{"id":13566,"date":"2018-04-07T14:52:50","date_gmt":"2018-04-07T09:22:50","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13566"},"modified":"2025-10-09T17:01:46","modified_gmt":"2025-10-09T11:31:46","slug":"cbse-sample-papers-class-12-chemistry","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/","title":{"rendered":"CBSE Class 12 Chemistry Sample Papers 2025"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Sample_paper_of_Chemistry_Class_12\" >Sample paper of Chemistry Class 12<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#CBSE_Class_12_Chemistry_Sample_Papers_2025_%E2%80%93_in_PDF\" >CBSE Class 12 Chemistry Sample Papers 2025 &#8211; in PDF<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Sample_Papers_Class_12_Chemistry_2024-25\" >Sample Papers Class 12 Chemistry 2024-25<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Solution\" >Solution<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Prepare_for_Exams_Effectively_with_myCBSEguide_App\" >Prepare for Exams Effectively with myCBSEguide App<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Sample_Papers_for_Class_12\" >Sample Papers for Class 12<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/#Download_Class_12_Sample_Papers_for_All_Subjects_from_myCBSEguide\" >Download Class 12 Sample Papers for All Subjects from myCBSEguide<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Sample_paper_of_Chemistry_Class_12\"><\/span>Sample paper of Chemistry Class 12<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Prepare effectively for your upcoming exams with <strong>CBSE <a href=\"https:\/\/mycbseguide.com\/cbse-sample-papers-class-12.html\">Class 12<\/a> Chemistry Sample Papers 2025<\/strong>, available for free download in <strong>PDF format<\/strong> on the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a> app and website.Download the <strong>CBSE Class 12 Chemistry Sample Papers 2025<\/strong> to practice effectively for your board exams. These sample papers are designed based on the latest <strong>CBSE marking scheme and blueprint<\/strong> for the 2025 board exams, ensuring that students are practicing with the most relevant and up-to-date materials. Each <strong>CBSE Class 12 Chemistry Sample Paper<\/strong> comes with detailed <strong>solutions<\/strong> to help students understand the best approach to answering questions. <strong>CBSE Sample Papers for Class 12 Chemistry 2025<\/strong> are created in line with the most recent exam pattern and blueprint.<\/p>\n<p>The <strong>CBSE Sample Papers for Class 12 Chemistry<\/strong> offer an excellent way for students to familiarize themselves with the types of questions likely to appear in the exams, while the included <strong>marking scheme<\/strong> and <strong>blueprint<\/strong> provide important insights into exam patterns. By downloading these sample papers, students can learn <strong>how to prepare for CBSE board exams<\/strong> and significantly improve their chances of scoring well. Accessing <strong>CBSE Chemistry Sample Papers Class 12<\/strong> through the <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> Website ensures that you have all the study material you need in one place. Get started today and practice <strong>CBSE Class 12 Chemistry Sample Papers<\/strong> with <strong>questions and answers<\/strong> to build confidence and master key concepts. Download the app now to get the most current and comprehensive study resources!<\/p>\n<p style=\"text-align: center;\"><strong>Sample Papers of Class 12 Chemistry 2025 with Solution<\/strong><\/p>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1267\/type\/2\">Download as PDF<\/a><\/strong><\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Class_12_Chemistry_Sample_Papers_2025_%E2%80%93_in_PDF\"><\/span>CBSE Class 12 Chemistry Sample Papers 2025 &#8211; in PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>myCBSEguide<\/strong> offers free access to <strong>CBSE Class 12 Chemistry Sample Papers 2025<\/strong> with solutions in <strong>PDF format<\/strong>. These sample papers are based on the latest <strong>CBSE syllabus<\/strong> and <strong>NCERT books<\/strong>, ensuring that students are practicing with the most current materials. It is essential for students to download and practice <strong>Class 12 Chemistry Sample Papers<\/strong> regularly to familiarize themselves with the <strong>marking scheme<\/strong> and <strong>blueprint<\/strong> outlined by CBSE. The <strong>CBSE Sample Papers<\/strong> for <strong>Class 12 Chemistry<\/strong> for the year 2025 follow the updated exam pattern, reflecting the latest changes in the syllabus and question types. Students can also download <strong>Class 12 Chemistry Sample Papers<\/strong> from the <strong>myCBSEguide app<\/strong> to practice on the go. To ensure the best preparation, students should always check the <strong>latest CBSE syllabus and marking scheme<\/strong> before attempting the sample papers. With <strong>myCBSEguide<\/strong>, students can access a wide range of <strong>CBSE Class 12 Chemistry Sample Papers<\/strong> for 2025, as well as papers for other subjects, designed to help them score higher in their board exams.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/12_chemistry_sample_paper.jpg\" alt=\"CBSE class 12 Chemistry Sample Papers 2024\" width=\"600\" height=\"300\" \/><\/p>\n<h3 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Sample_Papers_Class_12_Chemistry_2024-25\"><\/span>Sample Papers Class 12 Chemistry 2024-25<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: center;\"><strong>Class 12 &#8211; Chemistry<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<p><b>Maximum Marks: 70<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<p>Read the following instructions carefully.<\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>There are <strong>33<\/strong> questions in this question paper with internal choice.<\/li>\n<li>SECTION A consists of 16\u00a0multiple-choice questions carrying 1 mark each.<\/li>\n<li>SECTION B consists of 5 very short answer questions carrying 2 marks each.<\/li>\n<li>SECTION C consists of 7 short answer questions carrying 3 marks each.<\/li>\n<li>SECTION D consists of 2 case-based questions carrying 4 marks each.<\/li>\n<li>SECTION E consists of 3 long answer questions carrying 5 marks each.<\/li>\n<li><strong>All questions are compulsory.\u00a0<\/strong><\/li>\n<li><strong>Use of log tables and calculators is not allowed.<\/strong><\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section A<\/b><\/li>\n<li>Which of the following reactions is a halogenated exchange reaction:\n<div style=\"margin-left: 20px;\">a) <span class=\"math-tex\">{tex}R-OH\\;+\\;HCl\\;\\xrightarrow{ZnCl_2}\\;R-Cl\\;+\\;H_2O{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) <img decoding=\"async\" style=\"height: 45px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1709099449-6drcav.jpg\" \/><\/div>\n<div style=\"margin-left: 20px;\">c) <img decoding=\"async\" style=\"height: 77px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1709099863-ee7svd.jpg\" \/><\/div>\n<div style=\"margin-left: 20px;\">d) <span class=\"math-tex\">{tex}R-X\\;+\\;NaI\\;\\xrightarrow{Dry\\;acetone}\\;R-I\\;+\\;NaX{\/tex}<\/span><\/div>\n<\/li>\n<li>The most appropriate base pair in a double helix of DNA is\n<div style=\"margin-left: 20px;\">a) A \u2013 T<\/div>\n<div style=\"margin-left: 20px;\">b) A \u2013 C<\/div>\n<div style=\"margin-left: 20px;\">c) C \u2013 T<\/div>\n<div style=\"margin-left: 20px;\">d) T \u2013 G<\/div>\n<\/li>\n<li>Alcohols are easily dehydrated in the following order:\n<div style=\"margin-left: 20px;\">a) Secondary &gt; Tertiary&gt;Primary<\/div>\n<div style=\"margin-left: 20px;\">b) Primary&gt; Secondary &gt;Tertiary<\/div>\n<div style=\"margin-left: 20px;\">c) Tertiary &gt; Secondary &gt; Primary<\/div>\n<div style=\"margin-left: 20px;\">d) Secondary &gt; Primary&gt;Tertiary<\/div>\n<\/li>\n<li>Which of the following does not give Cannizaro reaction?\n<div style=\"margin-left: 20px;\">a) (CH<sub>3<\/sub>)<sub>3<\/sub> C &#8211; CHO<\/div>\n<div style=\"margin-left: 20px;\">b) (CH<sub>3<\/sub>)<sub>2<\/sub> CH &#8211;\u00a0CHO<\/div>\n<div style=\"margin-left: 20px;\">c) HCHO<\/div>\n<div style=\"margin-left: 20px;\">d) <img decoding=\"async\" style=\"height: 45px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1702014984-bnbc73.jpg\" \/><\/div>\n<\/li>\n<li>The chemical reaction in which reactants require high amount of activation energy are generally\n<div style=\"margin-left: 20px;\">a) Fast<\/div>\n<div style=\"margin-left: 20px;\">b) Instantaneous<\/div>\n<div style=\"margin-left: 20px;\">c) Slow<\/div>\n<div style=\"margin-left: 20px;\">d) Spontaneous<\/div>\n<\/li>\n<li>Match the items given in column I with that in column II:<br \/>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\">Column I<\/td>\n<td style=\"text-align: center;\">Column II<\/td>\n<\/tr>\n<tr>\n<td>(a) Hypertonic<\/td>\n<td>(i) NaCl.<\/td>\n<\/tr>\n<tr>\n<td>(b) Hypotonic<\/td>\n<td>(ii) Solution having higher osmotic pressure than other solution.<\/td>\n<\/tr>\n<tr>\n<td>(c) Isotonic<\/td>\n<td>(iii) Solution having lower osmotic pressure than other solution.<\/td>\n<\/tr>\n<tr>\n<td>(d) Electrolyte<\/td>\n<td>(iv) Solutions having same osmotic pressure.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div style=\"margin-left: 20px;\">a) (a) &#8211; (ii), (b) &#8211; (iv), (c) &#8211; (iii), (d) &#8211; (i)<\/div>\n<div style=\"margin-left: 20px;\">b) (a) &#8211; (i), (b) &#8211; (ii), (c) &#8211; (iii), (d) &#8211; (iv)<\/div>\n<div style=\"margin-left: 20px;\">c) (a) &#8211; (iv), (b) &#8211; (iii), (c) &#8211; (ii), (d) &#8211; (i)<\/div>\n<div style=\"margin-left: 20px;\">d) (a) &#8211; (ii), (b) &#8211; (iii), (c) &#8211; (iv), (d) &#8211; (i)<\/div>\n<\/li>\n<li>The conversion of an alkyl halide into an alcohol by aqueous NaOH is classified as\n<div style=\"margin-left: 20px;\">a) a dehydrohalogenation reaction<\/div>\n<div style=\"margin-left: 20px;\">b) a substitution reaction<\/div>\n<div style=\"margin-left: 20px;\">c) an addition reaction<\/div>\n<div style=\"margin-left: 20px;\">d) a dehydration reaction<\/div>\n<\/li>\n<li>The correct electronic configuration of copper atom is:\n<div style=\"margin-left: 20px;\">a) 3d<sup>10<\/sup>4s<sup>1<\/sup><\/div>\n<div style=\"margin-left: 20px;\">b) 3d<sup>5<\/sup>4s<sup>2<\/sup>4p<sup>4<\/sup><\/div>\n<div style=\"margin-left: 20px;\">c) 3d<sup>9<\/sup>4s<sup>2<\/sup><\/div>\n<div style=\"margin-left: 20px;\">d) 3d<sup>10<\/sup>4s<sup>2<\/sup><\/div>\n<\/li>\n<li>Which of the following rate laws is third order overall?\n<div style=\"margin-left: 20px;\">a) rate <span class=\"math-tex\">{tex} = K{\\left[ A \\right]^5}{\\left[ B \\right]^2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">b) rate <span class=\"math-tex\">{tex} = K\\left[ A \\right]{\\left[ B \\right]^2}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">c) rate <span class=\"math-tex\">{tex} = K{\\left[ A \\right]^3}{\\left[ B \\right]^3}{\/tex}<\/span><\/div>\n<div style=\"margin-left: 20px;\">d) rate\u00a0<span class=\"math-tex\">{tex} = K{\\left[ A \\right]^3}{\\left[ B \\right]^1}{\/tex}<\/span><\/div>\n<\/li>\n<li>Which one of the following is used for the separation and purification of aldehydes\n<div style=\"margin-left: 20px;\">a) hydrogen cyanide<\/div>\n<div style=\"margin-left: 20px;\">b) hydrogen sulphite<\/div>\n<div style=\"margin-left: 20px;\">c) Grignard reagent<\/div>\n<div style=\"margin-left: 20px;\">d) Sodium hydrogensulphite<\/div>\n<\/li>\n<li>By heating phenol with chloroform in alkali, it is converted into\n<div style=\"margin-left: 20px;\">a) Phenol benzoate<\/div>\n<div style=\"margin-left: 20px;\">b) Salicylic acid<\/div>\n<div style=\"margin-left: 20px;\">c) Anisole<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d) Salicylaldehyde<\/p>\n<p><strong>Download myCBSEguide App for Comprehensive Exam Preparation<\/strong>To enhance your exam preparation and practice more questions, download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\"><strong>myCBSEguide<\/strong><\/a>\u00a0app today. This app offers <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. The <strong>2025 Class 12 Chemistry sample papers<\/strong> are designed according to the latest CBSE syllabus and marking scheme. Whether you&#8217;re a student looking for quality study resources or a teacher aiming to create custom papers, myCBSEguide is your go-to solution for effective learning.In addition, teachers can use the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8<\/strong><\/a>\u00a0app to generate personalized exam papers with their own name, logo, and branding. With <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a>, teachers can create unique question papers for their students, making assessments more tailored and effective. <strong>Visit<a href=\"https:\/\/mycbseguide.com\/\"> myCBSEguide<\/a><\/strong> now for easy access to study materials, sample papers, and solutions, all in one place. Prepare confidently for your exams and ensure success with the best learning tools at your fingertips.<strong>\u00a0Class 12 Chemistry Sample Papers 2025<\/strong> <strong>CBSE <\/strong>provide detailed solutions to help students understand complex concepts.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li>Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is\n<div style=\"margin-left: 20px;\">a) Hoffmann Bromamide reaction<\/div>\n<div style=\"margin-left: 20px;\">b) Reaction with NH<sub>3<\/sub><\/div>\n<div style=\"margin-left: 20px;\">c) Gabriel phthalimide synthesis<\/div>\n<div style=\"margin-left: 20px;\">d) Sandmeyer reaction<\/div>\n<\/li>\n<li><strong>Assertion (A):\u00a0<\/strong>Uracil is present in DNA.<br \/>\n<strong>Reason (R): <\/strong>DNA undergoes replication.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong>\u00a0Pentan-2-one can be distinguished from pentan-3-one by iodoform test.<br \/>\n<strong>Reason (R):<\/strong>\u00a0Former is methyl ketone while the latter is not.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> (CH<sub>3<\/sub>)<sub>3<\/sub> C\u2013O\u2013CH<sub>3<\/sub> gives (CH<sub>3<\/sub>)<sub>3<\/sub> C\u2013I and CH<sub>3<\/sub>OH on treatment with HI.<br \/>\n<strong>Reason (R):<\/strong> The reaction occurs by S<sub>N<\/sub>1 mechanism.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> Phenols cannot be converted into esters by direct reaction with carboxylic acids.<br \/>\n<strong>Reason (R):<\/strong> Electron withdrawing groups increase the acidity of phenols.<\/p>\n<div style=\"margin-left: 20px;\">a) Both A and R are true and R is the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">b) Both A and R are true but R is not the correct explanation of A.<\/div>\n<div style=\"margin-left: 20px;\">c) A is true but R is false.<\/div>\n<div style=\"margin-left: 20px;\">d) A is false but R is true.<\/div>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section B<\/b><\/li>\n<li>Square planar complexes with a coordination number 4 exhibits geometrical isomerism whereas tetrahedral complexes do not why?<\/li>\n<li>Zinc salts are white while Cu<sup>2+<\/sup> salts are coloured. Why? [Atomic number of Zn = 30, Cu=29]<\/li>\n<li>Answer the following:\n<ol class=\"inner-group-question-ol\">\n<li>For a reaction R <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span>\u00a0P, half-life (t<sub>1\/2<\/sub>) is observed to be independent of the initial concentration of reactants. What is the order of reaction?<\/li>\n<li>Why is the probability of reaction with molecularity higher than three very rare?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molarity of <span class=\"math-tex\">{tex}9.8%{\/tex}<\/span>\u00a0(w\/W) solution of H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0if the density of the solution is <span class=\"math-tex\">{tex}1.02 \\;g\\; {mL}^{-1}{\/tex}<\/span>.(Molar mass of <span class=\"math-tex\">{tex}H_2S{O_4}{\/tex}<\/span>).\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Explain why aquatic species are more comfortable in cold water rather than in warm water.<\/li>\n<li>Write the structures of A, B and C in the following reactions:\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><img decoding=\"async\" style=\"height: 31px; width: 350px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1702278189-exfwp5.jpg\" \/><\/li>\n<li><span class=\"math-tex\">{tex}{\\mathrm{CH}}_3{\\mathrm{CH}}_2\\mathrm{Br}\\;\\xrightarrow{\\mathrm{KCN}}\\;\\mathrm A\\;\\xrightarrow{{\\mathrm{LiAlH}}_4}\\mathrm B\\xrightarrow[{0^{\\mathrm o}\\mathrm C}]{{\\mathrm{HNO}}_2}\\mathrm C{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Determine the value of equilibrium constant <span class=\"math-tex\">{tex}({K_c})\\,and\\,\\Delta {G^\\theta }{\/tex}<\/span> for the following reaction.<br \/>\n<span class=\"math-tex\">{tex}Ni(s) + 2A{g^ + }(aq) \\to N{i^{2 + }}(aq) + 2Ag(s){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{E^\\theta } = 1.05\\,V\\,(1\\,F = 96500\\,C\\,mo{l^{ &#8211; 1}}){\/tex}<\/span>Download myCBSEguide App for Effective Exam PreparationTo boost your exam preparation and practice more questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong>. This app offers comprehensive <strong>study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams. With <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a>, students can access a wide range of <strong>sample papers<\/strong>, <strong>solutions<\/strong>, and <strong>study notes<\/strong> tailored to their exam needs.In addition, <strong>teachers<\/strong> can enhance their teaching resources by using the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong> to create customized exam papers with their own <strong>name<\/strong> and <strong>logo<\/strong>. With <a href=\"https:\/\/examin8.com\/\"><strong>Examin8<\/strong><\/a>, teachers can create unique question papers for their students, making assessment more tailored and effective.Download the <strong>myCBSEguide<\/strong> app today to access a complete study package for all your exam preparation needs. From <strong>NCERT solutions<\/strong> to <strong>CBSE sample papers<\/strong>, everything you need for your academic success is available at your fingertips. For a well-rounded exam preparation, download <strong>CBSE Class 12 Chemistry Sample Papers 2025<\/strong> along with solutions and revision notes.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li>Consider the reaction:<br \/>\n2A + B <span class=\"math-tex\">{tex}\\to{\/tex}<\/span> C + D<br \/>\nFollowing results were obtained in experiments designed to study the rate of reaction:<\/p>\n<table style=\"width: 100%;\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\"><strong>Exp. No.<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>Initial concentration\u00a0(mol L<sup>-1 <\/sup>)\u00a0<span style=\"line-height: 1.6;\">[A]<\/span><\/strong><\/td>\n<td style=\"text-align: center;\"><strong>[B]<\/strong><\/td>\n<td style=\"text-align: center;\"><strong>Initial rate of formation\u00a0<span style=\"line-height: 1.6;\">[D] (m\/min)<\/span><\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">1.<\/td>\n<td style=\"text-align: center;\">0.10<\/td>\n<td style=\"text-align: center;\">0.10<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}1.5 \\times {10^{ &#8211; 3}}{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">2.<\/td>\n<td style=\"text-align: center;\">0.20<\/td>\n<td style=\"text-align: center;\">0.20<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}3.0 \\times {10^{ &#8211; 3}}{\/tex}<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">3.<\/td>\n<td style=\"text-align: center;\">0.20<\/td>\n<td style=\"text-align: center;\">0.40<\/td>\n<td style=\"text-align: center;\"><span class=\"math-tex\">{tex}6.0 \\times {10^{ &#8211; 3}}{\/tex}<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Write the rate law for the reaction.<\/li>\n<li>Calculate the value of rate constant for the reaction.<\/li>\n<li>Which of the following possible reaction mechanism is consistent with the rate law?\n<ol style=\"list-style-type: upper-roman;\" start=\"1\">\n<li><span class=\"math-tex\">{tex}{\\rm{A + B}} \\to {\\rm{C + E}}\\,{\\rm{(slow)}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A + E \\to D\\,\\,(fast){\/tex}<\/span><\/li>\n<li><span class=\"math-tex\">{tex}B \\to C + E\\,(slow){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A + E \\to F\\,\\,(fast){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}A + F \\to D\\,\\,(fast){\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>Name the following according to IUPAC system.\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><span class=\"math-tex\">{tex}C{H_3} &#8211; \\mathop C\\limits_{\\mathop |\\limits_{OH} } H &#8211; C{H_2} &#8211; C{H_3}{\/tex}<\/span><\/li>\n<li><img decoding=\"async\" style=\"width: 70px; height: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/iSUekNy.png\" alt=\"\" data-imgur-src=\"iSUekNy.png\" \/><\/li>\n<li><span class=\"math-tex\">{tex}C{H_3} &#8211; \\mathop C\\limits_{\\mathop |\\limits_{C{H_3}} }^{\\mathop |\\limits^{C{H_3}} } &#8211; C{H_2} &#8211; Cl{\/tex}<\/span><\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>How would you obtain the following.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Benzoquinone from phenol?<\/li>\n<li>2-methyl propan-2-ol from methyl magnesium bromide?<\/li>\n<li>Propan-2-ol from propene?<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Identify A, B and C in the sequence<br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/Rm5Xf1t.png\" alt=\"\" data-imgur-src=\"Rm5Xf1t.png\" \/><\/li>\n<li>Predict the products of the following reactions:\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li><img decoding=\"async\" style=\"width: 130px; height: 46px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/DAo0ReU.png\" alt=\"\" data-imgur-src=\"DAo0ReU.png\" \/><\/li>\n<li><img decoding=\"async\" style=\"width: 150px; height: 58px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/5wKTh8h.png\" alt=\"\" data-imgur-src=\"5wKTh8h.png\" \/><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the e.m.f. of the following cell at 298K:<br \/>\nFe(s) | Fe<sup>2+<\/sup> (0.001 M) || H<sup>+<\/sup> (0.01 M) | H<sub>2<\/sub>(g) (1 bar) | Pt(s)<br \/>\nGiven that <span class=\"math-tex\">{tex}E_{\\text {cell }}^{\\circ}{\/tex}<\/span> = +0.44V<br \/>\n[log 2 = 0.3010 log 3 = 0.4771 log 10 = 1]<\/li>\n<li>What are ambident nucleophiles? Explain with an example.<\/li>\n<li>In the button cell, widely used in watches, the following reaction takes place<br \/>\n<span class=\"math-tex\">{tex}Zn\\left( s \\right){\\rm{ }} + A{g_2}O\\left( s \\right){\\rm{ }} + {\\rm{ }}{H_2}O\\left( l \\right) {\/tex}<\/span> <span class=\"math-tex\">{tex}\\to Z{n^{2 + }}\\left( {aq} \\right) + 2Ag\\left( s \\right){\\rm{ }} + {\\rm{ }}20{H^ &#8211; }\\left( {aq} \\right){\/tex}<\/span> Given that <span class=\"math-tex\">{tex}E_{(Ag^+\/Ag)}^\\Theta = 0.80 V, E_{(Zn^{2+}\/Zn)}^\\Theta=-0.76\\;V{\/tex}<\/span><br \/>\nCalculate standard emf and standard free Gibbs energy of the cell.<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nThe unique behaviour of Cu, having a positive E\u00b0\u00a0accounts\u00a0for its inability\u00a0to liberate H<sub>2<\/sub> from\u00a0acids. Only oxidising acids (nitric and hot concentrated sulphuric) react with Cu, the acids being reduced.\u00a0The stability of the half-filled d sub-shell in Mn<sup>2+<\/sup> and the completely filled d<sup>10<\/sup> configuration in Zn<sup>2+<\/sup>\u00a0are related to their E\u00b0\u00a0values, whereas E\u00b0\u00a0for Ni is related to the highest negative <span class=\"math-tex\">{tex}\\Delta_{hyd}{\/tex}<\/span>H\u00b0.\u00a0An examination of the <span class=\"math-tex\">{tex}E^o_{(M^{3+}\/M^{2+})}{\/tex}<\/span> values the low value for Sc reflects the stability of Sc<sup>3+<\/sup> which has a noble gas configuration. The comparatively high value for Mn shows that Mn<sup>2+<\/sup>(d<sup>5<\/sup>) is particularly stable, whereas a comparatively low value for Fe shows the extra stability of Fe<sup>3+<\/sup>\u00a0(d<sup>5<\/sup>). The comparatively low value for V is related to the stability of V<sup>2+<\/sup>\u00a0(half-filled t<sub>2g<\/sub>\u00a0level).<br \/>\n<img decoding=\"async\" style=\"height: 197px; width: 250px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/agdWNFJ.png\" alt=\"\" \/><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Why Zn has high value for M<sup>3+<\/sup>\/M<sup>2+<\/sup>\u00a0Standard Electrode Potentials? (1)<\/li>\n<li>Transition metals, despite high E\u00b0 oxidation, are poor reducing agents. Justify. (1)<\/li>\n<li>Why is Cr<sup>2+<\/sup> reducing and Mn<sup>3+\u00a0<\/sup>oxidising when both Cr and Mn have d<sup>4<\/sup>\u00a0configuration? (2)<br \/>\n<strong>OR<\/strong><br \/>\nWhy Cu<sup>2+<\/sup> is more stable than Cu<sup>+<\/sup>? (2)<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nRaoult&#8217;s law for volatile liquids states that the partial vapour pressure of each component in the solution is directly proportional to its mole fraction, whereas for a non-volatile solute, it states that the vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. Two liquids A and B are mixed with each other to form a solution, the vapour phase consists of both components of the solution. Once the components in the solution have reached equilibrium, the total vapour pressure of the solution can be determined by combining Raoult&#8217;s law with Dalton&#8217;s law of partial pressures. If a non-volatile solute B is dissolved into a solvent A to form a solution, the vapour pressure of the solution will be lower than that of the pure solvent. The solutions which obey Raoult&#8217;s law over the entire range of concentration are ideal solutions, whereas the solutions for which vapour pressure is either higher or lower than that predicted by Raoult&#8217;s law are called non-ideal solutions. Non-ideal solutions are identified by determining the strength of the intermolecular forces between the different molecules in that particular solution. They can either show positive or negative deviation from Raoult&#8217;s law depending on whether the A &#8211; B interactions in solution are stronger or weaker than A\u00a0&#8211;\u00a0A and B\u00a0&#8211;\u00a0B interactions.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>20 mL of a liquid A was mixed with 20 mL of liquid B. The volume of resulting solution was found to be less than 40 mL. What do you conclude from the above data? (1)<\/li>\n<li>Which of the following show positive deviation from Raoult&#8217;s law? Carbon disulphide and Acetone; Phenol and Aniline; Ethanol and Acetone. (1)<\/li>\n<li>The vapour pressure of a solution of glucose in water is 750 mm Hg at 100<sup>o<\/sup>C. Calculate the mole fraction of solute. (2)<br \/>\n(Vapour pressure of water at 373 K = 760 mm Hg)<br \/>\n<strong>OR<\/strong><br \/>\nThe boiling point of solution increases when 1 mol of NaCl is added to 1 litre of water while addition of 1 mol of methanol to one litre of water decreases its boiling point. Explain the above observations. (2)<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section E<\/b><\/li>\n<li>Attempt any five of the following:\n<ol class=\"inner-group-question-ol\">\n<li>What type of protein is present in keratin?<\/li>\n<li>What happens when D-glucose is treated with the following reagents?\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>Br<sub>2<\/sub> water<\/li>\n<li>HCN<\/li>\n<\/ol>\n<\/li>\n<li>Differentiate between:\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Peptide linkage and Glycosidic linkage<\/li>\n<li>Nucleoside and Nucleotide<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>What is the difference between a nucleoside and nucleotide?<\/li>\n<li>What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?<\/li>\n<\/ol>\n<\/li>\n<li>What is the biological effect of denaturation of proteins?<\/li>\n<li>Write the full forms of DNA and RNA.<\/li>\n<li>Of the two bases named below, which one is present in RNA and which one is present in DNA?\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Thymine<\/li>\n<li>Uracil<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>For the complex, [Fe(CN)6]<sup>3-<\/sup>, write the hybridisation type, magnetic character and spin nature of the complex. (Atomic number of Fe = 26)<\/li>\n<li>Draw one of the geometrical isomers of the complex [Pt(en)<sub>2<\/sub>Cl<sub>2<\/sub>]<sup>2+<\/sup> which is optically active.<\/li>\n<\/ol>\n<\/li>\n<li>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>A metal complex having composition Cr(NH<sub>3<\/sub>)<sub>4<\/sub>Cl<sub>2<\/sub>Br has been isolated in two forms A and B. The form A reacts with AgNO<sub>3<\/sub> to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>State the hybridization of chromium in each of them.<\/li>\n<li>Calculate the magnetic moment (spin only value) of the isomer A.<\/li>\n<\/ol>\n<\/li>\n<li>Write IUPAC name and hybridization of the following complexes:\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>[Co(NH<sub>3<\/sub>)<sub>6<\/sub>]<sup>3+<\/sup><\/li>\n<li>[NiCl<sub>4<\/sub>]<sup>2-<\/sup><br \/>\n(Atomic number Ni = 28, Co = 27)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>State the reactions and reaction conditions for the following conversions :\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Benzene diazonium chloride to nitrobenzene.<\/li>\n<li>Aniline to benzene diazonium chloride.<\/li>\n<li>Ethylamine to methylamine.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>An aromatic compound &#8216;A&#8217; of molecular formula C<sub>7<\/sub>H<sub>6<\/sub>O<sub>2<\/sub> undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions.<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/M03UUR2.png\" alt=\"\" data-imgur-src=\"M03UUR2.png\" \/><\/p>\n<p><strong>Download the myCBSEguide App for Comprehensive Exam Preparation<\/strong><\/p>\n<p>For effective exam preparation and to practice a wide range of questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> today. This app provides <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>. Whether you&#8217;re studying for school exams or preparing for competitive tests, <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> Website has you covered with <strong>study notes<\/strong>, <strong>sample papers<\/strong>, and <strong>solutions<\/strong>.<\/p>\n<p>Teachers can also benefit from the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong>, which allows them to design custom <strong>question papers<\/strong> with their own <strong>name<\/strong> and <strong>logo<\/strong>, offering a personalized assessment experience for their students.<\/p>\n<p>Unlock the full potential of your exam preparation by downloading the <strong>myCBSEguide<\/strong> app now and gain access to high-quality resources to help you score better in your exams. With everything you need in one place, myCBSEguide is your ultimate study companion.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 12 &#8211; Chemistry<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ol style=\"padding-left: 20px;\">\n<li style=\"list-style-type: none;\">\n<ol style=\"padding-left: 20px;\">\n<li style=\"text-align: center; display: block;\"><b>Section A <\/b><\/li>\n<li>(d) <span class=\"math-tex\">{tex}R-X\\;+\\;NaI\\;\\xrightarrow{Dry\\;acetone}\\;R-I\\;+\\;NaX{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> <span class=\"math-tex\">{tex}R-X\\;+\\;NaI\\;\\xrightarrow{Dry\\;acetone}\\;R-I\\;+\\;NaX{\/tex}<\/span><\/li>\n<li>(a) A \u2013 T<br \/>\n<b>Explanation: <\/b> Two nucleic acid chains are wound about each other and held together by hydrogen bonds between pairs of bases. The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases. Adenine forms hydrogen bonds with thymine whereas cytosine forms hydrogen bonds with guanine. So base pair in a double helix of DNA is A-T.<\/li>\n<li>(c) Tertiary &gt; Secondary &gt; Primary<br \/>\n<b>Explanation: <\/b> The rate of dehydration is related to the ease of formation of Carbocation and the energy of the carbocation intermediate.<br \/>\nDehydration involves the formation of carbocation 1<sup>st<\/sup> and since tertiary carbocation is more stable than secondary and primary therefore 3\u00b0 is dehydrated first.<br \/>\nThe ease of formation of Carbocation is Tertiary&gt;Secondary&gt;Primary.<\/li>\n<li>(b) (CH<sub>3<\/sub>)<sub>2<\/sub> CH &#8211;\u00a0CHO<br \/>\n<b>Explanation: <\/b> (CH<sub>3<\/sub>)<sub>2<\/sub> CH &#8211;\u00a0CHO doesn&#8217;t give cannizaro reaction.<\/li>\n<li>(c) Slow<br \/>\n<b>Explanation: <\/b> As the energy of activation increases, more energy is required for reactants to complete the reaction hence chemical reaction become slow.<\/li>\n<li>(d) (a) &#8211; (ii), (b) &#8211; (iii), (c) &#8211; (iv), (d) &#8211; (i)<br \/>\n<b>Explanation: <\/b> (a) &#8211; (ii), (b) &#8211; (iii), (c) &#8211; (iv), (d) &#8211; (i)<\/li>\n<li>(b) a substitution reaction<br \/>\n<b>Explanation: <\/b> R &#8211; X\u00a0+ NaOH <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span>\u00a0R &#8211; OH +\u00a0NAX<\/li>\n<li>(d) 3d<sup>10<\/sup>4s<sup>2<\/sup><br \/>\n<b>Explanation: <\/b> 3d<sup>10<\/sup>4s<sup>2<\/sup><\/li>\n<li>(b) rate <span class=\"math-tex\">{tex} = K\\left[ A \\right]{\\left[ B \\right]^2}{\/tex}<\/span><br \/>\n<b>Explanation: <\/b> rate <span class=\"math-tex\">{tex} = K\\left[ A \\right]{\\left[ B \\right]^2}{\/tex}<\/span><br \/>\nsince the rate of a given reaction is first-order wrt A reactant and second-order wrt B reactant.<br \/>\norder of the reaction is the sum of powers of each reactant in rate law expression.<br \/>\nso, the order of reaction = 1+2 = 3 (a chemical\u00a0reaction\u00a0in which the rate of\u00a0reaction\u00a0is proportional to the concentration of each of three reacting molecules).<\/li>\n<li>(d) Sodium hydrogensulphite<br \/>\n<b>Explanation: <\/b> <sub><img decoding=\"async\" style=\"height: 110px; width: 300px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/luia1I9.png\" \/><\/sub><br \/>\nAldehydes and ketones form bisulphite addition product with NaHSO<sub>3<\/sub> while impurites does not\u00a0on hydrolysis we get pure aldehydes and ketones.<strong>Boost Your Exam Preparation with myCBSEguide App<\/strong>To enhance your exam preparation and practice a variety of questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> today. The app offers <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, making it the go-to resource for students aiming to excel in their exams. With a comprehensive collection of <strong>sample papers<\/strong>, <strong>solutions<\/strong>, and <strong>study notes<\/strong>, the myCBSEguide app ensures you&#8217;re well-prepared for your exams.Additionally, <strong>teachers<\/strong> can benefit from the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong>, which enables them to create personalized <strong>exam papers<\/strong> with their own <strong>name<\/strong> and <strong>logo<\/strong>.\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 This feature provides educators with an efficient way to design customized assessments for their students.Don&#8217;t miss out on the opportunity to streamline your study process.Visit<strong>\u00a0<a href=\"https:\/\/mycbseguide.com\/\">myCBSEguide<\/a><\/strong> <strong>Website<\/strong> and <strong><a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong> today and gain access to a wealth of resources designed to help you succeed in exams and improve your academic performance.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<ol style=\"padding-left: 20px;\">\n<li>(d) Salicylaldehyde<br \/>\n<b>Explanation: <\/b> Salicylaldehyde<\/li>\n<li>(c) Gabriel phthalimide synthesis<br \/>\n<b>Explanation: <\/b> Gabriel phthalimide synthesis is used to get primary amine is prepared\u00a0from alkyl halide without any change in the number of carbon atoms.<br \/>\n<img decoding=\"async\" style=\"width: 350px; height: 195px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/fr8qesR.png\" alt=\"\" data-imgur-src=\"fr8qesR.png\" \/><\/li>\n<li>(d) A is false but R is true.<br \/>\n<b>Explanation: <\/b> A is false but R is true.<\/li>\n<li>(a) Both A and R are true and R is the correct explanation of A.<br \/>\n<b>Explanation: <\/b> Both A and R are true and R is the correct explanation of A.<\/li>\n<li>(a) Both A and R are true and R is the correct explanation of A.<br \/>\n<b>Explanation: <\/b> (CH<sub>3<\/sub>)<sub>3<\/sub>-C-O-CH<sub>3<\/sub>\u00a0is an ether with two different alkyl groups, of which (CH<sub>3<\/sub>)<sub>3<\/sub>-C-I,\u00a0tertiary alkyl group, on reaction with hydrogen halide (HI) forms a tertiary halide.<br \/>\nThis occurs as the reaction is an S<sub>N<\/sub>1 reaction. The reaction involves the formation of a stable carbocation. If the ether has a primary alkyl group, then the reaction follows the S<sub>N<\/sub>2 mechanism.<\/li>\n<li>(b) Both A and R are true but R is not the correct explanation of A.<br \/>\n<b>Explanation: <\/b> Phenols cannot be converted into esters by direct reaction with carboxylic acids since phenols are less nucleophilic than alcohols.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li>The tetrahedral complexes do not show geometrical isomerism because the relative positions of the atoms with respect to each other will be the same. The square planar complexes on the other hand show geometrical isomerism because if same kind of ligands occupy positions adjacent to each other it is called cis-form and if these are opposite to each other it is called trans-form.<\/li>\n<li>In zinc salts, \u200b\u200b\u200b\u200b\u200b\u200bZn<sup>2+<\/sup>\u200b does not have unpaired electrons and has fully filled d orbital. Zn<sup>2+<\/sup> has configuration [Ar]3d<sup>10<\/sup> .Therefore, there is no possibility of undergoing d-d transitions whereas in Cu<sup>2+<\/sup>ions having electronic configuration [Ar]3d<sup>9<\/sup>, has unpaired electron which can undergo d-d transitions. Hence, zinc salts are white while\u00a0Cu<sup>2+<\/sup> salts are coloured.<\/li>\n<li>Answer the following:\n<ol class=\"inner-group-question-ol\">\n<li style=\"clear: both;\">For a reaction R\u00a0<span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> P, half-life (t<sub>1\/2<\/sub>)\u00a0is observed to be independent of the initial concentration of reactants. Thus, it follows first order reaction.<\/li>\n<li style=\"clear: both;\">As the number increases the chance of simultaneous collision decrease that means the probability of more than three molecules colliding simultaneously is very small. Therefore, the possibility of molecularity being three is very low.<\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\"><span class=\"math-tex\">{tex}M = \\frac{{{w_B}}}{{{M_B}}} \\times \\frac{{1000}}{{V(in{\\rm{ mL)}}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}98\\omega \/W{\/tex}<\/span>means 9.8g H<sub>2<\/sub>SO<sub>4<\/sub> is dissolved in 90.2g water.<br \/>\nMass of solution = 100g<br \/>\nDensity of solution = 1.02g mL<sup>-1<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a0Volume of solution\u00a0<span class=\"math-tex\">{tex} = \\frac{{mass\\;of\\;solution}}{{density\\;of\\;solution}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{\\rm{ = }}\\frac{{100}}{{1.02}} = 98.04mL{\/tex}<\/span><br \/>\nmass of the solute (H<sub>2<\/sub>SO<sub>4<\/sub>) w<sub>B\u00a0<\/sub>= 9.8g<br \/>\nmolar mass of sulphuric acid M<sub>B<\/sub> = 98g\/mol<br \/>\n<span class=\"math-tex\">{tex}{\\rm{M = }}\\frac{{9.8}}{{98}} \\times \\frac{{1000}}{{98.04}}{\/tex}<\/span><br \/>\nM = 1.02M<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Increase in temperature decreases the solubility of oxygen in the water. The amount of oxygen dissolved in water decreases at a higher temperature. As a result, it becomes more difficult to breathe less dissolved oxygen. Hence, the aquatic species are not comfortable in warm water.<\/li>\n<li>Structure of the compound\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><img decoding=\"async\" style=\"height: 68px; width: 320px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1702278514-2hxjdz.jpg\" \/><\/li>\n<li>A = CH<sub>3<\/sub>CH<sub>2<\/sub>CN B = CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>NH<sub>2<\/sub> C = CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub>OH<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b><\/li>\n<li>We have,<br \/>\n<span class=\"math-tex\">{tex}Ni(s) + 2A{g^ + }(aq) \\to N{i^{2 + }}(aq) + 2Ag(s){\/tex}<\/span><br \/>\nFor the reaction n = 2,<span class=\"math-tex\">{tex} E_{cell}^\\theta {\/tex}<\/span> = 1.05 V<br \/>\n<span class=\"math-tex\">{tex}\\Delta {G^\\theta } = &#8211; nF{E^\\theta }{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Delta {G^\\theta } = &#8211; 2 \\times 96500\\,C \\times 1.05\\,V{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Delta {G^\\theta } = &#8211; 202.65\\,kJ\\,mo{l^{ &#8211; 1}}{\/tex}<\/span><br \/>\nFor Equilibrium constant, we have,<br \/>\n<span class=\"math-tex\">{tex}\\Delta {G^\\theta } = &#8211; 2.303\\,RT\\,\\log \\,{K_c}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\log {K_c} = &#8211; \\frac{{\\Delta {G^\\Theta }}}{{2.303RT}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex} = &#8211; \\frac{{202650}}{{2.303 \\times 8.314 \\times 298}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{K_c} = {\\text{Antilog(35}}{\\text{.5161)}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}{K_c} = 3.284 \\times {10^{35}}{\/tex}<\/span>Enhance Your Exam Preparation with myCBSEguide AppFor effective exam preparation and ample practice, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> today. This app provides <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, offering a comprehensive suite of resources including <strong>sample papers<\/strong>, <strong>solutions<\/strong>, and <strong>study notes<\/strong>. With these materials, students can practice a variety of questions and strengthen their understanding of key concepts, ensuring they are exam-ready. In addition, <strong>teachers<\/strong> can use the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong> to create personalized <strong>exam papers<\/strong> with their own <strong>name<\/strong> and <strong>logo<\/strong>, allowing for customized assessments that cater to their teaching needs. Visit <a href=\"https:\/\/mycbseguide.com\/\"><strong>myCBSEguide<\/strong><\/a> <strong>Website<\/strong> and <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\"><strong>Examin8<\/strong><\/a> <strong>Website<\/strong> now and unlock the full potential of your study routine. Whether you&#8217;re preparing for <strong>board exams<\/strong> or competitive entrance exams, <strong>myCBSEguide<\/strong> offers the tools you need for success.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Let rate law is<br \/>\n<span class=\"math-tex\">{tex}Rate = k{[A]^x}{[B]^y}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}So, \\,1.5 \\times {10^{ &#8211; 3}} = k{[0.1]^x}{[0.1]^y}\\,&#8230;\\,(i){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}3.0 \\times {10^{ &#8211; 3}} = k{[0.2]^x}{[0.2]^y}\\,&#8230;\\,(ii){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}6.0 \\times {10^{ &#8211; 3}} = k{[0.2]^x}{[0.4]^y}\\,&#8230;\\,(iii){\/tex}<\/span><br \/>\nFrom eq.(ii) and (iii)<br \/>\n<span class=\"math-tex\">{tex}\\frac{{6 \\times {{10}^{ &#8211; 3}}}}{{3 \\times {{10}^{ &#8211; 3}}}} = \\frac{{k{{[0.2]}^x}{{[0.4]}^y}}}{{k{{[0.2]}^x}{{[0.2]}^y}}}{\/tex}<\/span><br \/>\n2<sup>y<\/sup>\u00a0= 2<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> y = 1<br \/>\nFrom eq.(i) and (ii)<br \/>\n<span class=\"math-tex\">{tex}\\frac{{3 \\times {{10}^{ &#8211; 3}}}}{{1.5 \\times {{10}^{ &#8211; 3}}}} = \\frac{{k{{[0.2]}^x}{{[0.2]}^1}}}{{k{{[0.1]}^x}{{[0.1]}^1}}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}2 = {2^x} \\times 2{\/tex}<\/span><br \/>\n2<sup>x<\/sup>\u00a0= 1<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span> x = 0<br \/>\nThus, the rate is given as Rate =k[B]<sup>1<\/sup><\/li>\n<li>Rate = k[B]\n<span class=\"math-tex\">{tex}k = \\frac{{Rate}}{{[B]}} = \\frac{{3 \\times {{10}^{ &#8211; 3}}}}{{0.2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\u00a0= 15 \\times {10^{ &#8211; 3}}\\,{\\min ^{ &#8211; 1}}{\/tex}<\/span><\/li>\n<li>B <span class=\"math-tex\">{tex}\\to{\/tex}<\/span> C + E (slow)\u00a0is the possible reaction with is consistent with the rate law i.e.,<br \/>\n<span class=\"math-tex\">{tex}Rate = k [B]^1.{\/tex}<\/span><br \/>\nHence, mechanism II is appropriate for the reaction.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><span class=\"math-tex\">{tex}\\mathop {\\mathop C\\limits^1 {H_3} &#8211; \\mathop {\\mathop C\\limits^2 }\\limits_{\\mathop |\\limits_{OH} } H &#8211; \\mathop C\\limits^3 {H_2} &#8211; \\mathop C\\limits^4 {H_3}}\\limits_{Butan &#8211; 2 &#8211; ol}{\/tex}<\/span><\/li>\n<li><img decoding=\"async\" style=\"width: 100px; height: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/K1WVe1v.png\" alt=\"\" data-imgur-src=\"K1WVe1v.png\" \/><\/li>\n<li><span class=\"math-tex\">{tex}\\mathop {\\mathop C\\limits^3 {H_3} &#8211; \\mathop {{}^2C}\\limits_{\\mathop |\\limits_{C{H_3}} }^{\\mathop |\\limits^{C{H_3}} } &#8211; \\mathop C\\limits^1 {H_2}Cl}\\limits_{2,2 &#8211; \\dim ethylchloropropane}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><img decoding=\"async\" style=\"width: 250px; height: 133px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/MQCzZYB.png\" alt=\"\" data-imgur-src=\"MQCzZYB.png\" \/><\/li>\n<li><span class=\"math-tex\">{tex}\\mathop {C{H_3}MgBr}\\limits_{\\mathop {Methyl\\,\\,magnesium}\\limits_{chloride} } + \\mathop {C{H_3} &#8211; \\mathop C\\limits_{\\mathop {||}\\limits_O } &#8211; C{H_3}}\\limits_{\\Pr opanone} \\to C{H_3} &#8211; \\mathop C\\limits_{\\mathop |\\limits_{C{H_3}} }^{\\mathop |\\limits^{OMgBr} } &#8211; C{H_3}\\xrightarrow[{ &#8211; MG(OH)Br}]{{{H_2}O}}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\mathop {C{H_3} &#8211; \\mathop C\\limits_{\\mathop |\\limits_{C{H_3}} }^{\\mathop |\\limits^{OH} } &#8211; C{H_3}}\\limits_{2 &#8211; methylpropan &#8211; 2 &#8211; ol} {\/tex}<\/span><\/li>\n<li><span class=\"math-tex\">{tex}\\mathop {C{H_2} = CH &#8211; C{H_3}}\\limits_{Propene} \\xrightarrow{{{H_2}O\/{H^ + }}}\\mathop {C{H_3} &#8211; \\mathop C\\limits_{\\mathop |\\limits_{OH} } H &#8211; C{H_3}}\\limits_{Propan &#8211; 2 &#8211; ol}{\/tex}<\/span><\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><img decoding=\"async\" style=\"width: 250px; height: 174px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/K9Xkpiz.png\" alt=\"\" data-imgur-src=\"K9Xkpiz.png\" \/>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li><img decoding=\"async\" style=\"width: 220px; height: 81px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/I0j0JpL.png\" alt=\"\" data-imgur-src=\"I0j0JpL.png\" \/><\/li>\n<li><img decoding=\"async\" style=\"width: 220px; height: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/wduZmlx.png\" alt=\"\" data-imgur-src=\"wduZmlx.png\" \/><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>According to the equation,<br \/>\nFe(s) + 2H<sup>+<\/sup> (aq) <span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span> Fe<sup>2+<\/sup> (aq) + H<sub>2<\/sub>(g)<br \/>\n<span class=\"math-tex\">{tex}E_{\\text {cell }}^{\\circ}=E_{\\text {cell }}^{\\circ}-E_{\\text {anode }}^{\\circ}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}E_{\\text {cell }}^{\\circ}{\/tex}<\/span> = 0 &#8211; (-0.44) V<br \/>\n<span class=\"math-tex\">{tex}E_{\\text {cell }}^{\\circ}{\/tex}<\/span> = + 0.44V<br \/>\nBy applying Nernst Equation<br \/>\nE<sub>cell<\/sub> = <span class=\"math-tex\">{tex}E_{\\text {cell }}^{\\circ}-\\frac{0.0591}{2} \\log \\frac{\\left[F e^{2+}\\right]}{\\left[H^{+}\\right]^2}{\/tex}<\/span><br \/>\nE<sub>cell <\/sub>= <span class=\"math-tex\">{tex}0.44 \\frac{-0.0591}{2} \\log \\frac{0.001}{1^2}{\/tex}<\/span><br \/>\nE<sub>cell <\/sub>= <span class=\"math-tex\">{tex}0.44-\\frac{0.0591}{2} \\log 10^{-3}{\/tex}<\/span><br \/>\nE<sub>cell <\/sub>= 0.44 + 0.089V<br \/>\nE<sub>cell <\/sub>= +0.53V<\/li>\n<li>Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.<br \/>\nFor example, nitrite ion is an ambident nucleophile.<br \/>\n<span class=\"math-tex\">{tex}\\left[ {{}^ &#8211; O &#8211; \\ddot N = O} \\right]{\/tex}<\/span><br \/>\nNitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/12\/sci\/ch10\/img67.png\" \/><\/li>\n<li>For the button cell, chemical reactions are:<br \/>\n<strong>At anode:<\/strong><br \/>\n<span class=\"math-tex\">{tex}Zn(s) \\to Z{n^{2 + }}(aq) + 2{e^ &#8211; }{\/tex}<\/span><br \/>\n<strong>At Cathode:<\/strong><br \/>\n<span class=\"math-tex\">{tex}A{g_2}O(s) + {H_2}O(l) + 2{e^ &#8211; } \\to 2Ag(s) + 2O{H^ &#8211; }(aq){\/tex}<\/span><br \/>\nComplete reaction is:<br \/>\n<span class=\"math-tex\">{tex}Zn\\left( s \\right){\\rm{ }} + A{g_2}O\\left( s \\right){\\rm{ }} + {\\rm{ }}{H_2}O\\left( l \\right){\/tex}<\/span> <span class=\"math-tex\">{tex} \\to Z{n^{2 + }}\\left( {aq} \\right) + 2Ag\\left( s \\right){\\rm{ }} + {\\rm{ }}20{H^ &#8211; }\\left( {aq} \\right){\/tex}<\/span>, for this reaction, n=2 moles of electrons.<br \/>\nNow,<br \/>\n<span class=\"math-tex\">{tex}E_{cell}^\\Theta = E_{A{g^ + }\/Ag}^\\Theta &#8211; E_{Z{n^{2 + }}\/Zn}^\\Theta {\/tex}<\/span><br \/>\n= 0.80 &#8211; (-0.76) = 1.56V<br \/>\nWe know that<br \/>\n<span class=\"math-tex\">{tex}{\\Delta _r}{G^\\Theta } = &#8211; nFE_{cell}^\\Theta {\/tex}<\/span><br \/>\nHere, n = 2<br \/>\nF = 96500 C mol<sup>-1<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\Delta G^\\Theta=-2\\times1.56\\;V\\times96000\\;C\\,\\;mol^{-1}{\/tex}<\/span><br \/>\n= -301080 J mol<sup>-1<\/sup><br \/>\n= -301.08 kJ mol<sup>-1<\/sup><br \/>\n<span class=\"math-tex\">{tex}\\therefore\\Delta G^\\Theta=-308.08\\;kJ\\;mol^{-1}{\/tex}<\/span><\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Due to the removal of an electron from the stable d<sup>10<\/sup>\u00a0configuration of Zn<sup>2+<\/sup>.<\/li>\n<li style=\"text-align: left;\">Transition metals despite having high E\u00b0 oxidation, are poor reducing agents because of their high heat of vaporization, high ionisation energies and low heats of hydration.<\/li>\n<li style=\"text-align: left;\">Cr<sup>2+<\/sup> is reducing as its configuration changes from d<sub>4<\/sub>\u00a0to d<sub>3<\/sub>, the having a half-filled t<sub>2g<\/sub> level. On the other hand, the change from Mn<sup>3+<\/sup> to Mn<sup>2+<\/sup> results in the half-filled (d<sub>5<\/sub>) configuration which has extra stability.<br \/>\n<strong>OR<\/strong><br \/>\nThe\u00a0Stability of Cu<sup>2+<\/sup> is more than Cu<sup>+<\/sup> as stability depends on the hydration energy of the ions when they bond to the water molecules. The Cu<sup>2+<\/sup> ion has a greater charge density than Cu<sup>+<\/sup> ion and thus forms much stronger bonds releasing more energy.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Solution shows a negative deviation from Raoult\u2019s law \/ A-A and B-B interactions are weaker than A-B interactions.<\/li>\n<li style=\"text-align: left;\">Carbon disulphide and acetone, Ethanol and acetone.<\/li>\n<li style=\"text-align: left;\">According to Raoult\u2019s law:<br \/>\np<sub>1<\/sub> = <span class=\"math-tex\">{tex}p_1^0 x_1{\/tex}<\/span> or x<sub>1<\/sub> = <span class=\"math-tex\">{tex}\\frac{p_1}{p_1^0}{\/tex}<\/span><br \/>\nx<sub>1<\/sub> = <span class=\"math-tex\">{tex}\\frac{750}{760}{\/tex}<\/span> = 0.987<br \/>\nx<sub>2<\/sub> = 1 &#8211; x<sub>1<\/sub><br \/>\n= 1 &#8211; 0.987 = 0.013<br \/>\n<strong>OR<\/strong><br \/>\nNaCl is a non-volatile solute, when it is added to water the vapour pressure decreases and hence boiling point increases.<br \/>\nMethanol is a volatile solute and its addition to water increases the total vapour pressure of the solution and hence boiling point decreases.<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section E <\/b><\/li>\n<li>Attempt any five of the following:\n<ol class=\"inner-group-question-ol\">\n<li style=\"clear: both;\">Fibrous Proteins\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li><img decoding=\"async\" style=\"height: 68px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1725943953-vvsmmg.jpg\" \/><\/li>\n<li><img decoding=\"async\" style=\"height: 79px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1725944035-sx9pxc.jpg\" \/><\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><strong>Peptide linkage: <\/strong>A linkage formed when two amino acids are joined through -CONH-\u00a0bond.<br \/>\nGlycosidic linkage: When two monosaccharides are joined through oxygen atom.<\/li>\n<li><strong>Nucleoside:<\/strong> Base + Sugar<br \/>\n<strong>Nucleotide: <\/strong>Base + Sugar + Phosphate<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>Nucleoside <span class=\"math-tex\">{tex}\\to{\/tex}<\/span> Nitrogenous base + Sugar whereas<br \/>\nNucleotide <span class=\"math-tex\">{tex}\\to{\/tex}<\/span>\u00a0Nitrogenous base + Sugar + Phosphate group.<\/li>\n<li>Thymine base, 2-Deoxyribose sugar, and a Phosphoric acid.<\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">On denaturation, protein globules unfold and unhelix gets uncoiled and protein looses its biological activity.<\/li>\n<li style=\"clear: both;\">DNA\u00a0<span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span>\u00a0Deoxyribonucleic acid<br \/>\nRNA\u00a0<span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span>\u00a0Ribonucleic acid<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Thymine is present in DNA.<\/li>\n<li>Uracil is present in RNA.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>In [Fe(CN)<sub>6<\/sub>]<sup>3-<\/sup> complex, Fe is present as Fe<sup>3+<\/sup>.<br \/>\nConfiguration of Fe= [Ar] 3d<sup>6<\/sup>4s<sup>2<\/sup>4p<sup>0<\/sup><br \/>\nOuter configuration of Fe<sup>3+<\/sup>= 3d<sup>5<\/sup><br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/3yFo5pF.png\" alt=\"\" data-imgur-src=\"3yFo5pF.png\" \/><br \/>\nCN<sup>&#8211;<\/sup> being strong field ligand, pair up the unpaired d-electrons. Thus, two 3d-orbitals are now available for CN<sup>&#8211; <\/sup>ions.<br \/>\n<span class=\"math-tex\">{tex}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}{\/tex}<\/span>\u00a0<img decoding=\"async\" style=\"width: 200px; height: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/XeFvSTp.png\" alt=\"\" data-imgur-src=\"XeFvSTp.png\" \/><br \/>\nSince one electron remains unpaired, the complex is paramagnetic.<br \/>\nMoreover, (n -1) d-orbitals are involved in bonding. So, it is an inner orbital or low spin complex.<\/li>\n<li>The complex, [Pt(en)<sub>2<\/sub>Cl<sub>2<\/sub>]<sup>2+<\/sup> contains + two symmetrical bidentate ligands, ethylenediamine (en) and exists in two geometrical isomers, as and trans. Trans isomer being symmetrical does not show optical isomerism and hence, this isomer is optically inactive. While cis being unsymmetrical shows optical isomerism.<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 205px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/xXmDIlM.png\" alt=\"\" data-imgur-src=\"xXmDIlM.png\" \/><br \/>\nHence, the structure of geometrical isomer (i.e. trans-isomer) of the complex [Pt(en)iCl2]2+ which is optically inactive is as follows:<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/4ImgjfX.png\" alt=\"\" data-imgur-src=\"4ImgjfX.png\" \/><br \/>\ntrans [Pt(en)<sub>2<\/sub>Cl<sup>2<\/sup>]<sup>2+<\/sup> isomer optically inactive (Superimposable mirror images)<\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li style=\"list-style-type: none;\">\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Hybridization of Cr in isomer A and B is d<sup>2<\/sup>sp<sup>3<\/sup>.<\/li>\n<li>Number of unpaired electrons in Cr<sup>3+<\/sup>(3d<sup>3<\/sup>) is 3<br \/>\nMagnetic moment <span class=\"math-tex\">{tex}=\\sqrt{n(n+2)}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}=\\sqrt{3(3+2)}{\/tex}<\/span><br \/>\n= 3.87 BM<br \/>\n(deduct half mark for wrong unit\/unit not written)<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>IUPAC : hexamine cobalt (III) ion. hybridisation : d<sup>2<\/sup>sp<sup>3<\/sup><\/li>\n<li>IUPAC : tetrachloridonickelate(II)ion. hybridisation : sp<sup>3<\/sup><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li><img decoding=\"async\" style=\"width: 300px; height: 99px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/rC8ZATN.png\" alt=\"\" data-imgur-src=\"rC8ZATN.png\" \/><\/li>\n<li><img decoding=\"async\" style=\"width: 250px; height: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/30c6FwX.png\" alt=\"\" data-imgur-src=\"30c6FwX.png\" \/><\/li>\n<li><span class=\"math-tex\">{tex}\\mathop {{C_2}{H_5}C{H_2}N{H_2}}\\limits_{\\begin{array}{*{20}{l}} {\\begin{array}{*{20}{l}} {Ethyla\\min e} \\end{array}} \\end{array}} {\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\xrightarrow{HNO_2}{\/tex}<\/span>\u00a0CH<sub>3<\/sub>CH<sub>2<\/sub>OH\u00a0<span class=\"math-tex\">{tex}\\xrightarrow[K_2Cr_2O_7\/ H_2SO_4]{}{\/tex}<\/span>\u00a0CH<sub>3<\/sub>CHOOH\u00a0<span class=\"math-tex\">{tex}\\xrightarrow[\\Delta]{NH_3}{\/tex}<\/span>\u00a0CH<sub>3<\/sub>CH<sub>2<\/sub>NOH<sub>2<\/sub>\u00a0<span class=\"math-tex\">{tex}\\xrightarrow{Br_2\/KOH}{\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}\\begin{array}{l}{\\mathrm{CH}_{3} \\mathrm{NH}_{2}} \\\\ {\\text { Methylamine }}\\end{array}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><img decoding=\"async\" style=\"width: 450px; height: 185px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/ekt5uuw.png\" alt=\"\" data-imgur-src=\"ekt5uuw.png\" \/><br \/>\n<img decoding=\"async\" style=\"width: 150px; height: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/EI6C5kG.png\" alt=\"\" data-imgur-src=\"EI6C5kG.png\" \/><img decoding=\"async\" style=\"width: 100px; height: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/HQEB5eV.png\" alt=\"\" data-imgur-src=\"HQEB5eV.png\" \/><img decoding=\"async\" style=\"width: 150px; height: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/GJmPGEi.png\" alt=\"\" data-imgur-src=\"GJmPGEi.png\" \/><br \/>\nD\u00a0is\u00a0<span class=\"math-tex\">{tex}\\begin{array}{c}{\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{CH}_{2} \\mathrm{NH}_{2}} \\\\ {\\quad \\text { Benzylamine }}\\end{array}{\/tex}<\/span>\u00a0<img decoding=\"async\" style=\"width: 150px; height: 111px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/VdLfbm9.png\" alt=\"\" data-imgur-src=\"VdLfbm9.png\" \/><br \/>\n<img decoding=\"async\" style=\"width: 500px; height: 205px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/u7KCbZJ.png\" alt=\"\" data-imgur-src=\"u7KCbZJ.png\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Prepare_for_Exams_Effectively_with_myCBSEguide_App\"><\/span><strong>Prepare for Exams Effectively with myCBSEguide App<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For comprehensive exam preparation and to practice a wide range of questions, <strong>download the <a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> today. This app provides <strong>complete study material<\/strong> for <strong>CBSE<\/strong>, <strong>NCERT<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA exams<\/strong>, making it the ultimate resource for students looking to excel in their exams. With a vast collection of <strong>sample papers<\/strong>, <strong>solutions<\/strong>, and <strong>study notes<\/strong>, myCBSEguide helps students to revise, practice, and master exam concepts efficiently. Get access to <strong>Class 12 Chemistry Sample Papers 2025<\/strong> and revise topics with chapter-wise practice tests.<\/p>\n<p>In addition, <strong>teachers<\/strong> can leverage the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> app<\/strong> to create custom <strong>exam papers<\/strong> with their <strong>name<\/strong> and <strong>logo<\/strong>, offering personalized assessments tailored to their students\u2019 needs.<\/p>\n<p>Don\u2019t miss out on these valuable resources. <strong>Visit <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a><\/strong> Website and<strong> <a href=\" https:\/\/examin8.com\/\">Examin8<\/a><\/strong> Website now for effective study support and enhanced exam preparation.<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><strong> Login to our <a href=\"https:\/\/mycbseguide.com\/dashboard\/\">student dashboard<\/a>.<\/strong><\/p>\n<p style=\"text-align: center;\"><b><strong><a class=\"button\" 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This offers a wide range of <strong>sample papers<\/strong> with detailed <strong>solutions<\/strong>, along with <strong>chapter-wise test papers<\/strong> for focused practice. Additionally, you\u2019ll find <strong>NCERT solutions<\/strong>, <strong>NCERT Exemplar solutions<\/strong>, <strong>quick revision notes<\/strong>, <strong>CBSE guess papers<\/strong>, and <strong>important question papers<\/strong> to help you prepare effectively. Improve your exam preparation by practicing <strong>Class 12 Chemistry Sample Papers 2025<\/strong> available for free download on myCBSEguide.<\/p>\n<p>All of these <strong>CBSE sample papers<\/strong> are available for <strong>free download<\/strong> through the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> app<\/strong> and website, making it the best study resource for CBSE students. Download now to enhance your exam preparation and boost your performance in <strong>Class 12 board exams<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Sample paper of Chemistry Class 12 Prepare effectively for your upcoming exams with CBSE Class 12 Chemistry Sample Papers 2025, available for free download in PDF format on the myCBSEguide app and website.Download the CBSE Class 12 Chemistry Sample Papers 2025 to practice effectively for your board exams. These sample papers are designed based on &#8230; <a title=\"CBSE Class 12 Chemistry Sample Papers 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-12-chemistry\/\" aria-label=\"More on CBSE Class 12 Chemistry Sample Papers 2025\">Read 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