{"id":172977,"date":"2022-03-14T17:32:33","date_gmt":"2022-03-14T12:02:33","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=172977"},"modified":"2022-03-19T15:02:13","modified_gmt":"2022-03-19T09:32:13","slug":"cbse-sample-papers-for-class-12-chemistry-term-2-set-1","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/cbse-sample-papers-for-class-12-chemistry-term-2-set-1\/","title":{"rendered":"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions"},"content":{"rendered":"<p>Students can access the\u00a0<a href=\"https:\/\/www.cbselabs.com\/cbse-sample-papers-class-12-chemistry\/\">CBSE Sample Papers for Class 12 Chemistry with Solutions<\/a> and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.<\/p>\n<h2>CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 1 with Solutions<\/h2>\n<p>Max. Marks: 35<br \/>\nTime: 2 Hours<\/p>\n<p><span style=\"color: #0000ff;\">General Instructions:<\/span><br \/>\nRead the following instructions carefully:<\/p>\n<ul>\n<li>There are 12 questions in this question paper with internal choice.<\/li>\n<li>Section A &#8211; Q. No. 1 to 3 are very short answer questions carrying 2 marks each.<\/li>\n<li>Section B &#8211; Q. No. 4 to 11 are short answer questions carrying 3 marks each.<\/li>\n<li>Section C- Q. No. 12 is case-based question carrying 5 marks.<\/li>\n<li>All questions are compulsory.<\/li>\n<li>Use of log tables and calculators is not allowed<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">SECTION A<\/span><\/p>\n<p>Question 1.<br \/>\nArrange the following in the increasing order of their property indicated (any 2):<br \/>\n(a) Benzoic acid, phenol, picric acid, salicylic acid (pka values).<br \/>\n(b) Acetaldehyde, acetone, methyl tert butyl ketone (reactivity towards NH<sub>2<\/sub>OH).<br \/>\n(c) Ethanol, ethanoic acid, benzoic acid (boiling point) [1 x 22]<br \/>\nAnswer:<br \/>\n(a) Picric acid p ka values (Picric acid) = 0.3<br \/>\np ka values (Salicylic acid) 2.97<br \/>\np ka values Phenol =10<br \/>\np ka values Benzoic acid = 4.20<br \/>\nThus, order is:<br \/>\nPicric acid &lt; salicylic acid &lt; Benzoic acid &lt; Phenol Methyl tert &#8211; Butyl ketone &lt; Acetone &lt; Acetaldehyde<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173082\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-1.png?resize=505%2C93&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 1\" width=\"505\" height=\"93\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-1.png?w=505&amp;ssl=1 505w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-1.png?resize=300%2C55&amp;ssl=1 300w\" sizes=\"(max-width: 505px) 100vw, 505px\" data-recalc-dims=\"1\" \/><br \/>\nWhen NH<sub>2<\/sub>OH reacts with a compound, the attacking species is a nucleophile OH- Therefore as the negative charge on compound increases, the reactivity with NH<sub>2<\/sub>OH decreases in above compounds +I effect increases and steric hindrance also increases in the same.<br \/>\n(c) Ethanol &lt;ethanoic acid &lt;benzoic acid (boiling point of carboxylic acids is higher than alcohols due<br \/>\nto extensive hydrogen bonding, boiling point increases with increase in molar mass).<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 2.<br \/>\nSolutions of two electrolytes A&#8217; and &#8216;B&#8217; are diluted. The Am of &#8216;B&#8217; increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. Graphically show the behaviour of A&#8217; and &#8216;B&#8217;. [1+1]<br \/>\nAnswer:<br \/>\nB is a strong electrolyte. The molar conductivity increases slowly with dilution as there is no increase in a number of ions on dilution as strong electrolytes are completely dissociated.<br \/>\nMolar conductivity \u03bb<sub>m<\/sub> = \\(\\frac{K}{C}\\) Where K is specific conductance C is molar concentrative of electrolyte.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173084\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-2.png?resize=258%2C227&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 2\" width=\"258\" height=\"227\" data-recalc-dims=\"1\" \/><br \/>\nElectrolyte B is strong because in B the number of ions remains same on dilution but only interionic attraction decreases, therefore, Am increases only 1.5 times while in case of weak electrolyte on dilution, number of constituent ions decreases.<\/p>\n<p>Question 3.<br \/>\nGive reasons to support the answer:<br \/>\n(a) Presence of alpha hydrogen in aldehydes and ketones is essential for aldol condensation.<br \/>\n(b) 3 -Hydroxy pentane-2-one shows positive Tollen&#8217;s test. [1&#215;2=2]<br \/>\nAnswer:<br \/>\n(a) The alpha hydrogen atoms are acidic in nature due to presence of electron-withdrawing carbonyl group. These can be easily removed by a base and the carbanion formed is resonance stabilized.<br \/>\n(b) Tollen&#8217;s reagent is a weak oxidizing agent not capable of breaking the C-C bond in ketones. Thus, ketones cannot be oxidized using Tollen&#8217;s reagent as it self gets reduced to Ag.<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">SECTION B<\/span><\/p>\n<p>Question 4.<br \/>\nAccount for the following:<br \/>\n(a) Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal conditions.<br \/>\n(b) N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature.<br \/>\n(c) Acylation of aniline is carried out in the presence of pyridine. [1&#215;3=3]<br \/>\nAnswer:<br \/>\n(a) In case of chlorobenzene, the C-Cl bond is quite difficult to break as it acquires a partial double bond character due to conjugation. So, under the normal conditions, ammonolysis of chlorobenzene does not yield aniline.<br \/>\n(b) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. Due to the presence of three hydrogen atoms, the intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Thus, butanamine boils at 350.8K while N-ethylethanamine boils at 329.3K.<\/p>\n<p>(c) During the acylation of aniline, stronger base pyridine is added. This done in order to remove the HC1 so formed during the reaction and to shift the equilibrium to the right-hand side.<br \/>\nOR<br \/>\nConvert the following:<br \/>\n(a) Phenol to N-phenylethanamide.<br \/>\n(b) Chloroethane to methenamine.<br \/>\n(c) Propanenitrile to ethanal. [1x 3=3]<br \/>\nAnswer:<br \/>\n(a) Phenol into N-phenylethylamine<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173089\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-3.png?resize=605%2C213&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 3\" width=\"605\" height=\"213\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-3.png?w=605&amp;ssl=1 605w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-3.png?resize=300%2C106&amp;ssl=1 300w\" sizes=\"(max-width: 605px) 100vw, 605px\" data-recalc-dims=\"1\" \/><br \/>\n(b) Chloroethane to methenamine<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173092\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-4.png?resize=747%2C76&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 4\" width=\"747\" height=\"76\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-4.png?w=747&amp;ssl=1 747w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-4.png?resize=300%2C31&amp;ssl=1 300w\" sizes=\"(max-width: 747px) 100vw, 747px\" data-recalc-dims=\"1\" \/><br \/>\n(c) Propanenitrile to ethanal<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173093\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-5.png?resize=525%2C108&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 5\" width=\"525\" height=\"108\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-5.png?w=525&amp;ssl=1 525w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-5.png?resize=300%2C62&amp;ssl=1 300w\" sizes=\"(max-width: 525px) 100vw, 525px\" data-recalc-dims=\"1\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 5.<br \/>\nAnswer the following questions:<br \/>\n(a) [Ni(H<sub>2<\/sub>O)<sub>6<\/sub>]<sup>2+<\/sup> (aq) is green in colour whereas [Ni(H<sub>2<\/sub>O)<sub>4<\/sub>(en)]<sup>2+<\/sup> (aq) is blue in colour, give reason in<br \/>\nsupport of your answer.<br \/>\n(b) Write the formula and hybridization of the following compound-tris(ethane-1,2_diamine) cobalt(III) sulphate. [1+2]<br \/>\nOR<br \/>\nIn a coordination entity, the electronic configuration of the central metal ion is t<sub>2<\/sub>g<sup>3<\/sup> e<sub>g<\/sub><sup>1<\/sup>.<br \/>\n(a) Is the coordination compound a high spin or low spin complex?<br \/>\n(b) Draw the crystal field splitting diagram for the above complex.<br \/>\nAnswer:<br \/>\n(a) The colour of coordination compound depends upon the type of ligand and d-d transition taking place. H<sub>2<\/sub>O is weak field ligand, which causes small splitting leading to the d-d transition corresponding to green colour, however, due to the presence of (en) which is strong field ligand, the splitting is increased. Due to the change in t<sub>2g<\/sub>&#8211; e<sub>g<\/sub> splitting, the colouration of the compound changes from green to blue.<\/p>\n<p>(b) Formula of the compound is [Co(H<sub>2<\/sub>NCH<sub>2<\/sub>CH<sub>2<\/sub>NH<sub>2<\/sub>)<sub>3<\/sub>]<sub>2<\/sub>(SO<sub>4<\/sub>)<sub>3 <\/sub>The hybridization of the compound is: d<sup>2<\/sup>sp<br \/>\nOR<br \/>\n(a) As the fourth electron enters one of the e<sub>g<\/sub> orbitais giving the configuration t<sub>2<\/sub>g<sup>3<\/sup>e<sup>1<\/sup><sub>g<\/sub>, which indicates<br \/>\n\u0394<sub>0<br \/>\n<\/sub>(b)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173096\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-6.png?resize=638%2C238&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 6\" width=\"638\" height=\"238\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-6.png?w=638&amp;ssl=1 638w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-6.png?resize=300%2C112&amp;ssl=1 300w\" sizes=\"(max-width: 638px) 100vw, 638px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nAccount for the following:<br \/>\n(a) Ti(IV) is more stable than the Ti (II) or Ti(III).<br \/>\n(b) In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number.<br \/>\n(c) Zinc is a comparatively a soft metal, iron and chromium are typically hard. [1&#215;3=31<br \/>\nAnswer:<br \/>\n(a) The electronic configuration of Ti is [Ar] 3d<sup>2<\/sup> 4s<sup>2<\/sup>. Ti (IV) is more stable than the Ti (Il) or Ti (III) because on losing 4 electrons Ti(IV) will become more stable as it will acquire the nearest noble gas configuration.<\/p>\n<p>(b) In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because as the new electron enters a d orbital each time<br \/>\nthe nuclear charge increases by unity. The shielding effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases and<br \/>\nthe ionic radius decreases.<\/p>\n<p>(c) Iron and chromium are typically hard metals because they have high enthalpy of atomization due to the presence of unpaired electrons, which accounts for their hardness. On the other hand, Zinc has low enthalpy of atomization as it has no unpaired electron. Hence zinc is comparatively a soft metal.<\/p>\n<p>Question 7.<br \/>\nAn alkene \u2018A\u2019 (Mol. formula C<sub>5<\/sub>, H<sub>10<\/sub>) on ozonolysis gives a mixture of two compounds \u2018B\u2019 and \u2018C\u2019. Compound \u2018B\u2019 gives positive Fehling\u2019s test and also forms iodoform on treatment with \u20182 and NaOH. Compound \u2018C\u2019 does not give Fehling\u2019s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C. [3]<br \/>\nAnswer:<br \/>\nCompound A is an alkene, on ozonolys\u00ees it will give carbonyl compounds. As both B and C have &gt;C = O<br \/>\ngroup, B gives positive Fehling\u2019s test so it is an aldehyde and it gives iodoform test so it has CH<sub>3<\/sub>C =O<br \/>\ngroup. This means the aldehyde is acetaldehyde C does not give Fehling\u2019s test, so it is a ketone. It gives<br \/>\npositive iodoform test so it is a methyl ketone means it has CH<sub>3<\/sub>C = O group<br \/>\nCompound A (C<sub>5<\/sub>H<sub>10<\/sub>) on ozonolysis gives B (CH<sub>3<\/sub>CHO) + C (CH<sub>3<\/sub>COR) So \u201cC\u201d is CH<sub>3<\/sub>COCH<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173098\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-7.png?resize=688%2C151&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 7\" width=\"688\" height=\"151\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-7.png?w=688&amp;ssl=1 688w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-7.png?resize=300%2C66&amp;ssl=1 300w\" sizes=\"(max-width: 688px) 100vw, 688px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 8.<br \/>\nObserve the figure given below and answer the questions that follow: [3]<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173100\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-8.png?resize=421%2C183&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 8\" width=\"421\" height=\"183\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-8.png?w=421&amp;ssl=1 421w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-8.png?resize=300%2C130&amp;ssl=1 300w\" sizes=\"(max-width: 421px) 100vw, 421px\" data-recalc-dims=\"1\" \/><br \/>\n(a) Which process is represented in the figure?<br \/>\n(b) What is the application of this process?<br \/>\n(c) Can the same process occur without applying electric field? Why is the electric field applied?<br \/>\nAnswer:<br \/>\n(a) Electrodialysis.<br \/>\n(b) Purification of colloidal solution<br \/>\n(c) Yes. Dialysis is a very slow process. To increase its speed, electric field is applied.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nWhat happens when reactions:<br \/>\n(a) N-ethylethanamine reacts with benzene sulphonyl chloride.<br \/>\n(b) Benzyl chloride is treated with ammonia followed by the reaction with Chioromethane.<br \/>\n(c) Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide. [1&#215;3]<br \/>\nOR<br \/>\n(a) Write the IUPAC name for the following organic compound:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173102\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-9.png?resize=231%2C103&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 9\" width=\"231\" height=\"103\" data-recalc-dims=\"1\" \/><br \/>\n(b) Complete the following:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173106\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-10.png?resize=646%2C52&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 10\" width=\"646\" height=\"52\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-10.png?w=646&amp;ssl=1 646w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-10.png?resize=300%2C24&amp;ssl=1 300w\" sizes=\"(max-width: 646px) 100vw, 646px\" data-recalc-dims=\"1\" \/><br \/>\n[1&#215;3]<br \/>\nAnswer:<br \/>\n(a) When N-ethylethanamine reacts with benzene sulphonyl chloride, N, N-.diethylbenzenesul sulphonamide is formed.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173110\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-11.png?resize=499%2C109&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 11\" width=\"499\" height=\"109\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-11.png?w=499&amp;ssl=1 499w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-11.png?resize=300%2C66&amp;ssl=1 300w\" sizes=\"(max-width: 499px) 100vw, 499px\" data-recalc-dims=\"1\" \/><br \/>\n(b) When benzyl chloride is treated with ammonia, Benzylamme is formed which on reaction with Chioromethane yields a secondary amine, N, N-dimethylbenzylamine.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173111\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-12.png?resize=685%2C54&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 12\" width=\"685\" height=\"54\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-12.png?w=685&amp;ssl=1 685w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-12.png?resize=300%2C24&amp;ssl=1 300w\" sizes=\"(max-width: 685px) 100vw, 685px\" data-recalc-dims=\"1\" \/><br \/>\n(c) When aniline reacts with chloroform in the presence of alcoholic potassium hydroxide, phenyl isocyanides or phenyl isonitrile is formed.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173112\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-13.png?resize=499%2C53&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 13\" width=\"499\" height=\"53\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-13.png?w=499&amp;ssl=1 499w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-13.png?resize=300%2C32&amp;ssl=1 300w\" sizes=\"(max-width: 499px) 100vw, 499px\" data-recalc-dims=\"1\" \/><br \/>\nOR<br \/>\n(a) N-Ethyl-N-methylbenzenamine or N-Ethyl-N-methylaniline<br \/>\n(b)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173114\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-14.png?resize=733%2C136&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 14\" width=\"733\" height=\"136\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-14.png?w=733&amp;ssl=1 733w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-14.png?resize=300%2C56&amp;ssl=1 300w\" sizes=\"(max-width: 733px) 100vw, 733px\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 10.<br \/>\nRepresent the cell in which the following reaction takes place. The value of E for the cell is 1.260 V. What<br \/>\nis the value of E<sub>cell<\/sub>?<br \/>\n2Al(s) + 3Cd<sup>2+<\/sup> (0.1M) \u2192 3Cd(s) + 2Al<sup>3+<\/sup> (0.01M) [3]<br \/>\nAnswer:<br \/>\nAl(s) \/Cd<sup>2+<\/sup> (0.1M) \/\/ Al (0.01M) \/Cd(s)<br \/>\n2Al(s) + 3Cd<sup>2+<\/sup> (0.1M) \u2192 3Cd (s) + 2Al<sup>3+<\/sup> (0.01M)<br \/>\nAl(s)\/Al<sup>3+<\/sup> (0.01 M)\/\/Cd<sup>2+<\/sup> (0.1 M)\/ Cd(s)<br \/>\nEcell = E<sup>0<\/sup>cell \\(\\frac{-0.059}{n} \\log \\frac{\\left[\\mathrm{Al}^{3+}\\right]^{2}}{\\left[\\mathrm{Cd}^{2+}\\right]^{3}}\\)<br \/>\nEcell = 1.26 \\(\\frac{-0.059}{6} \\log \\frac{(0.01)^{2}}{(0.1)^{3}}\\)<br \/>\n= 1.26 \\(\\frac{-0.059}{6}(-1)\\)<br \/>\n= 1.26 + 0.009 = 1.269 V<\/p>\n<p>Question 11.<br \/>\n(a) Why are fluorides of transition metals more stable in their higher oxidation state as compared to the lower oxidation state?<br \/>\n(b) Which one of the following would feel attraction when placed in magnetic field: CO<sup>2+<\/sup>,<br \/>\nAg<sup>+<\/sup> ,Ti<sup>4+<\/sup>, Zn<sup>2+<\/sup><br \/>\n(c) It has been observed that first ionization energy of 5d series of transition elements are higher than<br \/>\nthat of 3d and 4d series, explain why? [1 x 3 = 3]<br \/>\nOR<br \/>\nOn the basis of the figure given below, answer the following questions:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-173116\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Chemistry-Term-2-Set-1-with-Solutions-15.png?resize=295%2C271&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions 15\" width=\"295\" height=\"271\" data-recalc-dims=\"1\" \/><\/p>\n<p>(a) Why manganese has lower melting point than Chromium?<br \/>\n(b) Why do transition metals of 3d series have lower melting points as compared to 4d series?<br \/>\n(c) In the third transition series, identify and name the metal with the highest melting point [1&#215;3 = 3]<br \/>\nAnswer:<br \/>\n(a) The ability of fluorine to stabilize the highest oxidation state is attributed to the higher lattice energy or high bond enthalpy.<br \/>\n(b) Co<sup>2+<\/sup> has three unpaired electrons so it would be paramagnetic in nature, hence Co<sup>2+<\/sup> ion would be attracted to magnetic field.<br \/>\nThe transition elements of 5d series have intervening 4f orbitals. There is greater effective nuclear charge acting on outer valence electrons due to the weak shielding by 4f electrons.<br \/>\nHence, first ionisation energy of 5d series of transition elements are higher than that of 3d and 4d series.<br \/>\nOR<br \/>\nManganese has lower melting point as compared to chromium, because manganese has 5 free electrons in its d orbital whereas chromium has 6 free electrons in its d orbital. More the number of electrons, more it will take part in the metallic bonding and thus more energy is required in breaking the bonds and thus the melting point will be high. Therefore, manganese has lower melting point than chromium.<\/p>\n<p>There is much more frequent metal-metal bonding in compounds of the heavy transition metals i.e 4d and 5d series, which accounts for lower melting point of 3d series. Tungsten due to metallic bonding in tungsten. (M.P. = 3422\u00b0C).<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">SECTION C<\/span><\/p>\n<p>Question 12.<br \/>\nRead the passage given below and answer the questions that follow: Are there nuclear reactions going on in our bodies? There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix.<\/p>\n<p>The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1:1 for small nuclei (Z &lt; 20).<\/p>\n<p>Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food, you eat, and yes, in your body.<\/p>\n<p>The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive.<\/p>\n<p>In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaging chemicals.<\/p>\n<p>The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms.<\/p>\n<p>Slowly the carbon-14 atoms decay to nitrogen without being replaced so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first-order kinetics. It has a half-life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archaeologists can calculate when the person died.<\/p>\n<p>All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years.<\/p>\n<p>This is all because of the fact that nuclear reactions naturally occur in living organisms.<br \/>\n(a) Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6)<br \/>\n(b) Researchers have uncovered the youngest known dinosaur bone, dating around 65 million years ago. How was the age of this fossil estimated?<br \/>\n(c) Which are the two most common radioactive decays happening in human body?<br \/>\n(d) Suppose an organism has 20 g of Carbon -14 at its time of death. Approximately how much Carbon -14 remains after 10,320 years? (Given antilog 0.517 = 3.289)<br \/>\nOR<br \/>\n(d) Approximately how old is a fossil with 12 g of Carbon -14 if it initially possessed 32 g of Carbon -14? (Given log 2.667 = 0.4260) [1+1+1+2]<br \/>\nAnswer:<br \/>\n(a) Carbon-12 is stable because it contains 6 neutrons and 6 protons in its nucleus, due to this it never undergoes radioactive decay. On the other hand, Carbon-14 is unstable because it contains 6 protons and neutrons in its nucleus which makes it unstable and therefore, it undergoes radioactive decay with a half-life of about 5,730 years.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>(b) Age of fossils can be estimated by C-14 decay. All living organisms have C-14 which decays without being replaced back once the organism dies.<\/p>\n<p>(c) The most common natural radioactive isotopes in humans are carbon-14 which decays to stable nitrogen atom and potassium-40 which decays to stable calcium atom. Chemically, these isotopes behave exactly like stable carbon and potassium.<\/p>\n<p>(d) t = 2.303\/ k log (C<sub>0<\/sub>\/C<sub>t<\/sub>)<br \/>\nC<sub>0<\/sub> = 20<sub>g<\/sub>C<sub>t<\/sub> = ?<br \/>\nt = 10320 years k = 0.693\/6000 (half-life given in passage) substituting in equation:<br \/>\n10320 = 2.303 \/ (0.693\/6000) log 20\/ C<sub>t<\/sub><br \/>\n0.517 = log 20 \/ C<sub>t<\/sub> antilog (0.517) = 20\/C<sub>t<\/sub><br \/>\n3.289 = 20\/C<sub>t<\/sub><br \/>\nC<sub>t<\/sub>=6.17g<br \/>\nOR<br \/>\nt = 2.303\/ k log (C<sub>0<\/sub>\/C<sub>t<\/sub>)<br \/>\nC<sub>0<\/sub> = 32 g C<sub>t<\/sub> = 12<br \/>\nt = ? k = 0.693\/6000 (half life given in passage) substituting in equation:<br \/>\nt = 2.303 \/ (0.693\/6000) log 32\/12<br \/>\nt = 2.303 x (6000) \/0.693 log 2.667<br \/>\nt = 2.303 x 6000 x 0.4260 \/0.693<br \/>\n= 8494 years<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can access the\u00a0CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam. CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 1 with Solutions Max. Marks: 35 Time: 2 Hours General Instructions: Read the following &hellip;<\/p>\n","protected":false},"author":29,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[84914],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.2 (Yoast SEO v17.2) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions - CBSE Labs<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.cbselabs.com\/cbse-sample-papers-for-class-12-chemistry-term-2-set-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CBSE Sample Papers for Class 12 Chemistry Term 2 Set 1 with Solutions\" \/>\n<meta property=\"og:description\" content=\"Students can access the\u00a0CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam. 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CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 1 with Solutions Max. 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