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For the topic of chemistry, see the marking scheme and CBSE Sample paper questions with solutions below.
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CBSE Term 2 Class 12 Chemistry Sample Paper: Marking Scheme & Paper Pattern
There would be 12 questions in the question paper with internal choice for students.
The sections would be divided as follows:
SECTION A - would be very short answer questions carrying 2 marks each.
SECTION B - would be short answer questions carrying 3 marks each.
SECTION C- would be a case-based question carrying 5 marks.
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CBSE Term 2 Class 12th Chemistry Sample Paper With Answers:
Answer the following set of questions and find the correct answers below.
1. Arrange the following in the increasing order of their property indicated:
a. Benzoic acid, Phenol, Picric acid, Salicylic acid (PKA values).
b. Acetaldehyde, Acetone, Methyl tert butyl ketone (reactivity towards NH2OH).
c. Ethanol, ethanoic acid, benzoic acid (boiling point)
Ans. (a)Picric acid - salicylic acid - benzoic acid
(b)Methyl tert – butyl ketone - acetone- Acetaldehyde
2. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ 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’ and ‘B’.
Ans. B is a strong electrolyte. The molar conductivity increases slowly with dilution as there is no increase in the number of ions on dilution as strong electrolytes are completely dissociated. Draw the graph to support the answer.
Give reasons to support the answer:
Presence of Alpha hydrogen in aldehydes and ketones is essential for aldol condensation.
b. 3 –Hydroxy pentan-2-one shows positive Tollen’s test.
Ans. (a) The alpha hydrogen atoms are acidic in nature due to the presence of the electron-withdrawing carbonyl group. These can be easily removed by a base and the carbanion formed is resonance stabilized.
(b) Tollen’s reagent is a weak oxidizing agent not capable of breaking the C-C bond in ketones. Thus ketones cannot be oxidized using Tollen’s reagent itself gets reduced to Ag.
3. Account for the following:
a. Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal conditions.
b. N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature.
c. Acylation of aniline is carried out in the presence of pyridine.
Ans. a) In the 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 normal conditions, ammonolysis of chlorobenzene does not yield aniline.
b) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between the 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.
c) During the acylation of aniline, stronger base pyridine is added. This is done in order to remove the HCl so formed during the reaction and to shift the equilibrium to the right-hand side.
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