Class 12 Term 2 Chemistry Sample Paper
Chemistry Class 12 Term 2 Sample Paper: CBSE will conduct the Class 12 Term 2 Chemistry exam on 07th May 2022 (Saturday). Now the students must be buckling up to begin their revision for Chemistry Exam and solving the Class 12 Chemistry Sample Paper could be the best way to boos up yourself and enhance your confidence for the real examination. Are you searching for the Class 12 Chemistry Sample Paper to practice for your Chemistry Exam as per the new CBSE Exam Pattern? Then you are in the right place, as here we have provided the CBSE Class 12 Term 2 Chemistry Sample Paper that was issued by the CBSE on their official website as per the latest exam pattern.
Chemistry Sample Paper Class 12 Term 2 Exam
Central Board of Secondary Education has divided the Class 12 Chemistry Board Exam into two Terms. In Term 2 the exam is divided into three sections with 12 questions for which the exam pattern that has been followed is as follows-
1. Section A consists of 1 to 3 questions carrying 2 marks each [Very Short Answer Questions].
2. Section B consists of 4 to 11 questions carrying 3 marks each [Short Answer Questions].
3. Section C consists of 1 question carrying 5 marks [Case Based Question].
Chemistry Class 12 Term 2 Sample Paper PDF
The students who are doing their revision for the upcoming CBSE Class 12 Chemistry Exam must download the official sample paper PDF along with its solution from the below link. With this CBSE Class 12 Chemistry Sample Paper, it would be easier for the students to attempt the types of question that is going to be asked by CBSE in the Term 2 Board Exam. The best way is to solve the sample just like you are sitting in the examination hall and then analyse your attempt with the solution provided by CBSE and the link for the same is given below.
Chemistry Sample Paper Class 12 Term 2 with Solutions
Practice with the sample questions released by CBSE in Chemistry Sample Paper Class 12 Term 2 from the below section and prepare yourself for the examination.
Question 1- Arrange the following in the increasing order of their property indicated-
(a) Benzoic acid, Phenol, Picric acid, Salicylic acid (pka values).
Solution- Picric acid < salicylic acid < benzoic acid
(b) Acetaldehyde, Acetone, Methyl tert butyl ketone (reactivity towards NH2OH)
Solution- Methyl tert – butyl ketone < acetone< Acetaldehyde
(c) ethanol, ethanoic acid, benzoic acid (boiling point)
Solution- ethanol < ethanoic acid < benzoic acid (boiling point of carboxylic acids is higher than alcohols due to extensive hydrogen bonding, boiling point increases with increase in molar mass)
Question 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 behavior of ‘A’ and ‘B’.
Solution- B is a strong electrolyte. The molar conductivity increases slowly with dilution as there is no increase in number of ions on dilution as strong electrolytes are completely dissociated.
Question 3- Give reasons to support the answer:
(a) Presence of Alpha hydrogen in aldehydes and ketones is essential for aldol condensation
Solution- 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) 3 –Hydroxy pentane-2-one shows positive Tollen’s test.
Solution- 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.
Question 4- Account for the following:
(a) Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal conditions.
Solution- 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 the normal conditions, ammonolysis of chlorobenzene does not yield aniline.
(b) N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature.
Solution- 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.
(c) Acylation of aniline is carried out in the presence of pyridine.
Solution- During the acylation of aniline, stronger base pyridine is added. This done in order to remove the HCl so formed during the reaction and to shift the equilibrium to the right hand side.
Question 4- Convert the following:
(a) Phenol to N-phenylethanamide.
(b) Chloroethane to methenamine.
(c) Propanenitrile to ethanol.
Question 5- Answer the following questions:
(a) [Ni(H2O)6 ]²+ (aq) is green in colour whereas [Ni(H2O)4 (en)]²+ (aq)is blue in colour, give a reason in support of your answer.
Solution- The colour of the coordination compound depends upon the type of ligand and dd transition taking place. H2O is a weak field ligand, which causes small splitting, leading to the d-d transition corresponding green colour, however, due to the presence of ( en ) which is a strong field ligand, the splitting is increased. Due to the change in t2g -eg splitting the colouration of the compound changes from green to blue.
(b) Write the formula and hybridization of the following compound: tris(ethane-1,2–diamine) cobalt(III) sulphate
Solution- )Formula of the compound is [Co(H2NCH2CH2NH2 )3 ]2 (SO4 )3
The hybridisation of the compound is: d²sp³
Question 5- In a coordination entity, the electronic configuration of the central metal ion is t2g³ e^1
(a) Is the coordination compound a high spin or low spin complex?
Solution- As the fourth electron enters one of the eg orbitals giving the configuration t2g³ e^1 ,which indicates ∆o < P hence forms high spin complex.
(b) Draw the crystal field splitting diagram for the above complex.
Question 6- Account for the following:
(a) Ti(IV) is more stable than Ti (II) or Ti(III).
Solution- (a) Ti is having electronic configuration [Ar] 3d² 4s² . Ti (IV) is more stable as Ti^4+ acquires nearest noble gas configuration on loss of 4 e-.
(b) In the case of transition elements, ions of the same charge in a given series show a progressive decrease in radius with increasing atomic number.
Solution- In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. As the new electron enters a d orbital each time the 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 the ionic radius decreases.
(c) Zinc is a comparatively soft metal, iron and chromium are typically hard.
Solution- Iron and Chromium are having high enthalpy of atomization due to the presence of unpaired electrons, which accounts for their hardness. However, Zinc has low enthalpy of atomization as it has no unpaired electron. Hence zinc is comparatively a soft metal.
Question 7- An alkene ‘A’ (Mol. formula C5H10) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms an iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C.
Solution- Compound A is an alkene, on ozonolysis, it will give carbonyl compounds. As both B and C have >C=O group, B gives positive Fehling’s test so it is an aldehyde and it gives iodoform test so it is so it has CH3C=O group. This means the aldehyde is acetaldehyde C does not give Fehling’s test, so it is a ketone. It gives a positive iodoform test so it is a methyl ketone means it has CH3C=O group.
Compound A (C5H10) on ozonolysis gives B (CH3CHO) + C (CH3COR)
So “C” is CH3COCH3
Question 8- Observe the figure given below and answer the questions that follow:
(a) Which process is represented in the figure?
(b) What is the application of this process?
Solution- purification of colloidal solution
(c) Can the same process occur without applying an electric field? Why is the electric field applied?
Solution- Yes. Dialysis is a very slow process to increase its speed electric field is applied.
Question 9- What happens when reactions:
(a) N-ethylethanamine reacts with benzenesulphonyl chloride.
Solution- When N-ethylethanamine reacts with benzenesulphonyl chloride , N,N-diethylbenzenesulphonamide is formed.
(b) Benzylchloride is treated with ammonia followed by the reaction with Chloromethane.
Solution- When benzylchloride is treated with ammonia , Benzylamine is formed which on reation with Chloromethane yields a secondary amine , N-methylbenzylamine .
(c) Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide
Solution- When aniline reacts with chloroform in the presence of alcoholic potassium hydroxide , phenyl isocyanides or phenyl isonitrile is formed.
Question 9- (a) Write the IUPAC name for the following organic compound:
Solution- N-Ethyl-N-methylbenzenamine or N-Ethyl-N-ethylaniline
(b) Complete the following