CBSE GRADE 12 CHEMISTRY SET 1 Solutions & Answer Key

Q1 short answer 1 mark

Analysis shows that FeO has a non-stoichiometric composition with formula Fe0·95O. Give reason.

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FeO shows non-stoichiometric composition due to metal deficiency defect. Some Fe2+ ions are missing from the crystal lattice, and to maintain electrical neutrality, an equivalent number of Fe2+ ions are oxidized to Fe3+ ions, resulting in the formula Fe0.95O.

Q2 short answer 1 mark

CO (g) and H2 (g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions ?

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This reaction demonstrates the selectivity of a catalyst. Catalysts are highly selective in nature, meaning a specific catalyst can direct a reaction to yield a particular product from a given set of reactants.

Q3 short answer 1 mark

Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2].

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Coordination number of Pt is 6 (each 'en' is bidentate, providing 2 bonds x 2 = 4, plus 2 Cl- ligands). Oxidation state: Let x be the charge on Pt. x + 2(0) + 2(-1) = 0 => x = +2. So, coordination number is 6 and oxidation state is +2.

Q4 short answer 1 mark

Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why ?

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Benzyl chloride is more easily hydrolysed. In benzyl chloride, the C-Cl bond is weaker due to the formation of a stable benzyl carbocation intermediate. In chlorobenzene, the C-Cl bond acquires partial double bond character due to resonance, making it difficult to break.

Q5 short answer 1 mark

Write the IUPAC name of the following : CH3 - C - CH - CH3 | | C2H5 OH

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The IUPAC name is 3-methylpentan-2-ol. The longest chain contains 5 carbons, with an ethyl group at position 3 and a hydroxyl group at position 2.

Q6 short answer 2 marks

Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1·86 K kg mol–1)

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Moles of glucose = 60/180 = 1/3 mol. Molality (m) = (1/3) / 0.250 kg = 1.333 m. ΔTf = Kf * m = 1.86 * 1.333 = 2.48 K. Freezing point of solution = 273.15 - 2.48 = 270.67 K.

Q7 short answer 2 marks

For the reaction 2N2O5 (g)  4NO2 (g) + O2 (g), the rate of formation of NO2 (g) is 2·8  10–3 M s–1. Calculate the rate of disappearance of N2O5 (g).

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Rate = -(1/2) * d[N2O5]/dt = (1/4) * d[NO2]/dt. Given d[NO2]/dt = 2.8 * 10^-3 M s^-1. So, -(1/2) * d[N2O5]/dt = (1/4) * (2.8 * 10^-3). Therefore, rate of disappearance of N2O5 = -(d[N2O5]/dt) = (2/4) * 2.8 * 10^-3 = 1.4 * 10^-3 M s^-1.

Q8 short answer 2 marks

Among the hydrides of Group-15 elements, which have the (a) lowest boiling point ? (b) maximum basic character ? (c) highest bond angle ? (d) maximum reducing character ?

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(a) PH3 has the lowest boiling point due to lack of hydrogen bonding. (b) NH3 has the maximum basic character due to the small size of N and high electron density on the lone pair. (c) NH3 has the highest bond angle (107°). (d) BiH3 has the maximum reducing character due to the longest and weakest M-H bond.

Q9 short answer 2 marks

How do you convert the following ? (a) Ethanal to Propanone (b) Toluene to Benzoic acid OR Account for the following : (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction. (b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.

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(a) Ethanal to Propanone: Ethanal + CH3MgBr -> Propan-2-ol (on hydrolysis); Propan-2-ol + [O] -> Propanone. (b) Toluene to Benzoic acid: Toluene + alkaline KMnO4/H3O+ -> Benzoic acid. OR (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the -COOH group is strongly deactivating and the lone pair on the oxygen atoms of the carboxyl group attaches to the Lewis acid (catalyst), making the ring further electron-deficient. (b) 4-nitrobenzoic acid has a lower pKa because the -NO2 group is electron-withdrawing, which stabilizes the carboxylate ion conjugate base via the inductive and resonance effects, making the acid stronger.

Q10 short answer 2 marks

Complete and balance the following chemical equations : (a) Fe2+ + MnO4- + H+  (b) MnO4- + H2O + I– 

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(a) 5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O. (b) 2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3-.

Q11 short answer 3 marks

Give reasons for the following : (a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers. (b) Aquatic animals are more comfortable in cold water than in warm water. (c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.