CBSE Class 12 Chemistry Board Paper 5 — MCQs with Answers

8 solved questions from this paper

Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.

1. Q1Easy

   The rate of a reaction usually increases sharply with a rise in temperature mainly because:

   A.the fraction of molecules with energy equal to or greater than the activation energy increases sharply✓ Correct

   B.the activation energy of the reaction falls as the temperature rises

   C.the enthalpy change of the reaction becomes more negative

   D.the number of collisions per second becomes about a hundred times larger

   Answer: A. the fraction of molecules with energy equal to or greater than the activation energy increases sharply

   Explanation: Raising the temperature shifts the Maxwell–Boltzmann distribution so that many more molecules cross the activation energy threshold; near room temperature a 10 K rise roughly doubles that fraction. Activation energy is a property of the reaction path and is not changed by heating, and collision frequency rises only with the square root of temperature — far too little to account for the doubling of rate.

2. Q2Easy

   The IUPAC name of CH₃CH(Cl)CH₂CH₃ is:

   A.2-chlorobutane✓ Correct

   B.3-chlorobutane

   C.1-chloro-1-methylpropane

   D.2-chloro-2-methylpropane

   Answer: A. 2-chlorobutane

   Explanation: The longest chain has four carbons, so the parent is butane, and numbering is begun from the end that gives the substituent the lower locant — from the left the chlorine is on C2, from the right it would be on C3. Hence 2-chlorobutane. The name 1-chloro-1-methylpropane describes the same molecule but chooses a three-carbon chain, which breaks the rule of selecting the longest chain.

3. Q3Easy

   When an orange solution of potassium dichromate is made alkaline it turns yellow. The species responsible for the yellow colour is:

   A.Cr³⁺

   B.Cr₂O₇²⁻

   C.CrO₄²⁻✓ Correct

   D.CrO₂Cl₂

   Answer: C. CrO₄²⁻

   Explanation: Adding alkali shifts the equilibrium Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O to the right, and the yellow chromate ion is what is seen. Chromium stays in the +6 state throughout, so this is not a redox change; the green Cr³⁺ appears only when dichromate is REDUCED by a reagent such as Fe²⁺ in acid.

4. Q4Medium

   Which of the following pairs of liquids shows POSITIVE deviation from Raoult's law?

   A.chloroform and acetone

   B.phenol and aniline

   C.ethanol and acetone✓ Correct

   D.nitric acid and water

   Answer: C. ethanol and acetone

   Explanation: Positive deviation occurs when the new A–B attractions are WEAKER than the A–A and B–B attractions in the pure liquids, so more molecules escape and the vapour pressure exceeds the Raoult value. Adding acetone to ethanol breaks up the ethanol hydrogen-bonded network without replacing it with anything as strong. The other three pairs all form strong new hydrogen bonds between unlike molecules and therefore deviate negatively.

5. Q5Medium

   The number of geometrical isomers possible for the octahedral complex ion [Co(NH₃)₄Cl₂]⁺ is:

   A.3

   B.2✓ Correct

   C.4

   D.1

   Answer: B. 2

   Explanation: In an octahedron the two chloride ligands can be either adjacent, at 90° to each other, or opposite, at 180°, giving just the cis and trans forms. A third arrangement is not possible because all remaining positions are equivalent once the two chlorides are placed. Three isomers would arise for a complex of the type Ma₃b₃, which allows facial and meridional forms, not for Ma₄b₂.

6. Q6Medium

   During the discharge of a lead storage battery, the reaction taking place at the anode is:

   A.PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)

   B.PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq)

   C.Pb²⁺(aq) + 2e⁻ → Pb(s)

   D.Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻✓ Correct

   Answer: D. Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻

   Explanation: Oxidation always occurs at the anode, so the anode reaction must RELEASE electrons. The spongy lead plate is oxidised to lead(II) sulphate, giving up two electrons per lead atom. The half-equation involving PbO₂ consumes electrons and is the cathode reaction, while the reduction of PbSO₄ back to lead is what happens at that electrode during recharging, not discharge.

7. Q7Hard

   Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V, the standard Gibbs energy change for 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s), with F = 96500 C mol⁻¹, is:

   A.−22.2 kJ mol⁻¹

   B.−44.4 kJ mol⁻¹✓ Correct

   C.+44.4 kJ mol⁻¹

   D.−88.8 kJ mol⁻¹

   Answer: B. −44.4 kJ mol⁻¹

   Explanation: Fe³⁺ has the higher reduction potential, so it is reduced and iodide is oxidised: E°cell = 0.77 − 0.54 = 0.23 V. The equation as written transfers two electrons, so ΔG° = −nFE°cell = −2 × 96500 × 0.23 = −4.44 × 10⁴ J mol⁻¹. Taking n = 1 halves the magnitude to 22.2 kJ mol⁻¹, and a positive sign would contradict the fact that a positive cell potential means a spontaneous reaction.

8. Q8Hard

   For the reaction A + B → products, the rate doubles when [A] is doubled at constant [B], and becomes four times when [B] is doubled at constant [A]. The rate law is:

   A.rate = k[A]²[B]

   B.rate = k[A][B]

   C.rate = k[A][B]²✓ Correct

   D.rate = k[A]²[B]²

   Answer: C. rate = k[A][B]²

   Explanation: Doubling a concentration multiplies the rate by 2 raised to the order in that species. A factor of 2 for A means first order in A, while a factor of 4 = 2² for B means second order in B, giving rate = k[A][B]² and an overall order of three. Writing [A]²[B] reverses the two exponents and would predict a fourfold rise on doubling A, which the data contradict.

More Class 12 Chemistry board papers

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• Board Paper 5 — you are here

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