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ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2024-25: question 31(a)

From the official CBSE sample question paper for 2024-25, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerThe d- and f-Block Elements

a. A purple colour compound A, which is a strong oxidising agent and used for bleaching of wool, cotton, silk and other textile fibres was added to each of the three test tubes along with . It was followed by strong heating.

In which of the above test tubes; A,B or C:
(i) Violet vapours will be formed (1)
(ii) The bubbles of gas evolved will extinguish a burning matchstick. Write an equation for each of the above observations. (1)

b. A metal ion of the first transition series having configuration combines with three didentate ligands. Assuming :
(i) Draw the crystal field energy level diagram for the 3d orbital of this complex. (1)
(ii) What is the hybridisation of in this complex and why? (1)
(iii) Name the type of isomerism exhibited by this complex. (1)

Figure in the paper: Three test tubes labelled A, B, C containing: A - oxalic acid, B - PbS, C - KI (solid). See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Identify the purple oxidising agent first, then think which reagent gives violet vapours and which gives a non-supporter of combustion.

Hint 2

For the complex, three didentate ligands give six donor atoms. What geometry results?

Hint 3

If Δ₀ < P, do the electrons pair in t₂g or occupy e_g? What does that leave for hybridisation?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. A is (purple, strong oxidant, used to bleach textiles).
    a(i) Violet vapours () form in test tube C (KI).

    ½ mark
  2. ½ mark
  3. a(ii) The gas that puts out a burning matchstick is , formed in test tube A (oxalic acid).

    ½ mark
  4. ½ mark
  5. b(i) The complex is octahedral and high spin, since . The diagram shows the five degenerate 3d orbitals of the free ion split into: a lower set () at with one electron each, and an upper set () at with one electron each. The gap between them is . Configuration: , 5 unpaired electrons.

    1 mark
  6. b(ii) Hybridisation: **** (outer-orbital complex). Because , the five 3d electrons do not pair and occupy all five 3d orbitals. No inner 3d orbitals are free, so the metal uses and orbitals for the six donor atoms.

    1 mark
  7. b(iii) Optical isomerism. exists as two non-superimposable mirror images (d and l forms).

    1 mark

Answer: a(i) Test tube C: 10I⁻ + 2MnO₄⁻ + 16H⁺ → 5I₂ + 2Mn²⁺ + 8H₂O; a(ii) Test tube A: 5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 10CO₂ + 2Mn²⁺ + 8H₂O; b(i) t₂g³e_g² high-spin splitting; b(ii) sp³d² outer orbital; b(iii) optical isomerism

Where the marks are

  • Test tube C with balanced equation for I₂ formation (1 mark)
  • Test tube A with balanced equation for CO₂ formation (1 mark)
  • Octahedral splitting diagram with t₂g³e_g² (1 mark)
  • sp³d² with reason: no pairing since Δ₀ < P (1 mark)
  • Optical isomerism (1 mark)

Revise the chapter

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.