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Answer the following… (Class 12 Chemistry 2026, Q32b)

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ChemistryCBSE Class 12 · Board exam 2026, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2026: question 32(b)

Set in the CBSE Class 12 board exam of 2026, paper 56/1/1, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerThe d- and f-Block Elements

Answer the following questions :
(i) Name a member of the lanthanoid series
(I) which exhibits +4 oxidation state
(II) which exhibits +2 oxidation state.
(ii) Why transition metals act as good catalyst ?
(iii) Why Cr has higher melting point than Mn ?
(iv) What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/1/1.

Hints

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

Hint 1

For (i), look for lanthanoid ions that reach , or configurations.

Hint 2

For (iii), compare how many unpaired d and s electrons each metal can use for metallic bonding.

Hint 3

For (iv), think about which brown solid and which gas form as Mn(+7) is reduced on standing.

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. (i)(I) Cerium (Ce) shows the +4 oxidation state. Ce has the stable noble-gas configuration .

    ½ mark

  2. (i)(II) Europium (Eu) shows the +2 oxidation state. Eu has the stable half-filled configuration . Ytterbium (Yb, ) is also accepted.

    ½ mark

  3. (ii) Transition metals act as good catalysts for these reasons:

    • They can show variable (multiple) oxidation states, so they easily form and break down unstable intermediates.
    • They can form complexes with the reactants.
    • Their large surface area adsorbs reactant molecules, which weakens the bonds in the reactants and lowers the activation energy.
    1 mark

  4. (iii) Cr () has six unpaired electrons. All of them take part in metallic bonding, so its interatomic metallic bonds are strong. Mn () has a stable half-filled subshell that holds its electrons tightly, so fewer electrons are available for metallic bonding. The weaker metallic bonding in Mn gives it a lower melting point than Cr.

    1 mark

  5. (iv) An acidified KMnO solution slowly decomposes on standing. It forms a brown precipitate of MnO and gives off oxygen:

    1 mark

  6. In this reaction Mn is reduced from +7 to +4, and oxygen is oxidised from −2 to 0. It is therefore a redox reaction (a self-decomposition in which MnO is reduced).

    1 mark

Answer: (i) (I) Cerium; (II) Europium (or Ytterbium). (ii) Variable oxidation states, complex formation and large surface area. (iii) Greater number of electrons (3d + 4s) in metallic bonding in Cr. (iv) ; redox (reduction) reaction.

Where the marks are

  • Names Ce for the +4 state (½ mark)
  • Names Eu or Yb for the +2 state (½ mark)
  • Gives variable oxidation states / complex formation / large surface area as the reason for catalytic activity (1 mark)
  • Explains that more (n−1)d and ns electrons take part in metallic bonding in Cr (1 mark)
  • Writes the balanced decomposition equation of acidified MnO (1 mark)
  • Calls it a redox/reduction reaction (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from The d- and f-Block Elements, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 56/1/1: every question, by section.

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.