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(a) Why is chemistry of… (Class 12 Chemistry 2023, Q35)

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ChemistryCBSE Class 12 · Board exam 2023, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2023: question 35

Set in the CBSE Class 12 board exam of 2023, paper 56/5/1, section E, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCompetency-basedThe d- and f-Block Elements

(a) Why is chemistry of actinoids complicated as compared to lanthanoids ?
(b) Complete the following reaction and justify that it is a disproportionation reaction :
.
(c) The given graph shows the trends in melting points of transition metals :
Explain the reason why Cr has highest melting point and manganese (Mn) a lower melting point.

Figure in the paper: Graph of melting point (M.p./10^3 K, y-axis from 1 to 4) versus atomic number (x-axis) for transition metals in three series. Three curves: W (solid line, peaking near 3.6), Mo (dotted, peaking near 2.9), and a dashed line for the first series starting at Ti, rising to a peak at Cr (about 2.1), dropping sharply to a minimum at Mn (about 1.5), then rising slightly and falling. All curves fall toward about 1.2 at right. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 56/5/1.

Hints

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

Hint 1

For (a), compare how many oxidation states actinoids show, and think about their nuclear stability.

Hint 2

For (b), find the oxidation state of Mn in MnO4²⁻. Which Mn species in higher and lower oxidation states are common in acid?

Hint 3

For (c), relate melting point to the number of unpaired d-electrons available for metallic bonding.

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) Actinoids show a much wider range of oxidation states than lanthanoids, because the 5f, 6d and 7s levels are close in energy. Most actinoids are also radioactive, and many have short half-lives, so their study is difficult. Hence their chemistry is more complicated.

    1 mark

  2. (b) Completed and balanced equation:

    1 mark

  3. Justification: the oxidation state of Mn in is +6.

    • It rises to +7 in (oxidation).
    • It falls to +4 in (reduction).

    The same species is simultaneously oxidised and reduced, so this is a disproportionation reaction.

    1 mark

  4. (c) Cr () has six unpaired electrons, the maximum possible in the series. All of these can take part in interatomic metallic bonding, giving very strong metal–metal bonds and hence the highest melting point in the first series.

    1 mark

  5. Mn () has a stable half-filled configuration. Its d-electrons are held tightly and take little part in metallic bonding, so the metallic bonds are weaker and the melting point drops, as shown by the dip in the graph.

    1 mark

Answer: (a) Wide range of oxidation states and radioactivity of actinoids; (b) 3MnO4²⁻ + 4H⁺ → 2MnO4⁻ + MnO2 + 2H2O, Mn goes from +6 to +7 and +4; (c) Cr has the maximum unpaired electrons and strongest metallic bonding; Mn's stable half-filled d⁵ gives weaker metallic bonding

Where the marks are

  • Wide range of oxidation states / radioactive nature of actinoids (1 mark)
  • Products MnO4⁻ and MnO2 with correct balanced equation (1 mark)
  • Mn +6 → +7 (oxidation) and +6 → +4 (reduction) simultaneously (1 mark)
  • Cr: greater number of unpaired electrons, strong metal–metal bonding (1 mark)
  • Mn: stable half-filled d⁵, weaker metallic bonding (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 paper2023 board paper, set 56/5/1: 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.