(i) What is meant by lanthanoid contraction ? (ii) Why do transition metals form coloured compounds ? (iii) Why are E^oM²+/M values for Mn and Zn more negative than expected ? (iv) Which is the most stable oxidation state of Cu and why ? (v) Why is Ce^4+ in aqueous solution a good oxidising agent ?

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

CBSE Class 12 Chemistry 2026: question 32(b)

Set in the CBSE Class 12 board exam of 2026, paper 56/5/1, 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

(i) What is meant by lanthanoid contraction ?
(ii) Why do transition metals form coloured compounds ?
(iii) Why are $E_{M^{2 +} / M}^{o}$ values for Mn and Zn more negative than expected ?
(iv) Which is the most stable oxidation state of Cu and why ?
(v) Why is $C e^{4 +}$ in aqueous solution a good oxidising agent ?

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/5/1.

Hints

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

Hint 1

Think about how poorly 4f electrons shield the nucleus, and what the configurations of Mn²⁺ and Zn²⁺ look like.

Hint 2

For colour, think about electron jumps between split d orbitals. For copper, compare the hydration energies of Cu⁺ and Cu²⁺.

Hint 3

For Ce⁴⁺, ask which oxidation state is most common for lanthanoids and whether Ce⁴⁺ wants to gain or lose an electron.

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.

(i) Lanthanoid contraction is the steady decrease in the atomic and ionic radii ( $M^{3 +}$ ) of the lanthanoids from La to Lu as atomic number increases. It happens because electrons enter the 4f orbitals before 5d. The 4f electrons shield one another poorly from the rising nuclear charge, so the effective nuclear charge increases and the size shrinks.
1 mark
(ii) Most transition metal ions have partly filled d orbitals with unpaired electrons. In a compound the ligands split the d orbitals into sets of slightly different energy. An electron absorbs visible light and jumps from a lower to a higher d level (a d–d transition). The complementary colour of the absorbed light is the colour we see.
1 mark
(iii) Mn: $M n^{2 +}$ is $3 d^{5}$ , a half-filled configuration with extra stability. This favours forming $M n^{2 +}$ from Mn, so $E_{M n^{2 +} / M n}^{o}$ ( $- 1.18$ V) is more negative than the trend predicts.
½ mark
Zn: $Z n^{2 +}$ is $3 d^{10}$ , a completely filled and stable configuration. This also favours oxidation of Zn, so $E_{Z n^{2 +} / Z n}^{o}$ ( $- 0.76$ V) is more negative than expected.
½ mark
(iv) The most stable oxidation state of copper (in aqueous solution) is +2, i.e. $C u^{2 +}$ .
½ mark
Reason: $C u^{2 +}$ is small and highly charged, so its hydration enthalpy ( $Δ_{h y d} H$ ) is much more negative than that of $C u^{+}$ . This more than makes up for the second ionisation enthalpy of Cu. Hence $C u^{+}$ disproportionates in water: $2 C u^{+} \to C u^{2 +} + C u$ .
½ mark
(v) The common and most stable oxidation state of cerium (and of all lanthanoids) is +3. $C e^{4 +}$ therefore readily accepts an electron to revert to $C e^{3 +}$ : $C e^{4 +} + e^{-} \to C e^{3 +}$ , $E^{o} = + 1.74$ V. Because it is so easily reduced, $C e^{4 +}$ is a good oxidising agent.
1 mark

Answer: (i) Gradual decrease in size of lanthanoids from La to Lu due to poor shielding by 4f electrons; (ii) d–d transitions of unpaired electrons absorbing visible light; (iii) Mn²⁺ is half-filled d⁵ and Zn²⁺ is fully filled d¹⁰, both extra stable; (iv) +2 (Cu²⁺), because of its high hydration enthalpy; (v) Ce⁴⁺ readily changes to the stable +3 state.

Where the marks are

Defines lanthanoid contraction as the regular decrease in atomic/ionic radii from La to Lu due to filling of 4f orbitals (1 mark)
Explains colour by d–d transition of unpaired electrons in partly filled d orbitals (1 mark)
Gives half-filled d⁵ stability of Mn²⁺ (½ mark) and fully filled d¹⁰ stability of Zn²⁺ (½ mark)
States Cu²⁺ as the most stable state (½ mark) with reason: high negative hydration enthalpy (½ mark)
States Ce⁴⁺ tends to revert to the common +3 state, so it is easily reduced (1 mark)

Revise the chapter

Chapter: The d- and f-Block Elements (notes, formulas, where marks are lost and every past question)
Topic: Magnetic Properties and Colour of Transition Metal Ions

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 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2026: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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