Attempt any five of the following : (a) Ce(III) is easily oxidised to Ce(IV). Comment. (b) E°(Mn²+/Mn) is –1.18 V. Why is this value highly negative in comparison to neighbouring d block elements ? (c) Which element of 3d series has lowest enthalpy of atomisation and why ? (d) What happens when sodium chromate is acidified ? (e) Zn, Cd and Hg are soft metals. Why ? (f) Why is permanganate titration not carried out in the presence of HCl ? (g) The lower oxides of transition metals are basic whereas the highest are amphoteric/acidic. Give reason.

ChemistryCBSE Class 12 · Board exam 2024, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2024: question 31

Set in the CBSE Class 12 board exam of 2024, paper 56/5/1, section E, question 31, for 5 marks. Try it first; the hints and our step-by-step solution are below.

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

Attempt any five of the following :

(a)
Ce(III) is easily oxidised to Ce(IV). Comment.
1 mark
(b)
$E^{\circ} (M n^{2 +} /Mn)$ is –1.18 V. Why is this value highly negative in comparison to neighbouring d block elements ?
1 mark
(c)
Which element of 3d series has lowest enthalpy of atomisation and why ?
1 mark
(d)
What happens when sodium chromate is acidified ?
1 mark
(e)
Zn, Cd and Hg are soft metals. Why ?
1 mark
(f)
Why is permanganate titration not carried out in the presence of HCl ?
1 mark
(g)
The lower oxides of transition metals are basic whereas the highest are amphoteric/acidic. Give reason.
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/5/1.

Hints

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

Hint 1

Look at the electron configurations: which ions reach f⁰, d⁵ or d¹⁰?

Hint 2

Unpaired d electrons strengthen metallic bonding; a higher oxidation state increases covalent character.

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.

(a) Ce(III) is $4 f^{1}$ , while Ce(IV) is $4 f^{0}$ , the stable noble gas configuration. Ce(III) therefore readily loses one more electron to become Ce(IV).
1 mark
(b) $M n^{2 +}$ has a half-filled $3 d^{5}$ configuration with extra stability, so reducing $M n^{2 +}$ to Mn is difficult. Hence $E^{\circ}$ is much more negative than for its neighbours.
1 mark
(c) Zinc has the lowest enthalpy of atomisation. Its $3 d^{10}$ orbitals are completely filled, so no unpaired d electrons take part in metallic bonding and the metallic bonding is weak.
1 mark
(d) Acidified sodium chromate changes to sodium dichromate, and the colour changes from yellow to orange:
$2 C r O_{4}^{2 -} + 2 H^{+} \to C r_{2} O_{7}^{2 -} + H_{2} O$
1 mark
(e) Zn, Cd and Hg have completely filled $(n - 1) d^{10}$ orbitals with no unpaired electrons, so metallic bonding is weak and the metals are soft.
1 mark
(f) $K M n O_{4}$ oxidises HCl to $C l_{2}$ , so part of the oxidant is used up by the acid and the titration gives wrong results:
$2 M n O_{4}^{-} + 10 C l^{-} + 16 H^{+} \to 2 M n^{2 +} + 5 C l_{2} + 8 H_{2} O$
(This is an additional part; any five of (a)–(g) earn 5 marks.)
(g) In lower oxidation states the metal oxides are ionic, so they are basic (e.g. MnO). As the oxidation state rises, ionic character falls and covalent character increases, so the highest oxides are amphoteric or acidic (e.g. $M n_{2} O_{7}$ is acidic). (Additional part.)

Answer: (a) Ce⁴⁺ has stable 4f⁰; (b) stable d⁵ Mn²⁺; (c) Zn – filled d¹⁰, weak metallic bonding; (d) forms dichromate, yellow → orange; (e) filled d orbitals, weak metallic bonding; (f) HCl oxidised to Cl₂; (g) covalent character increases with oxidation state

Where the marks are

Each correct reason, any five parts (1 × 5 marks)

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 paper2024 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 31

Hints

Step-by-step solution

Revise the chapter

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