The group 14 elements A and B have the first ionisation enthalpy values of 708 and $715 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions $\mathrm{A}^{2+} \mathrm{B}^{4+}$ respectively is

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · p-Block Elements

The group 14 elements A and B have the first ionisation enthalpy values of 708 and

715 kJ mol^{- 1}

respectively. The above values are lowest among their group members. The nature of their ions

A^{2 +} B^{4 +}

respectively is

(A)both reducing
(B)both oxidising
(C)reducing and oxidising✓ Correct
(D)oxidising and reducing

Answer: (C)

Solution

Within group 14 the two members of lowest first ionisation enthalpy are the heaviest, tin and lead, so A is tin and B is lead. Tin in the

+ 2

state readily loses two further electrons to reach the stable

+ 4

state, making

S n^{2 +}

a reducing ion. For lead the inert-pair effect makes

+ 2

the stable state, so

P b^{4 +}

tends to gain electrons and is an oxidising ion. Thus

A^{2 +}

is reducing and

B^{4 +}

is oxidising.

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option

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