JEE Main 2026 — Jan 23, Shift 2 — Chemistry solution

JEE Main2026Jan 23, Shift 2Chemistry

It is noticed that $P b^{2 +}$ is more stable than $P b^{4 +}$ but $S n^{2 +}$ is less stable than $S n^{4 +}$ . Observe the following reactions

$P b O_{2} + P b \to 2 P b O$ ; $Δ_{r} G^{\circ}$ (1)

$S n O_{2} + S n \to 2 S n O$ ; $Δ_{r} G^{\circ}$ (2)

Identify the correct set from the following.

A
$Δ_{r} G^{\circ} (1) > 0$ ; $Δ_{r} G^{\circ} (2) < 0$
B
$Δ_{r} G^{\circ} (1) < 0$ ; $Δ_{r} G^{\circ} (2) < 0$
C
$Δ_{r} G^{\circ} (1) < 0$ ; $Δ_{r} G^{\circ} (2) > 0$
D
$Δ_{r} G^{\circ} (1) > 0$ ; $Δ_{r} G^{\circ} (2) > 0$

Solution

Because $P b^{2 +}$ is more stable than $P b^{4 +}$ , the reduction $P b O_{2} + P b \to 2 P b O$ moves the system to a more stable oxidation state, so it is spontaneous: $Δ_{r} G^{\circ} (1) < 0$ .

For tin, $S n^{2 +}$ is the less stable state, so the analogous reaction $S n O_{2} + S n \to 2 S n O$ produces a less stable species and is non-spontaneous: $Δ_{r} G^{\circ} (2) > 0$ .

$∴ Δ_{r} G^{\circ} (1) < 0$ and $Δ_{r} G^{\circ} (2) > 0$ .

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