ChemistryJEE Main 23 Jan 2026, Shift 2·Q2·MCQ·Easy–moderate

p-Block (13–14) · Inert pair effect

It is noticed that $\ce{Pb^{2+}}$ is more stable than $\ce{Pb^{4+}}$ but $\ce{Sn^{2+}}$ is less stable than $\ce{Sn^{4+}}$ . Observe the following reactions

$\ce{PbO2 + Pb -> 2PbO}$ ; $\Delta_\mathrm{r}G^\circ$ (1)

$\ce{SnO2 + Sn -> 2SnO}$ ; $\Delta_\mathrm{r}G^\circ$ (2)

Identify the correct set from the following.

(1) $\Delta_\mathrm{r}G^\circ(1) > 0$ ; $\Delta_\mathrm{r}G^\circ(2) < 0$
(2) $\Delta_\mathrm{r}G^\circ(1) < 0$ ; $\Delta_\mathrm{r}G^\circ(2) < 0$
(3) $\Delta_\mathrm{r}G^\circ(1) < 0$ ; $\Delta_\mathrm{r}G^\circ(2) > 0$
(4) $\Delta_\mathrm{r}G^\circ(1) > 0$ ; $\Delta_\mathrm{r}G^\circ(2) > 0$

We corrected a printing error in the original paper

Source printed “ $\ce{Pb^{2+}}$ is more stable than $\ce{Pb^{2+}}$ ” (second charge wrong; the intended $\ce{Pb^{4+}}$ follows from the $\ce{Sn^{2+}}$ / $\ce{Sn^{4+}}$ comparison).

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See the full 23 Jan 2026, Shift 2 paper