p Block Elements Part 3

Concept:Relative stability of $+2$ and $+4$ oxidation states determines whether a comproportionation reaction is spontaneous.Chemical Equation / Formula:$\mathrm{PbO_2} + \mathrm{Pb} \longrightarrow 2\mathrm{PbO}$$\mathrm{SnO_2} + \mathrm{Sn} \longrightarrow 2\mathrm{SnO}$Explanation:In the first reaction, Pb changes from $+4$ (in $\mathrm{PbO_2}$) and $0$ (in elemental Pb) to $+2$ (in $\mathrm{PbO}$).Since $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$, the reaction is spontaneous.Thus, $\Delta_{\mathrm{r}} G^{\circ}(1) < 0$.In the second reaction, Sn changes from $+4$ (in $\mathrm{SnO_2}$) and $0$ (in Sn) to $+2$ (in $\mathrm{SnO}$).But $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$, so the reaction is non‑spontaneous.Thus, $\Delta_{\mathrm{r}} G^{\circ}(2) > 0$.Shortcut:If the lower oxidation state is more stable, comproportionation is spontaneous (ΔG < 0); if less stable, comproportionation is non‑spontaneous (ΔG > 0).Answer:Option B: $\Delta_{r} G^{\circ}(1) < 0 ; \Delta_{r} G^{\circ}(2) > 0$