Q16 / 75 · JEE Main 2026 (23 January Shift 2) · Chemistry
It is noticed that $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$ but $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$. Observe the following reactions.
$\mathrm{PbO}_{2}+\mathrm{Pb} \rightarrow 2 \mathrm{PbO} ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)$
$\mathrm{SnO}_{2}+\mathrm{Sn} \rightarrow 2 \mathrm{SnO} ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)$
Identify the correct set from the following
$\mathrm{PbO}_{2}+\mathrm{Pb} \rightarrow 2 \mathrm{PbO} ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)$
$\mathrm{SnO}_{2}+\mathrm{Sn} \rightarrow 2 \mathrm{SnO} ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)$
Identify the correct set from the following
- A$\Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)<0 ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)<0$
- B$\Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)>0 ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)>0$
- Correct answerC$\Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)<0 ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)>0$
- D$\Delta_{\mathrm{r}} \mathrm{G}^{\circ}(1)>0 ; \Delta_{\mathrm{r}} \mathrm{G}^{\circ}(2)<0$
Solution
The stability of oxidation states in Group 14 is governed by the inert pair effect, which increases down the group.
For Lead ($Pb$), the $+2$ oxidation state is more stable than the $+4$ state ($Pb^{2+} > Pb^{4+}$).
In the reaction $PbO_2 + Pb \rightarrow 2PbO$, $Pb^{4+}$ is being reduced to $Pb^{2+}$ and $Pb^0$ is being oxidized to $Pb^{2+}$. Since the product contains the more stable oxidation state, the reaction is spontaneous, meaning $\Delta_r G^\circ(1) < 0$.
For Tin ($Sn$), the $+4$ oxidation state is more stable than the $+2$ state ($Sn^{4+} > Sn^{2+}$).
In the reaction $SnO_2 + Sn \rightarrow 2SnO$, the stable $Sn^{4+}$ is being converted into the less stable $Sn^{2+}$. This process is non-spontaneous, meaning $\Delta_r G^\circ(2) > 0$.
Thus, the correct set is $\Delta_r G^\circ(1) < 0$ and $\Delta_r G^\circ(2) > 0$.
For Lead ($Pb$), the $+2$ oxidation state is more stable than the $+4$ state ($Pb^{2+} > Pb^{4+}$).
In the reaction $PbO_2 + Pb \rightarrow 2PbO$, $Pb^{4+}$ is being reduced to $Pb^{2+}$ and $Pb^0$ is being oxidized to $Pb^{2+}$. Since the product contains the more stable oxidation state, the reaction is spontaneous, meaning $\Delta_r G^\circ(1) < 0$.
For Tin ($Sn$), the $+4$ oxidation state is more stable than the $+2$ state ($Sn^{4+} > Sn^{2+}$).
In the reaction $SnO_2 + Sn \rightarrow 2SnO$, the stable $Sn^{4+}$ is being converted into the less stable $Sn^{2+}$. This process is non-spontaneous, meaning $\Delta_r G^\circ(2) > 0$.
Thus, the correct set is $\Delta_r G^\circ(1) < 0$ and $\Delta_r G^\circ(2) > 0$.
Hence, the correct answer is option: 'C'.