JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqmedium

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

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

Options

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

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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 $\Delta_r G^\circ(2) > 0$ .

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Options

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

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

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

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

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 ( $Pb^{2+} > Pb^{4+}$ ).

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 $\Delta_r G^\circ(2) > 0$ .

JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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Solution

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JEE Main 2026 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

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