It is noticed that Pb^2+ is more stable than Pb^4+ but Sn^2+ is l… — Solution | JEE Chemistry

MCQEasyJEE 2026Group 14 Elements

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

PbO_2 + Pb \to 2PbO ; \Delta_rG^\circ(1)

SnO_2 + Sn \to 2SnO ; \Delta_rG^\circ(2)

Identify the correct set from the following

Answer

Correct answer:B

Standard Method

Given: $Pb^{2+}$ is more stable than $Pb^{4+}$ , whereas $Sn^{4+}$ is more stable than $Sn^{2+}$ .

Find: The correct signs of $\Delta_r G^\circ(1)$ and $\Delta_r G^\circ(2)$ .

The stability of the lower oxidation state $+2$ increases down group $14$ due to the inert pair effect.

For reaction $1$ :

Pb(+4) + Pb(0) \to Pb(+2)

Since $Pb^{2+}$ is more stable than $Pb^{4+}$ , the reaction proceeds forward spontaneously. Therefore,

\Delta_r G^\circ(1) < 0

For reaction $2$ :

Sn(+4) + Sn(0) \to Sn(+2)

Since $Sn^{4+}$ is more stable than $Sn^{2+}$ , the forward reaction is non-spontaneous. Therefore,

\Delta_r G^\circ(2) > 0

Thus the correct set is $\Delta_r G^\circ(1) < 0$ and $\Delta_r G^\circ(2) > 0$ . The correct option is B.

Assuming that both $Pb$ and $Sn$ behave identically because they belong to the same group is incorrect. Down the group, the inert pair effect becomes stronger, so $Pb^{2+}$ is stabilized much more than $Sn^{2+}$ . Always compare the relative stability of oxidation states for each element separately.
Equating stability directly with a positive $\Delta G^\circ$ is wrong. A reaction leading to the more stable oxidation state is thermodynamically favored, so its forward direction has $\Delta G^\circ < 0$ . First decide spontaneity, then assign the sign of $\Delta G^\circ$ .
Reading the second reaction as favorable toward $Sn^{2+}$ formation is a conceptual error. Since $Sn^{4+}$ is more stable than $Sn^{2+}$ , the forward conversion toward $Sn(+2)$ is not favored. Use oxidation-state stability, not only the written reaction form.

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.