Question

Consider the following changes : $\text{SnO}_2 \rightarrow \text{SnO}$ $\Delta G_1^o$ and $\text{PbO}_2 \rightarrow \text{PbO}$ $\Delta G_2^o$. Select the correct option

• A. $\Delta G_1^o>0$, $\Delta G_2^o>0$
• B. $\Delta G_1^o>0$, $\Delta G_2^o<0$
• C. $\Delta G_1^o<0$, $\Delta G_2^o<0$
• D. $\Delta G_1^o<0$, $\Delta G_2^o>0$

Hint:

The Inert Pair Effect causes the stability of the lower oxidation state (Group number $-2$) to increase down Groups 13-16. For Group 14, $\text{Sn}^{4+}$ is stable, while $\text{Pb}^{2+}$ is highly stable. Spontaneous reactions have $\Delta G^o<0$.

Answer 1

The Correct Option is B

Step 1: Both reactions involve reduction from +4 to +2 oxidation state.
Step 2: Due to inert pair effect, stability of +2 state increases down the group.
Step 3: Tin ($Sn$): +4 state is more stable than +2.
So reduction is non-spontaneous. \[ \Delta G_1^o>0 \] Step 4: Lead ($Pb$): +2 state is more stable due to inert pair effect.
So reduction is spontaneous. \[ \Delta G_2^o<0 \] Hence, correct option is (2).