Oxidizing Power and Stability of Lead and Tin Ions

To determine the correctness of the given statements, we analyze the periodic trends and stability of oxidation states for Group 14 elements ($\text{C}$, $\text{Si}$, $\text{Ge}$, $\text{Sn}$, $\text{Pb}$):

1. Analysis of STATEMENT-1:

   • Group 14 elements exhibit oxidation states of $+2$ and $+4$.
   • Due to the inert pair effect, the relativistic contraction of the inner $s$-orbitals causes the outermost $ns^2$ electron pair to remain non-bonding or "inert" as we move down the group.
   • Consequently, the stability of the lower oxidation state ($+2$) increases down the group, while the stability of the higher oxidation state ($+4$) decreases:
     $\text{Stability of }+2\text{ state: } \text{Ge}^{2+} < \text{Sn}^{2+} < \text{Pb}^{2+}$
     $\text{Stability of }+4\text{ state: } \text{Ge}^{4+} > \text{Sn}^{4+} > \text{Pb}^{4+}$
   • Since $\text{Pb}^{2+}$ is far more stable than $\text{Pb}^{4+}$, compounds of $\text{Pb}^{4+}$ readily undergo reduction to form $\text{Pb}^{2+}$:
     $\text{Pb}^{4+} + 2e^- \longrightarrow \text{Pb}^{2+} \quad (E^\circ = +1.69\text{ V})$
     making $\text{Pb}^{4+}$ a very strong oxidizing agent.
   • In contrast, for tin, $\text{Sn}^{4+}$ is more stable than $\text{Sn}^{2+}$ ($E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15\text{ V}$), so $\text{Sn}^{4+}$ compounds are not strong oxidizing agents (in fact, $\text{Sn}^{2+}$ acts as a reducing agent).
   • Therefore, $\text{Pb}^{4+}$ compounds are stronger oxidizing agents than $\text{Sn}^{4+}$ compounds.
   • STATEMENT-1 is True.

2. Analysis of STATEMENT-2:

   • The inert pair effect explains why the lower oxidation state ($+2$) becomes progressively more stable than the higher oxidation state ($+4$) for heavier elements like lead ($\text{Pb}$).
   • The statement claims that higher oxidation states are more stable for heavier members, which is incorrect.
   • STATEMENT-2 is False.

Thus, STATEMENT-1 is True and STATEMENT-2 is False.

Correct Option: C