It is because of inability of $n{s}^2$ electrons of the valence shell to participate in bonding that

Solution & Explanation

\textbf{Answer:} (A) $\mathrm{Sn}{}^{2+}$ is reducing while $\mathrm{Pb}{}^{4+}$ is oxidising. \textbf{Solution:} The inability of the $n{s}^2$ electrons to participate in bonding down a group of heavier p-block elements is the inert pair effect. Because of this, $\mathrm{Pb}(\text{II})$ is more stable than $\mathrm{Pb}(\text{IV})$, so $\mathrm{Pb}{}^{4+}$ is oxidising (readily reduced to $\mathrm{Pb}{}^{2+}$). In tin, $\mathrm{Sn}(\text{IV})$ is more stable than $\mathrm{Sn}(\text{II})$, so $\mathrm{Sn}{}^{2+}$ is reducing (readily oxidised to $\mathrm{Sn}{}^{4+}$).