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:} Due to the inert pair effect, the inability of the $n{s}^2$ electrons to participate in bonding increases down the group. Hence the +2 state becomes more stable for $\mathrm{Pb}$ and the +4 state more stable for $\mathrm{Sn}$. So $\mathrm{Pb}(\mathrm{IV})$ is easily reduced to $\mathrm{Pb}(\mathrm{II})$, i.e. $\mathrm{Pb}{}^{4+}$ is an oxidising agent, and $\mathrm{Sn}(\mathrm{II})$ is easily oxidised to $\mathrm{Sn}(\mathrm{IV})$, i.e. $\mathrm{Sn}{}^{2+}$ is a reducing agent. Therefore $\mathrm{Sn}{}^{2+}$ is reducing while $\mathrm{Pb}{}^{4+}$ is oxidising.