Question

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

• A. $S{n}^{2+}$ is oxidizing while $P{b}^{2+}$ is reducing
• B. $S{n}^{2+}$ and $P{b}^{2+}$ are both oxidizing and reducing
• C. $S{n}^{4+}$ is reducing while $P{b}^{4+}$is oxidizing
• D. $S{n}^{2+}$ is reducing while $P{b}^{4+}$is oxidizing

Answer 1

The correct option is C

$S{n}^{4+}$ is reducing while $P{b}^{4+}$is oxidizing

As we go down the group, the outermost s electrons of the $n{s}^{2}n{p}^2$ configuration tend not to participate in bond formation. Hence, the stability if +4 oxidation state decreases and the stability of the +2 oxidation state increases (as we descend). This is called inert pair effect.

This implies that $P{b}^{4+}$ would want to grab two electrons and become the more stable $P{b}^{2+}$ and itself get reduced. Hence $P{b}^{4+}$ is oxidizing. Likewise, for tin, $S{n}^{4+}$ is more stable than $S{n}^{2+}$ and hence, the latter ( as in - $S{n}^{2+}$) is reducing.