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

Answer 1

The correct option is A $S{n}^{2+}$ is reducing while $P{b}^{4+}$ is oxidising

Inability of $n{s}^2$ electrons of the valence shell to participate in bonding on moving down the group in heavier p-block elements is called inert pair effect.
As a result, $Pb\left(II\right)$ is more stable than $Pb\left(IV\right)$
$Sn\left(IV\right)$ is more stable than $Sn\left(II\right)$
$\therefore$ $Pb\left(IV\right)$ is easily reduced to $Pb\left(II\right)$
$\therefore$ $Pb\left(IV\right)$ acts as an oxidising agent
$Sn\left(II\right)$ is easily oxidised to $Sn\left(IV\right)$
$\therefore$ $Sn\left(II\right)$ acts as a reducing agent.