Question

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

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

Answer 1

Answer: (D)

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, $Pb(II)$ is more stable than $Pb(IV)$

$Sn(IV)$ is more stable than $Sn(II)$

$\therefore Pb(IV)$ is easily reduced to $Pb(II)$

$\therefore Pb(IV)$ is oxidising agent

$Sn(II)$ is easily oxidised to $Sn(IV)$

$\therefore Sn(II)$ is reducing agent