Question

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

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

Answer 1

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

$S{n}^{2+}$ is reducing while $P{b}^{4+}$ is oxidising.

$P{b}^{2+}$ compounds are more stable than $P{b}^{4+}$.

$P{b}^{4+}$ compounds are easily reduced to stable $P{b}^{2+}$ compounds.

$S{n}^{2+}$ can be oxidise to $S{n}^{4+}$ making it reducing agent.