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 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 while $$P{b^{4 + }}$$ is oxidising
Answer :
$$S{n^{4 + }}$$ is reducing while $$P{b^{4 + }}$$ is oxidising
Solution :
The inability of $$n{s^2}$$ electrons of the valence shell to participate in bonding is called as inert pair effect. Due to this effect, the lower oxidation state becomes more stable on descending the group. Thus, $$S{n^{2 + }}$$ is a reducing agent while $$P{b^{4 + }}$$ act as an oxidising agent.