Question

In view of ${△}_r{G}^{\circ }$ for the following reactions
$Pb{O}_2+Pb\to 2PbO,{△}_r{G}^{\circ }<0$
$Sn{O}_2+Sn\to 2SnO,{△}_r{G}^{\circ }>0$
Which oxidation state is more characteristic for lead and tin?

• A. For lead $+2$, for tin $+2$
• B. For lead $+4$, for tin $+4$
• C. For lead $+2$, for tin $+4$
• D. For lead $+4$, for tin $+2$

Answer 1

The correct option is C For lead $+2$, for tin $+4$

(i) For the reaction:
$Pb{O}_2+Pb\to 2PbO,△G^{\circ }<0$ i.e. $(-)ve$, means $PvO\left(P{b}^{2+}\right)$ is more stable hence it has more characteristic oxidation state.
(ii) For the reaction:
$Sn{O}_2+Sn\to 2SnO,△G^{\circ }>0$ i.e. $(+)ve$, means $Sn{O}_2\left(S{n}^{4+}\right)$ is more stable than $S{n}^{2+}\left(SnO\right)$ hence is more characteristic oxidation state.
Therefore, characteristic oxidation states for lead and tin are respectively $+2$ and $+4$.