Question

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

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

Answer 1

Answer: (D)

(i) For the reaction:

$Pb{O}_2+Pb⟶2PbO,\Delta G^{\circ }<0$ i.e. $(-)ve$

means $PbO\left(P{b}^{2+}\right)$ is more stable hence it has more characteristic oxidation state.

(ii) For the reaction:

$Sn{O}_2+Sn\to 2SnO,\Delta G^{\circ }>0$ i.e. $(+)ve$

means $Sn{O}_2\left(S{n}^{4+}\right)$ is more stable than $S{n}^{2+}$ (SnO) hence is more characteristic oxidation state.

Therefore, characteristic oxidation states for lead and tin are respectively $+2$ and $+4$.