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$PbO_2 \rightarrow PbO$, $\Delta G_{298} \lt 0$; $SnO_2 \rightarrow SnO$, $\Delta G_{298} \gt 0$. Most probable oxidation states of Pb and Sn will be :
A
$Pb^{4+}$, $Sn^{4+}$
B
$Pb^{4+}$, $Sn^{2+}$
C
$Pb^{2+}$, $Sn^{2+}$
D
$Pb^{2+}$, $Sn^{4+}$
Detailed Solution
$PbO_2 \rightarrow PbO$ has $\Delta G \lt 0$, so the change from +4 to +2 is spontaneous for lead. So +2 is the more stable state of Pb.
$SnO_2 \rightarrow SnO$ has $\Delta G \gt 0$, so the change from +4 to +2 is not spontaneous for tin. So +4 is the more stable state of Sn.
This is due to the inert pair effect, which increases down group 14: the $6s^2$ pair of Pb is reluctant to take part in bonding.
So the most probable oxidation states are $Pb^{2+}$ and $Sn^{4+}$.
$SnO_2 \rightarrow SnO$ has $\Delta G \gt 0$, so the change from +4 to +2 is not spontaneous for tin. So +4 is the more stable state of Sn.
This is due to the inert pair effect, which increases down group 14: the $6s^2$ pair of Pb is reluctant to take part in bonding.
So the most probable oxidation states are $Pb^{2+}$ and $Sn^{4+}$.