(A) $Sn ^{2+}$ is less stable than $Sn ^{4+}$,
therefore $SnCl _{2} \cdot 2 H _{2} O$ is a Lewis acid.
(B) $SnO _{2}$ is amphoteric, so it will dissolve in $KOH$ solution forming $K _{2} SnO _{3}$ or $K _{2}\left[ Sn ( OH )_{6}\right]$.
$SnO _{2}+2 KOH +2 H _{2} O \longrightarrow K _{2}\left[ Sn ( OH )_{6}\right]$
(C) $PbCl _{2}$ will form a complex with $HCl$
$PbCl _{2}( s )+2 HCl \longrightarrow H _{2}\left[ PbCl _{4}\right]$
So, we may assume that now this complex will dissociate as:
$H _{2}\left[ PbCl _{2}\right] \longrightarrow 2 H ^{+}+\left[ PbCl _{4}\right]^{2-}$
So, considering this possibility, option (C) should be incorrect.
(D) $Pb_3O_4$ reacts with conc.$HNO_3$ to form $PbO_2$ but the reaction is not a redox reaction.
$\overset{(+4)}{PbO_2}.\overset{(+2)}{2PbO} + \underset{\text{Conc.}}{4HNO_3} \rightarrow \overset{(+4)}{PbO_2}\downarrow + \overset{(+2)}{2Pb}(NO_3)_2 + 6H_2O$
or
$Pb_3O_4$
So, option (D) is incorrect.