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(D) . $SnCl_2 \cdot 2 H_2O$ is a reducing agent because $Sn^{2+}$ is easily oxidized to $Sn^{4+}$.$B$. $SnO_2$ is amphoteric and reacts with $KOH$ to

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$A, B, C$

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(D) . $SnCl_2 \cdot 2 H_2O$ is a reducing agent because $Sn^{2+}$ is easily oxidized to $Sn^{4+}$.$B$. $SnO_2$ is amphoteric and reacts with $KOH$ to form the stannate complex: $SnO_2 + 2 KOH + 2 H_2O \longrightarrow K_2[Sn(OH)_6]$.$C$. $PbCl_2$ is slightly soluble in water. In the presence of $HCl$,it exists in equilibrium as $Pb^{2+}$ and $Cl^-$ ions,though it can also form complex ions like $[PbCl_4]^{2-}$ in excess $HCl$.$D$. The reaction $Pb_3O_4 + 4 HNO_3 \longrightarrow PbO_2 + 2 Pb(NO_3)_2 + 2 H_2O$ is a disproportionation-like reaction where $Pb_3O_4$ (which is $2PbO \cdot PbO_2$) reacts with acid; it is not a redox reaction as the oxidation states of $Pb$ remain $+2$ and $+4$ throughout.

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