Properties and Reactions of Compounds of Tin and Lead

To determine the correct statement(s), we analyze each option individually:

1. Option (A): $\text{SnCl}_2 \cdot 2\text{H}_2\text{O}$ functions as a reducing agent.

   • In $\text{SnCl}_2 \cdot 2\text{H}_2\text{O}$, the oxidation state of tin is $+2$.
   • Group 14 elements exhibit $+2$ and $+4$ oxidation states. For tin ($\text{Sn}$), the $+4$ oxidation state is more stable than the $+2$ oxidation state due to a weak inert pair effect.
   • Therefore, $\text{Sn}^{2+}$ readily oxidizes to $\text{Sn}^{4+}$ by releasing two electrons: $\text{Sn}^{2+} \longrightarrow \text{Sn}^{4+} + 2e^-$
   • Since it undergoes oxidation easily, $\text{SnCl}_2 \cdot 2\text{H}_2\text{O}$ acts as a reducing agent.
   • Thus, Option (A) is correct.

2. Option (B): $\text{SnO}_2$ reacts with $\text{KOH}$ to form the complex $\text{K}_2[\text{Sn(OH)}_6]$.

   • $\text{SnO}_2$ is an amphoteric oxide. When treated with a strong base like potassium hydroxide ($\text{KOH}$), it dissolves to form the hexahydroxostannate(IV) complex: $\text{SnO}_2 + 2\text{KOH} + 2\text{H}_2\text{O} \longrightarrow \text{K}_2[\text{Sn(OH)}_6]$
   • Thus, Option (B) is correct.

3. Option (C): An aqueous solution of $\text{PbCl}_2$ in $\text{HCl}$ contains $\text{Pb}^{2+}$ and $\text{Cl}^-$ species.

   • $\text{PbCl}_2$ dissolves in concentrated $\text{HCl}$ due to the formation of a soluble chlorocomplex, rather than simple dissociation into free $\text{Pb}^{2+}$ ions: $\text{PbCl}_2\text{(s)} + 2\text{Cl}^-\text{(aq)} \longrightarrow [\text{PbCl}_4]^{2-}\text{(aq)}$
   • Thus, the solution primarily contains tetrachloroplumbate(II) complex ions, $[\text{PbCl}_4]^{2-}$, instead of free $\text{Pb}^{2+}$ ions.
   • Thus, Option (C) is incorrect.

4. Option (D): The reaction of $\text{Pb}_3\text{O}_4$ with hot dilute nitric acid producing $\text{PbO}_2$ is a redox process.

   • $\text{Pb}_3\text{O}_4$ is a mixed oxide containing two parts lead(II) oxide and one part lead(IV) oxide, represented as $2\text{PbO} \cdot \text{PbO}_2$.
   • When $\text{Pb}_3\text{O}_4$ reacts with hot dilute $\text{HNO}_3$, only the basic oxide $\text{PbO}$ reacts to form lead(II) nitrate, leaving the acidic $\text{PbO}_2$ unreacted: $\text{Pb}_3\text{O}_4 + 4\text{HNO}_3 \longrightarrow 2\text{Pb(NO}_3)_2 + \text{PbO}_2 + 2\text{H}_2\text{O}$
   • Oxidation states of $\text{Pb}$:
     • In $\text{PbO}$ ($+2$) $\longrightarrow$ $\text{Pb(NO}_3)_2$ ($+2$)
     • In $\text{PbO}_2$ ($+4$) $\longrightarrow$ $\text{PbO}_2$ ($+4$)
   • Since no element undergoes a change in oxidation state, this is an acid-base reaction, not a redox reaction.
   • Thus, Option (D) is incorrect.

Correct Options: A, B