Choose the correct statement(s) among the following
- $SnCl _{2} .2 H _{2} O$ is a reducing agent.
- $SnO _{2}$ reacts with $KOH$ to form $K _{2}\left[ Sn ( OH )_{6}\right]$.
- $A$ solution of $PbCl _{2}$ in $HCl$ contains $Pb ^{2+}$ and $Cl ^{-}$ions.
- The reaction of $Pb _{3} O _{4}$ with hot dilute nitric acid to give $PbO _{2}$ is a redox reaction.
The Correct Option is A, B
Solution and Explanation
(A) \(\text{SnCl}_2 \cdot 2\text{H}_2\text{O}\) acts as a reducing agent since \(Sn^{2+}\) tends to convert into \(Sn^{4+}\). This is evident in reactions where \(SnO_2\) reacts with KOH to form \(\text{K}_2\text{SnO}_3\) and water, or where SnO2 reacts with a base to form the amphoteric compound \(\text{K}_2[\text{Sn}(\text{OH})_6]\)
(B) When testing for first group cations (such as \(Pb^{2+}\)), they form insoluble chlorides (e.g., PbCl2) with HCl. However, PbCl2 is slightly soluble in water, and thus, lead ions may not be completely precipitated when hydrochloric acid is added. The remaining \(Pb^{2+}\) ions can then be quantitatively precipitated with H2S in an acidic medium. Therefore, the filtrate of the first group contains a solution of PbCl2 in HCl, consisting of \(Pb^{2+}\) and \(Cl^-\) ions. However, in the presence of concentrated HCl or excess HCl, \(\text{H}_2[\text{PbCl}_4]\) can be formed. Additionally, reactions such as Pb3O4 with HNO3 can produce PbO2, Pb(NO3)2, and water, or PbO.PbO2 in a non-redox reaction.
Therefore, options A, B be correct answers.
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