Question

Choose the correct statement(s) among the following

• A. $SnCl _{2} .2 H _{2} O$ is a reducing agent.
• B. $SnO _{2}$ reacts with $KOH$ to form $K _{2}\left[ Sn ( OH )_{6}\right]$.
• C. $A$ solution of $PbCl _{2}$ in $HCl$ contains $Pb ^{2+}$ and $Cl ^{-}$ions.
• D. The reaction of $Pb _{3} O _{4}$ with hot dilute nitric acid to give $PbO _{2}$ is a redox reaction.

Answer 1

The Correct Option is A, B

(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.