Question

Species "X" is dissolved in \( \mathrm{H_2SO_4} \) and reacts with \( \mathrm{SO_2} \) to give a green color solution. The species "X" is:

• A. \( \mathrm{KMnO_4} \)
• B. \( \mathrm{K_2Cr_2O_7} \)
• C. \( \mathrm{Pb(CH_3COO)_2} \)
• D. \( \mathrm{KI} \)

Hint:

When \( \mathrm{SO_2} \) reacts with oxidizing agents like \( \mathrm{K_2Cr_2O_7} \), the chromium species are reduced to \( \mathrm{Cr^{3+}} \), which forms a green-colored solution.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The reaction involves the reduction of an oxidizing agent by sulfur dioxide (\(SO_2\), a reducing agent).
The key observation is the formation of a green color.
This is a characteristic color change associated with the reduction of Chromium(VI) to Chromium(III).
\(Cr^{6+}\) (orange/yellow) \(\longrightarrow\) \(Cr^{3+}\) (green) upon reduction in acidic medium.
Step 2: Detailed Explanation:
When Potassium dichromate (\(K_2Cr_2O_7\)) is dissolved in sulfuric acid, the dichromate ion (\(Cr_2O_7^{2-}\)) is present in the orange/yellow form.
\(SO_2\) acts as a reducing agent and reduces \(Cr^{6+}\) to \(Cr^{3+}\).
The overall reaction is:
\[ K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O \]
In ionic form:
\[ Cr_2O_7^{2-} \text{ (orange)} + 3SO_2 + 2H^{+} \rightarrow 2Cr^{3+} \text{ (green)} + 3SO_4^{2-} + H_2O \]
The formation of \(Cr^{3+}\) gives the characteristic green color.
Why not the other options?
\(KMnO_4\): When reduced by \(SO_2\) in acidic medium, the purple color disappears and the solution turns colorless (formation of \(Mn^{2+}\) which is almost colorless in dilute solution).
\(Pb(CH_3COO)_2\): Lead acetate does not react with \(SO_2\) to give a green solution.
\(KI\): Reaction with \(SO_2\) does not produce a green color.
Step 3: Final Answer:
The specie ``X'' is K\(_2\)Cr\(_2\)O\(_7\).