Question

A paper dipped in a dil. $H_2SO_4$ solution of 'X' upon treatment with $SO_2$ gas turns into green. The compound 'X' is :

• A. KI-starch
• B. $KMnO_4$
• C. $Pb(CH_3COO)_2$
• D. $K_2Cr_2O_7$

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
$SO_2$ is a strong reducing agent. It reacts with oxidizing agents, often resulting in a characteristic color change.
Step 2: Detailed Explanation:
Acidified Potassium Dichromate ($K_2Cr_2O_7$) is orange in color. When it reacts with sulfur dioxide ($SO_2$), the dichromate ion is reduced to the chromic ion ($Cr^{3+}$), which is green.
Reaction:
\[ Cr_2O_7^{2-} (\text{orange}) + 3SO_2 + 2H^+ \to 2Cr^{3+} (\text{green}) + 3SO_4^{2-} + H_2O \]
Step 3: Final Answer:
The compound X is $K_2Cr_2O_7$.