During the qualitative analysis of \( SO_{3}^{-2}\) using dilute H2SO4, SO2 gas evolved which turns K2Cr2O7 solution (acidified H2SO4)
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In qualitative analysis, dichromate ions (Cr2O72–) are commonly used as oxidizing agents. Reduction of these ions often results in a color change, which can help identify the reducing species.
- Green
- Black
- Blue
- Red
The Correct Option is A
Solution and Explanation
Step 1: Reaction Involved
When \(\text{SO}_3^{2-}\) reacts with dilute \(\text{H}_2\text{SO}_4\), \(\text{SO}_2\) gas is evolved. The \(\text{SO}_2\) gas reduces the dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) to \(\text{Cr}^{3+}\), which is green in color. The reaction is as follows:
\[\text{Cr}_2\text{O}_7^{2-} + \text{SO}_3^{2-} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-}.\]
Step 2: Color Change
- Dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is orange in color.
- After reduction, \(\text{Cr}^{3+}\) ions form, which are green in color.
Conclusion: The solution turns green due to the formation of \(\text{Cr}^{3+}\). Therefore, the correct answer is \((3)\) Green.
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