Identification of Compound X Turning Green with Sulfur Dioxide

When filter paper dipped in an acidified solution of potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) in dilute $\text{H}_2\text{SO}_4$ is exposed to sulfur dioxide ($\text{SO}_2$) gas, a redox reaction occurs.

Potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), which is orange in color, acts as a strong oxidizing agent and oxidizes $\text{SO}_2$ to sulfate ions ($\text{SO}_4^{2-}$). In this reaction, the dichromate ions ($\text{Cr}_2\text{O}_7^{2-}$) are reduced to chromium(III) ions ($\text{Cr}^{3+}$), which impart a characteristic green color to the solution.

The ionic equation for the reaction is: $\text{Cr}_2\text{O}_7^{2-} (\text{aq, orange}) + 3\text{SO}_2 (\text{g}) + 2\text{H}^+ (\text{aq}) \rightarrow 2\text{Cr}^{3+} (\text{aq, green}) + 3\text{SO}_4^{2-} (\text{aq}) + \text{H}_2\text{O} (\text{l})$

The complete molecular chemical equation is: $\text{K}_2\text{Cr}_2\text{O}_7 + 3\text{SO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$

The formation of green chromium(III) sulfate, $\text{Cr}_2(\text{SO}_4)_3$, causes the paper to turn green.

Therefore, the compound '$X$' is $\text{K}_2\text{Cr}_2\text{O}_7$.

Correct Option: D ($\text{K}_2\text{Cr}_2\text{O}_7$)