During the qualitative analysis of `SO_(3)^(2-)` using dilute `H_(2)SO_(4),SO_(2)` gas is evolved which turns `K_(2)Cr_(2)O_(7)` solution (acidified with dilue `H_(2)SO_(4)` ) :

Options

A. red
B. blue
C. green
D. black

Answer

To solve the problem, we need to analyze the reactions occurring when `SO_(3)^(2-)` (sulfite ion) is treated with dilute `H_(2)SO_(4)` (sulfuric acid) and how the resulting `SO_(2)` gas interacts with `K_(2)Cr_(2)O_(7)` (potassium dichromate).

### Step-by-Step Solution:

1. **Identify the Reaction**: When `SO_(3)^(2-)` is treated with dilute `H_(2)SO_(4)`, it undergoes oxidation to form `SO_(2)` gas. The reaction can be represented as:
\[ SO_3^{2-} + H_2SO_4 \rightarrow SO_2 + HSO_4^{-} \]

2. **Evolution of SO2**: The `SO_(2)` gas that is evolved can now react with the potassium dichromate solution. Potassium dichromate in acidic medium is orange due to the presence of chromium in the +6 oxidation state.

3. **Reaction of SO2 with K2Cr2O7**: The `SO_(2)` gas acts as a reducing agent. It reduces the chromium from the +6 oxidation state in `K_(2)Cr_(2)O_(7)` to the +3 oxidation state. The reaction can be represented as:
\[ K_2Cr_2O_7 + SO_2 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O \]

4. **Color Change**: The orange solution of potassium dichromate changes color due to the reduction of chromium. The +3 oxidation state of chromium (in chromium(III) sulfate) gives a green color to the solution. Thus, the solution changes from orange to green.

5. **Final Answer**: Therefore, the color of the solution after the reaction is green.