By passing` SO_2` in the solution of `K_2 Cr_2 O_7`, it turns green due to

Options

A. `K_2 Cr O_4`
B. `CrSO_4`
C. `Cr_2 (SO_4)_3`
D. `Cr_2 S_3`

Explanation

To solve the question of why the solution of potassium dichromate (K₂Cr₂O₇) turns green when sulfur dioxide (SO₂) is passed through it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have potassium dichromate (K₂Cr₂O₇) and sulfur dioxide (SO₂) as our main reactants.

2. **Understand the Reaction**:
- When sulfur dioxide is introduced into an acidified solution of potassium dichromate, a redox reaction occurs.

3. **Oxidation and Reduction**:
- In this reaction, sulfur dioxide (SO₂) acts as a reducing agent. It gets oxidized to sulfuric acid (H₂SO₄).
- Simultaneously, potassium dichromate (which is orange in color due to the presence of Cr⁶⁺ ions) is reduced to chromic sulfate (Cr₂(SO₄)₃), which contains Cr³⁺ ions.

4. **Color Change**:
- The reduction of Cr⁶⁺ (orange) to Cr³⁺ (green) is what causes the color change in the solution. The solution transitions from orange to green as the reaction proceeds.

5. **Final Products**:
- The overall reaction can be summarized as:
\[ K_2Cr_2O_7 + SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O \]

6. **Conclusion**:
- The solution turns green due to the reduction of chromium from the +6 oxidation state (orange) to the +3 oxidation state (green).