When `SO_2` gas is passed through acidified `K_2Cr_2O_7` solution, the colour of the solution changes to:

Options

A. Red
B. Black
C. Orange
D. Green

Answer

To solve the problem, we need to understand the reaction that occurs when sulfur dioxide (SO₂) gas is passed through an acidified potassium dichromate (K₂Cr₂O₇) solution.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have sulfur dioxide (SO₂) and acidified potassium dichromate (K₂Cr₂O₇). The potassium dichromate is in an acidic medium, which is typically sulfuric acid (H₂SO₄).

2. **Understand the Oxidation States**:
- In K₂Cr₂O₇, chromium (Cr) is in the +6 oxidation state.
- In the product chromium(III) sulfate (Cr₂(SO₄)₃), chromium is in the +3 oxidation state.
- In SO₂, sulfur (S) is in the +4 oxidation state, and when it is oxidized to sulfuric acid (H₂SO₄), sulfur is in the +6 oxidation state.

3. **Write the Balanced Reaction**:
- The reaction can be summarized as follows:
\[ SO_2 + K_2Cr_2O_7 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O \]
- Here, SO₂ is oxidized to H₂SO₄, and K₂Cr₂O₇ is reduced to Cr₂(SO₄)₃.

4. **Color Change Observation**:
- Potassium dichromate (K₂Cr₂O₇) is orange in color due to the presence of Cr⁶⁺ ions.
- When it is reduced to Cr³⁺ ions in chromium(III) sulfate (Cr₂(SO₄)₃), the solution turns green.

5. **Conclusion**:
- Therefore, when SO₂ gas is passed through acidified K₂Cr₂O₇ solution, the color of the solution changes from orange to green.

### Final Answer:
The color of the solution changes to green.

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