Which one of the following statements is correct when `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution?

Options

A. Green `Cr_(2)(SO_(4))` is formed
B. The solution turns blue
C. The solution is decolourized
D. `SO_(2)` is reduced

Answer

To solve the question regarding the reaction of sulfur dioxide (SO₂) with acidified potassium dichromate (K₂Cr₂O₇), we can break down the process step by step:

### Step 1: Understand the Reactants
- **Sulfur Dioxide (SO₂)**: A colorless gas that can act as a reducing agent.
- **Potassium Dichromate (K₂Cr₂O₇)**: An orange compound that acts as an oxidizing agent when acidified.

### Step 2: Identify the Reaction Conditions
- The potassium dichromate is acidified, which typically means that sulfuric acid (H₂SO₄) is added. This provides an acidic environment necessary for the reaction.

### Step 3: Determine the Reaction
- When SO₂ is passed through the acidified K₂Cr₂O₇ solution, SO₂ is oxidized to sulfuric acid (H₂SO₄).
- Simultaneously, the dichromate ion (Cr₂O₇²⁻) is reduced to chromium(III) ions (Cr³⁺), which results in a color change.

### Step 4: Color Change Observation
- The orange color of the dichromate solution changes to green due to the formation of chromium(III) sulfate (Cr₂(SO₄)₃), which is green in color.

### Step 5: Conclusion
- The correct statement regarding the reaction is that the solution turns green due to the formation of Cr₂(SO₄)₃ from the reduction of K₂Cr₂O₇ and the oxidation of SO₂ to H₂SO₄.

### Final Answer
- The correct statement is: **Green Cr₂(SO₄)₃ is formed.**

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