When `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution

Options

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

Explanation

To solve the problem of what happens when sulfur dioxide (SO₂) is passed through acidified potassium dichromate (K₂Cr₂O₇) solution, we can break it down into a series of steps:

### Step 1: Understanding the Reactants
- **Potassium Dichromate (K₂Cr₂O₇)**: This is an orange-colored compound that contains chromium in the +6 oxidation state.
- **Sulfur Dioxide (SO₂)**: This is a colorless gas that can act as a reducing agent.

### Step 2: Reaction Conditions
- The potassium dichromate solution is acidified, typically with sulfuric acid (H₂SO₄). This acidic environment is necessary for the reaction to proceed.

### Step 3: Writing the Chemical Equation
- When SO₂ is passed through the acidified K₂Cr₂O₇ solution, a redox reaction occurs. The balanced chemical equation for this reaction is:

\[ K_2Cr_2O_7 + H_2SO_4 + SO_2 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + H_2O \]

### Step 4: Observing Color Change
- Initially, the solution is orange due to the presence of chromium in the +6 oxidation state. As SO₂ is introduced, it reduces the chromium from +6 to +3 oxidation state, resulting in a color change from orange to green.

### Step 5: Identifying the Products
- The main product formed is chromium(III) sulfate (Cr₂(SO₄)₃), which gives the solution a green color. Potassium sulfate (K₂SO₄) and water (H₂O) are also formed as by-products.

### Step 6: Conclusion
- The key observations are:
- The solution changes color from orange to green.
- Chromium is reduced from +6 to +3 oxidation state.
- Green chromium sulfate is formed.

### Final Answer
The correct conclusion is that **green chromium sulfate is formed** when sulfur dioxide is passed through acidified potassium dichromate solution.

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