Acidified `K_(2)Cr_(2)O_(7)` solution turns green when `Na_(2)SO_(3)` is added to it. This is due to the formation of

Options

A. `CrO_(4)^(2-)`
B. `Cr_(2)(SO_(3))_(3)`
C. `CrSO_(4)`
D. `Cr_(2)(SO_(4))_(3)`

Explanation

To solve the question, we need to analyze the reaction that occurs when acidified potassium dichromate (K₂Cr₂O₇) is mixed with sodium sulfite (Na₂SO₃).

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are acidified potassium dichromate (K₂Cr₂O₇) and sodium sulfite (Na₂SO₃).

2. **Understand the Role of Potassium Dichromate**:
- Potassium dichromate is a strong oxidizing agent. In acidic conditions, it can oxidize sulfite ions (SO₃²⁻) to sulfate ions (SO₄²⁻).

3. **Write the Balanced Chemical Reaction**:
- The balanced reaction can be written as follows:
\[ K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O \]
- In this reaction, the potassium dichromate is reduced, and the sulfite is oxidized.

4. **Identify the Product Responsible for the Color Change**:
- The product formed from the reduction of potassium dichromate is chromium(III) sulfate (Cr₂(SO₄)₃), which is responsible for the green color observed in the solution.

5. **Conclusion**:
- Therefore, the acidified K₂Cr₂O₇ solution turns green due to the formation of chromium(III) sulfate (Cr₂(SO₄)₃).

### Final Answer:
The solution turns green due to the formation of **Cr₂(SO₄)₃**.

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