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

Options

A. `Cr_(2)(SO_(4))_(3)`
B. `CrO_(4)^(2-)`
C. `Cr_(2)(SO_3)_3`
D. `CrSO_(4)`

Explanation

To solve the question regarding the color change of acidified \( K_2Cr_2O_7 \) solution when \( Na_2SO_3 \) is added, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
The reactants in this scenario are potassium dichromate (\( K_2Cr_2O_7 \)) and sodium sulfite (\( Na_2SO_3 \)). The solution is acidified, which means sulfuric acid (\( H_2SO_4 \)) is also present.

2. **Understand the Reaction:**
When \( K_2Cr_2O_7 \) is mixed with \( H_2SO_4 \), it dissociates to form chromium ions (\( Cr_2O_7^{2-} \)) and sulfate ions (\( SO_4^{2-} \)). The dichromate ion is orange in color.

3. **Addition of Sodium Sulfite:**
The sodium sulfite reacts with the dichromate ion. The sulfite ion (\( SO_3^{2-} \)) acts as a reducing agent and reduces the dichromate ion to chromium ions (\( Cr^{3+} \)), which are green in color.

4. **Write the Balanced Reaction:**
The overall reaction can be represented as:
\[ K_2Cr_2O_7 + 3 Na_2SO_3 + 4 H_2SO_4 \rightarrow 2 Cr_2(SO_4)_3 + 3 Na_2SO_4 + 4 H_2O \]
In this reaction, the \( Cr_2O_7^{2-} \) ions are reduced to \( Cr^{3+} \) ions, which results in the formation of chromium sulfate (\( Cr_2(SO_4)_3 \)).

5. **Conclusion:**
The green color observed in the solution is due to the formation of chromium sulfate (\( Cr_2(SO_4)_3 \)), which contains \( Cr^{3+} \) ions.

### Final Answer:
The acidified \( K_2Cr_2O_7 \) solution turns green due to the formation of chromium sulfate, \( Cr_2(SO_4)_3 \).

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