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 regarding the color change of acidified \( K_2Cr_2O_7 \) when \( Na_2SO_3 \) is added, we need to analyze the chemical reactions involved and the changes in oxidation states.

### Step-by-Step Solution:

1. **Identify the Reactants**:
We start with acidified potassium dichromate, \( K_2Cr_2O_7 \), which is typically acidified with sulfuric acid \( H_2SO_4 \), and sodium sulfite, \( Na_2SO_3 \).

2. **Write the Reaction**:
The reaction can be written as:
\[ K_2Cr_2O_7 + H_2SO_4 + Na_2SO_3 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + Na_2SO_4 + H_2O \]

3. **Balance the Reaction**:
Balancing the equation gives us:
\[ K_2Cr_2O_7 + 4 H_2SO_4 + 3 Na_2SO_3 \rightarrow Cr_2(SO_4)_3 + K_2SO_4 + 3 Na_2SO_4 + 4 H_2O \]

4. **Identify the Oxidation States**:
- In \( K_2Cr_2O_7 \), the oxidation state of chromium (Cr) is +6.
- In \( Cr_2(SO_4)_3 \), the oxidation state of chromium is +3.

5. **Determine the Changes**:
The chromium in \( K_2Cr_2O_7 \) is reduced from +6 to +3, indicating that it is acting as an oxidizing agent. Meanwhile, \( Na_2SO_3 \) is oxidized to \( Na_2SO_4 \).

6. **Color Change Explanation**:
The reduction of \( K_2Cr_2O_7 \) to \( Cr_2(SO_4)_3 \) results in a color change from orange (the color of \( K_2Cr_2O_7 \)) to green (the color of \( Cr_2(SO_4)_3 \)).

7. **Conclusion**:
Therefore, the solution turns green due to the formation of \( Cr_2(SO_4)_3 \) as a result of the reduction of \( K_2Cr_2O_7 \).

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