When `H_(2)S` is passed through acidified `K_(2)Cr_(2)O_(7)` solution, the solution turns :

Options

A. yellow
B. blue
C. green
D. white

Answer

To solve the question of what happens when \( H_2S \) is passed through acidified \( K_2Cr_2O_7 \) solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are \( H_2S \) (hydrogen sulfide) and \( K_2Cr_2O_7 \) (potassium dichromate) in an acidic medium.

2. **Understand the Reaction**: In acidic conditions, \( K_2Cr_2O_7 \) acts as a strong oxidizing agent. It oxidizes \( H_2S \) to elemental sulfur (\( S \)).

3. **Write the Balanced Chemical Equation**:
\[ 3H_2S + K_2Cr_2O_7 + 4H_2SO_4 \rightarrow K_2SO_4 + 3S + Cr_2(SO_4)_3 + 7H_2O \]
In this reaction, \( H_2S \) is oxidized to sulfur, and the dichromate ion (\( Cr_2O_7^{2-} \)) is reduced.

4. **Identify the Color Change**: The dichromate ion (\( Cr_2O_7^{2-} \)) is orange in color. When it is reduced to chromium ion (\( Cr^{3+} \)), the solution changes to green.

5. **Conclusion**: Therefore, when \( H_2S \) is passed through acidified \( K_2Cr_2O_7 \), the solution turns green.

### Final Answer:
The solution turns **green** when \( H_2S \) is passed through acidified \( K_2Cr_2O_7 \) solution.

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