Colour of `I_2` solution is discharged , when solution of 'X' is added . 'X' is

Options

A. `H_2SO_4`
B. `Na_2SO_4`
C. `Na_2S_2O_3`
D. `S_8`

Answer

To solve the question regarding the color of the \( I_2 \) solution being discharged when a solution of 'X' is added, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Color of \( I_2 \)**:
- The iodine solution (\( I_2 \)) has a characteristic violet color due to the presence of iodine in its molecular form, where the oxidation state of iodine is 0.

2. **Identifying the Reaction**:
- When a reducing agent is added to the iodine solution, it can reduce iodine from its oxidation state of 0 to a lower oxidation state, specifically -1, which corresponds to iodide ions (\( I^- \)). This reduction process leads to the discharge of the violet color, resulting in a colorless solution.

3. **Evaluating Possible Candidates for 'X'**:
- We need to identify which of the given options can act as a reducing agent to reduce \( I_2 \) to \( I^- \). The options provided are:
- \( H_2SO_4 \)
- \( Na_2SO_4 \)
- \( Na_2SO_3 \)
- \( S_8 \)

4. **Analyzing Each Option**:
- **\( H_2SO_4 \)**: This is a strong acid and does not act as a reducing agent for iodine.
- **\( Na_2SO_4 \)**: This is a neutral salt and does not have reducing properties.
- **\( Na_2SO_3 \)**: Sodium sulfite can act as a reducing agent. It can reduce \( I_2 \) to \( I^- \) and thus discharge the color.
- **\( S_8 \)**: Elemental sulfur does not act as a reducing agent in this context.

5. **Conclusion**:
- The only viable candidate that can reduce iodine and discharge its color is \( Na_2SO_3 \). Therefore, the solution 'X' is \( Na_2SO_3 \).

### Final Answer:
The solution 'X' is \( Na_2SO_3 \).