A violet colour is obtained on adding `Cl_(2)` water in solution of potassium halide in presence of chloroform and on adding excess of `Cl_(2)` water, violet colour disappears and colourless solution appears. The test shows the presence of

Options

A. Iodide ion
B. Bromide ion
C. Chloride ion
D. Iodide and bromide ion

Answer

To solve the question, we need to analyze the reactions taking place when chlorine water is added to a solution of potassium halide in the presence of chloroform.

### Step-by-Step Solution:

1. **Understanding the Initial Reaction**:
- When chlorine water (Cl2 in water) is added to a solution of potassium halide (for example, potassium iodide, KI), a reaction occurs where free iodine (I2) is liberated.
- The reaction can be represented as:
\[ 2KI + Cl2 \rightarrow 2KCl + I2 \]
- This indicates that iodine is being produced from the potassium iodide.

2. **Formation of Violet Color**:
- The liberated iodine (I2) has a brown color in aqueous solution. However, when it is shaken with chloroform (CHCl3), it forms a violet solution. This is due to the solubility of iodine in chloroform.
- The violet color indicates the presence of iodine in the solution.

3. **Adding Excess Chlorine Water**:
- When excess chlorine water is added to the solution, the violet color disappears, indicating that the iodine is being oxidized or reacted further.
- The excess chlorine reacts with the iodine, leading to the formation of iodide ions (I-), which are colorless in solution.
- The reaction can be summarized as:
\[ I2 + Cl2 \rightarrow 2ICl \]
- This reaction shows that iodine is being converted to iodide ions, which do not impart any color to the solution.

4. **Conclusion**:
- The presence of the violet color initially indicates the presence of iodine (I2), and the disappearance of the violet color upon adding excess chlorine indicates that iodine is being converted to iodide ions (I-).
- Therefore, the test shows the presence of iodide ions.

### Final Answer:
The test shows the presence of **iodide ions (I-)**.