Which of the following halogen disproportionates in water?

Options

A. `F_(2)`
B. `Cl_(2)`
C. `I_2`
D. all three

Answer

To determine which halogen disproportionates in water, we need to analyze the reactions of different halogens with water and check for the occurrence of disproportionation. Disproportionation is a reaction in which a single substance is both oxidized and reduced, resulting in two different products.

### Step-by-Step Solution:

1. **Understanding Disproportionation:**
- Disproportionation occurs when an element in a single oxidation state is both oxidized and reduced to form two different products.

2. **Analyzing Fluorine (F2):**
- Fluorine is a very strong oxidizing agent.
- When F2 reacts with water, the reaction can be represented as:
\[ F_2 + H_2O \rightarrow O_2 + HF \]
- In this reaction, F2 is reduced to HF, but it does not undergo disproportionation because it does not change oxidation states (it only forms HF).

3. **Analyzing Chlorine (Cl2):**
- Chlorine has a moderate oxidizing ability.
- The reaction of Cl2 with water is:
\[ Cl_2 + H_2O \rightarrow HCl + HOCl \]
- Here, chlorine changes from 0 (in Cl2) to -1 (in HCl) and +1 (in HOCl).
- This indicates that Cl2 is both oxidized and reduced, confirming that it undergoes disproportionation.

4. **Analyzing Iodine (I2):**
- Iodine is a weak oxidizing agent and does not react readily with water.
- Even if it does react, it does not lead to disproportionation because it does not change oxidation states significantly.

5. **Conclusion:**
- Among the halogens analyzed, only chlorine (Cl2) undergoes disproportionation in water.

### Final Answer:
The halogen that disproportionates in water is **Cl2**.