Reducing halogen is

Options

A. F
B. Cl
C. Br
D. I

Answer

To determine which halogen is a reducing halogen among Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I), we need to analyze their oxidation states and tendencies to act as reducing agents.

### Step-by-Step Solution:

1. **Understanding Reducing Agents**:
- A reducing agent is a substance that donates electrons to another substance, thereby reducing it while itself gets oxidized. In the context of halogens, we need to identify which halogen can undergo oxidation (increase in oxidation state) while reducing another species.

2. **Oxidation States of Halogens**:
- The oxidation state of halogens in their elemental form (e.g., F2, Cl2, Br2, I2) is 0. When they gain an electron, they form halide ions (F-, Cl-, Br-, I-) with an oxidation state of -1.
- The reaction can be summarized as:
- \( \text{X}_2 + 2e^- \rightarrow 2\text{X}^- \) (where X is a halogen)

3. **Analyzing Each Halogen**:
- **Fluorine (F)**:
- It has the highest electronegativity and is a strong oxidizing agent. It is unlikely to act as a reducing agent because it prefers to gain electrons rather than lose them.
- **Chlorine (Cl)**:
- Chlorine is also a strong oxidizing agent, but less so than fluorine. It can be reduced but does not have a strong tendency to oxidize itself.
- **Bromine (Br)**:
- Bromine has a moderate tendency to act as a reducing agent compared to chlorine and fluorine but is still not the strongest among the halogens.
- **Iodine (I)**:
- Iodine has the largest atomic size among the halogens, leading to weaker bonds in I2. This allows it to more readily lose electrons and thus act as a reducing agent.

4. **Conclusion**:
- Among the halogens, Iodine (I) has the highest tendency to act as a reducing agent because it can easily undergo oxidation (from I- to I2) while reducing other substances. Therefore, the correct answer is **Iodine (I)**.

### Final Answer:
**Iodine (I) is the reducing halogen.**