Among `Cl^-, Br^-, I^-`, the correct order for being oxidise to dihalogen is

Options

A. `I^(-) gt Cl^(-) gt Br^(-)`
B. `Cl^(-) gt Br^(-) gt I^(-)`
C. `I^(-) gt Br^(-) gt Cl^(-)`
D. `Br^(-) gt I^(-) gt Cl^(-)`

Answer

To determine the correct order of the halides \( Cl^- \), \( Br^- \), and \( I^- \) for being oxidized to their respective dihalogens \( Cl_2 \), \( Br_2 \), and \( I_2 \), we need to consider their ability to lose electrons, which is related to their reducing strength.

### Step-by-Step Solution:

1. **Understanding the Halides**:
- We have three halides: \( Cl^- \), \( Br^- \), and \( I^- \).
- These halides can be oxidized to their respective dihalogens: \( Cl_2 \), \( Br_2 \), and \( I_2 \).

2. **Oxidizing Strength**:
- The oxidizing strength of these halides is inversely related to their reducing strength. A stronger reducing agent will be more easily oxidized.
- The order of reducing strength for halides is: \( I^- > Br^- > Cl^- \).
- This means \( I^- \) is the strongest reducing agent and will be oxidized first, followed by \( Br^- \), and then \( Cl^- \).

3. **Conclusion**:
- Therefore, the correct order for being oxidized to dihalogen is:
\[ I^- > Br^- > Cl^- \]
- In terms of dihalogen formation, this translates to:
\[ I_2 > Br_2 > Cl_2 \]

### Final Answer:
The correct order for being oxidized to dihalogen is \( I^- > Br^- > Cl^- \).