When `I_(2)` is passed through `KCl, KF` and `KBr` solution

Options

A. `Cl_(2)` and `Br_(2)` are evolved
B. `Cl_(2)` is evolved
C. `Cl_(2), Br_(2)` and `F_(2)` are evolved
D. None of the above

Explanation

To solve the question regarding the behavior of iodine (I₂) when passed through solutions of KCl, KF, and KBr, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reactivity Series**:
- Iodine (I₂) is less reactive than chlorine (Cl₂), fluorine (F₂), and bromine (Br₂). This is due to its lower electronegativity and oxidizing power compared to these halogens.

2. **Identify the Nature of the Reaction**:
- In a displacement reaction, a more reactive halogen can displace a less reactive halogen from its compound.

3. **Analyze Each Solution**:
- **KCl (Potassium Chloride)**: Chlorine is more reactive than iodine. Therefore, iodine cannot displace chlorine from KCl.
- **KF (Potassium Fluoride)**: Fluorine is the most reactive halogen. Iodine cannot displace fluorine from KF.
- **KBr (Potassium Bromide)**: Bromine is more reactive than iodine. Thus, iodine cannot displace bromine from KBr.

4. **Conclusion**:
- Since iodine cannot displace chlorine, fluorine, or bromine from their respective potassium salts, no reaction occurs when I₂ is passed through KCl, KF, or KBr solutions.

5. **Final Answer**:
- There will be no reaction in any of the cases, and the correct answer is that iodine does not displace any of the halogens in these solutions.