When `I_2` is passed through `KCI, KF and KBr` solutions 

Options

A. `Cl_2` and `Br_2` are evolved
B. `Cl_2, Br_2` and `F_2` are evolved
C. `Cl_2` is evolved
D. None of these

Answer

To solve the question, we need to analyze the behavior of iodine (I₂) when it is passed through solutions of potassium chloride (KCl), potassium fluoride (KF), and potassium bromide (KBr).

### Step-by-Step Solution:

1. **Understanding Halogen Reactivity**:
- The reactivity of halogens decreases down the group in the periodic table. This means that fluorine (F) is the most reactive, followed by chlorine (Cl), bromine (Br), and iodine (I) is the least reactive.

2. **Displacement Reactions**:
- A more reactive halogen can displace a less reactive halogen from its salt solution. For example, if we have a solution of KBr, a more reactive halogen like chlorine can displace bromine.

3. **Analyzing the Solutions**:
- When I₂ is passed through KCl, KBr, and KF:
- **KCl**: Iodine cannot displace chlorine because iodine is less reactive than chlorine.
- **KBr**: Iodine cannot displace bromine because iodine is less reactive than bromine.
- **KF**: Iodine cannot displace fluorine because iodine is less reactive than fluorine.

4. **Conclusion**:
- Since iodine is less reactive than all the halogens present in the solutions (Cl, Br, and F), it cannot displace any of them from their respective potassium salts. Therefore, no reaction occurs when I₂ is passed through KCl, KF, or KBr solutions.

5. **Final Answer**:
- The correct option is that iodine does not displace any halogens from their salts, which leads us to conclude that the answer is **none of the above** (Option D).