Which one of the following is most unstable 

Options

A. `BrF`
B. `ClF`
C. `BrCl`
D. `IF`

Answer

To determine which compound is the most unstable among the given options, we will analyze the stability of each compound based on the principles of orbital overlap and atomic size.

### Step-by-Step Solution:

1. **Identify the Compounds**: We need to look at the compounds provided in the question. Let's assume they are:
- Iodine Fluoride (IF)
- Bromine Fluoride (BrF)
- Chlorine Fluoride (ClF)
- Bromine Chloride (BrCl)

2. **Understand Orbital Overlap**: The stability of a compound often depends on the effective overlap of atomic orbitals. A good overlap leads to a stable bond, while poor overlap can result in instability.

3. **Analyze Iodine Fluoride (IF)**:
- Iodine has an outer electron configuration of 5S² 5P⁵, while fluorine has 2S² 2P⁵.
- The size difference between iodine (5th period) and fluorine (2nd period) is significant. The large size of iodine leads to poor overlap with the small orbitals of fluorine.

4. **Conclusion on IF**: Due to the poor overlap between the orbitals of iodine and fluorine, IF is unstable. It tends to decompose into iodine and IF₅ (iodine pentafluoride).

5. **Compare with Other Compounds**:
- **Bromine Fluoride (BrF)** and **Chlorine Fluoride (ClF)**: These compounds have better orbital overlap because bromine and chlorine are closer in size to fluorine compared to iodine. Thus, they are more stable than IF.
- **Bromine Chloride (BrCl)**: Similar reasoning applies; bromine and chlorine are also of comparable size, leading to better stability.

6. **Final Answer**: Based on the analysis, Iodine Fluoride (IF) is the most unstable compound among the options provided.