Which one is the strongest bond

Options

A. `Br - F`
B. `F - F`
C. `Cl - F`
D. `Br - Cl`

Explanation

To determine which bond is the strongest among the given options (Br-F, F-F, Cl-F, Br-Cl), we need to analyze the electronegativity of the halogens involved in each bond. Here’s a step-by-step solution:

### Step 1: Identify the Electronegativity of Halogens
The electronegativity values of the halogens are as follows:
- Fluorine (F) - highest electronegativity
- Chlorine (Cl) - lower than F
- Bromine (Br) - lower than Cl
- Iodine (I) - lowest electronegativity

### Step 2: Analyze Each Bond
1. **F-F Bond**:
- Both atoms are fluorine, so the electronegativity difference is 0. This bond is relatively weak due to the lack of difference in electronegativity.

2. **Cl-F Bond**:
- Chlorine has a lower electronegativity than fluorine, but there is still a significant difference. This bond is stronger than F-F.

3. **Br-F Bond**:
- Bromine has an even lower electronegativity than chlorine. The electronegativity difference between Br and F is larger than that of Cl and F, which suggests that this bond is stronger than Cl-F.

4. **Br-Cl Bond**:
- Both bromine and chlorine have relatively low electronegativities compared to fluorine, and the difference between them is not as significant as that between Br and F or Cl and F.

### Step 3: Compare the Bonds
- The strongest bond will be the one with the largest electronegativity difference.
- The electronegativity differences are:
- F-F: 0
- Cl-F: Significant but less than Br-F
- Br-F: Largest difference
- Br-Cl: Smaller difference than Cl-F and Br-F

### Conclusion
The strongest bond among the given options is **Br-F** due to the maximum electronegativity difference.

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