Which bond is expected to be the least polar?

Options

A. O-F
B. P-F
C. Si-N
D. B-F

Explanation

To determine which bond is expected to be the least polar, we need to analyze the electronegativity differences between the atoms in each bond. The polarity of a bond is directly related to the difference in electronegativity between the two bonded atoms. A smaller difference in electronegativity indicates a less polar bond.

### Step-by-Step Solution:

1. **Understand Polarity**:
- Polarity in a bond arises from the difference in electronegativity between the two atoms involved in the bond. The greater the difference, the more polar the bond.

2. **Identify the Bonds**:
- The bonds given in the question are:
1. Oxygen and Fluorine (O-F)
2. Phosphorus and Fluorine (P-F)
3. Silicon and Nitrogen (Si-N)
4. Boron and Fluorine (B-F)

3. **Determine Electronegativity Values**:
- Look up the electronegativity values for each element:
- Oxygen (O): 3.44
- Fluorine (F): 3.98
- Phosphorus (P): 2.19
- Silicon (Si): 1.90
- Nitrogen (N): 3.04
- Boron (B): 2.04

4. **Calculate Electronegativity Differences**:
- O-F: |3.44 - 3.98| = 0.54
- P-F: |2.19 - 3.98| = 1.79
- Si-N: |1.90 - 3.04| = 1.14
- B-F: |2.04 - 3.98| = 1.94

5. **Analyze the Differences**:
- The calculated differences are:
- O-F: 0.54
- P-F: 1.79
- Si-N: 1.14
- B-F: 1.94
- The smallest difference is for the O-F bond.

6. **Conclusion**:
- Since the O-F bond has the smallest electronegativity difference (0.54), it is expected to be the least polar bond among the given options.

### Final Answer:
The bond that is expected to be the least polar is the **Oxygen and Fluorine (O-F) bond**.