Which of the following bonds is most polar ?

Options

A. `Cl-Cl`
B. `N-F`
C. `C-F`
D. `O-F`

Explanation

To determine which bond is the most polar among the given options, we need to analyze the electronegativity differences of the bonds in question. Here's a step-by-step solution:

### Step 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 in electronegativity, the more polar the bond.

### Step 2: Identify the Bonds
The given bonds to analyze are:
1. Cl-Cl (Chlorine-Chlorine)
2. N-F (Nitrogen-Fluorine)
3. C-F (Carbon-Fluorine)
4. O-F (Oxygen-Fluorine)

### Step 3: Determine Electronegativity Values
Using the Pauling scale, we can assign the following electronegativity values:
- Chlorine (Cl): 3.16
- Nitrogen (N): 3.04
- Carbon (C): 2.55
- Oxygen (O): 3.44
- Fluorine (F): 3.98

### Step 4: Calculate Electronegativity Differences
Now, we will calculate the electronegativity differences for each bond:
1. **Cl-Cl**: |3.16 - 3.16| = 0 (nonpolar)
2. **N-F**: |3.04 - 3.98| = 0.94
3. **C-F**: |2.55 - 3.98| = 1.43
4. **O-F**: |3.44 - 3.98| = 0.54

### Step 5: Compare the Differences
Now we compare the electronegativity differences:
- Cl-Cl: 0
- N-F: 0.94
- C-F: 1.43
- O-F: 0.54

### Step 6: Identify the Most Polar Bond
The bond with the highest electronegativity difference is C-F (1.43). Therefore, the most polar bond among the given options is the Carbon-Fluorine bond.

### Conclusion
The answer is that the **Carbon-Fluorine (C-F) bond is the most polar bond**.