KF has NaCl structure. The edge length of its unit cell has been found to be 537.6 pm The distance between `K^+` in and `F^-` in KF is

Options

A. 537.6 pm
B. 1075.2 pm
C. 268.5 pm
D. None of these

Answer

To find the distance between \( K^+ \) and \( F^- \) in KF, which has a NaCl-type structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure**:
- KF crystallizes in a NaCl-type structure, which means it has a face-centered cubic (FCC) lattice. In this structure, the \( K^+ \) ions are located at the corners of the cube and at the face centers, while the \( F^- \) ions occupy the edge centers.

2. **Identifying the Edge Length**:
- The edge length \( a \) of the unit cell is given as 537.6 pm (picometers).

3. **Calculating the Distance**:
- In the NaCl structure, the distance between the \( K^+ \) and \( F^- \) ions can be determined using the relationship between the edge length and the ionic radii. The distance between \( K^+ \) and \( F^- \) is equal to half the edge length:
\[ \text{Distance} = \frac{a}{2} \]
- Substituting the value of \( a \):
\[ \text{Distance} = \frac{537.6 \, \text{pm}}{2} = 268.8 \, \text{pm} \]

4. **Final Result**:
- The distance between \( K^+ \) and \( F^- \) in KF is approximately 268.8 pm.

### Answer:
The distance between \( K^+ \) and \( F^- \) in KF is **268.8 pm**.