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

Options

A. 26.88 pm
B. 268.8 pm
C. 2688 pm
D. Unpredictable

Answer

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

### Step-by-Step Solution:

1. **Understand the Structure**:
- KF has a NaCl-type structure, which means it has a face-centered cubic (FCC) arrangement of anions (F\(^-\)) with cations (K\(^+\)) occupying the octahedral voids.

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

3. **Determine the Relationship**:
- In a NaCl-type structure, the distance between the cation and anion can be determined using the formula:
\[ \text{Distance} = r_{K^+} + r_{F^-} \]
- Here, \( r_{K^+} \) is the radius of the potassium ion and \( r_{F^-} \) is the radius of the fluoride ion.

4. **Find the Distance**:
- The distance between \( K^+ \) and \( F^- \) ions in the unit cell can also be expressed in terms of the edge length \( a \):
\[ \text{Distance} = \frac{a}{2} \]
- This is because in the FCC structure, the \( K^+ \) ion is located at the center of the edge, and the \( F^- \) ions are located at the corners.

5. **Calculate the Distance**:
- Substitute the value of \( a \):
\[ \text{Distance} = \frac{537.6 \text{ pm}}{2} = 268.8 \text{ pm} \]

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

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