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 the potassium ion (K⁺) and the fluoride ion (F⁻) 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, meaning it has a face-centered cubic (FCC) arrangement for the anions (F⁻) and octahedral voids for the cations (K⁺).
- In this structure, K⁺ ions occupy the octahedral voids, which are located at the body center and the edge centers of the unit cell.

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

3. **Determine the Relationship Between Edge Length and Ionic Radii**:
- In a NaCl-type structure, the distance between the cation and anion can be expressed in terms of the ionic radii (r) of K⁺ and F⁻.
- The relationship is given by the equation:
\[ a = 4r \]
- Here, \( r \) is the radius of the fluoride ion (F⁻). The distance between K⁺ and F⁻ is \( 2r \).

4. **Calculate the Distance Between K⁺ and F⁻**:
- From the equation \( a = 4r \), we can express \( r \) as:
\[ r = \frac{a}{4} \]
- Therefore, the distance \( d \) between K⁺ and F⁻ can be calculated as:
\[ d = 2r = 2 \left(\frac{a}{4}\right) = \frac{a}{2} \]

5. **Substitute the Given Edge Length**:
- Now, substituting the value of \( a \):
\[ d = \frac{537.6 \text{ pm}}{2} = 268.8 \text{ pm} \]

### Final Answer:
The distance between K⁺ and F⁻ in KF is **268.8 pm**.

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