Potassium fluoride has `NaCI` type structure .What is the distance between `K' and F` ions if cell edge is a cm?

Options

A. 2a cm
B. a/2 cm
C. 4a cm
D. a/4 cm

Answer

To solve the problem of finding the distance between potassium (K⁺) and fluoride (F⁻) ions in potassium fluoride (KF) which has an NaCl type structure, follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**: Potassium fluoride (KF) crystallizes in a face-centered cubic (FCC) lattice structure similar to sodium chloride (NaCl). In this structure, each potassium ion (K⁺) is surrounded by six fluoride ions (F⁻) and vice versa.

2. **Identify the Cell Edge Length**: The problem states that the cell edge length (a) is given as 1 cm.

3. **Determine the Ion Positions**: In the NaCl type structure, the K⁺ ions occupy the corners of the cube, and the F⁻ ions occupy the face centers. The distance we want to find is between a corner K⁺ ion and a face-centered F⁻ ion.

4. **Calculate the Distance**: The distance between K⁺ and F⁻ ions can be calculated as half the length of the edge of the unit cell. This is because the K⁺ ion at the corner and the F⁻ ion at the face center are separated by half the edge length of the cube.

\[ \text{Distance} = \frac{a}{2} \]

5. **Substitute the Value of a**: Given that the edge length \( a = 1 \, \text{cm} \):

\[ \text{Distance} = \frac{1 \, \text{cm}}{2} = 0.5 \, \text{cm} \]

### Final Answer:
The distance between K⁺ and F⁻ ions is **0.5 cm**.