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`

Explanation

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

### Step 1: Understand the NaCl Structure
In the NaCl structure, the anions (in this case, fluoride ions, F⁻) are located at the corners of the cubic unit cell, while the cations (potassium ions, K⁺) are located at the center of the edges of the cube.

### Step 2: Define the Edge Length
We are given that the edge length of the unit cell (A) is 1 cm.

### Step 3: Determine the Position of Ions
In the NaCl structure:
- The fluoride ions (F⁻) are at the corners of the cube.
- The potassium ions (K⁺) are located at the center of the edges.

### Step 4: Calculate the Distance Between K⁺ and F⁻
The distance between the potassium ion and the fluoride ion can be calculated as follows:
- The distance from the center of the edge (where K⁺ is located) to the corner (where F⁻ is located) is half the edge length (A/2).

### Step 5: Conclusion
Thus, the distance between K⁺ and F⁻ ions is:
\[ \text{Distance} = \frac{A}{2} \]

Given that \( A = 1 \, \text{cm} \):
\[ \text{Distance} = \frac{1 \, \text{cm}}{2} = 0.5 \, \text{cm} \]

### Final Answer
The distance between potassium and fluoride ions is \( 0.5 \, \text{cm} \).

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