If the radius of `K^(+)` and `F^(-)` are 133 pm and 136 pm respectively, the distance between `K^(+)` and `F^(-)` in KF is......... pm 

Options

A. 269
B. 134.5
C. 136
D. 3

Answer

To find the distance between the potassium ion (K⁺) and the fluoride ion (F⁻) in potassium fluoride (KF), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Radii of Ions**:
- The radius of the potassium ion (K⁺) is given as 133 pm.
- The radius of the fluoride ion (F⁻) is given as 136 pm.

2. **Understand the Arrangement**:
- In the crystal structure of potassium fluoride, K⁺ ions are located in octahedral voids, while F⁻ ions are located in cubic close-packed (CCP) sites.

3. **Calculate the Distance Between Ions**:
- The shortest distance \( d \) between the K⁺ and F⁻ ions can be calculated by adding their radii together:
\[ d = r_{K^+} + r_{F^-} \]
- Substituting the values:
\[ d = 133 \, \text{pm} + 136 \, \text{pm} \]
- Performing the addition:
\[ d = 269 \, \text{pm} \]

4. **Conclusion**:
- The distance between the K⁺ and F⁻ ions in potassium fluoride (KF) is 269 pm.

### Final Answer:
The distance between K⁺ and F⁻ in KF is **269 pm**.

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