A solid compound PQ has CsCI structure. If the radius of cation is
120 pm then find radius of anion ?

Options

A. 88 pm
B. 140 pm
C. 164 pm
D. 120 pm

Explanation

To solve the problem of finding the radius of the anion in a solid compound PQ with a CsCl structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**: The CsCl structure indicates that the cation (Cs⁺) occupies the cubic voids in a body-centered cubic (BCC) lattice, while the anion (Cl⁻) occupies the corners of the cube.

2. **Identify the Given Data**:
- Radius of the cation (R⁺) = 120 pm.

3. **Use the Radius Ratio**:
- For the CsCl structure, the radius ratio of cation to anion is given by the formula:
\[ \frac{R^+}{R^-} = 0.732 \]
- Here, \(R^+\) is the radius of the cation and \(R^-\) is the radius of the anion.

4. **Set Up the Equation**:
- Substitute the known value of \(R^+\) into the radius ratio formula:
\[ \frac{120 \text{ pm}}{R^-} = 0.732 \]

5. **Solve for the Radius of the Anion (R⁻)**:
- Rearranging the equation gives:
\[ R^- = \frac{120 \text{ pm}}{0.732} \]
- Now, calculate \(R^-\):
\[ R^- \approx \frac{120}{0.732} \approx 163.9 \text{ pm} \]

6. **Round the Result**:
- Rounding 163.9 pm gives approximately:
\[ R^- \approx 164 \text{ pm} \]

### Final Answer:
The radius of the anion (R⁻) is approximately **164 pm**.