A solid `A^(+)B^(-)` has NaCl close packed structure. The radius of the cation when the radius of the anion is 250 pm is

Options

A. 103.5 pm
B. 207 pm
C. 69 pm
D. 276 pm

Explanation

To solve the problem of finding the radius of the cation \( A^{+} \) in a solid \( A^{+}B^{-} \) with a NaCl close-packed structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- The radius of the anion \( B^{-} \) is given as \( R_{B} = 250 \) pm.

2. **Understand the Structure:**
- The NaCl structure consists of cations occupying octahedral voids in a face-centered cubic (FCC) lattice formed by anions.

3. **Use the Radius Ratio Rule:**
- The radius ratio rule states that the ratio of the radius of the cation \( R_{A} \) to the radius of the anion \( R_{B} \) can be expressed as:
\[ \frac{R_{A}}{R_{B}} = r \]
- For an octahedral void, the radius ratio \( r \) is approximately \( 0.414 \).

4. **Rearrange the Formula:**
- To find the radius of the cation \( R_{A} \), we can rearrange the formula:
\[ R_{A} = r \times R_{B} \]

5. **Substitute the Values:**
- Substitute the known values into the equation:
\[ R_{A} = 0.414 \times 250 \text{ pm} \]

6. **Calculate the Radius of the Cation:**
- Perform the multiplication:
\[ R_{A} = 0.414 \times 250 = 103.5 \text{ pm} \]

7. **Final Answer:**
- The radius of the cation \( A^{+} \) is \( 103.5 \) pm.

### Summary:
The radius of the cation \( A^{+} \) in the solid \( A^{+}B^{-} \) with NaCl structure, given that the radius of the anion \( B^{-} \) is 250 pm, is \( 103.5 \) pm.