A solid AB has the NaCL structure, If radius of cation `A^(+)` is 120 pm, calculate the maximum possible value of the radius of the anion `B^(-)`

Options

A. 120 pm
B. 240 pm
C. 290 pm
D. 360 pm

Explanation

To solve the problem of finding the maximum possible radius of the anion \( B^{-} \) in a solid \( AB \) with an NaCl structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The radius of the cation \( A^{+} \) is given as \( 120 \, \text{pm} \).
- The structure is NaCl, which has a coordination number of 6.

2. **Understand the Radius Ratio**:
- For the NaCl structure, the radius ratio \( R \) (which is the ratio of the radius of the cation to the radius of the anion) is given by:
\[ R = \frac{r_{A^{+}}}{r_{B^{-}}} \]
- The radius ratio for a coordination number of 6 is between \( 0.414 \) and \( 0.732 \).

3. **Calculate the Maximum Radius of the Anion**:
- To find the maximum possible radius of the anion \( B^{-} \), we will use the minimum radius ratio \( 0.414 \):
\[ R = 0.414 = \frac{r_{A^{+}}}{r_{B^{-}}} \]
- Rearranging the formula to find \( r_{B^{-}} \):
\[ r_{B^{-}} = \frac{r_{A^{+}}}{0.414} \]
- Substitute the value of \( r_{A^{+}} \):
\[ r_{B^{-}} = \frac{120 \, \text{pm}}{0.414} \]

4. **Perform the Calculation**:
- Calculate \( r_{B^{-}} \):
\[ r_{B^{-}} = \frac{120}{0.414} \approx 289.86 \, \text{pm} \]
- Rounding this value gives:
\[ r_{B^{-}} \approx 290 \, \text{pm} \]

5. **Conclusion**:
- The maximum possible value of the radius of the anion \( B^{-} \) is approximately \( 290 \, \text{pm} \).

### Final Answer:
The maximum possible value of the radius of the anion \( B^{-} \) is **290 pm**.