A solid AB has NaCl structure. If the radius of cation `A^(+)` is 170 pm, calculate the maximum possible radius of the anion.

Explanation

To solve the problem of finding the maximum possible radius of the anion in a solid AB with NaCl structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data**:
- Radius of cation \( A^+ \) (denoted as \( R^+ \)) = 170 pm.

2. **Understand the Structure**:
- The solid AB has an NaCl type structure, which means it has a face-centered cubic (FCC) arrangement.
- In NaCl structure, the coordination number is 6 for both cations and anions.

3. **Use the Radius Ratio**:
- For a coordination number of 6, the radius ratio \( \frac{R^+}{R^-} \) (where \( R^- \) is the radius of the anion) is approximately 0.414.

4. **Set Up the Equation**:
- From the radius ratio, we can write the equation:
\[ \frac{R^+}{R^-} = 0.414 \]
- Rearranging gives:
\[ R^- = \frac{R^+}{0.414} \]

5. **Substitute the Known Value**:
- Substitute \( R^+ = 170 \, \text{pm} \):
\[ R^- = \frac{170 \, \text{pm}}{0.414} \]

6. **Calculate the Radius of the Anion**:
- Performing the calculation:
\[ R^- \approx \frac{170}{0.414} \approx 410.6 \, \text{pm} \]

7. **Final Result**:
- The maximum possible radius of the anion \( R^- \) is approximately **410.6 pm**.

### Summary:
The maximum possible radius of the anion in the solid AB with NaCl structure is **410.6 pm**.