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

Options

A. `410.6` pm
B. `397.4` pm
C. `210.9` pm
D. `347.9` pm.

Answer

To solve the problem, we need to calculate the maximum possible radius of the anion in a solid AB that has the NaCl structure, given the radius of the cation \( A^+ \) is 170 pm.

### Step-by-Step Solution:

1. **Understand the NaCl Structure**:
- The NaCl structure is a type of ionic crystal structure where cations and anions are arranged in a face-centered cubic (FCC) lattice. In this structure, each cation is surrounded by six anions and vice versa, giving a coordination number of 6.

2. **Use the Radius Ratio Rule**:
- In ionic compounds, the stability of the structure can be analyzed using the radius ratio of the cation to the anion. For a coordination number of 6, the radius ratio (\( r_A/r_B \)) should be approximately 0.414, where \( r_A \) is the radius of the cation and \( r_B \) is the radius of the anion.

3. **Set Up the Equation**:
- Given:
- Radius of cation \( r_A = 170 \) pm
- Radius ratio for coordination number 6 \( \frac{r_A}{r_B} = 0.414 \)
- We can express the radius of the anion \( r_B \) in terms of \( r_A \):
\[ \frac{r_A}{r_B} = 0.414 \implies r_B = \frac{r_A}{0.414} \]

4. **Substitute the Known Values**:
- Substitute \( r_A = 170 \) pm into the equation:
\[ r_B = \frac{170 \text{ pm}}{0.414} \]

5. **Calculate \( r_B \)**:
- Performing the calculation:
\[ r_B = \frac{170}{0.414} \approx 410.6 \text{ pm} \]

6. **Conclusion**:
- The maximum possible radius of the anion \( B^{-} \) is approximately \( 410.6 \) pm.

### Final Answer:
The maximum possible radius of the anion \( B^{-} \) is \( 410.6 \) pm.

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