A solid `A^(o+)B^(ɵ)` has `NaCl`-type close-packed structure. If the radius of the cation is `90` pm, calculate the probable range of the radius of the anion.

Explanation

To solve the problem, we need to find the probable range of the radius of the anion (R_a) given the radius of the cation (R_c) and the NaCl-type close-packed structure.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Radius of the cation (R_c) = 90 pm
- Structure type = NaCl-type close-packed structure (which has a coordination number of 6)

2. **Understand the Relationship in NaCl Structure:**
- In a NaCl-type structure, the ratio of the radius of the cation to the radius of the anion is given by:
\[ \frac{R_c}{R_a} = 0.414 \quad \text{(for the upper limit)} \]
\[ \frac{R_c}{R_a} = 0.732 \quad \text{(for the lower limit)} \]

3. **Calculate the Upper Limit for R_a:**
- Using the upper limit ratio:
\[ R_a \geq \frac{R_c}{0.414} \]
- Substitute the value of R_c:
\[ R_a \geq \frac{90 \text{ pm}}{0.414} \approx 217.39 \text{ pm} \]

4. **Calculate the Lower Limit for R_a:**
- Using the lower limit ratio:
\[ R_a \leq \frac{R_c}{0.732} \]
- Substitute the value of R_c:
\[ R_a \leq \frac{90 \text{ pm}}{0.732} \approx 122.95 \text{ pm} \]

5. **Combine the Results:**
- From the calculations, we find:
\[ 122.95 \text{ pm} < R_a < 217.39 \text{ pm} \]
- Therefore, the probable range of the radius of the anion (R_a) is:
\[ R_a \in (122.95 \text{ pm}, 217.39 \text{ pm}) \]

### Final Answer:
The probable range of the radius of the anion is approximately \( R_a \in (122.95 \text{ pm}, 217.39 \text{ pm}) \).