In `A^(+)B^(-)` ionic compound, radii of `A^(+)andB^(-)` ions are 180 pm and 187 pm respectively. The crystal structure of this compound will be:

Options

A. NaCl type
B. CsCl type
C. ZnS type
D. Similar to diamond

Answer

To determine the crystal structure of the ionic compound \( A^{+}B^{-} \) given the radii of the cation \( A^{+} \) and anion \( B^{-} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ionic Radii:**
- The radius of the cation \( A^{+} \) is given as 180 pm.
- The radius of the anion \( B^{-} \) is given as 187 pm.

2. **Calculate the Radius Ratio:**
- The radius ratio \( r \) is calculated using the formula:
\[ r = \frac{r_{cation}}{r_{anion}} = \frac{180 \, \text{pm}}{187 \, \text{pm}} \]
- Performing the calculation:
\[ r = \frac{180}{187} \approx 0.962 \]

3. **Apply the Radius Ratio Rule:**
- According to the radius ratio rule, we can determine the coordination number based on the calculated radius ratio:
- If \( 0.732 < r < 1 \), the coordination number is 8, indicating a body-centered cubic (BCC) structure.

4. **Determine the Crystal Structure:**
- Since the calculated radius ratio \( r \approx 0.962 \) falls within the range for a coordination number of 8, we conclude that the crystal structure of the compound \( A^{+}B^{-} \) is body-centered cubic (BCC).

5. **Final Conclusion:**
- The crystal structure of the ionic compound \( A^{+}B^{-} \) is BCC with a coordination number of 8.