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 \( A^+B^- \) based on the given ionic radii, we can follow these steps:

### Step 1: Identify the given data
- The radius of the cation \( A^+ \) is \( 180 \, \text{pm} \).
- The radius of the anion \( B^- \) is \( 187 \, \text{pm} \).

### Step 2: Calculate the ratio of the radii
To find the ratio of the radii of the cation to the anion, we use the formula:
\[ \text{Ratio} = \frac{\text{Radius of cation}}{\text{Radius of anion}} = \frac{r_A}{r_B} \]
Substituting the values:
\[ \text{Ratio} = \frac{180 \, \text{pm}}{187 \, \text{pm}} \approx 0.96 \]

### Step 3: Analyze the ratio
The calculated ratio \( 0.96 \) falls within the range of \( 0.732 \) to \( 1.000 \). This range indicates that the coordination number for this type of ionic compound is typically \( 8 \).

### Step 4: Determine the crystal structure
Based on the coordination number of \( 8 \) and the ratio of the ionic radii, the compound can be classified as having a structure similar to \( \text{CsCl} \) type. In this structure, each cation is surrounded by 8 anions and vice versa.

### Conclusion
Thus, the crystal structure of the ionic compound \( A^+B^- \) is similar to the \( \text{CsCl} \) type.

### Final Answer
The crystal structure of the ionic compound \( A^+B^- \) is \( \text{CsCl} \) type.

---