The ionic radii of `A^(+)` and `B^(-)` ions are `0.98xx10^(-10)m` and `1.81xx10^(-10)m`.The coordination number of each ion in `AB` is :

Options

A. 4
B. 8
C. 2
D. 6

Explanation

To find the coordination number of the ions \( A^+ \) and \( B^- \) in the compound \( AB \), we can follow these steps:

### Step 1: Identify the given ionic radii
The ionic radii provided are:
- Radius of \( A^+ \) (cation) = \( 0.98 \times 10^{-10} \, \text{m} \)
- Radius of \( B^- \) (anion) = \( 1.81 \times 10^{-10} \, \text{m} \)

### Step 2: Calculate the radius ratio
The radius ratio is calculated using the formula:
\[ \text{Radius Ratio} = \frac{r_{A^+}}{r_{B^-}} = \frac{0.98 \times 10^{-10} \, \text{m}}{1.81 \times 10^{-10} \, \text{m}} \]

### Step 3: Perform the calculation
Calculating the radius ratio:
\[ \text{Radius Ratio} = \frac{0.98}{1.81} \approx 0.541 \]

### Step 4: Determine the coordination number using the radius ratio
Using the radius ratio, we can refer to the coordination number table:
- For a radius ratio of \( 0.414 \) to \( 0.732 \), the coordination number is \( 6 \) with an octahedral arrangement.

### Conclusion
Thus, the coordination number of both \( A^+ \) and \( B^- \) ions in the compound \( AB \) is \( 6 \).

### Summary
- The coordination number for \( A^+ \) is \( 6 \).
- The coordination number for \( B^- \) is \( 6 \).

---