The ionic radii of `Rb^(+)` and `Br^(-)` ions are 147 pm and and 195 pm resectively. Deduce the possible C.N of `Rb^(+)` ions in RbBr

Explanation

To deduce the possible coordination number (C.N.) of Rb^(+) ions in RbBr, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ionic Radii**:
- The ionic radius of Rb^(+) is given as 147 pm.
- The ionic radius of Br^(-) is given as 195 pm.

2. **Calculate the Radius Ratio**:
- The radius ratio (r) is calculated using the formula:
\[ r = \frac{r_{cation}}{r_{anion}} = \frac{r_{Rb^{+}}}{r_{Br^{-}}} \]
- Substituting the values:
\[ r = \frac{147 \, \text{pm}}{195 \, \text{pm}} \approx 0.753 \]

3. **Determine the Coordination Number**:
- The coordination number can be inferred from the radius ratio. Based on empirical data:
- If \( r < 0.225 \), C.N. = 4
- If \( 0.225 < r < 0.414 \), C.N. = 6
- If \( 0.414 < r < 0.732 \), C.N. = 8
- If \( 0.732 < r < 1.0 \), C.N. = 12
- Since our calculated radius ratio \( r = 0.753 \) falls between 0.732 and 1.0, we can conclude that the coordination number is 8.

4. **Final Conclusion**:
- Therefore, the possible coordination number of Rb^(+) ions in RbBr is **8**.