The coordination number of `(Sr^(2+))`=113 pm and `r(F^(-))`=136 pm, is

Options

A. 6
B. 8
C. 12
D. 4

Explanation

To find the coordination number of \( \text{Sr}^{2+} \) and \( \text{F}^- \) based on their ionic radii, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data**:
- Radius of \( \text{Sr}^{2+} \) (cation) = 113 pm
- Radius of \( \text{F}^- \) (anion) = 136 pm

2. **Calculate the radius ratio**:
- The radius ratio \( r \) is calculated using the formula:
\[ r = \frac{r_{\text{cation}}}{r_{\text{anion}}} \]
- Substituting the values:
\[ r = \frac{113 \, \text{pm}}{136 \, \text{pm}} \approx 0.830 \]

3. **Determine the coordination number using the radius ratio**:
- The radius ratio \( r \) falls within specific ranges that correspond to different coordination numbers. For \( r \) values:
- \( 0.732 < r < 1 \) corresponds to a coordination number of 8.
- Since \( 0.830 \) lies within this range, we conclude that the coordination number is 8.

### Final Answer:
The coordination number of \( \text{Sr}^{2+} \) is **8**.

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