The radius of the bromide ion is 0.182 nm. According to the radius ratio rule the largest cation that can fit in tetrahedral hole has radius

Options

A. `7.53times10^(-2)nm`
B. `1.82times10^(-1)nm`
C. `7.53times10^(-1)nm`
D. `1.82times10^(-2)nm`

Explanation

To find the largest cation that can fit in a tetrahedral hole according to the radius ratio rule, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- The radius of the bromide ion (Br⁻) is given as \( r_{Br^-} = 0.182 \, \text{nm} \).

2. **Understand the radius ratio rule for tetrahedral holes:**
- The radius ratio rule states that the radius of the cation (\( r_{cation} \)) that can fit into a tetrahedral hole is related to the radius of the anion (\( r_{anion} \)) by the formula:
\[ \text{Radius ratio} = \frac{r_{cation}}{r_{anion}} \]
- For tetrahedral holes, the radius ratio typically ranges from approximately 0.225 to 0.414.

3. **Determine the maximum radius of the cation:**
- To find the largest cation that can fit in the tetrahedral hole, we will use the upper limit of the radius ratio, which is 0.414.
- Thus, we can express the radius of the cation as:
\[ r_{cation} = 0.414 \times r_{Br^-} \]

4. **Substitute the value of \( r_{Br^-} \):**
- Now, substituting the value of the bromide ion radius:
\[ r_{cation} = 0.414 \times 0.182 \, \text{nm} \]

5. **Calculate the radius of the cation:**
- Performing the multiplication:
\[ r_{cation} = 0.414 \times 0.182 = 0.075468 \, \text{nm} \]
- This can also be expressed in scientific notation:
\[ r_{cation} \approx 7.55 \times 10^{-2} \, \text{nm} \]

6. **Conclusion:**
- Therefore, the largest cation that can fit in the tetrahedral hole has a radius of approximately \( 0.0755 \, \text{nm} \) or \( 7.55 \times 10^{-2} \, \text{nm} \).