If the radius of the bromide ion is 0.182 nm, how large a cation can fit in each of the tetrahedral holes ?

Explanation

To determine the maximum size of a cation that can fit into a tetrahedral hole when given the radius of the bromide ion, we can follow these steps:

### Step 1: Understand the relationship between the cation and anion sizes in a tetrahedral void.
In a tetrahedral void, the size of the cation (R+) can be related to the size of the anion (R-) using the formula:
\[ \frac{R^+}{R^-} = k \]
where \( k \) is a constant that varies between 0.225 and 0.414 depending on the specific ionic sizes.

### Step 2: Identify the radius of the bromide ion.
Given that the radius of the bromide ion (R-) is:
\[ R^- = 0.182 \, \text{nm} \]

### Step 3: Use the maximum value of k for the calculation.
To find the maximum size of the cation, we will use the upper limit of k, which is 0.414. Thus, we can set up the equation:
\[ R^+ = k \times R^- \]

### Step 4: Substitute the values into the equation.
Substituting the values we have:
\[ R^+ = 0.414 \times 0.182 \, \text{nm} \]

### Step 5: Perform the calculation.
Now, calculate the value:
\[ R^+ = 0.414 \times 0.182 = 0.075468 \, \text{nm} \]

### Step 6: Round the result to an appropriate number of significant figures.
Rounding the result, we find:
\[ R^+ \approx 0.0755 \, \text{nm} \]

### Conclusion:
The maximum size of a cation that can fit into the tetrahedral holes is approximately:
\[ R^+ \approx 0.0755 \, \text{nm} \]