If the radius of anion is `0.20 nm`, the maximum radius of cation which can be filleed in respective voids is correctly matched in

Options

A. `r_(o+) = 0.0828 nm` for tetrahedral void
B. `r_(o+) = 0.045 nm` for triangular void
C. `r_(o+) = 0.1464 nm` for tetrahedral void
D. None of the above.

Explanation

To solve the problem of determining the maximum radius of cation that can be filled in respective voids given the radius of anion as 0.20 nm, we will follow these steps:

### Step 1: Understand the radius ratio for different voids
The maximum radius ratios for different types of voids are:
- **Octahedral void**: \( r^+ / r^- \leq 0.732 \)
- **Tetrahedral void**: \( r^+ / r^- \leq 0.414 \)
- **Triangular void**: \( r^+ / r^- \leq 0.225 \)

Where \( r^+ \) is the radius of the cation and \( r^- \) is the radius of the anion.

### Step 2: Given data
We are given that the radius of the anion \( r^- = 0.20 \) nm.

### Step 3: Calculate the maximum radius of cation for each type of void
1. **For Octahedral void**:
\[ r^+ / r^- = 0.732 \implies r^+ = 0.732 \times r^- = 0.732 \times 0.20 \text{ nm} = 0.1464 \text{ nm} \]

2. **For Tetrahedral void**:
\[ r^+ / r^- = 0.414 \implies r^+ = 0.414 \times r^- = 0.414 \times 0.20 \text{ nm} = 0.0828 \text{ nm} \]

3. **For Triangular void**:
\[ r^+ / r^- = 0.225 \implies r^+ = 0.225 \times r^- = 0.225 \times 0.20 \text{ nm} = 0.045 \text{ nm} \]

### Step 4: Summary of results
- For **Octahedral void**, the maximum radius of cation \( r^+ = 0.1464 \) nm.
- For **Tetrahedral void**, the maximum radius of cation \( r^+ = 0.0828 \) nm.
- For **Triangular void**, the maximum radius of cation \( r^+ = 0.045 \) nm.

### Step 5: Conclusion
The maximum radius of cation that can be filled in respective voids is:
- \( 0.1464 \) nm for octahedral void,
- \( 0.0828 \) nm for tetrahedral void,
- \( 0.045 \) nm for triangular void.