`Br^-` ions form a close packed structure. If the radius of `Br^–` ions is 195 pm, calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation `A^+` having a radius of 82 pm be shipped into be octahedral hole of the crystal `A^+ Br^–` ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We need to find the radius of the cation that fits into the tetrahedral hole given the radius of the `Br^-` ion, which is 195 pm. We also need to determine if a cation `A^+` with a radius of 82 pm can fit into the octahedral hole of the crystal `A^+ Br^-`.

### Step 2: Use the Radius Ratio Rule
The radius ratio rule helps us determine the size of ions that can fit into different types of voids in a crystal lattice.

- For tetrahedral holes, the radius ratio (r_cation/r_anion) should be between 0.225 and 0.414.
- For octahedral holes, the radius ratio should be between 0.414 and 0.732.

### Step 3: Calculate the Maximum Radius for Tetrahedral Hole
Given:
- Radius of `Br^-` (r_anion) = 195 pm

Using the upper limit of the tetrahedral hole radius ratio (0.414):
\[ r_{cation} = r_{anion} \times 0.414 \]
\[ r_{cation} = 195 \, \text{pm} \times 0.414 \]
\[ r_{cation} = 80.73 \, \text{pm} \]

### Step 4: Determine if `A^+` Can Fit into the Octahedral Hole
Now we need to check if the cation `A^+` with a radius of 82 pm can fit into the octahedral hole.

Since the maximum radius for tetrahedral holes is 80.73 pm, and the radius of `A^+` is 82 pm, we can conclude:
- Since 82 pm > 80.73 pm, `A^+` cannot fit into the tetrahedral hole.

### Step 5: Check if `A^+` Fits into the Octahedral Hole
Since `A^+` has a radius of 82 pm, we check if it fits into the octahedral hole:
- The radius ratio for octahedral holes is between 0.414 and 0.732. Since 82 pm is greater than 80.73 pm, it will fit into the octahedral hole.

### Final Conclusion
- The radius of the cation that just fits into the tetrahedral hole is **80.73 pm**.
- The cation `A^+` with a radius of **82 pm** can indeed fit into the octahedral hole.