Given: Radius of anion Br^{–} (r^{–} ) = 195 pm

Radius of cation ( Ar^{+} ) = 82 pm

To find:

i. The radius of the cation that just fits into the tetrahedral hole (r^{+} ) = ?

ii. Whether the cation A^{+} having a radius of 82 pm can be slipped into the octahedral hole of the crystal (A^{+} Br^{–} ) = ?

Formula:

Radius ratio = Radius of the cation/Radius of the anion

Calculation:

i. For Limiting value for for tetrahedral hole is 0.225 – 0.414

From formula, Radius of the tetrahedral hole

ii. For cation A+ with radius = 82 pm From formula,

As it lies in the range 0.414 – 0.732, hence the cation A^{+ } can be slipped into the octahedral hole of the crystal A^{+} Br^{−}.