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−.
