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# $Br^−$ ion forms 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 slipped into the octahedral hole of the crystal $A^+ Br^-$?

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Radius of the cations just filling into the tetrahedral hole= Radius of the tetrahedral hole =$0.225 \times 195$
$\Rightarrow 43.875$ pm
For cation $A^+$ with radius = 82 pm
Radius ratio $\large\frac{r^+}{r^-}=\frac{82}{195}$$=0.4205$
As it lies in the range 0.414 – 0.732, hence the cation $A^+$ can be slipped into the octahedral hole of the crystal $A^+ Br^-$