`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^–` ?

Explanation

(i) Radius of the cation just fitting into the tetrahedral hole= Radius of the tetrahedral hole
`= 0.225 xx r_(Br^(-)) = 0.225 xx 195 = 43.875 ` pm
(ii) For the cation ` A^(+)` with radius = 82 pm.
Radius ratio `= (r_(+))/(r_(-)) = (82"pm")/(195" pm") = 0.4205`
As it lies in the range 0.414 - 0.732 , hence tha catio ` A^(+)` can be slipped into octahedral hole of the crystal ` A^(+)Br^(-)`