`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

Radius of the cation just fitting into the tetrahedral hole
= Radius of tetrahedral hole = `0.225 xx r_(Br^(-))`
`=0.255 xx 195 = 43.875` pm
For the cation `A^(+)` with radius , 82 pm
Radius ratio = `(r_+)/(r_-) = (82 p m)/(195 p m) = 0.4205`
Since the radius ratio `(r_(+)//r_(-))` lies in the range 0.414 - 0.732, hence the cation `A^(+)` can be slipped into octahedral hole of the crystal `A^(+)Br^(-)`.