(a) If the close packed cations in an AB type solid have a radius of 75 pm, what would be the maximum and minimum sizes of the anions filling the voids ?
(b) `Br^(-)` ions form a close packed structure. If the radius of `Br^(-)` ion 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^(-)` ?

Explanation

(a) For close packed AB type solid, `((r^(+))/(r^(-)))=0.414` to 0.732
`therefore` Minimum value of `r^(-) = (r^(+))/(0.732)=(75)/(0.732)` pm
= 102.5 pm
Maximum value of `r^(-) = (r^(+))/(0.414)=(75)/(0.414)= 181.2` pm
(b) Radius of cation just fitting into the tetrachedral void.
= Radius of tetrahedral void.
`= 0.225xx r_(Br^(-))`
`= 0.225xx195=43.875` pm
For the cation `A^(+)` with radius = 82 pm
Radius ratio `(r^(+))/(r^(-))=(82)/(195)=0.4205`
As it lies in the range 0.414 to 0.732, hence the cation `A^(+)` can be slipped into the octahedral hole.