Question

`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^-` ?

Answer 1

(i)Radius of the cation just fitting into the tetrahedral hole=Radius of the tetrahedral hole `=0.255xxr_(Br^-)=0.225xx195`=43.875 pm
(ii)For the cation `A^+` with radius =82 pm
Radius ratio =`r_+/r_(-)="82pm"/"185pm"`=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^(-)`
Note. In case(ii), radius of octahedral void in which the cation can be fitted exactly =`0.414xxr_(Br^-) =0.414 xx195"pm"`=80.73 pm
This is the minimum size of the octahedral void in which if the cation is placed, it will touch all the anions and the anions also touch each other . However , if cation bigger than the above size is slipped into the octahedral void, cation will continue to touch all the anions but anion-anion contact will vanish. The arrangement remains octahedral upto the maximum size of the octahedral void viz . `0.732xxr_(Br^-)`
Hence, in such cases , we apply radius ratio rules to find the range of the void, for a particular arrangement