A solid `A^(o+)B^(ɵ)` has `NaCl`-type close-packed structure. If the anion has a radius of `250` pm, what should be the ideal radius for the cation? Can a cation `C^(o+)` having radius of `180` pm be slipped into the tetrahedral site of the crystal `A^(o+)B^(ɵ)`? Give reason for your answer.

Explanation

Step I. Calculation of Ideal radius of cation,
Since the solid AB has `Na^(+)Cl^(-)` structure, it is octahedral in nature. The ideal radius ratio for octahedral site is `0.414`. In this case, the cation `A^(+)` ion will touch the anions `B^(-)` and the arrangement will be closed packed. If the value 0.732 be considered, then the cation will be bigger in size and teh anions will not touch it. The arrangement will be less close packed and rather unstable.
`A^(+)//B^(-)=0.414" and "B^(-)="250 pm"`
therefore" "A^(+)=0.414xx250=103.5"pm"`
Thus, the ideal radius of the cation `(A^(+))` = 103.5pm
Setp II. Predicting whether cation 'C' can be slipped into the tetrahedral site
The ideal radius ratius ratio for tetrahedral site or void `(A^(+)//B^(-))=0.225`
Radius of anion `(B^(-))` = 250pm
Radius of cation `(A^(+))=0.225xx250=56.25"pm"`
This shows that the cation 'C' with radius 180 pm is bigger in size and cannot be slipped into the tetrahedral site.