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.

Options

A. Radius of octahedral void `103.4` pm
B. Radius of tetrahedral void `56.25` pm
C. Radius of `Cl^(-)` is larger than the size of the tetrahedral site , it cannot be slipped into it .
D. Radius of `Cl^(-)` is larger than the size of the tetrahedral site , it can be slipped into it .

Answer

A, B, C

Explanation

AS `A^(+)B^(-)` hs NaCl structure, `A^(+)` ions will be present in the octahedral voids , Ideal radius of the cation will be equal to the radius of the octahedral void because in that case it will touch the anions and the arrangment will be closely packed.

Hence, Radius of the octahedral void=`r_(A)`.
=0.414 `r_(B^(-))=0.414 250=103.4` pm
Radius of the tetrahedral site=0.225.
`r_(B)=0.225 250 pm=56.25`pm
As teh radius of the cation `C^(+)` (180pm) is large than the size of the tetrahedral site, it cannot be slipped into it.