Question

A solid `A^+B^-` has NaCl type close packed structure .If the anion has a radius of 241.5 pm , what should be the ideal radius of the cation ? Can a cation `C^+` having radius of 50 pm be fitted into the tetrahedral hole of the crystal `A^+ B^-` ?

Answer 1

As `A^+ B^-` has 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 arrangement will be close packed. Hence, Radius of the octahedral void =`r_(A^+)=0.414 xxr_(B^-)=0.414xx241.5` pm =100.0 pm
Radius of the tetrahedral void =`0.225 xxr_(B^-)` =0.225 x 241.5 pm =54.3 pm
As the radius of the cation `C^+` (50 pm) is smaller than the size of the tetrahedral void, it can be placed into the tetrahedral void (but not exactly fitted into it)