Question
The ionic radius of CI– ion is 181 pm. Consider the closest packed structure in which all anions are just touching:
(i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.
Solution
given:
Radius = 181 pm
thus,
Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
Cation having radius 74.934 pm will just fit into octahedral voids.
Radius of tetrahedral void
= 0.225 r = 0.225 x 181 pm = 40.725 pm
Cation having radius 40.725 pm will just fit into tetrahedral void.
Radius = 181 pm
thus,
Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
Cation having radius 74.934 pm will just fit into octahedral voids.
Radius of tetrahedral void
= 0.225 r = 0.225 x 181 pm = 40.725 pm
Cation having radius 40.725 pm will just fit into tetrahedral void.
