Question

The edge of the LiCl unit cell is 514 pm. Assuming that the Li+ ions fit in the octahedral holes of the closest packed Cl- ions, calculate the ionic radii for the Li+ and Cl- ions and compare them to the tabulated values of 60 pm and 181 pm respectively and discuss the significance of any discrepancies.

Answer 1

In ccp aarangement Cl- ions occupy at edges and at face center. Li+ ions occupy at edge center

along face diagional we have 2 half Cl- ions and 1 full Cl- ions

hence diagonal length = 4 x r    ( where r is radius of Cl-)

        1.414 a = 4r               ( where a = edge length of cell , face diagonal = 1.414 a )

1.414 x 514 pm = 4 r

r = 181.7 pm

So along one edge of unit cell we have 2 ( half Cl-) ions + 1 Li+ ion

edge length = 2 ( radius of Cl-) + 2 ( Li+ radius)

514 = 2( 181.7) + 2Li+ radius

Li+ radius = 75.3 ppm

Hence based on tabulated values we observe that radius of Cl- measure is nearly same as in tabulated

while Li+ calculated has shoen some difference with tabulated value