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**

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