Question

The unit cube length for `LiCl` (`NaCl` structure) is `5.14 Å`. Assuming anion-anion contact, calculate the ionic radius for chloride ion.

Answer 1

In a face-centred lattice, anions touch each other along the face diagonal of the cube.
`4r_(Cl^(-))=sqrt(2)a`
`r_(Cl^(-))=(sqrt(2))/(4)a`
`=(sqrt(2))/(4)xx5.14 = 1.82 Ã…`
Alternative : Distance between `Li^(+)` and `Cl^(-)` ions
`=(5.14)/(2)=2.57 Ã…`
Thus, distance between two chloride ions
`=sqrt((2.57)^(2)+(2.57)^(2))`
`=3.63 Ã…`
Hence,
radius of `Cl^(-)=(3.63)/(2)=1.82 Ã…`