Question

A solid AB has CsCl- type structure. Edge length of
the unit cell is 4.04 A'. The distance of closest
approach between A and B is​

Answer 1

Explanation:

In cscl structure,the distance of closest approach between the cation and anion is equal to half of the body diagonal of a cube

body diagonal=√3xedge length

=√3x4.04

=6.998A'

the distance of closest approach=1/2x body diagonal

=1/2x6.998

=3.449A'