A solid AB has CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of closest approach between A^+ and B^- ions?

A solid AB has CsCl type structure. The edge length of the unit cell i

1.	A solid AB has CsCl type structure. The edge length of the unit cell is 404 pm. Calculate the distance of closest approach between A^+ and B^- ions?
Answer» SOLUTION :Distance of closest APPROACH is EQUAL to the distance between the nearest neighbours (d). As CSCL has BCC lattice, `d=sqrt3/2a=1.732/2xx404` pm =349.9 pm