A solid AB has CsCl typestructure. The edge length of the unit cell is404 pm.Calculate the distance of closest approach between A^(+)and B^(-)ions ?

A solid AB has CsCl typestructure. The edge length of the unit cell is

1.	A solid AB has CsCl typestructure. The edge length of the unit cell is404 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 latttic ., `d = sqrt3/ 2 a = 1.732/2 XX 404"pm"= 349.9 ` pm