Question

CsCl crystallizes in a cubic cell that has a Cl- at each corner and Cs at the centre of the unit cell.if the radius of of Cs is 1.69Å and rcl - = 1.81Å, what is the edge length of unit cell?

Answer 1

we know for simple cubic lattice, rc+ra=a 
1.69+1.81=3.5A

Answer 2

Answer : The edge length of unit cell is, $4.0414\AA$

Solution :

From the given information, we conclude that the  CsCl has BCC (body centered cubic unit cell) structure.

The relation between the edge length and radius of unit cell.

$2(r_{Cs^+}+r_{Cl^-})=\sqrt{3}\times a$

where,

a = edge length of unit cell

$r_{Cs^+}$ = radius of cesium cation = $1.69\AA$

$r_{Cl^-}$ = radius of chloride anion = $1.81\AA$

Now put all the given values in the above relation, we get the edge length of the unit cell.

$2(1.69\AA+1.81\AA)=\sqrt{3}\times a$

$a=4.0414\AA$

Therefore, the edge length of unit cell is, $4.0414\AA$