Kcl crystallizes in an fcc unit cell. if the cl– ion radius in kcl is 181 pm, then what is the radius of the k+ ion in kcl if the edge length of the kcl unit cell is 628 pm? assume that the ions touch along the edge of the unit cell.

133 pm

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 628 pm

Also,  

For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,  

.................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

Thus,  

 ...................2

Given that:

To find,  

Using 1 and 2 , we get:

Size of the potassium ion = 133 pm