Question

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.

Answer 1

Answer:

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,  

$\frac {r^+}{r^-}=0.731$ .................1

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

Thus,  

$r^++r^-=\frac {a}{2}$  ...................2

Given that:

$Cl^-\ (r^-) = 181\ pm$

To find,  

$K^+\ (r^+)$

Using 1 and 2 , we get:

$1.731\ r^+=0.731\times \frac {628}{2}$

Size of the potassium ion = 133 pm