Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
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,
[tex]\frac {r^+}{r^-}=0.731[/tex] .................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
[tex]r^++r^-=\frac {a}{2}[/tex] ...................2
Given that:
[tex]Cl^-\ (r^-) = 1.82\ \dot{A}[/tex]
To find,
[tex]K^+\ (r^+) = ? \dot{A}[/tex]
Using 1 and 2 , we get:
[tex]1.731\ r^+=0.731\times \frac {6.28}{2}[/tex]
Size of the potassium ion = 1.33 Å
