The edge length of the unit cell of kcl (nacl-like structure, fcc) is 6.28 å. assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. the radius of the chloride ion is 1.82 å.

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,

.................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 = 1.33 Å

a electronegativity decreases beacuse the size of nucleus is bigger.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2