The face-centered cubic unit cell describing the structure of an ionic compound has an edge length of 207 pm. The lattice points of the unit cell are occupied by anions, and the cations are small enough so that the anions are in contact along the diagonal of the face of the unit cell. What is the ionic radius of the anion (in picometers)

73.2 pm

Explanation:

"The anions make contact along the diagonal of the face of the unit cell, so the diagonal of the face is equal to 4 ionic anion radii. By letting r represent the ionic radius of the anion, d represent the diagonal of the face of the unit cell, and x represent the edge length, then d=4r and

d^2=x^2+x^2=  2x^2

By substituting 4r for d, the second equation becomes

(4r)^2=(2)x^2

Solving this equation for r shows that

r=(x–√2)/4

Since the edge length x is 207 pm, the ionic radius of the anion is

r= ((207 pm) X )/4=73.2 pm

Notice that the formulae used to describe atomic radii include constants with exact values. Therefore, these values do not constrain the number of significant figures."

false

because it does not have the potential to be entirely used up