Metallic gold crystallizes in the fcc lattice.
The length of the cubic unit cell is `a = 4.242 A`.
a. What is the closest distance between gold atoms?
b. How many "nearest neighbours" does each gold atom have at the distance calculated in `(a)`?
(c) What is the density of gold? (Aw of Au `= 197.0 g mol^(-1)`)
d. Prove that the packing factor for gold is `0.74`.

Explanation

In fc c, `Z(eff) = 4`.
For fcc, atomic radius `(r) = (a) /(2sqrt2)`
a. Closest distance two Au atoms `= 2r`
`= 2 xx (a)/(2sqrt2) = (a)/(sqrt2) = (4.242)/(sqrt2) = (4.242)/(1.414) = 3.0 Å`

As the coordination number of fc c `= 12`.
So there are `12` nearest neighbours in all, the number expected for a close-packed structure.
c. Density `= rho = (Z_(eff) xx Aw)/(N_(A) xx a^(3))`
`= (4 xx 197 g)/(6 xx 10^(23) xx (3.0 xx 10^(-8) cm )^(3))`
`= 19.4 g cm^(-3)`
d. Packing factor for fcc lattice `= (sqrt2)/(6) pi = 0.7404`