Finding the atomic packing factor (APF) of Sodium Chloride and other FCC ionic compounds

Question

For a piece of coursework I am doing, I need to calculate the atomic packing factor of some ionic compounds. I have had no formal teaching in this area, so what I know comes from information I have found on the internet.

I take that the atomic packing factor is given by: $$ APF = \frac{V_\text{atoms}}{V_\text{unit cell}} $$ Taking a unit cell of $\ce{NaCl}$, we get: $$ APF = \frac{2\pi(r_\ce{Cl}^3 + r_\ce{Na}^3)}{3(r_\ce{Cl} + r_\ce{Na})^3} $$ Is this right? I took the side length of the unit cell to be $r_\ce{Cl} + r_\ce{Na}$, and reasoned that 1/8 of each ion is actually "inside" the unit cell, so there is essentially 1/2 of a chloride ion and 1/2 of a sodium ion "within" the cell. $r_\ce{Cl}$ is the ionic radius of chloride, and $r_\ce{Na}$ is that of sodium.

Would this also hold for all alkali metal-halides with a FCC crystal structure? For instance, for a compound with anion $\ce{A}$ and cation $\ce{C}$:

$$ APF = \frac{2\pi(r_\ce{A}^3 + r_\ce{C}^3)}{3(r_\ce{A} + r_\ce{C})^3} $$

Finally, is this a good measure of packing efficiency? As I mentioned, I have little experience in this topic, so I would be grateful if anyone knew of a better measure.

Answer 1

NaCl/ rocksalt lattice is a FCC structure

1. The length of the cubic for FCC is taken as: a
2. Volume of lattice is: a^3
3. Based on 1 atom +2 half atoms, the diagonal length is derived as Sqrt(2a^2 )= a sqrt 2
4. The diagonal length is made up of 2 atoms or total length is equals to 4r, where r is the radius of atom. Hence, r = (a sqrt 2) /4 OR a= 4r/ sqrt 2

APF of FCC typical value is 68%

Formula: Vatom/ Vlattice . Vatom = 4/3 PI (r)^3 Vlattice= a^3 OR (4r/ sqrt 2)^3

I might be wrong...

For Rocksalt structure APF: [4/3 *PI *((r)Na + (r)Cl)] / [(4 ((r)Na + (r)Cl))/ sqrt2]^3