# How to find the atomic packing factor of Sodium Chloride (and other FCC ionic compounds in general)?

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

Would this also hold for all alkali metal-halides with a FCC crystal structure? i.e for a compound with anion A and cation C:

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

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.

• What you call a unit cell is actually 1/8 of it. (Not that it matters much, though.) – Ivan Neretin Jan 22 at 13:35
• @IvanNeretin I wasn't sure, thanks for clarifying that. Does my formula look right otherwise? And will it hold for all alkali metal halides with a FCC structure? – J. Barker Jan 22 at 17:13
• Yeah, well, looks about right. – Ivan Neretin Jan 22 at 19:17