1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ What you call a unit cell is actually 1/8 of it. (Not that it matters much, though.) $\endgroup$ – Ivan Neretin Jan 22 at 13:35
  • $\begingroup$ @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? $\endgroup$ – J. Barker Jan 22 at 17:13
  • $\begingroup$ Yeah, well, looks about right. $\endgroup$ – Ivan Neretin Jan 22 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.