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

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.

• What you call a unit cell is actually 1/8 of it. (Not that it matters much, though.) Jan 22, 2021 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? Jan 22, 2021 at 17:13
• Yeah, well, looks about right. Jan 22, 2021 at 19:17

Coming to question 1, that is, is your calculation correct?

Quoting from Wikipedia,

In solid sodium chloride, each ion is surrounded by six ions of the opposite charge as expected on electrostatic grounds. The surrounding ions are located at the vertices of a regular octahedron. In the language of close-packing, the larger chloride ions (167 pm in size) are arranged in a cubic array whereas the smaller sodium ions (116 pm) fill all the cubic gaps (octahedral voids) between them. This same basic structure is found in many other compounds and is commonly known as the $$\ce{NaCl}$$ structure or rock salt crystal structure. It can be represented as a face-centered cubic (fcc) lattice with a two-atom basis or as two interpenetrating face centered cubic lattices. The first atom is located at each lattice point, and the second atom is located halfway between lattice points along the fcc unit cell edge.

Also refer this LibreTexts article.

What does this mean?

For starters, $$\ce{NaCl}$$ is an FCC structure, with $$\ce{Na+}$$ occupying the octahedral voids, and $$\ce{Cl-}$$ occupying the FCC lattice.

Edge length in FCC unit cell = $$2\sqrt{2}r$$.

You get unit cell volume from here.

There are 4 octahedral voids (OV) in a unit cell, and thus 4 $$\ce{Na+}$$ ions.

Take them as whole spheres, and find their volumes.

$$V_{\ce{Na+}}={4 \times \frac{4 \pi {r_{\ce{Na+}}}^3}{3}}$$

Number of chloride ions = 8 corners + 6 face centres = $$8\times\frac{1}{8}+6\times \frac{1}{2} = 4$$ (we are multiplying the various positions with their individual contributions.)

$$V_{\ce{Cl-}}={4 \times \frac{4 \pi {r_{\ce{Cl-}}}^3}{3}}$$

Now, volume occupied by the ions = $$V_{\ce{ions}}=V_{\ce{Na+}}+V_{\ce{Cl-}}=\frac{16\pi}{3}\times ({{r_{\ce{Na+}}}^3}+{r_{\ce{Cl-}}}^3)$$

Volume of unit cell = $$V_{\text{uc}} = (2\sqrt{2}r)^3=16\sqrt{2}r^3$$,

Assuming radius of the ions is the same (that's how I was taught, it's a simplification for the calculation),

The correct result is:

$$\text{Packing fraction}=\frac{V_{\ce{ions}}}{V_{\text{uc}}}=74\text{%}$$

The $$74\text{%}$$ number is standard for all FCC unit cells.

So you can refer to this for the correct calculation.

Question 2, will this hold true for all alkali metal halides with an FCC structure?

Of course! It holds true for all alkali metal halides with an FCC structure.

The only exception is $$\ce{CsCl}$$ which has a non-closed packed structure type, but is commonly mistaken to be body-centered cubic (BCC structure).