For a given packing (hcp, fcc, bcc), which ions occupy the corners and which occupy the faces or centers?

I recently came across the packing of atoms in unit cells. I wanted to know if there was a way to tell where the atoms in the unit cell are just by knowing the compound and its packing (hcp, fcc, etc.).

For example, if given the information that lithium iodide packs in fcc, how would I know where the lithium ions are and where the iodine ions are? Which would occupy the corners, or will both will share equally the corners and faces? Is there anything I could use to deduce where the atoms will be?

Another example, if NaCl exists in fcc where will the cations and anions be? Looking at its structure the chloride anions are on the edge centers. Can such forms be deduced or is memorizing the only way forward?

• In my opinion, no techniques exist to explain why a given compound packs in fcc or bcc, or even hcp. It may even change from one form to another one if the temperature is changed Oct 10 '21 at 13:01
• Many of the pretty pictures you find are not of primitive unit cells, so the full symmetries may not be obvious. Also, even primitive unit cells are not unique - they can be made into many different shapes and centered on different features of the atom basis. and still be primitive Oct 11 '21 at 19:17

The actual answer is that it doesn't matter. For many of the 1:1 solid-state structures, either the cations or the anions may be considered to be at the vertices (i.e. corners) of the unit cell. By symmetry, both representations are entirely equivalent.

To see why this is the case, it is helpful to look at a 2D analogue first. The following could be considered to be a 2D version of the cesium chloride (CsCl) structure.* The blue and orange dots represent cations and anions respectively (or the other way round; it doesn't matter, the point is that they're different things).

From the way I've drawn the unit cells (i.e. the black lines joining the dots), it appears that the blue dots should be at the vertices and the orange dot in the centre of the unit cell. It's obvious that this is a valid unit cell, as you can repeat it as many times as you wish to generate the full crystal structure.

However, I could just as easily have drawn this:

Again, this is a valid unit cell; but this time, the orange dots are at the vertices and the blue dot is in the centre.

This shows that there is no difference between the two representations of the unit cell. Both of them can be repeated (and translated) to form exactly the same crystal structure. That should already suggest to you the answer to the 3D case in your question.

Let's look at the 3D CsCl structure. You can draw it out in 3D and try to convince yourself again using the argument above. However, I think it's easier to use the 2D depiction on the left, where the numbers indicate the positions of the atoms (lattice points) in the third dimension, as a fraction of the unit cell length in that dimension. So, 0 and 1 indicates that there is a blue atom at the front and at the back, whereas 0.5 indicates the orange atom right in the middle of the unit cell.

What we need to do is to extend this unit cell a couple of times in each dimension. The first arrow shows us how to extend the unit cell in the third dimension, which has been collapsed: essentially, we need to add 1 to every number (which corresponds to adding a new unit cell behind the current one). The second arrow is just a tiling in the other two dimensions.

I'm now going to redraw the unit cell:

Now, all we need to do is to subtract 0.5 from every number (this is OK because it's just shifting the zero, i.e. moving our "point of view" backwards / forwards) and we get the same CsCl unit cell but with the atoms swapped round.

A similar exercise may convince you of the NaCl / rock salt case.†

Footnotes

* That's similar to the body-centred cubic structure, but is not the same: in the bcc structure the atom in the middle is the same as the one at the edges. Thanks to Karsten Theis for pointing that out.

† Similarly, the NaCl structure is similar to the face-centred cubic, but fcc has the same atom at all the lattice points.

• This was great! Blew my mind. I see how it works for the fcc case, depending on how it's viewed either can occupy the voids. Oct 10 '21 at 13:42
• Technically, BCC implies the same atom on the vertices and the center of the unit cell. Statements like "lithium iodide packs in fcc" means that lithium ions form a FCC lattice and iodide atoms form a FCC lattice, shifted by half a cell edge. What you say about having a choice of where to place the origin is absolutely correct. Oct 10 '21 at 14:13
• @KarstenTheis Indeed! I'm a bit rusty on this. I'll reword. It's CsCl and rock salt structure, IIRC. Thanks. Oct 10 '21 at 14:15
• Yes, (I had to look it up too), the CsCl structure is primitive cubic but if you squint and tilt your head and imagine anions and cations to be the same, it reminds you of BCC. Beautiful figures, nice colors, hand drawn and still so neat. Oct 10 '21 at 14:18
• @KarstenTheis thanks, the trick is to have an iPad, draw one unit cell, then copy-paste ;-) Oct 10 '21 at 14:22

Orthocresol’s answer is good at highlighting why it doesn’t make sense to distinguish between a rock-salt structure and an *anti-rock-salt structure: in the most basic cubic AB salt structures, the positions of cations and anions are symmetry-equivalent and thus either can be used to construct the unit cell and packing with the other sitting in the appropriate void.

This unfortunately changes rather quickly once you leave octahedral voids in the face and body-centred cubic structures. Trivially, this is the case with the fluorite structure ($$\ce{CaF2}$$ structure) in which calcium forms a face-centred cube and fluoride ions occupy all tetrahedral voids; as there are twice as many fluoride ions one cannot simply switch positions and arrive at the same structure. Instead, the opposite structure would be primitive-cubic with half of the cubes occupied. This is quite a mouthful which is why the shorthand anti-fluorite exists.

The same is true for hexagonal closest packing structures. While all closest packings have $$n$$ octahedral voids for $$n$$ atoms (and $$2n$$ tetrahedral voids), the octahedral voids of hcp do not form another hcp. For example in the nickel arsenide structure – hcp structure of the arsenic atoms with all octahedral voids occupied by nickel – if one were to take the nickel atoms as a lattice (as the image on Wikipedia does), the lattice turns out to be primitive hexagonal with every other central position occupied.

Things would get even more terrifying in a hexagonal structure with all tetrahedral voids occupied but gratifyingly such an $$\ce{AB2}$$ ionic structure does not exist in nature. The next-closest ist the wurtzit structure wherein half of the tetrahedral voids are occupied which, thankfully, forms interchangeable hcp structures. Unfortunately, other possibilities for occupying half of the tetrahedral voids exist in the patinum(II) sulfide and lead(II) oxide structures which feature square-planar platinum and square pyramidal lead atoms, respectively.

In these cases, where the different ions taken individually do not form the same crystal packing, one structure is the standard structure and the other is the anti structure. For example, in the fluorite structure of $$\ce{CaF2}$$, the calcium cations form an fcc substructure. Related structures in which the cation also forms the fcc substructure are known as fluorite structures. In compounds such as $$\ce{Na2O}$$ the positions of cations and anions are reversed: it is now the oxide anions that form the fcc substructure and the sodium cations occupying the tetrahedral voids. As this is opposite from the original fluorite, it is known as the anti-fluorite structure.

Similarly, anti-nickelarsenide, anti-platinumsulfide and anti-lead-oxide structures can be determined.

As can be seen when comparing the fluorite and the nickelarsenide structures, it is not always the same atom type that forms the lattice. In fluorite, the cations form the cubic lattice while the anions occupy the tetrahedral voids. In nickelarsenide, the more electronegative partner (although probably not strictly an anion) forms the hexagonal structure while the electropositive partner occupies the octahedral voids. So unfortunately, these have to be learnt by the student or researcher individually.

So far, this answer has barely scraped the surface. Take for example the spinel structure which contrasts with inverse spinels; notwithstanding various almost-spinels etc.