Oxidation state of ions in a unit cell

Given that $\ce{Al2O3}$ is face-centered cubic (cubic-closest packed) with respect to oxide anions, I am instructed to find the oxidation state of each aluminum ion "fragment" in the unit cell. Let's suppose that the picture above is of $\ce{Al2O3}$ and not $\ce{NaCl}$. The larger oxide ions are green; the smaller aluminum cations are purple.

There are a total of 4 oxide ions in a unit cell, and a total of 3 aluminum ions in the unit cell.

Given that 4 oxide ions impart a charge of -8, and that a unit cell must be neutral, then each of the "three" aluminum ions must have an oxidation state of +8/3 in the unit cell, correct?

I cannot find any textbook discussing the oxidation states of atoms in unit cells ... so I am at a loss at whether my work is correct or not.

• Well, as you should guess, you aren't. – Mithoron Nov 12 '15 at 11:28
• Are you saying I'm wrong? – Dissenter Nov 12 '15 at 17:54
• Single atoms can't have fractional ox. state. There would have to be 3 at. 8+ cluster which makes no sense. If your studying sth new and get results which contradict basic knowledge then you've done sth wrong (or learning quantum mech. :)) – Mithoron Nov 12 '15 at 19:12
• Well first I don't believe that Al2O3 is face centered cubic. In the example you show for NaCl there are four "whole" NaCl molecules per unit cell. For sodium four 1/8 "atoms" and one 1/2 atom on top and bottom layers. The middle layer has a "whole" Na in the center plus four 1/4 atoms. – MaxW Nov 12 '15 at 22:11

This is a really weird question.

Let's start with the general case for a FCC lattice, you'd count up the oxygen atoms per unit cell. In the $\ce{NaCl}$ example, as noted in the comment, there are four $\ce{Cl-}$ per unit cell (6 on the faces, 8 on the corners). Then you'd count up the number of Al atoms per unit cell and divide up the charge as you note above.

In the case of "fractional" oxidation state, it implies that there's a mixture of oxidation states.

Let's say, hypothetically, you had 3 cations and need a +8 charge to balance out the anions. That tells me there are two +3 cations and a +3 cation somewhere. The average is +8/3 but I need to assign formal charges as integers.

But...

Normal $\ce{Al2O3}$ is not an FCC lattice. It's triclinic. Here, you would do the same thing - count up the number of Al and O atoms per unit cell, considering the fractional volume of the atoms inside the unit cell. It's a bit trickier, since the volumes aren't as obvious as the FCC case.

This question is about $\gamma$-alumina Notice that the oxygen atoms actually form an FCC lattice. I have an oxygen at each corner of the unit cell, and one in the middle of each face. So that's 4 oxygen atoms per unit cell = +8 charge from the anions.

How many Al atoms are there? There are 5 Al atoms completely in the unit cell, plus one per edge (12, each counting 1/4 = 3) so a total of 8 Al atoms!

So that's an average oxidation state of +1 per Al...

• I will point out that for gamma alumina, the formula is not necessarily $\ce{Al2O3}$. – Geoff Hutchison Dec 9 '15 at 21:08
• @Geoffhutchinson - is there a reason the last picture seems to have ions in and out of the cube? – Dissenter Dec 9 '15 at 21:36
• @Dissenter - the oxygen atoms are on the faces and corners. – Geoff Hutchison Dec 9 '15 at 21:37
• @Geoffhutchinson - thank you for the explanation! I posted the corrected question - my professor changed it from Al2O3 to an iron-based ionic lattice. chemistry.stackexchange.com/questions/41950/… Does this change things? Or is +8/3rds wrong? – Dissenter Dec 9 '15 at 21:38