# Aluminium isopropoxide electronic structure

From Wikipedia:

A tetrameric structure of the crystalline material was verified by NMR spectroscopy and X-ray crystallography. The species is described by the formula $$\ce{Al[(μ-O-i-Pr)2Al(O-i-Pr)2]3}$$. The unique central $$\ce{Al}$$ is octahedral, and three other $$\ce{Al}$$ centers adopt tetrahedral geometry. The idealised point group symmetry is $$D_3$$.

I can understand that the 3 external $$\ce{Al}$$ atoms can be (inaccurately?) described as assuming an $${sp^3}$$ hybridization. With 4 bonds per $$\ce{Al}$$ atom, the valence shell is complete, following the octet rule. This too may deserve some additional attention, since $$\ce{Al}$$ formally has only 3 electrons available for bonding, yet here it forms 4 bonds. At least one of these is dative, I assume?

I'm less clear on the central $$\ce{Al}$$ atom. My initial instinct was to invoke the $${sp^3d^2}$$ hybridization. But this answer re $$\ce{SF6}$$ suggests that may not be the case. There are clearly similarities to the $$\ce{SF6}$$ case, but surely, also some differences? After all, $$\ce{Al}$$ is not $$\ce{S}$$.

In short, what is the electronic structure of the four $$\ce{Al}$$ centers?

• It might be easier to consider all of the Al atoms as Al3+ actions and all the propoxide groups as anions. Each O has three lone pairs that can interact with the metal Jan 21 at 0:49
• Yes, that's true. Each $\ce{O}$ atom has one valence electron used up in the $\ce{O-C}$ bond, and each $\ce{Al}$ donates one electron to each of 3 $\ce{O}$ atoms, completing their valence shells. The oxygens should all be equivalent, so we can probably think of all of the $\ce{O-Al}$ bonds as dative. Jan 21 at 6:55
• More accurately, the $\ce{O}$ should be almost equivalent. For example, some of them are bridging, while others are not. Jan 21 at 7:05