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 $D_3$.

Aluminum isopropoxide structure

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?

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    $\begingroup$ 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 $\endgroup$ – Andrew Jan 21 at 0:49
  • $\begingroup$ 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. $\endgroup$ – MichaelK Jan 21 at 6:55
  • $\begingroup$ More accurately, the $\ce{O}$ should be almost equivalent. For example, some of them are bridging, while others are not. $\endgroup$ – MichaelK Jan 21 at 7:05

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