I came up with this problem today and I can't find the answer elsewhere:
How does the oxygen atom bond to the four zinc atoms? To be more specific, what would be the molecular orbital diagram for the Zn4O cluster
Does this have an obvious answer?
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1$\begingroup$ I guess my question to you is why not? An oxygen in liquid water can be coordinated simultaneously to up to four hydrogens. Isn't this case exactly analogous? $\endgroup$– ZheCommented Jan 30, 2017 at 23:58
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$\begingroup$ Does it have an obvious answer? I have no experience of these types of clusters but I strongly suspect the answer is no. $\endgroup$– bonCommented Feb 3, 2017 at 17:32
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1 Answer
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As Zhe suggested -- it's just a tetrahedral bonding-interaction of oxygen. In the case of $\ce{H2O}$, oxygen has two lone pairs and two bonds to $\ce{H}$; in $\ce{[OZn4]^{6+}}$, those lone pairs are interacting with Zn atoms (strongly, because the $\ce{Zn}$ atoms bear a +2 charge).
Per the MOF-5 $\ce{H2}$ storage paper:
...inorganic $\ce{[OZn4]^{6+}}$ groups are joined to an octahedral array of $\ce{[O2C-C6H4-CO2]^{2–}}$ (1,4-benzenedicarboxylate, BDC) groups to form a robust and highly porous cubic framework.
Per the original synthesis:
A small amount of hydrogen peroxide was added to the reaction mixture in order to facilitate the formation of $\ce{O^{2-}}$ expected at the centre of the secondary building unit.
Thus the central oxygen has a full octet, able to supply 2 electrons to each $\ce{Zn^{2+}}$ in the coordination complex.
