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 clusterMOF-5 structure

Does this have an obvious answer?

  • 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? $\endgroup$ – Zhe Jan 30 '17 at 23:58
  • $\begingroup$ Does it have an obvious answer? I have no experience of these types of clusters but I strongly suspect the answer is no. $\endgroup$ – bon Feb 3 '17 at 17:32

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.


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