Consider either of the small, neutral molecules containing two Zn atoms below. If I were to determine the oxidation state of Zn from this molecule, I would formally assume each Zn to be Zn(I) in order to balance the overall charge (if I give 2- for each O, 3+ for the Al, and 1+ for the H). I also assume that each Zn(I) has 1 unpaired electron as a result of this oxidation state. With this, I would assume that the molecule may have a triplet ground-state (putting aside the possibility of any broken-symmetry behavior for now).

The left and right images are the result of density functional theory geometry optimizations where I set the multiplicity to be 1 and 3, respectively. I find the (closed-shell) singlet structure is lower in energy than the triplet by 28.7 kcal/mol, in contrast with the prior discussion.

How can I reconcile these two conflicting notions? If I were to study reactions involving this molecule, do I assume the structure is a triplet because that's what the 1+ oxidation state on the Zn atoms would likely yield? Alternatively, do I assume that the structure is a singlet because that is what I calculated, even if that does not seem to agree with the oxidation state assumption?

If the appropriate approach is the former, then perhaps the oxidation states of each atom is not what one would formally assign. If the appropriate approach is the latter, then perhaps since Zn does not tend to be in the 1+ oxidation state, there is some unusual change in the electron configuration to stabilize this structure.

Computational details: this is computed using the M06-L functional, I have checked the stability of the wavefunctions, there is minimal spin contamination, both structures are local minima, and the trend holds true for other model chemistries.

Singlet state: enter image description here Triplet state: enter image description here

  • $\begingroup$ You don't see that both molecules have exactly same oxidation states? $\endgroup$
    – Mithoron
    Jul 26, 2017 at 11:58
  • $\begingroup$ @Mithoron I assume you mean both transition metals, not both molecules? I manually set both molecules to have 0 charge (of course, molecules don't have oxidation states). With regards to the Zn atoms, the oxidation state isn't something you can directly obtain from DFT. I don't necessarily know what you're asking. Typically one would likely associate di-Zn(I) with the triplet state and di-Zn(II) with the singlet state, but this is admittedly a simplistic approach. $\endgroup$
    – Argon
    Jul 26, 2017 at 14:42
  • $\begingroup$ I'm really not getting it. If there is a bond between Zn atoms or not, they have both +1 (if it's neutral molecule) so what problem with ox. state you have? $\endgroup$
    – Mithoron
    Jul 26, 2017 at 19:32
  • $\begingroup$ The oxidation state is not an input parameter for electronic structure programs. The overall charge and spin multiplicity are. I have set the charge to be neutral and the multiplicity to be 1 and 3. I am implicitly assuming that each Zn is Zn(I) in order for the net charge to be zero. Typically I would associate a triplet with a di-Zn(I) complex since each Zn(I) would formally be $[Ar] 4s^{1}3d^{10}$. It is clear from my calculations that, instead, a (closed-shell) singlet has lower energy, and it is not clear to me how to reconcile the Zn(I) oxidation state with a singlet spin state. $\endgroup$
    – Argon
    Jul 26, 2017 at 19:36
  • $\begingroup$ Well, that's the molecule with Zn-Zn bond - singlet, both Zn have +1 state, everythings OK! $\endgroup$
    – Mithoron
    Jul 26, 2017 at 19:40

1 Answer 1


First reflex: As long as the calculations are otherwise reliable, I don't think you can ignore the possibility of a broken-symmetry/multiconfigurational solution. If such a solution ends up even lower in energy than your closed-shell singlet, I think you have your answer right there.

Alternatively, if the closed-shell singlet ends up exhibiting a lower energy than a BS solution, perhaps this is some sort of cationic Zn analogue of a peroxo species? If this were the case, then the $+1$ oxidation state wouldn't be nearly so surprising. As you note, the non-$\ce{Zn}$ portion of the system would be $\ce{[AlO3H]^{2-}}$, and thus the species as a whole might be better thought of as $\ce{[Zn2][AlO3H]}$. I have no feel for how realistic this hypothesis is, though Wikipedia notes that organozinc(I) compounds have been prepared, with decamethyldizincocene being the first such compound reported (public domain image; click to enlarge):


  • $\begingroup$ Thanks for your input. I will try running the BS singlet and see. I have occasionally seen this "odd" behavior when I consider other transition metals aside from Zn. For instance, I'd expect a di-Co(I) structure to be a quintet, but I find the septet is lower. For di-Fe(I), I'd expect a septet but I get a nonet. For all three of these examples, I have not yet considered a BS solution (but intend to). I will try it for the di-Zn(I) case now and will report back here for others to reference. Thank you for your point about the "peroxo"-like behavior. I'll look into that as well. $\endgroup$
    – Argon
    Jul 26, 2017 at 4:30
  • $\begingroup$ @Argon Beware that DFT is notoriously unreliable for metal spin states, especially with transition metals. For rigorous work, AFAIK one really can't avoid running multireference calculations (best augmented with perturbation theory or MRCI) to confirm the DFT conclusions. For a main group system this small (I would think Zn would behave as main-group here), I'd think such calculations should be feasible even on modest hardware. $\endgroup$
    – hBy2Py
    Jul 26, 2017 at 4:34
  • $\begingroup$ Yeah, I've kept that in mind as well. It's part of the reason why I wondered if I should go with the anticipated multiplicity even if it is not the lowest energy structure obtained from DFT in case DFT is not correctly capturing the energetics. Many computational papers I've seen on complex bumetallic complexes simply state they model the assumed high-spin multiplicity associated with the $d^n$ count without clear justification that other spin states are higher in energy. I don't feel extremely comfortable with that though. I may need to consider something beyond DFT I suppose. $\endgroup$
    – Argon
    Jul 26, 2017 at 14:46
  • $\begingroup$ Also, for what it's worth, I have not yet been able to obtain a BS singlet solution using Gaussian. I read in the orbitals of the triplet state and mix the orbitals for my initial guess. This creates a BS singlet, but as the geometry optimization progresses (and it progresses toward the closed-shell singlet structure), it inevitably switches off the BS singlet potential energy surface and onto the closed-shell singlet surface. I haven't had luck with defining "fragments" to obtain the BS singlet either. We'll see if I can get it another way to compare the energies. $\endgroup$
    – Argon
    Jul 26, 2017 at 14:49
  • 1
    $\begingroup$ @Argon Neat! Glad to not have been completely off-base. :-) And yeah, in my (limited) experience with BS calculations, if there's a BS state that's unambiguously lower in energy than the closed-shell singlet, the convergence algorithms usually solve to it quite quickly. $\endgroup$
    – hBy2Py
    Jul 28, 2017 at 22:44

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