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In its guide to model antiferromagnetic couplig, Gaussian uses $\ce{Mn2O2(NH3)8}$ as a model compound.[1] It argues that manganese is in II oxidation high spin state; in other words, in both manganese are all d orbitals are singly occupied. Furthermore it creates a structure, where the metal centres are octahedral coordinated. The level of theory used is U-B3LYP/3-21G, which is in my opinion not enough to discuss these complexes in the first case.

I could not find any evidence of a structure proposed by Gaussian, the closest I have gotten is given as a cation $\ce{[Mn2O2(NH3)8]^z+}, (z\in2,3,4)$.[2]

I have reason to doubt the structure given in the tutorial,[3] as in the $\frac{5}{2}$ high spin case there are only four orbitals in manganese left which could accept bonding partners. Given that two of them are used by the bridging oxo ligands, only two ammonia should bond to the manganese centres. This would lead to the formula $\ce{Mn2O2(NH3)4}$. The situation changes once you start removing electrons or pair them up. Then the question changes to whether ammonia is a strong enough ligand to induce lower spin states.

Summarising: Is $\ce{Mn2O2(NH3)8}$ a reasonable structure?

For convenience, here is the structure as proposed by Gaussian (they do create a different one within the tutorial though):

di manganese di-mu-oxo complex appearing on the Gaussian website

References & Notes:

  1. Gaussian Website: http://gaussian.com/afc/
  2. Mcgrady, J. E.; Stranger, R. Redox-Induced Changes in the Geometry and Electronic Structure of Di-μ-oxo-Bridged Manganese Dimers. J. Am. Chem. Soc. 1997, 119 (36), 8512–8522 DOI: 10.1021/ja964360r.
  3. In my optimisations on a different level of theory I was unable to reproduce the structure.
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  • $\begingroup$ Looks reasonable. Mn (II) complexes have no problem with being high spin and hexacoordinate. $\endgroup$
    – Mithoron
    Dec 13, 2018 at 21:33
  • $\begingroup$ @Mithoron Actually I'd supposed it would be quasi tetrahedral. $\endgroup$ Dec 13, 2018 at 22:21
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    $\begingroup$ Nah, should be octahedral just like manganese(II) chloride and other complexes. Counting orbitals seems to be only confusing for this stuff. $\endgroup$
    – Mithoron
    Dec 13, 2018 at 22:49
  • $\begingroup$ @Mithoron I guess confusing enough to ask a question here. Need to look up the chloride though, seems interesting enough. $\endgroup$ Dec 13, 2018 at 22:57
  • $\begingroup$ What do you mean ‘four orbitals left’? $\endgroup$
    – Jan
    Dec 14, 2018 at 1:30

1 Answer 1

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The structure is extremely reasonable to the point where I would ask why any other di-oxido-bridged manganese(II) ammine complex would not show this structure if it were proposed. To describe why, I am going to be lazy and just copy my molecular orbital scheme of a general octahedral 3d-metal complex that I have used many times before.

Molecular orbital scheme of an octahedral complex
Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

As you can see, the four empty orbitals ($\mathrm{a_{1g}}$ and $\mathrm{t_{1u}}$) are indeed present and part of the bond-forming orbitals, but the fact that the $\mathrm{e_g}$ orbitals are also half-filled does not prevent them from participating, too. In fact, these orbitals turn into $\mathrm{e_g^*}$ orbitals in the resulting complex, formally antibonding. This, however, doesn’t matter, because there are still many more electrons in bonding orbitals ($\mathrm{t_{1u}, a_{1g}}$ and $\mathrm{e_g}$) that this small amount of lost energy is more than offset. Typically, for a high-spin $\mathrm{d^5}$ complex, the bond order would be somewhere around 5/6. Still well within the overall bonding range.

If one looks at the kinetic stability of various complexes, one will find that $\mathrm{d^3}$ and low-spin $\mathrm{d^6}$ often give very inert octahedral complexes as no antibonding orbitals (of the simplified scheme) are populated—but even a $\mathrm{d^{10}}$ or $\mathrm{d^0}$ metal will form these complexes, their structures just won’t be as long-lasting. (Zinc has been shown to equilibrate between tetra, penta and hexacoordinated complexes in aquaeous solution as I was told in my coordination chemistry lecture.)

So yes, this octahedral complex is perfectly fine.

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