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With highly oxidized metal centers, we expect that the hydroxido ligand's proton becomes acidic and will be lost to the surrounding media, creating an oxido ligand.

The oxo ligand is a $\pi$ basic ligand and we expect both $\sigma $ and $\pi$ donation into the metal center.

The metal's $\mathrm{t_{2g}}$ orbitals are lowered in energy by the presence of the oxido ligand's filled p orbital donating into the $\mathrm{t_{2g}}$. My question is regarding the symmetry of the ligand group orbitals. A p orbital does not have $\mathrm{t_{2g}}$ symmetry and as we know non-symmetry interactions are forbidden. Therefore I am struggling to understand the details of the oxido ligand.

My only explanation is that the complex's symmetry is reduced with the presence of an oxo ligand maybe from $O_\mathrm{h}$ to $C_4$ or similar. This would then change the symmetry of the $\mathrm{t_{2g}}$ orbitals on the metal center, but I really need some help explaining this!

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    $\begingroup$ maybe an example to convey your problem would be better. $\endgroup$ – RE60K Apr 22 '15 at 16:32
  • $\begingroup$ Fact is that $\pi$ orbitals of an oxo ligand in $z$ direction can combine with $d_{xz}$ and $d_{yz}$ orbitals of the metal center. The irreducible representations for the orbitals is another topic for which we need more information from you. $\endgroup$ – LLang Feb 2 '16 at 20:50
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Although chemists and chemistry students alike often indiscriminatorily use the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ descriptors for the d-orbitals of a d-block metal in coordination complexes, those designations are only truly correct in an octahedral environment. The vast majority of coordination complexes, however, are not octahedral but distorted octahedra, octrahedroids with non-symmetric ligands and others. This is also the case for oxido-ligand coordinated complexes that you are asking about.

To restrict the discussion to a single species (although it can easily be expanded to others) I will craft my answer explaining the situation in the vanadyl cation $\ce{VO^2+}$. I will discuss the pentaaquaoxidovanadium(IV) cation $\ce{[VO(H2O)5]^2+}$ as Ballhausen and Gray analysed.[1] I will label the direction of the oxido ligand $+z$ (in line with their scheme).

Constructing the molecular orbital scheme

The first important thing to note is that the complex indeed does not display octahedral $O_\mathrm{h}$ symmetry but $C_\mathrm{4v}$. That is great for us because it facilitates practically every step in the construction of molecular orbitals. A character table for the $C_\mathrm{4v}$ point group can be found in orthocresol’s collection (Note: MathJax-heavy link. Give it time to load). In the tables below I have given the different metal and ligand orbitals and their irreducible representations.

$$\textbf{Table 1:}\text{ irreducible representations of the metal orbitals in }\ce{[VO(H2O)5]^2+}\\ \begin{array}{cccc}\hline \text{irrep} & \text{s orbital} & \text{p orbital} & \text {d orbital} \\ \hline \mathrm{a_1} & \mathrm{4s} & \mathrm{4p}_z & \mathrm{3d}_{z^2}\\ \mathrm{a_2} & & & \\ \mathrm{b_1} & & & \mathrm{3d}_{x^2 - y^2}\\ \mathrm{b_2} & & & \mathrm{3d}_{xy}\\ \mathrm{e} & & \mathrm{4p}_x, \mathrm{4p}_y & \mathrm{3d}_{xz}, \mathrm{3d}_{yz} \\ \hline \end{array}$$

$$\textbf{Table 2:}\text{ irreducible representations of the ligand (group) orbitals in }\ce{[VO(H2O)5]^2+}\\ \begin{array}{cccc}\hline \text{irrep} & \ce{O^2-} & \ce{$eq${-}H2O} & \ce{$ax${-}H2O} \\ \hline \mathrm{a_1} & \mathrm{p}_z & {\sigma_1} \atop {\pi_1} & \mathrm{p}_z\\ \mathrm{a_2} & & \pi_2 & & \\ \mathrm{b_1} & & {\sigma_2} \atop {\pi_3} & \\ \mathrm{b_2} & & \pi_4 & \\ \mathrm{e} & \mathrm{p}_x, \mathrm{p}_y & {\sigma_3, \sigma_4}\atop{\pi_5, \pi_6, \pi_7, \pi_8} & \mathrm{p}_x, \mathrm{p}_y\\ \hline\end{array}$$

If required, I can supply images of the π and σ symmetric orbitals of the equatorial ligands once I am back at a PC that has ChemDraw (not before the 8th of January). It is not always fully obvious which orbitals interact with which ones; however, it is very obvious that the π2 ligand group orbital is nonbonding with respect to the $\ce{L\bond{->}M}$ bond.

In a second round, we need to consider the strength of interaction. The oxido ligand is much closer to the vanadium centre ($d(\ce{V=O}) \approx170~\mathrm{pm}; d(\ce{V\bond{<-}OH2}) \approx 230~\mathrm{pm}$) meaning that any interactions with that will be considerably stronger. (There are hints pointing towards an even more elongated axial aqua ligand, but that would be secondary.)

The resulting scheme you construct should show a strong interaction of the oxido ligands $\mathrm{e}$ orbitals ($\mathrm{p}_x$ and $\mathrm{p}_y$) and the metal’s $\mathrm{e}$ orbitals ($\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$). This can be interpreted as the oxido ligand donating both p-type lone pairs to vanadium; while the resulting bond is typically depicted as $\ce{V=O}$ a better depiction may be $\ce{{V^-}#O+}$. The other interactions of $\mathrm{e}$ orbitals are considerably weaker.

The orbital scheme as published

Ballhausen and Gray, who I mentioned previously, also included a quantitative scheme in their paper, albeit having omitted the ligand-metal π interactions with the exception of those of the oxido ligand.[1] Thus, their scheme is a little bit simpler than the one I would have drawn. The low-lying σ orbital is the oxido ligand’s; the orbitals labelled π are obviously the oxido ligand’s $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. Due to the lack of aqua-π orbitals, the $\mathrm{d}_{xy}$ orbital appears unmodified. The $\mathrm{e}$ orbitals are raised notably, as is the $\mathrm{b_1}$ orbital ($\mathrm{d}_{x^2-y^2}$) due to the ligand’s σ interactions. Finally, the most destabilised metal d orbital is $\mathrm{a_1}$ corresponding to the $\mathrm{d}_{z^2}$ orbital.

Orbital scheme by Ballhausen and Gray

The resulting scheme can be seen as an atypical Jahn-Teller distortion of an octahedral complex. In typical Jahn-Teller distortions, the $z$-ligands would be moved away resulting in a stabilisation of orbitals with $z$ contribution. Here, one $z$-ligand is moved closer resulting in a destabilisation of anything including a $z$ contribution.


Reference

[1]: C. J. Ballhausen, H. B. Gray, Inorg. Chem. 1962, 1, 111. DOI: 10.1021/ic50001a022.

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