I was recently reading about the transition metals and some important compounds of them. I came across a rather interesting diagram in my book... a tetrahedral structure for the chromate ion. The only tetrahedral structure I know arises from an sp3 hybridization of the central atom, but as far as chromium is concerned, I think that a d3s would be more practical(?).. But then again, which specific orbitals of 3d would hybridize to give a tetrahedral structure?(this is beyond me, i seriously do not understand the shapes of the d orbitals)

  • $\begingroup$ Do not use hybridization with transition metals. Correction: Best not to use it for anything at all. $\endgroup$ Aug 28, 2019 at 16:44
  • $\begingroup$ but then again, how do you figure out shapes of stuff without that? like... normal covalent compounds, then we have the coordination spheres, how do you predict shape of such stuff WITHOUT taking hybridization into account? $\endgroup$
    – Om Nom
    Aug 28, 2019 at 17:03
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    $\begingroup$ im kind of new to this molecular orbital theory stuff, will surely check it out $\endgroup$
    – Om Nom
    Aug 28, 2019 at 17:52
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    $\begingroup$ but then @OscarLanzi is it so that valence bond theory and VSEPR cannot explain everything? $\endgroup$
    – Om Nom
    Aug 28, 2019 at 18:02
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    $\begingroup$ VSEPR is a very good model, but you have to ignore the hybridisation postrationalisation, which came much later. It doesn't get everything right, but it allows prediction on very simple reasoning. It's good for a first guess approach. Valence bond theory is much often taught in a completely abridged (beyond the useful) manner. It's a full quantum chemical model. The exact treatment of a molecule with it results in the same exact wave function as with molecular orbital theory. Beyond that, there is currently no theory to explain everything, or as it is also sometimes called "World Formula". $\endgroup$ Aug 28, 2019 at 19:59

1 Answer 1


The d orbitals are commonly grouped into two groups because that is how they transform under both octahedral (including a g subscript) and tetrahedral (no g subscript) symmetry:

  • $\mathrm{t_{2(g)}}: \mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$
  • $\mathrm{e_{(g)}}: \mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$

In tetrahedral environments, the p orbitals also transform as $\mathrm t_2$ while in octahedral ones they transform as $\mathrm{t_{1u}}$. By analogy to main group tetrahedral molecules, it would be easy to assume that just the three $\mathrm t_2$ d orbitals take part in the formation of tetrahedral complexes (throw in one s orbital for completion of the four).

In reality, all orbitals that transform as $\mathrm t_2$ will be represented in the final complex in some way or another, so that whatever bonding and antibonding orbitals of $\mathrm t_2$ symmetry you end up with will always include some contribution from the original p orbitals. Furthermore, outside of hydride complexes the ligands will also have available orbitals for π-type bonds which will complicate the picture further. The final MO for chromate will look something like the following:

Figure 1: tetrahedral $\ce{[ML4]}$ complex with ligand π contributions.

The metal orbitals on the left are d, s, p from bottom to top. The ligand orbitals on the right are the oxygens’ p orbitals (all 12 of them).

Don’t worry, these schemes will come to haunt you at one point when studying chemistry.

  • $\begingroup$ Thank you... exactly what i was looking for.. $\endgroup$
    – Om Nom
    Aug 31, 2019 at 14:15

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