I was reading about hybridisation and came across a point saying that $\mathrm{d^3s}$ hybridisation involves only $\mathrm d_{xy}$, $\mathrm d_{yz}$, and $\mathrm d_{zx}$ orbitals and not $\mathrm d_{x^2-y^2}$ nor $\mathrm d_{z^2}$. Why is that so?

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    $\begingroup$ Please read chemistry.stackexchange.com/questions/76726/… $\endgroup$
    – Ian Bush
    Apr 13, 2021 at 10:48
  • $\begingroup$ What is d3s hybridisation and in which compound have you seen it? $\endgroup$
    – S R Maiti
    Apr 13, 2021 at 15:35
  • $\begingroup$ @NisargBhavsar [MnCl4]- has Mn(+3) which is $\mathrm{d^4}$. I don't think d3s hybridisation is possible, it has to be sp3. Besides, hybrdisation is not useful for transition metals complexes. $\endgroup$
    – S R Maiti
    Apr 14, 2021 at 8:37
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    $\begingroup$ @NisargBhavsar But CrCl4- has Cr(+3) which is d3 so only two d orbitals are free, and you can't get d3s. Hybrdisation theory is good for main group elements; for transition metal complexes, it's not accurate and often gives wrong predictions or explanations. It has been mostly superseded by MO theory. $\endgroup$
    – S R Maiti
    Apr 14, 2021 at 19:23
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    $\begingroup$ @HarryHolmes Yes, that’s the one exception, and that’s why I kept it as such in the original post. $\endgroup$ May 19, 2021 at 1:43

1 Answer 1


First off, I'll reiterate what's mentioned in the comments - that hybrid orbitals are not a particularly good approach for transition metal complexes. But with that said, if you are going to apply this approach to a tetrahedral transition metal complex, here's how it works. For simplicity, I'm going to assume that only sigma bonding is relevent. You can do the same analysis including pi bonding, but it's more complicated.

Short answer: only the off-axis d orbitals have the right symmetry to match the ligands in a tetrahedral point group.

Long answer: The mathematical basis for constructing both delocalized and localized hybrid bond orbitals comes from group theory. If you haven't gotten to group theory in your studies, this explanation might not make a lot of sense, but at least you'll have to starting point to look deeper.

A tetrahedral molecule with four identical ligands is said to be in point group Td, which is just an abbreviation of "tetrahedral". Every point group has symmetries associated with it based on what operations can be performed that return the original structure. In point group Td, these are named A1, A2, E, T1 and T2.

When we build molecular orbitals, we combine the orbitals on the four ligands and look for possible symmetry groups. Then we look at the symmetry of the orbitals on the metal center and match the metal orbitals and "ligand group orbitals" based on symmetry. (An alternative approach is to look at the symmetry transformation of the bond vectors, which is described here: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/01%3A_Chapters/1.09%3A_Td_Molecular_Orbitals. The result is the same.)

Making a long story short, the tetrahedral ligand group orbitals for a sigma-bonded complex fall into two symmetry groups. One orbital, in which all four ligand orbitals have the same phase, is A1. A second group of three orbitals in which the ligand orbitals are pairwise in opposite phases matches the T2 symmetry of the Td group.

On the metal center, our valence atomic orbitals are 4s, 3d and 4p. The 4s orbital matches the A1 symmetry, while the 4p and 3d (xy, yz and xz) match the T2 symmetry. The 3d(x2-y2) and 3d(z2) orbitals, on the contrary, match the E symmetry from group Td. Since the E symmetry doesn't match either of our ligand group orbitals, we can't use these two orbitals for bonding.

Since the 4s orbital is the only match for the A1 symmetry, we must involve that orbital in bonding. For the T2 symmetry, however, we have two choices. We can either use the three 4p orbitals and ultimately make four sp3 hybrids, or we can use the three T2 symmetry 3d orbitals and end up with d3s hybrids. The sp3 hybridization is much more familiar to most people because they have learned it for organic compounds such as methane, but that's not the right choice here. If we consider the relative energy levels of the orbitals on a transition metal, we see that 3d is lower than 4p and so is closer to the ligand orbitals (which are lower in energy than both). Furthermore, the off-axis d orbitals are pointed more directly at the ligands in a tetrahedral complex (which are also off-axis), so this intuitively makes more sense as well.

Finally, to your question about $\ce{CrCl2O2}$, that molecule belongs to point group C2v, so you can apply the analysis above to that point group to get the allowable hybridizations in the same manner as for the Td case.

  • $\begingroup$ +1 But I would also add that the choice of which 3 d-orbitals are used for hybridisation, is entirely dependent on the choice of axes. In a free atom, all 5 d-orbitals are degenerate. It is only when you bring in the ligands and choose the x y z axes that you get a differentiation. $\endgroup$
    – S R Maiti
    May 12, 2021 at 18:28
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    $\begingroup$ @ShoubhikRMaiti That's only half true. The orbitals are indeed degenerate in the absence of ligands, and the axes in that case are completely arbitrary, but as soon as we assign a point group, that designation is done with respect to specific axes related to the symmetry operations, so the axes are no longer arbitrary. And we have specific conventions for coordinate systems of molecules that are used in group theory, so proper application of convention will always yield the same three d orbitals in the T2 group. $\endgroup$
    – Andrew
    May 12, 2021 at 19:13
  • $\begingroup$ Yes, but hybridisation is not usually taught with symmetry arguments. My point is that, if someone started reading hybridisation and asked why those 3 d-orbitals are used, the answer should be that it's because of the choice of axes. The choice of axes is a convention but convention in itself is arbitrary. Any three perpendicular axes could be chosen, but one of them has been decided to be the right one and that's an arbitrary choice. $\endgroup$
    – S R Maiti
    May 13, 2021 at 16:38

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