Why are the filled orbitals s and px (in the T-shaped case) also part of the hybrid orbitals of the central atom in sp3d hybridization?

In other words, why should the hybrid orbital not be a p3d, for the seesaw case, and a p2d for the t-shaped case?*

*Considering the original eletronic distribution to be a s2 p4 and s2 p5 respectively.


2 Answers 2


This is a complicated issue, and there are many problems with what you are asking.

  1. Hypervalency (expanded octets) don't appear to actually exist. They can be a useful model. This Wikipedia article does a reasonably good job at looking at the issue: hypervalent molecule.

    Basically, the even though we can draw Lewis structures of hypervalent compounds, it doesn't mean that they exist as such.

    Bonding in sulfur hexafluoride.  via: By Officer781 - Own work, Public Domain, https://commons.wikimedia.org/w/index.php?curid=39439792

    Sulfur hexafluoride, for example appears to be a covalent-ionic hybrid.

    At the same time, Lewis drawings of hypervalent species such as sulfate ion are still commonly used in organic chemistry, biochemistry and biology.

  2. The next issue is hybridization. As others have mentioned, it has been shown that d-orbitals do not participate in hybridization [E. Magnusen, J. Am. Chem. Soc. 1990, 112 (22), 7940–7951]. As such, the sp3d and sp3d2 hybridizations are incorrect.

    One of the problems here is that instructors (and textbooks), especially at an introductory level will have a tendency to over-generalize the rules for Lewis diagrams and for hybridization.
    Hybridization needs to be invoked to rationalize the molecular geometries around small atoms (O,N,C, etc) with the geometry of the atomic orbitals in these atoms.

    In larger atoms however, we do not always need to invoke hybridization. For example, the $\ce{H-S-H}$ bond angle in $\ce{H_2S}$ is $92^\circ$. This corresponds well to simple overlap between the p orbitals of sulfur with the s orbital of hydrogen. No need to hybridize.

I know this doesn't actually answer your question, but I hope that it explains why your question is flawed.

  • $\begingroup$ I didn't knew about this fact of hybridization, as you said many instructors and textbooks teach us that way, and I was one of this cases. That said, could you bring some light to the problem of finding the hibryd orbital in a T-shaped molecule, we can say C$\ell$F$_3$. $\endgroup$
    – liuzp
    May 9, 2019 at 2:17
  • $\begingroup$ Although it's not the main point of your answer, your statement that H2S bonding involves only p orbitals is not correct. It is a common misconception. If you look at the MO's of XH2-type molecules, you'll see that there is a large s orbital contribution to the lowest energy bonding MO. $\endgroup$
    – Andrew
    May 9, 2019 at 11:18
  • $\begingroup$ @Andrew, interesting point. The reason I brought H2S up in particular is because, in spite of the s-orbital may contribution, from a qualitative point of view, we don't need to invoke hybridization to explain the observed geometry. At an introductory level, the kind of fine distinction you are pointing out is besides the point, and usually quite beyond the scope of the course and the student's abilities. VSEPR, VB and MO are all models of bonding, and all models have their limits. My point was that hybridization is usually over-generalized to situations where it doesn't need to be applied. $\endgroup$ May 9, 2019 at 19:44
  • 1
    $\begingroup$ @liuzp I am not an expert in bonding theories. However, $\ce{ClF3}$ is hypervalent when drawn in the traditional way, so the traditional models like Lewis diagrams, VSEPR and VB don't correspond with reality (even if we can draw "valid" Lewis diagrams). Bonding models are just that, models. They have their limits, and don't work in every situation. $\endgroup$ May 9, 2019 at 19:52

Firstly, hybridisation is absolutely not what you should be using to explain structure and bonding, period.

It has been proven conclusively that the d orbitals do not participate in bonding whatsoever.

  • 1
    $\begingroup$ There is a multitude of cases, where d orbitals do take place in bonding. Almost all transition metal complexes need d orbitals. You are generalising to the other extend of what is useful. On another note, I'd consider this more like a comment than an actual answer. $\endgroup$ May 8, 2019 at 15:33
  • $\begingroup$ Hybridisation should be used even less in transition metal complexes, though. $\endgroup$ May 9, 2019 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.