It's known that $\ce{sp^3d}$ hybridization has 5 lobes; 3 equatorial with $\ce{sp^2}$ lobes and 2 axial with $\ce{dp}$ lobes. Do we have such splitting in $\ce{sp^3d^2}$ and $\ce{sp^3d^3}$ hybridization states too?
What I think is $\ce{sp^3d^2}$ splits into $4 \ce{dsp^2}$ lobes and $2 \ce{dp}$ lobes.

Is this correct?

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    $\begingroup$ We seem to continually get questions about such out-dated (not to mention wrong) theories. chemistry.stackexchange.com/questions/54737/…, chemistry.stackexchange.com/questions/66698/…, chemistry.stackexchange.com/questions/75009/…, chemistry.stackexchange.com/questions/129358/… $\endgroup$
    – C_Lycoris
    Apr 27, 2021 at 3:03
  • $\begingroup$ @C_Lycoris Please don't use $$..$$ for links, you can just paste the links, and it will do the job. Using $$ just makes your comment appear large as each link is displayed on a new paragraph. Also, hybridization is still taught in high schools and the first year of university; I would be cautious before calling it out-dated and wrong. $\endgroup$
    – S R Maiti
    Apr 27, 2021 at 9:22
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    $\begingroup$ @ShoubhikRMaiti It makes it by far worse that this garbage, deprecated, outdated, wrong stuff is still taught at universities. It is wrong, there is no doubt about it. Absolutely wrong. Is as wrong as the flag earth theory. If here is not the place to mention this, we could pack up shop immediately, there would be absolutely no Use for this site. $\endgroup$ Apr 27, 2021 at 19:15
  • $\begingroup$ @Martin-マーチン I am sorry, but I have to disagree. If you look at any organic textbook, most of the explanations still invoke hybridisation theory. A lot of other outdated stuff is also taught in high schools and universities as a pedagogic tool (like wet tests of ions etc.). I agree that it's not completely right, but as long as its taught in some level, I feel that people should be able to ask questions about it here. $\endgroup$
    – S R Maiti
    Apr 28, 2021 at 8:40
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    $\begingroup$ @ShoubhikRMaiti You can disagree with me, I'm fine with that. I've never said that I want to disallow these questions, but I object vehemently to the notion that we have to be courteous about our replies. If the theory is wrong, we have to be able, no it is our duty, to point that out. Also, the is a huge difference between outdated methods and completely wrong theories. And by the way, hybridisation is not a theory, it is a descriptive mathematical tool. What is described in organic chemistry texts is often enough okay; however, anything in this question certainly is not. $\endgroup$ Apr 28, 2021 at 15:58

1 Answer 1


To answer this, consider the purpose of the mathematical exercise we call "hybridization". The goal is to carve up the total electron density of the molecule into discrete segments (bonding orbitals) that are localized mostly to the area between two nuclei and which have some type of symmetry (usually either sigma or pi) with respect to the axis connecting the two nuclei.

A final important criterion is that these hybrid orbitals should be consistent with our intuitive understanding of equivalent bonds. For example, the four C-H bonds in methane are indistinguishable. Thus, the four hybrid bonding orbitals should also be indistinguishable.

With that in mind, consider that the six metal-ligand bonds in an octahedral complex are also indistinguishable (assuming all six ligands are identical). It does not make sense, then, to use four dsp2 hybrid orbitals and two dp, since that is not match six indistinguishable bonds. Instead, we extend the methane example and make six identical sp3d2 hybrids.

With odd numbers of ligands like 5 or 7, the metal ligand bonds are not all equivalent, so it makes sense to have a separate hybridization construct for each type of metal-ligand bond. For example, in a trigonal bipyramid, the equatorial bonds form one group and the axial bonds another.

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    $\begingroup$ While this is mathematically probably correct, I do reject this answer as it enforces the delusion that there is actually any physical meaning in this. $\endgroup$ Apr 27, 2021 at 19:19
  • $\begingroup$ @Martin-マーチン What is "physical meaning"? I was careful to specify that there's no difference in the physical result (ie net electron probability density) of a hybrid orbital description vs canonical delocalized orbitals (which I assume have "physical meaning" for you?). Given the indistinguishability of electrons, I'm not sure that any model that places electrons in discrete orbitals has physical meaning. $\endgroup$
    – Andrew
    Apr 30, 2021 at 12:49
  • $\begingroup$ I'm not sure I want to dive down this rabbit hole, but I also cannot let this accusation stand. The keyword in my original comment should be any. Can canonical orbitals be interpreted in a physical meaningful way? Maybe, but that wasn't my point, and I gladly refer to Observability of Orbitals and Orbital Energies and Evidence of orbitals? for further discussion. My point is (was) that it is impossible to utilise spd/dsp hybrid orbitals in any physically meaningful way. $\endgroup$ May 2, 2021 at 18:05

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