A molecule in which the central atom is $sp^3d^2$ hybridized ($\ce{SF_6}$), for example, acquires an octahedral shape, which can be explained by assuming that the hybrid orbitals arrange themselves to minimize repulsion amongst themselves. An octahedron makes sense, it the only way to symmetrically arrange six hybrid orbitals on an atom. The case for an $sp^3$ hybridized atom is similar which from a similar logic can be explained to have a tetrahedral geometry.

But consider the case of $sp^3d$ hybrid orbitals which arrange themselves in a trigonal bypyramidal shape ($\ce{PCl5}$). This is a gross breach of the usual adherence to symmetry. Two bonds are at the axial location and three at the equatorial location. In fact, the axial bonds are slightly longer than the equatorial ones. The case for $sp^3d^3$ hybridization is similar ($\ce{IF7}$) with five orbitals crammed into the equatorial plane and two sticking out above and below it. Why are these arrangements preferred in favor of one with a simple symmetric arrangement of orbitals (which, I think, would minimize the repulsion)?


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    $\begingroup$ Which alternative "simple symmetric arrangement" do you have in mind? Do you know of possible structures for $\ce{PCl5}$ or $\ce{IF7}$ which are more symmetric than their trigonal bipyramidal and pentagonal bipyramidal geometries with $D_{3h}$ and $D_{5h}$ symmetry, respectively? You may also be interested in this page which discusses a way to calculate these predicted geometries without any direct inference to Chemistry or VSEPR theory. $\endgroup$ May 17, 2015 at 15:15
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    $\begingroup$ Hybridization involving d-orbitals is not an accurate description for these molecules, rather hypercoordinated (or hypervalent) bonding using only s- and p-orbitals provides a better explanation. For application of this concept to $\ce{SF6}$ see here. for $\ce{PCl5}$ see here and for $\ce{IF7}$ see here. $\endgroup$
    – ron
    May 17, 2015 at 15:21
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    $\begingroup$ @NicolauSakerNeto: As far as I am aware, the structure I have in mind does not have any common name. I think the best description might be an operational one. To take the example of $sp^3d^3$, tie seven balloons at their common centre. The structure they assume is similar to the one I have in mind. $\endgroup$
    – Gerard
    May 17, 2015 at 15:29
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    $\begingroup$ @NicolauSakerNeto: I think the answer might lie in the fact that only five platonic solids are possible. And as mentioned in the link you provided: "The vertices of the five Platonic solids give the only perfectly symmetrical distributions of points on the surface of a sphere. Therefore, if N positively charged particles are constrained to the surface of a sphere, and N is not equal to the number of vertices of a Platonic solid, then the particles must have an equilibrium configuration that is not perfectly symmetrical. " $\endgroup$
    – Gerard
    May 17, 2015 at 15:32
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    $\begingroup$ Lowest energy states don't always have the highest symmetry. For example, in the Thomson problem, where the goal is to find the points on a sphere where the Coulomb repulsion energy is minimized, the configurations for n=3 and n=5 points have two points of one type of symmetry on opposite poles and the other points of a different type of symmetry and forming a polygon around the equator. $\endgroup$ May 17, 2015 at 17:29

2 Answers 2


As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.

d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.

Conveniently, all the cases you presented in your question can be adequately explained using only s- and p-orbitals. See the links for $\ce{SF6}$, $\ce{PCl5}$ and $\ce{IF7}$ respectively. Many thanks to ron for providing them in his comments!


For a structure to be perfectly symmetric, it must satisfy this condition:

For any two bonds, the plane passing through them must divide the molecule into two symmetric halves.

Note that this condition is satisfied by trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and pentagonal bipyramidal geometry.

I'm going to show that there are no better structures satisfying this condition for sp$^3$d and sp$^3$d$^3$ hybridisation.

For sp$^3$d, we're going to first arrange any two bonds at some angle (unknown). Obviously, the two bonds form a plane ($3$D here): Step 1

Using the condition above, we have to arrange $3$ bonds around the plane, such that they are symmetric about it.

We cannot put $2$ bonds on one side and $1$ bond on the other side, as that would be unsymmetrical ($3$D here).


Therefore, for symmetry about the plane, we must keep a third bond on the plane. The instant we do that, we are left with only two options: trigonal bipyramidal and pentagonal planar. Of the two, trigonal bipyramidal is obviously a more stable structure (bond angles $90°$ and $120°$) than pentagonal planar (bond angle $72°$) ($3$D here).


When we proceed in a similar manner for sp$^3$d$^3$, we have two options: heptagonal planar (bond angle $51.43°$) and pentagonal bipyramidal (bond angles $72°$ and $90°$. Again, the pentagonal bipyramidal structure is more stable.

Thus, these are the only feasible structures that are symmetrical.


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