Why an asymmetric geometry with sp3d and sp3d3 hybridization?

Question

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)?

Answer 1

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!

Answer 2

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):

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.