# Why an asymmetric geometry with sp3d and sp3d3 hybridization?

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)?

• 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. – Nicolau Saker Neto May 17 '15 at 15:15
• 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. – ron May 17 '15 at 15:21
• @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. – Gerard May 17 '15 at 15:29
• @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. " – Gerard May 17 '15 at 15:32
• 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. – user1704042 May 17 '15 at 17:29

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!