The central atom has a hybridization of $\mathrm{sp^3d^3}$. Thus, its structure should be pentagonal bipyramidal.
Why is it not that but a distorted octahedron?
The central atom has a hybridization of $\mathrm{sp^3d^3}$. Thus, its structure should be pentagonal bipyramidal.
Why is it not that but a distorted octahedron?
This is one of the many reasons why hybridisation including d-orbitals fails for main-group elements.
Xenon in $\ce{XeF6}$ is not hybridised at all. Instead of invoking populated core d-orbitals or energetically removed d-orbitals (remember the aufbau principle: the next shell’s s-orbital has a lower energy than the d-orbitals you are proposing to include in hybridisation!) xenon just offers its three p-orbitals $\mathrm{p}_x, \mathrm{p}_y$ and $\mathrm{p}_z$ for four-electron-three-centre bonds. These 4e3c bonds can be understood using the following two mesomeric structures:
$$\ce{F^-\bond{...}Xe^+-F <-> F-Xe^+\bond{...}F-}$$
Each $\ce{Xe-F}$ bond has a bond order of ½, and for each fluorine there is another with a bond angle $\angle(\ce{F-Xe-F}) \approx 180^\circ$ as part of the same 4e3c bond.
Also note that this means that xenon’s lone pair is comfortably located in the $\mathrm{5s}$ orbital.
It has to do with the steric crowding of the ligands. Thus, the lone pair has very little room so it moves all over the molecule to get as far away as possible. The motion of the lone pair distorts the molecule from a perfect octahedron.