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?
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2$\begingroup$ related chemistry.stackexchange.com/questions/34073/… $\endgroup$ – Mithoron Dec 15 '16 at 17:52
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4$\begingroup$ Possible duplicate of What is hybridisation of XeF6 in solid state? $\endgroup$ – Mithoron Dec 15 '16 at 22:30
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$\begingroup$ NOT a duplicate of @Mithoron's second comment. The answers may be similar, but this question asks about the structure due to an assumed hybridization, whereas the linked "duplicate" asks about the hybridization itself. $\endgroup$ – hBy2Py Dec 16 '16 at 4:57
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$\begingroup$ While the question linked in @Mithoron's first comment does provide the symmetry group of $\ce{XeF6}$, it does not indicate the actual structure of the molecule. Further, it's closed. $\endgroup$ – hBy2Py Dec 16 '16 at 4:59
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$\begingroup$ >Why is it not that but a distorted octahedron? || It actually is, though for different reasons. $\endgroup$ – permeakra Apr 22 '18 at 19:07
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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.
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$\begingroup$ So, the molecules of $XeF_6$ are arranged in a lattice with a structure that has more similarity with transition metal halide salts with bridge-bonding, right? $\endgroup$ – Eashaan Godbole Dec 15 '16 at 18:27
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$\begingroup$ @EashaanGodbole Erm … what? I’m sorry, I don’t understand what you’re asking … $\endgroup$ – Jan Dec 15 '16 at 18:28
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$\begingroup$ @ Jan The bonding between xenon and the fluorides is 3c-4e like the titanium chloride salts...? $\endgroup$ – Eashaan Godbole Dec 17 '16 at 16:54
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$\begingroup$ @EashaanGodbole I’m not familiar with solid state ionic structures too much, but $\ce{TiCl3}$ is a typical ionic structure with closest packing of anions and cations in the octahedral voids. This doesn’t compare well to molecular structures as are present in xenon compounds which feature distinct $\ce{XeF_{2n}}$ molecules. $\endgroup$ – Jan Dec 17 '16 at 17:47
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1$\begingroup$ Correct to a first approximation; the 3c4e model doesn't fully explain XeF6, though, as it is not a static octahedral molecule. [Of course, as you said, there is no (or at most, little) d-orbital involvement.] $\endgroup$ – orthocresol♦ Sep 8 '17 at 14:35
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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.
