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In sulfur hexafluoride, there are two 3-centre 4-electron bonds, according to the book I am reading (Essential Trends in Inorganic Chemistry by D. P. M. Mingos). The other two fluorines are bonded to sulfur, by interacting with its sp-hybrid orbitals. Therefore, we should expect to observe two sets of bond lengths, one set for those involved in the 3-c 4-e bonds and one set for those involved in the bonds involving the hybrid orbitals. However, only one set of bond length (156.4 pm) is reported on Wikipedia. Why is this the case?

In comparison, we observe two different sets of bond lengths in sulfur tetrafluoride, one set of values for the axial bonds and another set of values for the equatorial bonds.

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    $\begingroup$ SF6 has symmetric bond angles, so it sort of makes sense that it also has symmetric bond lengths. Maybe the wavefunction is a rotationally symmetric superposition of the different orientations of the structure you describe. $\endgroup$ – Ben Crowell Nov 23 '17 at 23:27
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    $\begingroup$ Yes that’s precisely it: one resonance form is not sufficient to describe the structure of SF6. Alternatively, why is there only one C–O bond length in the carbonate anion even though the conventional Lewis structure has one double bond and two single bonds? The answer to that question is exactly the same as the answer to this, it just needs to be adapted to the present context. $\endgroup$ – orthocresol Nov 24 '17 at 0:38
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    $\begingroup$ @orthocresol Why is it that we can describe PCl5 and SF4 with just one resonance form and yet describe SF6 with multiple resonance forms? Does molecular symmetry determine the resonance forms? $\endgroup$ – Tan Yong Boon Nov 24 '17 at 0:45
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    $\begingroup$ Yes, exactly. In SF6 when you stick to one single resonance form, you have arbitrarily decided which atoms are participating in a single bond and which are involved in the 3c4e bond. However, because all atoms are symmetry equivalent, one resonance form cannot be sufficient. Same goes for carbonate. The follow up question is presumably why do we assume it is symmetrical? and truthfully I don’t know a full answer. There are cases where it is more stable to not be symmetrical, cf Jahn–Teller distortion, but SF6 is not one of them. $\endgroup$ – orthocresol Nov 24 '17 at 0:46
  • $\begingroup$ @orthocresol Do you know the answer to the follow-up question you proposed? If you do, please feel free to write an answer. I would think that the symmetry is justified by experimental data (i.e. exp. bond lengths and angles)? $\endgroup$ – Tan Yong Boon Nov 24 '17 at 0:50
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This is an old question and is largely answered in the comments, but here's a full answer for the sake of completeness.

There are really two questions here: 1) is the given bonding orbital description correct for SF6? 2) why is SF6 octahedral?

The first question is easier: no, your description is incomplete as you've described it, but perhaps correct as intended by the author (I don't have the textbook to see the exact wording). It is correct if you explain further that there are three possible representations with two $sp$ bonding orbitals and two 3c/4e bonds. The three different representations look identical except that the $sp$ orbitals are aligned on different Cartesian axes (ie one for $x$, one for $y$ and one for $z$). The true structure is a composite of these three, meaning that each S-F bond is 1/3 $sp$ and 2/3 3c/4e (which is essentially 1/2 of a $p$ orbital) which adds up to a total of 2/3 of a bond with a contribution of 1/6 from $s$ and 1/2 from $p$.

I suspect that the textbook uses this representation because it connects readily to earlier discussions of integer hybridization, i.e. "$sp$" orbitals are familiar and easy to visualize. And presumably the concept of a 3c/4e bond has also been previously introduced. Making a composite of familiar concepts is perhaps easier than introducing the idea of delocalized molecular orbitals, but in this case they are simple enough that it's worth going through them to show that the result is the same.

First, the $s$ orbital is spherically symmetric, so it necessarily interacts the same with all six identical ligands. The lowest energy molecular orbital is thus represented by the $s$ orbital on sulfur and the six fluoride ligand orbitals all in the same phase (images below). That's where the 1/6 $s$ comes from. The $p$ orbitals each are able to interact with two of the ligands, one on each end of orbital. The other four ligands do not contribute at all, because they are exactly on the nodal plane of the $p$ orbital. Since each $p$ interacts with two ligand orbitals, each S-F bond gets 1/2 $p$ to go with the 1/6 $s$, exactly the result we got from the composite of hybrids. Below are drawings of the four bonding molecular orbitals:

SF6 bonding MOs

Now on to the second question: why is SF6 octahedral if it only makes a total of 4 bonds?

Given that there is a lot of negative charge on the ligands and positive charge on S, a simplistic representation is to make the atoms solid charged spheres. A simple electrostatic potential energy minimization would then lead to an octahedral arrangement of the ligand spheres around the sulfur sphere. That arrangement minimizes the repulsive interactions between the ligands and maximizes the attractive interaction between ligand and sulfur.

Without a lot of empirical data about atomic interactions, though, we cannot rule out the possibility that there could be favorable bonding interactions with a strong distance dependence that would make it more favorable to, for example, push four of the ligands even closer to the sulfur and push the other two farther away. However, it turns out that the distance dependence of the bonding interaction is similar enough to the distance dependence of the electrostatic interaction that the geometric shape of the minimum energy conformation is the same - all six ligands at equal distance from the sulfur and spaced evenly around it.

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  • $\begingroup$ "The true structure is a composite of these three" Do you have anything to support this claim? Because I have never encountered this before when I was reading up the literature. $\endgroup$ – Tan Yong Boon Feb 3 at 23:05
  • $\begingroup$ That statement is supported by the MO diagrams, as I tried to explain. To be clear, I mean by "composite" that it is one structure that is the average of those three, not that it switches between them in any way. The average of those three is the symmetric octahedral structure that is well known for SF6, and the average of those orbitals is the same as the four well-established MOs of SF6 that I've shown. $\endgroup$ – Andrew Feb 4 at 1:23

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