The structure of xenon fluoride is a capped octahedron. The lone pair is stereochemically active, i.e. it will rotate about.

Structure of XeF6

The dipole moments of the axial fluorides will get cancelled, and so will those of the four equatorial fluorides will also cancel each other. Thus, overall, the molecule should be nonpolar.

But what effect will the stereochemically active lone pair have on the polarity? Will the polarity keep on changing?


1 Answer 1


First of all, the point group for $\ce{XeF6}$ is disputed. The latest I saw was $C_\mathrm{3v}$, not $O_\mathrm h$ but the $O_\mathrm h$ is so energetically close to its $C_\mathrm{3v}$ that it is likely in a perpetual state of transition between the two. As for polarity, due to this it is hard to think it would have a permanent dipole. One would assume it would be easy to induce one, though.

$\ce{XeF6}$ is not the typical molecule the different models were made for, so YMMV.

  • $\begingroup$ >First of all, the point group for XeF6 is C3v, not Oh as you have shown || AFAIK there is no clear consensus on it. $\endgroup$
    – permeakra
    Jun 19, 2017 at 15:57
  • $\begingroup$ I have no idea about point groups . Is there some other way to explain its polarity ? $\endgroup$
    – Excalibur
    Jun 19, 2017 at 16:54
  • $\begingroup$ @permeakra I accept that there is no clear consensus. It is somehow relevant though to mention that it is either highly transitory or doesn't allow for polarity. Should I rephrase? $\endgroup$
    – Stian
    Jun 19, 2017 at 20:47
  • $\begingroup$ Yes please rephrase. $\endgroup$
    – Excalibur
    Jun 20, 2017 at 1:36

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