Recently, in a class discussion, the following point was brought up:

Although the axial and equatorial bond lengths in $\ce {PCl5}$ are inequivalent, when we consider the time-averaged $\ce {P-Cl}$ bond length, there should only be one value sincethe molecule is symmetric and thus, all the chlorine atoms interact with the phosphorus atom to the same extent.

The idea of "time-averaged" means that we are considering the $\ce {P-Cl}$ bond over a long duration. This bond length varies over time since the atoms are constantly in motion. When this bond length is "time-averaged", it means that we are taking the average of the bond lengths measured at infinitely many instants in time.

The argument is that although at any particular instant, the molecule would adopt a configuration with axial bonds being longer than equatorial bonds, when the bonds are time-averaged, they should all be the same length.

This has caused much dispute among members of the class and it has also led me to consult a professor who is well-versed with quantum chemistry. The professor mentioned that the argument put forward is not valid because in $\ce {PCl5}$, there is a breaking of degeneracy due to perturbation. He cited the Jahn-Teller distortion as a situation similar to that in $\ce {PCl5}$, where degeneracy is broken.

I am very confused by this. What exactly is this perturbation? Why does it exist for $\ce {PCl5}$ and not for similar molecules, such as $\ce {SF6}$ (as seen in the comments for this question)?

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    $\begingroup$ The experimental data shows that the two axial bonds ($C_3$ axis) are significantly longer than the 3 equatorial ones i.e. outside expt. error different. So there is no time averaging going on here as that would predict similar bond lengths. Additionally, electronegative atoms preferentially fill the axial positions e.g $\ce{PCl3F2}$ indicating a difference in bonding between axial and equatorial positions. The proposed trigonal bipyramidal ($sp^3d$ bonding) is consistent with minimising electron pair repulsion; $sp^2$ along equatorial; $pd$ along axial bonds. $\endgroup$
    – porphyrin
    May 24, 2018 at 13:01
  • $\begingroup$ @porphyrin I am completely aware of the 3c4e bonding scheme involved and I understand the rationale behind the difference in bond lengths and thus, I am not looking for this in the answer that I require $\endgroup$ May 24, 2018 at 13:20
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    $\begingroup$ Rapid pseudorotation on the observational time scale is what makes the bonds euivalent. $\endgroup$
    – ron
    May 24, 2018 at 13:27
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    $\begingroup$ That being said, the pseudorotation appears rapid to some physical methods and slow to the others. $\endgroup$ May 24, 2018 at 13:31
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    $\begingroup$ The molecule will undergo a Berry rotation , but that does not make the bonds equivalent it effectively just swaps them around, rather like the effect of rotating the axes. The experimental evidence is that the two types of bonds have different lengths. A similar, but simpler, effect occurs in ammonia inversion or $\ce{CH2}$ wagging in cyclopentene or any number of molecules with small barriers to rotation/inversion. $\endgroup$
    – porphyrin
    May 24, 2018 at 13:43

1 Answer 1


I would guess he means a spontaneous symmetry breaking, in which case the "perturbation" is any old force from the environment. This can happen in physical systems: it's possible for a system of high symmetry to spontaneously fall to a lower-energy state of lower symmetry. Only the tiniest of perturbations is required, if you wait long enough. A familiar classical example is crystallization, in which below a certain temperature a system of atoms spontaneously loses complete translational symmetry even in the absence of an external potential. A quantum example is magnetization when a ferromagnet cools below the Curie temperature -- the spins lose rotational symmetry and pick some essentially random preferred direction, in response to the tiniest of perturbations from the environment. In the case of PCl5, one would assume, very roughly (and it would take detailed calculation to verify) that if 3 of the Cls pull in closer to the nucleus and 2 move further away, you get a better balance of nuclear-electron attraction and nuclear-nuclear and electron-electron repulsion than if all 5 remain at the same distance. So the system spontaneously breaks rotational symmetry, in response to any old tiny perturbation in the environment.

That said, in the absence of an external potential, the rotational quantum state of the molecule will be entirely symmetric, and you will have an equal chance of finding the long Cl-P-Cl axis pointing in any direction.

  • $\begingroup$ So if the PCl5 molecule isn't interacting with the environment, say it is in some vacuum, then it would adopt the symmetrical configuration? $\endgroup$ May 25, 2018 at 13:54
  • $\begingroup$ Why does the same not occur for SF6? $\endgroup$ May 25, 2018 at 13:55
  • $\begingroup$ The whole point of spontaneous symmetry breaking is that the perturbation required to achieve it becomes vanishingly small, so if you mean a real vacuum, such as interstellar space far form a nearby galaxy, no, because even in that lonely place there'd be a perturbation greater than strictly zero. If you mean a hypothetical empty universe -- who knows? There's certainly a theoretical answer, but you have to bear in mind our theories were inferred from experiment in a non-empty universe. They need not give the correct answer for other (untestable) situations. $\endgroup$ May 26, 2018 at 18:10
  • $\begingroup$ I don't know (why it doesn't happen for SF6). One could do deep calculations and maybe find an answer why the balance is better struck in that case by a purely symmetrical arrangement. We do know symmetry is strongly preferred in a symmetric environment, so much so that a case like PCl5 is a surprise that needs to be explained. Spontaneous symmetry breaking is rare, most of the time the symmetry of the ground state of a system matches the symmetry of its environment. $\endgroup$ May 26, 2018 at 18:13

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