37
$\begingroup$

I've always been a bit uncomfortable with the concept of more than two electrons in a single orbital-like region(probability-wise) which occurs in resonance.

This seems to disobey Pauli's exclusion principle.

Let's take benzene for example. It has a delocalized cloud of electrons. These all had the "original" (as in, if resonance didn't occur) quantum numbers $n=2,l=1,m=\text{(1,0,or -1; but it must be the same for all)},m_s=\pm\frac12$. But, due to delocalization, they're now all in one extra-large orbital-like region (by orbital-like, I mean that the pi cloud is similar to an orbital in the sense that it is also a probability cloud which defines the probabilities of finding one or more electrons at different coordinates).

Now, it's obvious that the original quantum numbers are insufficient here due to Pauli's exclusion principle. They may not even exist for these electrons.

The easiest thing to do would be to "propose" (not exactly propose, since I guess the Dirac equation can prove their existence) an extra quantum number, which can take twelve values. Or a combination of a few extra quantum numbers that have the same effect.

But this raises the question: do the electrons belonging to the pi cloud "retain" their "original" quantum numbers as well?

Basically I want to know how benzene and other resonating compounds are described quantum mechanically, with focus on quantum numbers. (links to papers are fine, probably necessary here--but please explain the gist of the link)

I'd prefer the answer in terms of simpler quantum mechanics, though I'm not averse to the idea of some extra reading.

$\endgroup$
1
  • 5
    $\begingroup$ I would simply says that for the molecule, the original orbitals are no longer adequate… the molecule has a different set of orbitals, which are the solutions to Schrödinger’s equation. These orbitals do not have more than two electrons each in them (or one if you consider spin-orbitals). $\endgroup$
    – F'x
    Commented May 8, 2012 at 9:33

3 Answers 3

37
$\begingroup$

To understand why the exclusion principle isn't violated in this system, you really need to shift from valence bond theory to molecular orbital theory. The $p_{z}$ orbitals in benzene combine according to the $D_{6h}$ symmetry of the molecule to generate a set of bonding molecular orbitals (MOs) (which are lower in energy than the isolated $p_{z}$ orbitals) and a set of antibonding molecular orbitals (which are higher in energy). Bonding orbitals are generated from sets of $p_{z}$ orbitals which are mostly or totally in phase, whereas the highest energy antibonding orbital of the set is composed of $p_{z}$ orbitals that are totally out of phase with respect to their adjacent neighbors.

Here, for instance, is the MO that is generated from the totally in-phase set (I calculated this at a spin-restricted RI-BP86/6-311G* level of theory in ORCA(1) and visualised the isosurfaces in VMD(2)):

enter image description here

Now, this MO is shared equally across all of the carbons, i.e. it is not localised. However it only contains 2 electrons and thus satisfies the exclusion principle. (F'x mentions spin-orbitals - this is where the orbitals are split on the basis of their spin, which entails a single electron per orbital. Benzene is a closed-shell singlet so we would expect the $\alpha$ and $\beta$ spin orbitals to be spatially indistinguishable.)

Not to sound like a walking advertisment for Housecroft and Sharpe, but H&S has a great visual introduction to MO theory.

P.S.

It was suggested in chat that I expand upon the generated orbitals - here's one of the two highest occupied molecular orbitals, from the same calculation.

enter image description here

As you can see, there's a big node along one of the mirror planes. This orbital is higher in energy than the earlier example because it's composed of two antiphase sets of 3 $p_{z}$ orbitals, so this orbital has antibonding character for two of the carbon-carbon pairs. It should be easy to extrapolate to the highest lying MO of this type - antibonding all the way around.


(1) Neese, F. ORCA – an ab initio, Density Functional and Semiempirical program package, Version 2.6. University of Bonn, 2008

(2) Humphrey, W., Dalke, A. and Schulten, K., 'VMD - Visual Molecular Dynamics', J. Molec. Graphics 1996, 14.1, 33-38.

$\endgroup$
3
  • 3
    $\begingroup$ This is the correct answer to this question. The six pi electrons occupy three different molecular orbitals of very different shapes. These are provided even by the very simple calculations of Hückel theory. $\endgroup$ Commented May 12, 2012 at 7:32
  • $\begingroup$ Could you perhaps post the ORCA input you used somewhere? Playing with ORCA is in my to-do list for a long time now and using this example I could just dive in. $\endgroup$
    – Ajasja
    Commented Jul 18, 2012 at 9:31
  • $\begingroup$ What is all the fuss? Pauli exclusion isn't violated, even in MO theory, because the two electrons have different quantum numbers (opposite spin) $\endgroup$
    – Eric Brown
    Commented May 31, 2013 at 13:33
20
$\begingroup$

The short answer to your question is that no, they don't all occupy the same orbital but yes, they occupy very similar regions in space. What tends to be confusing in these pictures is it sometimes unclear that the pictures are sometimes of the molecular orbitals and sometimes of the electron density, which is an observable quantity and quite another thing altogether.

The six pi electrons occupy three different molecular orbitals of very different shapes, as can be seen here, for example, and in Richard Terrett's answer. However, their corresponding probability densities can be superposed onto similar regions of real space. There is no conflict with Pauli exclusion because you can see (and calculate!) that the orbitals are orthogonal by virtue of having an appropriate number of nodes. While on average the electrons look like they are all in the same region of space, if you measure the positions at any time, no two electrons will ever occupy exactly the same point in space.

Your discussion of orbital and spin angular momenta is completely irrelevant. Angular momentum operators $L_z$ and $L^2$ commute with atomic Hamiltonians because of rotational and Runge-Lenz invariance; however, they do not commute with molecular Hamiltonians because molecules have neither of these symmetries. Therefore, molecular orbitals are not indexable by the same set of quantum numbers as can be done for atomic orbitals. This is not because of Pauli exclusion, but because the Hamiltonian, $L_z$ and $L^2$ no longer form a complete set of commuting observables. (Note that Pauli exclusion must also hold in atoms where this latter statement is true!) Nor is there any need to introduce new fictitious quantum numbers.

$\endgroup$
2
  • $\begingroup$ (and electrons of opposite spin in the same MO have different quantum numbers, so their Slater determinants don't vanish.) $\endgroup$
    – Eric Brown
    Commented May 31, 2013 at 13:48
  • $\begingroup$ Great answer! Here is also a more detailed image and explanation of the 3 different molecular orbital shapes that were mentioned in this answer: chemcomp.com/journal/molorbs.htm $\endgroup$
    – akhmed
    Commented Mar 20, 2018 at 23:39
17
$\begingroup$

I very much like the first answer, especially the superb visuals!

However, I will nonetheless add a simpler approach to answering your question, which is this: Pauli exclusion holds as long as there is something in space or in spin (spin being the pairs of electrons) that firmly distinguishes two "stable" states.

What that means is that your initial intuition is essentially correct if all the electron pairs shared exactly the same molecular orbital.

So the answer to your question in general terms is that electron pairs settle into molecule-sized orbitals (molecular orbitals) that are physically spread out over the entire ring, but which nonetheless have unique geometric features that keep them distinguished in real space. I would add that they are even more unique when represented in an abstract space called momentum space, where the geometric differences translate into actual distances between the electron pairs. These abstract distances in momentum space satisfy the Pauli exclusion principle just as well as physical separation in real space, so that even very similar looking orbitals can remain distinct under Pauli exclusion.

For more details and some easy visual analogies, you can read more in the addendum below.


The same general problem of how to distinguish electron pairs to meet Pauli exclusion exists in metals. In that case, the conduction electrons are free to drift over the entire macroscopic metal. The result is an even more extreme need than in benzene for the pairs to distinguish themselves firmly from each other, since for conduction electrons all of these gazillions [technical term] of pairs must find "unique places" in which they can maintain their unique identity. That's tough!

Now it turns out that these two problems of benzene electrons and metallic conduction electron pairs are related in a rather deep fashion, since a benzene ring can be interpreted (quite correctly) as a special tiny case of a one-dimensional metal. That means that if you understand what happens conceptually in metals, you are well on the way to understanding all forms of hybridization in organic molecules. So, let's take a quick look at how that works.

The situation for electron pairs in metals is most easily understood by imagining the pairs moving back and forth along the $x$ length of a very narrow, highly conductive, and unusually regular metallic wire. Next, borrow the $y$ and $z$ plane to represent the complex plane (complex phase of a wave function). The amplitude for finding an electron from the pair becomes the radius of the complex plain at each point along $x$. For stable electron pair states, these radii and their phases will exhibit a lot of continuity as you move their as along the $x$ length of the wire. They will also rotate at the same speed in the complex plane of $y$ and $z$.

In fact... if you calculate the stable or "stationary" states possible in such a configuration, you will find that they bear a striking resemblance both visually and mathematically to skip rope solutions, that is, to the ways that you can twirl a skip rope so that it maintains stable, unmoving (the "stationary" part) loops.

So let's look at that situation as a good analog solution to the quantum equations. Anyone can make one large loop, and that in fact represents the electron pair state with the lowest possible energy along the length of the wire, or the base state. Incidentally, the volume enclosed by the skip rope is proportional to the square of the radius, which as I mentioned earlier is the "amplitude" of the wave function. So... that means that if you simply look at the volume enclosed by the skip rope as it turns, you pretty much instantly know what the probability of finding an electron from the pair is. In the case of the base state, it means that the electron pair will tend to be found in the center of the wire, and almost never at the ends.

In chemistry such varying probabilities of finding electrons are usually represented as densities in ordinary space, without the benefit of the dynamics of the complex plane. And for good reason, since for anything except wires, trying to show the complex plane would result in five-dimensional spaces. Those are tough to visualize! Instead, the idea of a phase is attached to tell the relative positions of the rotating points in the unseen complex plane.

The skip rope in the one-dimensional wire makes the idea of phase a lot easier to visualized. For example, good skip-ropers can easily create double loop skip ropes. These correspond to the first excited state of the electron pair, one with a bit more energy added (you have to twirl the rope faster) and two distinct loops. Phase just means this for such a double loop: When one loop is towards you, the other is opposite of you. You can label one such loop positive and one negative, but of course it's all arbitrary, and really just a technique for roughly describing those invisible but rope-like complex plane dynamics.

Notice that the volume-equals-probability rule still applies to the double loop. That means that in the first excited state of the electron pair, the odd are good and roughly equal for finding it in the left loop or right loop, but close to zero at the ends and at the center. So, this really is a unique state for the excited electron pair, one that is quite distinct from the first single-loop or base state.

Now there's an interesting bit of mathematics, Fourier transforms, in which any very long sinusoidal shape can always be distinguished very sharply from any other sinusoidal, even if they only differ slightly in wavelength. People use this idea all the time, since in the case of radio waves (which are sinusoidal waves, mostly) it's called tuning a radio!

The uniqueness of long sinusoidal waves is relevant here because it turns out that Pauli exclusion is met equally well if two pairs of electrons are separated from each other in frequency space.

So, another way of saying all of this is that when electrons are "delocalized" either in bonds or in metals -- it really is the same idea -- they can achieve uniqueness by bumping around either in ordinary space or in a more abstract frequency space called momentum space. Both work equally well. The "extra quantum number" you speculated about in your question could be equated to this idea of additional separation in momentum space for example.

So, the benzene orbitals maintain Pauli exclusion uniqueness by finding specific stable solutions that are similar but geometrically distinct, and which are even more unique when interpreted in terms of frequencies or loop counts. Those equate to a quite different-looking, distance-like separations in the abstract space called momentum space.

So to finish up, what about that extreme case I mentioned of metals that have an electron or two for every atom in what may be a very large crystal?

That's a truly fascinating case, a sort of "benzene gone bad" scenario in which the need for Pauli exclusion uniqueness pushes the electron pairs into absolutely bizarre behaviors, ones very important to modern (and ancient) technologies: Some of the electrons are forced to get very hot. Now that's weird!

What happens is easy to see with the above background. The first pair of electrons in a silver wire is very, very cold, near absolute zero, moving so slowly that it forms only a single loop in the center of the wire. To maintain Pauli exclusion -- the momentum space version of it to be precise -- the next pair must look like a skip rope with two loops, and the third three, and so forth. You get the picture.

But each such loop has more energy, even if only by a little bit! So what happens when you have a gazillion such electron pairs all vying for uniqueness?

Someone gets pushed very, very high up in frequency, and thus also in energy, while other lurk in the deeper and colder regions of this pile of moving electrons. The result is almost like a sea in momentum space, and that is in fact what it is called: A Fermi sea, with cold electron pairs at the bottom, and some truly hot ones at the surface. Fermi seas are also referred to by the more common phrase of "conduction bands," meaning the range of energies that the electrons occupy in order to maintain Pauli uniqueness.

You see Fermi surfaces all the time, since it is only the hottest electrons in a metal that interact with photons and bounce them back at you. You are in effect looking directly at a hot Fermi surface every time you look into a silver or aluminum backed mirror.

How hot? Well, believe it or not, X-ray hot. A typical piece of ordinary shiny metal, say a fork or spoon, would kill you from X-ray exposure if all of its hot Fermi surface and near-surface electrons could suddenly fall back into a "ground state" of just one loop. They are tamed and prevented from doing so by all those other cooler electrons that refuse to let them fall to the bottom of the Fermi sea.

And with that I'll close. I realize the parts about metals are a bit remote from the original chemistry of benzene rings. But then again, as you get into longer and longer carbon chains such as the ones that carry off electrons from chlorophyll molecules, you may find that recognizing the "almost metallic" nature of multiple pairs and the way they interact in momentum space can be a good shorthand for understanding how some pretty complicated forms of large-molecule dynamics play out.

$\endgroup$
10
  • $\begingroup$ I knew it was you behind the long answer! I knew it! :D Excellent explanation, removes all lingering doubts from my mind. I guess my main confusion was that I assumed the electrons to all have the same region in their MOs, which was explained in the above answer--and that "region" includes momentum space. $\endgroup$ Commented May 9, 2012 at 2:52
  • $\begingroup$ Manishearth, yep, I am the master of overkill! I didn't realize it was you asking until I got to the very end. If I'd been paying attention to that, I might have kept my answer down to "Pauli exclusion works the same in both ordinary space and momentum space, so clear separation in momentum space allows orbitals to overlap significantly in real space." (To be accurate, it's the unique shapes in all six dimensions that tell the full story.) One the other hand, it really is a delightful topic about which to write, so I appreciate the excuse to talk about ways of visualizing such things. $\endgroup$ Commented May 9, 2012 at 22:25
  • 2
    $\begingroup$ This answer is mostly nonsensical. The six pi electrons occupy three different molecular orbitals of very different shapes. Momentum space has nothing to do with this. In fact, momentum space representations of molecular orbitals are effectively useless because of the distinct lack of spatial periodicity in molecular systems. $\endgroup$ Commented May 12, 2012 at 7:34
  • 1
    $\begingroup$ To be clear: By choice, I mostly answer questions from folks who are unlikely to understand how probability waves and Hilbert spaces work, but who clearly want to learn. If you hit them with barrages of new terms for which they have no mental framework, most will very quickly learn to stop asking. That's bad, since the world needs to get a lot more young people interested science. As for whether a couple of your comments may have rated as more personal than professional, the only answer that counts there is your assessment of yourself. I'm good, and have had far more smiles than frowns today. $\endgroup$ Commented May 13, 2012 at 4:12
  • 1
    $\begingroup$ I must agree with AcidFlask. This has elements of truth, but it is far from the simplest answer: electrons of opposite spin may exist in the same region of space because they have different quantum numbers. Also, MO $\endgroup$
    – Eric Brown
    Commented May 31, 2013 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.