Fewer bands than electrons

A photoelectron spectrum of a small molecule shows a handful of bands in the valence region, and the obvious question about it is how many there should be. Three plausible answers are available and two of them are wrong.

One per valence electron. No: electrons are removed in pairs from doubly occupied orbitals, and the two members of a pair are indistinguishable.

One per bond or lone pair. Also no, and this is the answer worth taking seriously because it is the one a Lewis structure suggests. Methane has four equivalent bonds and shows two bands.

One per occupied symmetry species. This is the right count, and it is computable from the geometry alone — by the reduction formula character tables and reduction builds, running on a group generated from the molecule’s own coordinates.

The count, computed #

The valence basis of a molecule like these is the central atom’s s and p functions plus one s function per hydrogen. Reducing each of those sets in the molecule’s own point group gives what they span, and the union is what the occupied orbitals can be built from.

Methane, Td. Carbon 2s spans A₁ and 2p spans T₂; the four hydrogen 1s functions span A₁ ⊕ T₂. So the valence basis is 2A₁ ⊕ 2T₂, eight functions, of which the bonding half — four orbitals in the species A₁ and T₂ — is occupied. Two species, two bands.

Ammonia, C3v. Nitrogen 2s spans A₁ and 2p spans A₁ ⊕ E; the three hydrogen 1s functions span A₁ ⊕ E. The occupied set is A₁, E and A₁ again — four orbitals in three species. Three bands.

Water, C2v. Oxygen 2s spans A₁ and 2p spans A₁ ⊕ B₁ ⊕ B₂; the two hydrogen 1s functions span A₁ ⊕ B₂. The four occupied orbitals fall into A₁, B₁, B₂ and A₁ — every one of them non-degenerate. Four bands.

Eight valence electrons in each case, four occupied orbitals in each case, and two, three and four bands. The difference is entirely degeneracy: methane’s T₂ is three orbitals at one energy, ammonia’s E is two, and water has none at all.

Which four of the eight basis combinations are the occupied ones is a separate question, and it is worth being explicit that the reduction does not answer it. The reduction says the valence basis spans 2A₁ ⊕ 2T₂; it does not say that the lower member of each pair is filled. That ordering comes from the bonding argument — a combination in phase with the central atom’s function lies below one out of phase with it — and it is an energy statement rather than a symmetry one.

What symmetry does guarantee is that the two A₁ combinations mix with each other and with nothing else, and likewise the two T₂ sets. So the occupied orbitals are some combination within each species, the count of species is fixed regardless, and the band count survives whatever the mixing turns out to be. That is why the prediction is robust: it uses only the part of the calculation that does not depend on the energies.

Ammonia’s three hydrogens in C₃ᵥ span a₁ ⊕ e: one fewer hydrogen than methane, a smaller group, and a degeneracy of two instead of three. Eight valence electrons, three bands — the count of bands follows the group and the count of electrons does not.

What the spectra show #

The measured valence ionisation energies are, in electronvolts:

Every band’s label is a species the computed valence basis spans, and that is checked: a band labelled with a species the basis does not contain would be refused rather than drawn.

The intensities are not computed here and neither are the energies. What is computed is the count and the labels, and that is a stronger claim than it sounds — it says how many distinguishable bands a molecule may show, and the number is small and specific.

The intensity pattern is the same count again #

Each band carries as many electrons as its species has dimensions, doubled, so the degeneracies predict a pattern of relative intensities with no cross-section anywhere in the argument.

Those are integer ratios and they are checkable by eye on a published spectrum. Methane’s three-to-one is the assignment’s original evidence: the band at $12.7$ eV is visibly the larger of the two, and no picture in which the four bonds are equivalent and independent produces a three-to-one anything.

The prediction is not exact in practice, because photoionisation cross-sections depend on the orbital’s character — an orbital with more s character on the heavy atom behaves differently from one with more p — and on the photon energy. What is exact is the electron count behind it, and the departures are second-order corrections to a leading term that is a pure integer.

What a fifth band would mean #

A count that says how many at most is a falsifiable statement, and it is worth spelling out what would falsify it.

If methane’s valence region showed three well-separated bands rather than two, the molecule would not be tetrahedral. There is no arrangement of parameters, no adjustment of energies and no refinement of the model that gets three bands out of Td with eight valence electrons: the basis spans two occupied species and that is the end of it.

That is the same logic two structures, two spectra uses on vibrational spectra, where a linear XY₂ and a bent one differ in band count rather than in band position, and it is why band counting settled several structural questions decades before anybody could compute an energy. A count is a discrete statement, and discrete statements survive experimental error.

The converse does not hold, and it is worth saying so. Observing two bands does not prove the molecule is tetrahedral, since bands can overlap, lie outside the photon energy available, or be too weak to see. The count bounds the number of distinguishable bands from above, and an experiment reporting fewer has an ordinary explanation.

The description that predicts the wrong number #

The four C–H bonds of methane are equivalent. Four equivalent localised bond orbitals can be constructed, and this collection constructs them explicitly: the localisation transformation applies the unitary transformation that takes the canonical a₁ and t₂ set to four equivalent localised bonds and demonstrates that the total density is unchanged to $10^{-16}$.

So there are two descriptions of the same wavefunction, related by a transformation that changes no observable. One of them has four identical orbitals and the other has two species. Which one does the spectrum count?

The spectrum counts the canonical set, and the reason is not that the localised description is wrong. It is that ionisation energy is not invariant under the transformation. Removing an electron leaves an ion, the ion has a definite energy, and the energies of the states reached by removing an electron from each of the four localised orbitals are not equal — they mix, and the eigenstates of the ion are the canonical combinations.

That is worth stating carefully because it is the sharpest available example of the site’s recurring theme. A change of basis leaves every observable unchanged; an ionisation energy is an observable; and the resolution is that the localised orbitals are not eigenstates of anything, so removing an electron from one is not a well-defined process producing a stationary state.

Methane’s canonical orbitals can be transformed to four equivalent localised bonds with the density computed both ways and found identical to 10⁻¹⁶. The transformation is exact and it is not a change of physics — which is why the four bonds a chemist draws and the two bands a spectrometer sees are both correct and are answers to different questions.

The total electron density along a line through the molecule is the same curve computed from the canonical set or from the localised one, to the last bit a double holds. That is the demonstration that the count of bands is a statement about the description and the count of electrons a statement about the molecule.

Why symmetry compresses a spectrum #

The pattern across the three molecules can be stated as a rule with a reason attached.

A group with degenerate species compresses a spectrum, because a degenerate set of orbitals is one energy and therefore one band. The more symmetric the molecule, the more degeneracy is available, and the fewer bands its spectrum can show from a fixed number of electrons.

That is a loss of information, and it runs against the intuition that a symmetric molecule is easier to study. A symmetric molecule is easier to predict and harder to measure things about: methane’s two bands are consistent with a wide range of pictures of its bonding, while water’s four bands, in four different symmetry species, pin down considerably more.

The same trade appears one field over. A spectrum that changes when only a mass does finds methane’s nine vibrations giving four frequencies and CH₂D₂’s giving nine, from the identical force field, for exactly this reason — and concludes that the least symmetric member of a family is the most informative and the hardest to assign.

C₂ᵥ has every species one-dimensional, which is what it means for a group to have no degeneracies available — and therefore why water shows one band per occupied orbital and methane does not. The number of bands is bounded above by the number of orbitals and below by the number of species, and both bounds come from the table.

What the count does not settle #

Three things the reduction has nothing to say about, each of which the spectrum does report.

Where the bands are. The energies above are measured. Computing them requires the total energies of a molecule and of its ion, which needs the electron-electron repulsion not evaluated here — the gap the smallest many-electron calculation works around by using a model small enough to solve exactly instead.

Whether Koopmans’ identification holds. The habit of reading an ionisation energy as minus an orbital energy is an approximation resting on a cancellation between two neglected errors of about an electronvolt each — the relaxation of the remaining electrons, which lowers the ion’s energy, and the correlation energy, which is larger in the molecule than in the ion. What a photoelectron spectrum measures works through what that cancellation is worth.

How intense each band is. The three-to-one ratio in methane’s spectrum follows from the degeneracies and is therefore a symmetry consequence, but the absolute cross-sections are not, and nothing here computes one.

What the vibrational structure on each band says. A band is not a line: removing an electron from a bonding orbital leaves an ion whose equilibrium geometry differs from the molecule’s, and the resulting vibrational progression is often the most informative part of the spectrum. Reading it is a Franck–Condon argument about overlaps between vibrational wavefunctions of two different electronic states, and neither state is computed here.

What the count does settle is a question that no amount of measurement answers on its own: how many bands there could be. A spectrum showing five valence bands for methane would mean the molecule is not tetrahedral, and that inference needs no energy, no intensity and no model of the bonding.

Water’s four bands, and the pair that is not one #

Water is the case where the count does the most work, and its consequence has an essay of its own.

Four bands in four different species means the four occupied valence orbitals are all inequivalent. Two of them — the b₁ at $12.6$ eV and the a₁ at $14.8$ — are the ones a Lewis structure calls “the two lone pairs”, and they differ in energy by more than two electronvolts and belong to different symmetry species.

So the two lone pairs of water are not a pair in any sense the spectrum recognises. Water’s lone pairs are not a pair makes that argument in full; what is added here is that the number four was fixed before any of the energies were looked at, by a reduction that takes a few lines.

Counting bands as a way of deciding a shape #

A band count that falls as the symmetry rises is a relation with two ends, and the useful direction is the one the essay does not run: given a count, decide the symmetry.

That is a real analytical procedure and it works for the same reason the essay’s series works. A molecule of higher symmetry has more degeneracies, degenerate orbitals ionise at one energy, and the spectrum has fewer bands than the electron count implies. Lower the symmetry and the degeneracies lift, each split producing a new band, and the count rises.

So two candidate structures for one formula, differing in point group, predict different band counts — and counting is easier and more robust than measuring positions, because it survives the calibration, the resolution and the assignment being imperfect.

The essay’s own three molecules are the cleanest demonstration of the mechanism. Eight valence electrons throughout; methane’s arrangement is the most symmetric and gives two bands; ammonia’s is less so and gives three; water’s is least and gives four. Nothing changed but the shape.

Two cautions belong with the procedure and both are the ordinary ones for a count.

Bands can overlap. Two distinct ionisations at similar energies appear as one broad feature, so an observed count is a lower bound on the true one, and a structure predicted to give five bands is not refuted by four being resolved.

And a count is not a unique inverse. Several point groups can give the same number of bands for the same electron count, so counting narrows the candidates rather than selecting one. It is at its strongest when the candidates are two — which is the situation an isomer question usually presents.

That combination is what makes the technique useful rather than decisive. It answers which of these two more often than which of all possible, and it does so with an integer, which is the one kind of spectroscopic evidence that does not depend on how well the instrument was calibrated.

There is a second count in the same spectrum that is worth separating from the first, because the two are often conflated. The number of bands counts distinct ionisation energies; the intensity of each band counts electrons. A degenerate level ionises at one energy and contributes proportionally more intensity, so methane’s two bands are not of equal height — the one from its threefold set carries three times the electrons of the other, and the spectrum shows it.

Reading both together recovers what the band count alone discards. Two bands in the ratio three to one is a considerably stronger statement about a structure than two bands, because it says not only that there are two distinct energies but that one of them is threefold degenerate — which is a symmetry statement with an integer in it, and there are not many point groups that can supply a threefold degeneracy.

So the procedure is at its strongest when the intensities are used as well as the positions, and the intensities are the half a band count throws away.

Who found it, and when #

Photoelectron spectroscopy of molecules dates from the early 1960s, developed independently by David Turner in Oxford and by Kai Siegbahn’s group in Uppsala, using the helium resonance line at $21.2$ eV for the valence region. Turner’s book with Baker, Baker and Brundle (1970) collected the spectra of most small molecules and is where the assignments above come from.

The group-theoretic count is much older than the experiment — it is the same reduction Hund and Mulliken were doing in the late 1920s — and the interest of the 1960s spectra was precisely that they tested it. A molecular orbital picture predicts the number of distinguishable ionisation energies and their degeneracy ratios; a localised-bond picture predicts a different number; and until there were spectra, the argument had no umpire. Hybridisation does not explain collects what else the localised picture is asked to carry and cannot.

Methane was the decisive case, and it is worth being clear about why. The two-band spectrum is not evidence that molecular orbital theory is right and localised bonds are wrong — the two are related by an exact transformation, so no experiment can prefer one. It is evidence about which quantity is being measured: the spectrum reports the ion’s stationary states, and those correspond to the canonical orbitals. That is a subtler conclusion than the one usually drawn, and it is the one the localisation calculation supports.

Still open: one reduction behind three kinds of counting #

The argument began by asking what a photoelectron band is, went on to what the band count says about water’s lone pairs, and now sets the count itself on the group.

Three fields now count something with the same reduction — vibrational bands, electronic transitions and ionisation bands — and in each case what is counted is a number of symmetry species rather than of physical objects. That the same reduction formula answers all three is the strongest available argument that the formula is about the molecule’s symmetry rather than about any of the three experiments.