H s on methane: a₁ ⊕ t₂

One of the figures on representations: Character tables generated from a molecule's own operations, bases reduced in them, and what a descent in symmetry does to a level.

16 essays draw this figure, each at the values its own argument needs rather than at the setting shown above. What each one uses it to show is below, in the words of its own caption.

Hybridisation does not explain

The reduction, with the arithmetic on the page. The row marked χ is the count of hydrogens each class of operation leaves in place, and the rows beneath it are the two representations that reproduce it. Nothing on this figure is an energy.

Carbon’s three 2p functions on the same molecule, reduced the same way. They span t₂ — one three-dimensional species and nothing else — so the three of them cannot be split by anything the molecule’s symmetry can do.

The two hydrogen 1s functions on water, reduced in C₂ᵥ. They span a₁ ⊕ b₂ — one symmetric under the twofold rotation and one antisymmetric — so two equivalent bonds again give two ionisation energies rather than one.

Why water is bent

The two hydrogen orbitals reduced. Both stay put under the identity and under reflection in the molecular plane; both are exchanged by the twofold rotation and by the perpendicular mirror. The character is 2, 0, 0, 2, and it reduces to a₁ ⊕ b₂ — one symmetric combination and one antisymmetric one, at different energies.

Character tables and reduction

Ammonia’s three hydrogen 1s functions reduced in C₃ᵥ, which is the same arithmetic on a smaller group. Three unmoved hydrogens under the identity, one under each threefold rotation, one under each mirror — and the reduction formula turns that tally into a₁ ⊕ e. Two species again, but with degeneracies one and two rather than one and three, because C₃ᵥ has no three-dimensional representation to offer.

The same molecule with a bigger basis on it: all three Cartesian displacements of each hydrogen rather than one s function apiece, spanning a₁ ⊕ e ⊕ t₁ ⊕ 2t₂. Twelve functions, five species, and the multiplicities still come out whole — which is the check, and it is a stronger one here because there are more ways for it to fail.

Methane’s four hydrogen 1s orbitals reduced, with the arithmetic on the page. The row marked χ is the count of hydrogens each class leaves in place; the rows beneath are the two representations that rebuild it. The answer is a₁ ⊕ t₂.

Site symmetry, and what it constrains

Sulfur hexafluoride’s six fluorines are a single orbit — the forty-eight operations of Oh carry any one onto any other — so all six sigma orbitals enter one reduction and span a₁g ⊕ eg ⊕ t₁u. Three species from one orbit.

A spectrum counts environments, not atoms

Sulfur hexafluoride’s six fluorine 1s functions reduced in its own group. They span a₁g ⊕ eg ⊕ t₁u — three species from six equivalent atoms, and the six sit in one orbit. The number of species is what a spectrum can resolve; the number of atoms is not.

The three-fluorine case for contrast: one orbit of three atoms spanning a₁′ ⊕ e′, so two species from three equivalent atoms. Reducing the same basis on a molecule with the same formula and a different shape gives a different answer, which is what makes the count structural evidence.

Degeneracy is a group theorem

Methane’s four hydrogen 1s functions reduced in Td: a₁ ⊕ t₂. The reduction is the sum over classes of the character times the class size, divided by the group order, and every multiplicity it returns must be a whole number — a requirement that is checked. Four functions have gone in and a one and a three have come out, which is the prediction that methane’s photoelectron spectrum shows two bands.

An infinite group, worked in a finite one

What is done instead. Carbon dioxide’s vibrations computed in D2h — the largest finite subgroup that keeps the molecular axis — with the name each species would carry in the real group beside it. The count must come out at 3N−5 rather than 3N−6, because rotation about the molecular axis moves no atom and is not a motion of the molecule at all.

The unsymmetrical case. Hydrogen cyanide is C∞v — no centre of inversion, no horizontal mirror — and is worked in C2v. Its four vibrations give three distinct frequencies, and every one of them is active in both experiments, because a group without a centre has representations that carry both linear and quadratic functions. Comparing this with carbon dioxide is two structures, two spectra’s argument in its cleanest form.

Carbonyl sulfide, worked the same way. Linear and unsymmetrical, so C∞v in C₂ᵥ: four vibrations from three atoms, three distinct frequencies, and every species carrying both a linear and a quadratic function because there is no centre of inversion to separate them. Compare the carbon dioxide figure above, where the same count splits into disjoint halves.

Six bonds and four orbitals

The six fluorine σ functions of SF₆ reduced in Oh. The character under each operation is the number of fluorines left in place by it, and the reduction gives a₁g ⊕ eg ⊕ t₁u — one non-degenerate combination, one doubly degenerate pair and one triply degenerate set.

The five fluorine σ functions of PF₅ reduced in D3h. Two copies of a₁′ appear, one for the axial pair and one for the equatorial set, and only one of them finds a partner in the phosphorus 3s.

XeF₄’s four fluorine σ functions in D4h: a₁g ⊕ b₁g ⊕ eu. The b₁g combination is the unmatched one, and it is the same shape as a dx²−y² orbital — which is what makes the d-orbital story so tempting here.

What a photoelectron spectrum measures

Methane’s four hydrogen 1s functions reduced in its own group. They span a₁ ⊕ t₂ — two species, in a one-to-three ratio — so the four bonding electrons cannot all be at one energy and a single band is symmetry-forbidden. This is a prediction from four coordinates, made before any energy is mentioned.

The other half of the same basis, for comparison: carbon’s three 2p functions in the same group span t₂ and nothing else — one species, threefold. So the t₂ channel has a carbon contribution and the a₁ channel does not, which is why the two bands are not two halves of one thing. The reduction is a count of what each operation leaves in place, and the multiplicities it returns have to be whole numbers; that requirement is the check, and a mistyped character anywhere in the table produces a fraction here.

And water’s, which is the case where the counting says something a picture denies. Oxygen’s three 2p functions in C₂ᵥ reduce to a₁ ⊕ b₁ ⊕ b₂ — three species, no two of them alike, none of them degenerate, because the group holds no two-dimensional representation at all. So water’s spectrum cannot show a degenerate valence band and ammonia’s can, and that difference was fixed before either was measured.

Water's lone pairs are not a pair

Oxygen’s 2s function reduced in the same group, for comparison: it spans a₁ and nothing else, because a spherical function on the unique atom is fixed by every operation there is. So the a₁ channel has two oxygen functions in it and the b₁ channel has one, which is why two of the four bands are a₁ and why the b₁ band is the one with no bonding character to lose.

Oxygen’s three 2p functions reduced in water’s group. They come out a₁ ⊕ b₁ ⊕ b₂ — three different species, no two of them the same — which is the crux: the p orbital perpendicular to the molecular plane is in a species of its own and can mix with nothing, while the two in the plane are in species that the hydrogens also span.

The hydrogens’ half of the story. Their two 1s functions span a₁ ⊕ b₂, so they can mix with oxygen orbitals of those two species and with nothing else. That single fact is what leaves the oxygen’s b₁ orbital alone and unmixed, and it is why one of water’s two “lone pairs” is pure and the other is not.

Hypervalency is about the ligands

The reduction that makes the count exact: six fluorine σ functions in Oh span a₁g ⊕ eg ⊕ t₁u. The eg pair finds nothing of the right symmetry on the sulfur, so it is non-bonding — two orbitals holding four electrons entirely on the fluorines, which is the same four electrons the three-centre picture puts on the ends.

The projector is unique, the basis is not

The reduction that says how many. Four functions, twenty-four operations, characters summed class by class and divided by the group order — and out come two whole numbers. Nothing here says what the combinations are.

Four is all that s and p can match

The number of ligand σ combinations that transform as one of the central atom’s s and p orbitals, against the number of ligands and lone pairs. The count follows the diagonal and then stops at four, because there are four orbitals. The vertical bars are what is left over: pairs with nowhere on the central atom to go.

The full census: what each molecule’s σ set spans, how many combinations are matched, how many of the four orbitals are held by lone pairs, what is left over, and the electron count. The last two columns are related by a factor of two in every row.

The reduction for the octahedral case, done from the operations: six fluorine σ functions spanning A₁g ⊕ Eg ⊕ T₁u, with the character under each class computed by counting which fluorines each operation leaves alone. The Eg pair is the part with nowhere to go.

The count is the population

The ligand σ set of chlorine trifluoride reduced in its own point group. Three combinations, two of which find a partner among the central atom’s four orbitals and one of which does not — and the one that does not is where two of the ten electrons go. Everything below is a measurement of where those two electrons actually sit.

The mean charge on a ligand against the orphan count divided by the ligand count, with no electronegativity in the model. Every molecule sits on the diagonal, to twelve decimal places, with nothing fitted.

Every molecule, with the charge the model puts on each ligand set. Chlorine trifluoride splits one and two; phosphorus pentafluoride splits three and two; the octahedral and square-planar species have a single set. In every case the mean over all the ligands is the identity above, whatever the split.

Two models that disagree about the shape

Where the orphan count comes from for the molecule in question: the five fluorine σ functions reduced in the molecule’s own group, against the four orbitals the phosphorus has to match them with. The count is what the identity divides; and the count itself moves with the arrangement.

A label that prices nothing

The reduction the whole census is built on, run on a smaller basis for the molecule that raised the question. Every count comes back a whole number by construction, and a species appearing twice is visible in the arithmetic long before it is a difficulty.