Hypervalency — where it appears
Named by 17 essays across 2 fields — each of them below, with the objects they name alongside it.
Hypervalency without d orbitals
Sulfur hexafluoride is not d²sp³ hybridised. The d orbitals are far too high in energy to contribute meaningfully, the bonding is three-centre four-electron, and the textbook account has been known to be wrong for fifty years.
Where two-centre bonding stops
A bond between two atoms is a special case, not the general one. Rings, clusters and metals are held together by orbitals spread over many centres, and the arithmetic that describes them is the arithmetic already used for benzene.
Three-centre bonding, computed
Three orbitals in a line and four electrons: a bonding level, a level with exactly zero amplitude on the central atom, and an empty antibonding one. The middle atom never exceeds an octet, and the ligands carry the charge — which is why every molecule that needs this arrangement has electronegative ligands.
Six bonds and four orbitals
The six fluorine σ functions of sulfur hexafluoride span a₁g ⊕ eg ⊕ t₁u. Sulfur's 3s and 3p supply a₁g and t₁u and nothing else, so four bonding orbitals hold twelve electrons across six bonds — a bond order of two thirds, computed from characters with no energy anywhere in it.
Hypervalency is about the ligands
The three-centre four-electron bond is offered as the reason sulfur hexafluoride needs no d orbitals. Read it forwards instead of backwards and it is a requirement rather than a permission — the arrangement puts half an electron onto each ligand before any electronegativity difference is applied, which is why the hypervalent compounds are fluorides and SH₆ is not a compound.
Hypervalency does not stop at three centres
The three-centre four-electron bond is written up everywhere as an arrangement peculiar to hypervalent molecules. It is the first member of a family — five centres and six electrons, seven and eight — and the family predicts alternating bond strengths that the polyiodide crystal structures have.
Four is all that s and p can match
Reduce the ligand σ set of ten molecules in each one's own point group and ask how many of its components transform as one of the central atom's four valence orbitals. The answer is never more than four — not by arrangement, in every geometry from linear to octahedral — and what is left over is n + L − 4, with exactly twice that many electrons in excess of an octet.
The count is the population
The census counted how many ligand combinations have no partner on the central atom and called the count n + L − 4. A σ-only model built from each molecule's own coordinates says what that count is worth: with no electronegativity difference anywhere, the mean charge on a ligand is minus the orphan count divided by the ligand count, exactly, in all ten cases. And the prediction the census made — that the charge grows with the orphan count — is refused by the divisor.
Two models that disagree about the shape
The identity says how much charge a hypervalent molecule's ligands must share and nothing about how. Working out which arrangement shares it most evenly puts the σ model and the repulsion model on one axis for the first time: the even sharer is the pentagonal plane, which is the arrangement the repulsion likes least — and along the interchange chemistry actually uses, one model sees 0.15 per cent of a change and the other sees 54.
The square that wastes an orbital
The orphan count that prices hypervalency was treated as a property of a molecule's composition — ligands plus lone pairs minus four. Run on twenty-six arrangements of the same ten molecules it is right for twenty-three and wrong for three, and all three are flat. A planar arrangement gives a main-group centre three usable orbitals rather than four, so the count is a property of the shape.
Expensive is not the same as unadopted
A counting formula right for twenty-three arrangements and wrong for three invites a reading: the failures are the arrangements nothing adopts, so the formula is reliable because chemistry stays away from where it breaks. Put the repulsion energy on the same axis and the reading fails — the most expensive arrangement in the census is one the formula gets right.
A count that changes at one point
Where does the orphan count step along a distortion, relative to where the energy's minimum sits? A rule that depends on an exact symmetry may have no answer for a real molecule. On the path from a tetrahedron to a square plane the count is the same at every angle up to 89.99° and changes only at 90° exactly — which is the energy's maximum, not its minimum, and a single geometry out of a continuum.
The group nobody wrote a table for
The orphan count is predicted to be constant along the Bailar twist, because the twist keeps D3 the whole way. The premise is exactly right — at every angle strictly between the prism and the octahedron the symmetry finder assembles six operations that hold to four parts in 10¹⁶. The conclusion cannot be tested here, because a count is a reduction, and the calculation's nineteen character tables did not include D3.
The count the table was hiding
The interior of the Bailar twist looks uncountable — exactly D3 at every angle, and D3 is a point group whose character table is rarely written out. Writing it takes nine lines and settles the prediction made for the path: the orphan count is two at every geometry on the path that can be answered, in three different groups, out of three different decompositions.
The leftover changes sides
A main-group centre brings four valence orbitals against six ligand combinations, so two are orphaned. A transition metal brings nine, so the arithmetic inverts and three metal orbitals are left instead — three at every geometry of the Bailar twist, out of decompositions that share no species. Run past the whole arrangement census, exactly one arrangement orphans anything at a metal, and it needs an f orbital to fix.
The gap found on purpose
A twist between two coordination arrangements passes through a point group with no character table among the twenty in use, and that was found by accident. Running every path between the arrangements — eighteen of them, two hundred and seventy geometries — finds exactly one more gap, at a group of order ten. It also finds something the search was not looking for: a group without a table had been reported as an infinite group.
Folding the ring does not give the orbital back
Three arrangements break the orphan count n + L − 4, all flat, and the reason given was flatness: the p orbital perpendicular to the ring has no ligand combination of its species. Fold the ring into an umbrella at any angle and that orbital becomes totally symmetric — and the count stays wrong by exactly one, for rings of four to eight. The ring offers one symmetric combination to a centre with two symmetric orbitals. Writing C₅ᵥ to see it also counts the pentagonal pyramid at last, and the formula holds there.
Named alongside it
The objects these essays reach for when they reach for this one.
Irreducible representationsPoint groupThree-centre bondingNon-bonding orbitalsReduction formulaElectron countModel limitSymmetry operationLone pairOctet ruleCharacter tableDegeneracy