The leftover changes sides

The Bailar twist, once its missing character table is written, has an orphan count of two at every answerable geometry, and the reason is not the one first given. Not the group is constant so the count is constant — the group is not constant, the path passes through three of them — but six ligand combinations minus four matched central orbitals, neither of which moves.

Then it named the case that inverts the arithmetic.

Nine valence orbitals against six σ combinations inverts the arithmetic — there are more central orbitals than ligand combinations, so the leftover is on the other side, and what the twist does to that number is not obviously constant.

It is constant, at three, and the more interesting result is what happens when the same nine orbitals are shown the whole arrangement census rather than one path.

What nine orbitals does to the bookkeeping #

A main-group centre has one s and three p, so four. A transition metal adds five d, so nine. Against six ligand σ combinations that is a surplus rather than a deficit, and the matching has three columns instead of two: how many combinations find a partner, how many do not, and how many central orbitals are left unused.

The identity is immediate and worth stating because it is not the finding. Matched plus orphaned is the ligand set’s dimension and matched plus spare is the centre’s, so

spare − orphan = 9 − n

at every arrangement, whatever the group. At six ligands that is three; at five, four; at four, five. Nothing in that is about chemistry, and it is checked here only because a matching done without multiplicity satisfies it by accident and would go unnoticed.

The content is in which of the two leftovers is zero, and that is where the metal and the main-group centre part company.

Along the twist #

Every ligand combination finds a metal partner at every geometry on the path. The orphan count is zero throughout and the spare count is three.

The species tell a different story from the number. At the prism the three spare orbitals are A₁′ ⊕ E′; through the D₃ interior they are A₁ ⊕ E; at the octahedron they are T₂g. Three decompositions with no label in common, and the last of them is not even the same shape as the others — a single three-dimensional species where the others are a one and a two.

That is the replacement rule holding in the case it was least likely to. Its statement is that the count is constant because the matched dimension is constant, and here the matched dimension is six at every geometry for a quite different reason: not because four central orbitals all find partners, but because six ligand combinations all do.

The six geometries with no answer are the same six found on the main-group twist, in the same two bands near the ends, and they are refused for the same reason — the symmetry finder’s own tolerance, not a missing character table. Adding five d orbitals to the centre changes nothing about how nearly symmetric a geometry is.

The t₂g set survives the twist #

There is a consequence here for hypervalency that is worth drawing out, because it is a statement about a rule rather than about a count.

At an octahedron the three spare orbitals are the t₂g set, and the eighteen-electron rule is built on them: nine orbitals below a gap, eighteen electrons, and the t₂g are the three that hold six of them without contributing to a metal–ligand bond.

The twist does not destroy that set. It becomes A₁ ⊕ E in the D₃ interior and A₁′ ⊕ E′ at the prism, and it stays three-dimensional at every angle. So a d⁶ centre has the same closed shell all the way along the path — the same count of non-bonding orbitals holding the same electrons — and whatever resists a Bailar twist in a real tris-chelate complex, it is not the electron count.

That is a genuinely useful negative. The twist is the standard mechanism for racemising an octahedral tris-chelate, whose two ends a repulsion census prices, its barriers are measured, and the natural first explanation for one is that something in the electronic structure is lost on the way. Nothing in the σ-matching arithmetic is.

One arrangement still refuses #

Run past the whole census — fifteen answerable arrangements from two ligands to six — the metal matches everything, at every one, with a single exception.

The planar hexagon orphans one combination. Its six σ functions span a set containing B₁u in D₆h, and no s, p or d function of a central atom transforms as B₁u in that group. The metal’s nine orbitals span A₁g ⊕ A₂u ⊕ E₁u ⊕ E₁g ⊕ E₂g, and B₁u is not among them at any multiplicity.

Reaching it would take an f orbital. That is the first case here where a valence set of s, p and d is demonstrably not enough for a symmetry match, and it arrives on the kind of arrangement that has been awkward throughout: the three arrangements a counting formula fails on are all flat, and this is the flat six.

The parallel is close enough to be worth stating exactly. At a main-group centre, a planar arrangement gives the centre three usable orbitals rather than four, because the fourth points out of the plane and finds nothing to match. At a metal, planarity does something related and not the same: the centre keeps all nine orbitals, and the ligand set acquires a component with the wrong behaviour under the horizontal mirror combined with the sixfold rotation. Flatness costs the centre an orbital in one case and costs the ligands a partner in the other, and both are consequences of the same missing third dimension.

The multiplicity is where a set-based matching would fail #

The D₃ interior deserves a paragraph on its own, because it contains the trap identified in the main-group case and it is worse here.

In D₃ the metal’s nine orbitals span A₁ twice — once from the s and once from a d — A₂ once, from a p, and E three times: once from the p pair and twice from the d shell. The six ligand combinations span A₁ once, A₂ once and E twice. Matching by label, as somebody reading two lists down a page would, finds every ligand species present among the metal’s and reports nothing left over: A₁ is there, A₂ is there, E is there.

The answer is three spare, and it is arrived at by subtracting multiplicities: two copies of A₁ on the metal against one in the ligand set leaves one spare, three copies of E against two leaves one more — which is two-dimensional — and the single A₂ matches exactly. One plus two is three. A set-based matching reports zero at every geometry in the interior and is wrong at all of them.

This is the same defect the main-group case shows running the other way — there, the orphan could not be identified by its label, because the leftover E was the second copy of a species that had also matched. Here the spare cannot be identified by its label for the identical reason. In both cases the interior of the path is where labels stop being sufficient, and in both cases the two ends are where a reader would check.

What the spare count is worth as a function of ligand count #

The census makes one more thing visible that a single path could not, and it is the arithmetic behind every electron-counting rule in the field.

Across the census the spare count runs 7, 6, 5, 4, 3 for two, three, four, five and six ligands — one fewer for each ligand added, without exception on the fourteen arrangements that orphan nothing. That is forced by the identity rather than measured: with the orphan count zero, spare is nine minus the ligand count and there is nothing else it could be.

Which is the eighteen-electron rule, before any energy. Nine valence orbitals hold eighteen electrons; n of them are used for σ bonds and 9 − n are left. A six-coordinate complex has three non-bonding orbitals holding six electrons and six bonding orbitals holding twelve, and eighteen is the sum. A four-coordinate one has five left over — which is where sixteen comes from once one of the five is pushed above the gap, and the gap that does the pushing is an energy rather than a count.

So the surplus arithmetic is not a curiosity produced by giving a centre more orbitals. It is the counting rule the applied field is built on, arriving from the direction the hypervalency argument approaches everything: reduce the ligand set, match by species, and see what has no partner. The deficit case gives hypervalency and a charge on the ligands; the surplus case gives eighteen electrons and a non-bonding set. One matching, two regimes, and which one a centre is in is decided by whether nine minus n is positive.

The rule’s exceptions are then visible in the same terms. A count fails where the matching fails, and the matching fails where a ligand combination has no partner — which is the planar hexagon here and, at a main-group centre, every flat arrangement in the census.

What was computed, and how #

The path, the geometries, the symmetry search and the arrangement census are those of the main-group case, unchanged, so this extends that measurement rather than a neighbouring one. The one addition is a d shell on the centre, spanned by the character of the l = 2 representation under each operation — $\sin(5\theta/2)/\sin(\theta/2)$ for a rotation by θ, which is what a rotation does to five functions. It is five at the identity, one at a two-fold rotation and minus one at a three-fold one, and those three numbers are the whole of what the matching needs from the metal’s side at every geometry on the twist.

The matching is done with multiplicity, which matters here in a way it did not before: a species appearing twice among the metal’s nine and once in the ligand set must leave one copy spare rather than none, and a matching that ignores multiplicity would report zero and be wrong at every geometry in the D₃ interior — the trap of reading a count off a label, where E appears twice on the metal.

Six results are checked on the twist and five on the census. The bookkeeping balances at every geometry, on both sides. The orphan count is zero along the whole path, so the leftover is the metal’s. The spare count is one number across three groups. And the species of the spare set do change — a constant count with a constant decomposition would be a much weaker finding, and the change is what says the number is not a label.

On the census: the identity $\text{spare} - \text{orphan} = 9 - n$ holds at every arrangement. Exactly one arrangement orphans anything — stated as a count rather than as none, because the answer turned out not to be none. And the one is the planar hexagon, with a one-dimensional orphan.

Where the model stops #

Symmetry only. Nothing here is an energy. A spare orbital is one no ligand σ combination can interact with by symmetry, which says it is non-bonding in a σ-only model and says nothing about where it sits or what a π interaction would do to it. The ligand-field essays are where the energies are, and they agree that t₂g is the non-bonding set at an octahedron for reasons symmetry alone does not supply.

σ only. A real ligand set has π functions, and adding them changes which species are spanned — which is exactly how a π donor or acceptor reaches the t₂g set. The spare count here is a σ count and would be smaller with π in.

Nine orbitals is a choice. A first-row transition metal’s 4s, 4p and 3d are not degenerate and not equally available, and treating them as one valence set is the same idealisation the eighteen-electron rule makes. The arithmetic is exact for the set it is given.

And the census is fifteen arrangements. Two of its seventeen are linear and are reported as such rather than worked in a finite stand-in. A larger census would probably find more B₁u-like cases, and the honest statement is that one arrangement in this set refuses rather than that flat six-coordination is the only case that can.

The generalisation #

The transferable point is about which side of a matching a surplus lands on, and what changes when it moves.

Hypervalency is entirely about a deficit: too few central orbitals, a leftover on the ligands, and a charge that has to go somewhere — which is why every hypervalent compound is a fluoride. Inverting the surplus does not invert the story, it removes it. Three spare metal orbitals carry no charge anywhere, put no requirement on what the ligands are made of, and are the reason a transition metal’s chemistry is not constrained the way a main-group centre’s is. The same arithmetic, run with a different valence count, stops being about the same thing.

The second half is about the one exception and how it was found. The census was run to confirm a result already established on one path, which is the sort of run that usually produces a table nobody reads. It produced one row that does not fit, and the row is informative — it identifies the exact circumstance in which a nine-orbital valence set is not enough, and it names the orbital that would fix it.

A confirmation run over a wider set is worth doing precisely because it can fail somewhere the original could not. One path through six-coordinate geometries could never have found the planar hexagon; the census had to include arrangements nobody adopts. The same point appears from the other direction where the most expensive arrangement in a census turns out to be one the formula gets right and the reading that failures cluster where chemistry stays away was refuted.

Who found it, and when #

The Bailar twist is from 1958, the eighteen-electron rule is older, and the reduction of a σ set in a point group is standard. The path, the census and the main-group counts are original to these essays, and so are the nine-orbital matching, the constancy of the spare count across three decompositions and the planar hexagon’s orphan.

The metal case was flagged in advance as not obviously constant, rather than assumed to be. Being explicit that a prediction is not obvious is what makes running it worth an essay rather than a paragraph.

Still open: whether an f shell removes the last orphan #

The obvious open question is the f orbital. One arrangement in the census needs one, and no l = 3 shell has been spanned in any group here — though the same closed form gives the character of any angular momentum under any rotation, so nothing new is needed to do it. Adding f to the valence set and re-running the census would say whether the planar hexagon is the only case that needs it and whether anything then has no orphan at all, which would be a statement about how large a valence set has to be before symmetry stops being an obstacle.

The nearer question is one the main-group twist raised. This path went through a group with no character table among those in use, and that was found by accident. Enumerating the one-parameter paths between the arrangements the census actually contains, and asking which of them pass through groups not on the list, would turn one gap discovered by luck into a set of them discovered on purpose — and the census above now supplies both halves of the input.