One scale, from two centres to a cage

One question has been approached from several sides. Diborane’s bridge is one pair over three centres; the methyllithium tetramer’s pairs are one pair over four, and no mixing of them reduces it; a closo borane’s skeleton needs one pair more than it has corners. The tetramer essay ended by naming what would unify them: a single treatment producing all three, with the number of centres as an output rather than as a label.

The tool to do that is the localisation itself, which measures how many centres a pair sits on rather than assuming one. This essay runs it up the whole scale, and finds two things: the output is two numbers rather than one, and at the top of the scale it stops being an output at all. Neither is a failure of the idea — both are what happens when a label is replaced by a measurement.

Two numbers, and where they come apart #

A localised orbital has a population on each atom, and there are two obvious ways to say how many atoms it is on.

Count them. How many atoms carry more than a threshold’s worth of amplitude — five per cent, here.

Weight them. The participation number, one over the sum of the squared populations, which is a standard measure of how many things a distribution is spread over and which is two for a pair shared equally between two atoms.

They agree when the sharing is even and separate when it is not, and the separation is the interesting quantity.

Two separate ethenes are the control and give exactly two by both measures, which is the check that the localisation can find a two-centre bond when there is one. Benzene gives 2.842 on four atoms, which is the classical statement that it cannot be localised into three double bonds — the numerical form of what a hexagon’s shape is the frame’s doing says about the same molecule from the geometry. The hypervalent chains give participation numbers rising from 1.684 to 2.149 as they lengthen, on two to two and a half atoms — a pair that is less than two-centred, because in a chain one pair short of a full complement each pair leans on one bond and reaches weakly to the next.

What the scale actually orders #

Set out in order of how many atoms one pair is asked to hold, the participation numbers do not increase monotonically, and the reason is worth having.

The three-centre chain’s pairs have a participation number of 1.684 — the smallest on the scale — and it has fewer pairs than bonds. The tetramer’s have 2.540 and the six-vertex cage’s 2.200. Benzene, which nobody calls electron-deficient, has the largest at 2.842.

So the participation number is not a measure of electron deficiency. It is a measure of how spread out one pair is, and a delocalised system with enough electrons is spread out too.

What does track deficiency is the gap between the two counts. Benzene’s pairs sit on four atoms with a participation number of 2.842, a gap of 1.16, and are evenly spread over three of them plus a tail. The tetramer’s sit on four with 2.540, a gap of 1.46, and are one large share and three small. The distinction is between a pair spread over many atoms and a pair belonging to one atom and lent to several, which is what an electron-deficient bond is, and is why a cage needs one pair more than it has corners.

The cage that breaks its own symmetry #

The six-vertex borane skeleton has four occupied orbitals for six vertices, and every vertex is equivalent to every other. The localisation does not produce four equivalent orbitals.

It produces two that are ordinary two-centre bonds, participation number exactly 2.000, and two spread over five atoms with participation 2.400. That answer is reached from all twenty-four starting rotations tried, and the functional it reaches is exactly $11/6$.

That is not a defect. The occupied space is symmetric; a basis for it need not be, and the criterion being maximised has no reason to prefer a symmetric basis. What is worth noticing is that the unsymmetric answer is the unique maximum: it is not one of a set of equivalent broken-symmetry solutions that a symmetric average would restore, because twenty-four random starts all land on the same set of participation numbers.

The density does not move, and that is what makes any of this legitimate #

Every mixing here is orthogonal, so the density is untouched — and it is checked rather than assumed. The charge on every atom before and after the localisation agrees to $10^{-9}$ for every system on the scale, which is the property that makes a localised description a description rather than a different calculation.

That check is worth having for a reason beyond diligence. A localisation sweep applies rotations to pairs of orbitals, and writing one of the two new rows with the wrong original row is not a rotation at all — it is a mistake that leaves the orbitals looking entirely reasonable and moves the charge. It is an easy slip to make, and a density check is what catches it.

The system that has no answer #

The twelve-vertex cage is where the measurement stops working, and it stops in a way that is worth separating from failure.

Twenty-four starts reach ten distinct maxima. Their functionals span 0.0236 — from 3.7905 to 3.8141 — against a sweep that converges to $10^{-13}$. So these are not one maximum computed to different precision. They are different maxima, and the sweep lands on whichever one its starting point drains into.

The honest statement of that result has three parts, and only the first is usually made.

The answer is not unique. Ten distinct sets of participation numbers, and there is no reason to think ten is all of them.

The best found is a lower bound. 3.814131 is the largest functional twenty-four starts reached. Nothing here shows it is the global maximum, and a hundred starts might find a better one.

And twenty-four is a sample. A run that had used one start — which is what a localisation ordinarily does — would have reported one of these ten and would have looked exactly like the seven systems that do have an answer.

That last point is the practical one. The difference between a system with one localisation and a system with ten is invisible unless it is looked for, because a single sweep converges cleanly in both cases and reports a number with as many digits as anyone wants.

Why the small ones are unique and the large one is not #

The mechanism is a counting one. A localisation maximises over the orthogonal mixings of the occupied set, which is a space of $\binom{p}{2}$ dimensions for $p$ pairs: one for two pairs, six for four, twenty-eight for eight. The number of local maxima a smooth function on such a space has grows quickly with its dimension, and eight pairs is where it stops being small.

So the boundary is not chemical, and it is not a boundary of the method either — a bigger cage would be no harder to diagonalise and no easier to localise.

That distinction is worth having because the two failure modes look alike from outside. A calculation that cannot be done reports an error; a maximisation with many maxima reports a number. The same thing happens to a force-field search, and it happens to an unconstrained methane fit, which walks along a flat direction to a constant of −97,596 millidynes per ångström and reproduces every observed frequency on the way. It is not that a twelve-vertex cage is more delocalised than a six-vertex one — its participation numbers are smaller, 2.106 against 2.200. It is that the search space is larger, and the answer that a localisation gives is the answer to a maximisation rather than a property of the molecule.

That is the same shape of difficulty as a force field with more constants than a spectrum can determine, with one difference: there the family of answers all reproduce the data exactly, and here they do not — one of these maxima is the largest and the others are worse. The trouble is not that the criterion fails to choose; it is that finding what it chose is a search that can fail.

What the scale is for #

The tetramer essay asked for a single treatment producing diborane’s three centres, the tetramer’s four and a cage’s delocalised skeleton, with the count as an output. There is one now, and the thing it produces is not the tidy series that was hoped for.

The three systems do not lie on a monotone scale by either measure. By the count of atoms they run three, four, three; by participation number 1.68, 2.54, 2.11. What separates them is not how many atoms one pair touches but how unevenly it touches them, which is a shape rather than a count, and that quantity — the gap between the two counts — runs 0.32, 1.46, 1.02, which does order them in the way the chemistry suggests.

So the unification is real and its axis is not the one the label suggests. “Three-centre bond” and “four-centre bond” are counts of atoms; what makes them a family is a distribution rather than a count, and the family they belong to includes benzene, which nobody calls electron-deficient and which has the largest participation number on the scale.

The objects a localisation starts from are the canonical orbitals — one of benzene’s degenerate π pair is the standing example — and the occupied space they span is what every localised description in this essay is a different basis for. Nothing in the search changes that space; it only chooses coordinates in it.

What this cannot say #

The model is Hückel throughout. Every system here is a graph with one orbital per site and no repulsion, so the “pairs” are pairs of electrons in a one-electron model. The Pipek–Mezey criterion applied to a real calculation’s orbitals would give different numbers and the same structure of difficulty.

One criterion. When a second one is tried, Boys and Pipek–Mezey agree to three or four figures on every system here, because the disagreements they are known for are between σ and π on the same atom, and a model with one orbital per site cannot have them. So the multiplicity found here is a property of one functional’s landscape.

The threshold is arbitrary. Five per cent is where “carries amplitude” is drawn, and a different threshold moves the atom counts by one here and there. The participation number has no threshold in it, which is one reason for reporting both.

The threshold and the participation number are two summaries of one distribution. Neither is the distribution. A pair with populations 0.7, 0.1, 0.1, 0.1 and a pair with 0.55, 0.35, 0.05, 0.05 have participation numbers of 1.92 and 2.13 and both sit on four atoms, and no single number distinguishes what a chemist would draw differently.

And twenty-four starts is a number chosen for cost. It is enough to distinguish one maximum from ten and is not enough to enumerate the second.

What was checked #

The control, exactly. Two separate ethenes give 2.000000000 by both measures, and their gap is zero to $10^{-9}$.

The tetramer’s four atoms and its participation number below three, and the gap between them above one whole centre.

Both kinds present. Some systems here have the two measures agreeing to within a quarter of a centre and some have them more than one apart, because a table in which every row showed a gap would be evidence of a bug rather than of chemistry.

Seven systems with a single localisation from every start, and exactly one without — named, so that the check would fail if a different system became multi-valued.

And the six-vertex cage’s functional at exactly $11/6$, with its four pairs taking exactly two distinct values, which is the check that the symmetry breaking is real and reproducible rather than a run-to-run artefact.

The gap between the two measures is a polarity #

Two numbers for the same pair — four centres and a participation of 2.54 — look like a disagreement to be reconciled. They are not: they measure different things, and the difference between them is itself a quantity worth reading.

A count of centres asks how many atoms carry a non-negligible share. A participation number is the reciprocal of the sum of the squared shares, which is an effective count: it weights each atom by how much it holds, so an atom with a small share contributes little to it. Spread a pair evenly over four atoms and both measures return four. Concentrate most of it on one and the count stays at four while the participation falls.

So the gap is a measure of inequality, and for a pair spread over one carbon and three lithiums the arithmetic inverts to give the shares.

Writing $w$ for the carbon’s share and $(1-w)/3$ for each lithium’s, the participation is $1/\bigl(w^2 + (1-w)^2/3\bigr)$, and setting that equal to 2.54 gives

$$w = 0.578, \qquad \text{each lithium} = 0.141.$$

The pair is 58 per cent on the carbon and 14 per cent on each metal.

That is not an abstract characterisation. It is methyllithium’s entire chemistry stated as a number: the compound is used as a source of a carbanion, its reactions are those of a nucleophilic methyl group, and the reason is that the bonding pair sits mostly on the carbon while remaining genuinely shared with three lithiums rather than transferred to any of them.

It also makes the two measures worth reporting together rather than choosing between. The count says how many atoms are involved and the participation says how evenly, and a system where the two agree is one whose pair is shared equally — which for a homonuclear cage it should be, and for a bond between two elements of very different electronegativity it should not.

That gives the scale a second dimension. Running along it from a two-centre bond to a cage measures how many centres a pair holds together; comparing the two measures at each point measures how fairly. A borane cage is many centres shared evenly; the methyllithium tetramer is four centres shared unevenly; and an ionic solid, at the far end, would be one centre with three spectators.

One caution belongs with the inversion. Recovering the shares from a participation number required assuming the shape of the distribution — one large share and three equal small ones — which the tetramer’s symmetry supplies and which nothing supplies in general. A participation number is one number and a distribution of shares is several, so the inversion is possible only where a symmetry has reduced the several to one. For a system with four inequivalent atoms the same participation number is consistent with many distributions, and the honest output there is the number itself rather than the shares it might have come from.

Still open: how many maxima, and what they are #

The natural open question is the enumeration declined here. Ten maxima found from twenty-four starts is a sample of an unknown population, and the population is finite and countable: the number of distinct localisations of a given occupied space is a property of the space and could be estimated by running the search until new answers stop appearing, which is the ordinary way of deciding when any search has run dry.

The nearer question is what the multiple maxima are. Ten answers with functionals within six parts in a thousand of each other are ten descriptions of one molecule that a chemist would draw differently — and whether they are related by a symmetry of the cage, or are genuinely inequivalent, is a question the coordinates would settle. If they are symmetry-related then the multiplicity is an artefact of the search and the true answer is their average; if they are not, then a twelve-vertex borane has several inequivalent localised descriptions, and choosing between them is a choice nobody has been making explicitly. That is the same shape of thing as a resonance energy measured from somewhere: a number that is quoted as a property and is a property of a decision.