What one pair can hold together

The phrase delocalised bonding carries an implication that is worth testing: that spreading electrons over more atoms makes each bond weaker, and therefore that a delocalised system is held together less firmly per bond and perhaps in total.

Half of that is right and half is not, and the arithmetic separating them is short enough to do exactly. Where two-centre bonding stops asks when a pair has to be shared among more than two atoms; this essay asks what the sharing costs.

One pair, any ring #

Put two π electrons into a ring of $n$ carbons. They occupy the lowest level, whose eigenvector has the same coefficient $1/\sqrt{n}$ on every atom — the totally symmetric combination.

The bond order between neighbouring atoms is then $2 \times (1/\sqrt{n})(1/\sqrt{n}) = 2/n$, and there are $n$ bonds, so the total is exactly $2$.

The total does not move. One pair supplies a total bond order of two, whether it holds three atoms together or eight, and the per-link figure falls exactly as $1/n$ because that is how the total is divided.

The energy behaves the same way, and more surprisingly. A ring’s lowest level is at $\alpha + 2\beta$ for every ring size, since the closed form $E_k = \alpha + 2\beta\cos(2\pi k/n)$ gives $2\beta$ at $k = 0$ regardless of $n$. So two electrons in a ring have a π energy of $4\beta$ at $n = 3$, at $n = 8$, and in the limit.

Spreading a pair over more centres in a ring therefore costs nothing at all in total binding, and gains nothing.

A chain is different, and the difference is the ends #

Rings are the special case. Doing the same thing to an open chain gives a total bond order that rises toward two rather than sitting at it.

The chain’s lowest level is $2\beta\cos(\pi/(n+1))$, which is $1.0000$, $1.4142$, $1.6180$ and $1.8019$ for those four sizes and approaches $2$ from below. So a pair in a long chain eventually binds as strongly as a pair in a ring, and never quite reaches it.

The reason is the ends. A ring has no boundary and its lowest state is perfectly uniform; a chain’s lowest state has to fall to zero just outside each end, so it is not uniform, and the bond orders near the ends are visibly smaller than those in the middle — $0.194$ against $0.543$ for the six-atom chain.

That difference is the whole content of “surface effects”. A ring is a chain that has had its ends removed, and the cost of having ends falls as the chain lengthens because the two disturbed regions are a smaller fraction of it. The end is the hardest place to bind computes the same distinction from the other direction, finding an end site’s binding threshold converging on exactly one β while a middle site’s converges on zero.

Where the pair puts its charge #

The eigenvector supplies a second set of numbers, and they follow the same pattern with one exception worth noting.

In a ring, the pair’s charge is $2/n$ on every atom by symmetry — $0.667$ each for three atoms, $0.333$ each for six, $0.250$ each for eight. Perfectly even, because the ring’s operations permute the atoms among themselves and the lowest orbital is totally symmetric.

In a chain it is not. Allyl’s three atoms carry $0.500$, $1.000$, $0.500$; butadiene’s four carry $0.276$, $0.724$, $0.724$, $0.276$; hexatriene’s six carry $0.108$, $0.349$, $0.543$, $0.543$, $0.349$, $0.108$.

The middle of a chain carries five times what its ends do at six atoms, and the disparity grows with length. So the pair is not merely less well bound in a chain than in a ring — it is unevenly distributed as well, and the two facts are the same fact: the standing wave has to vanish beyond the ends, and a wave that is not uniform is not extracting the maximum from its links.

Coordination, counted from the levels #

There is a second exact statement in this neighbourhood, and it is the one that carries the argument out of molecules.

The mean square of the level distribution is the mean coordination number, exactly, for any structure and at any size. The reason is that $\mathrm{Tr}(A^2)/n$ counts closed walks of length two, and a closed walk of length two is a step to a neighbour and back — so the sum counts each atom’s neighbours.

Computed on a ring of sixty it gives $2.0000000$ and on a chain of sixty $1.9667$, which is $2(n-1)/n$: the chain’s two end atoms have one neighbour each rather than two, and the deficit is exactly the two missing bonds divided among sixty sites.

The second moment of a level distribution is exactly the mean coordination, because it counts closed two-step walks — an identity rather than an approximation, and the tripwire behind it is that alternating the bonds changes the moment in a way the count predicts. Every total in this essay is a sum over the same spectra that identity is checked on.

That exact statement has a consequence for structures a chain model cannot build. The spread of the levels — the band width — goes as the square root of the coordination, since the second moment is the coordination and the moment is a squared width. Meanwhile the number of bonds per atom goes as the coordination itself.

So the binding available per atom grows as $\sqrt{Z}$ while the number of bonds it is divided among grows as $Z$, and the binding per bond falls as $1/\sqrt{Z}$. A twelve-coordinate close-packed metal’s individual bonds are about $0.41$ of the strength of a two-coordinate chain’s, while its atom is held about $2.4$ times as firmly.

That is the same accounting as the ring result, with coordination in place of ring size, and it is why metals combine high coordination numbers, weak individual bonds and large total cohesion. The scaling argument is stated here rather than computed, because every structure computed here has a coordination of two; what is computed is the moment relation the scaling rests on.

What the conservation law is #

The result generalises past rings and chains, and the statement is worth having in its general form because it is exact.

For any system, the total bond order summed over all links, for a single doubly occupied orbital with coefficients $c_i$, is

$$\sum_{\langle ij\rangle} 2 c_i c_j$$

and the corresponding energy is $2\beta$ times the same sum. So the total bond order and the π energy of one orbital are the same quantity in different units, and a claim about one is a claim about the other.

For a ring’s lowest orbital that sum is exactly $1$, giving a total bond order of $2$ and an energy of $2\beta$ per electron. For a chain’s it is $\cos(\pi/(n+1))$, always less. The maximum any orbital can achieve on a graph of maximum degree two is $1$, and the ring achieves it — which is what makes a closed loop the efficient arrangement for a pair.

That is also why a molecule with an electron-deficient centre reaches for a bridged structure. Diborane’s two bridging hydrogens are held by two pairs across four links rather than by four pairs across four links, and the bridged arrangement extracts the maximum total bond order those pairs can supply. Three-centre bonding, computed works that case out in full, including the non-bonding orbital whose central coefficient is exactly zero.

Where this becomes a metal #

Taking the ring to a large size is where this counting turns into the solid-state picture, and the transition is worth watching because nothing discontinuous happens.

At one pair, the ring’s energy is $4\beta$ at every size. Fill it properly — one electron per atom, which is what a half-filled band means — and the energy per site approaches a limit instead, and the individual bond orders fall.

The bond orders in a half-filled ring fall as roughly $1/n$ times a constant while the number of bonds grows as $n$, so the total stays finite. What the ring gains from being large is not more bonding per electron; it is a denser set of levels, the quantity a solid is a molecule that did not stop follows to two thousand atoms, and that is what changes the chemistry — a large ring has excitations available at arbitrarily small energy and a small one does not.

The band limit computes that convergence and what a metal actually is uses it. The point to carry from here is that the two properties come apart: the binding is roughly size-independent and the level spacing is not.

What a bond order is not #

Three cautions, since bond order is a quantity that invites over-reading.

It is a property of a model. The numbers above come from Hückel theory, which has one parameter per bond and no repulsion — the model whose reach Hückel theory and what it gets right sets out, and whose blindness to repulsion where molecular orbital theory dissociates measures. A bond order computed from a better wavefunction differs, and bond orders computed by different definitions differ from each other — Mulliken, Wiberg, Mayer and natural bond order are not the same quantity.

It is not proportional to a bond energy. The frequency is not the bond strength makes the corresponding point for vibrational frequencies, and the reason is the same: the energy involves the whole Hamiltonian, the bond order involves one matrix element.

It says nothing about breaking one bond. A total bond order of two spread over six links does not mean the ring survives losing one link with five sixths of its bonding intact; removing a link changes the whole eigenvalue problem. What the conservation says is about the intact system.

The two-centre case is the one everybody starts from: one pair between two atoms, bond order one, and a splitting computed from an overlap integral rather than from a graph. Everything in this essay is the same arithmetic with more than two sites, and the total the pair supplies is two in every one of them.

The refusal this counting has to survive #

A conservation law that holds for every case tested proves nothing unless something can break it, so the arithmetic was asked for a case where the total is not two — and the first candidate refused to break it.

Alternating the bond strengths does nothing. Giving a six-ring alternating interactions of $1.2$ and $0.8$ leaves every bond order at $0.3333$ and the total at exactly $2.0000$. The reason is worth having: every site still has one strong neighbour and one weak one, so every row of the matrix sums to the same $2.0$, and a matrix with equal row sums has the uniform vector as an eigenvector with that sum as its eigenvalue. The distortion that a chain cannot stay even is about changes the upper levels and the gap, and leaves the lowest state exactly where it was.

A heteroatom does break it. Pyrrole’s ring with one pair in it gives bond orders of $0.561$, $0.218$, $0.165$, $0.218$, $0.561$ and a total of $1.7226$. A diagonal entry makes the sites inequivalent, the lowest orbital concentrates on the deepest one, and the sum falls.

So the honest statement is narrower than the headline and sharper for it: a pair in a system whose sites are all equivalent supplies a total bond order of two, where equivalent means equal diagonal entries rather than equal bonds. Where the sites differ the pair concentrates and the total falls below two; the uniform case is the maximum, and unequal bonds between equal sites are not enough to move it.

That is the sort of correction worth recording rather than tidying away. The alternating ring was written down as the obvious counterexample, computed, and came back at $2.0000$ — which turned a headline into a theorem with a stated condition.

One pair divides as 1/n and a filled band does not #

The conservation has an immediate consequence for how strong an individual link is, and comparing it against what a real structure does is worth doing, because the two answers differ and the difference identifies what the one-pair result leaves out.

A single pair in a ring’s lowest orbital has equal amplitude on every site, so it supplies a bond order of $2/n$ to each of the $n$ links. Double the number of centres and every bond is worth half as much. The strength per link falls as the reciprocal of the coordination.

That is exact and it is not what a real structure does. The same quantity has been measured for a filled band — the binding a bond supplies at half filling, across structures with two, four and six neighbours per site — and it falls by a factor of about two across a threefold change in coordination, which is a reciprocal square root rather than a reciprocal.

The two results are not in conflict; they describe different situations, and the gap between them says what a band adds.

One pair occupies one orbital, the nodeless one, and that orbital contributes bonding across every link with the same sign. Nothing cancels, so the total is fixed and dividing it is all that happens.

A filled band occupies many orbitals, and the higher ones have nodes. An orbital with a node across a link contributes antibonding there, so the sum over occupied orbitals is a partial cancellation rather than a division — and the cancellation is what makes the total grow with $\sqrt{n}$ instead of staying fixed. The bond order per link then falls as $\sqrt{n}/n$, which is the measured behaviour.

So the two scalings bracket the situation and each is right where it applies.

A single pair is the electron-deficient limit, where the pair is scarce and the structure is trying to make it do as much work as possible. A bridged borane is that case, and the $1/n$ division is why a three-centre bond is an efficient use of a pair rather than a weak version of a two-centre one.

A filled band is the electron-rich limit, where there are enough pairs to occupy orbitals that partly undo one another, and the per-link strength falls more slowly because the total is growing.

The practical form of that is a rule about which limit a system is in. Count the pairs against the links: where they are scarce the strength per link falls steeply with coordination, and a structure gains by keeping its coordination low. Where they are plentiful it falls gently, and a structure gains by raising its coordination as far as packing allows — which is the difference between a molecular solid and a metal, arrived at from a conserved total and a count.

It also says which of the two limits the result is safe in. The conservation is exact for one pair and exact for any structure, so it is a theorem; the $1/n$ division that follows from it is exact only when the pair occupies the nodeless orbital, which is to say only when there is one pair. Quoting the division as a general rule about how bonding spreads over centres would be extending a theorem past the case it was proved in — and the measured square root is what that extension costs.

Who found it, and when #

The uniform lowest orbital of a ring is as old as Hückel’s 1931 solution, and the bond-order definition used here is Charles Coulson’s, from 1939 — introduced precisely so that a delocalised system could be described in the vocabulary of bonds.

The electron-deficient molecules that need multicentre counting are a slightly later story. Diborane’s structure was disputed into the 1940s, with an ethane-like arrangement defended for two decades before the bridged structure was established; Longuet-Higgins gave the three-centre account in 1949, and Lipscomb’s systematic treatment of the boranes followed in the 1950s and earned a Nobel Prize.

What is striking in hindsight is that the counting argument was available before the structures were. A pair supplying a total bond order of two, however many centres it spans, says immediately that a bridged structure is not a compromise but an efficient use of a scarce pair — and the boranes are exactly the compounds in which pairs are scarce.

From a pair to a metal #

The question of where two-centre bonding stops takes a ring to its limit, and the accounting here completes it: a pair supplies a fixed total, geometry decides how it is divided, and only a closed loop achieves the maximum.

That accounting is what makes the solid-state field’s arguments continuous with the molecular ones. A metal is not a molecule whose bonds have become weak in some new way; it is a molecule with many centres and few pairs, dividing a conserved total among a great many links — which is why its individual bonds are weak, its coordination numbers are high, and its total cohesion is nevertheless large.