Hypervalency does not stop at three centres

Three-centre four-electron bonding is how hypervalency is explained without d orbitals. Three atomic orbitals on a line make three molecular orbitals — bonding, non-bonding, antibonding — and four electrons fill the first two, leaving the pair in the non-bonding orbital sitting entirely on the two ends. The result is two half-strength bonds and half a negative charge on each end, and it accounts for xenon difluoride, the triiodide ion, and the axial bonds of a trigonal bipyramid — without any d orbitals being needed.

It is always presented as an arrangement, singular. It is not. It is the smallest member of a series, and the rest of the series exists.

The family #

Take a chain of $m$ identical centres holding $m + 1$ electrons: one pair more than the chain has bonds to fill. For $m = 3$ that is four electrons in two bonds, which is the familiar case. For $m = 5$ it is six electrons in four bonds, for $m = 7$ eight electrons in six.

Only odd $m$ works, and the reason is arithmetic: $m + 1$ has to be even for the electrons to pair, so $m$ has to be odd. An even chain with one pair too few is a different problem with no arrangement of whole pairs leaving exactly one bond over, and the calculation refuses it rather than computing it.

The charges alternate #

The population on each centre is the first thing to come out, and it has a pattern nobody puts in.

For three centres: 1.5, 1.0, 1.5. The ends carry half an electron more than a neutral atom and the middle carries exactly one. That is the standard result — half a negative charge on each end — and it appears with no electronegativity difference anywhere in the calculation, which is the whole reason it is worth computing rather than simply stating.

For five: 1.333, 1.000, 1.333, 1.000, 1.333. For seven: 1.25 and 1.00 alternating. For nine: 1.20 and 1.00.

The even-numbered centres carry exactly one electron, at every chain length. The extra pair distributes itself over the odd-numbered sites, and the share each of them gets is $1 + 2/(m+1)$ — which falls as the chain lengthens, because the same one extra pair is spread further.

The exactness of the 1.000 on the even sites is worth noticing. It is not a near-miss; it is exact, and it follows from the alternant symmetry of a chain — the same pairing theorem that puts every alternant hydrocarbon’s charges at unity in a neutral molecule. Here the chain is charged and the theorem still fixes half of the sites.

Where the 0.707 comes from #

The three-centre case has one number in it that surprises people and is worth deriving, because it recurs throughout the family.

Four electrons in a three-centre system look like two bonds’ worth of electrons spread over two bonds, so the naive expectation is a bond order of one half. The computed value is $1/\sqrt{2} = 0.707$.

The reason is that bond order is not a count of electrons per bond. It is a sum over occupied orbitals of the product of the two coefficients, weighted by occupancy, and the bonding orbital of a three-centre system has coefficients $\tfrac{1}{2}$, $\tfrac{1}{\sqrt{2}}$, $\tfrac{1}{2}$ rather than three equal ones. The central atom carries more of it, so the two links it makes are stronger than a naive split would give.

The non-bonding orbital contributes nothing to either bond order, because its coefficient on the central atom is exactly zero — which is also why the pair in it sits entirely on the ends and produces the charges.

So 0.707 is what happens when a pair is shared unevenly: the bond order goes as the geometric mean of the two shares rather than as their average, and a geometric mean of unequal numbers beats what a straight division suggests. That is a general fact about bond order and it is worth having, because the same effect appears at every chain length in the table above.

The orders alternate too, and that is the prediction #

Every bond in every chain is bonding — none has zero or negative order — so the chain is held together throughout. And every bond is weaker than a full bond, which is what one pair short means.

What is not obvious is the pattern. For five centres the orders are 0.789, 0.577, 0.577, 0.789: strong at the outside, weak in the middle. For seven: 0.816, 0.545, 0.653, 0.653, 0.545, 0.816. The alternation grows with the chain, from nothing at three to 0.211 at five and 0.296 at nine.

That is a testable statement about a real class of compounds.

The polyiodides #

Iodine forms a series of anions — I₃⁻, I₅⁻, I₇⁻, and longer chains in the solid state — and they are exactly this family. Their structures have been known for decades.

Short outside, long inside is the chain’s prediction and it is what the crystal shows.

The competing picture — a polyiodide as an iodide ion with iodine molecules attached — predicts the opposite. If I₅⁻ were I⁻ with two I₂ units stuck to it, the short bonds would be the I₂ units and the long ones the contacts to the central iodide, giving long–short–short–long. That is not what is measured.

Nothing about iodine is in the calculation. It is a chain of identical centres with one pair too few, and the alternation comes from the filling.

Where the alternation comes from #

The reason is in the shape of the highest occupied orbital, and it is worth following because it connects this to something apparently unrelated.

A chain of $m$ sites has orbitals with $0, 1, \dots, m-1$ nodes. Filling $(m+1)/2$ of them puts electrons in every orbital up to the one with $(m-1)/2$ nodes, and the topmost occupied orbital has its nodes distributed along the chain. Where a node falls between two sites, that pair contributes nothing to their bond order; where it does not, the contribution adds. This is the same reading of the coefficients that bond order from the eigenvectors sets out.

For an odd chain one pair short of full, the nodes of the highest occupied orbital fall in the middle rather than at the ends, and the result is the alternation.

That is the same mechanism as a chain that cannot stay even. There, an infinite half-filled chain lowers its energy by alternating its bond lengths, opening a gap at the Fermi level. Here a finite chain at a filling one pair short shows the alternation in its bond orders without needing to distort. The connection is the filling: in both cases the electron count picks out a pattern along the chain, and in both cases the pattern has a period of two sites.

The check that isolates the missing pair #

The alternation could conceivably be a property of being a chain rather than of the electron count, and the way to find out is to fill the chain completely.

A five-site chain with all ten electrons — one pair per orbital, every level occupied — has a much flatter set of bond orders. So the alternation is the missing pair’s doing, not the chain’s.

That is a small check and it is the sort worth writing, because the alternative explanation is exactly the one a sceptical reader would offer and it costs one extra calculation to rule out.

Counting the same thing two ways #

The chain gives a second reading of what hypervalency is, and it is worth setting beside the usual one.

The octet reading: a hypervalent atom is one with more than eight electrons around it, which is forbidden, so something must be wrong with the picture. The three-centre bond repairs it by giving the central atom two half-bonds instead of two bonds, so its share is four electrons rather than eight.

The chain reading: there is nothing special about the central atom at all. Every centre in the chain is in the same situation, the extra pair is delocalised over the odd sites, and the arrangement is a property of the chain’s electron count rather than of any one atom’s valence.

The second reading is better because it extends. Asking which atom in I₅⁻ is hypervalent has no good answer — the two odd-numbered inner iodines are equivalent by symmetry and both have two neighbours — and the octet bookkeeping has to be repeated at each of them. The chain simply counts one pair too few and reports the consequences.

That is the same shift hypervalency is about the ligands makes: hypervalency is a property of the system, and locating it on one atom is a bookkeeping convention rather than a physical statement.

Where the model stops #

Identical centres, no repulsion. Every site has the same energy and every bond the same resonance integral. Real polyiodides sit in crystals where the counter-ion pulls the chain out of symmetry, and the effect is large: the same I₃⁻ measures 2.90/2.90 in a large symmetric cation’s salt and 2.83/3.04 in a caesium salt. The chain describes the symmetric case and says nothing about the asymmetry.

Bond order is not bond length. The comparison above is a comparison of orderings, not of numbers. Turning a bond order into a length needs an empirical relation, and no such relation survives being applied across a factor of two in order.

One orbital per centre. The chain uses one p orbital per iodine, pointing along the chain. The other valence orbitals — the perpendicular p orbitals, the s — are assumed to be non-bonding and full, which is the same assumption the three-centre picture makes and the same assumption the argument against d orbitals rests on.

Closed shell only. Every chain here has an even number of electrons and a filled set of orbitals, so there is a gap and the ground state is a singlet. Nothing in the treatment would notice if a chain’s highest occupied and lowest unoccupied levels came close together, and for a long enough chain they do — which is where a one-electron model stops being adequate for the same reasons it stops being adequate anywhere else.

No σ framework. A polyiodide chain is bent in most of its crystal structures, sometimes strongly, and a linear chain model has nothing to say about the angles.

The other members of the family #

Three-centre four-electron systems are not only iodine, and the family reading suggests where else to look.

Xenon difluoride is the canonical three-centre case: F–Xe–F, four electrons in the axial system, and the two bonds are long and weak. Xenon tetrafluoride is two such systems at right angles sharing a central atom, which is why its geometry is square planar and not tetrahedral — the two three-centre axes are independent and each wants to be linear.

The axial bonds of a trigonal bipyramid are the same arrangement, which is why they are systematically longer than the equatorial ones: phosphorus pentafluoride’s axial bonds are 1.577 Å against 1.534 equatorial, and the difference is a bond order of about a half against about one. That is a three-centre system embedded in a molecule whose other bonds are ordinary, and it is the reason the two kinds of site in a five-coordinate structure differ in more than their angles.

The bifluoride ion, [F–H–F]⁻, is a three-centre four-electron system in which the central atom is a hydrogen with no p orbitals to spare — the strongest hydrogen bond known, symmetric, with the proton exactly in the middle.

What the family reading adds is that none of these is an exception to anything. They are all one pair short of what a two-centre description needs, and the consequences — long bonds, charge on the ends, and in the longer members an alternation — follow from the count.

What a bond order of two thirds means #

There is a persistent difficulty in reading any of these numbers, and it is worth naming because it applies to every bond order in the hypervalency argument.

A bond order is defined inside a model. Here it is $\sum_k n_k c_{ki} c_{kj}$, a sum over occupied orbitals of the product of two coefficients — a well-defined quantity in a one-orbital-per-atom description and nothing else. It is not the number of electron pairs between two atoms, it is not measurable, and it does not have a unit.

What can be done with it is comparison. Within one model, applied consistently, a larger bond order goes with a shorter and stronger bond, and the ordering is what the polyiodide comparison tests. Nothing above claims that a bond of order 0.577 is 57.7 per cent of anything.

That restraint costs something. It means the model cannot say how much longer the inner bonds of I₅⁻ should be than the outer ones — only that they should be longer. A relation converting order to length would supply the number and would be an empirical fit, calibrated on compounds, with no more standing than the fit it came from. The ordering is a prediction; a number would be a parametrisation.

Where the ordering is the whole content, it is worth saying so, and the polyiodide structures are a case where the ordering alone distinguishes two pictures that differ qualitatively. That is as much as a model with one parameter should be asked to do.

Why the family has only odd members #

The even chains are refused by the arithmetic, and the species that exist are not counterexamples — they are built out of the odd members.

The requirement is structural. A three-centre four-electron system works because its non-bonding level has a node at the middle atom, and a node at the middle requires there to be a middle: an odd number of centres. Five and seven have one; four and six do not, and a chain of them cannot produce the level that puts the charge on the ends.

The higher polyiodides duly turn out to be assemblies rather than chains. Their structures are described, throughout the crystallographic literature, as arrangements of three building blocks — iodide ions, iodine molecules, and triiodide ions — packed together and weakly linked. An eight-iodine dianion is not an eight-centre bond; it is two triiodides with an iodine between them.

So the family is not odd chains plus some awkward even cases. It is odd chains, plus assemblies of odd chains, and the even stoichiometries are counts of atoms rather than counts of centres in a bond.

That is a stronger closure than the essay’s own hedge, and it costs nothing: the same node that puts the charge on the ligands is the reason there is no even member to look for.

Still open: the band limit, and the even chains #

The chain has an obvious limit and it can be reached from the other side too. Let $m$ grow and the discrete levels crowd into a band; the filling stays one pair short of half, which for a chain of $p$ orbitals is a nearly-half-filled band; and the alternation becomes a periodic distortion. That is the band limit with a specific occupancy, and the polyiodides are a finite piece of it — which is the sense in which a molecular hypervalent bond and a Peierls-distorted solid are two ends of one argument.

The other open question is the even chains, which the calculation refuses. An even chain with one pair short of full has an odd number of electrons, so it is a radical rather than a closed-shell anion, and the family does contain such species — I₄²⁻ and I₈²⁻ exist, and they are not simple chains but branched or bent arrangements. Working out why the even members escape the pattern by changing shape rather than by holding an unpaired electron is a question about geometry, and it needs a calculation of shape rather than of orbitals.