Molecular Geometry & Bonding
what the shape decides
The collection

Every essay — page 35

Page 35 of 36, continuing through the fields in the same order.

Orbitals Where the atoms go Bonding models What symmetry decides Beyond the octet What a spectrum settles When the molecule does not stop What the shape is for What is taught wrongly Series Named objects Orbitals Refutations Search

What is taught wrongly

The explanations that are confident, memorable and false — stated fairly and then tested against a calculation rather than an opinion.

A cheap number that predicts an expensive failure. Thirty-two systems. Along the bottom, how far the mean field's own symmetry breaking has moved between the reference and the target — a quantity available before any exact calculation. Up the side, how wrong the transferred correction turns out to be. They rank together at 0.902, and the open marks are the systems whose broken solution has collapsed entirely, which is where the diagnostic stops being a scale and becomes a warning.

The warning a cheap calculation gives

A correlation correction computed on one system and carried to another works until it does not, and nothing in the scheme says in advance which. The mean field's own symmetry breaking says: it collapses at a definite field, and the transfer fails where it goes. Across thirty-two systems the two rank together at 0.90.

7 figures The carriers a distortion was hiding. A half-filled ring of 40 — one of the 4m rings that carry exactly one pair of carriers at every temperature — allowed to distort. Cold, it alternates by 0.1232, opens a gap of 0.4927 and carries 3.6e-15 carriers rather than one pair. The alternation is undone continuously at kT = 0.1358, and the carrier count comes back as it goes.

The carriers a distortion was hiding

A half-filled ring of 4m carries exactly one pair of thermal carriers at every temperature, which is true only of a ring held rigid. Allowed to move, it does not carry them: it alternates, opens a gap of 0.4927, and carries none at all until a temperature that undoes the distortion.

6 figures The slope goes to a half, and a window fit stops short of it. The local slope of the alternation against the reduced temperature, between each neighbouring pair of points, on a ring of 40 at K = 1.6. It rises monotonically from 0.4115 to 0.5053 as the transition is approached, crossing a half at about a part in a thousand of the reduced temperature. The fitted 0.44 is the average of the left-hand end of this curve; the exponent is one half, which is what a free energy analytic in one order parameter is obliged to give.

The exponent was the window's

A fit over the last decade before a distortion vanishes gives an exponent of 0.44, and running it on larger rings should say whether the number belongs to the transition or to a forty-site ring. It belongs to neither. The local slope runs to 0.5020 as the transition is approached, and 0.44 is what a fit over that particular decade returns — on every ring size and every stiffness, because the whole curve is one curve.

7 figures A fourth data point, and 3 negative electronegativities. Each element's electronegativity from three points on its energy curve, the cubic coefficient a fourth point adds, and what the fourth point leaves. The cubic coefficient is one sixth of a second difference of the ionisation series, so it is largest where that series has a kink — and the alkali metals, whose second electron comes out of a closed shell, are pushed to Li -7.91, Na -3.42, K -1.49 eV. The last column is where the two roots of the fixed-point equation collide, beyond which the atom has no solution at all.

Where a closed form stops being one

What happens when the quadratic energy is not enough? A cubic makes the equalisation condition a quadratic with two roots, and something has to choose between them. The choice is easy and the finding is somewhere else — a cubic fitted through the dication gives lithium an electronegativity of −7.907 eV, a capacity of 0.073 of an electron, and a molecule of two alkali metals no solution at all.

6 figures The pair's regime is not a property of the pair. How much the pair's bonding responds to its own overlap — the ratio of what it is bonded by at an overlap of 0.4 to what it is bonded by at 0.1 — with and without a third orbital coupled to both. Alone it is 11.83, which is the regime in which bonding tracks overlap. With a third orbital present it falls to 1.46, 0.92, 0.74 — and two of those are below one, meaning a fourfold increase in the overlap between the two atoms buys them less bonding rather than more. The coupling comes from the overlap by the Wolfsberg–Helmholz rule with K = 1.75, which is fitted rather than derived. Every stabilisation here inherits that; the shape of the curve against separation does not, because K is a constant.

A regime that belongs to the neighbours

Two orbitals at the same energy are bonded in proportion to their overlap and two far apart are barely bonded at all — two regimes, and the natural question is whether the regime is a property of the pair. It is not. Put a third orbital beside them and the pair's response to its own overlap falls from twelvefold to less than one: more overlap buys less bonding.

5 figures The capacity, against whether the anion exists. Each atom at its electron affinity and the capacity the cubic model gives it — the largest amount of electron the model says it can accept. A negative affinity is an anion that is not bound, which is the statement that the true capacity at the integer is zero. The three atoms with negative affinities are beryllium, magnesium and nitrogen, and they have the largest capacities in the set: two of them infinite and the third 31.5. The three smallest capacities all belong to atoms whose anions are bound.

A capacity that is largest where there is none

A cubic through four charge states implies a largest amount of electron an atom can accept, and it is natural to ask whether the idea survives its own model. It does not. The three atoms whose anions are not bound — beryllium, magnesium and nitrogen — have the largest capacities in the set, two of them infinite; the three smallest capacities all belong to atoms whose anions are bound. And among the fourteen where the test cannot bite, the ordering is sensible.

6 figures How much stronger a fundamental is than a satellite, against the repulsion. The weakest fundamental divided by the strongest satellite, for a six-site ring and chain at every filling from a third to a half, against the on-site repulsion. Below the line at two the two kinds of line cannot be told apart by their height. The half-filled systems cross it and the third-filled ones do not — not at any repulsion up to sixty-four times the hopping, where the third-filled ring is still at 8.3.

A satellite that never loses its place

The repulsion at which a satellite stops being tellable from a fundamental orders exactly with the one-electron gap across four systems. Changing the gap by the filling instead is the sharper test, and the ordering does not survive it: a six-site chain has a larger gap at half filling and a smaller boundary. Below half filling there is no boundary at all, at any repulsion up to sixty-four times the hopping.

6 figures The alternation a spring buys, three ways. The alternation against the elastic constant, on a logarithmic axis. The middle line solves (2/π)(K − E)/(1 − δ²) = K for the complete elliptic integrals; the lower one is the exponential form every account of a Peierls distortion quotes, which is its own asymptote and is 6.5 per cent low at K = 1.2; the upper one is a ring of 40, which leaves the infinite chain as the spring stiffens because a smaller alternation is a longer coherence length.

The amplitude the collapse left behind

Five Peierls curves became one curve when each was divided by the alternation its ring settles at cold, so that amplitude is the whole of what distinguished them — and it was five golden-section searches over diagonalisations with no formula anywhere. It has one, exactly, as a sum of square roots; and writing it down says that one of the five rings was never measuring a long chain.

6 figures Where the broken solution appears. The mean field's spin polarisation against the on-site repulsion, at half filling and no site-energy modulation, for three systems. Two of them are symmetric below a threshold and polarised above it — 1.672 for a chain of four and 2.355 for a ring of six. The third is polarised at every repulsion tested, because its half-filled shell is degenerate and the symmetric solution is unstable however small the repulsion is.

The half of the square a ring of four cannot show

There is a warning that says in advance whether a transferred correction will hold: the mean field's own symmetry breaking, which collapses at a definite site-energy modulation and takes the transfer with it. The other axis of the same square has a threshold too — on the other side — and the ring of four it was all measured on is the one system with no threshold to find.

7 figures A pair with no overlap, and a third orbital swept past it. The three levels of a trio in which the two outer orbitals have exactly no overlap with each other, as the third orbital's energy is swept. The middle line is at -13.6 at every point — the antisymmetric combination of the two, which has no partner of its own symmetry and cannot mix with anything. The other two move, so the pair is split by an orbital it has no direct contact through.

A bond order between atoms that do not interact

A diatomic held where its overlap changes sign has no interaction between its two orbitals at all — which is what the sign change of its overlap means. Put a third orbital beside it and the pair is still split, one line sits exactly at the free-atom energy at every third-orbital energy, and the bond order between the two runs to −0.9999. Three measures of the same bond disagree completely.

6 figures One geometry, four electron counts, four answers. The bond order between two orbitals with exactly no overlap and no resonance integral, as the third orbital's energy is swept, at every count the trio can hold. With none it is identically zero. With two it is positive and rises past one. With four it is negative and reaches -0.954. With six it is a horizontal line — the third orbital's energy stops mattering entirely.

A filled shell is not an empty statement

A bond order of −0.954 between two orbitals with no overlap and no resonance integral invites the prediction that at six electrons — every level occupied, the sum over a complete set — it would be exactly zero. It is exactly one seventh, and the reason is that a complete set in a non-orthogonal basis sums to the inverse of the overlap matrix, which has entries where the overlap has none.

6 figures What happens to the level that was exact. The three levels of the trio as one of the two outer orbitals is raised. At zero detuning the middle one sits at -13.6 exactly — it is the antisymmetric combination, and nothing of its symmetry exists for it to mix with. The moment the two are made inequivalent that statement is gone: the level leaves linearly, and the other two barely move by comparison.

A symmetry holds or it does not

One level of a three-orbital trio sits at the free-atom energy exactly, at every third-orbital energy, because the antisymmetric combination of the pair has nothing of its own symmetry to mix with. Detuning one of the two by a twentieth of an electron volt moves it by half of that — first order, immediately, with no protected regime at all.

6 figures