Molecular Geometry & Bonding
what the shape decides
The collection

Every essay — page 18

Page 18 of 42, 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

Orbitals

A one-electron wavefunction, drawn as a contour surface at a level somebody chose. Nodes, signs, and what the picture is a picture of.

Every usable window, with three centres and with four. For three pairs of arrangements on a line — every centre alike, the heavy centres inside, the heavy centres outside — the stretch of accuracy line over which each basis size is usable at every separation, across three decades of line. Adding a centre takes the covered share from 44% to 58%, from 27% to 54%, and from 67% down to 60%. The four-centre uniform chain is the one arrangement with no usable basis size at a kilocalorie a mole.

The overshoot was one arrangement

A fourth fragment leaves four three-body terms out of a pairwise counterpoise assembly as well as the four-body one, so the window in which the assembly is usable was expected to narrow. Asked of three arrangements that each gain one centre, it widens twice and narrows once, the uniform chain loses its answer at a kilocalorie a mole altogether, and the overshoot every earlier calculation reported turns out to belong to the arrangement with the heavy centre inside.

7 figures The usable stretches, assembled from pairs and from triples. For each four-centre arrangement, the stretches of accuracy line with a usable basis size when each fragment's correction is assembled from pairs, and when the three-body increments are added to it, labelled with the sizes usable there. The covered share rises from 58% to 67% (1-1-1-1), 54% to 75% (1-2-2-1), 60% to 80% (2-1-1-2), and stretches where more than one basis size is usable appear where there were none.

A repair that costs more than the whole

Four fragments are the first system in which a counterpoise correction can be assembled from something between pairs and the whole. Adding the three-body increments leaves what is still missing below a kilocalorie a mole on every cell, widens the usable range for every arrangement and turns a single usable basis size into two — and under a cubic model of cost it is dearer than the full calculation it stands in for until the cluster has eleven fragments. Keeping only the consecutive triples, which pays from five, works for two arrangements and does worse than pairs for the third.

7 figures The error falls along the sum of the two exponents. The ionic model's relative error on each of the six checkable separations, against the sum of the two ions' Slater exponents, with the least-squares line through all six. The rank correlation is −0.986 and only 4 of the 720 possible orderings of six points do as well, where the count of third-row ions it replaces is matched by 24. The three pairs with one third-row ion, which the count could not tell apart, fall in the order the line runs.

The sum of the exponents, not the softer ion

The row of the periodic table sorted an ionic model's errors into three groups, and a count that takes three values can say nothing inside a group. Made continuous, the variable the proposed mechanism names — how diffuse the softer ion is — carries no information: an oxide's 2p and a chloride's 3p have the same exponent to a hundredth. The sum of the two exponents carries nearly all of it, orders the middle group, and, asked about that group without having seen it, predicts its spread at twice the size.

7 figures Two conditions, and one factor that nearly meets both. The mean error of each group of pairs as the third-row p exponents alone are contracted, with the second-row pairs untouched by construction. One factor has to bring both other groups onto them. The pairs with one third-row ion arrive at 1.376 and potassium chloride at 1.346, 2.2 per cent apart — a test that could have produced two factors nowhere near each other, and did not.

One contraction for two conditions

The ionic model's errors run with the row of the periodic table, and the test proposed for that — stiffen the repulsion and watch for second-row pairs moving out and third-row pairs moving in — produces exactly that pattern from a repulsion that knows nothing about shells. The test that can fail contracts the third-row shells alone and asks one factor to bring two different groups of pairs onto the second-row ones. The two factors needed are 1.35 and 1.38.

6 figures Three ions with the same shell, and three different contractions. Each pair with one third-row ion, its error plotted against a contraction of that ion's p exponents alone. Sodium chloride's error falls to the second-row pairs' mean when chloride is contracted by 1.302, potassium fluoride's when potassium is contracted by 1.387, and calcium oxide's when calcium is contracted by 1.441. The single factor that contracts every third-row shell at once, 1.376, is drawn faint: it sits between the three, and the three span 10.7 per cent.

Three contractions for one shell

One contraction of the third-row p shells removed the row pattern from an ionic model's errors, with two conditions met by factors 2.2 per cent apart, and left the order of three pairs untouched. Taken ion by ion, chloride needs 1.302, potassium 1.387 and calcium 1.441 — the pairs' own order — and potassium chloride, fitted on nothing, lands among the second-row pairs. The single factor's two conditions agreed because both were averages of these three.

6 figures The same count in both pictures, and no rule between them. Every hydrogenic orbital with a radial node, drawn twice: its position nodes on the left axis and its momentum nodes on the right, at the same nuclear charge. The counts are identical and exact — n − l − 1 in each — because the two radial functions are polynomials of the same degree. The positions are unrelated: a node three quarters of the way out in one picture is not three quarters of the way out, or anywhere in particular, in the other.

The nodes in the other variable

An orbital has n − l − 1 radial nodes, and it has exactly that many in momentum too — the two radial functions are polynomials of the same degree. Nothing pairs one node with another: they are zeros of two different classical families. What is exact is the product over all of them, which is a ratio of factorials and does not depend on the nuclear charge at all.

7 figures A bond is a cosine in momentum space. The two factors a two-centre bonding orbital's momentum density is made of. One is the atomic momentum density, unchanged by the bond and falling as the eighth power of the momentum. The other is cos²(q·R/2), the interference between the two centres, whose period is fixed by the bond length and by nothing else. Everything that distinguishes a bond from two atoms in this picture is that cosine — and it has zeros where the atomic factor has none.

Oblate in the picture nobody draws

Every drawing of a σ bond shows a density stretched along the bond, and the position-space calculation agrees: the second moment along the axis is twice the one across it. In momentum the same orbital is flattened in the same direction, because the interference between the two centres cuts the distribution off at π over the bond length — and that cut-off is a zero a measurement could find.

7 figures An antibonding occupation fills the zero and moves the minimum. The profile along the bond near π/R, per electron, on a logarithmic scale, for five antibonding occupations of the same two orbitals. With nothing in the antibonding orbital the profile is exactly zero at π/R = 1.573. Two hundredths of an electron leave a minimum at 1.608, which reads the separation as 1.954 bohr instead of 1.997. At 0.104 the minimum becomes a flat shoulder, and the Heitler–London bond, at 0.236, has none.

The zero belongs to one determinant

A bonding orbital's momentum profile along the bond is exactly zero at π/R, and that zero reads a bond length with nothing fitted. It is a property of putting every electron into that one orbital. Any antibonding occupation fills it in linearly and drags the minimum outward, a tenth of an electron erases it, and the valence-bond wavefunction built from the same two functions never has one at any separation.

7 figures The zero at π/R follows parity, not bonding. The bonding and antibonding combinations of three atomic functions on nitrogen, each along the bond and each normalised to its own largest value, against momentum in units of π/R. For 2s the bonding combination is zero at π/R and the antibonding one is not. For 2p along the bond it is the other way round: the σ bond carries a sine, is zero at the origin and near its largest at π/R, and the antibonding combination carries the cosine. For 2p across the bond the π bond carries the cosine again. The factor is a cosine exactly when the orbital's inversion parity matches the atomic function's.

The zero is a parity, not a bond

A hydrogen-like σ bond has a momentum profile along its axis that vanishes at π/R, and it is natural to read that zero as a bond's signature. Built from 2p functions pointing along the axis, the σ bond carries a sine instead and sits at 83 per cent of its peak there. Which factor an orbital carries is decided by whether its inversion parity matches its atom's, and bonding has nothing to do with it.

6 figures The consecutive triples fail for a uniform chain at every length. The share of three decades of accuracy line over which some basis size is usable, against the number of fragments from three to eight, for three arrangements and three ways of assembling the correction. On the uniform chain the consecutive-triple assembly covers less than pairs alone at every length from four, and the shortfall grows. On the heavy-inside chain it covers exactly what every triple covers. On the heavy-outside chain it covers more than every triple from five fragments on.

Length did not rescue the consecutive triples

On four fragments, keeping only the consecutive triples of a counterpoise assembly worked for two arrangements and did worse than pairs for the uniform chain, and a short chain was the obvious excuse. Carried to eight fragments the excuse fails: the uniform chain's consecutive assembly settles at 48 per cent of the line against 68 for pairs. And the heavy-outside chain turns the lesson over — from five fragments every triple together covers less than the consecutive ones alone.

6 figures One zero becomes three, because the average couples them. What the interference factor becomes when the molecule is averaged over every orientation, one curve per angular kind. A directional profile along the bond carries cos(qR) for every orbital, so every cosine combination is exactly zero at q = π/R. The average replaces it by j₀(pR) for an s combination, j₀ − 2j₂ for a pσ one and j₀ + j₂ for a pπ one — because the cross term multiplies the interference by the orbital's own angular density and the average of the product is not the product of the averages. Their first zeros are at 0.66π, π and 1.43π.

The pair that cancels only at zero overlap

A directional Compton profile along a bond is exactly zero at π/R for every cosine combination, which makes the depth there a count of parity mismatches. A gas measurement averages over orientations, and the question was whether the count survives. It does not, for a reason nothing in the directional picture shows: the bonding and antibonding combinations of one atomic function are normalised by 2 + 2S and 2 − 2S, so a filled pair leaves a residue proportional to its overlap — and in nitrogen that residue is the largest single term, from a pair whose imbalance is zero.

6 figures The minimum fills in, and it fills in before any charge moves. The σ profile along the bond for three pairs of atoms, each with equal coefficients on the two atoms — no charge transfer at all. The homonuclear pair has an exact zero at π/R because its two atomic contributions are identical and cancel. The other two do not, because their radial functions differ: the cancellation at π/R needs cₐ gₐ = cᵇ gᵇ everywhere, and two Slater functions of different exponents are nowhere proportional.

A floor no charge transfer explains

A σ bond between two identical atoms has a momentum profile that is exactly zero at π/R, because the two atomic contributions are identical and cancel. Two different atoms cannot cancel there at any coefficients, so the depth of the minimum becomes a measure of polarity — and it is one, monotone in the charge imbalance in the direction chemistry moves charge. What it is not is a measure with a zero: it has a floor set by how far the two radial functions are from proportional, and B–N's floor is deeper than C–O's although C–O is the more polar bond.

6 figures