The collection

Every essay — page 26

Page 26 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 a spectrum settles

A spectrum is a list of positions, and how many there can be is decided by the shape before any of them is measured. Counting them, and reading a structure back out.

H₂O: 3 distinct modes. The displacement of every atom in 3 normal modes of H₂O, drawn from the eigenvectors of the mass-weighted Hessian. Under each is the internal coordinate with the largest share of the motion and how large that share is; where no coordinate holds nine tenths, the mode is not a motion of one bond or one angle and is labelled so.

The six motions that are not modes

Three N minus six is quoted in every textbook and the six are almost never computed. Diagonalising a mass-weighted Hessian gives six eigenvalues at arithmetic noise, and projecting their vectors onto the three translations and three rotations written down from the geometry alone accounts for every one of them.

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Mutual exclusion across every group here. The 20 groups with tables here, each with its highest rotation order, whether it has a centre of inversion, which of its representations carry a coordinate, which carry a product of coordinates, and whether any carries both. Every group with a centre excludes, which is a theorem. 1 group without a centre excludes as well — D5h — so the rule does not run backwards, and the counterexample needs a fivefold axis.

Mutual exclusion does not prove a centre

A centrosymmetric molecule shows no band in both its infrared and its Raman spectrum. The rule is a theorem and its converse is read off as though it were part of it — but ferrocene in the gas phase has no centre of inversion and no coincidence either, and the reason is that a fivefold axis separates the coordinates from their products where a threefold or fourfold axis cannot.

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A microwave constant predicted from an infrared one. For four diatomics: the rotational constant and the stretching frequency, both measured, and the centrifugal distortion constant predicted from them by 4B³/ω² — then the constant a microwave spectroscopist fits to the line positions. The prediction and the fit agree within a few per cent across three orders of magnitude in the quantity, and nothing connects the two measurements except the assumption that the bond stretching under rotation is the same bond that vibrates.

The rotor that stretches

A rigid rotor's lines are evenly spaced, and a real molecule's are not — it pulls itself apart as it spins. How much is not a fitting parameter: it follows from the stretching frequency by one relation, and the prediction agrees with the measured constant to a few per cent across four molecules spanning three orders of magnitude.

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H₂O with 1→D: what each mode is made of. H₂O with 1→D. Each row is one distinct frequency and each column one internal coordinate; the bar is the share of the motion in that coordinate. A mode whose largest share reaches nine tenths is a motion of one bond or one angle and is named for it. 3 of 3 here are.

When a mode becomes a bond stretch

Water's two stretching modes are each exactly half in one O–H bond and half in the other, which is why neither of them belongs to a bond. Change one hydrogen to deuterium and the same force field at the same geometry gives two modes that are 99.5 and 99.7 per cent in a single bond each. Nothing about the bonding changed; a mass did.

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ammonia's rotational levels, sorted by K. The rigid rotational levels of ammonia up to J = 4, each drawn at its computed energy and grouped by J. Within a group the levels are pushed apart by the second rotational constant, so it is plainly there in the level pattern.

The constant a spectrum cannot see

A symmetric top has two rotational constants and its microwave spectrum reports one of them. Not badly, not with difficulty: ammonia's A of 6.3406 wavenumbers appears in none of its lines at any J and any K, because the term it belongs to cancels exactly out of every transition. The molecule turns about that axis, the energy is real, and the measurement is blind to it.

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boron trifluoride at 1454 cm⁻¹, shared out four ways. One mode of boron trifluoride, with each internal coordinate's share of it computed by four conventions. Each convention is a bar in every group; the groups are the coordinates. A number quoted for this band without saying which bar it is has not said much.

How much of a band is a bond stretch

Boron trifluoride's 1454 cm⁻¹ band is 36.7 per cent B–F stretch, or 49.9, or 97.9, depending on which of four standard ways of sharing a mode out among internal coordinates is used. All four are defensible, all four sum to one, and the spread between them is sixty-one percentage points on one band of one molecule.

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water: every level up to J = 4, from a matrix. The rotational levels of water at κ = -0.4322, each J diagonalised in the symmetric-top basis. A symmetric top would show one level per K with everything above K = 0 doubly degenerate; here every degeneracy is split, and the size of each splitting is what the third constant is measured from.

The top that reports all three

A symmetric top hides one of its two rotational constants in every line of its spectrum. Break the symmetry and the hiding stops: for water, twenty-three of the twenty-five levels up to J = 4 move when A is changed, and the two that do not are the ground state and the one at B + C. There is no formula for any of them.

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Three bands of one spectrum, and the bond length behind each. Nitrogen's three photoelectron bands, drawn as the vibrational intensity distributions computed from the measured bond lengths and vibrational constants of the three states of the ion. Each band's lines add to one. The middle band is spread over five lines because the electron removed came out of a strongly bonding orbital and the bond lengthened by 77.22 thousandths of an ångström; the outer two keep 92 and 88 per cent of their strength in a single line.

The width of a band is a bond length

Nitrogen's three photoelectron bands are one sharp line, a progression of five, and a line with a shoulder. Computed from the measured bond lengths of the three states of the ion, the intensities come out at 0.917, 0.263 and 0.880 in the first line of each — because removing a weakly bonding electron lengthens the bond by 18.7 thousandths of an ångström, a strongly bonding one by 77.2, and an antibonding one shortens it by 23.7.

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Three parameters, two numbers, and a curve of answers. seven structures of formaldehyde, every one of which reproduces the measured rotational constants A and B exactly. The C=O length runs from 1 to 1.3 ångström, the C–H length from 1.56 down to 0.95, and the HCH angle from 74.58 to 164.47 degrees. The third constant is not a third number: for a planar molecule it is fixed by the other two, and it comes out at 1.14 for every member.

Three numbers is not a structure

Formaldehyde's rotational spectrum gives three constants, of which a planar molecule's are only two independent numbers, and its structure has three parameters. Seven structures are computed here that reproduce A and B to the last digit the solver carries: the C=O length runs from 1.000 to 1.300 ångström, the C–H length from 1.557 down to 0.952, and the HCH angle from 74.6 degrees to 164.5.

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A photoelectron band is a filter, and the group chooses the filter. The Huang–Rhys factor of every vibration of five molecules under a change of geometry that lengthens every bond alike — which is what removing an electron from a non-degenerate orbital does. On a logarithmic scale spanning sixteen decades, eight modes carry the whole of it and the rest sit on the floor at arithmetic noise. Which ones is decided by the point group: only a totally symmetric vibration can appear, whatever the size of the change.

A band is a filter on the modes

A photoelectron band's vibrational structure reports the frequencies of a few of the ion's vibrations and is silent about the rest, and which few is decided by the point group before any geometry is known. Under a change of shape that lengthens every bond alike, methane's totally symmetric stretch gets a Huang–Rhys factor of 0.905 and its other eight modes get between 10⁻²⁸ and 10⁻³⁵.

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Every sign, lost. Formaldehyde in its own principal axes. Open circles are the atoms where they are; filled ones are where Kraitchman's equations put them, from the change in the three moments when each atom in turn is made heavier. The two agree to 7.6e-8 ångström — the equations are an identity for a rigid structure — but they return the square of each coordinate, so the two hydrogens at b = ±0.9348 both come back at +0.9348 and land on the same point.

The coordinate an isotope reports

Kraitchman's equations return an atom's position from the change in the moments when that atom alone is made heavier, and for a rigid structure they are an identity — formaldehyde's four atoms come back to a part in ten million. What they return is the square of each coordinate, so both hydrogens at b = ±0.9348 come back at +0.9348; every out-of-plane coordinate comes back imaginary at a moment error of one part in a hundred thousand; and the famous error cancellation, measured at a factor of thirteen, still leaves the answer two and a half times worse than a direct fit.

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Three lines, then a hundred. The exact removal spectrum of a 6-site Hubbard ring at half filling: every final state of the ion, at the energy it costs to reach and with the intensity the matrix element gives it. With no repulsion there are 3 lines and they are the occupied orbital energies. At U = 8 there are 100, on a molecule with 6 orbitals — so the spectrum cannot be read as a list of orbital energies, because there are more bands in it than there are orbitals to name.

More bands than there are orbitals

A photoelectron spectrum is read as a list of orbital energies, one band per occupied orbital. Computed exactly for a six-orbital ring, it has three bands with no repulsion and a hundred with eight — and by then fifty-three per cent of the intensity is in lines that no orbital corresponds to. The total intensity is three at every repulsion, exactly, because that is a sum rule and not a fit.

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