When a mode becomes a bond stretch

Normal modes are not bond stretches is right. Water has two stretching frequencies and two O–H bonds, and it is almost irresistible to pair them off; computed, each mode is exactly half in one bond and half in the other, and neither frequency belongs to a bond at all.

That essay states its conclusion as a general one. This one is about the case where it stops being true, and about how little has to change for that to happen.

The change is one neutron on one of two hydrogens, and it takes the mixing from exactly a half to less than one part in two hundred. Nothing about the bonding has been touched, and the two O–H bonds are chemically the same bond they always were.

What “half in each bond” means, and why it is exact #

A mode’s composition is how much of its motion is in each internal coordinate: how much stretching of the first O–H, how much of the second, how much bending. The numbers are squared contributions and they sum to one.

For water they come out at 50.0 and 50.0 for both stretches, and the exactness is not an accident of the fit. The molecule has a twofold axis that exchanges the two hydrogens, so the two O–H coordinates are exchanged by a symmetry operation of the molecule. A normal mode has to belong to a symmetry species, which means it is either unchanged by the exchange or reversed by it — and either way its amplitude in the two coordinates has the same magnitude.

So symmetry forbids a localised stretch in water. There is no force field, however contrived, that gives a mode confined to one O–H bond, because the constraint is on the symmetry species rather than on the constants.

It is worth being clear about what is and is not being said. The two stretching modes are not two bond stretches that happen to have mixed; they are the symmetric and antisymmetric combinations of the two bond stretches, and the combinations are what the molecule has. A bond stretch of one O–H is a perfectly well-defined motion and it is not a stationary state of the vibrating molecule: excite it and the energy runs into the other bond and back.

The time it takes to do that is set by the coupling between the two coordinates, which the force field contains as a stretch–stretch constant. For water that constant is small, so the sloshing is slow compared with the vibration itself — and it is nonetheless total, because slow and complete are different things.

Substituting both hydrogens changes nothing structural #

This is the control the argument needs. Replacing both hydrogens moves the frequencies a long way and leaves the compositions exactly alone, because the twofold axis is still there: two deuteriums are exchanged by it as readily as two hydrogens.

So a change of mass by itself does not delocalise or localise anything. What matters is whether the change is the same on both sides.

Substituting one hydrogen localises both stretches #

Nothing about the electrons has changed. The force field is the same four constants; the geometry is the same; the two O–H bonds are chemically identical because a nucleus’s mass has no bearing on the electronic structure at all. What has gone is the symmetry: the twofold axis exchanged two identical nuclei and now exchanges a hydrogen with a deuterium, which is not an operation of anything.

With the constraint removed, the two coordinates are free to sort themselves out by frequency, and they differ enormously. An O–H oscillator and an O–D oscillator have very different natural frequencies — a factor of about 1.4 — and two oscillators that far apart in frequency barely mix however strongly they are coupled. That is a general fact about coupled oscillators and it is the same arithmetic that decides whether two orbitals interact: the mixing goes as the coupling divided by the energy separation, and a large separation kills it.

Drawn as motions rather than as compositions the two cases look as different as the numbers say. A spectrum counts environments, not atoms draws both sets: HOD’s middle motion moves the deuterium and leaves the hydrogen nearly still, its highest moves the hydrogen and leaves the deuterium nearly still, while both of H₂O’s stretching pictures move both hydrogens by the same amount and neither shows a bond stretching on its own.

The frequencies are a prediction #

The force field used above was fitted to H₂O and D₂O and never saw HOD. So the frequencies it gives for HOD — 1,445, 2,824 and 3,890 wavenumbers — are predictions rather than fits, and they can be compared with what is measured.

They are close: the harmonic values from experiment are about 2,824 and 3,890 for the two stretches. That agreement is a check on the whole construction, and it is a different check from the usual ones, because it involves a molecule the fit had no knowledge of.

The isotope shift is arithmetic draws the substitution the fit did see — H₂O against D₂O, with the ratio of each pair of frequencies — and none of the three ratios is the √2 the rule of thumb predicts.

It is worth noticing which quantity the prediction is a prediction of. The fit determined four force constants from six measured frequencies — three from H₂O and three from D₂O — which is the underdetermination of a force field: three frequencies cannot fix four constants, and a second isotopologue is what makes the problem solvable at all.

So HOD is a third isotopologue used as a test rather than as data, and it is the strictest test available, because it is the one whose symmetry differs from both of the others.

The bend does not move #

One row of the tables has been passed over and deserves a paragraph, because it is a control the argument did not have to pass and does.

The bending mode is 99.8 per cent in the angle coordinate in H₂O, 99.9 per cent in D₂O and 99.8 per cent in HOD. It is localised in all three, and it was localised before any symmetry was broken.

The reason is the one the rule above gives: there is only one bending coordinate, so there is nothing for it to mix with. Localisation is not a property a mode acquires by being simple; it is what happens when a coordinate has no partner at a comparable frequency, and a molecule with one angle has a bend that is a bend in every isotopologue.

That makes the bend a useful control on the whole calculation. Whatever the substitution does to the stretches, it should leave the bend’s composition where it was — and it does, to a tenth of a per cent.

Why this is the sharpest available demonstration #

The claim that a normal mode is a property of the whole molecule rather than of a bond is easy to state and hard to make vivid, because in most molecules there is nothing to compare it against.

Here there is. Three molecules, one force field, one geometry, one set of electrons. The only thing that varies between them is which nuclei are which isotope, and the compositions go from exactly shared, to exactly shared, to almost perfectly localised. Any account of the delocalisation that appealed to the bonding would have to explain why the bonding changed when a neutron was added, and it did not.

The general statement the three cases support is stronger than the familiar one:

A normal mode is shared between equivalent coordinates because they are equivalent, and equivalence is a statement about the nuclei’s masses as well as about the electrons.

Which is to say the delocalisation is a symmetry effect, and symmetry in vibrational problems is symmetry of the mass-weighted Hessian rather than of the molecule’s electronic structure.

How far apart two coordinates have to be #

The mixing between two coupled oscillators goes as the coupling divided by their separation in frequency, so the question of how localised a mode is has an arithmetic answer rather than a qualitative one.

For water the two O–H coordinates are identical in frequency — a separation of zero — and the mixing is total whatever the coupling, which is the symmetry statement arriving as a limit. For HOD the O–H and O–D coordinates differ by about 1,100 wavenumbers, and the stretch–stretch coupling in this force field corresponds to a few tens; the ratio is small, and the modes are localised to within half a per cent.

That gives a usable rule. Two internal coordinates mix strongly if their natural frequencies differ by less than the coupling between them and hardly at all if they differ by much more, and everything in between is a matter of degree. Symmetry is the case where the difference is exactly zero and the answer is exactly half.

The two numbers are worth stating side by side because the comparison is the argument. HOD’s two stretches are 1,066 wavenumbers apart, which is what makes them mix so little. H₂O’s are 110 apart — a tenth of that — and are nonetheless exactly half and half, because there the separation is not what decides anything. Symmetry does, and symmetry does not care how close two frequencies are.

Group frequencies, and what this says about them #

Group frequencies, and where they stop is about when a frequency can be attributed to a fragment — a carbonyl band near 1,700 wavenumbers in every ketone — and it finds that four of twenty-one distinct frequencies in its set are localised in a single internal coordinate.

HOD’s two stretches join that list and are the most localised members of it. What puts them there is the same mechanism the essay identifies: a coordinate whose natural frequency is far from every other coordinate’s stays to itself. In a ketone that isolation comes from a stiff bond between heavy atoms; here it comes from a light nucleus being replaced by a heavy one on one side only.

Carried further, the count of distinct frequencies rises and then falls again: CH₄ has four, CH₃D six, CH₂D₂ nine, CHD₃ six, CD₄ four. Symmetry is highest at the two ends and lowest in the middle, and so is the number of things a spectrum can resolve — which is the same statement as the localisation, counted rather than measured.

So an isotopic substitution is a way of manufacturing a group frequency, and it is used that way — replacing one hydrogen in a molecule to make a single bond’s stretch visible on its own is a standard trick, and this is the arithmetic of why it works.

What has not changed, and it is worth checking #

Two things are invariant across the three isotopologues and both are worth stating, because the essay’s whole claim is that one thing changed and one thing only.

The force constants are identical by construction — the same four numbers, in the same units, describing the same electronic structure — and that is not an approximation being made here. Within the fixed-nuclei approximation the potential energy surface is a property of the electrons, and isotopes differ only in the mass that moves on it.

The number of modes is three throughout: three atoms, non-linear, 3N − 6 = 3. The six motions that are not modes is the essay about the other six, and they are six for all three isotopologues too.

What changed is the point group of the isotopically substituted molecule, from C₂ᵥ to Cₛ, and every consequence in this essay follows from that one change.

The symmetry that matters is not the one on the label #

There is a general point here about what “the symmetry of a molecule” means in a vibrational problem, and it is easy to state wrongly.

HOD’s electronic structure has a mirror plane exchanging the two hydrogen positions, because the electrons cannot tell a proton from a deuteron: the potential energy surface of HOD is the same function as the potential energy surface of H₂O. What HOD does not have is that symmetry in its mass-weighted Hessian, because the mass weighting divides each coordinate by a different square root.

So there are two symmetries in play and they part company under isotopic substitution. The electronic one governs the force constants and every property of the surface. The mass-weighted one governs the normal modes and the spectrum.

That distinction is why a spectrum counts environments, not atoms can be true of one kind of spectroscopy and a different count be right for another, and it is why the point group written at the top of a table is not always the group the question in hand is about.

What a chemist takes from this #

The practical reading is short and is worth separating from the theory.

If a stretching frequency is wanted for one particular bond in a molecule that has two equivalent ones, substituting an isotope into one of them will produce it — and the frequency that comes out is close to what the bond on its own would give, because the mode has become that bond’s motion.

If a pair of frequencies is being assigned to two bonds in a symmetric molecule, the assignment is wrong before it begins, and no amount of care about which band is which will fix it.

And if a force field is being fitted, the isotopologue that constrains it best is the one whose symmetry is lowest, because that is the one whose modes are least determined by symmetry and most determined by the constants. HOD is a better datum than D₂O for exactly that reason, and it is the one that was left out of the fit here so that it could be used as a test.

That is worth following, because it says the inequality above is incomplete rather than wrong. It compares two oscillators of the same kind — two stretches — and ammonia’s N–D stretch is well separated from its two N–H stretches by exactly the eight hundred wavenumbers that localise methane’s. What it is not separated from is the bending coordinates, and ammonia’s force field couples a stretch to an angle where methane’s very nearly does not: the N–D mode puts as much of itself into the three angles as into the bond it is named for.

So breaking the symmetry is necessary and not sufficient. A localised stretch needs isolation from every other coordinate the force field couples it to, and a pyramidal molecule’s umbrella coordinate is a coordinate that couples to stretches. Water works because it has one angle and a stretch–bend constant small enough to ignore; methane works because its stretches and bends barely talk; ammonia does not work, and the essay’s rule of thumb would have predicted that it would.

The inequality that decides it, and the 99.7 per cent it predicts #

The finding — complete delocalisation in one molecule and 99.5 per cent localisation in the other — is a large change from a small cause, and it follows from one inequality that can be evaluated on the numbers already in hand.

Two oscillators coupled by an amount $V$ and separated in frequency by $\Delta$ mix through an angle $\theta$ with

$$\tan 2\theta = \frac{2V}{\Delta}.$$

At $\Delta = 0$ the angle is 45° and the mixing is complete — two identical bonds, two exactly half-and-half modes, which is water. As $\Delta$ grows past $2V$ the angle collapses and each mode retreats onto its own oscillator.

Both quantities are available. The coupling is half the splitting between water’s two stretches, 99 wavenumbers, so $V \approx 50$. The detuning is the difference between an O–H and an O–D oscillator, which is roughly 3,700 against 2,700 — about 1,000 wavenumbers.

Putting those in: $\tan 2\theta = 99/980$, so $\theta = 2.9°$, and the fraction of the mode sitting in the wrong bond is $\sin^2\theta = 0.0025$.

The prediction is 99.75 per cent, and the computed values are 99.5 and 99.7.

That is worth more than the agreement, because the same inequality accounts for three separate findings about normal modes. A mode is delocalised when the detuning is small against the coupling and localised when it is large, and the detuning can be produced three ways: by nothing, in a symmetric molecule; by a mass, which is the case here; and by anharmonicity as the vibrational levels are climbed. One expression, three mechanisms, and the answer in each case is a ratio of two numbers a spectrum supplies.

It also says what would be needed to make a symmetric molecule’s modes local, which is a detuning of order fifty wavenumbers — a very small perturbation. Substituting one hydrogen with deuterium supplies twenty times that, which is why the change is not gradual: the molecule goes from one regime to the other in a single substitution, with nothing in between to observe.

What is left #

The compositions here are for a molecule with two equivalent bonds, which is the simplest case where the effect exists. In a molecule with three or more equivalent coordinates the intermediate cases are richer: substituting one of methane’s four hydrogens leaves three still equivalent, so the modes localise partially rather than completely, and a spectrum that changes when only a mass does is the essay about the count of bands that results.

There is also a case met from the other end that is rarely connected to this one. Water’s lone pairs are not a pair makes the same structural argument about orbitals rather than modes: two equivalent lone pairs is a description, the canonical orbitals are what have energies, and a photoelectron spectrum sees the second. The vibrational version is the same theorem about a different matrix, and the isotopic substitution here has no orbital counterpart at all — which is the cleanest possible statement that the two symmetries are different symmetries.

And the whole calculation is harmonic. A real O–H stretch is anharmonic enough that the fundamental and the harmonic frequency differ by about 150 wavenumbers, and anharmonicity couples modes in a way this model has none of. What that does to a composition is a question a harmonic model cannot ask, and the answer is not obviously small for a coordinate as anharmonic as an O–H stretch.