A spectrum that changes when only a mass does

A vibrational spectrum is usually read as a report on the bonding: which bonds are stiff, which atoms are heavy, where the groups absorb. Isotopic substitution is then treated as a controlled experiment on the second of those, since replacing hydrogen with deuterium changes a mass and leaves the electronic structure alone — a cleaner intervention than comparing two molecules, which is what two structures, two spectra has to do, and a different question from the one how many frequencies, not how many modes asks about a single molecule.

It changes something else as well, and the something else changes the number of lines rather than their positions.

The operations a mass pattern permits #

A symmetry operation belongs to a molecule’s point group if it maps the structure onto itself — which requires it to send each atom to an atom of the same kind. Two nuclei of different mass are not the same kind for this purpose, even though they are the same element and sit in identical positions.

So the operations surviving a substitution can be counted exactly, with no geometry involved: take each of the parent’s operations, read off the permutation it performs on the atoms, and keep it if every atom is sent to one of equal mass. That is arithmetic on a list of masses.

For methane the answers are

The count of surviving operations always divides twenty-four, which it must, since the survivors are a subgroup. And the number of distinct frequencies moves in the opposite direction from the number of operations: the least symmetric member of the family has the most lines.

Naming the subgroup without being told #

The subgroup is not looked up. Its conjugacy classes are computed by conjugating every surviving operation by every other, the resulting class structure is compared against the thirteen character tables in use, and the match is by order and by class shape — how many classes there are and, for each, its kind, its rotation order and its size.

That comparison is checked to be unambiguous rather than assumed to be: if two tabulated groups ever shared a shape, that would be reported instead of the first match being taken. Among the thirteen, none do.

Six operations forming two classes of sizes $1$, $2$, $3$ with a threefold rotation among them is C3v. Four operations in four classes of size one, with a twofold rotation and two mirrors, is C2v. Those are the answers the calculation returns, and they are the groups a chemist would name by inspection — which is the point of asking a computation to do it rather than doing it by hand.

Where the extra lines come from #

Nine vibrations is $3N - 6$ with $N = 5$, and it holds for every row: substitution cannot change how many degrees of freedom a molecule has.

What changes is how they are grouped. Reducing the vibrational representation in each subgroup gives

• CH₄ and CD₄: a₁ ⊕ e ⊕ 2t₂ — one non-degenerate, one doubly, two triply. Four species, four frequencies, nine modes.
• CH₃D and CHD₃: 3a₁ ⊕ 3e — three non-degenerate and three doubly. Six frequencies, nine modes.
• CH₂D₂: 4a₁ ⊕ a₂ ⊕ 2b₁ ⊕ 2b₂ — every species one-dimensional. Nine frequencies, nine modes.

The degeneracies are the whole of the difference, and which degeneracies a group permits is fixed before any molecule is chosen, as degeneracy is a group theorem establishes. A triply degenerate mode is three motions at one frequency; break the symmetry that makes them equivalent and they separate. C2v has no degenerate species at all, so a C2v molecule can never have two vibrations at the same frequency except by accident.

Which operations are lost, and which survive #

The surviving sets are worth naming, because the naming is what the class-shape match reproduces.

CH₃D keeps the identity, the two rotations about the C–D axis, and the three mirrors containing it — six operations. Everything else in Td moves the deuterium onto a hydrogen, and eighteen operations go. The threefold axis that survives is the one through the substituted position, which is why the substituted position is the one the remaining symmetry is organised around.

CH₂D₂ keeps the identity, one twofold rotation, and two mirrors — four operations. The twofold axis is the one bisecting both the H–C–H and the D–C–D angles, and each mirror contains one of those pairs. Nothing with threefold character can survive, since a threefold rotation about any C–H direction sends one hydrogen onto a deuterium.

CD₄ keeps everything, because a uniform mass pattern is as symmetric as no substitution at all.

The counts $24$, $6$, $4$, $6$, $24$ are therefore not a smooth function of how much deuterium is present. Substituting one hydrogen removes three quarters of the group; substituting two removes another third; substituting the third restores the threefold axis about the remaining hydrogen; and substituting the fourth restores the lot. Symmetry is not a quantity that erodes with disturbance, which is the same lesson descent in symmetry draws from a different starting point.

The maximum a five-atom molecule can reach #

CH₂D₂’s nine distinct frequencies are the largest number the family can produce, and the bound is easy to state: a molecule with $N$ atoms has $3N - 6$ vibrations, and a group with no degenerate species gives each of them its own frequency. Nine modes, nine frequencies, and no room for more.

That makes the least symmetric isotopologue the most informative one for anybody trying to determine a force field, since every mode reports separately. It is also the hardest to assign, since nine lines have to be attributed to nine species rather than four lines to four. The trade between the two is a permanent feature of the subject: symmetry compresses a spectrum, which makes it easier to read and less able to constrain.

The two routes that have to agree #

The frequency counts above were derived from the group. They can also be obtained by a route with nothing in common with it, and the agreement is what makes the argument worth trusting.

The second route is the mass-weighted Hessian. Internal coordinates are generated from the geometry, a quadratic potential is written in them, second derivatives are taken by central differences, the result is mass-weighted with the substituted masses and diagonalised. Counting distinct eigenvalues gives $4$, $6$, $9$, $6$, $4$ down the family.

Four, six, nine, six, four, from a matrix of second derivatives that has never heard of a point group; and four, six, nine, six, four from a character sum over permutations that has never seen a force constant. The equality is checked row by row, and every row agrees.

That check is stronger than it looks, because the two routes fail differently. A force field with an error in it still diagonalises, still gives nine positive frequencies, and still satisfies the trace relations — exactly that happens with a force field giving two symmetry-equivalent bonds different constants, which passes every test but the species one — the same blindness normal modes are not bond stretches exploits from the other direction. Meanwhile a group-theoretic count that had lost an operation would give the right number of modes and the wrong number of frequencies.

What the infrared sees, and what the Raman sees #

Counting frequencies is one question and counting bands is another, since a frequency appears in a spectrum only if the corresponding species carries the right kind of function.

In Td only t₂ transforms as a translation, so only t₂ is infrared active: methane has two infrared fundamentals. All four species transform as quadratics or as the totally symmetric function, so all four are Raman active.

In C3v both a₁ and e are infrared active, so CH₃D has six. In C2v every species but a₂ carries a translation, so CH₂D₂ has eight infrared bands and nine Raman ones, with one mode visible only to the Raman experiment.

Two infrared bands become eight, with no force constant changed anywhere. The molecule that absorbs at two frequencies and the molecule that absorbs at eight have the same potential energy surface.

None of the five has a silent mode, and that is a computed result rather than a general rule: what an absence proves collects the molecules in this collection that do have modes invisible to both experiments, and benzene has nine of them.

Why the mixed isotopologues are the useful ones #

The mass-only view of substitution says that a heavier atom moves more slowly, so every frequency involving it falls. That is true and it is the content of the isotope shift is arithmetic, where the shifts are computed from the masses alone.

The symmetry view adds what the mass-only view cannot see, and the two together explain why partially substituted molecules earn their place in a spectroscopist’s work.

Fully substituted molecules test the force field. CD₄ has the same symmetry as CH₄ and its four frequencies are related to methane’s by the mass change alone, so agreement is a check on the constants.

Partially substituted molecules test the assignment. CH₃D’s six frequencies come from methane’s four by a splitting pattern that the group fixes: a t₂ must become a₁ ⊕ e, and the two products must be those particular linear combinations. A force field that reproduced all four of methane’s frequencies and got the mode shapes wrong would produce the wrong splitting, and the error would be visible in a way it is not in the parent.

That is why the product-rule check is worth having: it is the one test the force constants cancel out of entirely, so it examines the mass-weighting and the geometry alone.

Substituting one hydrogen localises motion that was spread over four equivalent bonds, which is the same phenomenon HOD shows in three modes rather than nine. The mode shapes change because the mass matrix does, and the symmetry change is what makes the count change with them.

The same argument on a second molecule #

Ammonia gives the pattern again with a different parent group, which is the check that nothing above is peculiar to a tetrahedron.

Ammonia is C3v with six operations and $3N - 6 = 6$ vibrations in species 2a₁ ⊕ 2e. Substituting one hydrogen leaves two operations — the identity and the mirror through the unique hydrogen — so the group is Cs, all six species are one-dimensional, and six frequencies appear where four did.

The numbers differ and the mechanism is identical: degeneracy is a group property, substitution removes group elements, and every degeneracy that loses its supporting operations resolves.

C₃ᵥ’s character table is generated by closing ammonia’s found operations under multiplication and sorting them into classes, and it is the table the mixed methane isotopologues’ vibrations are sorted into. Nothing about it knows that the molecule it came from was ammonia.

Drawn as displacement fields, CH₃D’s modes show the deuterium moving visibly less in the high-frequency ones, and the threefold symmetry that survives is visible in the shapes. The pictures and the line count are the same fact — a symmetry the substitution left standing.

Where the model stops #

The force field is fitted and harmonic. It was fitted to methane’s and CD₄’s eighteen frequencies together and reproduces them to $0.40$ per cent. Nothing here is an ab initio calculation of anything.

Real spectra have more lines than these counts. Overtones, combination bands, Fermi resonances and rotational structure all add features that a harmonic normal-mode count does not predict, and the counts above are counts of fundamentals.

The mass ratio used is exact and the reduced masses are not the whole story. Anharmonicity differs between isotopologues, so measured ratios depart from computed ones by amounts that grow with the frequency.

None of that touches the argument. The band counts are decided by which operations survive, that decision is exact, and the anharmonic corrections shift lines rather than creating or destroying species.

The same effect used on purpose #

A spectrum changing when only a mass does is presented here as a surprise, and it is the basis of the most widely used assignment technique in vibrational spectroscopy — because the change is not arbitrary and what it changes says which atoms move in which band.

The rule is simple and follows from the arithmetic. A mode’s frequency depends on the masses of the atoms that move in it, weighted by how far each moves. So substituting one atom shifts a band by an amount proportional to how much that atom participates: a mode dominated by the substituted atom shifts by nearly the full mass factor, and a mode in which it barely moves barely shifts at all.

So comparing two isotopologues’ spectra assigns bands to motions, with no calculation, no force field and no symmetry argument.

The procedure is used constantly and in two forms.

Substitute and watch which bands move. A band that shifts by the full amount when a hydrogen is replaced by deuterium is a hydrogen motion; one that does not shift involves the heavy atoms. That distinguishes an O–H stretch from a C=O stretch without knowing either frequency in advance.

Substitute at one position out of several. In a molecule with two chemically distinct hydrogens, deuterating one of them and not the other shifts the bands belonging to that position and leaves the others — which assigns bands to positions rather than merely to elements, and is how a complicated spectrum is taken apart.

The reason this works so well is the one the numbers here demonstrate. Nothing else changes. The potential energy surface is identical, so every force constant is the same, and the entire difference between the two spectra is the masses. That is a controlled experiment of a kind chemistry rarely offers: one variable changed exactly, everything else held exactly.

It also explains why the technique is trusted where a calculation is not. A computed assignment depends on a force field, and a force field has been shown to be underdetermined by its own spectrum. An isotopic assignment depends on the masses, which are known to more figures than anything else in the problem.

There is one limit worth stating, because it is where the method stops being a labelling exercise and starts needing the arithmetic. Two modes that shift by similar amounts are not distinguished by the shift, and modes mix: a substitution that changes the masses also changes which combinations of coordinates the modes are built from, so a band can shift because its own atom moved less rather than because that atom is lighter.

The remedy is the sum rule the whole treatment rests on. The product of all the frequencies of one isotopologue, divided by the product for the other, is fixed by the masses and the geometry alone — no force constants enter. So a proposed assignment can be checked against a number the molecule cannot argue with, and an assignment that satisfies every individual shift but fails the product is wrong somewhere it did not look wrong.

Who found it, and when #

Isotopic substitution as a tool for vibrational assignment dates from the 1930s, when deuterium became available in quantity following Harold Urey’s isolation of it in 1931. The Teller–Redlich product rule — a relation between the frequencies of two isotopologues that the force constants cancel out of — is from 1934 and 1935, published independently by Edward Teller and Otto Redlich, and it remains the sharpest check available on an assignment.

The symmetry half of the argument is older than the spectra it explains. Group-theoretic vibrational analysis was worked out in the late 1920s and early 1930s, principally by Eugene Wigner and by Placzek for the Raman case, and it was the first place in chemistry where a representation-theoretic count made a prediction nobody could reach by other means.

What is unusual about the isotopologue case is that it separates the two cleanly. Everything about the potential is shared between the rows of the table above, and everything that differs is the mass pattern and its group — so a phenomenon usually described as an interaction of structure and dynamics can be seen with the structure held perfectly still.

Still open: whether isotopologues fix a force field #

Frequencies have been counted against modes, spectra against structures, and environments against atoms. This counts the same molecule against itself.

What it leaves is a question that can be posed here and not answered: whether the frequencies of a molecule and of its isotopologues determine the force field. The force field is not in the spectrum shows that one molecule’s spectrum does not, and that the isotopologue does discriminate among fields the parent cannot — so the counting argument here and the underdetermination argument there are the two halves of what substitution is for.