Vibrational (IR) spectroscopy

Result

Vibrational (IR) spectroscopy

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Statement

The harmonic oscillator and vibrational selection rules.

Why it matters

rotational-spectroscopy already established the quantised-energy-level, resonance-absorption logic used throughout this spectroscopy unit, for the lowest-energy motion a molecule can undergo. Vibrational (infrared) spectroscopy applies that same logic one level up the energy hierarchy, to bond stretching and bending, and is one of the single most widely used routine techniques for identifying functional groups, making it a load-bearing companion to uv-vis-electronic and nmr-chemical-shift within the unit's overall structure-elucidation toolkit.

Because a bond's vibrational frequency depends predictably on both its stiffness and the mass of the atoms it joins, characteristic "group frequencies" recur at roughly the same wavenumber across many different molecules, which is what makes a routine IR spectrum such a fast, practical diagnostic for the presence or absence of specific functional groups.

Hypotheses

The harmonic oscillator model: the bond's potential energy is approximated as a simple parabola, \(V(x)=\tfrac12kx^2\), around the equilibrium bond length.This is a good approximation only near the bottom of the true, anharmonic potential well; it cannot describe bond dissociation at high vibrational energy, since a parabola, unlike the true potential, never flattens out or allows the bond to break. A vibration is IR-active only if it produces a change in the molecule's dipole moment as the bond stretches or bends.A vibration that leaves the dipole moment completely unchanged — the symmetric stretch of a homonuclear diatomic, or certain symmetric modes of more complex, centrosymmetric molecules — is IR-inactive regardless of how energetically favourable that vibration is. The true molecular potential is anharmonic; overtone transitions (\(\Delta v=\pm2,\pm3,\ldots\)), strictly forbidden under the harmonic selection rule, appear weakly precisely because of this anharmonicity, and bond dissociation itself is possible at sufficiently high vibrational energy, which the harmonic model cannot describe at all by construction.

Proof

E_v=\left(v+\tfrac12\right)h\nu_{\text{vib}}, \qquad v=0,1,2,\ldots

Solving the harmonic-oscillator Schrödinger equation gives evenly spaced vibrational energy levels with a non-zero zero-point energy \(\tfrac12h\nu_{\text{vib}}\) present even in the lowest state \(v=0\), a general quantum-mechanical consequence of confinement. A

\nu_{\text{vib}}=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}

The classical vibration frequency from the harmonic model depends on the bond's force constant \(k\) (a measure of bond stiffness) and the reduced mass \(\mu\) of the two vibrating atoms. A

\Delta v=\pm1, \quad\text{and only if the vibration changes the dipole moment (Hypotheses).}

For the harmonic oscillator, allowed transitions satisfy \(\Delta v=\pm1\) exactly, and only occur at all for modes that change the molecular dipole moment; together these two conditions determine which vibrational modes of a given molecule appear in its IR spectrum. A

k\text{ increases with bond order}; \nu_{\text{vib}}\propto\sqrt{k/\mu} \Rightarrow \text{characteristic "group frequencies."}

Because \(k\) increases with bond order and \(\nu_{\text{vib}}\) scales with \(\sqrt{k/\mu}\), stronger and lighter-atom bonds vibrate at systematically higher wavenumber, the physical basis for transferable group frequencies (e.g. O-H, C-H, C=O, C≡N) recurring at roughly the same wavenumber across many different molecules. A

\text{The fingerprint region contains dense, largely molecule-specific combinations of coupled skeletal vibrations.}

The lower-wavenumber third of the mid-IR range is generally too complex to assign mode-by-mode, but its overall pattern is close to unique to a given compound, valuable for comparison against a reference spectrum even without full assignment. B

Result

\nu_{\text{vib}}=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}, \qquad E_v=\left(v+\tfrac12\right)h\nu_{\text{vib}}, \qquad \Delta v=\pm1

Reading. A bond's vibrational frequency is set jointly by its stiffness and the mass of the atoms it joins, and only certain, dipole-changing modes absorb IR radiation at all.

Scope. The harmonic model and its simple selection rule apply well near the bottom of the potential well, at low \(v\), and break down at higher vibrational energy where anharmonicity (Hypotheses) becomes significant.

Corollaries & converses

Group frequencies (Step 4) are the practical basis of routine IR structural identification, complementing rather than duplicating the electronic-transition information from uv-vis-electronic and the nuclear-environment information from nmr-chemical-shift.
Isotopic substitution changes \(\mu\) substantially without appreciably changing \(k\) (bonding is nearly unaffected by which isotope is present), exactly as in rotational-spectroscopy's identical use of isotopic substitution to probe structure independently of chemistry, predictably shifting the observed vibrational frequency.
The same resonance-absorption framework used here and in rotational-spectroscopy differs essentially only in the energy scale of the quantised motion probed, which is why a real vibration-rotation spectrum shows rotational fine structure built on top of a single vibrational transition.

Fails without

Drop the dipole-change requirement (Hypotheses): a vibrational mode that leaves the dipole moment unchanged, such as the symmetric stretch of carbon dioxide, remains completely IR-inactive no matter how strongly it is otherwise excited thermally or mechanically — it simply produces no absorption signal at all.
Treat the harmonic model as exact at high \(v\) (Hypotheses' \(t3\)): the true, anharmonic potential allows bond dissociation and produces weak overtone bands the strict harmonic selection rule forbids; a purely harmonic treatment cannot describe either effect and increasingly overestimates the true energy-level spacing as \(v\) grows.

Common errors

Assuming every vibrational mode of a molecule is automatically IR-active; the dipole-change requirement rules out several common, symmetric modes.
Confusing bond order with bond strength alone when predicting relative stretching wavenumber, without also accounting for the reduced mass \(\mu\) (e.g. C-H stretches appear at much higher wavenumber than C-C stretches, driven mostly by the much smaller \(\mu\)).
Attempting to assign every single fingerprint-region absorption to one isolated bond vibration, when many arise from strongly coupled combinations of several skeletal motions.
Forgetting the zero-point energy term in \(E_v\), and incorrectly treating \(v=0\) as a state of zero vibrational energy.

Discussion

Infrared radiation was first observed by William Herschel in 1800, in the course of measuring the heating effect of light beyond the red end of the visible spectrum; its systematic chemical application — quantised molecular vibrations, harmonic-oscillator interpretation, and functional-group correlation tables — developed much later, through the 20th century, alongside the broader development of quantum mechanics and, eventually, practical benchtop FTIR instrumentation.

Common misconception: that fingerprint-region IR absorptions can and should each be assigned to one specific, isolated bond vibration in the same way the higher-wavenumber group-frequency region often can. As Step 5 notes, fingerprint-region absorptions frequently result from strongly coupled, delocalised skeletal motions involving many bonds simultaneously, and the region's diagnostic value lies mainly in its overall pattern, not mode-by-mode assignment.

Worked examples

\text{HCl: } k\approx480\,\text{N/m}, \quad \mu=\frac{(1.008)(34.97)}{1.008+34.97}\,\text{u}\approx0.980\,\text{u}=1.63\times10^{-27}\,\text{kg}

Using standard, widely tabulated values for HCl's force constant and the reduced mass computed from its two atomic masses, Step 2's formula is ready to be evaluated numerically. A

\nu_{\text{vib}}=\frac{1}{2\pi}\sqrt{\frac{480}{1.63\times10^{-27}}}\approx8.6\times10^{13}\,\text{Hz} \;\Rightarrow\; \tilde\nu=\nu/c\approx2880\,\text{cm}^{-1}

Substituting into Step 2 and converting frequency to wavenumber gives a predicted fundamental stretch around \(2880\,\text{cm}^{-1}\), matching HCl's well-established, experimentally observed fundamental vibration near \(2886\,\text{cm}^{-1}\) closely. A

\tilde\nu(\text{H-Cl})\approx2880\,\text{cm}^{-1}, \quad\text{from } k\text{ and }\mu\text{ alone}

Reading. The harmonic-oscillator formula, given only a bond's force constant and the atomic masses involved, predicts a vibrational wavenumber matching the real, experimentally measured value closely.

Scope. The identical calculation applies to any diatomic with a known or estimated force constant.

Problems

Explain, without recalculating fully, why the C-D stretching wavenumber is lower than the corresponding C-H stretching wavenumber for the same carbon framework.
Solution
Deuterium substitution roughly doubles the reduced mass \(\mu\) while leaving the bond's force constant \(k\) essentially unchanged (bonding is nearly unaffected by nuclear mass). Since \(\nu_{\text{vib}}\propto1/\sqrt{\mu}\) (Step 2), a larger \(\mu\) gives a lower vibrational frequency, so C-D stretches appear at systematically lower wavenumber than the corresponding C-H stretch.
Explain why the symmetric stretch of \(\text{CO}_2\) is IR-inactive, referencing the Hypotheses.
Solution
In the symmetric stretch, both C=O bonds lengthen and shorten in phase, so the two bond dipoles' changes cancel exactly by the molecule's linear, centrosymmetric geometry, leaving the overall molecular dipole moment (zero throughout, for this non-polar linear molecule) completely unchanged. Since IR activity requires a changing dipole moment (Hypotheses), this mode produces no IR absorption at all, regardless of how strongly it is excited.
A C=O bond (higher bond order) and a C-C bond (lower bond order) are compared, with masses of the bonded atoms taken as roughly similar. Predict which shows the higher stretching wavenumber, and justify using Step 4.
Solution
The C=O bond, having the higher bond order, has the larger force constant \(k\) (a stiffer bond); with reduced masses assumed roughly comparable, \(\nu_{\text{vib}}\propto\sqrt{k/\mu}\) (Step 2) predicts the C=O stretch appears at the higher wavenumber of the two, consistent with the general group-frequency trend of Step 4.