Vibrational IR Spectroscopy: Principles

Hold a chemical bottle up to an infrared beam and the spectrum comes back scarred with dips. Each dip marks a bond rocking back and forth at a frequency the light supplied exactly. This is vibrational infrared (IR) spectroscopy, and it turns molecular motion into a readable fingerprint. The technique matters because it is fast, non-destructive, and universal. Every molecule with more than one atom vibrates, and most vibrations interact with infrared light. A chemist can identify a carbonyl, confirm a synthesis, or monitor a reaction in real time, often in under a minute. This article builds the physical picture of vibrational absorption. You will see why only some vibrations absorb IR light. You will learn the energy scale that places vibrations in the infrared. You will also learn how to estimate absorption frequencies for real bonds.

Key takeaways

IR absorption promotes a molecule between vibrational energy levels using photons in the 200 to 4000 cm⁻¹ range.
A vibration is IR-active only if it changes the molecular dipole moment during the motion.
Vibrational energy spacings, roughly 10⁻²⁰ J per quantum, are far larger than typical rotational spacings but far smaller than electronic gaps.
Bond stiffness and reduced mass together set the vibrational frequency, giving each functional group a characteristic wavenumber.
Symmetric, non-polar vibrations (like N≡N stretching) are IR-silent even though the bond clearly vibrates.
Key idea: A photon is absorbed only when its energy matches a vibrational spacing and the vibration itself shakes the dipole moment.

The core insight

Infrared absorption happens when an oscillating electric field couples to an oscillating molecular dipole, so a vibration must both match the photon's energy and change the dipole moment during its motion for absorption to occur at all.

The physical picture

A diatomic molecule is not a rigid dumbbell. The bond behaves like a spring, stretching and compressing about an equilibrium length. Quantum mechanics restricts this motion to a discrete ladder of vibrational energy levels, evenly spaced (to a first approximation) by an amount set by the bond's stiffness and the masses of the atoms involved. This ladder, and the harmonic-oscillator model that generates it, is developed in full in the next article. Here we only need one fact: transitions between adjacent rungs of that ladder correspond to photon energies that fall squarely in the infrared region of the electromagnetic spectrum, roughly 2.5 to 25 micrometers in wavelength, or 400 to 4000 cm⁻¹ in wavenumber. But energy matching alone does not guarantee absorption. Light is an oscillating electric field. For a molecule to absorb a photon and jump between vibrational states, that oscillating field must be able to do work on the molecule. It does so by pushing and pulling on the molecule's dipole moment. If the vibration itself causes the dipole moment to oscillate at the same frequency as the incoming light, the field and the molecule can exchange energy resonantly. If the vibration leaves the dipole moment completely unchanged, no such coupling exists, and the transition is said to be IR-forbidden, regardless of how much energy the photon carries. This is why homonuclear diatomics like N₂, O₂, and Cl₂ are invisible in IR spectroscopy. Their bonds vibrate exactly as vigorously as any polar bond, but stretching a symmetric, non-polar bond leaves the dipole moment at zero throughout the motion. There is nothing for the light to grab onto. Contrast this with carbon monoxide, CO, whose vibration modulates a genuine (if small) permanent dipole moment, making it strongly IR-active and observable even in the tenuous atmosphere of Mars. For polyatomic molecules the picture generalizes: each independent way the atoms can vibrate is called a normal mode, and each normal mode is checked separately for whether it changes the dipole moment. Carbon dioxide is instructive here. Its symmetric stretch, in which both oxygens move outward together, leaves the linear molecule's dipole moment at zero by symmetry: IR-silent. Its asymmetric stretch and its bending modes both break that symmetry and create a fluctuating dipole: IR-active. This is precisely why CO₂ is a potent greenhouse gas through its bend and asymmetric stretch, despite its symmetric stretch being spectroscopically dark.

Figure

Dipole moment during a stretching vibration: HCl versus N₂

The second useful picture is the energy-scale ladder that places vibrational transitions among their spectroscopic neighbors. Rotational transitions are the closest in the microwave and far-infrared; electronic transitions sit far above in the visible and ultraviolet.

Figure

Energy scale of molecular transitions

The core relations

\Delta E_{vib} = h\nu = hc\tilde{\nu}

where $\tilde{\nu}$ is the vibrational wavenumber in cm⁻¹, $h$ is Planck's constant, and $c$ is the speed of light.

\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}

where $k$ is the bond force constant (N/m) and $\mu$ is the reduced mass of the vibrating pair.

\mu = \frac{m_1 m_2}{m_1 + m_2}

where $m_1, m_2$ are the masses of the two atoms joined by the bond.

\text{IR active} \iff \left(\frac{\partial \mu_{dipole}}{\partial Q}\right)_{Q=0} \neq 0

where $Q$ is the normal-mode coordinate and $\mu_{dipole}$ is the molecular dipole moment.

Derivation AFrom bond stiffness to an infrared wavenumber

Goal. Show that a bond modeled as a classical spring, combined with the quantum condition for absorption, predicts wavenumbers that land in the observed infrared window, and derive the working formula chemists use every day.

Step 1: model the bond as a Hookean spring. Treat two bonded atoms as point masses connected by a spring with force constant $k$ . Classical mechanics for two masses connected by a spring reduces to the motion of a single effective mass, the reduced mass $\mu$ , oscillating about the equilibrium bond length. The classical angular frequency of that oscillator is

\omega = \sqrt{\frac{k}{\mu}}

Step 2: convert to a frequency and then a wavenumber. Since $\omega = 2\pi\nu$ , the ordinary vibrational frequency is

\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}

Spectroscopists prefer wavenumber $\tilde{\nu} = \nu/c$ , giving

\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}

Step 3: connect to quantized energy levels. Quantum treatment of this same spring (the harmonic oscillator, developed fully in the next article) shows the allowed vibrational energies are evenly spaced by $h\nu$ , so the photon energy needed for the fundamental transition is exactly

\Delta E = h\nu = hc\tilde{\nu}

The classical spring frequency and the quantum energy spacing turn out to be governed by the same $\sqrt{k/\mu}$ dependence, which is why the simple mechanical model already predicts the right spectral region.

Step 4: plug in realistic numbers to check the region. Typical single-bond force constants run 400 to 500 N/m; typical light-atom reduced masses run roughly 1 to 15 u ( $1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$ ). Taking $k = 500\ \text{N/m}$ and $\mu = 1.63\times10^{-27}\ \text{kg}$ (an H–Cl-like reduced mass).

\tilde{\nu} = \frac{1}{2\pi (3.00\times10^{10}\ \text{cm/s})}\sqrt{\frac{500\ \text{N/m}}{1.63\times10^{-27}\ \text{kg}}}

\tilde{\nu} \approx 2990\ \text{cm}^{-1}

This lands right in the mid-infrared, consistent with real HCl absorption near 2886 cm⁻¹.

Step 5: state the selection requirement. The above gives the energy condition. A second, independent condition governs whether the transition is observed at all: the dipole moment must change during the motion. Expanding the dipole moment in a Taylor series in the normal coordinate $Q$ about equilibrium,

\mu_{dipole}(Q) \approx \mu_{dipole}(0) + \left(\frac{\partial \mu_{dipole}}{\partial Q}\right)_0 Q + \cdots

the oscillating electric field of light couples to the linear term. If that derivative vanishes by symmetry, as it does for the stretch of a homonuclear diatomic, there is no term through which the field can drive the transition regardless of how well $\Delta E$ matches $hc\tilde{\nu}$ .

\boxed{\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \quad \text{observed only if} \quad \left(\frac{\partial \mu_{dipole}}{\partial Q}\right)_0 \neq 0}

Worked examples

Example 1: Predicting the C=O stretch wavenumber of carbon monoxide. Carbon monoxide has a force constant of $k = 1860\ \text{N/m}$ (a typical value for a strong double/triple-character bond). Compute the expected fundamental stretching wavenumber. Step 1 — reduced mass. Carbon: $m_C = 12.00\ \text{u}$ . Oxygen: $m_O = 16.00\ \text{u}$ .

\mu = \frac{(12.00)(16.00)}{12.00+16.00}\ \text{u} = \frac{192.0}{28.00}\ \text{u} = 6.857\ \text{u}

Convert to kilograms using $1\ \text{u} = 1.6605\times10^{-27}\ \text{kg}$ :

\mu = 6.857 \times 1.6605\times10^{-27}\ \text{kg} = 1.1388\times10^{-26}\ \text{kg}

Step 2 — apply the wavenumber formula.

\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} = \frac{1}{2\pi (2.998\times10^{10}\ \text{cm/s})}\sqrt{\frac{1860\ \text{N/m}}{1.1388\times10^{-26}\ \text{kg}}}

Inside the square root: $1860 / 1.1388\times10^{-26} = 1.633\times10^{29}\ \text{s}^{-2}$ , so the square root is $4.041\times10^{14}\ \text{s}^{-1}$ .

\tilde{\nu} = \frac{4.041\times10^{14}\ \text{s}^{-1}}{1.884\times10^{11}\ \text{cm/s}} = 2145\ \text{cm}^{-1}

Step 3 — sanity check. The experimental CO fundamental is 2143 cm⁻¹, essentially an exact match, confirming both the force constant estimate and the model. This wavenumber corresponds to a photon energy of

\Delta E = hc\tilde{\nu} = (6.626\times10^{-34}\ \text{J s})(2.998\times10^{10}\ \text{cm/s})(2145\ \text{cm}^{-1}) = 4.26\times10^{-20}\ \text{J}

per molecule, or 25.6 kJ/mol — small compared to a bond energy (~1070 kJ/mol for CO) but far larger than $k_BT$ at room temperature (4.1×10⁻²¹ J), which is exactly why most molecules sit in the vibrational ground state at 298 K, as later discussed for Boltzmann populations. Since CO carries a small permanent dipole (about 0.11 D) that changes as the bond stretches, this mode is IR-active, and the prediction can be checked directly against a real gas-phase IR spectrum.

Example 2: Distinguishing IR activity in the normal modes of CO₂. Carbon dioxide, O=C=O, is linear with three atoms, so it has $3N - 5 = 4$ vibrational normal modes: the symmetric stretch, the asymmetric stretch, and a doubly degenerate bend (two perpendicular bending motions of equal energy). Task: determine which modes are IR-active, and use the observed asymmetric-stretch wavenumber (2349 cm⁻¹) to estimate the effective C=O force constant, then compare with the fundamental force constant obtained from treating CO₂ as an isolated diatomic-like oscillator for that mode. Step 1 — symmetry analysis of the dipole moment. In the symmetric stretch, both C=O bonds lengthen and shorten in phase; the molecule remains linear and symmetric at every instant, so the two bond dipoles always cancel. The derivative of the total dipole moment with respect to this coordinate is zero: IR-inactive. (It is, however, Raman-active — a preview of the complementary technique covered elsewhere.) In the asymmetric stretch, one C=O bond lengthens while the other shortens. The molecule's instantaneous dipole moment no longer cancels, oscillating along the molecular axis: IR-active. Observed at 2349 cm⁻¹, this is in fact the strongest CO₂ absorption and the one exploited by non-dispersive infrared CO₂ gas sensors. In the bending mode, the molecule flexes away from linearity, producing a net dipole moment perpendicular to the former axis: IR-active. Observed near 667 cm⁻¹, this is the mode primarily responsible for CO₂'s absorption of outgoing terrestrial infrared radiation in the atmosphere. Step 2 — estimate an effective force constant from the asymmetric stretch. Treat one C=O oscillator using the reduced mass of carbon against oxygen, $\mu = 1.1388\times10^{-26}$ kg (as in Example 1), purely as an approximation (the true motion is more complex because both bonds move and couple to each other, but this comparison illustrates the reasoning).

k = \mu (2\pi c \tilde{\nu})^2 = (1.1388\times10^{-26}\ \text{kg})\left[2\pi(2.998\times10^{10}\ \text{cm/s})(2349\ \text{cm}^{-1})\right]^2

Computing the bracketed angular frequency: $2\pi \times 2.998\times10^{10} \times 2349 = 4.425\times10^{14}\ \text{s}^{-1}$ . Squaring gives $1.958\times10^{29}\ \text{s}^{-2}$ .

k = (1.1388\times10^{-26}\ \text{kg})(1.958\times10^{29}\ \text{s}^{-2}) = 2230\ \text{N/m}

Step 3 — sanity check. This effective force constant (2230 N/m) is noticeably larger than the isolated CO force constant (1860 N/m from Example 1), consistent with the C=O bonds in CO₂ carrying somewhat more double-bond character and being stiffer than in free carbon monoxide. The qualitative ordering (bend, softest, lowest wavenumber; asymmetric stretch, stiffest, highest wavenumber) matches intuition: bending motions distort weak angular restoring forces, while bond stretches fight the strong radial force constant directly.

Limiting cases & common traps

Situation	What actually happens
Homonuclear diatomic (N₂, O₂, H₂, Cl₂)	Vibrates normally but is IR-silent; zero dipole derivative by symmetry
Heteronuclear diatomic (HCl, CO, NO)	Always IR-active if it has any permanent dipole that changes with bond length
Centrosymmetric mode of a polyatomic (CO₂ symmetric stretch)	Can be IR-silent even though other modes of the same molecule are active
Very stiff bond, light atoms (C–H, ~3000 cm⁻¹)	High wavenumber, since $\tilde\nu \propto \sqrt{k/\mu}$ and small $\mu$ pushes frequency up
Weak bond, heavy atoms (C–I, ~500 cm⁻¹)	Low wavenumber, near the edge of typical mid-IR instruments
Overtone and combination bands	Weak extra peaks at roughly 2× or sums of fundamentals; foreshadows anharmonicity

The chemist behind it

William Weber Coblentz spent the early 1900s painstakingly recording the infrared spectra of hundreds of pure organic compounds using a rock-salt prism spectrometer he built largely by hand at the US National Bureau of Standards. His catalogs, published starting in 1905, first demonstrated that specific functional groups reliably absorbed at reproducible wavelengths — the empirical seed of the "characteristic group frequency" concept that underlies every modern IR functional-group chart. Coblentz worked without the quantum theory that would later explain his results; the harmonic-oscillator model and dipole-derivative selection rule came decades afterward. Yet his patient, systematic measurements gave the field the first large body of reliable data against which any later theory had to be checked.

Origins · 1 minWilliam W. Coblentz and how this was discovered

William W. Coblentz

1873-1962 · Pioneer of quantitative infrared spectroscopy at the US Bureau of Standards

Ask the tutor: How does anharmonicity change these predicted wavenumbers, and where do overtone bands actually show up in a real spectrum?

Why does infrared light specifically excite vibrations rather than rotations or electronic transitions?

Infrared photons carry energies of roughly 10⁻²¹ to 10⁻¹⁹ J, which happens to match typical vibrational level spacings almost exactly. Rotational spacings are one to three orders of magnitude smaller, matching microwave photons instead. Electronic spacings are one to two orders of magnitude larger, matching visible or ultraviolet photons. The match between photon energy and level spacing is what determines which type of transition a given spectral region can drive; nothing about "vibration" is special to infrared beyond this coincidence of energy scales.

Can a molecule with zero permanent dipole moment still show an IR spectrum?

Yes, as long as at least one of its vibrations creates a temporary dipole moment during the motion. Carbon dioxide has zero permanent dipole at equilibrium yet shows strong IR bands from its bending and asymmetric-stretching modes, both of which distort the molecule away from its symmetric equilibrium shape. Only vibrations that preserve the zero-dipole symmetry at every instant, like the symmetric stretch, remain IR-silent.

How is the force constant k actually measured if it is used to predict the spectrum?

In practice the logic runs backward from how it is presented pedagogically. Chemists measure the fundamental wavenumber experimentally with an IR spectrometer, then solve the same formula for $k$ given the known reduced mass. The force constant obtained this way is a genuinely useful, transferable descriptor of bond stiffness, and comparing $k$ values across related bonds (C–C, C=C, C≡C) gives a quantitative measure of bond order and strength.

Why do bonds to hydrogen, like C–H, N–H, and O–H, always show up above 2700 cm⁻¹?

Wavenumber scales as the square root of $k/\mu$ , and hydrogen's small mass dominates the reduced mass for any X–H bond regardless of what X is. Even though C–H, N–H, and O–H have different force constants, their reduced masses are all close to 1 u, keeping $\mu$ small and $\tilde\nu$ large. This is why the entire X–H stretching region clusters near 2700 to 3700 cm⁻¹, distinctly separated from heavier-atom stretches below 2000 cm⁻¹.

How does this topic connect to overtones and anharmonicity, covered next?

Everything above assumes a perfectly harmonic spring, giving evenly spaced levels and only fundamental transitions. Real bonds are anharmonic: they dissociate at large stretch, and their energy levels converge at higher quantum numbers. This produces weak overtone and combination bands at roughly (but not exactly) integer multiples of the fundamental wavenumber, and it slightly shifts the fundamental itself. The next article develops this quantitative correction explicitly.