Lesson 5 · 30 min
Infrared spectroscopy: molecular vibrations
Bonds are springs and infrared spectra record their vibrations. Why does CO₂ absorb infrared while N₂ does not? Why are symmetric stretches invisible? And how do group frequencies reveal what a molecule contains?
In this lesson
- Link a vibration’s wavenumber to bond stiffness and atomic masses with the harmonic model ν̃ = (1/2πc)√(k/μ)
- Count a molecule’s vibrations (3N − 6 or 3N − 5) and describe stretches and bends
- Apply the selection rule — only vibrations that change the dipole moment absorb infrared — and explain IR-silent symmetric stretches and the complementary Raman spectrum
- Use group-frequency regions and the fingerprint region to identify functional groups and carbonate, sulfate and nitrate
N₂ and O₂ make up 99 % of the air and are completely transparent to the infrared radiation the Earth gives off; CO₂, at about 0.04 %, and a little water vapour absorb a great deal of it — that is why there is a greenhouse effect. The same physics lets chemists “see” the skeleton of a molecule: an infrared spectrum records how a molecule vibrates. But why does N₂ not absorb while CO₂ does? Neither has a permanent dipole moment.
Predict first
Which of these gases absorbs infrared light?
Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.
Bonds are springs
Treat the bond between two atoms as a spring: the stiffer the bond (the larger its force constant k) and the lighter the atoms (the smaller the reduced mass μ), the faster it vibrates. Infrared spectra are plotted in wavenumber ν̃ (waves per centimetre, cm⁻¹), which is proportional to frequency and energy; wavelength follows from λ/µm = 10⁴/(ν̃/cm⁻¹). The mid-infrared runs from about 4000 to 400 cm⁻¹ (2.5–25 µm).
- vibrational wavenumber, cm⁻¹
- speed of light (in cm/s)
- force constant of the bond, N/m
- reduced mass
A molecule of N atoms has 3N degrees of freedom; subtract 3 translations and 3 rotations (only 2 for a linear molecule) and 3N − 6 (3N − 5 if linear) normal vibrations remain: 3 for H₂O, 4 for CO₂ (its bend is doubly degenerate), 9 for CH₄. Vibrations are stretches (symmetric or asymmetric), which change bond lengths, and bends, which change angles.
Selection rule: a vibration absorbs infrared light only if it changes the dipole moment; it appears in the Raman spectrum only if it changes the polarisability. For molecules with a centre of symmetry, such as CO₂, no vibration is both IR and Raman active (the rule of mutual exclusion), so the two spectra complement each other. Highly symmetric ions behave the same way: the symmetric stretches ν₁ of CO₃²⁻, NO₃⁻ and SO₄²⁻ are invisible in the infrared yet are the strongest lines of their Raman spectra.
| Ion | ν₁ sym. stretch | ν₂ | ν₃ asym. stretch | ν₄ |
|---|---|---|---|---|
| 1063 (R) | 879 (IR) | 1415 (IR, R) | 680 (IR, R) | |
| 1050 (R) | 831 (IR) | 1390 (IR, R) | 720 (IR, R) | |
| 983 (R) | 450 (R) | 1105 (IR, R) | 611 (IR, R) |
Although a vibration involves the whole molecule, X–H, C=O and C≡N vibrations stay largely confined to one bond and appear at similar wavenumbers in every molecule — group frequencies: O–H 3200–3600 cm⁻¹ (broad and rounded when hydrogen-bonded), N–H 3300–3500, C–H 2850–3100 (above 3000 for alkenes and aromatics), C≡N and C≡C 2100–2260, C=O 1670–1780 (ketones about 1715), C=C 1600–1680 and C–O 1000–1300. Below 1500 cm⁻¹ lies the fingerprint region, vibrations of the whole skeleton that differ for every compound and are matched against reference spectra.
Worked example
Force constants of C=O and C–O
A ketone’s C=O stretch is at 1715 cm⁻¹ and an alcohol’s C–O stretch near 1100 cm⁻¹. Treat each as an isolated diatomic oscillator (m_C = 12.000 u, m_O = 15.995 u) and estimate both force constants.
Concept check
The infrared spectrum of calcite (solid calcium carbonate) shows nothing at 1086 cm⁻¹ — yet that is the strongest line in its Raman spectrum (the same vibration of free CO₃²⁻ in solution is at 1063 cm⁻¹). Why?
Try the calculation
H³⁵Cl absorbs at 2886 cm⁻¹. With the same force constant, where does D³⁵Cl absorb? (Atomic masses: H 1.0078 u, D 2.0141 u, ³⁵Cl 34.969 u.)
Key takeaways
- ν̃ = (1/2πc)√(k/μ): stiffer bonds and lighter atoms vibrate at higher wavenumbers; isotopic substitution shifts bands as √μ.
- An N-atom molecule has 3N − 6 (linear 3N − 5) normal vibrations: stretches and bends.
- Only vibrations that change the dipole moment absorb infrared; those that change the polarisability appear in the Raman spectrum; with a centre of symmetry the two are mutually exclusive.
- Group frequencies (O–H, N–H, C–H, C=O …) identify functional groups; the fingerprint region below 1500 cm⁻¹ identifies the whole molecule.
Data and references
- T. Shimanouchi, Tables of Molecular Vibrational Frequencies, Consolidated Volume I, NSRDS-NBS 39 (1972), via the NIST Chemistry WebBook.
- K. Nakamoto, Infrared and Raman Spectra of Inorganic and Coordination Compounds, Part A, 6th ed., Wiley (2009).
- S. Gunasekaran, G. Anbalagan, S. Pandi, “Raman and infrared spectra of carbonates of calcite structure”, J. Raman Spectrosc. 37, 892 (2006) (calcite ν₁ near 1086 cm⁻¹, Raman only).
- K. P. Huber, G. Herzberg, Constants of Diatomic Molecules (1979), via the NIST Chemistry WebBook (HCl ωₑ, ωₑxₑ).
- D. L. Pavia, G. M. Lampman, G. S. Kriz, J. R. Vyvyan, Introduction to Spectroscopy, 5th ed., Cengage (2015), ch. 2.
- P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed., OUP (2018), topics 11C–11D.