Lesson 7 · 40 min

Counting vibrations with group theory: infrared and Raman

Of a molecule’s 3N − 6 vibrations, which absorb infrared and which appear in the Raman spectrum? Turn symmetry operations into characters and reduce them: without solving a single equation of motion you can predict how many bands to expect — and tell cis from trans isomers.

In this lesson

  • Write the characters of Γ₃N as (atoms left in place) × (±1 + 2 cos θ)
  • Reduce Γ₃N into irreducible representations with the reduction formula, and remove translations and rotations to get Γ_vib
  • Use the character table to decide which vibrations are infrared and Raman active, and apply the rule of mutual exclusion
  • Tell cis from trans isomers by counting their stretching bands

Methane has 5 atoms and 9 vibrational modes, yet its gas-phase infrared spectrum shows only two fundamental bands, at 3019 and 1306 cm⁻¹. The other two vibrations, at 2917 and 1534 cm⁻¹, appear only in the Raman spectrum. Sulfur hexafluoride has 15 vibrational modes but only 2 infrared fundamentals and 3 Raman lines — and no line appears in both. What decides which vibrations are “visible”? The answer lies entirely in the molecule’s symmetry, and it can be worked out without solving a single equation of motion.

Predict first

How many infrared fundamental bands do methane’s (TdT_d) 9 vibrational modes give?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Characters: symmetry operations as numbers

Give every atom three small arrows (displacements along x, y and z) — 3N in all for an N-atom molecule. A symmetry operation turns these arrows into combinations of one another and can be written as a 3N × 3N matrix; group theory needs only its character, the sum of its diagonal elements. That is easy to find: an atom moved elsewhere contributes only zeros to the diagonal, and an atom left in place contributes the trace of a 3 × 3 rotation matrix, ±1+2cos⁡θ\pm 1 + 2\cos\theta (+ for proper rotations, − for reflections, inversion and improper rotations). So

  • E: 3 per unmoved atom; C2C_2: −1; C3C_3: 0; C4C_4: 1;
  • σ: 1; i: −3; S4S_4: −1; S6S_6: 0.

Write out Γ₃N class by class and split it with the reduction formula into the irreducible representations of the character table. Three of them are translations (transforming like x, y and z), three are rotations (like RxR_x, RyR_y, RzR_z; a linear molecule has only two), and what is left is the vibrations, Γ_vib.

χ3N(R)=Nunmoved(R) (±1+2cos⁡θ),ni=1h∑Rχi(R) χ3N(R)\chi_{3N}(R) = N_\text{unmoved}(R)\,(\pm 1 + 2\cos\theta),\qquad n_i = \frac{1}{h}\sum_R \chi_i(R)\,\chi_{3N}(R)
The character of Γ₃N, and the reduction formula: how many times irreducible representation i occurs. The sum runs over all h operations of the group (operations in one class share a character, so multiply by the class size).
NunmovedN_\text{unmoved}
number of atoms that R leaves in place
θ\theta
rotation angle (0° for a reflection, 180° for inversion)
hh
order of the group
χi(R)\chi_i(R)
character of irreducible representation i

Worked example

The three vibrations of water

Water belongs to C2vC_{2v} and lies in the yz plane. Find Γ_vib and decide which vibrations are infrared and Raman active.

Selection rules: which vibrations are seen

To absorb infrared light a vibration must change the dipole moment, which is a vector — so only vibrations in the same irreducible representation as x, y or z are infrared active. Raman scattering needs a change in polarisability, which transforms like the quadratic functions x2x^2, xyxy …, so only vibrations in the same representation as a quadratic function are Raman active.

In a molecule with a centre of inversion, x, y and z are all u (they change sign under inversion) and the quadratic functions are all g, so no vibration can appear in both spectra — the rule of mutual exclusion. The symmetric stretch of CO₂ is seen only in the Raman spectrum (the Fermi-resonance doublet at 1285 and 1388 cm⁻¹); the antisymmetric stretch and the bend only in the infrared. Conversely, if several infrared and Raman lines coincide, the molecule cannot have a centre of inversion.

Normal modes · the dipole rule · group frequencies

Infrared spectra: molecules in motion

COO
This vibration leaves the dipole moment unchanged: infrared cannot see it; only Raman spectroscopy can.

Normal modes (4 = 3N − 5)

In the Raman spectrum it splits into 1285 and 1388 cm⁻¹ (Fermi resonance).

ν1 13334000350030002500200015001000500Wavenumber ν̃ / cm⁻¹020406080100Transmittance / %
Computed: gas-phase fundamentals for H₂O, CO₂, SO₂, NH₃ and CH₄ (Shimanouchi/NIST) and typical free-ion values for CO₃²⁻, NO₃⁻ and SO₄²⁻ (Nakamoto). For planar and linear species the IR activity is computed from the dipole change Σqᵢdᵢ of the vibration; for the rest it comes from the symmetry tables. Band positions are measured; heights are coarse intensity classes and widths and shapes are illustrative. The displacement arrows show the symmetry of each mode, not exact normal coordinates, and the animation is exaggerated.
See mutual exclusion at work: CO₂’s symmetric stretch (Σg⁺) gives no infrared absorption, while the antisymmetric stretch and the bend (Σu⁺, Πu) do. Then switch to SO₂ — no centre of inversion, and all three vibrations are infrared active.

Worked example

Telling cisplatin from transplatin by infrared

The anticancer drug cisplatin is cis\textit{cis}-[PtClX2(NHX3)X2]\ce{[PtCl2(NH3)2]}; its trans isomer is inactive. Considering only the two Pt–Cl stretching vibrations (treating NH₃ as a point), how many infrared bands does each show?

Concept check

Which statement about the vibrations of SF₆ (OhO_h) is true?

Try the calculation

Trigonal planar BF₃ (D3hD_{3h}) has Γ_vib = A₁′ + A₂″ + 2E′. How many infrared fundamental bands does it show?

Key takeaways

  • The characters of Γ₃N are (atoms left in place) × (±1 + 2 cos θ); reduce, then remove translations and rotations to get Γ_vib.
  • Vibrations transforming like x, y or z are infrared active; like a quadratic function, Raman active; with a centre of inversion, never both.
  • Degenerate vibrations count once and inactive ones give no band: CH₄’s 9 modes give only 2 infrared fundamentals.
  • Counting stretching bands distinguishes isomers: cis-[PtCl₂(NH₃)₂] shows two, trans one.
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Data and references

  1. F. A. Cotton, Chemical Applications of Group Theory, 3rd ed., Wiley, 1990, chapters 4 and 10 and Appendix IIA (character tables).
  2. K. Nakamoto, Infrared and Raman Spectra of Inorganic and Coordination Compounds, Part A, 6th ed., Wiley, 2009, chapters 1–2.
  3. T. Shimanouchi, Tables of Molecular Vibrational Frequencies, Consolidated Volume I, NSRDS-NBS 39 (1972), via the NIST Chemistry WebBook.
  4. R. S. Mulliken, “Report on notation for the spectra of polyatomic molecules”, J. Chem. Phys. 23, 1997 (1955).