Symmetry, Point Groups, and Normal Modes of Vibration

A polyatomic molecule looks at first like an intractable mechanical problem: $3N$ nuclei coupled by every bond and bond angle, all pulling on one another at once. The Born-Oppenheimer approximation cuts the first knot — it fixes the nuclei on a single electronic potential energy surface, and near a stable geometry that surface is quadratic. The coupled tangle then collapses, exactly as the harmonic-oscillator article promised in one dimension, into a set of independent oscillators: the normal modes, each a collective in-phase displacement pattern ringing at its own frequency. Any real vibration is a chord struck from these fundamental notes. The second knot is cut by symmetry. The equilibrium geometry is left unchanged by a set of rotations, reflections, inversions, and improper rotations that close under composition into the molecule's point group. Because the vibrational Hamiltonian commutes with every one of them, the normal coordinates must transform as basis functions of the group's irreducible representations — so group theory hands each mode a symmetry label ( $A_1$ , $B_2$ , $E_u$ , $\Pi_u$ ) and its degeneracies before any equation of motion is solved. The same labels decide spectroscopy: a mode absorbs infrared light only if its irrep carries a dipole component $x,y,z$ , and scatters Raman light only if it carries a polarizability component $x^2,xy,\dots$ . This article makes that machinery concrete — counting modes by geometry, reducing the Cartesian representation, and reading IR and Raman activity off the character table for water, carbon dioxide, ammonia, and boron trifluoride.

Key takeaways

A molecule of $N$ atoms has $3N$ degrees of freedom; remove 3 translations and 3 rotations (2 if linear) and $3N-6$ (or $3N-5$ ) normal modes remain.
The symmetry operations form a point group; the $3N$ Cartesian displacements give a reducible $\Gamma_{\mathrm{tot}}$ that decomposes into irreducible representations, one labelling each mode's symmetry species.
The character of $\Gamma_{\mathrm{tot}}$ is pure geometry: only atoms an operation leaves in place contribute, each with a fixed factor $1+2\cos\theta$ (proper) or $-1+2\cos\theta$ (improper).
A mode is infrared active when its irrep spans a translation ( $x,y,z$ ) and Raman active when it spans a quadratic ( $x^2,xy,\dots$ ); the totally symmetric mode is always Raman active.
Rule of mutual exclusion: in a centrosymmetric molecule no vibration is both IR and Raman active, since $g$ quadratics and $u$ linears never share an irrep.
Key idea: normal coordinates transform as irreps of the point group, so IR activity requires the irrep to span a linear coordinate $(x,y,z)$ and Raman activity a quadratic one $(x^2,xy,\dots)$ — the spectrum is fixed by symmetry before a single frequency is computed.

The core insight

Normal coordinates transform as irreducible representations of the molecular point group; a mode is infrared active if and only if its irrep spans a linear coordinate $(x,y,z)$ and Raman active if and only if it spans a quadratic one $(x^2,xy,\dots)$ .

Where this lives today

The reduction procedure is quietly built into every modern quantum-chemistry package: to compute vibrational frequencies a program first detects the point group, then block-diagonalises the Hessian by irrep — turning one large eigenvalue problem into several small ones and labelling each frequency with its species automatically. The same labels drive the assignment of measured spectra: symmetry tells the analyst how many fundamentals to expect in the IR and Raman and which should coincide, so a mismatch flags a wrong assignment, an overtone, or a wrong structure. Mutual exclusion remains a front-line diagnostic — the absence of any IR–Raman coincidence is still the cleanest signature of a centre of inversion, used to sort centrosymmetric from non-centrosymmetric complexes and clusters. The polarization refinement is a working tool too: a totally symmetric mode gives a polarized Raman line ( $\rho<3/4$ ) while every other mode is depolarized ( $\rho=3/4$ ), so rotating a polarizer assigns species directly — routine in resonance-Raman of biomolecules and catalysts. And the labels feed forward: a non-totally-symmetric mode can lend intensity to a forbidden electronic transition through vibronic (Herzberg–Teller) coupling, shaping the band envelopes the Franck–Condon principle otherwise governs. Symmetry, set once at equilibrium, propagates through the whole spectroscopy of the molecule.

The physical picture

Imagine the atoms strung together by springs and shaken at exactly the right frequency. Then every atom moves in phase, all passing through equilibrium at the same instant and reaching their extremes together — a single rigid-looking pattern of arrows that grows and shrinks in place. That synchronized pattern is a normal mode, its frequency one of the molecule's fundamentals; a generic thermal wobble is a superposition of these modes, the way a plucked string mixes its harmonics. Symmetry sorts the modes by how they behave in a mirror. For each mode, ask what every symmetry operation does to it. A pattern left unchanged is totally symmetric, labelled $A_1$ ; one that flips sign under some reflection or rotation gets a different label; and two motions that a rotation carries into each other must share a frequency — a degeneracy, bookkept by a two- or three-dimensional irrep such as $E$ or $T$ . The first figure shows the three normal modes of water with their species and measured frequencies.

Figure

The three normal modes of water in point group C₂ᵥ. Symmetric stretch (A₁, 3657 cm⁻¹) and bend (A₁, 1595 cm⁻¹) are totally symmetric; the antisymmetric stretch (B₂, 3756 cm⁻¹) flips sign under C₂. All three are both infrared and Raman active. Arrows are instantaneous nuclear displacements.

To count the modes you never touch a force constant — you count atoms. An operation that carries an atom to a different site puts nothing on the diagonal of the transformation matrix, so only atoms left in place contribute to that operation's character, each weighted by a fixed number set by whether the operation is a rotation, reflection, inversion, or improper rotation. Reduce the resulting characters against the character table, strike out the pieces that are rigid translation and rotation, and what remains is the exact list of vibrational species. Whether each species couples to light is a question of what oscillates: a changing dipole (a vector, like $x,y,z$ ) absorbs infrared, a changing polarizability (a tensor, like $x^2,xy,\dots$ ) scatters Raman. In a centrosymmetric molecule these two demands can never be met by the same mode, as the second figure shows for carbon dioxide.

Figure

The four vibrations of linear centrosymmetric CO₂ in D∞ₕ. The symmetric stretch (Σg⁺, even under inversion) is Raman-active only; the antisymmetric stretch (Σu⁺) and doubly degenerate bend (Πu, odd under inversion) are infrared-active only — no band in both spectra, the rule of mutual exclusion.

The core equations

The starting point is a head count. Every atom carries three Cartesian degrees of freedom, the whole molecule can translate and rotate as a rigid body, and whatever is left over is internal vibration — fixing the mode number before any symmetry is invoked.

3N = N_{\mathrm{trans}} + N_{\mathrm{rot}} + N_{\mathrm{vib}} \qquad (1)

where: $N$ — number of atoms; $N_{\mathrm{trans}}=3$ — rigid translations; $N_{\mathrm{rot}}$ — rigid rotations, $3$ (nonlinear) or $2$ (linear); $N_{\mathrm{vib}}$ — vibrational normal modes. Rearranged, $N_{\mathrm{vib}}=3N-6$ or $3N-5$ : a linear molecule gets one extra vibration because rotation about its own axis moves no nucleus. Those vibrations decouple because of the potential's shape near equilibrium. Expand the Born-Oppenheimer surface to second order; the linear term vanishes at a minimum, leaving a quadratic form whose matrix of second derivatives — the Hessian — governs the motion.

V(\mathbf{q}) \approx V_0 + \tfrac{1}{2}\sum_{i,j} K_{ij}\, q_i q_j,\qquad K_{ij}=\left.\frac{\partial^2 V}{\partial q_i \partial q_j}\right|_0 \qquad (2)

where: $q_i=\sqrt{m_i}\,\Delta x_i$ — mass-weighted displacement; $K_{ij}$ — element of the mass-weighted force-constant (Hessian) matrix; $V_0$ — the potential minimum. Diagonalising the real symmetric $K$ yields the normal coordinates $Q_k$ , each an independent oscillator $\ddot Q_k=-\omega_k^2 Q_k$ . Symmetry enters through how the $3N$ displacement vectors transform under each operation $R$ . The character — the trace of that matrix — factorises: a shifted atom gives an off-diagonal block and hence nothing, so only fixed atoms count, each carrying the same $3\times3$ vector character.

\chi_{\mathrm{tot}}(R) = N_{\mathrm{unshifted}}(R)\times \chi_{xyz}(R),\qquad \chi_{xyz}=\pm 1 + 2\cos\theta \qquad (3)

where: $\chi_{\mathrm{tot}}(R)$ — character of the Cartesian representation $\Gamma_{\mathrm{tot}}$ ; $N_{\mathrm{unshifted}}(R)$ — atoms left in place by $R$ ; $\chi_{xyz}$ is $+1+2\cos\theta$ for a proper rotation $C_n(\theta)$ , $-1+2\cos\theta$ for an improper $S_n(\theta)$ . Special cases: $E\to3$ , a mirror $\sigma\to+1$ , inversion $i\to-3$ . The great orthogonality theorem then tells you how many times each irrep $\Gamma$ hides inside $\Gamma_{\mathrm{tot}}$ — a weighted overlap of the two characters summed over all operations.

n_\Gamma = \frac{1}{h}\sum_{R} g(R)\,\chi_{\mathrm{tot}}(R)\,\chi_\Gamma(R)^{*} \qquad (4)

where: $n_\Gamma$ — multiplicity of $\Gamma$ in $\Gamma_{\mathrm{tot}}$ ; $h$ — order of the group; $g(R)$ — size of the class of $R$ ; $\chi_\Gamma(R)$ — character of $\Gamma$ from the table. The $n_\Gamma$ are non-negative integers, and $\sum_\Gamma n_\Gamma\,\dim\Gamma = 3N$ is a built-in check. The last step strips off the rigid-body motions: the table flags which irreps carry the translations $(x,y,z)$ and rotations $(R_x,R_y,R_z)$ , and you subtract those multiplicities.

\Gamma_{\mathrm{vib}} = \Gamma_{\mathrm{tot}} \ominus \Gamma_{\mathrm{trans}} \ominus \Gamma_{\mathrm{rot}} \qquad (5)

where: $\Gamma_{\mathrm{trans}}$ — irreps spanned by $(x,y,z)$ ; $\Gamma_{\mathrm{rot}}$ — irreps spanned by $(R_x,R_y,R_z)$ ; $\ominus$ — subtraction of multiplicities. The survivors assign every one of the $3N-6$ (or $3N-5$ ) modes a definite species. Activity reads off the same table. An infrared transition from the totally symmetric ground state requires the mode's irrep to contain a dipole component — one of the linear functions $x,y,z$ .

\text{IR active} \iff \Gamma(Q_k)=\Gamma(x)\ \text{or}\ \Gamma(y)\ \text{or}\ \Gamma(z) \qquad (6)

where: $\Gamma(x,y,z)$ — irreps carrying the dipole components. The moment $\langle Q_k|\hat\mu_\alpha|0\rangle$ survives only if $\Gamma(Q_k)\otimes\Gamma(\alpha)$ contains the totally symmetric irrep — i.e. when the mode and $\alpha$ share a species. Raman works the same way through the polarizability tensor, whose components transform as quadratics — a second, independent selection rule.

\text{Raman active} \iff \Gamma(Q_k)\supset\Gamma(\alpha_{ij}),\quad \alpha_{ij}\in\{x^2,y^2,z^2,xy,xz,yz\} \qquad (7)

where: $\alpha_{ij}$ — components of the polarizability tensor. Since the totally symmetric irrep always contains $x^2+y^2+z^2$ , the symmetric-stretch mode of any molecule is guaranteed Raman active, with a strongly polarized line.

Derivation AFrom coupled springs to independent normal coordinates

The goal is to show why a molecule of $N$ mutually bonded atoms — a system whose equations of motion couple every coordinate to every other — nevertheless behaves as $3N-6$ independent one-dimensional oscillators. The engine is the diagonalisation of one real symmetric matrix, the mass-weighted Hessian, and the payoff is the object symmetry then classifies: the normal coordinate $Q_k$ , a fixed displacement pattern that oscillates at one frequency $\omega_k$ and never exchanges energy with any other.

Step 1. Write Newton's law with the quadratic potential of equation (2): $\ddot q_i = -\sum_j K_{ij}\,q_j$ . Why mass-weight: absorbing $\sqrt{m_i}$ into the coordinates leaves the kinetic energy as $\tfrac12\sum\dot q_i^2$ with no mass matrix, so the remaining dynamical matrix $K$ is real symmetric — the object with a complete orthonormal eigenbasis.

Step 2. Seek collective solutions $q_i(t)=a_i\cos(\omega t+\varphi)$ ; substituting turns the coupled ODEs into $K\mathbf a = \omega^2\mathbf a$ . Why: a normal mode has every atom sharing one frequency and phase, so its pattern is time-independent up to a scalar — exactly the ansatz that reduces dynamics to linear algebra.

Step 3. Being real symmetric, $K$ has $3N$ real eigenvalues $\omega_k^2$ and an orthogonal $L$ with $L^{\mathsf T}KL=\mathrm{diag}(\omega_k^2)$ ; the normal coordinates $Q_k=\sum_i L_{ik}\,q_i$ obey the decoupled $\ddot Q_k=-\omega_k^2 Q_k$ . Why it matters: $L$ diagonalises kinetic and potential energy at once, so the total energy is a sum of independent oscillator energies — modes never exchange energy.

Step 4. Six eigenvalues (five if linear) come out at $\omega_k^2=0$ : the rigid translations and rotations, which cost no potential energy. Discarding them leaves the $3N-6$ (or $3N-5$ ) vibrations. Why the zeros appear: translating or rotating the molecule changes no bond length or angle, so $V$ is flat there and the curvature — hence frequency — is zero; equation (5) removes them by symmetry label.

Go deeper · Derivation B · 2 minReducing the Cartesian representation of water in C₂ᵥ

Derivation BReducing the Cartesian representation of water in C₂ᵥ

We run the full machine on water to extract its three vibrational species and their activity from symmetry alone. Water is bent, point group $C_{2v}$ with operations $\{E,\,C_2,\,\sigma_v(xz),\,\sigma_v'(yz)\}$ and order $h=4$ ; place it in the $yz$ plane with $C_2$ along $z$ . The result $\Gamma_{\mathrm{vib}}=2A_1\oplus B_2$ names the symmetric stretch, bend, and antisymmetric stretch, and predicts — correctly — that all three appear in both the infrared and Raman spectra.

Step 1. Count unshifted atoms: $E$ leaves all $3$ ; $C_2$ (about $z$ ) leaves only the on-axis oxygen, $1$ ; $\sigma_v(xz)$ is perpendicular to the molecular plane and holds only oxygen, $1$ ; $\sigma_v'(yz)$ is the molecular plane and holds all $3$ . Why only these count: swapping the two hydrogens moves them to new sites, contributing off-diagonal blocks to the $9\times9$ matrix and nothing to its trace.

Step 2. Attach $\chi_{xyz}$ from equation (3): $E\to3$ ; $C_2(\theta=\pi)\to 1+2\cos\pi=-1$ ; each mirror $\to +1$ . Multiplying gives $\chi_{\mathrm{tot}}$ : $E{:}\,9,\ C_2{:}\,{-}1,\ \sigma_v(xz){:}\,1,\ \sigma_v'(yz){:}\,3$ . Why the product: each retained atom carries a full copy of the $3\times3$ vector block, so the trace is fixed-atom count times block trace.

Step 3. Reduce with equation (4), $h=4$ , all classes size one: $n_{A_1}=\tfrac14(9-1+1+3)=3$ , $n_{A_2}=\tfrac14(9-1-1-3)=1$ , $n_{B_1}=\tfrac14(9+1+1-3)=2$ , $n_{B_2}=\tfrac14(9+1-1+3)=3$ . So $\Gamma_{\mathrm{tot}}=3A_1\oplus A_2\oplus 2B_1\oplus 3B_2$ , dimensions summing to $9=3N$ . Why check the sum: any slip in the characters almost always breaks the $\sum n_\Gamma\dim\Gamma=3N$ balance — a free guard.

Step 4. Remove rigid motions from the $C_{2v}$ table: $z\to A_1$ , $x\to B_1$ , $y\to B_2$ so $\Gamma_{\mathrm{trans}}=A_1\oplus B_1\oplus B_2$ ; $R_z\to A_2$ , $R_y\to B_1$ , $R_x\to B_2$ so $\Gamma_{\mathrm{rot}}=A_2\oplus B_1\oplus B_2$ . Subtracting, $\Gamma_{\mathrm{vib}}=2A_1\oplus B_2$ — exactly $3N-6=3$ . Why these irreps: the table lists the linear and quadratic functions each species carries; translations and rotations are literally the $(x,y,z)$ and $(R_x,R_y,R_z)$ rows.

Step 5. Read activity: $A_1$ carries $z$ (IR) and $x^2,y^2,z^2$ (Raman); $B_2$ carries $y$ (IR) and $yz$ (Raman). Every mode spans both a linear and a quadratic, so all three are IR and Raman active — matching bands at $3657$ , $1595$ , $3756\ \mathrm{cm^{-1}}$ . Why no exclusion: water has no inversion centre, so nothing forbids one species carrying both a linear and a quadratic function.

Go deeper · Derivation C · 2 minCO₂ and the rule of mutual exclusion

Derivation CCO₂ and the rule of mutual exclusion

This derivation classifies the four vibrations of carbon dioxide and proves that a centre of inversion splits the vibrational spectrum cleanly in two. $\mathrm{CO_2}$ is linear and centrosymmetric, point group $D_{\infty h}$ with the inversion centre on carbon; being linear it has $3N-5=4$ vibrations. The outcome $\Gamma_{\mathrm{vib}}=\Sigma_g^{+}\oplus\Sigma_u^{+}\oplus\Pi_u$ splits into a Raman-only set and an IR-only set with not a single band in common — the diagnostic that first pinned down the molecule's straight, symmetric shape.

Step 1. Fix the count: $3N-5=3(3)-5=4$ — a symmetric stretch, an antisymmetric stretch, and a doubly degenerate bend. Why the bend is degenerate: its two perpendicular bending planes are carried into each other by rotation about the axis, so symmetry forces one shared frequency — a single line for two motions, the two-dimensional irrep $\Pi_u$ .

Step 2. Assign parity under inversion $i$ . The symmetric stretch (oxygens move oppositely, carbon fixed) maps onto itself, so it is even, $\Sigma_g^{+}$ ; the antisymmetric stretch and the bend reverse, so both are odd, $\Sigma_u^{+}$ and $\Pi_u$ . Why parity is the whole story: with an inversion centre, $g$ / $u$ is a good label — and it is exactly the label the linear and quadratic operators carry too.

Step 3. Linear (IR) functions: $z\to\Sigma_u^{+}$ , $(x,y)\to\Pi_u$ . So the antisymmetric stretch ( $2349\ \mathrm{cm^{-1}}$ ) and the bend ( $667\ \mathrm{cm^{-1}}$ ) are IR active — both $u$ , both build an oscillating dipole. Why the symmetric stretch is IR-silent: it stays centrosymmetric throughout, so its dipole never changes and the infrared field has nothing to grip.

Step 4. Quadratic (Raman) functions: $x^2+y^2$ and $z^2\to\Sigma_g^{+}$ . Only the symmetric stretch is Raman active, near $1333\ \mathrm{cm^{-1}}$ — in practice a Fermi-resonance dyad at $1285$ and $1388\ \mathrm{cm^{-1}}$ from accidental degeneracy with the bend overtone. Why the split is harmless: Fermi resonance mixes two $\Sigma_g^{+}$ levels but keeps both $g$ , hence Raman-only.

Step 5. Combine: $g$ modes carry only quadratics (Raman), $u$ modes only linears (IR), and no irrep is both, so $\Gamma_g\cap\Gamma_u=\varnothing$ . No vibration is both IR and Raman active — the rule of mutual exclusion. Why it generalises: the argument used only the inversion centre, so it holds for benzene, ethylene, octahedral $\mathrm{SF_6}$ , and every centrosymmetric molecule.

Limiting cases

The mode count and activity rules take sharp limiting forms by geometry and symmetry class:

Case	Result	Meaning
Nonlinear molecule	$N_{\mathrm{vib}}=3N-6$	Three translations and three rotations are removed from the $3N$ Cartesian degrees of freedom.
Linear molecule	$N_{\mathrm{vib}}=3N-5$	Rotation about the molecular axis moves no nucleus, so only two rotations are removed — leaving one extra vibration, a doubly degenerate bend.
Centrosymmetric molecule	$\Gamma_g\cap\Gamma_u=\varnothing$	Rule of mutual exclusion: $g$ modes are Raman active (quadratics are even), $u$ modes are IR active (linears are odd), and none is both.
Totally symmetric mode	$\Gamma=A_1\ (\text{or }A_{1g},\ \Sigma_g^{+})$	Always Raman active, since the totally symmetric irrep contains $x^2+y^2+z^2$ ; its Raman line is strongly polarized ( $\rho<3/4$ ).
Degenerate mode ( $E$ , $T$ , $\Pi$ )	one frequency, $\dim\Gamma$ motions	Counts once in the spectrum but occupies two or three independent normal coordinates locked to the same $\omega$ .

Common trapAn IR-inactive mode is not a real vibration

It is a genuine normal mode that merely fails to couple to an oscillating dipole. The symmetric stretch of $\mathrm{CO_2}$ is a real motion at $\approx 1333\ \mathrm{cm^{-1}}$ , plain in the Raman spectrum, but invisible in the infrared because the molecule stays centrosymmetric and its dipole never changes. "Silent" in one technique means "look in the other," not "does not exist." Two related traps: a degenerate mode ( $E$ , $T$ , $\Pi$ ) is one frequency but several motions — count it once among observed bands, yet use its full dimension when balancing $\sum n_\Gamma\dim\Gamma=3N$ ; and "IR active" and "Raman active" are independent tests, so a mode can pass one, both, or neither.

Worked examples

1 · Vibrations of ammonia, NH₃ (confidence builder)

Ammonia is a trigonal pyramid of point group $C_{3v}$ , operations $\{E,\,2C_3,\,3\sigma_v\}$ , order $h=6$ . Classify its normal modes, count the distinct fundamental frequencies, and state which are IR and which Raman active.

Step 1. With $N=4$ , $3N-6=6$ . Unshifted atoms: $E{:}\,4$ ; each $C_3$ leaves only the on-axis nitrogen, $1$ ; each $\sigma_v$ holds nitrogen and one hydrogen, $2$ . Vector characters $\chi_{xyz}$ : $E\to3$ , $C_3(120°)\to 1+2\cos120°=0$ , $\sigma_v\to+1$ . Products: $\chi_{\mathrm{tot}}=(12,\,0,\,2)$ .

Step 2. Reduce with equation (4), class sizes $g=(1,2,3)$ and characters ( $A_1{:}\,1,1,1$ ; $A_2{:}\,1,1,-1$ ; $E{:}\,2,-1,0$ ): $n_{A_1}=\tfrac16(12+0+6)=3$ , $n_{A_2}=\tfrac16(12-0-6)=1$ , $n_{E}=\tfrac16(24+0+0)=4$ . So $\Gamma_{\mathrm{tot}}=3A_1\oplus A_2\oplus 4E$ , dimensions summing to $3+1+8=12=3N$ . Check passes.

Step 3. Remove rigid motions: $z\to A_1$ , $(x,y)\to E$ give $\Gamma_{\mathrm{trans}}=A_1\oplus E$ ; $R_z\to A_2$ , $(R_x,R_y)\to E$ give $\Gamma_{\mathrm{rot}}=A_2\oplus E$ . Subtracting leaves $\boxed{\Gamma_{\mathrm{vib}}=2A_1\oplus 2E}$ .

Step 4. Two $A_1$ plus two $E$ modes give $2\cdot1+2\cdot2=6$ states ( $=3N-6$ ) but only four distinct frequencies. In $C_{3v}$ , $A_1$ carries $z$ (IR) and $z^2,\,x^2+y^2$ (Raman); $E$ carries $(x,y)$ (IR) and quadratics (Raman) — so all four are both IR and Raman active: the symmetric stretch and umbrella bend ( $A_1$ , $\approx 3337$ and $950\ \mathrm{cm^{-1}}$ ) and the degenerate stretch and bend ( $E$ , $\approx 3444$ and $1627\ \mathrm{cm^{-1}}$ ). With no inversion centre, nothing forbids coincidence.

2 · Planar versus pyramidal XY₃: telling geometry from band counts

A trihalide $\mathrm{XY_3}$ could be trigonal planar ( $D_{3h}$ , like $\mathrm{BF_3}$ ) or trigonal pyramidal ( $C_{3v}$ , like $\mathrm{NH_3}$ ). From symmetry alone, predict how the IR and Raman band patterns differ so an experiment can decide the shape.

Step 1. The pyramid is Example 1: $\Gamma_{\mathrm{vib}}(C_{3v})=2A_1\oplus 2E$ , four frequencies, all IR active and all Raman active — total coincidence.

Step 2. For the plane, reduce in $D_{3h}$ ( $E,\,2C_3,\,3C_2,\,\sigma_h,\,2S_3,\,3\sigma_v$ , $h=12$ ). Unshifted counts $(4,1,2,4,1,2)$ times vector characters $(3,0,-1,1,-2,1)$ give $\chi_{\mathrm{tot}}=(12,0,-2,4,-2,2)$ . Reducing yields $\Gamma_{\mathrm{tot}}=A_1'\oplus A_2'\oplus 3E'\oplus 2A_2''\oplus E''$ ; removing $\Gamma_{\mathrm{trans}}=A_2''\oplus E'$ and $\Gamma_{\mathrm{rot}}=A_2'\oplus E''$ leaves $\boxed{\Gamma_{\mathrm{vib}}(D_{3h})=A_1'\oplus 2E'\oplus A_2''}$ — again four frequencies.

Step 3. Activity in $D_{3h}$ : $A_1'$ carries quadratics but no linear — Raman only (symmetric stretch, $\approx 888\ \mathrm{cm^{-1}}$ ); $A_2''$ carries $z$ but no quadratic — IR only (out-of-plane bend, $\approx 719\ \mathrm{cm^{-1}}$ ); the two $E'$ carry both — both active ( $\approx 1450$ and $480\ \mathrm{cm^{-1}}$ ).

Step 4. The verdict: the pyramid shows four bands coincident in IR and Raman; the plane shows a Raman-only symmetric stretch and an IR-only out-of-plane bend. A symmetric stretch bright in Raman yet dark in the infrared is the fingerprint of planarity — the non-coincidence argument that, pushed to a full inversion centre, becomes the rule of mutual exclusion. Group theory settles planar-versus-pyramidal by counting bands, before any ab-initio frequency.

Physics of the portrait

Molecular symmetry is where nineteenth-century crystallography, the group theory of the 1920s, and twentieth-century spectroscopy fused into a single predictive tool — one that reads a whole vibrational spectrum off a character table.

Origins · 1 minGerhard Herzberg and how this was discovered

Gerhard Herzberg

1904 to 1999, Hamburg to Ottawa · molecular spectroscopy and the group theory of vibrations

Gerhard Herzberg turned the abstract machinery of point groups into the everyday working language of chemists and spectroscopists. Trained as a physicist in Germany, he fled the Nazi regime in 1935 and built, at the National Research Council of Canada in Ottawa, the pre-eminent molecular-spectroscopy laboratory of its era. His treatise Molecular Spectra and Molecular Structure — especially Volume II, Infrared and Raman Spectra of Polyatomic Molecules (1945) — codified exactly the procedure of this article: enumerate the point group, reduce the $3N$ Cartesian representation, subtract translation and rotation, read activity from the character table. Generations learned to classify normal modes from its tables. He was also a superb experimentalist, detecting spectra of free radicals and molecular ions others thought unobservable and finding $\mathrm{H_2}$ in planetary and stellar atmospheres, work that won him the 1971 Nobel Prize in Chemistry. The synthesis he championed — symmetry first, dynamics second — is why a student today can predict that $\mathrm{CO_2}$ shows two IR bands and one Raman band without writing a single force constant.

History

1830

— Hessel enumerates the 32 crystallographic point groups

1928

— Raman and Krishnan report the inelastic scattering of light, the Raman effect

1930

— Wigner analyses the normal vibrations of symmetric molecules by group representation theory

1945

— Herzberg's Infrared and Raman Spectra of Polyatomic Molecules codifies the reduction procedure

1963

— Cotton's Chemical Applications of Group Theory makes the method standard for chemists

Going further

Texts. Herzberg, Molecular Spectra and Molecular Structure II — the founding reference · Cotton, Chemical Applications of Group Theory, 3rd ed. — reduction procedure and character tables · Bunker and Jensen, Molecular Symmetry and Spectroscopy · Atkins and Friedman, Molecular Quantum Mechanics, 5th ed. · Demtröder, Laser Spectroscopy, Vol. 1 · Wilson, Decius and Cross, Molecular Vibrations — the classic on the normal-coordinate analysis of Derivation 1.

Provenance. The degree-of-freedom counting, the character formula $\chi_{\mathrm{tot}}=N_{\mathrm{unshifted}}\times(\pm1+2\cos\theta)$ , and the reduction formula (a corollary of the great orthogonality theorem) are standard, following Cotton and Herzberg. A verifier should reproduce the four reductions of this article: $\mathrm{H_2O}$ $\to 2A_1\oplus B_2$ , $\mathrm{CO_2}$ $\to \Sigma_g^{+}\oplus\Sigma_u^{+}\oplus\Pi_u$ , $\mathrm{NH_3}$ $\to 2A_1\oplus 2E$ , $\mathrm{BF_3}$ $\to A_1'\oplus 2E'\oplus A_2''$ . Fundamentals quoted are standard textbook values: water $3657/1595/3756$ ; $\mathrm{CO_2}$ $667/1333/2349\ \mathrm{cm^{-1}}$ (symmetric stretch as the Fermi dyad $1285/1388$ ); ammonia $3337/950/3444/1627$ ; $\mathrm{BF_3}$ $\approx 888/719/1450/480\ \mathrm{cm^{-1}}$ (Herzberg; Demtröder).

Prerequisites. The Harmonic Oscillator Model of Vibration · The Born-Oppenheimer Approximation · The Chemical Bond and Molecular Potential Energy Curves · Selection Rules for Molecular Transitions. Leads to. Vibration-Rotation Spectra and P, Q, R Branches · The Franck-Condon Principle · Electronic Transitions in Molecules.

Ask the tutor: Give me any molecule and I will walk you through its point group, reduce the Cartesian representation step by step, and predict exactly how many infrared and Raman bands you should see.

Why is the mode count 3N−6 for most molecules but 3N−5 for linear ones?

Each atom has three Cartesian degrees of freedom, so a molecule starts with $3N$ . Three describe rigid translation and three tumbling; neither stretches a bond, leaving $3N-6$ . A linear molecule is the exception: rotation about its own axis moves no nucleus — the nuclei lie on the axis — so it is not a genuine rotation. Only two rotations exist, one fewer is subtracted, and the count becomes $3N-5$ . That extra vibration is the doubly degenerate bend, which is why $\mathrm{CO_2}$ has four modes rather than three.

How do I find the number of unshifted atoms quickly?

Picture each operation acting on the ball-and-stick model and ask which atoms end up exactly where they started — no coordinates needed. For $E$ , all $N$ ; for a rotation axis, only atoms on the axis; for a mirror, only atoms in the plane; for inversion, at most the atom at the centre; for $S_n$ , only an atom at its centre. Multiply each count by the vector character ( $+1+2\cos\theta$ proper, $-1+2\cos\theta$ improper) to build $\chi_{\mathrm{tot}}$ . Any atom the operation moves contributes zero — it sits off-diagonal in the matrix.

Why is the totally symmetric mode always Raman active?

Raman activity requires the mode's irrep to contain a quadratic — $x^2$ , $z^2$ , $xy$ , and so on. The totally symmetric irrep ( $A_1$ , $A_{1g}$ , or $\Sigma_g^{+}$ ) is by definition unchanged by every operation, and $x^2+y^2+z^2$ is likewise invariant, being just the squared distance from the origin. So the totally symmetric irrep always carries that quadratic, guaranteeing the mode couples to the polarizability and scatters Raman light. As a bonus its line is polarized ( $\rho<3/4$ ), a handy way to spot the symmetric mode in a crowded spectrum.

If a mode is infrared inactive, is it still a real vibration?

Yes — activity is about coupling to light, not existence. Every one of the $3N-6$ normal coordinates is a genuine oscillation with a definite frequency. "Infrared inactive" only means that motion makes no oscillating dipole, so the infrared field has nothing to push on and the band is absent from the IR spectrum. The symmetric stretch of $\mathrm{CO_2}$ is the classic case: a real vibration near $1333\ \mathrm{cm^{-1}}$ , plain in Raman, yet silent in the infrared because the molecule stays centrosymmetric. Always look in the complementary technique before calling a mode absent.

How can group theory predict activity without solving the equations of motion?

Because activity is a yes/no question of whether an integral vanishes, and symmetry alone answers it. A transition rate goes as $|\langle Q_k|\hat O|0\rangle|^2$ , with $\hat O$ the dipole (IR) or polarizability (Raman). The ground state is totally symmetric, so the integral is non-zero only if $\Gamma(Q_k)\otimes\Gamma(\hat O)$ contains the totally symmetric irrep — which reduces to " $Q_k$ and $\hat O$ share a species." That test uses only the character table, never a frequency or force constant. What it cannot give you is how strong the band is; intensity depends on the dynamics.

What does mutual exclusion actually let me conclude in the lab?

If the IR and Raman spectra of an unknown share not one band frequency, that is strong evidence of a centre of inversion. With an inversion centre, dipole components ( $x,y,z$ ) are odd ( $u$ ) and polarizability components (quadratics) are even ( $g$ ), so no vibration can be both IR and Raman active. Complete non-coincidence thus rules out any geometry lacking that centre. This is how $\mathrm{CO_2}$ was confirmed linear and centrosymmetric rather than bent, and it routinely separates a centrosymmetric trans isomer from its cis partner, or a planar from a pyramidal skeleton.