Infrared Activity from Character Tables

Introduction

Infrared absorption occurs when light transfers energy into a vibrational motion through interaction with the molecular electric dipole. A polar bond alone does not ensure that every normal mode produces an IR band. The collective displacement must change an appropriate component of the whole molecule's dipole moment. Character tables offer a symmetry test: a fundamental vibration can be IR active if its symmetry matches x, y or z, the components of a polar vector.

Core explanation

Let Q be a normal coordinate measured from the equilibrium geometry. Expand one dipole component near equilibrium as μk(Q) ≈ μk(0) + (∂μk/∂Q)₀ Q + higher-order terms, where k is x, y or z. In the harmonic, electric-dipole approximation, the fundamental vibrational transition intensity depends on the square of the relevant dipole derivative. If the derivative is exactly zero by symmetry, the fundamental is IR-forbidden in that ideal approximation. A nonzero permanent dipole μk(0) is not the criterion; it is the change during the vibration that matters.

The symmetry rule follows from the transition integral. The vibrational ground state is totally symmetric in the harmonic model. The one-quantum excited state for mode Q transforms like Q. The dipole component transforms like x, y or z. The product of ground state, dipole and excited state must contain the totally symmetric species. Therefore Γ(Q) must match at least one linear-coordinate species for a simple fundamental IR transition. If several coordinates share the same species, symmetry may permit more than one polarisation component.

For water in the stated C₂v convention, z is A₁, x is B₁ and y is B₂. Its normal-mode representation is 2A₁ + B₂. Both A₁ modes can change a z-directed dipole component by symmetry, and the B₂ mode can change a y-directed component. Thus all three fundamentals are IR allowed under the ideal molecular symmetry. Their intensities are not necessarily equal, because each depends on the actual dipole derivative and vibrational eigenvector.

Carbon dioxide gives an important contrast. Ideal linear CO₂ has zero permanent dipole and inversion symmetry, yet some of its vibrations are IR active. Its asymmetric stretch changes the molecular dipole and is IR allowed. The doubly degenerate bend also produces a transient dipole perpendicular to the axis. The symmetric stretch retains zero dipole change at first order and is IR inactive in the ideal isolated molecule. This example refutes the notion that a molecule must have a permanent dipole to absorb IR radiation.

An IR-allowed mode need not produce a strong band. Symmetry only says the derivative is not forced to zero. Chemical bond polarity, charge redistribution and cancellation among moving atoms affect its magnitude. A forbidden ideal fundamental can also appear weakly when anharmonicity, isotopic substitution, solvent effects or vibronic coupling changes the strict selection-rule conditions. One should state the assumed ideal geometry and approximation when interpreting an absent or weak feature.

The group-theory workflow depends on the previous displacement analysis. First derive Γvib, then compare its irrep species to x, y and z in the same character-table convention. A table copied with switched axes can relabel B₁ and B₂, but the physical polarisation prediction remains invariant if all labels are transformed consistently. Numerical frequencies still require force constants or experiment.

Step-by-step reasoning

Classify the molecule's equilibrium geometry and derive or obtain each normal-mode symmetry. Read the x, y and z species from the corresponding character table. Mark a mode IR allowed if it matches at least one coordinate species, then identify the possible light polarisation. If an experimental band is weak, distinguish small allowed intensity from symmetry-forbidden intensity activated by perturbations.

Visual explanation

Draw water's symmetric stretch with both O–H bonds changing together and an arrow along z for dipole change. Draw its antisymmetric stretch with one bond lengthening while the other shortens and an arrow along y. For CO₂, draw the symmetric stretch with equal opposite bond changes that cancel dipole variation, then the asymmetric stretch with an unbalanced arrow.

Real-world analogy

A pair of people pulling opposite ends of a rope may make large motions while the rope's centre does not shift; another coordinated motion can move the centre. IR activity similarly depends on a net changing dipole, not simply how visibly atoms move. The analogy is incomplete because dipole change involves electron redistribution as well as nuclear positions.

Real-world example

Infrared spectra help identify functional groups because characteristic bond vibrations occur in predictable frequency ranges, but symmetry and molecular geometry govern which collective motions appear. CO₂'s strong asymmetric-stretch and bend absorption is environmentally important despite the molecule's lack of a permanent dipole. Its symmetric stretch is prominent through a different spectroscopic mechanism discussed with Raman activity.

Why?

Why does x, y or z appear in the selection rule? The electric dipole is a polar vector. Its components transform exactly like Cartesian coordinates under point-group operations. A vibrational coordinate of matching symmetry can couple linearly to a dipole component, allowing a first-order transition moment.

Common misconception

IR activity is not equivalent to having polar bonds or a permanent molecular dipole. A mode of a polar molecule may be IR inactive if its dipole changes cancel, while a nonpolar molecule may have IR-active asymmetric vibrations. Also, symmetry allowance says nothing exact about the band frequency and only a qualitative possibility about intensity.

Worked example

Use C₂v water with Γvib = 2A₁ + B₂. The table lists z as A₁ and y as B₂. Each A₁ vibration can have (∂μz/∂Q)₀ nonzero, while the B₂ vibration can have (∂μy/∂Q)₀ nonzero. Therefore all three normal modes are symmetry-allowed as IR fundamentals. The actual derivative for each mode must be calculated or measured to compare intensities.

Quick check

1. Must a molecule have a permanent dipole to show an IR-active vibration? Answer: No. The vibration must change a dipole component; ideal CO₂ has IR-active asymmetric stretch and bend despite zero equilibrium dipole. 2. Which C₂v water mode species can couple to a y-directed dipole in the stated axes? Answer: B₂, because y transforms as B₂.

Exam focus

State the derivative criterion and the character-table shortcut. Derive mode species before applying it and keep the axis convention consistent. Qualify results as symmetry allowed or forbidden in the electric-dipole fundamental approximation, not as guaranteed strong or absent under every physical condition.

Advanced insight

Transition intensity is proportional to the squared transition-dipole matrix element. In a harmonic approximation, the leading fundamental contribution is controlled by the linear dipole derivative, but overtones and combination bands can gain intensity from higher-order electrical or mechanical anharmonicity. These mechanisms explain why simple selection rules are most exact for an idealised first-order model.

Summary

A normal vibration is IR allowed when its symmetry matches a Cartesian dipole component x, y or z, so that a dipole derivative can be nonzero. Water's 2A₁ + B₂ modes are all allowed in its stated C₂v axes. CO₂ shows that permanent polarity is unnecessary: collective motion and dipole change determine activity. Symmetry does not give frequency or intensity magnitude.

Practice questions

1. Why is ideal CO₂'s symmetric stretch IR inactive in the simplest model? Answer: Equal changes of its two equivalent C–O bonds do not create a first-order net dipole change, and inversion symmetry forbids the electric-dipole fundamental. 2. A normal mode transforms as A₂ in C₂v, while x, y and z are B₁, B₂ and A₁. Is its fundamental IR allowed? Answer: No in the ideal electric-dipole harmonic approximation, because A₂ matches no dipole coordinate. Perturbations can still lend weak intensity. 3. Two modes have the same symmetry and both are IR allowed, but one band is much weaker. Does that contradict group theory? Answer: No. Group theory only removes exact symmetry zeros; actual dipole derivatives can differ greatly between modes.