Vibrational Spectroscopy: Force Constants and Anharmonicity

The harmonic oscillator

Model the bond as a spring obeying Hooke's law. The vibrational energy levels are

with the frequency set by the force constant k and the reduced mass:

Two predictions follow directly, and both are heavily used in interpretation. A stiffer bond — higher k, meaning a stronger or higher-order bond — absorbs at higher wavenumber. Lighter atoms — smaller μ — also absorb at higher wavenumber. This is why C≡C appears above C=C, which appears above C–C, and why C–H appears far above C–Cl.

Zero-point energy

Setting v = 0 leaves E = ½hν, not zero. A molecule retains vibrational energy even at absolute zero, and this is a real physical effect, not a mathematical artefact — it is required by the uncertainty principle, since a stationary bond of exactly fixed length would specify both position and momentum precisely.

Its most useful consequence is the isotope effect: substituting deuterium for hydrogen increases μ, lowers ν, and therefore lowers the zero-point energy. Because breaking the bond then requires more energy, C–D bonds break more slowly than C–H bonds — the kinetic isotope effect, which is the standard evidence for a rate-determining bond cleavage.

Selection rules

• Δv = ±1 for a harmonic oscillator.
• The dipole moment must change during the vibration for infrared activity.

Anharmonicity

Real bonds do not obey Hooke's law indefinitely. Stretch far enough and the bond breaks; compress and the nuclei repel sharply. The Morse potential captures this, and the corrected energy levels are

where x_e is the anharmonicity constant. Three consequences follow, all examinable:

• Levels converge as v increases, rather than staying evenly spaced, and eventually reach the dissociation limit. A harmonic oscillator has infinitely many equally spaced levels and can never dissociate, which is its most serious failure.
• Overtones become weakly allowed. Transitions with Δv = 2 or 3 appear faintly, at slightly less than two or three times the fundamental frequency — not exactly, because of the convergence.
• Hot bands appear at higher temperature, from molecules already in v = 1 absorbing to v = 2. These occur at slightly lower wavenumber than the fundamental.

Vibrational modes of polyatomics

The difference is that a linear molecule has only two rotational degrees of freedom rather than three, since rotation about the molecular axis moves nothing. Being asked to count modes for a given molecule is routine, and the linear-versus-bent distinction is the whole question.

The rule of mutual exclusion

In a molecule with a centre of symmetry, no vibration is both infrared and Raman active. So observing a band in both spectra proves the molecule is not centrosymmetric — a powerful structural conclusion from a simple comparison, and one of the most useful applications of vibrational spectroscopy in structure determination.

Frequently asked questions

Why does a stronger bond absorb at higher wavenumber?

Because the force constant is larger, and frequency goes as the square root of the force constant. It is the stiffness rather than the bond energy directly, though the two correlate well.

Why are overtones weak?

Because they are forbidden in the harmonic approximation and become allowed only through anharmonicity, which is a small correction. Their intensity reflects how small.

What does the dissociation energy have to do with the spectrum?

The converging levels approach a limit, and extrapolating the spacing to zero gives the dissociation energy — the Birge–Sponer method, which appears as a graphical question.

Why is the harmonic model still used if it is wrong?

Because it is accurate near the bottom of the well, where most molecules sit at ordinary temperatures, and it gives the correct relationship between frequency, force constant and mass. Anharmonicity is a correction, not a replacement.