Rotational Spectroscopy: Measuring a Bond Length From a Spectrum

The rigid rotor model

Treat a diatomic molecule as two masses joined by a rigid rod. Solving the Schrödinger equation for that system gives quantised rotational energies:

where B is the rotational constant, inversely proportional to the moment of inertia:

Here μ is the reduced mass and r the bond length. A heavier molecule or a longer bond means a larger moment of inertia and therefore more closely spaced levels.

The selection rules

• ΔJ = ±1. Only adjacent levels are connected.
• The molecule must have a permanent dipole moment. A homonuclear diatomic has none, so it shows no pure rotational spectrum at all.

Why the lines are equally spaced

The transition from J to J+1 has energy

So transitions occur at 2B, 4B, 6B and so on. Successive lines are separated by exactly 2B, which is what makes the spectrum so easy to interpret: measure any gap between adjacent lines, halve it, and you have B.

Getting the bond length — worked chain

1. Measure the spacing between adjacent lines. That equals 2B.
2. Halve it to obtain B.
3. Rearrange B = h/(8π²Ic) to find I.
4. Compute the reduced mass μ = m₁m₂/(m₁ + m₂), converting atomic masses to kilograms.
5. Solve I = μr² for r.

The arithmetic errors in this chain are almost always unit errors: forgetting to convert atomic mass units to kilograms, or leaving B in wavenumbers where the formula expects consistent SI units. Write the units alongside each step.

Intensity, and why the strongest line is not the first

Line intensity depends on how many molecules occupy the starting level, which is governed by two competing factors. The Boltzmann factor decreases with J, since higher levels cost more energy. But the degeneracy of level J is 2J+1, which increases with J.

Their product peaks at some intermediate J, so the most intense line is not the lowest-frequency one. Being asked to explain the intensity distribution, or to find the most populated level, is a standard question, and the answer requires both factors.

Centrifugal distortion

Real molecules are not rigid. At high J the rapid rotation stretches the bond, increasing the moment of inertia and lowering the energy slightly below the rigid-rotor prediction. The corrected expression adds a term in J²(J+1)².

Experimentally, the line spacing therefore decreases slightly at high J rather than remaining exactly 2B. Noticing that in supplied data and attributing it to centrifugal distortion is a higher-order question, and treating it as experimental error is the wrong answer.

Polyatomic molecules, briefly

Molecules are classified by their three principal moments of inertia: linear, spherical top with all three equal, symmetric top with two equal, and asymmetric top with all three different. The spectra become progressively more complex, and asymmetric tops require computational analysis rather than a simple formula.

Frequently asked questions

Why do homonuclear diatomics show no microwave spectrum?

Because they have no permanent dipole moment, so rotation produces no oscillating dipole for the radiation to couple to. They can still be studied by rotational Raman spectroscopy.

What does isotopic substitution do to the spectrum?

It changes the reduced mass without changing the bond length, so B changes predictably. Comparing spectra of two isotopologues is a standard way to confirm an assignment, and it is a common numerical.

Why is the J = 0 level not affected by temperature the way others are?

Its degeneracy is one and its energy is zero, so its population falls simply as other levels become accessible. The competition between degeneracy and Boltzmann factor is what shifts the intensity maximum upward.

How precise are bond lengths from this method?

Very precise — microwave spectroscopy gives some of the most accurate bond lengths available, which is why it is the reference method for small molecules.