Rotational spectroscopy

Result

Rotational spectroscopy

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Statement

Bond lengths from the rigid-rotor microwave spectrum.

Why it matters

vibrational-ir-spectroscopy and uv-vis-electronic both probe molecular structure by resonant absorption of light, but at successively higher energy scales (bond vibration, then electronic structure); rotational spectroscopy sits at the bottom of that hierarchy, probing the lowest-energy motion a molecule can undergo — end-over-end tumbling — and is, for suitable small molecules, the single most precise experimental route to bond lengths available in all of chemistry.

Because the technique reads bond length directly from a quantised energy-level spacing rather than from any indirect model, it provides an essential experimental benchmark against which more approximate structural methods (electron diffraction, X-ray crystallography, computed equilibrium geometries) are routinely checked, and it underlies the identification of molecules in interstellar space by radio and microwave astronomy, where rotational transitions of simple molecules such as carbon monoxide are readily observed.

Hypotheses

The molecule is treated as a rigid rotor: a fixed bond length \(r\), unaffected by the rotation itself.Real bonds stretch slightly under centrifugal force as rotation speeds up (higher \(J\)), and vibration and rotation are not perfectly separable in a real molecule; the rigid-rotor model neglects both effects, giving energy levels that are exactly evenly spaced rather than very slightly compressed at high \(J\) (centrifugal distortion). The molecule possesses a permanent electric dipole moment.A pure rotational spectrum arises from the rotating dipole's oscillating electric field interacting with incident radiation; a molecule with no permanent dipole (any homonuclear diatomic such as \(\text{N}_2\) or \(\text{O}_2\), or any other centrosymmetric molecule) has no such handle for the radiation to couple to and shows no pure rotational absorption spectrum at all, regardless of how it is illuminated. The simple, evenly-spaced line spectrum derived here holds cleanly only for a linear (or symmetric-top) rotor; asymmetric-top molecules (most polyatomic molecules of general shape) have three distinct moments of inertia and a correspondingly more complex rotational spectrum, beyond this page's rigid linear-rotor treatment.

Proof

E_J = BJ(J+1), \qquad B=\frac{h}{8\pi^2cI}, \qquad J=0,1,2,\ldots

Solving the rigid-rotor Schrödinger equation for a diatomic molecule gives quantised rotational energy levels indexed by the rotational quantum number \(J\), with the rotational constant \(B\) (in wavenumber units) fixed entirely by the molecule's moment of inertia \(I\). B

I=\mu r^2, \qquad \mu=\frac{m_1 m_2}{m_1+m_2}

For a diatomic, the moment of inertia reduces to the reduced mass \(\mu\) of the two atoms times the square of the bond length \(r\); since \(\mu\) is known directly from atomic masses, measuring \(I\) (via \(B\), Step 1) determines \(r\) uniquely. A

\Delta J=\pm1, \qquad \text{requires a permanent dipole moment (Hypotheses)}

The rotational selection rule follows from evaluating the transition dipole moment between rotational states: it is non-zero only between adjacent \(J\) levels, and only at all if the molecule has a permanent dipole moment for the rotating charge distribution to couple to incident radiation. A

\tilde\nu_{J\to J+1} = E_{J+1}-E_J = 2B(J+1)

Applying the selection rule to the energy-level formula (Step 1) gives absorption lines at \(2B, 4B, 6B,\ldots\) for \(J=0\to1,\,1\to2,\,2\to3,\ldots\) — a series of perfectly evenly spaced lines separated by exactly \(2B\), the single most distinctive signature of a rigid-rotor spectrum. A

\text{Measure the line spacing } 2B \Rightarrow B \Rightarrow I \Rightarrow r

Working the chain of Steps 1–4 backwards from an observed spectrum: the measured spacing gives \(B\) directly, Step 1 converts \(B\) to \(I\), and Step 2 converts \(I\) to the bond length \(r\) once the reduced mass is known — the practical measurement procedure underlying this entire result. A

Result

B=\frac{h}{8\pi^2cI}, \qquad \tilde\nu_{J\to J+1}=2B(J+1)

Reading. A rigid diatomic rotor's absorption spectrum is a ladder of perfectly evenly spaced lines, and the single spacing value \(2B\) encodes the molecule's moment of inertia, and hence its bond length, essentially exactly.

Scope. Requires a permanent dipole moment (Hypotheses) and applies most cleanly to small, rigid, gas-phase linear or symmetric-top molecules; breaks down at high \(J\) (centrifugal distortion stretches the bond) and does not apply in this simple form to general asymmetric-top polyatomics.

Corollaries & converses

Isotopic substitution changes the reduced mass \(\mu\) (Step 2) but leaves the bond length \(r\) essentially unchanged, since bonding is an electronic property to which nuclear mass is nearly irrelevant; comparing the rotational spectra of two isotopologues of the same molecule is therefore a standard, independent check on a measured bond length and a precise route to isotopic mass ratios.
vibrational-ir-spectroscopy's higher-resolution spectra show rotational fine structure (P and R branches) superimposed on the vibrational transition, since a real molecule rotates and vibrates simultaneously; the spacing within those branches is set by essentially the same \(B\) derived here.
Converse: a molecule showing no pure rotational absorption spectrum at all, despite being a genuine gas-phase species, is strong evidence that it has no permanent dipole moment (Hypotheses), exactly the situation for homonuclear diatomics, which must instead be studied by rotational Raman spectroscopy.

Fails without

Drop the rigid-rotor assumption (Hypotheses): at high \(J\), centrifugal distortion stretches the bond, and the strictly evenly-spaced line spacing of Step 4 becomes progressively compressed relative to the ideal rigid-rotor prediction, requiring an additional correction term to fit the real spectrum.
Drop the permanent-dipole requirement: a homonuclear diatomic such as \(\text{N}_2\) shows no pure rotational absorption spectrum at all, regardless of how it is illuminated, since there is no oscillating dipole for the radiation field to couple to.

Common errors

Assuming any diatomic molecule shows a pure rotational absorption spectrum — homonuclear diatomics such as \(\text{N}_2\) and \(\text{O}_2\), lacking a permanent dipole, show none at all (Hypotheses).
Reading the observed line spacing directly as \(B\) rather than \(2B\) (Step 4), giving a rotational constant, and hence a bond length, that is a factor of two wrong.
Confusing the rotational quantum number \(J\) with the vibrational quantum number \(v\) used in vibrational-ir-spectroscopy — the two describe entirely different types of molecular motion with very different energy-level spacings.
Forgetting that \(\mu\) is the reduced mass of the two atoms, not the mass of either atom alone or their simple sum, when converting a measured \(I\) into a bond length.

Discussion

Practical microwave spectroscopy developed rapidly after the Second World War, drawing directly on microwave-generation technology originally developed for radar; rotational transitions of small molecules typically fall in the microwave region of the electromagnetic spectrum, at far lower photon energy than vibrational (infrared) or electronic (ultraviolet-visible) transitions, exactly the energy-scale ordering that runs through beer-lambert-law's companion spectroscopic results in this unit.

Rotational spectroscopy is also the primary tool of radio and microwave astronomy for identifying molecules in cold interstellar gas clouds, where electronic and vibrational transitions are far too energetic to be populated at the prevailing low temperatures, but rotational transitions of simple, dipolar molecules such as carbon monoxide remain readily excited and observable, making CO's rotational lines a standard tracer used to map molecular gas throughout the galaxy.

Common misconception: that a larger molecule always has a larger rotational constant \(B\). Since \(B\propto1/I\), and \(I\) grows with both mass and the square of bond length, heavier or more extended molecules in fact have systematically smaller \(B\) and more closely spaced rotational lines, not larger ones.

Worked examples

\text{HCl: } B\approx10.59\ \text{cm}^{-1}\ (\text{a standard, widely tabulated value})

Using \(\mu(\text{H},{}^{35}\text{Cl})=\dfrac{(1.008)(34.97)}{1.008+34.97}\,\text{u}\approx0.9796\,\text{u}=1.627\times10^{-27}\,\text{kg}\), Step 1's formula \(B=h/(8\pi^2cI)\) converts the measured spacing between HCl's microwave absorption lines into this rotational constant. A

I=\frac{h}{8\pi^2cB}, \qquad r=\sqrt{I/\mu}

Rearranging Step 1 and substituting \(B=10.59\,\text{cm}^{-1}\) gives \(I\approx2.65\times10^{-47}\,\text{kg m}^2\); dividing by \(\mu\) and taking the square root gives \(r\approx1.275\times10^{-10}\,\text{m}=127.5\,\text{pm}\), matching the well-established bond length of HCl closely. A

r(\text{H-Cl})\approx127.5\,\text{pm}, \quad \text{from } B\approx10.59\,\text{cm}^{-1}\text{ alone}

Reading. A single measured spectroscopic quantity, the rotational constant, is converted through two purely algebraic steps into a bond length accurate to well within a picometre.

Scope. The identical two-step procedure (spacing \(\to B\to I\to r\)) applies to any rigid diatomic showing a resolved pure rotational spectrum.

Problems

Given a diatomic molecule's rotational absorption lines spaced \(2B=1.688\,\text{cm}^{-1}\) apart, find \(B\) and identify which two transitions (in terms of \(J\)) are adjacent to each other with exactly this spacing.
Solution
\(B=0.844\,\text{cm}^{-1}\). By Step 4, every adjacent pair of lines in the series \(J=0\to1,\,1\to2,\,2\to3,\ldots\) is separated by exactly \(2B\); any such adjacent pair (e.g. the \(J=0\to1\) and \(J=1\to2\) transitions) shows this spacing, since the whole series is evenly spaced by construction.
Explain, using Step 2's formula \(I=\mu r^2\), why substituting deuterium for hydrogen in HCl (giving DCl) shifts the rotational spectrum to smaller line spacing, even though the bond length is essentially unchanged.
Solution
Deuterium substitution roughly doubles \(\mu\) (D is about twice as massive as H, while Cl's mass and the bond's electronic structure, and hence \(r\), are essentially unaffected). Since \(I=\mu r^2\) grows in direct proportion to \(\mu\) at fixed \(r\), and \(B=h/(8\pi^2cI)\) is inversely proportional to \(I\), \(B\) for DCl is roughly half that of HCl, and the line spacing \(2B\) (Step 4) shrinks correspondingly.
A gas sample shows no pure rotational absorption spectrum at all, despite clearly containing diatomic molecules (confirmed by another technique). Suggest the most likely explanation.
Solution
By the Hypotheses, a pure rotational absorption spectrum requires a permanent dipole moment; the most likely explanation is that the diatomic is homonuclear (such as \(\text{N}_2\), \(\text{O}_2\), or \(\text{H}_2\)), which has no permanent dipole by symmetry and is therefore microwave-inactive, regardless of how strongly it might absorb in other spectral regions or techniques.