University Physics IV · Molecular and Solid-State Physics · 12.4

Vibrational & Rotation-Vibration Spectra

An infrared band is a measuring instrument in disguise. This topic teaches you to turn one into numbers: the line spacing gives a bond length, the origin gives a force constant, the overtone gives the anharmonicity — and the missing central line tells you the molecule had to rotate as it stretched.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Near the bottom of any bonding curve, U(R) is a parabola: expand about Rₑ, the linear term dies at the minimum, and what survives is ½k(R−Rₑ)² with k the curvature there. That single move buys the whole harmonic ladder, Eᵥ = (v+½)ħω with ω = √(k/μ) — evenly spaced rungs, a Δv = ±1 rule, and a floor at ½ħω that no cooling removes. The cost is that a parabola never dissociates, while the real curve flattens toward Dₑ.

Morse repairs it: rungs shrink by 2ω̃ₑ xₑ per step, the ladder terminates, and the same anharmonicity mixes the states so that weak Δv = ±2 overtones appear a hundred wavenumbers below twice the fundamental. Meanwhile no vibrational jump in a gas happens alone. The photon carries one unit of angular momentum, and a Σ-state diatomic has no electronic angular momentum to absorb it, so ΔJ = ±1 is compulsory: the band splits into a P branch and an R branch about an origin that carries no line at all.

Because the v = 1 bond is longer, B̃₁ < B̃₀, and the two branches drift at different rates — a nuisance if you wanted one number, a gift if you want two.

Simple definition: A vibrational spectrum is the set of transitions between the quantised stretching levels of a bond, Eᵥ = (v+½)ħω; in a gas each one must be accompanied by a rotational change, so it appears as a band of lines rather than a single line.
Example: H³⁵Cl absorbs its fundamental near 2886 cm⁻¹, but not as one line: an R branch sits above and a P branch below, lines about 21 cm⁻¹ apart, with a 41 cm⁻¹ hole exactly where the origin is.

Harmonic ladder from the curvature of U(R)Eᵥ = (v + ½)ħω, ω = √(k/μ), k = d²U/dR² at Rₑ

Rungs are equally spaced, and the lowest sits ½ħω above the well bottom — 0.185 eV for HCl.

v = 0, 1, 2, …; k in N m⁻¹; μ = m₁m₂/(m₁+m₂) in kg

Reduced mass and the isotope shiftμ = m₁m₂/(m₁+m₂), ω̃ ∝ μ^(−½)

k belongs to the electrons and is shared, so a 0.15% mass change drops the H³⁷Cl origin by 2.2 cm⁻¹.

H³⁵Cl μ = 0.9796 u, H³⁷Cl μ = 0.9811 u; 1 u = 1.6605×10⁻²⁷ kg

Morse levels and the closing ladderG(v) = ω̃ₑ(v+½) − ω̃ₑ xₑ(v+½)², ΔG = ω̃ₑ − 2ω̃ₑ xₑ(v+1)

Spacings shrink by 2ω̃ₑ xₑ per rung, so the ladder ends near v = 27 instead of running forever.

All in cm⁻¹; for H³⁵Cl ω̃ₑ = 2990 cm⁻¹ and ω̃ₑ xₑ = 51.9 cm⁻¹

Fundamental, overtone, and well depthν̃₁₀ = ω̃ₑ − 2ω̃ₑ xₑ, ν̃₂₀ = 2ω̃ₑ − 6ω̃ₑ xₑ, D̃ₑ = ω̃ₑ²/4ω̃ₑ xₑ

Two measured origins fix both constants; the Morse Dₑ then overshoots the true 4.62 eV by about 16%.

H³⁵Cl measures 2885.9 and 5668.0 cm⁻¹, giving D̃ₑ ≈ 4.31×10⁴ cm⁻¹

P and R branch line positionsν̃ = ν̃₀ + (B̃₁ + B̃₀)m + (B̃₁ − B̃₀)m², m = J″+1 (R), −J″ (P)

One formula covers both branches; the m² term alone makes R crowd together and P spread apart.

m ≠ 0; B̃ᵥ = h/(8π²cμ⟨R⁻²⟩ᵥ) in cm⁻¹, and B̃ᵥ = B̃ₑ − α̃ₑ(v+½)

Combination differencesR(J−1) − P(J+1) = 4B̃₀(J+½), R(J) − P(J) = 4B̃₁(J+½)

Separates the two rotational constants using measured lines only — the origin itself carries none.

Two lines sharing an upper level isolate B̃₀; sharing a lower level isolates B̃₁

Why a bond behaves like a spring near the bottom

Expand the electronic energy curve about its minimum: U(R) = U(Rₑ) + U′(Rₑ)(R−Rₑ) + ½U″(Rₑ)(R−Rₑ)² + … . The linear term vanishes because Rₑ is a minimum, so the leading behaviour is a parabola with spring constant k = U″(Rₑ) — the curvature of the well, in N m⁻¹. Solve that oscillator with the reduced mass μ = m₁m₂/(m₁+m₂) and you get Eᵥ = (v+½)ħω with ω = √(k/μ). For H³⁵Cl, ω̃ₑ = 2990 cm⁻¹ gives ω = 2πcω̃ₑ = 5.63 × 10¹⁴ rad s⁻¹, and with μ = 0.9796 u = 1.627 × 10⁻²⁷ kg, k = μω² = 516 N m⁻¹. Note what k is not: it is not the bond strength. F₂ has almost the same curvature, k ≈ 470 N m⁻¹, but a well only 1.66 eV deep against HCl's 4.62 eV. Stiffness is a local property of the bottom; depth is a global property of the whole curve, and the harmonic model, which has no top, cannot tell them apart.

The zero-point rung you cannot remove

Set v = 0 and the energy is ½ħω, not zero. For H³⁵Cl that is 1495 cm⁻¹, or 0.185 eV, and it is why two dissociation energies are quoted for every molecule: Dₑ is measured from the bottom of the well, D₀ from the v = 0 level, and D₀ = Dₑ − ½ħω. HCl's 4.62 eV and 4.43 eV differ by exactly that 0.19 eV. The same rung sets what a room-temperature gas can do. At 300 K, kT is worth 208.5 cm⁻¹, so the v = 1 to v = 0 population ratio is exp(−2886/208.5) = e^(−13.8) ≈ 1 × 10⁻⁶. Fundamental-band absorption therefore starts essentially from v = 0 alone, which is why the line intensities across a band track rotational populations and not vibrational ones. And because ω ∝ μ^(−½), deuterating a bond lowers the zero-point rung by roughly 30%, leaving a C–D bond measurably harder to break than C–H — the kinetic isotope effect, read straight off the ladder.

Anharmonicity closes the ladder and turns on overtones

A parabola never dissociates. Morse's curve U(R) = Dₑ[1 − e^{(−a(R−Rₑ)})]² does, and it is still exactly soluble: G(v) = ω̃ₑ(v+½) − ω̃ₑ xₑ(v+½)², so successive spacings ΔG = ω̃ₑ − 2ω̃ₑ xₑ(v+1) shrink by a fixed 2ω̃ₑ xₑ per rung. For H³⁵Cl, with ω̃ₑ = 2989.7 and ω̃ₑ xₑ = 51.9 cm⁻¹, the ladder runs 2886.2, 2782.4, 2678.6 cm⁻¹ … and closes at v = 27. Extrapolating those spacings linearly to zero is the Birge–Sponer route to D₀. Anharmonicity also breaks the selection rule. Δv = ±1 rested on two things: harmonic eigenfunctions, for which ⟨v′|x|v⟩ vanishes unless v′ = v ± 1, and a dipole expanded only to first order, μ(R) ≈ μₑ + μ′x. Morse eigenstates are mixtures, and the μ″x²/2 term connects v to v ± 2, so a weak first overtone appears — for HCl at 5668 cm⁻¹, roughly 1% as strong as the fundamental and 104 cm⁻¹ below twice it.

Every vibrational jump drags a rotation with it

A photon carries one unit of angular momentum and it has to go somewhere. In a ¹Σ diatomic there is no electronic angular momentum to absorb it, so the rotational quantum number must change: ΔJ = ±1, with ΔJ = 0 strictly forbidden. Every vibrational transition therefore arrives as a band. Adding rigid-rotor energy to each vibrational level gives ν̃ = ν̃₀ + B̃₁J′(J′+1) − B̃₀J″(J″+1). Taking J″ = 0 → J′ = 1 puts the first R line at ν̃₀ + 2B̃₁; taking J″ = 1 → J′ = 0 puts the first P line at ν̃₀ − 2B̃₀. For H³⁵Cl those are 2906.2 and 2865.0 cm⁻¹, a gap of 2(B̃₀ + B̃₁) = 41.2 cm⁻¹ where neighbouring lines are only about 20.6 cm⁻¹ apart. The band origin is thus a fitted number, never a measured line. The missing Q branch is a fact about this molecule, not about light: NO, with a ²Π ground state, does show a Q branch, and so do the bending modes of CO₂ and HCN.

B̃₁ < B̃₀, and the two branches drift apart

The rotational constant is not a constant. B̃ᵥ = h/(8π²cμ⟨R⁻²⟩ᵥ), and because the outer wall of an anharmonic well is soft, raising v raises ⟨R⟩ and lowers ⟨R⁻²⟩. Empirically B̃ᵥ = B̃ₑ − α̃ₑ(v+½); for H³⁵Cl, B̃ₑ = 10.593 and α̃ₑ = 0.302 cm⁻¹, so B̃₀ = 10.44 and B̃₁ = 10.14 cm⁻¹. Write both branches with one running index, m = J″+1 in R and m = −J″ in P: ν̃ = ν̃₀ + (B̃₁+B̃₀)m + (B̃₁−B̃₀)m². The linear term sets the ≈ 2B̃ spacing; the quadratic term, negative because B̃₁ < B̃₀, pulls every line down by an amount growing as m². On the R side that fights the linear growth and the lines crowd — 19.68 cm⁻¹ between the first pair, 18.48 cm⁻¹ between the third and fourth. On the P side it reinforces, and spacings open from 21.48 to 22.08 cm⁻¹. Far enough out, at m ≈ 34 here, the R branch would stop and turn back into a band head; in electronic spectra, where ΔB̃ is far larger, heads appear within the first dozen lines.

Reading a bond length and a force constant out of the band

The extraction runs in a fixed order, and each step names its own cost. First, combination differences: R(J−1) − P(J+1) = 4B̃₀(J+½) uses two lines sharing an upper level, so it returns B̃₀ alone, while R(J) − P(J) = 4B̃₁(J+½) returns B̃₁ — neither needs the origin, which carries no line. Second, the bond length: B̃₀ = 10.44 cm⁻¹ with μ = 1.6266 × 10⁻²⁷ kg gives I = 2.681 × 10⁻⁴⁷ kg m² and R₀ = √(I/μ) = 128.4 pm, sensibly longer than Rₑ = 127.5 pm. Third, fit the m-polynomial for ν̃₀, then combine the fundamental with the overtone to split ω̃ₑ from ω̃ₑ xₑ; only then is k = μ(2πcω̃ₑ)² = 516 N m⁻¹ legitimate. What limits each number: the rigid rotor ignores centrifugal distortion, −D̃J²(J+1)² with D̃ ≈ 5.1 × 10⁻⁴ cm⁻¹, which already pushes the J = 10 level about 6 cm⁻¹ down — a third of a line spacing — so a fit run to high J with D̃ omitted biases B̃₀ and the bond length with it.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotational constant B₀ (v = 0)10.4 cm⁻¹

Bond stretch B₀ − B₁0.30 cm⁻¹

Temperature T300 K

Push the stretch from 0 to 0.6 cm⁻¹ and watch the R branch on the right crowd together while the P branch on the left spreads apart — that one-sided drift is B₁ < B₀, the bond longer in v = 1. Then raise T: the tall sticks move out to higher J″ while every line position stays put.

B₀ (v = 0)10.40 cm⁻¹

B₁ (v = 1)10.10 cm⁻¹

GAP AT ORIGIN41.00 cm⁻¹

BRIGHTEST J″3

Live interpretationB₀ (v = 0): 10.40 cm⁻¹. B₁ (v = 1): 10.10 cm⁻¹. GAP AT ORIGIN: 41.00 cm⁻¹. BRIGHTEST J″: 3

Catch the common trap

Explain before calculating.

The infrared fundamental of H³⁵Cl is quoted at 2885.9 cm⁻¹, yet no absorption line stands at that wavenumber. The two innermost lines sit at 2906.2 and 2865.0 cm⁻¹, while neighbouring lines within each branch are about 21 cm⁻¹ apart. What accounts for the doubled gap in the middle?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyH³⁵Cl has a harmonic vibrational constant ω̃ₑ = 2990 cm⁻¹ and a reduced mass μ = 0.9796 u. Find the force constant of the bond and the zero-point energy in electronvolts, then compare that zero-point energy with kT at 300 K.

Turn the wavenumber into an angular frequency: ω = 2πcω̃ₑ = 2π(2.998 × 10¹⁰ cm s⁻¹)(2990 cm⁻¹) = 5.632 × 10¹⁴ rad s⁻¹.
Put the reduced mass in kilograms: μ = 0.9796 × 1.6605 × 10⁻²⁷ kg = 1.6267 × 10⁻²⁷ kg. It is close to the proton mass because the chlorine barely moves.
Invert ω = √(k/μ): k = μω² = 1.6267 × 10⁻²⁷ × (5.632 × 10¹⁴)² = 1.6267 × 10⁻²⁷ × 3.172 × 10²⁹ = 516 N m⁻¹.
Zero-point energy is ½ħω = ½hcω̃ₑ = ½ × 2990 cm⁻¹ = 1495 cm⁻¹. With 1 cm⁻¹ = 1.2398 × 10⁻⁴ eV that is 0.1854 eV.
kT at 300 K is 0.02585 eV, so the zero-point energy is 7.2 kT. Vibration is frozen out at room temperature, unlike rotation, whose ≈ 21 cm⁻¹ steps sit far below the 208 cm⁻¹ that kT is worth.

Answerk = 5.16 × 10² N m⁻¹ and a zero-point energy of 0.185 eV — 7.2 times kT at 300 K, and exactly the gap between the well depth Dₑ = 4.62 eV and the measured dissociation energy D₀ = 4.43 eV.

MediumFor the H³⁵Cl fundamental, ν̃₀ = 2885.9 cm⁻¹, B̃₀ = 10.44 cm⁻¹ and B̃₁ = 10.14 cm⁻¹. Locate R(0), R(1), R(2), R(3), P(1), P(2) and P(3), then show that the central gap is twice a normal step while the R spacings shrink and the P spacings grow.

Line positions are ν̃ = ν̃₀ + B̃₁J′(J′+1) − B̃₀J″(J″+1), with J′ = J″ + 1 in the R branch and J′ = J″ − 1 in the P branch. ΔJ = 0 is not allowed, so no J″ produces a line at ν̃₀ itself.
R(0): J″ = 0 → J′ = 1, so ν̃ = 2885.9 + 10.14(2) − 0 = 2906.18 cm⁻¹. P(1): J″ = 1 → J′ = 0, so ν̃ = 2885.9 + 0 − 10.44(2) = 2865.02 cm⁻¹.
Central gap: 2906.18 − 2865.02 = 41.16 cm⁻¹, which is exactly 2(B̃₀ + B̃₁) = 2(20.58).
R(1) = 2885.9 + 10.14(6) − 10.44(2) = 2925.86; R(2) = 2885.9 + 10.14(12) − 10.44(6) = 2944.94; R(3) = 2885.9 + 10.14(20) − 10.44(12) = 2963.42. Spacings: 19.68, 19.08, 18.48 cm⁻¹.
P(2) = 2885.9 + 10.14(2) − 10.44(6) = 2843.54; P(3) = 2885.9 + 10.14(6) − 10.44(12) = 2821.46. Stepping down from P(1), the spacings are 21.48 then 22.08 cm⁻¹.
So an ordinary step is about 20.6 cm⁻¹, the central gap is twice it, and the two branches drift in opposite directions by 0.60 cm⁻¹ per line — that drift is 2(B̃₀ − B̃₁).

AnswerR(0) = 2906.18 cm⁻¹, P(1) = 2865.02 cm⁻¹, a central gap of 41.16 cm⁻¹ = 2(B̃₀ + B̃₁), twice the ordinary step. R spacings fall 19.68 → 18.48 cm⁻¹ while P spacings rise 21.48 → 22.08 cm⁻¹.

HardThe H³⁵Cl fundamental band origin is 2885.9 cm⁻¹ and the first overtone origin is 5668.0 cm⁻¹. Treating the ladder as Morse, find ω̃ₑ and ω̃ₑ xₑ, estimate the well depth from D̃ₑ = ω̃ₑ²/4ω̃ₑ xₑ, compare it with the spectroscopic 4.62 eV, and say how many bound vibrational levels the fitted curve holds.

Morse origins: ν̃(0→1) = ω̃ₑ − 2ω̃ₑ xₑ and ν̃(0→2) = 2ω̃ₑ − 6ω̃ₑ xₑ. Two equations, two unknowns.
Double the first and subtract the second: 2(2885.9) − 5668.0 = 5771.8 − 5668.0 = 103.8 cm⁻¹, and that difference is 2ω̃ₑ xₑ, so ω̃ₑ xₑ = 51.9 cm⁻¹. The overtone falling 104 cm⁻¹ short of twice the fundamental is the whole measurement.
Back-substitute: ω̃ₑ = 2885.9 + 2(51.9) = 2989.7 cm⁻¹, which is 3.6% above the observed fundamental. Using 2885.9 as ω̃ₑ instead would give k = 481 rather than 516 N m⁻¹.
Well depth: D̃ₑ = ω̃ₑ²/(4ω̃ₑ xₑ) = 2989.7²/(4 × 51.9) = 8.9383 × 10⁶/207.6 = 4.3055 × 10⁴ cm⁻¹. At 1.2398 × 10⁻⁴ eV per cm⁻¹ that is 5.34 eV.
Against the spectroscopic Dₑ = 4.62 eV the estimate is 16% high, because the real curve falls to its asymptote faster than a Morse exponential does; a linear Birge–Sponer extrapolation overshoots for the same reason.
Spacings ΔG = ω̃ₑ − 2ω̃ₑ xₑ(v+1) vanish when v + 1 = 2989.7/103.8 = 28.8, so the last bound rung is v = 27 and the ladder holds 28 levels — again generous, since HCl really supports about twenty.

Answerω̃ₑ = 2989.7 cm⁻¹, ω̃ₑ xₑ = 51.9 cm⁻¹, D̃ₑ = 4.31 × 10⁴ cm⁻¹ = 5.34 eV against the true 4.62 eV (16% high), and the ladder closes after v = 27, so 28 bound levels.