Vibrational and rotation-vibration spectra

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.

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.

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 2885.9, 2782.1, 2678.3 cm⁻¹ … and closes at v = 28. 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.

A photon carries one unit of angular momentum, so the triangle rule allows ΔJ = 0 or ±1 and nothing more; parity removes the middle option. In a ¹Σ diatomic the rotational levels alternate in parity with J, exactly as Y_JM does, while the dipole operator is odd, so ⟨J|cos θ|J⟩ = 0 and 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 whose Λ-doubling gives every J both parities, does show a Q branch, and so do the bending modes of CO₂ and HCN.

The rotational constant is not a constant. B̃ᵥ = h⟨R⁻²⟩ᵥ/(8π²cμ), 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.

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.