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13.7 Vibrational and Electronic Spectra: Anharmonicity and Franck–Condon

★★★ 2021–2025: 19 parts, 140 marks (Civil Services 2 parts / 15 m · IFoS 5 / 35 m · Geo-Scientist P-II 12 / 90 m). Geo-Scientist P-II asks a 10-mark question on it every year; isotope numericals and CO₂ modes repeat.

Syllabus lines. Civil Services: “Vibrational: diatomic molecules, linear triatomic molecules, specific frequencies of functional groups in polyatomic molecules. Electronic: singlet and triplet states”. IFoS: “Vibrational spectra: diatomic molecules, linear triatomic molecules, specific frequencies of functional groups in polyatomic molecules”. Geo-Scientist P-II: “Vibrational spectroscopy of diatomic molecules: simple harmonic oscillator model, selection rules and vibration spectra … Overtones, anharmonicity, normal mode analysis of polyatomic molecules”; “Franck–Condon principle and vibrational structure of electronic spectra. Bond dissociation and principle of determination of dissociation energy”; “spectrum of hydrogen atom”. Rotational and Raman spectra: 13.6.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

13.7.1 The harmonic oscillator ★★★

(13.7.1)
(13.7.2)
  • Zero-point energy: the lowest level lies ½hν above the potential minimum and stays even at 0 K. A molecule at rest at re would have Δx = 0 and Δp = 0, which the uncertainty principle forbids. In joules E₀ = ½hν; in cm⁻¹ ε₀ = ½ν (anharmonic: G(0) = ½ωe − ¼ωexe) IFoS 2024 Q5(b) 5m Geo 2023 Q1(m) 5m.SN
  • Selection rules: (i) gross rule: the dipole moment must change during the vibration, (∂μ/∂Q)₀ ≠ 0, so homonuclear H₂, N₂, O₂ give no IR spectrum; (ii) specific rule: Δv = ±1 (harmonic). All levels are equally spaced, so a harmonic diatomic shows one line at ν.SN
  • Force constant measures bond strength; mass sets the frequency. H₂⁺ 2322, D₂ 3118, H₂ 4400 cm⁻¹ give k = 159, 573, 570 N m⁻¹: H₂ and D₂ share one potential curve (same electrons), so the same k; H₂⁺ (bond order ½) has a much weaker bond Geo 2024 Q9 10m.SN

Solved PYQ Geo 2024 Q9

  1. k = 4π²c²ν²μ with c = 2.998 × 10¹⁰ cm s⁻¹, μ in kg (1 u = 1.6605 × 10⁻²⁷ kg).
  2. H₂: μ = 0.5 u = 8.30 × 10⁻²⁸ kg; k = 4π²(2.998 × 10¹⁰ × 4400)² × 8.30 × 10⁻²⁸ = 570 N m⁻¹.
  3. D₂: μ = 1.0 u → 573 N m⁻¹; H₂⁺: μ = 0.5 u → 159 N m⁻¹.
  4. Ratio ν(H₂)/ν(D₂) = 1.41 ≈ √2 = √(μD₂/μH₂): the isotope changes only the mass.

13.7.2 Isotope effects and zero-point energy ★★★

(13.7.3)
  • C–H → C–D: μ(C–H) = 12/13 u, μ(C–D) = 24/14 u; ν(C–D) = 2900 × √(7/13) = 2128 cm⁻¹, a fall of 772 cm⁻¹ IFoS 2022 Q5(e) 5m.SN
  • HCl → DCl (H = 1, D = 2, Cl = 35.5): μ(HCl) = 0.9726 u, μ(DCl) = 1.8933 u; ν(DCl) = 2890 × √(0.9726/1.8933) = 2071 cm⁻¹ (from k = 483 N m⁻¹ directly: 2081 cm⁻¹) Geo 2023 Q15(a) 10m.SN
  • H₂, HD, D₂: μ = 0.50, 0.67, 1.00 u, same k, so H₂ > HD > D₂ (4161, 3632, 2994 cm⁻¹). They have no dipole: the values come from Raman spectra Geo 2021 Q1(n) 5m.SN

Solved PYQ CSE 2023 Q6(a)(i) 10m IFoS 2025 Q8(b)(i) 10m

HCl + D₂ → DCl + HD; ν = 2885, 2990, 1990, 3627 cm⁻¹ (v = 0 states). Energy change?

  1. Electronic energies are unchanged by isotopes; only zero-point energies differ: ΔE₀ = ½hc[Σν(products) − Σν(reactants)].
  2. Σ products = 1990 + 3627 = 5617 cm⁻¹; Σ reactants = 2885 + 2990 = 5875 cm⁻¹; Δ = ½(5617 − 5875) = −129 cm⁻¹.
  3. Per molecule: −129 × 100 m⁻¹ × 6.626 × 10⁻³⁴ × 2.998 × 10⁸ = −2.56 × 10⁻²¹ J; × NA = −1.54 kJ mol⁻¹.
  4. Negative: energy is released (exothermic by 1.54 kJ mol⁻¹), because the products together hold less zero-point energy.SN

13.7.3 Anharmonicity, overtones and hot bands ★★★

020 00040 000internuclear distance r / pmenergy / cm−1100200300v = 0: zero-point energy G(0)dissociation limitDₑD0harmonicMorseD0 = Dₑ − G(0)
Fig. 13.7.1 Harmonic and Morse curves for HCl (computed: ω̃ₑ = 2990.9, ω̃ₑxₑ = 52.82 cm⁻¹).
Draw it in 60 s: a U-shaped well that flattens to a horizontal line on the right; dashed parabola inside it; levels closer near the top; arrows Dₑ from the bottom and D0 from v = 0.
(13.7.4)
harmonic: equal gapsanharmonic: gaps shrinkv = 0v = 0v = 1v = 1v = 2v = 2v = 3v = 3Δv = ±1 onlyfundamental1st overtonehot bandfundamental ωₑ(1 − 2xₑ) · 1st overtone 2ωₑ(1 − 3xₑ) · hot band ωₑ(1 − 4xₑ)
Fig. 13.7.2 Fundamental, first overtone and hot band.
Draw it in 60 s: two ladders: equal rungs on the left; on the right rungs closing up; arrows 0→1 (fundamental), 0→2 (overtone), dashed 1→2 (hot band).
LineTransitionPositionHow to tell it
Fundamental0 → 1ωe(1 − 2xe)strongest
First overtone0 → 22ωe(1 − 3xe)just under twice the fundamental; weak; hardly changes with T
Hot band1 → 2ωe(1 − 4xe)2ωexe below the fundamental; grows on heating
  • Real bonds are anharmonic: levels converge, Δv = ±2, ±3 become weakly allowed (overtones), and the bond breaks at De. HCl: 2886 (fundamental), 5668 (overtone) cm⁻¹ Geo 2022 Q15(b) 10m.SN
  • Hot-band population, I₂ at 300 K (ωe = 215 cm⁻¹, xe = 0.003): G(1) − G(0) = 215(1 − 2 × 0.003) = 213.7 cm⁻¹; N₁/N₀ = exp(−hc × 213.7/kBT) = e−1.025 = 0.359: the hot band is about 36 % of the fundamental (equal transition strengths assumed); it lies at 212.4 cm⁻¹ IFoS 2025 Q8(b)(ii) 10m.SN
  • De ≈ ωe²/(4ωexe) (Morse); D₀ = De − G(0). Plotting ΔG against v + 1 and finding the area under the line (Birge–Sponer) gives D₀.

13.7.4 Polyatomic molecules: normal modes and IR activity ★★★

OCOν1 sym. stretch1388 cm−1IR inactiveOCOν3 antisym. stretch2349 cm−1IR activeOCOν2 bend (two, degenerate)667 cm−1IR activeOHHν1 sym. stretch3657 cm−1 · IR activeOHHν2 bend1595 cm−1 · IR activeOHHν3 antisym. stretch3756 cm−1 · IR activeCO2 (linear): 3N − 5 = 4 modes · H2O (bent): 3N − 6 = 3 modes, all IR active
Fig. 13.7.3 Normal modes of CO₂ and H₂O.
Draw it in 60 s: three O–C–O sticks: arrows out–out (no dipole change), left–right–left, up–down–up; below, a bent H–O–H three times: both H out, both H in-and-up, one out one in.
  • Number of normal modes: 3N − 5 (linear), 3N − 6 (non-linear). CO₂: 4 modes but 3 fundamentals (the two bends are degenerate) Geo 2022 Q10 10m CSE 2023 Q8(c)(iii) 5m.SN
  • A mode is IR active if it changes the dipole moment. CO₂: ν₁ symmetric stretch (1388 cm⁻¹) inactive; ν₂ bend (667) and ν₃ antisymmetric stretch (2349) active. Observed IR: two strong bands; ν₁ shows only in the Raman (Fermi doublet 1285/1388); weak combination bands near 3716 and 3609 cm⁻¹.SN
  • Mutual exclusion: with a centre of symmetry no mode is both IR and Raman active. CS₂ (centrosymmetric): bend 397 and antisymmetric stretch 1535 cm⁻¹ IR active, symmetric stretch 658 cm⁻¹ inactive. OCS (no centre, polar): all three (520, 859, 2062 cm⁻¹) IR active Geo 2021 Q15(b) 5m.SN
  • Does it have an IR spectrum? A permanent dipole is not needed, only a dipole change in some mode: CH₃OH, CH₃NH₂, CO, CO₂ and CH₄ are all IR active (CH₄ and CO₂ through their asymmetric modes) IFoS 2024 Q8(a)(ii) 5m.SN

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