Spectroscopy: Atomic Terms, Rotational, Vibrational, Raman and Electronic Spectra, Line Shapes and NMR Principles
1. Atomic spectroscopy, Russell–Saunders coupling and term symbols
In light atoms the electrostatic repulsion between electrons is much larger than spin–orbit coupling, so the individual orbital angular momenta couple to a total L and the spins to a total S, and only then do L and S couple to J = L + S, L + S − 1, …, |L − S| — Russell–Saunders (LS) coupling. A term symbol is written ^(2S+1)L_J with L = 0, 1, 2, 3 written S, P, D, F. Filled subshells contribute nothing, and a subshell with n electrons gives the same terms as one with n holes. The number of microstates for n electrons in a subshell of 2(2l + 1) spin-orbitals is the binomial coefficient C(2(2l + 1), n): 15 for p², 20 for p³, 45 for d².
Hund’s rules pick the ground term: (1) maximum S; (2) for that S, maximum L; (3) J = |L − S| if the subshell is less than half full, J = L + S if more than half full. So carbon p² gives ³P, ¹D, ¹S with ground ³P₀; nitrogen p³ gives ⁴S, ²D, ²P with ground ⁴S₃/₂; oxygen p⁴ has ground ³P₂; Ti²⁺ d² has ground ³F₂. Selection rules follow from the transition-moment integral and the photon’s one unit of angular momentum: for one electron Δl = ±1 (the Laporte parity rule); for the atom ΔS = 0, ΔL = 0, ±1, ΔJ = 0, ±1 but J = 0 ↛ J = 0. Spin–orbit coupling splits a term into its J levels — the sodium D doublet at 589.0 and 589.6 nm is ²P₃/₂ → ²S₁/₂ and ²P₁/₂ → ²S₁/₂. For hydrogen the lines obey ν̃ = R_H(1/n₁² − 1/n₂²), with R_H = 109 677 cm⁻¹ (Lyman n₁ = 1, Balmer n₁ = 2).
2. Rotational spectra
A molecule shows a pure rotational (microwave) spectrum only if it has a permanent dipole moment — HCl and CO do, N₂, CO₂ and CH₄ do not. For a rigid diatomic F(J) = BJ(J + 1) with B = h/(8π²cI) and I = μr², and the selection rule ΔJ = ±1 gives absorption lines at ν̃ = 2B(J + 1) for J → J + 1: 2B, 4B, 6B, …, equally spaced by 2B. Measuring the spacing gives B, hence I and the bond length. Isotopic substitution changes μ and B but not r. At higher J the bond stretches — centrifugal distortion, F(J) = BJ(J + 1) − DJ²(J + 1)² — and the lines close up slightly. Line intensities follow the populations N_J ∝ (2J + 1)e^(−hcBJ(J+1)/kT), largest at J_max ≈ √(kT/2hcB) − ½.
Polyatomic rotors are classed by their three moments of inertia: linear (I_A = 0), spherical tops (all equal; CH₄, SF₆ — no microwave spectrum), symmetric tops (two equal; NH₃, CH₃Cl, energies F(J, K) = BJ(J + 1) + (A − B)K² for a prolate top) and asymmetric tops (all different; H₂O).
3. Vibrational, vibration–rotation and Raman spectra
A vibration is infrared active only if it changes the dipole moment. For a harmonic oscillator the selection rule is Δv = ±1, giving one line at ω̃. Real molecules are anharmonic, G(v) = (v + ½)ω̃ₑ − (v + ½)²ω̃ₑxₑ, so weak overtones (Δv = ±2, ±3) appear and the lines are not exact multiples: the fundamental 0 → 1 lies at ω̃ₑ(1 − 2xₑ), the first overtone 0 → 2 at 2ω̃ₑ(1 − 3xₑ) and the second at 3ω̃ₑ(1 − 4xₑ). The levels converge at v_max ≈ 1/(2xₑ) − ½, and the well depth is Dₑ ≈ ω̃ₑ/(4xₑ).
In the gas phase each vibrational transition carries rotational structure. With ΔJ = +1 the R branch lies above the band origin and with ΔJ = −1 the P branch below it; the Q branch (ΔJ = 0) is missing for a diatomic in a Σ state, leaving a gap of about 4B at the centre (NO, with electronic angular momentum, shows one). Because B in v = 1 is slightly smaller than in v = 0, R lines converge and P lines spread out.
Raman scattering is inelastic: most scattered light keeps the incident frequency (Rayleigh), a little is shifted down (Stokes) or up (anti-Stokes, weaker because it needs an excited initial state). A mode is Raman active if it changes the polarisability. The rotational Raman rule is ΔJ = 0, ±2, giving Stokes and anti-Stokes lines at 6B, 10B, 14B, … from the exciting line — first line at 6B, then a spacing of 4B — and homonuclear molecules such as N₂ and H₂, silent in the microwave, have rotational Raman spectra. The vibrational Raman rule is Δv = ±1, and in centrosymmetric molecules IR and Raman activity are mutually exclusive.
| Spectrum | Gross requirement | Specific rule | Line positions (diatomic) |
|---|---|---|---|
| Microwave | permanent dipole | ΔJ = ±1 | 2B, 4B, 6B … |
| Infrared | dipole changes in the vibration | Δv = ±1 (overtones weak) | ω̃ₑ(1 − 2xₑ) |
| Rotational Raman | anisotropic polarisability | ΔJ = 0, ±2 | 6B, 10B, 14B … |
| Vibrational Raman | polarisability changes | Δv = ±1 | same shift as the IR fundamental |
4. Electronic spectra, Beer–Lambert, Einstein coefficients and the Jablonski diagram
An electronic transition is fast compared with nuclear motion, so it is drawn vertically on the potential-energy curves — the Franck–Condon principle — and the intensity of each vibronic band is proportional to the square of the overlap of the two vibrational wave functions (the Franck–Condon factor). When the excited state has a longer bond, the most intense band is not the 0–0 band. The Beer–Lambert law states that absorbance A = log₁₀(I₀/I) = εcl, with ε the molar absorption (extinction) coefficient in L mol⁻¹ cm⁻¹, c in mol/L and l in cm; transmittance is T = I/I₀ = 10^(−A), so A = 1 passes 10% and A = 0.30 about 50%.
Einstein’s coefficients describe absorption (B₁₂), stimulated emission (B₂₁ = B₁₂ for equal degeneracies) and spontaneous emission (A₂₁), with A₂₁/B₂₁ = 8πhν³/c³: spontaneous emission grows as ν³, which is why it dominates in the UV and why lasers are harder to make at short wavelength. All three are proportional to |μ₁₂|², the square of the transition-moment integral μ₁₂ = ⟨ψ₁|μ̂|ψ₂⟩. The integrated absorption band measures the same quantity: the oscillator strength f = 4.32 × 10⁻⁹ ∫ε dν̃ (ε in L mol⁻¹ cm⁻¹, ν̃ in cm⁻¹), f ≈ 1 for a fully allowed transition, 10⁻⁵–10⁻³ for spin-allowed d–d bands and far less for spin-forbidden ones.
The Jablonski diagram maps what happens after absorption to S₁ or S₂. Vibrational relaxation (picoseconds) brings the molecule to the bottom of the excited state; internal conversion (IC) crosses non-radiatively to a lower state of the same multiplicity; fluorescence S₁ → S₀ is spin-allowed (lifetimes of nanoseconds) and lies at longer wavelength than absorption (the Stokes shift); intersystem crossing (ISC) S₁ → T₁ is spin-forbidden but helped by heavy atoms; phosphorescence T₁ → S₀ is slow (milliseconds to seconds). Kasha’s rule: emission occurs from the lowest excited state of a given multiplicity, whatever state was first excited.
5. Line broadening: Lorentzian and Gaussian shapes
No line is infinitely sharp. Lifetime (natural) broadening follows from the energy–time uncertainty: a state that decays with lifetime τ has a width δν ≈ 1/(2πτ), so a 1 ns lifetime gives about 160 MHz. It produces a Lorentzian profile, g(ν) ∝ 1/[(ν − ν₀)² + (Δ/2)²]. Collisional (pressure) broadening shortens the effective lifetime and is also Lorentzian, growing with pressure. Doppler broadening comes from the Maxwell distribution of molecular velocities along the line of sight and gives a Gaussian profile, g(ν) ∝ exp[−(ν − ν₀)²/2σ²], with a width proportional to ν₀√(T/m).
6. Principles of nuclear magnetic resonance
A nucleus of spin I has a magnetic moment μ = γħ√(I(I + 1)), where γ is the gyromagnetic (magnetogyric) ratio. In a field B₀ its 2I + 1 levels are E = −γħm_I B₀, so for a spin-½ nucleus (¹H, ¹³C, ¹⁹F, ³¹P) the two levels are split by ΔE = γħB₀ and resonance occurs at the Larmor frequency ν = γB₀/2π. For ¹H, γ = 2.675 × 10⁸ rad s⁻¹ T⁻¹, so a 9.4 T magnet gives 400 MHz. The energy gap is tiny compared with kT, so the population excess of the lower level is only of order 10⁻⁵ — NMR is intrinsically insensitive, and sensitivity rises with γ and B₀.
Electrons around a nucleus set up a small opposing field, so the local field is B₀(1 − σ), with σ the shielding constant. The chemical shift is defined relative to a reference (tetramethylsilane for ¹H and ¹³C) and made field-independent: δ = (ν − ν_ref)/ν_spectrometer × 10⁶ ppm. So 1 ppm is 400 Hz on a 400 MHz instrument and 600 Hz on a 600 MHz one. Deshielding by electronegative atoms, by the ring current of aromatic rings and by the anisotropy of C=O raises δ. Nuclei also couple through the bonding electrons: spin–spin coupling splits a signal, the coupling constant J (in Hz) is independent of field, and in a first-order spectrum n equivalent spin-½ neighbours give n + 1 lines in binomial intensities (1:1, 1:2:1, 1:3:3:1). Equivalent nuclei do not split one another.
Key takeaways
- LS coupling: terms ^(2S+1)L_J, Hund’s rules for the ground term (C ³P₀, N ⁴S₃/₂, O ³P₂); ΔS = 0, ΔL = 0, ±1, ΔJ = 0, ±1 (not 0 → 0), Δl = ±1.
- Microwave needs a permanent dipole: lines at 2B(J + 1), spacing 2B, B = h/8π²cI. Rotational Raman: ΔJ = ±2, first line 6B then 4B spacing.
- IR needs a changing dipole, Raman a changing polarisability; anharmonic fundamental ω̃ₑ(1 − 2xₑ), first overtone 2ω̃ₑ(1 − 3xₑ); P and R branches with no Q for Σ diatomics.
- A = εcl and T = 10^(−A); A/B = 8πhν³/c³; f = 4.32 × 10⁻⁹∫ε dν̃ ∝ |μ₁₂|²; Jablonski: IC, ISC, fluorescence (ns), phosphorescence (ms–s), Kasha’s rule.
- Lifetime and collisional broadening are Lorentzian (δν ≈ 1/2πτ), Doppler is Gaussian; NMR resonance at ν = γB₀/2π, δ in ppm is field-independent, J in Hz, and n equivalent neighbours give n + 1 lines.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the ground-state term symbol of the carbon atom (1s²2s²2p²)?
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Answer: A — ³P₀
p² gives ³P, ¹D and ¹S. Hund: maximum S picks ³P (L = 1, S = 1); the subshell is less than half full, so J = |L − S| = 0, giving ³P₀. ³P₂ is the ground term of oxygen, p⁴, which is more than half full.How many microstates arise from the p² electron configuration?
Numerical answer — type the value.
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Answer: 15
There are 6 p spin-orbitals and 2 electrons: C(6, 2) = 6!/(2!4!) = 15. Check against the terms: ³P (9) + ¹D (5) + ¹S (1) = 15. Squaring 6 counts the same electron twice and gives the wrong 36.The two sodium D lines at 589.0 and 589.6 nm arise because
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Answer: A — spin–orbit coupling splits the 3p ²P term into ²P₃/₂ and ²P₁/₂
For 3p¹, L = 1 and S = ½ give J = 3/2 and 1/2; the two levels differ by about 17 cm⁻¹, and each combines with ²S₁/₂. Hyperfine splitting from the nuclear spin is far smaller, and 3p → 3s obeys Δl = −1.H³⁵Cl has a bond length of 127.5 pm. Using atomic masses H = 1.008 and ³⁵Cl = 34.97 amu, what is its rotational constant B, in cm⁻¹, to one decimal place? (h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m/s, 1 amu = 1.6605 × 10⁻²⁷ kg)
Numerical answer — type the value.
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Answer: 10.6
μ = (1.008 × 34.97)/(35.978) = 0.9798 amu = 1.627 × 10⁻²⁷ kg; I = μr² = 1.627 × 10⁻²⁷ × (1.275 × 10⁻¹⁰)² = 2.645 × 10⁻⁴⁷ kg m². B = h/(8π²cI) = 1058 m⁻¹ = 10.58 cm⁻¹, i.e. 10.6. Forgetting to convert m⁻¹ to cm⁻¹ gives 1058.A diatomic molecule has B = 10.6 cm⁻¹. At what wavenumber, in cm⁻¹, does its J = 3 ← J = 2 rotational absorption line lie, to one decimal place?
Numerical answer — type the value.
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Answer: 63.6
For J → J + 1, ν̃ = 2B(J + 1); with J = 2 this is 6B = 6 × 10.6 = 63.6 cm⁻¹. Using the upper J (8B = 84.8) or the level energy 12B confuses the line with the level.For a diatomic molecule with B = 10.0 cm⁻¹ at 300 K, which rotational level J is the most populated? (Take kT/hc = 208.6 cm⁻¹ at 300 K.)
Numerical answer — type the value.
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Answer: 3
J_max = √(kT/2hcB) − ½ = √(208.6/20) − 0.5 = 3.23 − 0.5 = 2.73, and the nearest integer is J = 3. The (2J + 1) degeneracy pushes the maximum above J = 0 even though the Boltzmann factor alone always favours J = 0.A diatomic molecule has ω̃ₑ = 2990 cm⁻¹ and ω̃ₑxₑ = 52 cm⁻¹. What is the wavenumber of its fundamental (v = 0 → 1) vibrational band, in cm⁻¹?
Numerical answer — type the value.
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Answer: 2886
G(v) = (v + ½)ω̃ₑ − (v + ½)²ω̃ₑxₑ, so G(1) − G(0) = ω̃ₑ − 2ω̃ₑxₑ = 2990 − 104 = 2886 cm⁻¹. Subtracting ω̃ₑxₑ only once gives 2938; the first overtone would be 2ω̃ₑ − 6ω̃ₑxₑ = 5668 cm⁻¹.For the same molecule (ω̃ₑ = 2990 cm⁻¹, ω̃ₑxₑ = 52 cm⁻¹), at what wavenumber does the first overtone (v = 0 → 2) lie, in cm⁻¹?
Numerical answer — type the value.
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Answer: 5668
G(2) − G(0) = 2ω̃ₑ − 6ω̃ₑxₑ = 5980 − 312 = 5668 cm⁻¹. Doubling the fundamental (2 × 2886 = 5772) ignores the extra anharmonic shrinkage of the higher level, which is why overtones are never exact multiples.In the rotational Raman spectrum of a linear molecule with B = 2.0 cm⁻¹, how far from the exciting line, in cm⁻¹, is the first Stokes line?
Numerical answer — type the value.
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Answer: 12
The rule ΔJ = +2 from J = 0 gives F(2) − F(0) = 6B = 12 cm⁻¹; later lines follow at 10B, 14B …, a spacing of 4B = 8 cm⁻¹. Answering 2B = 4 cm⁻¹ applies the microwave rule ΔJ = ±1 to a Raman spectrum.Which of the following molecules show a pure rotational (microwave) absorption spectrum?
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Answer: A — HCl; C — CO
Microwave absorption needs a permanent dipole: HCl and CO have one. CO₂ is linear and centrosymmetric and CH₄ is a spherical top, both with zero dipole, so neither absorbs, though CO₂ has a rotational Raman spectrum.A solution of a dye with ε = 1.5 × 10⁴ L mol⁻¹ cm⁻¹ at 2.0 × 10⁻⁵ mol/L is measured in a 1.0 cm cell. What percentage of the incident light is transmitted, to the nearest whole number?
Numerical answer — type the value.
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Answer: 50
A = εcl = 1.5 × 10⁴ × 2.0 × 10⁻⁵ × 1.0 = 0.30, and T = 10^(−A) = 10^(−0.30) = 0.501, i.e. 50%. Using e^(−A) instead of 10^(−A) gives 74%, the natural-log slip.An absorption band is approximated as a rectangle of height ε = 1.0 × 10⁴ L mol⁻¹ cm⁻¹ and width 4000 cm⁻¹. Using f = 4.32 × 10⁻⁹ ∫ε dν̃, what is the oscillator strength, to two decimal places?
Numerical answer — type the value.
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Answer: 0.17
∫ε dν̃ = 1.0 × 10⁴ × 4000 = 4.0 × 10⁷ L mol⁻¹ cm⁻², so f = 4.32 × 10⁻⁹ × 4.0 × 10⁷ = 0.173, i.e. 0.17: a strongly allowed π → π* band. Using the peak ε without the width gives a meaningless 4 × 10⁻⁵.Which statements about the Jablonski diagram are correct?
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Answer: A — Fluorescence is S₁ → S₀ emission and is spin-allowed; B — Intersystem crossing changes the spin multiplicity
Fluorescence S₁ → S₀ keeps the spin (ns lifetimes) and ISC S₁ → T₁ changes it. Phosphorescence T₁ → S₀ is spin-forbidden and therefore slow (ms to s), and internal conversion is non-radiative, not radiative.An excited state has a lifetime of 1.0 ns. Estimate its lifetime-broadened linewidth δν = 1/(2πτ), in MHz, to the nearest whole number.
Numerical answer — type the value.
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Answer: 159
δν = 1/(2π × 1.0 × 10⁻⁹ s) = 1.59 × 10⁸ Hz = 159 MHz. Omitting the 2π gives 1000 MHz; this broadening is Lorentzian and does not depend on temperature or molecular mass.A Gaussian and a Lorentzian line have the same position and the same full width at half maximum. Compared with the Gaussian, the Lorentzian
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Answer: A — has much more intensity far out in the wings
A Lorentzian falls as 1/(ν − ν₀)², a Gaussian as exp[−(ν − ν₀)²], so the Lorentzian keeps long wings. Lorentzian shapes come from lifetime and collisions; the Doppler profile is the Gaussian one, and neither curve reaches exactly zero.On a 400 MHz spectrometer, two ¹H signals appear at δ 1.20 and δ 1.70 ppm. How far apart are they, in Hz?
Numerical answer — type the value.
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Answer: 200
Δδ = 0.50 ppm, and 1 ppm = 400 Hz at 400 MHz, so the separation is 0.50 × 400 = 200 Hz. The same pair would be 300 Hz apart at 600 MHz, whereas a coupling constant J stays the same in Hz at any field.What is the ¹H Larmor frequency in a field of 9.4 T, in MHz, to the nearest whole number? (γ(¹H) = 2.675 × 10⁸ rad s⁻¹ T⁻¹)
Numerical answer — type the value.
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Answer: 400
ν = γB₀/2π = 2.675 × 10⁸ × 9.4/(2π) = 4.002 × 10⁸ Hz = 400 MHz. Forgetting the 2π gives the angular frequency, 2.51 × 10⁹ rad/s, which is not the frequency the instrument is named for.