CSIR-NET Rotational and Vibrational Spectroscopy — Bond Lengths and Force Constants

By Aniket Bhardwaj · 19 September 2026 · CSIR-NET Chemistry

Structure elucidation from IR, NMR and mass spectra is a pattern-recognition skill. This article is about the other half of spectroscopy — the quantitative half, where a microwave spectrum gives you a bond length to three decimal places and an infrared band gives you a force constant in newtons per metre. These questions are pure physical chemistry, the formulae are few, and the marks are reliable once you have practised the unit handling. Every number below is computed step by step.

Rotation: the rigid rotor

E_J = B J(J+1) (in cm⁻¹) · B = h / (8π² c I) · I = μ r² · μ = m₁m₂ / (m₁ + m₂)

J — rotational quantum number, 0, 1, 2, … ; each level is (2J+1)-fold degenerate.
B — rotational constant. Expressed in cm⁻¹ when c is in cm s⁻¹; in Hz if c is omitted.
I — moment of inertia, kg m². μ — reduced mass, kg (convert from u with 1 u = 1.66054 × 10⁻²⁷ kg).
Selection rule ΔJ = ±1, and the molecule must have a permanent dipole moment. So HCl and CO give microwave spectra; N₂, O₂ and CO₂ do not.

The transition J → J+1 appears at ν̃ = 2B(J+1), so the pure rotational spectrum is a ladder of lines separated by 2B. That even spacing is the fingerprint of a rigid rotor and the fastest way to extract B from a printed spectrum.

Worked example 1 — bond length of ¹²C¹⁶O from B = 1.9313 cm⁻¹.

Step 1 — reduced mass.
μ = (12.000 × 15.9949) ÷ (12.000 + 15.9949) = 191.939 ÷ 27.9949 = 6.8562 u.
In kg: 6.8562 × 1.66054 × 10⁻²⁷ = 1.13850 × 10⁻²⁶ kg.

Step 2 — moment of inertia. Rearranging B = h/(8π²cI):
8π² = 78.9568.
Denominator = 78.9568 × (2.99792 × 10¹⁰ cm s⁻¹) × 1.9313 = 4.5715 × 10¹².
I = (6.62607 × 10⁻³⁴) ÷ (4.5715 × 10¹²) = 1.4494 × 10⁻⁴⁶ kg m².

Step 3 — bond length.
r = √(I/μ) = √[(1.4494 × 10⁻⁴⁶) ÷ (1.13850 × 10⁻²⁶)] = √(1.2731 × 10⁻²⁰) = 1.1283 × 10⁻¹⁰ m = 112.8 pm.

Cross-check. The accepted equilibrium bond length of carbon monoxide is about 112.8 pm, so the arithmetic and the unit conversions are both sound. Note that c had to be in cm s⁻¹ because B was in cm⁻¹ — mixing that up is the single most common way to be out by a factor of 100.

Line spacing: 2B = 3.86 cm⁻¹, which is where in the far infrared or microwave you would look for these lines.

Two refinements the paper likes

Centrifugal distortion. A real bond stretches as the molecule spins faster, so I grows and the levels fall slightly below the rigid-rotor prediction:

E_J = B J(J+1) − D J²(J+1)², with D ≪ B

The observable consequence is that the line spacing slowly decreases at high J rather than staying exactly 2B. A question that says "the spacing is not quite constant" is asking for this term.

Which line is strongest? The intensity of a rotational line depends on the population of the starting level, which is a product of a rising degeneracy (2J+1) and a falling Boltzmann factor. The maximum occurs at

J_max ≈ √[ kT / (2hcB) ] − ½

Worked example 2 — most populated level of CO at 300 K.

kT = (1.380649 × 10⁻²³) × 300 = 4.1419 × 10⁻²¹ J.
hcB = (6.62607 × 10⁻³⁴)(2.99792 × 10¹⁰)(1.9313) = 3.8364 × 10⁻²³ J, so 2hcB = 7.6729 × 10⁻²³ J.
Ratio = 4.1419 × 10⁻²¹ ÷ 7.6729 × 10⁻²³ = 53.98.
J_max = √53.98 − 0.5 = 7.35 − 0.5 = 6.85, so J = 7 is the most populated level.

This is why a rotational band peaks in the middle rather than at J = 0: the ground level has the largest Boltzmann factor but the smallest degeneracy.

Vibration: the harmonic oscillator

E_v = (v + ½) h ν · ν̃ = (1 / 2πc) √(k/μ) · k = 4π² c² ν̃² μ

v — vibrational quantum number; note the zero-point energy ½hν, which does not vanish at v = 0.
k — force constant, N m⁻¹; a measure of bond stiffness, not bond strength, though the two usually run together.
Selection rule Δv = ±1 for a harmonic oscillator, and the vibration must change the dipole moment for it to be infrared active.

Worked example 3 — force constant of H³⁵Cl from ω_e = 2990 cm⁻¹.

Step 1 — reduced mass.
μ = (1.00783 × 34.96885) ÷ (1.00783 + 34.96885) = 35.2427 ÷ 35.9767 = 0.97960 u
= 0.97960 × 1.66054 × 10⁻²⁷ = 1.62666 × 10⁻²⁷ kg.

Step 2 — frequency in s⁻¹.
ν = c ν̃ = (2.99792 × 10¹⁰) × 2990 = 8.96378 × 10¹³ s⁻¹.

Step 3 — force constant.
k = 4π² ν² μ = 39.4784 × (8.96378 × 10¹³)² × (1.62666 × 10⁻²⁷)
= 39.4784 × (8.0349 × 10²⁷) × (1.62666 × 10⁻²⁷)
= 317.21 × 1.62666 = 516 N m⁻¹.

That agrees with the standard quoted value for the H–Cl bond, which is the check worth making before you move on.

Worked example 4 — the isotope shift, D³⁵Cl. The force constant is a property of the electronic structure and is essentially unchanged by isotopic substitution, so only μ changes:

μ(DCl) = (2.01410 × 34.96885) ÷ (2.01410 + 34.96885) = 70.4308 ÷ 36.9830 = 1.90441 u.
ν̃(DCl)/ν̃(HCl) = √(μ_HCl/μ_DCl) = √(0.97960 ÷ 1.90441) = √0.51439 = 0.71721.
ν̃(DCl) = 2990 × 0.71721 = 2145 cm⁻¹.

A near-halving of the wavenumber on going from H to D is the standard qualitative result, and it is why C–H stretches are deliberately shifted out of the way by deuteration in mechanistic work.

Anharmonicity — where overtones come from

A real potential is not a parabola: it flattens out towards dissociation. The Morse approximation gives

G(v) = ω_e(v + ½) − ω_ex_e(v + ½)²
Fundamental (0→1): ν̃ = ω_e − 2ω_ex_e
First overtone (0→2): ν̃ = 2ω_e − 6ω_ex_e

Two consequences follow immediately. Levels get closer together as v rises, and Δv = ±2, ±3 transitions become weakly allowed — which is why overtones exist at all, and why the first overtone appears at slightly less than twice the fundamental rather than exactly twice.

Worked example 5 — HCl with ω_e = 2990.9 cm⁻¹ and ω_ex_e = 52.8 cm⁻¹.

Fundamental = 2990.9 − 2(52.8) = 2990.9 − 105.6 = 2885.3 cm⁻¹, matching the observed HCl fundamental near 2886 cm⁻¹.
First overtone = 2(2990.9) − 6(52.8) = 5981.8 − 316.8 = 5665.0 cm⁻¹ — clearly less than 2 × 2885.3 = 5770.6.

Dissociation energy, and an honest caveat. The Morse model gives D_e = ω_e² ÷ (4 ω_ex_e) = (2990.9)² ÷ (4 × 52.8) = 8 945 483 ÷ 211.2 = 42 356 cm⁻¹. Using 1 cm⁻¹ = 11.9627 J mol⁻¹, that is about 507 kJ mol⁻¹. The experimentally accepted dissociation energy of HCl is appreciably lower than this. The discrepancy is not an arithmetic error — the Morse formula systematically overestimates D_e for real molecules, because the true potential deviates from the Morse shape at large separation. Quote the method with the number, and say that it is an upper estimate.

Rovibrational spectra: P, Q and R branches

In the gas phase, a vibrational transition is accompanied by a change in J, which splits the band into branches.

Branch	ΔJ	Position relative to the band centre	When it appears
P	−1	Lower wavenumber	Always, for a diatomic
Q	0	At the band centre	Only when the molecule has vibrational angular momentum — e.g. a bending mode; absent in a diatomic stretch
R	+1	Higher wavenumber	Always, for a diatomic

The lines within each branch are spaced by roughly 2B, and there is a gap of about 4B across the missing centre where the Q branch would be. Being asked to explain the "missing central line" of the HCl gas-phase infrared band is a standard question, and the answer is simply that ΔJ = 0 is forbidden for that transition.

Raman, and how many modes there are

Raman scattering depends on a change in polarisability, not dipole moment, so it sees vibrations that the infrared cannot. Selection rules are ΔJ = 0, ±2 for rotation (giving lines spaced by 4B) and Δv = ±1 for vibration.

Rule of mutual exclusion: in a molecule with a centre of symmetry, no vibration is both infrared and Raman active.

CO₂ is the textbook illustration. It is linear with three atoms, so it has 3N − 5 = 4 vibrational modes: a symmetric stretch (Raman active, infrared inactive, because the dipole stays zero), an antisymmetric stretch (infrared active) and a doubly degenerate bend (infrared active). Water, being bent, has 3N − 6 = 3 modes and all three are active in both techniques, because it has no centre of symmetry. Benzene, with 12 atoms and non-linear, has 3(12) − 6 = 30 vibrational modes.

Stokes lines (the molecule gains energy) are more intense than anti-Stokes lines (it loses energy) simply because the lower level is more populated at ordinary temperatures — a Boltzmann argument, not a selection rule.

Mistakes that cost marks

Using c in m s⁻¹ with B in cm⁻¹. Match the units first; this error changes the answer by a factor of 100 and is by far the commonest.
Using molar masses instead of atomic masses in μ. Divide by N_A, or work in u and convert with 1.66054 × 10⁻²⁷ kg per u.
Forgetting the dipole requirement. Homonuclear diatomics have no pure rotational or infrared vibrational spectrum — but they are Raman active.
Assuming the strongest rotational line is J = 0. Degeneracy pushes the maximum out to J_max.
Treating the first overtone as exactly twice the fundamental. That is only true for a perfect harmonic oscillator, in which case overtones would not appear at all.
Assuming a force constant changes on isotopic substitution. It does not, to a very good approximation — only μ does.
Using 3N − 6 for a linear molecule. Linear molecules have only two rotational degrees of freedom, so it is 3N − 5.
Expecting a Q branch in a diatomic stretch. It is absent, and the gap is the evidence.

Where this appears in the paper

Sub-topic	Typical question form
Rigid rotor	Bond length from B, or B from a line spacing
Populations	Most intense rotational line; effect of temperature
Harmonic oscillator	Force constant from a wavenumber, or the reverse
Isotope effects	Predict ν̃ for the deuterated or ¹³C-substituted species
Anharmonicity	Fundamental and overtone positions; dissociation energy estimate
Band structure	Explain P, Q, R branches and the missing central line
Activity rules	Which modes are infrared or Raman active; mutual exclusion
Mode counting	3N − 6 versus 3N − 5 for a stated molecule

Use this as a checklist of sub-topics rather than a forecast. How many questions appear and how they are weighted is set by the official notification for the session you are sitting, and that document is the only reliable source.

Every calculation above needs the same four numbers. Planck's constant, the speed of light, the Boltzmann constant and the atomic mass unit appear in all five worked examples, and one mistyped exponent ruins the answer. The Scientific Constants tool lists them with their exact accepted values and units, so you can copy them correctly instead of trusting memory.

Open the Scientific Constants Tool →

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