Lesson 6 · 40 min

Rotation and vibration: the infrared band of a gas

The infrared absorption of gaseous HCl is not one peak but a comb of fine lines with a gap in the middle. Rotational levels, a selection rule and the Boltzmann distribution explain its shape; analyse a set of real lines with combination differences and you can measure the bond length to a thousandth of an ångström.

In this lesson

Explain the P and R branches of a gas-phase infrared band, and the gap between them, with rigid-rotor levels and the selection rule ΔJ = ±1
Use the Boltzmann distribution to say which lines are strongest and how temperature reshapes the band
Describe how vibration–rotation coupling, centrifugal distortion and isotopes shift the lines
Obtain rotational constants and a bond length from measured lines with combination differences

Put a tube of hydrogen chloride gas into a Fourier-transform infrared spectrometer (FTIR). By the last lesson, HCl has a single vibration and should give one absorption near 2886 cm⁻¹. What comes out instead is a row of evenly spaced lines like the teeth of a comb, with a gap in the middle; raise the resolution and every line splits in two, heights in the ratio of about 3 : 1. The comb records the rotation of the molecules — and from it the H–Cl bond length can be measured to a thousandth of an ångström.

Predict first

When HCl jumps from v = 0 to v = 1, its rotational quantum number J must change by ±1 at the same time. What appears exactly at the band origin ν̃₀, where the pure vibrational transition would be?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

The strongest line of the band A weak line Nothing at all Two overlapping lines

Rotational levels

Treat a diatomic molecule as a dumb-bell and its energy of rotation about the centre of mass is quantised too. A rigid rotor has levels $F(J) = BJ(J+1)$ , J = 0, 1, 2, …, each 2J + 1-fold degenerate. The rotational constant B is inversely proportional to the moment of inertia $I = \mu r^2$ : the lighter the atoms and the shorter the bond, the larger B and the sparser the levels. HCl contains the lightest atom, hydrogen, and has B ≈ 10.4 cm⁻¹, one of the largest rotational constants there is. Neighbouring levels are 2B(J + 1) apart, wider and wider as J climbs.

F(J) = BJ(J+1), \qquad B = \frac{h}{8\pi^2 c\, \mu r^2}, \qquad \mu = \frac{m_1 m_2}{m_1 + m_2}

B in cm⁻¹. With the constants collected,

r^2 = 16.8576 / (\mu B)

for μ in u and r in Å.

$F(J)$: rotational term, cm⁻¹
$J$: rotational quantum number
$B$: rotational constant
$\mu$: reduced mass
$r$: bond length

In a vibrational transition J changes at the same time. If the upper and lower vibrational levels had the same B, the lines of the R and P branches would lie at

R(J) = \tilde\nu_0 + 2B(J+1), \qquad P(J) = \tilde\nu_0 - 2BJ

J is the rotational quantum number of the lower level. Both branches are evenly spaced by 2B, and each stops 2B short of the band origin, so R(0) and P(1) are 4B apart.

Worked example

A first bond length from the gap

For $\ce{H^{35}Cl}$ , R(0) is at 2906.25 cm⁻¹ and P(1) at 2865.10 cm⁻¹; the reduced mass is μ = 0.97959 u. Estimate B and the bond length.

Which lines are strongest

How strong a line is depends mostly on how many molecules sit in its lower level. The degeneracy 2J + 1 grows with J while the Boltzmann factor shrinks with energy, and their product peaks in between: at room temperature the most populated level of HCl is J = 3, so both branches rise and then fall. Heat the gas and the maximum moves to higher J and the band broadens — which is how astronomers and combustion scientists read a “rotational temperature” off gases they cannot touch.

\frac{N_J}{N} \propto (2J+1)\, e^{-hcBJ(J+1)/kT}, \qquad J_\text{max} \approx \sqrt{\frac{kT}{2hcB}} - \frac{1}{2}

hc/k = 1.4388 cm K. The nearest integer is the most populated level.

Try the calculation

HCl has B₀ = 10.44 cm⁻¹. Which rotational level J is most populated at 296 K?

Vibration–rotation coupling, centrifugal distortion and isotopes

Look closely at the spacing: the R branch crowds together further out, while the P branch spreads apart. The vibration is not harmonic — a molecule in v = 1 has a longer average bond, so B₁ = 10.14 cm⁻¹ is smaller than B₀ = 10.44 cm⁻¹. On top of that, the faster a molecule spins, the more the bond is stretched by centrifugal force, lowering the levels by $-D[J(J+1)]^2$ : centrifugal distortion. D is only about 5 × 10⁻⁴ cm⁻¹ for HCl, but by J = 10 it has pulled the level down by 6 cm⁻¹.

Isotopes change the masses, not the bond. $\ce{H^{37}Cl}$ has a slightly larger reduced mass, so it vibrates and rotates a little more slowly: its band sits about 2.1 cm⁻¹ lower, and every line carries a “shadow” about a third as tall (the ³⁷Cl : ³⁵Cl abundance ratio). Replace H by D and the reduced mass almost doubles: the DCl band moves to about 2091 cm⁻¹ and its lines are half as far apart.

Combination differences: from lines to a bond length

R(J − 1) and P(J + 1) end on the same upper level (v = 1, J), so their difference contains only two rotational levels of the lower state; likewise R(J) − P(J) contains only the upper state. Such differences, which cancel one level completely, are combination differences. For a rigid rotor each equals 4B(J + ½): plotted against J + ½ it is a straight line through the origin with slope 4B.

The data below come from a real measurement: in 2021 researchers in Finland recorded the first infrared lines of the radioactive isotopologue $\ce{H^{36}Cl}$ . Chlorine-36 forms in reactor materials when neutrons hit ³⁵Cl impurities; with a half-life of about 300 000 years, and a chance of escaping as H³⁶Cl, it needs spectroscopic methods that can monitor it on site. They made the sample the textbook way, by dripping concentrated sulfuric acid onto ³⁶Cl-containing NaCl in the cell. Fit these lines with the rigid rotor first, then look at the residuals.

R(J-1) - P(J+1) = 4B_0\left(J+\tfrac{1}{2}\right), \qquad R(J) - P(J) = 4B_1\left(J+\tfrac{1}{2}\right)

With centrifugal distortion, the difference divided by (J + ½) is 4B − 8D[(J + ½)² + ¾], still a straight line: the intercept gives B and the slope D.

Least squares · measured lines

From lines to a bond length: combination differences

Data

Fit

Lower level v = 0: R(J−1) − P(J+1)

Upper level v = 1: R(J) − P(J)

Results
Quantity	Value
B₀ (v = 0)	10.3233 ± 0.0149 cm⁻¹
B₁ (v = 1)	10.0003 ± 0.0168 cm⁻¹
D₀	not fitted (rigid rotor)
αₑ = B₀ − B₁	0.3229 cm⁻¹
Bₑ = B₀ + αₑ/2	10.4847 cm⁻¹
Bond length rₑ	1.2806 Å
Band origin ν̃₀ ≈ ½[R(0) + P(1)] + B₀ − B₁	2884.91 cm⁻¹

Compare: Huber and Herzberg give rₑ = 1.27455 Å for H³⁵Cl. The equilibrium bond length does not depend on the isotope; only the rotational constants change with the reduced mass.

Line list (cm⁻¹)

P(10) 2651.0675
P(9) 2676.8101
P(8) 2702.0639
P(7) 2726.8168
P(6) 2751.0527
P(5) 2774.7574
P(4) 2797.9207
P(3) 2820.5297
P(2) 2842.5712
P(1) 2864.0247
R(0) 2905.1493
R(1) 2924.7790
R(2) 2943.7809
R(3) 2962.1397
R(4) 2979.8392
R(5) 2996.8709
R(6) 3013.2251
R(7) 3028.8887
R(8) 3043.8510
R(9) 3058.1013
R(10) 3071.6279

The rigid-rotor line looks like a good fit, yet its residuals bend into an arc — the sign that the model is missing something. Switch to the plot that includes centrifugal distortion: the residuals scatter at random, B₀ rises by about 0.1 cm⁻¹ and the bond length comes back to 1.2746 Å. Then switch to the H³⁵Cl model data: different B, same bond length.

Try the calculation

For $\ce{H^{36}Cl}$ , R(2) is at 2943.7809 cm⁻¹ and P(4) at 2797.9207 cm⁻¹. Treating the molecule as a rigid rotor, find B₀ from this pair.

Concept check

The lines of the CO fundamental are only about 3.8 cm⁻¹ apart, against about 21 cm⁻¹ for HCl. Why, mainly?

The CO bond is shorter than HCl’s CO’s reduced mass is about 7 times HCl’s CO has a triple bond, stronger than HCl’s CO is less polar than HCl

Model limits What the model computes, and what it leaves out

Line positions are computed from the molecular constants compiled by Huber and Herzberg, taken to higher order; they predict the independently measured H³⁶Cl lines to better than 0.02 cm⁻¹. For deuterium, the small failure of the Born–Oppenheimer approximation shifts the prediction by a few tenths of a cm⁻¹. Intensities include only isotopic abundance, the Hönl–London factors and the Boltzmann factor, leaving out the slight asymmetry between R and P intensities (the Herman–Wallis effect); the line shape is the instrument resolution alone, without pressure or Doppler broadening. ¹Σ molecules such as HCl have no Q branch; molecules with electronic angular momentum such as NO, and the 667 cm⁻¹ bending vibration of CO₂, do show a ΔJ = 0 Q branch.

Key takeaways

Rigid-rotor levels F(J) = BJ(J+1) with B = h/(8π²cμr²); ΔJ = ±1 gives R and P branches spaced by about 2B, with no line at the band origin.
Line strengths follow the population of the lower level, rising then falling; at room temperature HCl peaks at J = 3, and heating broadens the band.
B₁ < B₀ crowds the R branch and spreads the P branch; centrifugal distortion D lowers the high-J levels; isotopes change masses, not the bond length.
The combination differences R(J−1) − P(J+1) and R(J) − P(J) contain only the lower and upper level; the measured H³⁶Cl lines give rₑ = 1.2746 Å.

Progress is stored only in this browser and never uploaded.

Data and references

K. P. Huber and G. Herzberg, Constants of Diatomic Molecules, Van Nostrand Reinhold, 1979, as served by the NIST Chemistry WebBook (HCl and CO ground states, including the footnoted higher-order terms).
S. Larnimaa, M. Vainio and V. Ulvila, “Infrared spectroscopy of radioactive hydrogen chloride H³⁶Cl”, J. Quant. Spectrosc. Radiat. Transfer (2022), doi:10.1016/j.jqsrt.2021.107984; arXiv:2108.13031, Table 2.
G. Herzberg, Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules, 2nd ed., Van Nostrand, 1950, chapter III (combination differences).
M. Wang et al., “The AME 2020 atomic mass evaluation (II)”, Chinese Phys. C 45, 030003 (2021).