分子 Molecular Spectroscopy · Laboratory experiment

Rotation–vibration spectrum of HCl gas (FTIR)

Aim: to record the fundamental (v = 0→1) and first overtone (v = 0→2) bands of gaseous HCl with a Fourier-transform infrared spectrometer, assign the P- and R-branch lines of H³⁵Cl and H³⁷Cl, and determine the rotational constants, the equilibrium bond length, the vibrational constants and the force constant.

FTIR spectrometer

Interferogram (detector signal against OPD)

Transmission spectrum

Click on a line (an absorption minimum) to mark it. Drag to zoom; double-click to reset. A doublet is H³⁵Cl (stronger) plus H³⁷Cl (24%, about 2 cm⁻¹ lower).

Observation table: line positions and assignments

m = J″+1 for the R branch, m = −J″ for the P branch

Graph: ν̃(m) for H³⁵Cl, v = 0→1

Results

Theory

The term values of a vibrating rotor (an anharmonic oscillator with rotation–vibration coupling) are

$$ T(v,J) = \omega_e(v+\tfrac12) - \omega_ex_e(v+\tfrac12)^2 + B_vJ(J+1) - D J^2(J+1)^2,\qquad B_v = B_e - \alpha_e(v+\tfrac12) .$$

With the selection rule ΔJ = ±1 (there is no Q branch for a ¹Σ molecule), the R (ΔJ = +1) and P (ΔJ = −1) lines combine into a single formula in \(m\):

$$ \tilde\nu(m) = \tilde\nu_0 + (B_1+B_0)\,m + (B_1-B_0)\,m^2 - 4D\,m^3,\qquad m = J''+1\ (\text{R}),\ m=-J''\ (\text{P}).$$

A cubic fit therefore gives \(\tilde\nu_0\), \(B_0\), \(B_1\) and D. Then \(\alpha_e = B_0 - B_1\), \(B_e = B_0 + \alpha_e/2\), and

$$ r_e = \sqrt{\frac{h}{8\pi^2 c\,\mu B_e}},\qquad \mu = \frac{m_Hm_{Cl}}{m_H+m_{Cl}} .$$

The band origins of the fundamental and the overtone give the vibrational constants (neglecting \(\omega_ey_e\)) and the force constant:

$$ \tilde\nu_{01} = \omega_e - 2\omega_ex_e,\quad \tilde\nu_{02} = 2\omega_e - 6\omega_ex_e,\qquad k = 4\pi^2c^2\mu\,\omega_e^2,\qquad D_e \approx \frac{\omega_e^2}{4\omega_ex_e}\ (\text{Morse}).$$

Intensities

Line strength ∝ (Hönl–London factor) × (thermal population of J″), with population ∝ \((2J+1)e^{-hcB_0J(J+1)/kT}\). The maximum lies at \(J_{max} \approx \sqrt{kT/2hcB} - \tfrac12\). Raise the temperature and watch the branches widen.

Procedure

  1. Record the background (empty cell), fill the cell with HCl (about 40 Torr), and record the sample. The simulator shows the ratio, i.e. the transmittance.
  2. Choose a resolution of 0.5 cm⁻¹ or better, so the ³⁵Cl/³⁷Cl doublets (Δν̃ ≈ 2 cm⁻¹) are resolved.
  3. Mark every H³⁵Cl line and assign m. Note the gap at the band centre: m = 0 is missing.
  4. Fit the cubic. Repeat for the overtone near 5668 cm⁻¹ (use a long path, since the overtone is about 50 times weaker).

Things to notice

  • Spacing shrinks in the R branch and grows in the P branch, because B₁ < B₀ (the vibrationally excited bond is longer on average).
  • Isotope shift: ν̃₀(37)/ν̃₀(35) ≈ √(μ₃₅/μ₃₇) = 0.99924.
  • Saturation: at high pressure or low resolution, strong lines bottom out. Apparent depths no longer scale with concentration, the same failure of Beer's law met in the Beer–Lambert experiment.

Viva questions

Why is there no Q branch?

For a ¹Σ diatomic, ΔJ = 0 is forbidden in a vibrational transition, because the photon carries one unit of angular momentum and there is no electronic angular momentum to take it up.

Why are the lines not equally spaced?

B₁ ≠ B₀ (rotation–vibration interaction), together with centrifugal distortion D.

What is the FTIR advantage?

Fellgett (multiplex: all frequencies measured at once), Jacquinot (throughput: no slit), and Connes (accurate wavenumbers from a HeNe reference laser).