Questions: Rotational Quantum Numbers and Energy Levels

Question 1

A student argues that because rotational energy levels in a diatomic molecule are unevenly spaced, the absorption lines in a microwave spectrum should also be unevenly spaced. What is wrong with this reasoning?

• A. The student is correct — lines are unevenly spaced because the levels are unevenly spaced
• B. Although levels are unevenly spaced, the ΔJ = ±1 selection rule produces transitions at 2B, 4B, 6B, ... which are separated by a constant 2B — so lines are evenly spaced
• C. Rotational energy levels are actually evenly spaced, so the student's premise is wrong
• D. Microwave spectra do not consist of discrete lines, so spacing is irrelevant

Answer: (B)

The energy levels E_J = BJ(J+1) are indeed unevenly spaced — the gap between J and J+1 is 2B(J+1), which grows with J. But each transition J → J+1 appears at frequency 2B(J+1), giving lines at 2B, 4B, 6B, 8B, ... These are separated from each other by exactly 2B. So the spectral lines are evenly spaced even though the underlying energy levels are not. This uniform 2B spacing is the experimental signature of a rigid rotor and is exactly what allows B to be extracted from the spectrum.

Question 2

Two isotopes, H³⁵Cl and H³⁷Cl, both appear in a microwave rotational spectrum. Which will have the larger rotational constant B, and what does this imply about their spectral line spacing?

• A. H³⁷Cl has larger B because the heavier chlorine isotope makes the molecule more rigid
• B. Both isotopes have the same B because the bond length does not change with isotope substitution
• C. H³⁵Cl has larger B because its smaller reduced mass gives a smaller moment of inertia, making B = ℏ²/2I larger
• D. H³⁵Cl has larger B because heavier isotopes always have lower rotational constants

Answer: (C)

B = ℏ²/(2I) and I = μr², where μ is the reduced mass. H³⁵Cl has a slightly smaller reduced mass than H³⁷Cl (because ³⁵Cl is lighter), so its moment of inertia is smaller and B is larger. Larger B means more widely spaced spectral lines. This is why isotope substitution shifts rotational line positions — a useful analytical technique. The bond length r is essentially unchanged between isotopologues, so the difference in B traces directly to the reduced mass.

Question 3

The rotational energy levels of a diatomic rigid rotor are equally spaced.

• T. True
• F. False

Answer: False

False. The energy levels are E_J = BJ(J+1), which gives gaps of 2B, 4B, 6B, ... between successive levels. The gap between J and J+1 is 2B(J+1), which increases with J. Equal spacing would imply E_J ∝ J, as in a harmonic oscillator. The J(J+1) dependence is a hallmark of angular momentum quantization and leads to the equally spaced spectral lines (despite unequally spaced levels) via the ΔJ = ±1 selection rule.

Question 4

Molecular nitrogen (N₂) does not produce a pure rotational microwave spectrum because it lacks a permanent electric dipole moment.

• T. True
• F. False

Answer: True

True. To absorb microwave radiation, a molecule must have a permanent dipole moment that can interact with the oscillating electric field of the photon. N₂ is a homonuclear diatomic — its electron distribution is symmetric and there is no permanent dipole. The same applies to O₂, H₂, and Cl₂. Heteronuclear diatomics like HCl, CO, and HF do have permanent dipoles and show rich rotational spectra. The distinction is critical: microwave spectroscopy is inherently limited to polar molecules.

Question 5

Why do successive absorption lines in a pure rotational spectrum appear at equally spaced frequencies, and what quantity does this spacing directly measure?

Model answer: The ΔJ = ±1 selection rule means absorption lines occur at transitions J → J+1, with energies 2B, 4B, 6B, ... The spacing between consecutive lines is always 2B, making the spectrum a ladder of lines uniformly separated by 2B. This constant spacing directly measures the rotational constant B = ℏ²/(2I), from which the moment of inertia I and ultimately the bond length can be determined.

The uniform 2B spacing follows because each successive transition energy (J → J+1 vs. J+1 → J+2) differs by exactly 2B: [2B(J+2)] − [2B(J+1)] = 2B. This elegant regularity means that a single measurement of the line spacing immediately yields B, and B encodes the molecular geometry through I = μr². The ability to extract bond lengths to sub-picometer precision from this simple pattern is one of the triumphs of microwave spectroscopy.