Questions: Vibrational Energy Levels and Selection Rules

Question 1

A chemist records the IR spectrum of pure N₂ gas and observes no absorption in the fundamental stretching region. The most likely explanation is:

• A. N₂ has no vibrational modes because the triple bond is too strong to vibrate
• B. Stretching the N≡N bond does not change the molecular dipole moment, so the IR selection rule is violated
• C. The N≡N stretching frequency falls outside the standard IR range
• D. N₂ absorbs only in the microwave region due to its small moment of inertia

Answer: (B)

IR activity requires two conditions: Δv = ±1 AND a change in dipole moment during the vibration. N₂ is homonuclear — it has zero dipole moment by symmetry at any bond length. Stretching the bond doesn't change this, so the dipole-moment condition is never satisfied. N₂ does vibrate (it has zero-point energy and normal modes), but it cannot interact with infrared radiation regardless of frequency. This is why atmospheric N₂ doesn't absorb IR radiation despite being the dominant gas — it's IR-invisible.

Question 2

For a harmonic oscillator, the energy spacing between vibrational levels v=1 and v=2 compared to the spacing between v=0 and v=1 is:

• A. Larger, because higher vibrational states are more energetic
• B. Smaller, because quantum numbers increase but the ladder slows
• C. Identical — spacing is constant at ℏω regardless of quantum number
• D. Temperature-dependent — spacing increases at higher temperatures

Answer: (C)

In the harmonic oscillator, E_v = ℏω(v + ½), so the spacing ΔE = ℏω is the same between every pair of adjacent levels. This uniform spacing is a defining property of the harmonic potential. Real molecules are anharmonic, which makes spacing decrease with increasing v — but the question specifies the harmonic oscillator model, where spacing is strictly constant. This constant spacing is also why the selection rule Δv = ±1 gives a single fundamental absorption: all allowed transitions occur at the same frequency ω.

Question 3

A vibrational mode that obeys the Δv = ±1 selection rule will still not absorb infrared radiation if the vibration does not change the molecular dipole moment.

• T. True
• F. False

Answer: True

True. IR absorption requires both conditions simultaneously: the quantum selection rule Δv = ±1 (from the transition dipole moment integral over wavefunctions) AND a nonzero dipole moment derivative with respect to the normal coordinate. The dipole condition is the physical requirement that the oscillating electric field of the photon can couple to an oscillating charge distribution in the molecule. A symmetric stretching mode in CO₂, for example, satisfies Δv = ±1 but is IR-inactive because it preserves the molecular symmetry and produces no net dipole change.

Question 4

Hot bands in an IR spectrum appear at higher frequency than the fundamental transition because molecules in excited vibrational states vibrate faster.

• T. True
• F. False

Answer: False

False. Hot bands appear at slightly lower frequency than the fundamental. They arise from transitions between thermally populated excited states (e.g., v=1 → v=2). In anharmonic potentials, the energy level spacing decreases with increasing v — the levels are closer together higher up. So the v=1→v=2 transition has a smaller ΔE than the v=0→v=1 fundamental, producing absorption at lower frequency. Their intensity increases with temperature as higher v states become more populated, making them a diagnostic for thermal vibrational excitation.

Question 5

Explain why overtone transitions (Δv = ±2) appear in real molecular spectra even though they are forbidden for a harmonic oscillator.

Model answer: In a perfect harmonic oscillator the wavefunction integrals for Δv = ±2 vanish exactly, making overtones strictly forbidden. Real molecules have anharmonic potentials — the true potential energy curve deviates from a parabola. Anharmonicity mixes the pure harmonic wavefunctions, giving each vibrational state a small admixture of other quantum numbers. This mixing makes the transition dipole moment integral for Δv = ±2 (and ±3 etc.) small but nonzero, allowing weak overtone absorption at approximately twice the fundamental frequency.

The key point is that the harmonic oscillator selection rule is exact only for a perfectly parabolic potential. Any anharmonicity is a perturbation that relaxes the rule. Overtone intensities decrease rapidly with increasing Δv because the anharmonic mixing coefficients are small — the v=0→v=2 overtone is typically 10–100 times weaker than the fundamental, and v=0→v=3 weaker still.