Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A3

A polar diatomic molecule can absorb microwaves through its permanent electric dipole moment. In the linear rigid rotor approximation,The electric-dipole selection rule is $\Delta J=+1$ in absorption. It givesso adjacent lines are separated by $2B_0$. ThereforeAssign a line near $m\times 21.999\mathrm{GHz}$ to $J=m-1\to m$ for positive integer $m$. The first member is $0\to 1$, the next $1\to 2$, and so on; if the recorded window misses the lowest lines, their absolute frequencies fix the integer offset. At high temperature many rotational states are populated, so the strongest line need not be $0\to 1$.

This is the leading rotational pattern, not a claim that real ZnF lacks spin or hyperfine structure. Fine splittings and centrifugal distortion can perturb the ideal lines; they do not explain the three systematically displaced progressions as economically as vibrational excitation.

The two extra progressions are rotational transitions within thermally populated $v=1$ and $v=2$ vibrational levels. Isotopic purity rules out attributing them to different isotope masses. Their rotational constants areThe Boltzmann distribution makes each higher vibrational manifold less populated, so its rotational spectrum is weaker. A Morse oscillator explores larger separations as it is excited: the bond's vibrationally averaged moment of inertia increases and the rotational spacing decreases. More precisely, $B_v$ involves an average of the inverse square separation; it is not simply the inverse square of the mean separation.

The observed decrement is constant to the given precision:This fits the leading vibration-rotation interaction law $B_v=B_e-{\alpha }_e(v+\frac{1}{2})$.

An equilibrium bond length requires $B_e$, not merely the ground-vibrational-state constant. Extrapolating the linear vibration-rotation interaction law givesUsing integer isotope masses, the reduced mass isFor two nuclei $I_e=\mu r_e^2$, soUsing $B_0$ instead gives $r_0\simeq 177.1$ pm, an effective vibrational separation. The small difference is precisely why the weaker vibrational progressions matter when the target is equilibrium geometry. Higher terms in $B_v$ and the rigid-rotor approximation limit this estimate's precision.

A vibrational fundamental is the transition $v=0\to 1$. A first vibrational overtone is $v=0\to 2$. A hot band starts from a thermally populated vibrationally excited level, such as $v=1\to 2$.

For a strictly harmonic potential and a dipole linear in displacement, the allowed transitions have $\Delta v=\pm 1$. A fundamental and a $1\to 2$ hot band then share the same frequency, while the overtone has zero intensity. Anharmonicity separates the adjacent gaps and can give overtone intensity; nonlinear displacement dependence of the dipole can also contribute. A hot band is defined by its lower state, not merely by being weak.

The strong band is $0\to 1$, while the weak band is the first vibrational overtone $0\to 2$. Let $w={\tilde{\omega }}_e$ and $q={\tilde{\omega }}_e{x}_e$. Subtracting the Morse oscillator levels givesTwice the fundamental is $881.2\mathrm{c}{\mathrm{m}}^{-1}$, slightly above the overtone because the higher vibrational gap is smaller. Eliminate $w$:soThe implied dimensionless anharmonicity constant is $x_e\simeq 0.004164$. The overtone's weakness follows from its being forbidden in the leading harmonic/linear-dipole model, rather than a lack of population in its lower state.

The relevant allowed absorption is $1\to 2$. Under the stipulated population-only intensity model, the vibrational levels are nondegenerate andSet this ratio to $0.10$ and solve:The energy in the exponential is the measured $0\to 1$ gap, not the harmonic constant $444.3\mathrm{c}{\mathrm{m}}^{-1}$. At this temperature, $N_2/N_0\simeq 0.0102$, so considering the first thermally excited manifold is sensible. Actual band intensity ratios also involve transition moments and rotational structure; the calculation follows the question's simplifying assumption.

Anharmonicity places the $1\to 2$ hot band atonly $3.7\mathrm{c}{\mathrm{m}}^{-1}$ below the fundamental. It can be detected with adequate spectral resolution and signal-to-noise; it is not guaranteed to appear as a separate peak in an unresolved band contour. The predicted ten-percent population-based intensity is appreciable, but overlapping rotational branches and linewidth can hide it beneath the stronger fundamental. No instrumental resolution or linewidth is supplied, so an unconditional assertion of visibility or invisibility would go beyond the information given.