Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A3

The resolved lines arise from simultaneous changes of vibration and rotation in hydrogen cyanide. A vibrational fundamental connects many populated lower rotational levels to different upper rotational levels, giving the fine structure of a rovibrational spectrum.

The prominent central Q branch, together with flanking P branch and R branch lines, identifies a perpendicular band. For linear HCN, this is the doubly degenerate bending fundamental, not either stretching fundamental. Bending creates a transition-dipole component perpendicular to the molecular axis; its vibrational angular momentum permits $\Delta J=0$ as well as $\Delta J=\pm 1$.

The two perpendicular bending coordinates share one vibrational frequency. Their degeneracy is different from the series of resolved rotational transitions superimposed on that fundamental. The stretches have their electric dipole transition moment along the molecular axis and would give the usual parallel-band P/R pattern without this strong central Q feature.

Let $J$ label the lower rotational state and write the elementary band model asHere $B^{\prime \prime },B^{\prime }$ are lower and upper rotational constants, in inverse centimetres; a constant term associated with upper-state vibrational angular momentum can be included in ${\tilde{\nu }}_{\ast }$. The P branch has $J^{\prime }=J-1$, and the R branch has $J^{\prime }=J+1$.

In this model the six printed labels are assigned asThe low-frequency series steps away from the central origin towards increasing lower $J$. The first two high-frequency R positions are approximately $714.96$ and $717.93{\mathrm{cm}}^{-1}$, so the labels at $720.91$ and $723.90$ correspond to $R(2)$ and $R(3)$; they are not the first two rotational transitions.

There is a physical limitation to taking this schematic model literally. A strict $l=0\to |l|=1$ bending band has $J^{\prime }\ge 1$, so a genuine $P(1)$ transition to $J^{\prime }=0$ is forbidden. The printed $709.04$ feature has the formal $P(1)$ position of the elementary model. The numerical analysis below follows the printed line sequence; it does not infer a physically allowed $J^{\prime }=0$ bending level from that feature.

Use rovibrational spectroscopy combination differences so that the band origin cancels. With the assignments above, $R(2)$ and $P(4)$ end at the same upper level $J^{\prime }=3$, givingLikewise $R(3)$ and $P(3)$ start at the same lower level $J=3$, soOnly the printed wavenumbers are needed:Thus $B^{\prime }-B^{\prime \prime }\simeq 0.00429{\mathrm{cm}}^{-1}$. The rotational constants differ only slightly, so the average moment of inertia changes little on vibrational excitation. The upper value is slightly larger, corresponding to a slightly smaller effective moment of inertia. This sign is possible for a bending excitation; the intuition that excitation must always lengthen a diatomic bond is not a general rule for a polyatomic bend.

The rounded printed values are internally consistent: fitting all six positions to the same elementary model gives approximately ${\tilde{\nu }}_{\ast }=711.9945{\mathrm{cm}}^{-1}$, $B^{\prime \prime }=1.47797{\mathrm{cm}}^{-1}$ and $B^{\prime }=1.48203{\mathrm{cm}}^{-1}$, with residuals below $0.002{\mathrm{cm}}^{-1}$. The displayed combination-difference values have more digits than their physical accuracy; the meaningful scale is about $1.48{\mathrm{cm}}^{-1}$.

For a linear polyatomic molecule, the first-order polyatomic vibration-rotation interaction iswhere $d_i$ is the degeneracy of normal mode $i$. The bending mode has $d_2=2$, so the two measured states implyThe two stretching zero-point contributions are not supplied. The true all-mode equilibrium constant is therefore not uniquely determined by this band alone. Neglecting those unknown stretching corrections gives the bend-only estimateUsing $B=h/(8{\pi }^{2}cI)$ and converting inverse centimetres to inverse metres,If the intended course approximation instead treats the observed vibration as a single nondegenerate oscillator with $B_v=B_e-{\alpha }_e(v+1/2)$, its extrapolation givesThat is a useful effective one-coordinate estimate, but it should not be identified with an exact equilibrium constant of the doubly degenerate polyatomic bend. Both estimates give $I\simeq 1.90\times 10^{-46}\mathrm{kg}{\mathrm{m}}^2$ at the accuracy warranted here.

The expanded feature resolves the Q branch: each line has $\Delta J=0$ but starts from a different lower rotational level. In the simple rigid rotor band model,Because the upper rotational constant is larger, the lines extend towards higher wavenumber. Their adjacent spacings areso they crowd together at the low-frequency edge and become increasingly separated. A strictly resolved bending-band analysis may distinguish the upper parity component sampled by Q lines from that sampled by P/R lines; it is not necessary to impose identical high-resolution constants on every branch in this sketch.

The intensity envelope reflects rotational state populations and the rotational transition strength. For this perpendicular Q branch, the usual rotational line-strength factor gives the approximate dependenceAt small $J$, the increasing magnetic degeneracy makes the line intensity rise; at large $J$, Boltzmann suppression dominates. The maximum is therefore at an intermediate $J$. This is a population envelope across distinct transitions, rather than one intrinsically broad vibrational line.

Count the successive Q branch lines from the low-frequency edge, taking $J=1$ as the first allowed bend transition; the first lines are crowded and partly unresolved. The maximum near $712.55{\mathrm{cm}}^{-1}$ is approximately $J=12$. Treating $J$ as continuous in the intensity envelope from part (i),HenceUsing $hc/k_{\mathrm{B}}=1.43878\mathrm{K}\mathrm{cm}$, $B^{\prime \prime }=1.47786{\mathrm{cm}}^{-1}$ and $J_{\max }\simeq 12$ givesA useful check avoids locating the exact maximum. The plotted $J\simeq 20$ line has roughly $0.70$ of the $J\simeq 12$ peak height. Thusgives about $659\mathrm{K}$. An appropriate graphical estimate is $T\simeq 6.6\times 10^2\mathrm{K}$, with precision limited by line counting, line overlap and reading the plotted intensities. There is no basis for reporting a temperature to the nearest kelvin from this figure.