Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section B / 21

The coarse spacing is about $2\tilde{B}$, so $\tilde{B}\simeq 0.25\mathrm{c}{\mathrm{m}}^{-1}$. The two groups near 1.5 and $2.0\mathrm{c}{\mathrm{m}}^{-1}$ therefore have $J=2\to 3$ and $J=3\to 4$, respectively. For positive ${\tilde{D}}_{JK}$, increasing $K^2$ moves a component to lower wavenumber. The assignments areWithin one group, successive separations scale as $(K+1{)}^2-K^2=1,3,5,\dots$. The observed gaps follow that pattern to their printed precision. There are $J+1$ distinct $|K|$ components because $|K|\le J$ in the lower state; the $+K$ and $-K$ members remain coincident without an applied field. A high-resolution feature here is therefore not simply assigned to an arbitrary magnetic sublevel.

The $K=0$ lines give two equations independent of ${\tilde{D}}_{JK}$:Solving yields $\tilde{B}\simeq 0.25022\mathrm{c}{\mathrm{m}}^{-1}$ and ${\tilde{D}}_J\simeq 2.1\times 10^{-7}\mathrm{c}{\mathrm{m}}^{-1}$. The small distortion is obtained by subtracting close numbers, so its precision is considerably worse than that of $\tilde{B}$.

For components in a fixed group,For example the $J=2,K=2$ displacement gives ${\tilde{D}}_{JK}=0.00008/24$, while $J=3,K=3$ gives $0.00024/72$. Both give ${\tilde{D}}_{JK}\simeq 3.3\times 10^{-6}\mathrm{c}{\mathrm{m}}^{-1}$; the smaller gaps are consistent after rounding. The pure rotational transitions have $\Delta K=0$, so the $(\tilde{A}-\tilde{B})K^2$ and ${\tilde{D}}_K{K}^4$ terms cancel. Neither $\tilde{A}$ nor ${\tilde{D}}_K$ can be determined from these zero-field lines.

A spectral component shifts by the difference between its upper- and lower-level Stark shifts. With $\Delta J=+1$, $\Delta K=0$ and $\Delta M=q$, the supplied energy formula givesFor radiation polarized parallel to the static field, $q=0$, soThus the assigned $K=0$ feature has no first-order splitting, whereas the $K\ne 0$ features acquire symmetrically arranged magnetic-sublevel components with spacings proportional to $|K|$. With perpendicular radiation, $q=\pm 1$ gives the corresponding different allowed pattern. Recording field-dependent spectra with specified polarization would test both the J values and the K sequence; using an individual level's shift alone would not predict the observed line positions. Higher-order shifts of the nominally unsplit feature are outside this first-order test.

The largest applied field givesFor parallel polarization, adjacent allowed M values in a $J,K$ component are separated byAlready at $K=1$ this is $2.24\times 10^{-4}\mathrm{c}{\mathrm{m}}^{-1}$ for $J=2$ and $8.96\times 10^{-5}\mathrm{c}{\mathrm{m}}^{-1}$ for $J=3$. Both exceed the stated $2\times 10^{-5}\mathrm{c}{\mathrm{m}}^{-1}$ resolution, and larger $|K|$ gives larger first-order spacings. The spectrometer should resolve suitable Stark components, provided their physical linewidths and signal-to-noise permit it. This predicts resolvable splittings rather than incorrectly comparing the maximum level shift directly with instrumental resolution.

A regularly repeated set of sharp features is characteristic of the unresolved Q subbranches of a perpendicular rovibrational band. The transition dipole is perpendicular to the symmetry axis, giving $\Delta K=\pm 1$; for the Q branch $\Delta J=0$. Holding the rotational constants fixed between vibrational states cancels the $BJ(J+1)$ term, leavingMany J transitions therefore coincide within each K-changing Q subbranch in this approximation. The different K subbranches are separated byThe P and R branches have $\Delta J=-1$ and $+1$, respectively. Their extra J-dependent terms spread many transitions across a broad background at modest resolution. A parallel band, with $\Delta K=0$, would not generate this repeated Q-subbranch spacing under the stated assumptions.

The subbranch spacing gives $\tilde{A}-\tilde{B}=4.9\mathrm{c}{\mathrm{m}}^{-1}$, hence $\tilde{A}\simeq 5.15\mathrm{c}{\mathrm{m}}^{-1}$. Rotation about the C–I symmetry axis has momentC and I lie on this axis, so only the three hydrogens contribute to $I_a$. If $\theta$ is the I–C–H angle, their perpendicular distances are $r_{\mathrm{CH}}\sin \theta$. Taking integer hydrogen mass givesThe molecular arrangement selects the obtuse solution, so $\theta \simeq 108^{\circ }$. The supplementary acute angle describes the opposite axial direction, not the bond angle to iodine. This is a rigid-geometry estimate based on an effective rotational constant and the stipulated bond length, rather than a precision equilibrium-structure measurement.