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 / b

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.