 Solution

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.