Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section B / 21 / c

$I_{C_4}\simeq 2.43\times 10^{-45}\mathrm{kg}{\mathrm{m}}^2=146.6m_{\mathrm{u}}{\mathring{\mathrm{A}}}^{,2}$. The parallel-band progression gives $2\tilde{B}=0.42{\mathrm{cm}}^{-1}$, so $\tilde{B}=0.210{\mathrm{cm}}^{-1}$. The perpendicular Q-feature spacing givesFor an oblate symmetric top, $I_{C_4}>I_{\perp }$, hence $\tilde{C}<\tilde{B}$ and $\tilde{C}=0.115{\mathrm{cm}}^{-1}$. The rotational constant relation then givesThe estimated constants neglect centrifugal distortion and the change of rotational constants in the excited vibration.

$I_{C_4}=4m_B{R}^2(1-\cos \theta )$; the central atom makes no contribution. Let $\rho$ be the radius of the square of outer atoms and $h_A$ the central atom's height above its plane. Then $R^2={\rho }^2+h_A^2$. Adjacent bond vectors have perpendicular in-plane components, soThe centre of mass lies on the fourfold axis, even when $m_A\ne m_B$. Thus the central atom and the shift of origin along the axis contribute no perpendicular lever arm. The four outer atoms each contribute $m_B{\rho }^2$, givingThe result concerns the smallest adjacent bond angle; the angle between opposite outer atoms is a different angle in a pyramid.

With integer fluorine mass $m_B=19m_{\mathrm{u}}$ and $R=1.5\mathring{\mathrm{A}}$,This angle is below the planar limit $90^{\circ }$, consistent with the proposed pyramid. The sulfur mass does not enter this axial moment of inertia.