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

The nine normal coordinates transform as $A_{1g}+B_{1g}+B_{2g}+A_{2u}+B_{2u}+2E_u$. Place the molecule in the $xy$ plane, with its four outer atoms on the $x$ and $y$ axes, as specified for the vertical reflection planes. In the class orderthe fifteen Cartesian displacements have characters of a representationFor each operation this character is the number of fixed atoms times the trace of its polar-vector matrix. Subtract translations $A_{2u}+E_u$ and rotations $A_{2g}+E_g$, givingMolecular character reduction in the D4h molecular point group yields the stated decomposition. Its dimensions sum to $1+1+1+1+1+2(2)=9=3(5)-6$, providing an independent degree-of-freedom check.

An infrared fundamental must transform like a component of the electric dipole moment; a Raman spectroscopy fundamental must transform like a component of the symmetric polarizability tensor. For this centrosymmetric D4h molecular point group, the resulting classification is:

Here parallel and perpendicular refer to the principal fourfold axis. The table obeys infrared and Raman mutual exclusion.

Three separated infrared fundamentals: one parallel $A_{2u}$ band and two perpendicular $E_u$ bands. Each $E_u$ frequency belongs to a doubly degenerate mode, but degeneracy does not create two distinct band origins.

The square-planar rotor is an oblate symmetric top. A parallel band has $\Delta K=0$ and $\Delta J=0,\pm 1$, producing P and R progressions with approximate spacing $2\tilde{B}$ and a central Q contribution from $K\ne 0$; the $K=0$ parallel Q branch is absent. At modest resolution many nearly coincident $K$ components merge.

A perpendicular band has $\Delta K=\pm 1$ and $\Delta J=0,\pm 1$. Its Q subbranches produce a characteristic series of sharper features on the broader unresolved P/R envelope. Neglecting changes of rotational constants on vibration, successive perpendicular Q features are separated by approximately $2|\tilde{B}-\tilde{C}|$. These are rotational contours rather than extra normal-mode fundamentals.

All the listed modes are Raman active; only $A_1$ and $E$ are infrared active. For the C4v molecular point group, with vertical planes through the outer atoms:

There is no inversion centre, so infrared and Raman mutual exclusion does not apply.

In the stipulated ideal, fully resolved limit, yes: square-planar geometry gives three infrared fundamentals, whereas square-pyramidal geometry gives four: two parallel $A_1$ bands and two perpendicular $E$ bands. The extra parallel fundamental is the distinguishing feature. In a real incomplete spectrum, a weak band, overlap or accidental degeneracy could hide that difference, so absence of a fourth observed band would not alone prove planarity.

$I_{C_4}\simeq 2.43\times 10^{-45}\mathrm{kg}{\mathrm{m}}^2=146.6m_{\mathrm{u}}{\mathring{\mathrm{A}}}^{,2}$. 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.

The inferred dipole moment is approximately $0.626\mathrm{D}$. With the static field parallel to the radiation electric field, the microwave selection rule is $\Delta M_J=0$. A fully resolved three-component first-order rotational Stark effect pattern naturally identifies the $J=1\to 2$, $|K|=1$ transition: the three initial $M_J=-1,0,1$ values give the three components. $K=0$ would have no first-order shift.

Subtract the lower-state field shift from the upper-state field shift, rather than equating a transition shift to one level's shift:For $J=1$ and $|K|=1$, neighbouring components are separated by $\mu E_{\mathrm{field}}/(3h)$. HenceThis assignment assumes the stated splitting resolves all allowed first-order components, rather than three accidentally unresolved groups of a higher-$J$ transition.