Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A4

Let $Q_k$ be a vibrational normal mode coordinate with symmetry ${\Gamma }_k$. The harmonic vibrational ground state is totally symmetric; the one-quantum state transforms as $Q_k$. Integration over a symmetry-invariant domain annihilates any integrand lacking the totally symmetric representation. Thus a fundamental requiresfor some Cartesian component $\alpha$. For the one-dimensional irreducible representations used here, this means ${\Gamma }_k=\Gamma (x)$, $\Gamma (y)$ or $\Gamma (z)$. Equivalently, expanding the electric dipole moment givesthe derivative can be nonzero only if $Q_k$ and ${\mu }_{\alpha }$ have compatible symmetry. This is the symmetry criterion for vibrational infrared activity; it permits a transition without guaranteeing a large intensity.

With $z$ along C=O and the molecule in the $xz$ plane, formaldehyde has $E$, $C_2(z)$, ${\sigma }_{xz}$ and ${\sigma }_{yz}$. It belongs to the C2v point group, $\boxed{C_{2v}}$. Keeping this axis convention is essential when assigning the labels $B_1$ and $B_2$.

For operations ordered as $(E,C_2(z),{\sigma }_{xz},{\sigma }_{yz})$, the characters of the twelve Cartesian displacements areOnly atoms left fixed contribute: all four for $E$ and ${\sigma }_{xz}$, but only C and O for $C_2$ and ${\sigma }_{yz}$. Multiplying their counts by the Cartesian traces $3,-1,1,1$ gives these numbers. Subtractto obtain ${\chi }_{\mathrm{vib}}=(6,0,4,2)$. The character table reduction givesThe dimensions sum to $6=3N-6$, as required. The $B_2$ mode is the out-of-plane wag with this axis convention.

In this C2v point group convention, $z\sim A_1$, $x\sim B_1$, and $y\sim B_2$. Therefore all six normal modes have symmetry-allowed infrared fundamentals: $\boxed{6\text{ IR features}}$, absent accidental coincidences or vanishing intensities.

For Raman spectroscopy, examine the polarizability tensor: $x^2,y^2,z^2\sim A_1$, $xz\sim B_1$, and $yz\sim B_2$; $xy\sim A_2$ is also Raman-allowed but no $A_2$ vibration occurs here. All six vibrations are therefore Raman-active too, giving $\boxed{6\text{ Raman features and }6\text{ IR/Raman coincidences}}$. The centrosymmetric mutual-exclusion rule does not apply to $C_{2v}$ formaldehyde.

The C=O length is invariant under every operation, so its stretch is $A_1$. If $s_L,s_R$ are the two C–H bond-length changes, their sum and difference giveHence $\boxed{{\Gamma }_{\mathrm{stretch}}=2A_1+B_1}$. These are stretching symmetry coordinates; the two $A_1$ motions can mix in actual normal modes, and mass-weighted atomic motions need not have the schematic arrow lengths below.

There are three stretching fundamentals, although four stretching-region features are listed. The low feature at $1746\mathrm{c}{\mathrm{m}}^{-1}$ is predominantly C=O stretch ($A_1$). The two features at $2783$ and $2843\mathrm{c}{\mathrm{m}}^{-1}$ are the symmetric and antisymmetric C–H stretches; their order is not determined by this mode count alone.

The $3482\mathrm{c}{\mathrm{m}}^{-1}$ feature is consistent with the first C=O vibrational overtone, close to $2(1746)=3492\mathrm{c}{\mathrm{m}}^{-1}$ and shifted downward by anharmonicity. Its symmetry is $A_1\otimes A_1=A_1$, so symmetry permits the weak anharmonic transition. It is not a fourth stretching fundamental. The combination-band evidence in the next part distinguishes the two C–H assignments.

The out-of-plane displacements in the wag transform as $y$, hence as $B_2$. A one-quantum combination of two distinct normal modes transforms as the product of their irreducible representations. The observed feature satisfiesA combination of the wag with an $A_1$ stretch has symmetry $B_2\otimes A_1=B_2$ and can absorb through ${\mu }_y$. In contrast, $B_2\otimes B_1=A_2$ has no electric-dipole component in $C_{2v}$ and is forbidden in the stated symmetry model. The missing feature near $4010=1167+2843\mathrm{c}{\mathrm{m}}^{-1}$ therefore identifies the $2843$ stretch as $B_1$.

The completed assignments areThe allowed combination obtains intensity through anharmonic/dipole-coupling terms; a purely separable harmonic model with a strictly linear dipole has no combination bands. The argument uses both the observed combination and the selection rule: absence of a feature by itself would not generally prove a forbidden symmetry.^[ref]