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

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.