Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section A / 29

Use the axes actually printed: the two hydrogens lie in the $xz$ plane, $z$ is the principal twofold axis, and $y$ is perpendicular to that plane. For a Cartesian-displacement group representation, only atoms left at their own sites contribute to the trace; their vector components must also be transformed. In the order $(E,C_2(z),\sigma (xz),\sigma (yz))$, identity fixes six components, the twofold rotation exchanges the hydrogens, the $xz$ reflection fixes both hydrogens with vector matrix $\mathrm{diag}(1,-1,1)$, and the $yz$ reflection exchanges them. ThereforeThe nonzero mirror character of a representation is $2(1-1+1)=2$, not six.

The C2v molecular point group has four one-dimensional irreducible representations. In the operation order used above, the character table rows are $A_1=(1,1,1,1)$, $A_2=(1,1,-1,-1)$, $B_1=(1,-1,1,-1)$ and $B_2=(1,-1,-1,1)$. Their character inner products with ${\chi }_{\Gamma }=(6,0,2,0)$ giveHence $\Gamma =2A_1\oplus A_2\oplus 2B_1\oplus B_2$, whose dimensions sum to six. The labels $B_1,B_2$ refer to the source axes; swapping $x$ and $y$ swaps these names.