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 / 28

For the character of a localized orbital basis, take the trace of the transformation matrix: both permutation of centres and signs of atomic orbitals matter. The two metal centres lie on the principal axis and remain fixed under rotation about $z$. At either metal the $(p_x,p_y)$ basis has the matrixThe full four-dimensional group representation is block diagonal with two such blocks. Consequently $\chi (R)=4\cos \theta$, and for the $60^{\circ }$ rotation of the D6h molecular point group, $\chi (C_6)=4\cos (\pi /3)=2$.

A $180^{\circ }$ rotation leaves both metal centres fixed but reverses each transverse p orbital: $p_x\mapsto -p_x$ and $p_y\mapsto -p_y$. Its matrix is $-I_4$, so its character of a representation is $\chi (C_6^3)=-4$, also obtained from $4\cos \pi$.

Choose transverse directions parallel and perpendicular to the vertical mirror plane. At each metal the parallel p orbital is unchanged, whereas the perpendicular p orbital changes sign. Thus each two-dimensional block has trace $1-1=0$. The total character of a representation is $\chi ({\sigma }_d)=0$. A mirror plane passing through an orbital centre does not imply that every orbital on that centre contributes $+1$.