Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A5

For a displacement group representation, an atom contributes only if the operation leaves its site fixed. Its three Cartesian displacements then contribute the trace of the polar-vector transformation.

Take $z$ perpendicular to the ring and an in-plane $x$ axis through opposite carbon atoms. To avoid confusing the principal twofold rotation with the in-plane axes, write $C_2(z)$ for the former, $C_{2a}$ for an in-plane axis through atoms and $C_{2b}$ for one through opposite bond midpoints. The carbon displacement characters in the D6h molecular point group areA through-atom half-turn fixes two carbon sites and has Cartesian trace $-1$, giving $-2$. The molecular-plane reflection fixes all six carbons with trace $+1$, giving $6$. A through-atom vertical mirror fixes two sites with trace $+1$, giving $2$.

Using molecular character reduction, the 18-dimensional carbon-only result isThe six one-dimensional terms contribute six dimensions and the six copies of two-dimensional representations contribute twelve, totaling eighteen. The distinction between carbon-only and whole-molecule displacements is needed in part (ii).

Each hydrogen lies on the radial line through its attached carbon. Consequently every molecular symmetry operation fixes or permutes the hydrogen sites exactly as it does the carbon sites, and the Cartesian trace is the same. Thus ${\Gamma }_H={\Gamma }_C$ and the complete benzene displacement group representation is ${\Gamma }_{3N}=2{\Gamma }_C$, of dimension $36$.

Subtract translations $A_{2u}+E_{1u}$ and rotations $A_{2g}+E_{1g}$. The full set of normal modes isThe one-dimensional components contribute $2+1+2+1+2+2=10$ coordinates. The two-dimensional components contribute $2(1+4+3+2)=20$, giving $30=3(12)-6$ vibrational degrees of freedom. There are twenty distinct symmetry-labelled fundamentals before accidental coincidences, because each doubly degenerate pair has one frequency.

Subtracting translations and rotations directly from the carbon-only result would instead give twelve carbon-skeleton coordinates. That is a six-site mechanical model, not all the normal modes of ${\mathrm{C}}_6{\mathrm{H}}_6$. The hydrogen contribution is required for the actual molecule.

The transition strength depends on $|R_{ji}{|}^2$, so a necessary condition is that at least one Cartesian electric dipole transition moment is nonzero. An integral can be nonzero only if its integrand contains the totally symmetric representation:The harmonic vibrational ground state is totally symmetric, and a one-quantum state of normal mode $k$ transforms as ${\Gamma }^{(k)}$. For the real irreducible representations here, the fundamental is therefore symmetry-allowed precisely whenEquivalently, expand the electric dipole moment near equilibrium:The fundamental has $⟨1_k|Q_k|0_k⟩\ne 0$, so vibrational infrared activity requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving $A_{2u}$ and $E_{1u}$ as the IR-active species. Symmetry permission does not guarantee a large band intensity.

From the complete normal mode result, the vibrational infrared activity belongs toThe $A_{2u}$ fundamental couples to the out-of-plane dipole component and each $E_{1u}$ fundamental to the two in-plane components. Counting one frequency for each degenerate pair, the expected number is $1+3=4$ fundamental bands, containing seven vibrational coordinates.

This is the symmetry count for the full molecule, including hydrogen displacements. The experimental compilation likewise lists the out-of-plane $A_{2u}$ fundamental and three $E_{1u}$ fundamentals; anharmonic resonances and weak intensity borrowing can produce additional observed features beyond the ideal fundamental count.^[ref]

The electronic transition starts from $A_{1g}$, so the electric-dipole test reduces to whether the final electronic irreducible representation occurs among $A_{2u}+E_{1u}$. The excited $E_{1u}$ state matches the $(x,y)$ dipole components and is therefore electric-dipole allowed.

The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is electric-dipole forbidden at the equilibrium geometry. Both excited states are singlets, so the difference is spatial symmetry rather than the spin selection rule. Inversion parity alone does not decide the result: both final states are ungerade, but only the matching vector representation is allowed.

The overall vibrational ground state is invariant under every molecular symmetry operation, so it transforms as $A_{1g}$. Combining it with the electronic ground state givesFor a degenerate harmonic normal mode, the zero-quantum wavefunction depends on the invariant sum $Q_x^2+Q_y^2$, so the ground vibrational symmetry remains totally symmetric even when excited vibrational states form multiplets.

One quantum in an $E_{2g}$ normal mode gives an $E_{2g}$ vibrational pair. The electronic factor is $B_{2u}$, henceThis follows by multiplying the characters of the two group representations. In particular, multiplication of $E_2$ by $B_2$ produces $E_1$ in the hexagonal group, and the inversion parities multiply as $u\times g=u$. The resulting vibronic state is twofold degenerate.

The initial vibronic state is $A_{1g}$ and the final one is $E_{1u}$, so an in-plane dipole component satisfiesThe product contains the totally symmetric representation. The combined electronic–vibrational transition is therefore symmetry-allowed, with an in-plane electric dipole transition moment.

A nonzero intensity still needs vibronic coupling. In a coordinate-independent separation, the forbidden electronic matrix element would remain zero. Its first derivative with respect to the $E_{2g}$ coordinate can instead be nonzero, and the vibrational factor $⟨1|Q|0⟩$ supplies the required one-quantum transition. This is vibrational activation of an electronic transition and explains why the band can be weak.

A coordinate $Q_s$ of an $A_{1g}$ normal mode is unchanged by every molecular symmetry operation. Every harmonic-oscillator excitation $H_n(Q_s)e^{-Q_s^2/2}$ therefore also has symmetry $A_{1g}$, including odd values of $n$. Consequentlyfor any number of quanta in that totally symmetric coordinate. The same electric-dipole permission survives every $A_{1g}$ excitation.

This statement concerns symmetry. Actual progression intensities depend on vibrational inner products, equilibrium displacements and the coordinate dependence of the electric dipole transition moment; a permitted line need not have measurable intensity for every vibrational quantum number.