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The alternating short and long bonds preserve fourfold, but not eightfold, rotations. Including the molecular plane and inversion symmetry, the planar structure belongs to $D_{4h}$, the D4h molecular point group.

Use the printed axes: the short bonds are $1$–$2$, $3$–$4$, $5$–$6$ and $7$–$8$, and the $x$ and $y$ axes pass through the midpoints of opposite short bonds. Interchanging short and long bonds is not a molecular symmetry operation of this geometry.

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 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.

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.

There are four internal C–N stretching coordinates. A stretch length is unchanged in sign when a molecular symmetry operation carries a bond to an equivalent bond, so these coordinates have the same permutation group representation as the four ligand sigma functions.

For the square,Only $E_u\sim (x,y)$ has vibrational infrared activity. The $A_{1g}$ and $B_{1g}$ stretches have Raman vibrational activity. Thus the ideal square gives one IR feature and two Raman features, with no common fundamental. The two components of $E_u$ have one frequency, not two different lines.

For the tetrahedron,The $T_2$ stretch is IR active, while both $A_1$ and $T_2$ are Raman active. Thus the tetrahedron also gives one IR feature and two Raman features, but one feature, $T_2$, is shared between the spectra. Its three components likewise describe three independent displacements at one frequency.

The separate line counts alone do not distinguish the geometries. Comparing their frequencies does: the tetrahedral IR C–N feature coincides with one Raman C–N feature, whereas the square has no symmetry-enforced coincidence. This assumes resolved fundamentals and excludes accidental equality of unrelated frequencies.

Work with $M=(H-\alpha I)/\beta$. Its nearest-neighbour ring entries are $1$, and its diagonal entries are $1$ on the nitrogen sites and $0$ on the carbon sites. Because $M$ commutes with each molecular symmetry operation, it only mixes symmetry-adapted linear combinations with the same irreducible representation.

In the bases $(N_{+},C_0)$ and $(N_{-},C_y)$, the two nontrivial matrix blocks areTheir characteristic polynomials are $(1-\lambda {)}^2-2$ and ${\lambda }^2-3$. The remaining single-function blocks are $M_{B_{2g}}=1$ and $M_{A_u}=-1$. Thus, ordered from lowest to highest physical energy because $\beta <0$, the molecular orbitals areAll six levels are nondegenerate in this parametrization. The right-hand ladder of the diagram in part (ii) shows these levels and occupations. The occupied $B_{2g}$ level is the highest occupied molecular orbital, and the unoccupied $B_{1u}$ level immediately above it is the lowest unoccupied molecular orbital.

Pyrazine belongs to $D_{2h}$, the D2h point group. With the printed axes, its three twofold axes are $x,y,z$; it also has an inversion centre and the three coordinate mirror planes. The nitrogen atoms occupy sites 1 and 4 on the $y$ axis. Interchanging carbon and nitrogen is not a molecular symmetry operation, so the sixfold benzene symmetry is lost.

The molecular symmetry operations must be closed under composition. Write the proposed nonidentity operations as $R=C_2$ and $S=\sigma$. Their product $RS$ cannot equal any proposed member:Each conclusion contradicts the distinct operations stipulated. Therefore at least a fourth symmetry operation must exist. For a mirror plane containing the twofold axis, the product generates another mirror plane; for a plane perpendicular to it, the product is inversion. A tilted plane may generate still more operations, but cannot evade group closure.

A symmetry-closed orbital basis, often called a complete basis in this context, is a linearly independent set whose span is carried into itself by every group operation. No transformed basis function requires a function outside that span. This is completeness for the chosen group representation, not a claim that a small atomic-orbital set spans the full electronic Hilbert space.

A representation matrix $D(g)$ specifies the linear transformation produced by a molecular symmetry operation $g$. With the column convention,Its character of a representation is the matrix trace,The character is unchanged by a simultaneous change of basis and is constant on a conjugacy class, which is why character tables have one column per class.

An ordinary molecular point group consists of the identity, proper rotations $C_n$, and, when present, reflections $\sigma$, inversion $i$, and improper rotations $S_n$. Each operation leaves the nuclear geometry and chemical labels equivalent, so only identical nuclei can be interchanged. Translations are not elements of the point group of a finite isolated molecule. The molecular symmetry operations determine how electronic and vibrational coordinates transform.

An orbital set whose span is invariant under every molecular symmetry operation. Completeness here means closure of the chosen representation space; it does not mean that a finite orbital set spans the entire electronic Hilbert space.