1,4-Benzoquinone 2026-10-20
A planar conjugated molecule with two opposite carbonyl groups and molecular point group . A simple eight-site Hückel method model uses one perpendicular orbital on each of its six carbon and two oxygen atoms.
Bond-length alternation 2026-10-20
Alternation between shorter and longer chemical bonds along a conjugated system. In the Hückel method, unequal nearest-neighbour couplings split the frontier molecular orbitals. An electronic energy gain must be compared at a stated constraint on the mean coupling and against the geometric distortion cost.
Delocalization energy 2026-10-20
The change in energy on allowing a specified localized orbital reference to delocalize across a conjugated system. In a Hückel method calculation, the reference must use the same one-centre parameters and bond types as the delocalized molecule. With the signed convention , stabilization is negative; its positive magnitude is often also called the delocalization energy, so the sign convention must be stated.
Hückel orbital charge density 2026-10-20
In an orthonormal Hückel method basis, an occupied molecular orbital with occupancy contributes electrons to site . Adding one electron to a fixed orbital changes the site population by . A coefficient sign indicates orbital phase, not charge sign. A net atomic charge also needs a reference population and treatment of the electrons outside the model.
Hückel square-to-trapezium distortion 2026-10-20
For a four-site cycle with site energy , three resonance integrals and one weakened integral , , put . In the independent-electron Hückel method, filling the two or four lowest spin orbitals gives total electronic energies and . At the square limit , these become and . The weakened-edge distortion raises the two-electron sum but lowers the four-electron sum. Near the latter change is , so its first-order term cancels. This electronic comparison omits structural repulsion and other restoring forces, and does not determine the open-shell spin at the square limit.
Past exam of the chemistry A course of the University of Cambridge 2021 Paper 2 A10 b ii Solution 2026-10-20
Differentiating the tight binding dispersion produced by the Hückel method givesHenceUnlike a free particle, the band effective mass depends on wavevector and can change sign. At an inflection point , the inverse mass vanishes and the curvature-based mass diverges; a quadratic approximation then ceases to describe the dispersion locally. For the usual convention, the bottom at has positive mass and the top at has negative electron mass. The constant has no effect on curvature.
Past exam of the chemistry A course of the University of Cambridge 2022 Paper 1 A4 b ii Solution 2026-10-20
Use the four supplied opposite-site combinations in the order . Their Hückel method matrix isThe negative corner is essential. For example, includes the long bonds – and –, each with opposite coefficient signs, giving .
Past exam of the chemistry A course of the University of Cambridge 2022 Paper 1 A4 b i Solution 2026-10-20
Write and use the orthonormal basis of the Hückel method. The four normalized one-dimensional symmetry-adapted linear combinations areFor example, the mirror plane exchanges sites , , , , without changing the sign of an individual orbital. The all-positive combination is unchanged by every vertical mirror but changes sign under , giving . The alternating combination changes sign under both sorts of vertical mirror, as required for . The two remaining sign patterns distinguish from in the printed axes. The signs refer to the upper lobes; the lower lobes have opposite quantum phases, and neither colour denotes an electric charge.
For a normalized coefficient vector , the energy isEvery site has one short-bond and one long-bond neighbour. Equivalently, apply the Hückel method matrix to any coefficient and read off its two neighbouring signs. The results areWith negative resonance integrals and , the and levels are occupied, while and are unoccupied in the closed-shell alternating-bond state.
Past exam of the chemistry A course of the University of Cambridge 2022 Paper 1 A4 c i 1 Solution 2026-10-20
Fill four spatial molecular orbitals with the eight electrons. For negative and , the occupied energies are , , and the pair at . ThereforeFor equal integrals, , this becomesAt equal bond lengths, the last two electrons occupy the frontier degeneracy at , so their distribution within that degenerate space does not change the bare Hückel method energy.
Past exam of the chemistry A course of the University of Cambridge 2023 Paper 1 A5 a ii Solution 2026-10-20
Filling the three bonding molecular orbitals with two electrons each givesThree isolated ethene-like bonds, evaluated with the same Hückel method parameters, have . Thus the signed delocalization energy isEquivalently, the positive stabilization is .
Benzene has a planar cyclic conjugated system with six electrons, satisfying the count with . All three bonding molecular orbitals are filled and the antibonding levels are empty, giving the closed-shell stabilization associated with aromaticity. The conclusion uses the continuous cyclic overlap as well as the electron count.
Past exam of the chemistry A course of the University of Cambridge 2023 Paper 1 A5 a i Solution 2026-10-20
Use the printed numbering: site 1 is at the top on , sites 2 and 3 are on the right, site 4 is at the bottom, and sites 5 and 6 are on the left; is perpendicular to the ring. Let be consistently phased, orthonormal atomic orbitals at those sites.
In the Hückel method, and , including the 6–1 bond. Set , with the usual . The cyclic nearest-neighbour matrix has eigenvalues . One real, normalized orthonormal eigenbasis isEach row means . The three lowest molecular orbitals hold the six electrons. Any real orthogonal mixing within either degenerate pair is an equally valid choice of molecular orbitals; the coefficient vectors specify exactly which sketches are used here.
Original benzene pi-orbital energy diagram and six real orbital phase sketches in the printed atom numbering
. Details. The coloured circles show the signs and relative magnitudes of the lobes above the molecular plane. The lobes below it have the opposite signs; colours are quantum phases, not electric charges.
Past exam of the chemistry A course of the University of Cambridge 2023 Paper 1 A5 b ii Solution 2026-10-20
The nitrogen pair and the four-carbon set are separately invariant under the D2h point group. Construct symmetry-adapted linear combinations within these two sets; mixing equivalent representations across the sets is left for the Hückel method calculation.
For the nitrogen atomic orbitals, normalized combinations areFor the carbon atomic orbitals, takeThese six functions are orthonormal under the zero-overlap assumption. The suffixes describe coefficient patterns across the carbon sites; the sign acquired by each underlying function must also be included when assigning an irreducible representation. In particular, reflection through the molecular plane changes the sign of every function, while inversion sends to . Ignoring that orbital sign would give incorrect labels.
The complete group representation is thereforeThe diagram shows all six ring symmetry-adapted linear combinations and the resulting energy ordering used in part (iii).
Past exam of the chemistry A course of the University of Cambridge 2023 Paper 1 A5 b iv Solution 2026-10-20
The six electrons give the Hückel method energyThe appropriate localized reference contains two C=N bonds and one C=C bond. Three C=C bonds would incorrectly replace the nitrogen one-centre parameters. For one isolated C=N pair, the dimensionless matrix iswhose eigenvalues are . Since , the occupied bonding molecular orbital has energy . Two doubly occupied C=N bonds and one doubly occupied C=C bond consequently giveTherefore the signed delocalization energy isThe positive stabilization is , smaller than the benzene value in the stated model but still stabilizing.
Pyrazine is aromatic: its planar cyclic conjugated system contains six electrons and has a filled set of three bonding molecular orbitals. Each ring nitrogen contributes one electron; its in-plane lone pair is outside this system. The heteroatoms change orbital energies and coefficients without interrupting cyclic conjugation or the count. Aromaticity does not require the stabilization to equal benzene's.
Past exam of the chemistry A course of the University of Cambridge 2025 Paper 1 A5 c Solution 2026-10-20
Set , with the usual bonding convention . In the Hückel method, the dimensionless matrix has diagonal entries one on oxygen sites 1 and 8, zero on carbon, and unit entries between bonded neighbours. In the bases and above, its blocks areThe one-dimensional and blocks have and , respectively. The characteristic polynomial of the block factors asso its three molecular orbital energies areFor the block the matrix trace is zero. Thus its three values sum to zero, and the pair supplied in the paper gives the remaining one as . Directly, its secular equation is , whose roots are approximately , and . The supplied pair is approximate, so from its trace and from the exact block should not be mistaken for an additional physical splitting.
All eight energies, ordered from low to high for , areThe eight electrons fill the four lowest molecular orbitals. The equality at is an accidental degeneracy of this parameterized model, not a multidimensional representation of .
Original Hückel method energy-level diagram for neutral para-quinone, with eight electrons and its unoccupied level
. Details. and . The Hückel method gives benzene's occupied frontier level as and its unoccupied frontier level as . ThereforeUsing the Koopmans' theorem approximation for the ionization energy, , gives and the stated parameters. This calibration neglects orbital relaxation, electron correlation and any distinction between a measured optical excitation and an independent-particle orbital gap.
The model predicts . For a uniform ten-site cyclic Hückel method model,Ten pi electrons fill with two electrons per orbital. The Homo is the degenerate pair, soThis is the prediction for the stipulated planar conjugated ring. It does not assert that an unconstrained all-cis ten-membered ring actually adopts a strain-free planar geometry.
An effective one-electron Hückel method operator may be writtenThe first term is one-electron kinetic energy. The effective site potentials represent attraction to the nuclei and any averaged screening implicit in the chosen parametrization. They replace the explicit many-electron interaction; they do not contain a coordinate-dependent two-electron operator .
For normalized hydrogen-centred orbitals, symmetry givesThe Hückel Coulomb parameter is an on-site one-electron energy, not the electron–electron Coulomb repulsion integral. The Hückel resonance integral couples the sites; with the usual same-phase atomic-orbital convention it is negative and stabilizes the bonding combination. For zero overlap, the two energies are .
Number the four sites of the cyclobutadienyl dianion cyclically and apply the Hückel method. The orthonormal pi basis givesA convenient real orthonormal eigenbasis isThus one bonding and two degenerate nonbonding orbitals are filled, and the antibonding orbital is empty. There are six pi electrons, including the two added by the dianion charge. Rotating the two occupied nonbonding orbitals within their degenerate subspace changes their individual sketches but not the total density matrix.
Use a common Coulomb integral on each site, a resonance integral on each adjacent pair, zero interaction across a diagonal, and the identity overlap matrix. In site order the Hückel method givesApplying it to the four symmetry-adapted linear combinations givesFor the conventional bonding sign , the all-in-phase combination is lowest and the alternating combination is highest. The combinations have canceling nearest-neighbour interactions and are nonbonding within this model.
Treat the site atomic orbitals as orthonormal, as in the Hückel method. A normalized set of symmetry-adapted linear combinations isThe first combination is unchanged by every operation; the alternating combination has the transformation signs of for the printed axes. The last two transform together like . The squared coefficients sum to one in each case, and their pairwise inner products vanish. Any orthogonal rotation of the two combinations is an equally valid basis for that degenerate subspace.





