Chemical Bonding and Group Theory: VB and MO Theory, Hückel Theory, Point Groups and Selection Rules
1. Born–Oppenheimer, valence-bond theory and hybridisation
Nuclei are thousands of times heavier than electrons and move far more slowly, so in the Born–Oppenheimer approximation the electronic Schrödinger equation is solved with the nuclei held fixed. Repeating this for many nuclear arrangements gives the potential-energy curve or surface on which the nuclei then vibrate and rotate; bond lengths, force constants and dissociation energies are all read from it. The approximation fails where two electronic surfaces come close, as at conical intersections in photochemistry.
The Heitler–London valence-bond treatment of H₂ pairs one electron on each atom: ψ_VB = 1s_A(1)1s_B(2) + 1s_A(2)1s_B(1), multiplied by the singlet spin function. It gives binding because of the exchange (resonance) integral, and it omits the ionic terms H⁺H⁻ entirely; adding them with a small coefficient improves the energy. Hybridisation mixes orbitals on one atom to point along bonds: sp (linear, 180°), sp² (trigonal, 120°), sp³ (tetrahedral, 109.47°), dsp² (square planar), sp³d (trigonal bipyramidal) and sp³d² (octahedral). For two equivalent s–p hybrids with fractional s-character α the interorbital angle obeys cos θ = −α/(1 − α), so more s-character means a wider angle — Bent’s rule: more electronegative substituents take orbitals richer in p-character.
2. LCAO-MO theory: H₂⁺, H₂ and diatomic molecules
In the LCAO-MO method a molecular orbital is a linear combination of atomic orbitals, and the variational principle fixes the coefficients through a secular determinant. For H₂⁺, ψ± = (1s_A ± 1s_B)/√(2(1 ± S)) with energies E± = (α ± β)/(1 ± S), where α is the Coulomb integral, β the (negative) resonance integral and S the overlap. Because of the 1 ± S denominators the antibonding orbital is raised more than the bonding orbital is lowered, so filling both bonding and antibonding orbitals (He₂) is net repulsive. H₂ puts two paired electrons in σ1s; the simple MO function contains ionic and covalent terms in equal weight, the opposite fault to the VB function, and configuration interaction repairs both.
For second-row homonuclear diatomics the valence MOs are σ2s, σ2s, σ2p, π2p (×2), π2p (×2) and σ*2p. s–p mixing pushes σ2p above π2p for Li₂ to N₂, while for O₂ and F₂ the order is σ2p below π2p. Bond order = ½(bonding − antibonding electrons). O₂ has two unpaired electrons in the degenerate π* pair and is paramagnetic, which the Lewis structure cannot explain; B₂ is paramagnetic for the same reason in its π2p pair; C₂ and N₂ are diamagnetic. In heteronuclear molecules the more electronegative atom’s orbitals lie lower and dominate the bonding MOs: in CO (isoelectronic with N₂) the HOMO is a σ orbital concentrated on carbon, which is why CO binds metals through C; NO has one electron in π* and bond order 2.5; in HF the bonding σ is mostly F 2p.
| Species | Valence electrons | Bond order | Unpaired electrons |
|---|---|---|---|
| B₂ | 6 | 1 | 2 |
| C₂ | 8 | 2 | 0 |
| N₂ | 10 | 3 | 0 |
| N₂⁺ | 9 | 2.5 | 1 |
| O₂⁺ | 11 | 2.5 | 1 |
| O₂ | 12 | 2 | 2 |
| O₂⁻ | 13 | 1.5 | 1 |
| O₂²⁻ | 14 | 1 | 0 |
| F₂ | 14 | 1 | 0 |
| NO | 11 | 2.5 | 1 |
| CO | 10 | 3 | 0 |
3. Hückel theory of conjugated π systems
Hückel theory treats only the π electrons, one p orbital per carbon, and makes drastic simplifications: every Hᵢᵢ = α; Hᵢⱼ = β for bonded neighbours and 0 otherwise; Sᵢⱼ = 0 for i ≠ j. Writing x = (α − E)/β, the secular determinant has x on the diagonal and 1 for each bonded pair. Both α and β are negative, so E = α + kβ with k > 0 is bonding. For ethylene E = α ± β. For a linear chain of N atoms the roots are E_k = α + 2β cos(kπ/(N + 1)), and for a ring E_k = α + 2β cos(2πk/N).
| System | Orbital energies, x | Total π energy | Delocalisation energy |
|---|---|---|---|
| Ethylene (2 e) | +1, −1 | 2α + 2β | 0 |
| Allyl cation (2 e) | +1.414, 0, −1.414 | 2α + 2.828β | 0.828β |
| Allyl radical (3 e) | +1.414, 0, −1.414 | 3α + 2.828β | 0.828β |
| Butadiene (4 e) | ±1.618, ±0.618 | 4α + 4.472β | 0.472β |
| Cyclobutadiene (4 e) | +2, 0, 0, −2 | 4α + 4β | 0 |
| Benzene (6 e) | +2, +1, +1, −1, −1, −2 | 6α + 8β | 2β |
The delocalisation energy is the π energy minus that of the same number of isolated ethylenic bonds (2β per bond). Benzene’s 2β is the Hückel measure of aromatic stabilisation. The ring formula explains Hückel’s 4n + 2 rule: a ring has one lowest orbital and then degenerate pairs, so 4n + 2 electrons fill a closed shell, while 4n electrons half-fill a degenerate pair — cyclobutadiene is predicted to be a diradical with no delocalisation energy, i.e. antiaromatic. The coefficients also give π-bond orders and charge densities: in butadiene the C1–C2 π bond order is 0.894 and C2–C3 is 0.447, matching the short–long–short bond pattern.
4. Symmetry elements, groups and point-group classification
A symmetry operation moves a molecule into an indistinguishable orientation; the symmetry element is the point, line or plane about which it acts. There are five kinds: the identity E; a proper rotation Cₙ by 360°/n; a reflection σ (σ_h perpendicular to the principal axis, σ_v containing it, σ_d containing it and bisecting two C₂ axes); inversion i through a centre; and an improper rotation Sₙ, rotation by 360°/n followed by reflection in the perpendicular plane (S₁ = σ, S₂ = i). The operations of a molecule form a group: the product of any two is a member (closure), multiplication is associative, E is present, and every operation has an inverse. The order h is the number of operations. A multiplication table lists every product, and operations that are conjugate (Y = X⁻¹AX for some X in the group) form a class — in C₃v the two C₃ rotations are one class and the three σ_v another.
| Point group | Order h | Examples |
|---|---|---|
| C₁ | 1 | CHFClBr |
| Cₛ | 2 | HOCl, NOCl |
| Cᵢ | 2 | meso-1,2-dibromo-1,2-dichloroethane (anti) |
| C₂ | 2 | H₂O₂ (skew), 1,3-dichloroallene |
| C₂v | 4 | H₂O, SO₂, CH₂Cl₂, cis-N₂F₂ |
| C₃v | 6 | NH₃, CHCl₃, PCl₃ |
| C₂h | 4 | trans-N₂F₂, trans-CHCl=CHCl |
| D₂d | 8 | allene, cyclooctatetraene (tub) |
| D₃h | 12 | BF₃, PCl₅, eclipsed ethane |
| D₃d | 12 | staggered ethane, chair cyclohexane |
| D₄h | 16 | XeF₄ and [PtCl₄]²⁻ |
| D₆h | 24 | benzene |
| T_d | 24 | CH₄, [NiCl₄]²⁻ |
| O_h | 48 | SF₆, [Co(NH₃)₆]³⁺ |
| C∞v / D∞h | ∞ | HCl, CO / CO₂, N₂ |
Classification follows a fixed sequence: look first for a linear or high-symmetry (T_d, O_h, I_h) molecule; otherwise find the principal axis Cₙ; ask whether there are n C₂ axes perpendicular to it (the D groups) or not (C and S groups); then look for σ_h (Dₙh, Cₙh), for n σ_d or σ_v (Dₙd, Cₙv), or for an S₂ₙ alone. Two consequences are examined repeatedly: a molecule is chiral if and only if it has no improper axis Sₙ (no σ, no i, no S₄ …), and it can have a dipole moment only in the groups C₁, Cₛ, Cₙ and Cₙv.
5. Representations, character tables and the reduction formula
A set of matrices that multiply like the group’s operations is a representation. Using any set of vectors as a basis (atomic displacements, orbitals, bonds) gives a reducible representation Γ, which can be broken into irreducible representations — the smallest blocks, listed with their characters (traces) in the character table. The number of irreducible representations equals the number of classes, and Σ dᵢ² = h. Mulliken labels: A and B are one-dimensional (symmetric or antisymmetric to the principal rotation), E two- and T three-dimensional; subscript g/u marks symmetric/antisymmetric under inversion, 1/2 under a C₂ or σ_v, and ′/″ under σ_h.
| C₂v | E | C₂(z) | σ_v(xz) | σ_v′(yz) | Functions |
|---|---|---|---|---|---|
| A₁ | 1 | 1 | 1 | 1 | z, x², y², z² |
| A₂ | 1 | 1 | −1 | −1 | R_z, xy |
| B₁ | 1 | −1 | 1 | −1 | x, R_y, xz |
| B₂ | 1 | −1 | −1 | 1 | y, Rₓ, yz |
The reduction formula gives how many times each irreducible representation i occurs in Γ: nᵢ = (1/h) Σ_R g(R) χ_Γ(R) χᵢ(R), summed over classes with g(R) operations each. For the 3N displacements of H₂O the characters are found from the number of atoms left in place by each operation times the contribution per unshifted atom (E: 3, C₂: −1, σ: +1): Γ₃N = (9, −1, 3, 1). Reduction gives Γ₃N = 3A₁ + A₂ + 3B₁ + 2B₂. Removing translations (A₁ + B₁ + B₂, the functions x, y, z) and rotations (A₂ + B₁ + B₂) leaves Γ_vib = 2A₁ + B₁: two totally symmetric modes (symmetric stretch and bend) and one antisymmetric stretch. With the molecule placed in the yz plane instead, the last label is B₂; the physics is unchanged.
6. Selection rules, normal modes, SALCs and hybrid orbitals
A transition between states ψᵢ and ψ_f is allowed only if the transition-moment integral ∫ψᵢ* μ̂ ψ_f dτ is non-zero, which requires the direct product Γᵢ ⊗ Γ_μ ⊗ Γ_f to contain the totally symmetric representation. For a fundamental vibration from the totally symmetric ground state this becomes simple: a mode is IR active if it transforms as x, y or z and Raman active if it transforms as a quadratic function (x², xy, …). In a molecule with a centre of inversion the x, y, z functions are u and the quadratic ones g, so no mode is both — the rule of mutual exclusion. CO₂ (D∞h) has 3N − 5 = 4 modes: the symmetric stretch Σg⁺ (Raman only), the antisymmetric stretch Σu⁺ and the doubly degenerate bend Πu (IR only). For electronic transitions in a centrosymmetric complex, g → g d–d bands are forbidden by the same argument — the Laporte rule.
Internal coordinates (bond stretches, angle bends) used as the basis give the symmetry of the stretching and bending modes directly: the two O–H stretches of water span A₁ + B₁, the bend A₁. A symmetry-adapted linear combination (SALC) is a combination of equivalent basis functions that transforms as one irreducible representation, obtained with the projection operator; only orbitals of the same symmetry can combine into MOs. The same logic constructs hybrids: the three σ bonds of BF₃ span A₁′ + E′ in D₃h, matched by s (A₁′) and (pₓ, p_y) (E′) — sp²; the four σ bonds of CH₄ span A₁ + T₂ in T_d, matched by s + (pₓ, p_y, p_z) — sp³ — or, in principle, s + (d_xy, d_xz, d_yz) — sd³.
Key takeaways
- Born–Oppenheimer separates electronic from nuclear motion; VB pairs electrons between atoms, LCAO-MO spreads them over the molecule, and E± = (α ± β)/(1 ± S) makes antibonding raise more than bonding lowers.
- Bond order = ½(bonding − antibonding); s–p mixing puts π2p below σ2p up to N₂; O₂ and B₂ are paramagnetic, C₂ and N₂ diamagnetic.
- Hückel: E = α + xβ; butadiene delocalisation 0.472β, benzene 2β, cyclobutadiene 0 — the 4n + 2 rule follows from the ring energies α + 2β cos(2πk/N).
- Point groups follow a fixed flowchart; chiral means no Sₙ of any order, and a dipole needs C₁, Cₛ, Cₙ or Cₙv.
- nᵢ = (1/h)Σ g χ_Γ χᵢ reduces representations; IR activity needs x, y, z symmetry and Raman activity a quadratic function, with mutual exclusion in centrosymmetric molecules.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
What is the bond order of the superoxide ion O₂⁻ in molecular-orbital theory? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 1.5
O₂⁻ has 13 valence electrons: σ2s² σ2s² σ2p² π2p⁴ π2p³, so bonding 8 and antibonding 5, bond order = (8 − 5)/2 = 1.5. O₂ itself is 2 and the peroxide O₂²⁻ is 1; adding electrons to π* lowers the order.How many unpaired electrons does the B₂ molecule have in its ground state, according to MO theory with s–p mixing?
Numerical answer — type the value.
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Answer: 2
B₂ has 6 valence electrons: σ2s² σ*2s² then π2p², because s–p mixing lifts σ2p above the π2p pair. By Hund’s rule the two π electrons occupy the two degenerate orbitals singly, so B₂ is paramagnetic with 2 unpaired electrons. Without s–p mixing σ2p² would give 0.Which of the following species are paramagnetic?
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Answer: A — O₂; C — N₂⁺
O₂ has two unpaired π* electrons and N₂⁺ (9 valence electrons) one unpaired σ2p electron, so both are paramagnetic. C₂ ends with a filled π2p⁴ pair and CO is isoelectronic with N₂, both with every electron paired.In the LCAO treatment of H₂⁺ with overlap S included, E± = (α ± β)/(1 ± S). What does this imply?
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Answer: A — The antibonding orbital is destabilised more than the bonding orbital is stabilised
E₊ − α = (β − αS)/(1 + S) and E₋ − α = −(β − αS)/(1 − S); the smaller denominator 1 − S makes the antibonding shift larger. That is why He₂, with both levels full, is unbound. Equal splitting holds only when S is set to zero, as Hückel theory does.Two equivalent s–p hybrid orbitals on one atom each have one-third s-character. What is the angle between them, in degrees?
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Answer: 120
cos θ = −α/(1 − α) = −(1/3)/(2/3) = −0.5, so θ = 120°: sp² hybrids. The same formula with α = 1/4 gives cos θ = −1/3 and the tetrahedral 109.47°; with α = 1/2 it gives 180°.Using Hückel theory, what is the π delocalisation energy of 1,3-butadiene, in units of |β|, to two decimal places?
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Answer: 0.47
For a four-atom chain x = 2cos(kπ/5) = ±1.618, ±0.618. Four electrons fill the two bonding levels: E_π = 2(α + 1.618β) + 2(α + 0.618β) = 4α + 4.472β. Two isolated ethylenes give 4α + 4β, so the delocalisation energy is 0.472β, i.e. 0.47|β|.In Hückel theory the total π-electron energy of benzene is written 6α + nβ. What is n?
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Answer: 8
The ring energies α + 2β cos(2πk/6) are α + 2β, α + β (×2), α − β (×2) and α − 2β. Six electrons fill the three bonding orbitals: 2(α + 2β) + 4(α + β) = 6α + 8β. Subtracting three ethylenes (6α + 6β) gives the delocalisation energy 2β, which is often confused with n.What is the Hückel delocalisation energy of the allyl cation (two π electrons), in units of |β|, to two decimal places?
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Answer: 0.83
For three atoms x = 2cos(kπ/4) = +1.414, 0, −1.414. Both electrons go into α + 1.414β, giving 2α + 2.828β. One localised double bond gives 2α + 2β, so the delocalisation energy is 0.828β, i.e. 0.83|β|. The radical and anion have the same value because the extra electrons enter the non-bonding level α.What is the point group of ammonia, NH₃?
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Answer: A — C₃v
NH₃ has one C₃ axis through N and three σ_v planes each containing an N–H bond, with no perpendicular C₂ axes and no σ_h — C₃v, order 6. D₃h would need a planar molecule like BF₃.What is the point group of ethane in its staggered conformation?
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Answer: A — D₃d
Staggered ethane has a C₃ axis along C–C, three perpendicular C₂ axes, three dihedral planes σ_d, a centre of inversion and an S₆ axis, but no σ_h — D₃d. The eclipsed conformer has σ_h and is D₃h.Which of the following molecules are chiral?
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Answer: A — 1,3-Dichloroallene, ClHC=C=CHCl; B — Bromochlorofluoromethane, CHFClBr
1,3-Dichloroallene (C₂) and CHFClBr (C₁) have no improper axis of any kind, so they are chiral. CH₂Cl₂ (C₂v) has mirror planes and trans-ClCH=CHCl (C₂h) has σ_h and i, so both are achiral.Which of the following are symmetry elements of the BF₃ molecule (D₃h)?
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Answer: A — A C₃ axis; C — A horizontal mirror plane σ_h; D — An S₃ improper axis
D₃h contains E, 2C₃, 3C₂, σ_h, 2S₃ and 3σ_v, order 12; S₃ exists because C₃ and σ_h both do. A trigonal planar molecule has no centre of inversion — inverting one B–F bond sends F to an empty position.What is the order of the point group D₃h?
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Answer: 12
D₃h = {E, 2C₃, 3C₂, σ_h, 2S₃, 3σ_v}: 1 + 2 + 3 + 1 + 2 + 3 = 12 operations. The pure rotation subgroup D₃ has order 6, which is the answer if the improper operations are forgotten.For H₂O (C₂v, molecule in the xz plane) the reducible representation of the 3N Cartesian displacements has characters χ(E) = 9, χ(C₂) = −1, χ(σ_v(xz)) = 3, χ(σ_v′(yz)) = 1. How many times does A₁ occur in it?
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Answer: 3
n(A₁) = (1/4)[9(1) + (−1)(1) + 3(1) + 1(1)] = 12/4 = 3. The full reduction is 3A₁ + A₂ + 3B₁ + 2B₂; removing translations and rotations leaves Γ_vib = 2A₁ + B₁, so 2 is the number of A₁ vibrations, not of A₁ in Γ₃N.How many normal modes of CO₂ are infrared active, counting each component of a degenerate mode separately?
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Answer: 3
CO₂ has 3N − 5 = 4 modes: the symmetric stretch Σg⁺ (Raman only), the antisymmetric stretch Σu⁺ and the doubly degenerate bend Πu. The last two change the dipole moment, giving 1 + 2 = 3 IR-active modes but only two IR bands (about 2349 and 667 cm⁻¹).The rule of mutual exclusion (no fundamental both IR and Raman active) applies to molecules that have
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Answer: A — a centre of inversion
With i present, x, y, z transform as u and the quadratic functions as g, so a mode cannot belong to both sets. H₂O has a C₂ axis and mirror planes yet its modes are active in both IR and Raman, which rules out those options.In D₃h, the three B–F σ bonds of BF₃ span A₁′ + E′. Which set of boron orbitals can form these σ bonds?
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Answer: A — s, pₓ and p_y (sp²)
In the D₃h table s transforms as A₁′ and (pₓ, p_y) together as E′, while p_z is A₂″. So the σ framework matches s + pₓ + p_y, the sp² set; p_z is left for π bonding with the fluorine lone pairs.