§ C-21 Molecular Orbitals (LCAO–MO): Bonding σ/π, Antibonding σ*/π* & Bond Order (Grade 10–11 · Quantum Chemistry)
Wavefunction interference (ψ_A ± ψ_B), bonding σ/π & antibonding σ*/π* orbitals, bond order, and paramagnetism of O₂. An atomic orbital is monocentric, whereas a molecular orbital embraces all nuclei of the molecule as a single quantum wave.
Theory Summary & Scientific Mechanism
The Molecular Orbital (LCAO–MO) theory treats valence electrons as delocalized over the entire molecule via linear combinations of atomic orbitals ψ_A and ψ_B. Constructive interference (ψ₊ = c(ψ_A + ψ_B)) accumulates electron density between the nuclei, forming a lower-energy bonding MO (σ_g, πᵤ). Destructive interference (ψ₋ = c(ψ_A - ψ_B)) creates a nodal plane between the nuclei, forming a higher-energy antibonding MO (σᵤ^*, π_g^*).
Chemical bond order is defined as B = (1)/(2)(N_bond - Nₐnti). For He₂, B = (1)/(2)(2 - 2) = 0 (unstable), whereas for dinitrogen N₂, B = (1)/(2)(8 - 2) = 3.0 (D₀ = 945 kJ/mol).
MO theory triumphantly explains why liquid and gaseous dioxygen O₂ is paramagnetic: of its 12 valence electrons, the highest two occupy the degenerate antibonding π₂p,x^* and π₂p,y^* orbitals singly with parallel spins (Hund’s rule, S = 1, μₛ = 2.83 μ_B), giving B = 2.0.
Everyday Analogy
Picture two water waves meeting: crest-on-crest builds electron density between the nuclei to glue the atoms together (bonding MO σ, π), whereas crest-on-trough cancels out at a nodal plane and pushes the nuclei apart (antibonding MO σ^*, π^*).
Interactive Question & Derivation
In Molecular Orbital (MO) theory, how is the bond order B defined, and why does the oxygen molecule O₂ exhibit paramagnetism (attraction into a magnetic field)?
- A: Bond order is B = (1)/(2)(N_b - Nₐ) (B = 2 for O₂), and O₂ is paramagnetic because two unpaired electrons occupy the two degenerate antibonding π₂p^* orbitals according to Hund’s rule
- B: Bond order equals the sum of all valence electrons (N_b + Nₐ), and O₂ is diamagnetic with all spins paired
- C: Antibonding σ^* and π^* orbitals have lower energy than atomic orbitals and stabilize the molecule without nodes
Study Hint: Calculate bond order B = (1)/(2)(N_b - Nₐ) and apply Hund’s rule to the last two electrons of O₂ in the degenerate π^*₂p orbitals.
Proof & Explanation: Constructive interference of atomic wavefunctions (ψ₊ = c(ψ_A + ψ_B)) builds electron density between nuclei, forming a lower-energy bonding MO (σ, π). Destructive interference (ψ₋ = c(ψ_A - ψ_B)) creates a nodal plane between nuclei and a higher-energy antibonding MO (σ^*, π^*). In O₂ (12 valence electrons), the last two electrons singly occupy the degenerate π₂p,x^* and π₂p,y^* orbitals with parallel spins (S = 1), yielding bond order B = (1)/(2)(8 - 4) = 2.0 and strong paramagnetism.
Key Equations & 3D Simulation Parameters
ψ_± = (ψ_A ± ψ_B)/(√(2(1 ± S))), B = (N_bond - Nₐnti)/(2), μₛ = √(n(n+2)) μ_B
- Diatomic species (1=H₂, 2=He₂⁺, 3=N₂, 4=O₂, 5=F₂): 1 … 5 (default: 4)
- Internuclear distance R: 0.7 … 2.2 Å (default: 1.21 Å)
- Displayed MO (1=σ_g bonding, 2=σ*ᵤ, 3=πᵤ, 4=π*_g): 1 … 4 (default: 1)
- Resonance overlap integral |β|: 1 … 4.5 eV (default: 2.5 eV)