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Molecular Orbitals — H₂, O₂, N₂ Topic #17

§1 Interactive Simulation

Three.js r128
Molecule
H₂
Bond Order
1.0
Bonding e⁻
2
Antibonding e⁻
0
Unpaired e⁻
0
Magnetic
Diamag.
⚙ Controls
Internuclear Distance R (Å)
0.50.744.0
Overlap coefficient S
00.591
Speed
0.25×1×4×
Show phase colors
Show nodal plane
Show spin arrows

§2 The Idea, Step by Step

From two clouds in step to a shared electron glue

Hold two slinky springs side by side and shake them with the same rhythm. In step, the waves pile up and grow taller; out of step, they cancel and leave a dead spot. Electron clouds do exactly this when two atoms drift close enough to touch. Each atom brings its own cloud (an atomic orbital), and the two clouds blend into brand-new, shared clouds that belong to the whole molecule — molecular orbitals. Combine them in phase and the clouds reinforce between the nuclei; combine them out of phase and a gap (a node) opens up between the nuclei.

That single in-phase / out-of-phase split is the whole story. The in-phase blend, the bonding orbital, packs extra negative charge between the two positive nuclei and pulls them together — it sits lower in energy than the lone atoms, so electrons are happy to live there. The out-of-phase blend, the antibonding orbital (written with a star, $\sigma^*$), starves the middle and sits higher in energy, prying the atoms apart. Now just count electrons, drop them into the lowest orbitals first, and read off the bond order: $\text{BO} = \tfrac{1}{2}(n_b - n_a)$, bonding minus antibonding electrons over two. Hydrogen has two electrons; both fit the bonding $\sigma_{1s}$, giving $\text{BO}=\tfrac{1}{2}(2-0)=1$ — one bond, a real molecule. Helium has four; the extra pair is forced into $\sigma_{1s}^*$, so $\text{BO}=\tfrac{1}{2}(2-2)=0$ and He$_2$ simply falls apart.

The precise statement writes each MO as a linear combination of atomic orbitals, $\psi_\pm = \frac{1}{\sqrt{2(1\pm S)}}(\phi_A \pm \phi_B)$, where the overlap integral $S$ measures how much the two clouds share space. Their energies are $E_\pm=\frac{\alpha\pm\beta}{1\pm S}$, and because of the $1\pm S$ in the bottom the antibonding level is pushed up more than the bonding level is pulled down — that asymmetry is why filling both cancels a bond rather than leaving it neutral. Head-on overlap makes cylinder-shaped $\sigma$ orbitals; side-on overlap makes $\pi$ orbitals with a nodal plane through the bond axis. The famous payoff: $\text{O}_2$ must place its last two electrons singly into two equal-energy $\pi^*$ orbitals (Hund's rule), leaving two unpaired spins — so oxygen is paramagnetic, a fact MO theory predicts and the older dot-structure picture cannot. In the sim, the molecule menu sets the electron count and the $R$ slider sets how far apart the nuclei sit, which controls the overlap $S$.

Try this in the sim above Pick He₂ from the molecule menu and watch the Bond Order readout fall to 0 as the antibonding pair cancels the bonding pair — then switch to H₂ and see it jump back to 1. Next choose O₂ and open the Magnetism mode to see the two unpaired $\pi^*$ electrons that make it stick to a magnet. Finally drag the Internuclear Distance R slider toward 4 Å and watch the Overlap S graph collapse toward zero — pull the atoms far apart and the shared cloud, and the bond, disappears.

§3 Equation Derivation

LCAO-MO Theory & Bond Order
\[\psi_{\pm} = \frac{1}{\sqrt{2(1\pm S)}}(\phi_A \pm \phi_B)\] \[\text{Bond Order} = \frac{n_b - n_a}{2}\]
SymbolMeaningUnit
\(\psi_+\)Bonding MO (σ or π)—
\(\psi_-\)Antibonding MO (σ* or π*)—
\(\phi_A, \phi_B\)Atomic orbitals on atoms A and B—
\(S\)Overlap integral = ∫φ_A φ_B dVdimensionless
\(n_b\)Electrons in bonding MOscount
\(n_a\)Electrons in antibonding MOscount
Step-by-Step Derivation
Step 1 — LCAO Ansatz MO = linear combination of AOs: \(\psi = c_A\phi_A + c_B\phi_B\). For homonuclear diatomics, symmetry requires |cA| = |cB|, giving bonding (+ combination) and antibonding (− combination).
Step 2 — Normalization with overlap S \(\int|\psi_+|^2 dV = 1 \Rightarrow c = \frac{1}{\sqrt{2(1+S)}}\) for bonding, \(\frac{1}{\sqrt{2(1-S)}}\) for antibonding. S > 0 means bonding MO is lower energy than either AO.
Step 3 — Energy of bonding MO \(E_+ = \frac{\alpha + \beta}{1+S}\) where α = Coulomb integral (AO energy), β = resonance integral (< 0). Bonding MO energy < α (stabilized). Antibonding: \(E_- = \frac{\alpha - \beta}{1-S}\) > α (destabilized).
Step 4 — MO configurations and bond order Fill MOs from lowest energy upward (Aufbau for MOs). Bond order = (bonding − antibonding)/2.
H₂: (σ1s)²: BO = 1. He₂: (σ1s)²(σ*1s)²: BO = 0 (doesn't exist). N₂: BO = 3 (triple bond).
Step 5 — O₂ paramagnetism O₂ MO config: (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)²
The two π*2p electrons go into two degenerate orbitals with parallel spins (Hund's rule) → 2 unpaired electrons → O₂ is paramagnetic! This was a triumph of MO theory (VB theory predicted wrongly diamagnetic O₂).
Step 6 — σ vs π bonding σ bonds: AOs overlap head-on (s+s, s+pz, pz+pz). Cylindrically symmetric around bond axis. π bonds: AOs overlap side-on (px+px, py+py). Have one nodal plane containing the bond axis. π* is higher energy than σ* for the same principal quantum number.
Worked Example — N₂ Bond Order

N₂ MO filling: σ1s² σ*1s² σ2s² σ*2s² π2p⁴ σ2p² — total 14 electrons

n_b = 2+2+4+2 = 10 (counting σ1s,σ2s,π2p,σ2p). n_a = 2+2 = 4 (σ*1s,σ*2s)

Bond Order = (10−4)/2 = 3 ✓ (consistent with N≡N triple bond, shortest diatomic, 945 kJ/mol bond energy)

Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., §10D "Molecular orbital theory" | Housecroft & Sharpe — Inorganic Chemistry, 5th Ed., §2.3

§4 Frequently Asked Questions

🔬 SimulationWhat is each mode showing?▼
The MO Diagram shows the energy level diagram with AOs on the sides and MOs in the center, with electrons filled by Aufbau. σ Orbital renders the bonding σ1s (egg-shaped density between nuclei) and the σ*1s (density on outsides, node between). π Orbital shows the π2p bonding (two lobes above/below) and π*2p (four lobes with two nodal planes). Bond Formation animates two H atoms approaching and merging into an H₂ molecule. Magnetism shows which molecules are paramagnetic (unpaired e⁻) and why. Key: MO theory is superior to VB because it correctly predicts O₂ is paramagnetic — something Lewis structures and hybridization cannot do.
🌍 Real LifeWhere does MO theory matter in real life?▼
The paramagnetism of O₂ is exploited in liquid oxygen (used as rocket fuel oxidizer) — liquid O₂ is attracted to magnets. MRI contrast agents work because unpaired d-electrons (predicted by MO/crystal field theory) create magnetic moments. Organic semiconductors in OLED displays (phones, TVs) are designed by tuning HOMO-LUMO gaps using MO theory. Transition metal catalysts for industrial hydrogen production and pharmaceutical synthesis are optimized using MO-based computational chemistry. Even the green/blue colors of Cu(II) complexes arise from MO transitions. Key: MO theory underlies rational drug design, materials science, and catalysis — the HOMO-LUMO gap is one of the most important chemical concepts.
💡 Non-ObviousWhy does He₂ not exist but He₂⁺ does?▼
He₂: (σ1s)²(σ*1s)² → BO = (2−2)/2 = 0 → no bond, doesn't exist under normal conditions. He₂⁺ has one fewer electron: (σ1s)²(σ*1s)¹ → BO = (2−1)/2 = 0.5 → weakly bonded, detected spectroscopically. The lesson: even a fractional bond order means stabilization relative to separate atoms. H₂⁺ (BO = 0.5) has a bond energy of 256 kJ/mol and bond length 1.06 Å — shorter than most single bonds. Half bonds are real and detectable. Key: Bond order 0.5 is still a real, measurable bond — MO theory quantifies bonding beyond the integer bonds of Lewis structures.
🧮 MathematicalHow do you calculate bond order for O₂?▼
O₂ has 16 electrons. Fill MOs: σ1s² σ*1s² σ2s² σ*2s² σ2p² π2p⁴ π*2p². Bonding electrons: σ1s(2) + σ2s(2) + σ2p(2) + π2p(4) = 10. Antibonding electrons: σ*1s(2) + σ*2s(2) + π*2p(2) = 6. Bond order = (10−6)/2 = 2 ✓ (consistent with O=O double bond). The two π*2p electrons occupy the two degenerate π* orbitals singly (Hund's rule) → 2 unpaired electrons → paramagnetic. Key: O₂ bond order = 2 from MO theory — this matches the experimental bond energy (498 kJ/mol) and length (121 pm).
🧪 ConceptualWhy is the MO energy ordering different for N₂ vs O₂?▼
For molecules up to N₂, the σ2p orbital is higher in energy than the π2p orbitals. From O₂ onwards, σ2p is lower than π2p. This reversal occurs because for lighter elements (Li₂ to N₂), the 2s and 2p orbitals are closer in energy — their mixing (s-p mixing) pushes σ2p above π2p. For O, F: the energy gap between 2s and 2p is larger, so mixing is less significant and σ2p drops below π2p. This affects the MO filling order and explains why N₂ has a σ2p filled last while O₂ has π* filled last. Key: s-p mixing causes the MO energy reversal between N₂ and O₂ — it's not arbitrary but has a clear quantum mechanical cause.
🌌 DeepWhat is the connection between MO theory and aromaticity?▼
Benzene's 6 π electrons fill three bonding π MOs formed by combining the 6 carbon p orbitals. The lowest MO has no nodes (like 1s-type); the next two are degenerate with one node; all three are fully occupied → no net angular momentum, extra stability = aromaticity. Hückel's rule (4n+2 π electrons for aromaticity) follows directly from the MO energy pattern of cyclic conjugated systems. This MO-based understanding of aromaticity explains why benzene is unusually stable, does substitution rather than addition, and has unique ring current NMR shifts. Key: MO theory explains aromaticity quantitatively — Hückel's rule is a direct consequence of π MO filling patterns in cyclic conjugated molecules.

Reference: LibreTexts Chemistry — Molecular Orbital Theory https://chem.libretexts.org/Bookshelves/General_Chemistry/Map:_Chemistry_-_The_Central_Science/9:_Molecular_Geometry_and_Bonding_Theories | MIT OCW 5.111 Lectures 14-16

§5 Common Misconceptions

❌ Misconception: "Antibonding orbitals are just empty orbitals that don't contribute to bonding."
✅ Correction: Antibonding orbitals actively destabilize the molecule when occupied — each electron in an antibonding orbital contributes −½ to the bond order and raises the total energy by more than a bonding orbital lowers it (because |β/(1−S)| > |β/(1+S)|). This is why He₂ (BO=0) doesn't form: the antibonding pair exactly cancels the bonding pair. Empty antibonding orbitals (LUMO) are also crucial as the site of nucleophilic attack in reactions and as electron acceptors in coordination chemistry.
📖 Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., §10D.2: "The hydrogen molecule-ion" — antibonding destabilization quantified
❌ Misconception: "O₂ has a double bond, so it must be diamagnetic (like most organic double bonds)."
✅ Correction: This was actually believed before MO theory was developed — Linus Pauling's early VB treatment predicted diamagnetic O₂. But liquid O₂ is visibly attracted to a magnetic field (paramagnetic), which VB could not explain. MO theory predicts two unpaired electrons in degenerate π*2p orbitals — exactly consistent with the observed paramagnetism. This was a landmark validation of MO theory over simple VB theory for understanding electronic structure.
📖 Reference: Housecroft & Sharpe — Inorganic Chemistry, 5th Ed., §2.3.4: "Dioxygen, O₂" — historical VB vs MO comparison
❌ Misconception: "Bonding MOs always have electron density between the nuclei, and antibonding MOs always have density away from the nucleus."
✅ Correction: Bonding MOs have increased electron density between nuclei (constructive interference), which is correct. Antibonding MOs have a nodal plane between the nuclei and electron density concentrated on the outer sides — but they still have significant electron density around both nuclei. The key distinction is the node between atoms (absent in bonding, present in antibonding) and whether the electron density stabilizes (bonds) or destabilizes (antibonds) the internuclear region.
📖 Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., Figure 10D.3: "Electron density in σ and σ* MOs"
❌ Misconception: "LCAO-MO just means combining orbitals — you can combine any atomic orbitals to get MOs."
✅ Correction: Only atomic orbitals of similar energy and correct symmetry can combine to form MOs. Two AOs can overlap only if: (1) they have similar energy (within ~10 eV), (2) they have the correct symmetry match (e.g., s+s, pz+pz for σ; px+px for π), and (3) their overlap integral S is nonzero. For example, an s orbital on one atom cannot form a π bond with a p orbital on another atom (wrong symmetry — integral = 0 by symmetry). These selection rules are essential for understanding which orbital combinations are meaningful.
📖 Reference: Atkins & de Paula — Physical Chemistry, 11th Ed., §10D.1: "Molecular orbital theory: the hydrogen molecule-ion"
❌ Misconception: "N₂ has a triple bond because it has three lone pairs and three shared pairs — same as from Lewis structure."
✅ Correction: While Lewis structures correctly count a triple bond in N₂, they cannot explain why N₂ is so unreactive (bond energy 945 kJ/mol) or predict its diamagnetism. MO theory provides the deeper explanation: the three bonds (one σ2p + two π2p) arise from three bonding MOs with no corresponding occupied antibonding MOs of the same type. The MO picture also explains why N₂ is a better ligand than CO despite similar structure, through backbonding into π* orbitals.
📖 Reference: Housecroft & Sharpe — Inorganic Chemistry, 5th Ed., §2.3.3: "Dinitrogen, N₂"

Section 4 reference: Taber, K.S. — Chemical Misconceptions (RSC, 2002) | Galloway, K.R. et al. — Students' reasoning about molecular orbital diagrams. Chem. Educ. Res. Pract. 2017, 18, 353