Chemical Bonding II: Valence Bond & Molecular Orbital Theory

Why Lewis Structures and VSEPR Are Not Enough

Lewis structures and VSEPR theory are excellent for predicting overall molecular shape and estimating polarity, but they treat bonds as simple dots and lines and cannot explain finer details of bonding behavior. They cannot account for why some bonds involve a cylindrically symmetric overlap while others involve sideways overlap, why rotation is hindered around double bonds but free around single bonds, or why a molecule like O₂ is experimentally paramagnetic despite Lewis theory pairing all its electrons. Two more sophisticated quantum-mechanical models — valence bond (VB) theory and molecular orbital (MO) theory — fill these gaps by describing bonding in terms of atomic and molecular orbitals.

• •Lewis/VSEPR: fast, good for overall shape and polarity, but no orbital detail.
• •Valence bond theory: retains atomic orbital character, explains hybridization and σ/π bonds.
• •Molecular orbital theory: builds new orbitals delocalized over the whole molecule, explains magnetism and some bond orders Lewis theory gets wrong.

Valence Bond Theory and Orbital Overlap

Valence bond theory describes a covalent bond as forming when atomic orbitals on two atoms overlap, allowing a pair of electrons with opposite spins to occupy the shared region between the nuclei. The degree of overlap correlates with bond strength: greater orbital overlap generally produces a stronger, shorter bond. Unlike molecular orbital theory, VB theory keeps the electrons largely localized between specific pairs of atoms, which matches the localized bonds drawn in Lewis structures and makes VB theory intuitive for predicting geometry through hybridization.

• •A covalent bond forms only when orbital overlap places significant electron density between two nuclei.
• •The two shared electrons in a bond must have opposite (paired) spins, consistent with the Pauli exclusion principle.
• •Greater overlap → stronger bond → generally shorter bond length.

Sigma and Pi Bonds

A sigma (σ) bond results from head-on (end-to-end) orbital overlap along the internuclear axis, producing electron density that is cylindrically symmetric around that axis; every single bond is a σ bond, and it is always the first bond formed between any two atoms. A pi (π) bond results from sideways (parallel) overlap of unhybridized p orbitals above and below (or in front of and behind) the internuclear axis, producing two lobes of electron density, one above and one below the bond axis, with a node along the axis itself. Because π overlap is less effective than head-on σ overlap, π bonds are generally weaker than σ bonds, and a molecule can never have a π bond without first having a σ bond between the same two atoms.

• •Single bond = 1 σ bond only.
• •Double bond = 1 σ bond + 1 π bond.
• •Triple bond = 1 σ bond + 2 π bonds (using two mutually perpendicular unhybridized p orbitals).
• •π bonds restrict rotation: rotating around a π bond would break the sideways overlap, so double and triple bonds are rigid, while single (σ-only) bonds rotate freely.

Hybridization of Atomic Orbitals

Hybridization is a mathematical mixing of an atom's standard atomic orbitals (s, p, and sometimes d) into a new set of degenerate hybrid orbitals oriented to match the observed molecular geometry predicted by VSEPR. The type of hybridization needed on a central atom is determined directly by counting its total electron domains (the same count used for VSEPR), since each domain requires one hybrid orbital pointed toward a bonded atom or a lone pair. Any p orbitals not used in hybridization remain as pure, unhybridized p orbitals available to form π bonds.

• •2 domains → sp hybridization → linear arrangement, 180°.
• •3 domains → sp² hybridization → trigonal planar arrangement, 120°; one unhybridized p orbital remains for π bonding.
• •4 domains → sp³ hybridization → tetrahedral arrangement, 109.5°; no p orbitals remain for π bonding.
• •5 domains → sp³d hybridization → trigonal bipyramidal arrangement (requires an available d orbital, period 3+).
• •6 domains → sp³d² hybridization → octahedral arrangement (requires an available d orbital, period 3+).
• •The number of hybrid orbitals produced always equals the number of atomic orbitals mixed (conservation of orbitals).

Multiple Bonds in Valence Bond Terms

Multiple bonds are best understood as a combination of one σ bond, formed from hybrid orbital overlap, plus one or more π bonds, formed from leftover unhybridized p orbitals. In ethylene (C₂H₄), each carbon is sp² hybridized, using its three sp² orbitals for three σ bonds (two to H, one to the other C), while the remaining unhybridized p orbital on each carbon overlaps sideways to form the π bond of the C=C double bond. In acetylene (C₂H₂), each carbon is sp hybridized, using its two sp orbitals for two σ bonds (to H and to the other C), while two unhybridized, mutually perpendicular p orbitals on each carbon form two π bonds, together making the C≡C triple bond.

• •Ethylene C₂H₄: each C is sp², C=C bond is 1σ + 1π, molecule is planar, no rotation about C=C.
• •Acetylene C₂H₂: each C is sp, C≡C bond is 1σ + 2π, molecule is linear.
• •Restricted rotation about π bonds is the structural basis for cis/trans (E/Z) isomerism.

Molecular Orbital Theory: LCAO

Molecular orbital (MO) theory takes a fundamentally different approach from VB theory by treating electrons as delocalized over the entire molecule rather than localized between two atoms. Atomic orbitals on different atoms are mathematically combined via the linear combination of atomic orbitals (LCAO) method to generate an equal number of new molecular orbitals. When two atomic orbitals combine in phase (constructive interference), they produce a lower-energy bonding molecular orbital with increased electron density between the nuclei; when they combine out of phase (destructive interference), they produce a higher-energy antibonding molecular orbital (marked with an asterisk, e.g., σ*) with a node of zero electron density between the nuclei.

• •Number of MOs formed = number of atomic orbitals combined (orbitals are conserved).
• •Bonding MOs are lower in energy than the original atomic orbitals and stabilize the molecule.
• •Antibonding MOs (σ*, π*) are higher in energy and destabilize the molecule if occupied.
• •σ MOs form from head-on overlap (of s or end-on p orbitals); π MOs form from sideways overlap of p orbitals, each producing both a bonding and an antibonding version.

MO Diagrams for Period-2 Homonuclear Diatomics

For period-2 diatomic molecules, the 2s orbitals combine to form σ2s and σ2s* MOs, and the 2p orbitals combine to form σ2p, two degenerate π2p, two degenerate π2p*, and σ2p* MOs, filled according to the Aufbau principle, Pauli exclusion, and Hund's rule. For B₂, C₂, and N₂, s–p mixing (interaction between the σ2s and σ2p orbitals of similar energy) pushes the σ2p orbital above the π2p orbitals in energy, so the correct filling order is σ2s, σ2s*, π2p (×2, degenerate), σ2p, π2p* (×2), σ2p*. For O₂, F₂, and Ne₂, the 2s–2p energy gap is large enough that s–p mixing is negligible, restoring the more 'expected' order: σ2s, σ2s*, σ2p, π2p (×2), π2p* (×2), σ2p*.

• •Memory aid: 'Before O, p before s-crossover' — B₂, C₂, N₂ have π2p filled before σ2p.
• •N₂ (σ2p filled last, no unpaired electrons in that filling order) is diamagnetic with bond order 3, consistent with its Lewis triple bond.
• •O₂, F₂, Ne₂ follow σ2p before π2p, matching simpler 'textbook' MO ordering.
• •This reordering only matters for period-2 homonuclear diatomics; it is a required, specific piece of exam knowledge.

Bond Order, Stability, and the Paramagnetism of O₂

Bond order in MO theory is calculated from the number of electrons in bonding versus antibonding orbitals, and it correlates directly with bond length (higher bond order, shorter bond) and bond energy (higher bond order, stronger bond); a bond order of zero means the molecule does not form at all. Molecular orbital theory's most celebrated success is correctly predicting that O₂ is paramagnetic (attracted into a magnetic field due to unpaired electrons): filling the two degenerate π2p* orbitals with one electron each (per Hund's rule) leaves two unpaired electrons, exactly matching experimental observation, something Lewis theory (which pairs all valence electrons in O=O) fails to predict.

• •Bond order = (electrons in bonding MOs − electrons in antibonding MOs) / 2.
• •Higher bond order → shorter bond length and higher bond dissociation energy.
• •A species is paramagnetic if it has any unpaired electrons in its MO diagram; diamagnetic if all electrons are paired.
• •O₂ has bond order 2 with 2 unpaired π2p* electrons — paramagnetic, matching experiment; this is MO theory's classic triumph over simple Lewis theory.

MO Treatment of Diatomic Ions and He Species

Removing or adding electrons to a neutral diatomic changes its MO electron configuration and therefore its bond order, bond length, and stability, and MO diagrams predict these changes reliably even for species with no simple Lewis structure. Removing an electron from a bonding MO (as in forming N₂⁺ from N₂, or O₂⁺ from O₂) lowers bond order and weakens/lengthens the bond, while removing an electron from an antibonding MO (also possible depending on which orbital is highest occupied) can actually raise bond order. He₂ has equal electrons in σ1s and σ1s*, giving a bond order of zero, which is why He₂ does not exist as a stable molecule; He₂⁺, however, has one fewer antibonding electron, giving a bond order of ½ and a weak but real bond, detected spectroscopically.

• •N₂ (bond order 3) → N₂⁺ (bond order 2.5): removing a bonding σ2p electron weakens the bond.
• •O₂ (bond order 2) → O₂⁺ (bond order 2.5): removing an antibonding π2p* electron strengthens the bond and shortens it relative to O₂.
• •O₂ (bond order 2) → O₂⁻ (bond order 1.5): adding an antibonding electron weakens/lengthens the bond.
• •He₂: bond order = (2−2)/2 = 0, no stable molecule; He₂⁺: bond order = (2−1)/2 = 0.5, a real, weakly bound ion.

MO Electron-Count Reference Table

This table collects the valence electron counts, resulting bond orders, and magnetic behavior for the key homonuclear diatomic species and ions covered on the exam, which is the fastest way to review before test day.

• •B₂: 6 valence e⁻, bond order 1, paramagnetic (2 unpaired e⁻ in degenerate π2p).
• •C₂: 8 valence e⁻, bond order 2, diamagnetic.
• •N₂: 10 valence e⁻, bond order 3, diamagnetic — very strong, short bond.
• •N₂⁺: 9 valence e⁻, bond order 2.5, paramagnetic.
• •O₂: 12 valence e⁻, bond order 2, paramagnetic (2 unpaired e⁻ in π2p*).
• •O₂⁺: 11 valence e⁻, bond order 2.5, paramagnetic.
• •O₂⁻: 13 valence e⁻, bond order 1.5, paramagnetic.
• •F₂: 14 valence e⁻, bond order 1, diamagnetic.
• •Ne₂: 16 valence e⁻, bond order 0, does not exist as a stable molecule.
• •He₂: 4 electrons, bond order 0, does not exist; He₂⁺: 3 electrons, bond order 0.5, exists weakly.

Comparing the Three Bonding Models

Each bonding model trades off simplicity for explanatory power, and knowing which to reach for is itself an exam skill. Lewis structures with VSEPR are fastest for predicting overall shape, polarity, and formal charge, and should be the default first step for any species. Valence bond theory (hybridization, σ/π bonds) is the right tool when a question asks about hybridization, orbital overlap, or rotation around a bond. Molecular orbital theory is required whenever a question asks about magnetism, bond order that doesn't match a simple Lewis structure, or the stability/existence of diatomic ions and unusual species like He₂ or O₂⁺.

• •Use Lewis/VSEPR for: shape, bond angles, polarity, resonance, formal charge.
• •Use valence bond theory for: hybridization, σ vs. π character, rotational rigidity of double/triple bonds.
• •Use MO theory for: magnetism, bond order of ions/odd-electron species, relative stability of related diatomics.

Determine the hybridization of the central atom and the σ/π composition of all bonds in CO₂.

1. 1Draw the Lewis structure: O=C=O, with carbon as the central atom bonded to two oxygens by double bonds and no lone pairs on C.
2. 2Count electron domains on carbon: 2 double-bonded groups = 2 electron domains (each multiple bond counts once).
3. 32 domains corresponds to sp hybridization on carbon, consistent with CO₂'s linear (180°) geometry.
4. 4Each C=O double bond consists of 1 σ bond, formed from overlap of a carbon sp orbital with an oxygen orbital, plus 1 π bond, formed from sideways overlap of an unhybridized carbon p orbital with an oxygen p orbital.
5. 5Since carbon has two unhybridized p orbitals (mutually perpendicular) remaining after sp hybridization uses only one s and one p orbital, each is used to form one π bond, one to each oxygen.
6. 6Total bonding: 2 σ bonds + 2 π bonds around carbon, matching the two C=O double bonds.

Carbon in CO₂ is sp hybridized (linear); each C=O bond is composed of 1 σ + 1 π bond, giving 2 σ and 2 π bonds total.

Construct the MO electron configuration for N₂ and calculate its bond order and magnetism.

1. 1N₂ has 2 × 5 = 10 valence electrons to place in period-2 MOs.
2. 2Since N₂ is B₂/C₂/N₂-type (s–p mixing applies), use filling order: σ2s, σ2s*, π2p(×2), σ2p, π2p*(×2), σ2p*.
3. 3Fill: σ2s² σ2s*² π2p⁴ (2 electrons in each of the 2 degenerate π2p orbitals) σ2p² — total electrons placed = 2+2+4+2 = 10 ✓.
4. 4Bonding electrons: σ2s(2) + π2p(4) + σ2p(2) = 8. Antibonding electrons: σ2s*(2) = 2.
5. 5Bond order = (8 − 2)/2 = 3.
6. 6All electrons are paired (no unpaired electrons in any orbital) → N₂ is diamagnetic.

N₂: bond order = 3, diamagnetic — consistent with its very strong, short N≡N triple bond.

Construct the MO electron configuration for O₂ and explain its magnetic behavior.

1. 1O₂ has 2 × 6 = 12 valence electrons.
2. 2O₂ follows the O₂/F₂/Ne₂ filling order (no significant s–p mixing): σ2s, σ2s*, σ2p, π2p(×2), π2p*(×2), σ2p*.
3. 3Fill: σ2s² σ2s*² σ2p² π2p⁴ π2p*² — total placed = 2+2+2+4+2 = 12 ✓.
4. 4The final 2 electrons go into the two degenerate π2p* orbitals; by Hund's rule, they occupy separate orbitals with parallel spins, leaving 2 unpaired electrons.
5. 5Bonding electrons: σ2s(2)+σ2p(2)+π2p(4) = 8. Antibonding electrons: σ2s*(2)+π2p*(2) = 4.
6. 6Bond order = (8−4)/2 = 2.
7. 7Because there are 2 unpaired electrons, O₂ is paramagnetic — confirmed experimentally by liquid O₂ being attracted to a magnet, something simple Lewis theory cannot explain.

O₂: bond order = 2, paramagnetic due to 2 unpaired electrons in the degenerate π2p* orbitals.

Predict how the bond order and bond length change upon forming O₂⁺ from O₂.

1. 1Start from O₂'s MO configuration: bonding electrons = 8, antibonding electrons = 4, bond order = 2.
2. 2Forming O₂⁺ removes one electron from the highest-occupied MO, which for O₂ is the antibonding π2p* set.
3. 3New antibonding electron count = 4 − 1 = 3; bonding electrons remain 8.
4. 4New bond order = (8 − 3)/2 = 2.5.
5. 5Since bond order increased from 2 to 2.5, the O–O bond in O₂⁺ is shorter and stronger than in neutral O₂.
6. 6O₂⁺ retains one unpaired electron in the π2p* set, so it remains paramagnetic.

O₂⁺ has bond order 2.5 (up from 2 in O₂), giving a shorter, stronger O–O bond; it remains paramagnetic.

Explain why He₂ does not exist as a stable molecule but He₂⁺ does, using MO theory.

1. 1Each He atom contributes 2 electrons (1s²), so He₂ has 4 total electrons to place in σ1s and σ1s* MOs.
2. 2Fill He₂: σ1s² σ1s*² — 2 bonding and 2 antibonding electrons.
3. 3Bond order for He₂ = (2 − 2)/2 = 0, meaning there is no net stabilization from bonding — He₂ does not form as a stable molecule.
4. 4He₂⁺ has one fewer electron (3 total): σ1s² σ1s*¹.
5. 5Bond order for He₂⁺ = (2 − 1)/2 = 0.5, a nonzero, net-stabilizing bond order.
6. 6A bond order greater than zero means a real, if weak, bond exists, so He₂⁺ is experimentally observed while He₂ is not.

He₂: bond order 0, does not exist. He₂⁺: bond order 0.5, a real (weakly bound) diatomic ion.

Q1. Which bonding model correctly predicts that O₂ is paramagnetic?

Q2. A central atom with 4 total electron domains is best described by which hybridization?

Q3. A triple bond between two carbon atoms consists of which combination of bonds?

Q4. Why is rotation about a C=C double bond restricted compared to a C–C single bond?

Q5. For which set of homonuclear diatomics does s–p mixing raise σ2p above π2p in energy?

Q6. What is the bond order of the diatomic ion N₂⁺, formed by removing one electron from N₂'s highest occupied MO (σ2p)?

Q7. Which statement about He₂ is correct according to molecular orbital theory?

Q8. Ethylene, C₂H₄, has each carbon hybridized as sp². What happens to the leftover unhybridized p orbital on each carbon?

Q9. Which molecule/atom arrangement requires sp³d² hybridization on the central atom?

Q10. A σ bond is characterized by which of the following?

Q11. Compared to O₂, how does the bond order of O₂⁻ change, and why?

Q12. Which model should be used to determine whether a molecule has hybridized orbitals and how many σ vs π bonds it contains?

Q13. What is the correct MO filling order for F₂?

Q14. How many total molecular orbitals form when two atoms each contribute one 2s and three 2p atomic orbitals (8 atomic orbitals total)?

Q15. Acetylene, C₂H₂, has linear geometry with each carbon sp hybridized. How many π bonds are present in the C≡C triple bond?