Hybridization and Bonding: Advanced Bonding Models

Beyond Basic Hybridisation

While orbital hybridisation explains geometry, this guide explores how hybridisation determines bond strength, bond length, delocalisation, and the fascinating chemistry of aromatic systems.

Learning Goals: By the end of this guide, you should be able to:

1. Relate hybridisation to bond length and strength.
2. Explain delocalisation and conjugation using orbital models.
3. Describe bonding in benzene and its consequences.
4. Predict molecular properties from bonding analysis.

Bond Length and Bond Strength

As hybridisation changes from $sp^3 → sp^2 → sp$, the s-character increases:

Why? $s$ orbitals are closer to the nucleus than $p$ orbitals. Higher s-character means the hybrid orbital is held closer→ shorter, stronger bonds.

Data: C–H Bond Lengths

Delocalisation

Delocalisation occurs when $\pi$ electrons are not confined between two atoms but are spread over three or more atoms. This happens when multiple $sp^2$ atoms are adjacent, allowing their $p$ orbitals to overlap continuously.

Conditions for Delocalisation

1. Adjacent atoms must all be $sp^2$ (or $sp$).
2. Their unhybridised $p$ orbitals must be parallel (same plane).
3. No $sp^3$ atom breaks the chain.

Consequences

• Stabilisation: Delocalised systems are lower in energy (more stable).
• Equal bond lengths: In benzene, all C–C bonds are the same length (1.39 Å) — intermediate between single (1.54 Å) and double (1.34 Å).
• Planarity: Delocalisation requires all involved atoms to be coplanar.

Benzene: The Perfect Example

The Kekulé Model (Incomplete)

Kekulé proposed that benzene alternates single and double bonds. This predicts:

• Alternating short/long C–C bonds → Wrong (all are 1.39 Å)
• Typical alkene reactivity → Wrong (benzene resists addition, favours substitution)

The Delocalised Model (Correct)

Each carbon in benzene is $sp^2$, with one unhybridised $p$ orbital. The six $p$ orbitals overlap to form a continuous ring of electron density above and below the plane — a delocalised $\pi$ system.

Conjugation

A conjugated system has alternating single and double bonds (or a lone pair adjacent to a $\pi$ bond), allowing continuous $p$ orbital overlap.

Examples:

• Buta-1,3-diene: $CH_2=CH-CH=CH_2$ (4 conjugated carbons)
• Carbonyl with adjacent double bond: $C=C-C=O$
• Lone pair conjugation: $-NH_2$ next to $C=O$ (amide bond)

Effects of Conjugation

• Shorter single bonds: The "single" bond in a conjugated system gains partial $\pi$ character.
• UV absorption shifts: Extended conjugation absorbs longer wavelengths → this is why many dyes and pigments are conjugated molecules.
• Increased stability: Delocalisation energy lowers the overall energy of the system.

Worked Examples

Example 1: Bond Length Comparison

Question: Which C–C bond is shorter: ethane or ethene?

Solution: Ethane has $sp^3$ C–C (1.54 Å). Ethene has $sp^2$ C=C (1.34 Å). The double bond in ethene is shorter because: (a) it has a $\pi$ bond providing extra attraction, and (b) $sp^2$ hybrid orbitals have more s-character, bringing the carbons closer.

Example 2: Explaining Benzene Stability

Question: The enthalpy of hydrogenation of cyclohexene is –120 kJ/mol. Predict and compare with benzene.

Solution: If benzene were "cyclohexatriene" (3 double bonds): expected = $3 \times (-120) = -360$ kJ/mol. Actual: –208 kJ/mol. The difference (152 kJ/mol) is the resonance stabilisation energy — proof that delocalisation stabilises benzene.

Common Mistakes

1. Drawing benzene with alternating single/double bonds — While Kekulé structures are accepted for mechanism diagrams, always state that the bonds are actually equal due to delocalisation.

2. Confusing conjugation with aromaticity — All aromatic compounds are conjugated, but not all conjugated systems are aromatic. Aromaticity requires a planar, cyclic, fully conjugated system with $4n+2$ $\pi$ electrons (Hückel's rule).

3. Forgetting lone pairs can conjugate — A nitrogen lone pair adjacent to a $C=O$ can delocalise into the $\pi$ system, as in amides.

Exam Tips

• For benzene questions: mention equal bond lengths, resistance to addition, and delocalisation energy as evidence.
• When comparing bond lengths: more s-character = shorter, and $\pi$ bonding = shorter.
• Know Hückel's rule for aromaticity: $4n+2$ $\pi$ electrons (n = 0, 1, 2...), so 2, 6, 10, 14 electrons.

Related Topics

• Orbital Hybridisation — The foundational theory behind $sp$, $sp^2$, $sp^3$.
• Molecular Coplanarity — Why delocalised systems must be planar.
• VSEPR Theory — Geometry from electron pair repulsion.