Orbital Hybridisation: sp, sp², and sp³ Explained

Why Do We Need Hybridisation?

VSEPR theory tells us that methane ($CH_4$) is tetrahedral with $109.5°$ bond angles. But carbon's electron configuration is $1s^2\ 2s^2\ 2p^2$ — it has one filled $2s$ orbital and two half-filled $2p$ orbitals. That should give at most 2 bonds at $90°$. How does carbon form 4 equivalent bonds at $109.5°$?

The answer is hybridisation: atomic orbitals mix to form new, equivalent hybrid orbitals that match the observed geometry.

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

1. Explain why hybridisation is necessary.
2. Describe the three common hybridisation types ($sp^3$, $sp^2$, $sp$).
3. Link hybridisation to molecular geometry and bond angles.
4. Distinguish between sigma ($\sigma$) and pi ($\pi$) bonds.
5. Determine the hybridisation of any central atom.

The Three Hybridisation Types

$sp^3$ Hybridisation

Mixing: 1 $s$ orbital + 3 $p$ orbitals → 4 equivalent $sp^3$ hybrid orbitals

Each $sp^3$ orbital holds one bonding pair or one lone pair. In methane, all 4 are bonding; in water, 2 are bonding and 2 are lone pairs.

$sp^2$ Hybridisation

Mixing: 1 $s$ orbital + 2 $p$ orbitals → 3 equivalent $sp^2$ hybrid orbitals. The remaining unhybridised $p$ orbital is perpendicular to the plane and forms a $\pi$ bond.

$sp$ Hybridisation

Mixing: 1 $s$ orbital + 1 $p$ orbital → 2 equivalent $sp$ hybrid orbitals. Two unhybridised $p$ orbitals remain for $\pi$ bonds.

Summary Comparison

Sigma (σ) and Pi (π) Bonds

Understanding the difference is essential:

Counting σ and π Bonds

This is a common exam question:

Rule: Every bond has exactly one σ bond. The rest are π bonds.

How to Determine Hybridisation

Quick Method: Count Electron Domains

The number of electron domains (bonding pairs + lone pairs) on the central atom equals the number of hybrid orbitals needed:

Remember: Double and triple bonds each count as one domain for hybridisation purposes.

Worked Examples

Example 1: Hybridisation of Carbon in Ethene ($C_2H_4$)

Each carbon forms: 2 C–H bonds + 1 C=C bond = 3 bonding domains (the double bond is one domain).

3 domains → $sp^2$ hybridisation → trigonal planar geometry → $120°$ angles.

The remaining unhybridised $p$ orbital on each carbon overlaps side-on to form the π bond in the double bond.

Example 2: Hybridisation of Nitrogen in Ammonia ($NH_3$)

Nitrogen has: 3 N–H bonds + 1 lone pair = 4 electron domains.

4 domains → $sp^3$ hybridisation → tetrahedral electron geometry → trigonal pyramidal molecular shape → $\approx 107°$ angles (compressed by the lone pair).

Example 3: How Many σ and π Bonds in $CH_3CHCHCH_2$?

Count systematically:

• C–H single bonds: each has 1 σ → there are 6 C–H bonds → 6 σ
• C–C single bonds: 2 σ
• C=C double bond: 1 σ + 1 π → 1 σ and 1 π

Total: 9 σ bonds and 1 π bond.

Common Mistakes

1. Confusing hybridisation with VSEPR — VSEPR predicts shape from electron pair repulsion. Hybridisation explains why orbitals have those orientations at the quantum level. They give the same answer but approach it from different directions.

2. Forgetting lone pairs count as domains — In water ($H_2O$), oxygen has 4 electron domains (2 bonding + 2 lone pairs), so it's $sp^3$, not $sp^2$.

3. Saying "π bonds are weaker than σ" — This is generally true, but the statement needs context. A C=C double bond (1 σ + 1 π) is stronger overall than a C–C single bond (1 σ). It's the individual π bond that's weaker than the individual σ bond.

4. Treating every bond type separately — A double bond counts as one domain for both VSEPR and hybridisation. Don't count the σ and π separately.

5. Mixing up "hybridisation of orbitals" and "hybridisation of atoms" — It's the atom that is hybridised. Orbitals are the things being mixed. Say "carbon is $sp^2$ hybridised," not "$p$ orbital is hybridised."

Exam Tips (A-Level / AP / IB)

• Quick determination: Count bonding groups + lone pairs on the central atom. 2 = $sp$, 3 = $sp^2$, 4 = $sp^3$.
• For "describe the bonding" questions, always state: the type of hybridisation, the geometry, and which bonds are σ and which are π.
• When asked about restricted rotation (e.g., in alkenes), explain it via the π bond: "The side-on overlap of $p$ orbitals above and below the bond axis prevents free rotation."
• Know that benzene has all $sp^2$ carbons with delocalised π electrons forming a ring above and below the plane.

Frequently Asked Questions

Why do orbitals hybridise?

Hybridisation occurs because the resulting hybrid orbitals are lower in energy than the separated atomic orbitals when they form bonds. The energy released from forming stronger, more directional bonds drives the mixing process.

Is hybridisation a real physical process?

Hybridisation is a mathematical model that explains observed molecular geometries. Electrons don't literally "mix" — rather, the hybrid orbital model accurately predicts the shapes and energies we observe experimentally.

Can atoms other than carbon hybridise?

Yes. Any atom that forms covalent bonds can hybridise. Nitrogen in $NH_3$ is $sp^3$, oxygen in $H_2O$ is $sp^3$, boron in $BF_3$ is $sp^2$, beryllium in $BeCl_2$ is $sp$.

What is the hybridisation in benzene?

Each carbon in benzene is $sp^2$ hybridised. Each carbon uses three $sp^2$ orbitals for σ bonds (two to adjacent carbons, one to hydrogen). The remaining unhybridised $p$ orbital on each carbon overlaps with neighbours to form a delocalised π electron ring.

Related Topics

• VSEPR Theory — Predict molecular shapes using electron pair repulsion, which aligns with hybridisation predictions.
• Chemical Bonds — Foundational bonding concepts before diving into orbital theory.
• Hybridization and Bonding — Explore more advanced hybridisation concepts and bonding types.