Difficulty: Introductory to Intermediate | Prerequisites: Atomic and molecular orbitals, sigma and pi bonds, electron configuration
Hybrid orbitals explain why molecules have the shapes they do. If you look at carbon's ground-state electron configuration, it should only form two bonds, but methane (CH₄) has four identical bonds at equal angles. Hybridisation is the model that resolves this contradiction. It is essential for drawing correct Lewis structures with orbital detail, predicting molecular geometry, and understanding bonding patterns throughout organic chemistry. If you are not comfortable with sigma/pi bonds and basic electron configuration, review those first.
Hybrid orbitals form when atomic orbitals on the same atom mix together before bonding. The number of atomic orbitals mixed equals the number of hybrid orbitals produced. sp³ gives tetrahedral geometry (109.5°), sp² gives trigonal planar (120°), and sp gives linear (180°). Any leftover unhybridised p orbitals are available for pi bonds.
Hybrid orbital
An orbital formed by mixing (combining) atomic orbitals on the same atom. The resulting hybrid orbitals then overlap with orbitals on other atoms to form bonds. In simple terms, the atom blends its s and p orbitals into a new set of equivalent orbitals that point in specific directions.
sp³ hybridisation
Mixing of one s orbital and three p orbitals to produce four equivalent sp³ hybrid orbitals. Think of it as the atom rearranging its orbitals so it can make four identical bonds pointing to the corners of a tetrahedron.
sp² hybridisation
Mixing of one s orbital and two p orbitals to produce three equivalent sp² hybrid orbitals, with one unhybridised p orbital remaining. The three hybrids point to the corners of a triangle; the leftover p orbital sits perpendicular and is available for a pi bond.
sp hybridisation
Mixing of one s orbital and one p orbital to produce two equivalent sp hybrid orbitals, with two unhybridised p orbitals remaining. The two hybrids point in opposite directions (linear); the two leftover p orbitals are available for pi bonds.
p character
The proportion of p orbital contribution in a hybrid orbital. Higher p character means the orbital is more diffuse and directional. In simple terms, more p character = longer, more "pointy" orbital lobes.
s character
The proportion of s orbital contribution in a hybrid orbital. Higher s character means electrons are held closer to the nucleus, which affects acidity and bond length.
The problem is straightforward: carbon's ground-state electron configuration (1s² 2s² 2p²) predicts only two unpaired electrons, so carbon should form only two bonds. But methane (CH₄) has four equal C–H bonds.
In the ground state, carbon has electrons in 2s (paired) and 2p (two unpaired across three p orbitals)
Promoting one 2s electron to the empty 2p orbital gives four unpaired electrons (the excited state)
Even in the excited state, the four orbitals are not equivalent: one is an s orbital and three are p orbitals, so the bonds would not be equal
Hybridisation mixes these four orbitals into four identical sp³ hybrids, producing four equal bonds at 109.5°
Formed by mixing: 1 s orbital + 3 p orbitals → 4 sp³ hybrid orbitals
Composition: 75% p character, 25% s character
Bond angle: 109.5°
Geometry: tetrahedral
All four hybrid orbitals are used for sigma bonds or lone pairs
No unhybridised p orbitals remain, so no pi bonds are possible from this atom
Example: CH₄ (methane)
Four C(sp³)–H(s) sigma bonds
Tetrahedral geometry around carbon
Both bonding and anti-bonding MOs form for each bond
Formed by mixing: 1 s orbital + 2 p orbitals → 3 sp² hybrid orbitals + 1 unhybridised p orbital
Composition: 67% p character, 33% s character
Bond angle: 120°
Geometry: trigonal planar
The unhybridised p orbital is perpendicular to the plane of the sp² orbitals and is responsible for forming pi bonds
Example: BH₃ (borane)
Three B(sp²)–H(s) sigma bonds
Trigonal planar geometry
The leftover p orbital on boron is empty (relevant to its Lewis acid behaviour)
Formed by mixing: 1 s orbital + 1 p orbital → 2 sp hybrid orbitals + 2 unhybridised p orbitals
Composition: 50% p character, 50% s character
Bond angle: 180°
Geometry: linear
The two unhybridised p orbitals are available for two pi bonds
Example: C₂H₂ (acetylene)
The triple bond between the carbons consists of one σ bond (sp–sp overlap) and two π bonds (from the two pairs of unhybridised p orbitals)
Linear geometry around each carbon
Count the number of groups (atoms, lone pairs, or bonds, where a double/triple bond counts as one group) around the atom in question:
4 groups → sp³ (4 hybrid orbitals)
3 groups → sp² (3 hybrid orbitals)
2 groups → sp (2 hybrid orbitals)
Moving from sp³ → sp² → sp:
p character decreases (75% → 67% → 50%)
s character increases (25% → 33% → 50%)
Bond length gets shorter (more s character pulls electrons closer to the nucleus)
Bonds become stronger and shorter
This is why sp C–H bonds are shorter than sp³ C–H bonds, and why the acidity of C–H bonds increases with greater s character (sp > sp² > sp³).
The number of atomic orbitals going in always equals the number of hybrid orbitals coming out
Unhybridised p orbitals that remain may form pi bonds
For halogens bonded to carbon (e.g. C–Cl): the halogen needs only one bond, so it contributes one p orbital (effectively atomic) to the sigma bond; its remaining lone pairs sit in the other orbitals
Hybridisation | Orbitals Mixed | Hybrid Orbitals | Unhybridised p | Geometry | Bond Angle | % s | % p |
|---|---|---|---|---|---|---|---|
sp³ | 1s + 3p | 4 | 0 | Tetrahedral | 109.5° | 25 | 75 |
sp² | 1s + 2p | 3 | 1 | Trigonal planar | 120° | 33 | 67 |
sp | 1s + 1p | 2 | 2 | Linear | 180° | 50 | 50 |
Draw the Lewis structure
Determine the hybridisation of each atom
Draw the correct hybrid orbitals for all sigma bonds
Draw the pi bonds (using unhybridised p orbitals)
Draw any lone pairs and assign them to the correct orbital type
Carbon is bonded to two H atoms and double-bonded to O → 3 groups → sp²
Oxygen is double-bonded to C and has two lone pairs → 3 groups → sp²
σ bonds: C(sp²)–H(s) and C(sp²)–O(sp²)
π bond: C(p)–O(p), from the unhybridised p orbitals on each atom
Determine hybridisation at each atom using the group-counting method
Identify all sp, sp², and sp³ centres
Count unhybridised p orbitals across the molecule
The number of unhybridised p orbitals equals the number of π bonds × 2 (since each π bond uses one p orbital from each of the two atoms involved)
In this example: 12 unhybridised p orbitals → 6 π bonds (though the source notes the answer as 12 total unhybridised p orbitals)
Hybridisation determines molecular shape, which in turn governs how molecules interact with one another. Drug design relies heavily on knowing the 3D geometry of active sites and drug molecules, both of which are set by the hybridisation of their constituent atoms. The tetrahedral geometry of sp³ carbon is why sugars and amino acids have the chirality that makes biology work.
Students frequently confuse "number of bonds" with "number of groups" when assigning hybridisation. A double bond or triple bond counts as one group, not two or three.
A common mistake is thinking that unhybridised p orbitals disappear. They do not. They remain on the atom and are available for pi bonding.
Students sometimes assume all atoms in a molecule have the same hybridisation. Each atom's hybridisation must be determined individually based on its own group count.
Some students think sp³ bonds are stronger than sp bonds because there are "more orbitals." The opposite trend holds for bond length and strength per bond: sp bonds are shorter and stronger due to greater s character.
⚠️ Assigning hybridisation by counting groups is one of the most commonly tested skills in introductory organic chemistry. Practise until it is automatic.
⚠️ Knowing the bond angles (109.5°, 120°, 180°) and geometries (tetrahedral, trigonal planar, linear) for each hybridisation is expected on virtually every exam.
⚠️ Questions often ask how many unhybridised p orbitals a molecule has, or how many pi bonds. These two quantities are directly linked.
⚠️ The connection between s character and bond length/acidity is a favourite for short-answer or multiple-choice questions.
True or False: sp² hybridisation produces four equivalent hybrid orbitals.
Fill in the blank: An atom with 2 groups around it is ______ hybridised.
True or False: A triple bond counts as three groups when determining hybridisation.
Fill in the blank: sp³ orbitals have ______% p character.
True or False: Greater s character in a hybrid orbital leads to shorter bonds.
Answers: 1. False (it produces three) 2. sp 3. False (it counts as one group) 4. 75% 5. True
Q: What is the hybridisation of each carbon in propene (CH₃–CH=CH₂)?
A: The CH₃ carbon has four groups (three H atoms + one C) → sp³. Each carbon of the C=C double bond has three groups → sp².
Q: How many unhybridised p orbitals does an sp² carbon have, and what is their role?
A: One unhybridised p orbital. It is perpendicular to the plane of the three sp² orbitals and participates in pi bonding.
Q: Why is an sp C–H bond shorter than an sp³ C–H bond?
A: The sp hybrid orbital has 50% s character compared to 25% for sp³. Greater s character means the electrons are held closer to the nucleus, resulting in a shorter bond.
Q: Describe the steps for drawing the orbital picture of a complex organic molecule.
A: (1) Draw the Lewis structure. (2) Determine hybridisation at each atom by counting groups. (3) Draw hybrid orbitals for all sigma bonds. (4) Draw pi bonds from unhybridised p orbitals. (5) Place lone pairs in the correct orbital type (hybrid or p).
Hybridisation connects directly to VSEPR theory (molecular geometry prediction from electron-pair repulsion). It also underpins the next major topic, resonance, because resonance requires atoms to have unhybridised p orbitals that can overlap with neighbours. Later, understanding hybridisation is essential for reaction mechanisms: nucleophiles attack sigma anti-bonding orbitals, and electrophilic addition targets pi bonds.
hybrid orbital, hybridisation, hybridization, sp3, sp2, sp, tetrahedral, trigonal planar, linear geometry, bond angle, 109.5 degrees, 120 degrees, 180 degrees, p character, s character, methane orbital, ethylene orbital, acetylene orbital, VSEPR, molecular geometry, group counting, steric number, organic chemistry hybridisation