Difficulty: Introductory to Intermediate | Prerequisites: Atomic orbitals (s, p, d), electron configuration, Lewis structures, VSEPR basics.
Hybridization is the foundational model that explains why carbon (and other atoms) form the shapes they do. If you have covered atomic orbitals and electron configurations but molecular geometry still feels hand-wavy, this is the section that ties the two together. It sits right at the start of organic chemistry because every bond you draw from here on depends on understanding which orbitals mixed and what shape that produces. You should already be comfortable with s and p orbitals and basic Lewis structures before reading on.
Atoms mix (hybridise) their s and p atomic orbitals to form new, equal-energy hybrid orbitals that overlap more effectively and produce stronger bonds. The three common types are sp³ (four orbitals, tetrahedral, 109.5°), sp² (three orbitals, trigonal planar, 120°), and sp (two orbitals, linear, 180°). Counting the number of bonded atoms plus lone pairs on a central atom tells you which hybridisation it uses.
Hybridization (orbital hybridisation)
The mixing of atomic orbitals on a single atom to form new, equivalent hybrid orbitals. Think of it as blending different-shaped rooms into identical rooms so every bond gets the same amount of space.
sp³ hybridization
One s orbital combines with three p orbitals to produce four identical sp³ hybrid orbitals. In simple terms, this is why methane is a perfect tetrahedron, not a flat cross.
sp² hybridization
One s orbital combines with two p orbitals to produce three identical sp² hybrid orbitals, leaving one unhybridised p orbital. Think of it as the reason double-bonded carbons sit in a flat triangle.
sp hybridization
One s orbital combines with one p orbital to produce two identical sp hybrid orbitals, leaving two unhybridised p orbitals. This is the geometry behind triple bonds: everything lines up in a straight line.
Bond angle
The angle formed between two adjacent bonds on the same atom. Each hybridisation state has a characteristic bond angle: 109.5°, 120°, or 180°.
s character / p character
The fractional contribution of s or p atomic orbitals to a hybrid orbital. More s character means the orbital is held closer to the nucleus and electrons in it are lower in energy.
Sigma (σ) bond
A bond formed by head-on overlap of orbitals along the internuclear axis. Every single bond is a sigma bond; every double or triple bond contains exactly one sigma bond with the rest being pi bonds.
Tetrahedral geometry
A molecular shape where four groups point toward the corners of a tetrahedron, with bond angles of 109.5°. In simple terms, it is the shape you get when four balloons are tied together and push apart equally.
Trigonal planar geometry
A molecular shape where three groups lie in the same plane, separated by 120° angles.
Linear geometry
A molecular shape where two groups sit on opposite sides of the central atom, separated by 180°.
Atoms hybridise their orbitals because the resulting compounds are more stable.
Hybrid orbitals are more directional than pure atomic orbitals, which means they overlap more effectively with neighbouring atoms.
Greater overlap produces stronger bonds.
The process also minimises electron repulsion by spacing bonding pairs as far apart as possible.
What mixes: one 2s orbital + three 2p orbitals → four equivalent sp³ orbitals.
Orbital composition: 25% s character, 75% p character.
Geometry: tetrahedral.
Bond angle: 109.5°.
Electron configuration change (carbon example):
Ground state: 1s² 2s² 2p² (two unpaired electrons in the 2p sub-shell).
After hybridisation: 1s², then four sp³ orbitals each holding one electron, ready to form four equivalent bonds.
Classic example: methane (CH₄). Carbon forms four identical C–H sigma bonds pointing toward the corners of a tetrahedron.
Determining sp³ in practice: count the steric number (bonded atoms + lone pairs). If the total is 4, the atom is sp³. Lone pairs count as occupied hybrid orbitals.
Example from the source notes: oxygen in ⁻OH has one bond and three lone pairs (steric number = 4), so it is sp³ and adopts a tetrahedral electron geometry.
What mixes: one 2s orbital + two 2p orbitals → three equivalent sp² orbitals.
What stays behind: one unhybridised 2p orbital, perpendicular to the plane of the sp² orbitals. This leftover p orbital is available for pi (π) bonding.
Orbital composition: 33% s character, 67% p character.
Geometry: trigonal planar.
Bond angle: 120°.
Electron configuration change (carbon example):
Ground state: 1s² 2s² 2p².
After hybridisation: 1s², three sp² orbitals each with one electron, plus one unhybridised 2p orbital with one electron.
Classic example: aluminium trihydride (AlH₃). Aluminium uses three sp² hybrid orbitals to bond with three hydrogen atoms in a flat, triangular arrangement.
Determining sp² in practice: steric number = 3 (three regions of electron density around the central atom). If a carbon has formed three bonds and no lone pairs, it is sp².
Example from the source notes: a carbocation (H₂C⁺–H) has three bonds and no lone pairs, giving sp² hybridisation and a trigonal planar shape.
What mixes: one 2s orbital + one 2p orbital → two equivalent sp orbitals.
What stays behind: two unhybridised 2p orbitals, both perpendicular to each other and to the sp axis. These two p orbitals are available for two pi bonds.
Orbital composition: 50% s character, 50% p character.
Geometry: linear.
Bond angle: 180°.
Electron configuration change (carbon example):
Ground state: 1s² 2s² 2p².
After hybridisation: 1s², two sp orbitals each with one electron, plus two unhybridised 2p orbitals each with one electron.
Classic example: acetylene (HC≡CH). Each carbon uses two sp hybrid orbitals (one for the C–C sigma bond, one for a C–H bond), and the two leftover p orbitals on each carbon form the two pi bonds of the triple bond.
Determining sp in practice: steric number = 2 (two regions of electron density). If a carbon forms two bonds (with any multiplicity) and no lone pairs, it is sp.
Example from the source notes: the right-hand carbon in H–C≡C–H forms two bonds and has a linear shape, confirming sp hybridisation.
Steric number rule
Steric number = (number of bonded atoms) + (number of lone pairs on the central atom). This single count determines hybridisation: 4 = sp³, 3 = sp², 2 = sp.
s character and p character
For any spⁿ hybrid set, s character = 1/(n+1) and p character = n/(n+1).
Hybridization | Orbitals mixed | Hybrid orbitals formed | s character | p character | Bond angle | Geometry | Unhybridised p orbitals |
|---|---|---|---|---|---|---|---|
sp³ | 1s + 3p | 4 | 25% | 75% | 109.5° | Tetrahedral | 0 |
sp² | 1s + 2p | 3 | 33% | 67% | 120° | Trigonal planar | 1 |
sp | 1s + 1p | 2 | 50% | 50% | 180° | Linear | 2 |
Key relationship to remember: as s character increases (sp³ → sp² → sp), the hybrid orbital holds electrons closer to the nucleus, bonds become shorter and stronger, and the bond angle widens.
Diamond is a giant lattice of sp³ carbon atoms, each bonded tetrahedrally to four neighbours, which is why it is so hard. Graphite, by contrast, is built from sheets of sp² carbon, and the leftover p orbitals delocalise across the plane, giving graphite its electrical conductivity and slippery feel.
The rigidity of a double bond (sp²) versus the free rotation of a single bond (sp³) is the reason cis and trans isomers exist in fats, pharmaceuticals, and polymers. Understanding hybridisation is the first step to predicting whether a molecule can rotate or is locked into a particular shape.
Students often think hybridisation is something that happens physically to an atom over time. It does not. Hybridisation is a mathematical model that explains the observed bond angles and strengths.
Students often forget to count lone pairs when determining hybridisation. A nitrogen with three bonds and one lone pair has a steric number of 4 and is sp³, not sp².
Students sometimes believe that a double bond means the atom is sp³ because "there are more electrons." The number of bonds to different atoms (sigma bonds) and lone pairs determines hybridisation, not the total number of electrons or bond order.
Students often confuse the geometry of the electron groups with the shape of the molecule. An sp³ oxygen with two bonds and two lone pairs has tetrahedral electron geometry but a bent molecular shape.
⚠️ Expect questions that give you a Lewis structure and ask you to identify the hybridisation of a specific atom. The method is always: count bonded atoms + lone pairs = steric number.
⚠️ Bond angles are a favourite multiple-choice distractor. Know the three angles cold: 109.5°, 120°, 180°.
⚠️ Questions about unhybridised p orbitals test whether you understand what is left over after hybridisation. sp² leaves one, sp leaves two. These leftover orbitals form pi bonds.
⚠️ Charged species (carbocations, carbanions, hydroxide) appear frequently. The charge does not change the counting method: bonds + lone pairs = steric number.
⚠️ s character and p character percentages are commonly asked as fill-in-the-blank or matching questions.
True or False: An sp² hybridised atom has two unhybridised p orbitals remaining. (False, it has one.)
Fill in the blank: The bond angle in a tetrahedral (sp³) molecule is ____. (109.5°)
True or False: Lone pairs count when determining hybridisation. (True.)
Fill in the blank: sp hybrid orbitals have ____% s character and ____% p character. (50%, 50%)
True or False: An atom with a steric number of 3 is sp³ hybridised. (False, it is sp².)
Q: What is the hybridisation of carbon in methane (CH₄), and what bond angle does this predict?
A: sp³. Carbon has four bonded atoms and no lone pairs (steric number = 4), giving a tetrahedral geometry with 109.5° bond angles.
Q: In a carbocation such as CH₃⁺, what is the hybridisation of the central carbon, and what is the molecular geometry?
A: sp². The carbon has three bonds and no lone pairs (steric number = 3), so it is trigonal planar with 120° bond angles. It has one empty, unhybridised p orbital.
Q: Determine the hybridisation of the oxygen atom in hydroxide ion (⁻OH).
A: sp³. Oxygen has one bond and three lone pairs (steric number = 4), so it uses sp³ hybrid orbitals. The electron geometry is tetrahedral, though the molecular shape is just a single bond.
Q: In acetylene (HC≡CH), what is the hybridisation of each carbon, and how many unhybridised p orbitals does each carbon have?
A: Each carbon is sp hybridised (steric number = 2, linear geometry, 180°). Each carbon retains two unhybridised p orbitals, which form the two pi bonds of the triple bond.
Q: An sp² hybrid orbital has what percentage of s character and p character?
A: 33% s character and 67% p character (one s orbital divided among three hybrid orbitals gives 1/3 s; two p orbitals divided among three gives 2/3 p).
Q: Why are hybrid orbitals more effective at forming bonds than pure atomic orbitals?
A: Hybrid orbitals are more directional than pure s or p orbitals, so they overlap more effectively with orbitals on adjacent atoms. Greater overlap produces stronger, more stable bonds.
Hybridisation connects directly to VSEPR theory. VSEPR predicts the shape from electron-pair repulsion; hybridisation explains why the orbitals adopt that shape in the first place. If you are comfortable with one, the other should click quickly.
This topic also feeds into molecular orbital (MO) theory later in the course. MO theory is a more complete picture of bonding, but hybridisation remains the practical shortcut for predicting geometry in organic chemistry.
When you reach functional groups (alcohols, carbonyls, alkynes), you will use hybridisation to predict bond angles, reactivity, and acidity. For instance, the acidity of a C–H bond increases with s character (sp > sp² > sp³) because the electrons in the resulting anion are held closer to the nucleus.
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