Difficulty: Introductory | Prerequisites: General chemistry bonding, Lewis structures, VSEPR basics
Every carbon atom in an organic molecule uses one of three hybridisation states (sp, sp², or sp³), and you can work out which one by counting the number of regions of electron density around it. The hybridisation determines the geometry, bond angles, and which orbitals overlap to form sigma and pi bonds. Once you can assign hybridisation confidently, drawing a full orbital picture of any small molecule is a matter of matching each bond and lone pair to the right orbital.
sp³ hybridisation
The carbon mixes one s orbital and three p orbitals to produce four equivalent sp³ hybrid orbitals arranged in a tetrahedron (bond angle ~109.5°). All four bonds are sigma bonds.
Think of it as: four equal arms pointing as far apart as possible, like the legs of a camera tripod plus one arm straight up.
sp² hybridisation
The carbon mixes one s orbital and two p orbitals to produce three equivalent sp² hybrids in a trigonal planar arrangement (bond angle ~120°). One unhybridised p orbital remains perpendicular to the plane and forms a pi bond.
In simple terms, this means the carbon is flat with three bonds in a plane and one p orbital sticking above and below for a pi bond.
sp hybridisation
The carbon mixes one s orbital and one p orbital to produce two sp hybrids arranged linearly (bond angle 180°). Two unhybridised p orbitals remain, each available for a pi bond.
Think of it as: the carbon points in two opposite directions, with two leftover p orbitals that can each form a pi bond.
Sigma (σ) bond
A bond formed by head-on (end-to-end) overlap of orbitals along the internuclear axis. Every single bond is a sigma bond. In double and triple bonds, one of the bonds is always sigma.
In simple terms, this is the first and strongest bond between two atoms, formed by orbitals pointing directly at each other.
Pi (π) bond
A bond formed by side-by-side (lateral) overlap of unhybridised p orbitals above and below (or in front of and behind) the internuclear axis.
Think of it as: the "extra" bond in a double or triple bond, formed by p orbitals overlapping sideways rather than head-on.
Lone pair
A pair of valence electrons on an atom that is not shared in a bond. Lone pairs occupy hybrid orbitals (or unhybridised p orbitals, depending on context) and count as a region of electron density when determining hybridisation.
In simple terms, electrons that belong to one atom only and are not involved in bonding, but they still take up space and affect geometry.
Count the steric number (number of electron-density regions) around the carbon. Each single bond, double bond, triple bond, or lone pair counts as one region.
4 regions of electron density = sp³ (tetrahedral, ~109.5°)
3 regions of electron density = sp² (trigonal planar, ~120°)
2 regions of electron density = sp (linear, 180°)
A double bond counts as one region, not two. A triple bond also counts as one region. This is the single most important rule for getting hybridisation right.
Structure (a): HC≡C–C(=O)–C(=CH)–CH₃
The HC≡C carbon at the end: two regions (triple bond + H) = sp
The C≡C carbon bonded to the carbonyl: two regions (triple bond + single bond) = sp
The C=O carbon: three regions (double bond to O + two single bonds) = sp²
The C=CH carbon: three regions (double bond + two single bonds) = sp²
The CH₃ carbon: four regions (three C–H bonds + one C–C bond) = sp³
Structure (b): Acetic anhydride type
Carbonyl carbons (C=O): three regions each = sp²
CH₃ carbons: four regions each = sp³
Structure (c): Benzamide type (ring + C(=O)NH₂)
All six ring carbons: three regions each = sp²
The carbonyl carbon (C=O): three regions = sp²
Any CH₃ or CH₂ side-chain carbons: four regions = sp³
Acetaldehyde has two carbons: the methyl carbon (sp³) and the carbonyl carbon (sp²).
Methyl carbon (CH₃, sp³):
Four sp³ hybrid orbitals, each forming a sigma bond
Three sp³–s(1s) sigma bonds to hydrogen atoms
One sp³–sp² sigma bond to the carbonyl carbon
Carbonyl carbon (C=O, sp²):
Three sp² hybrid orbitals in a trigonal plane
One sp²–sp³ sigma bond to the methyl carbon
One sp²–s(1s) sigma bond to hydrogen
One sp²–sp² sigma bond to oxygen (the first bond of the C=O)
One unhybridised p–p pi bond to oxygen (the second bond of the C=O), formed by lateral overlap above and below the molecular plane
Oxygen (sp²):
One sp² orbital forms the sigma bond to carbon
Two sp² orbitals hold lone pairs
One unhybridised p orbital forms the pi bond to carbon
Every single bond = 1 sigma
Every double bond = 1 sigma + 1 pi
Every triple bond = 1 sigma + 2 pi
For acetaldehyde (CH₃CHO): 6 sigma bonds total (3 C–H on methyl, 1 C–C, 1 C–H on carbonyl carbon, 1 C–O sigma) and 1 pi bond (C=O).
Students often count each bond in a double bond as a separate region of electron density. A double bond is one region, not two. The same goes for triple bonds.
Students sometimes forget that lone pairs count as regions of electron density when assigning hybridisation. On oxygen in a carbonyl, the two lone pairs each occupy a hybrid orbital.
A common error is labelling all carbons in a ring as sp³. If the ring has alternating double bonds (as in benzene), those carbons are sp².
Students frequently confuse sigma and pi bonds. The first bond between any two atoms is always sigma. Pi bonds are the "extra" bonds in double and triple bonds, formed by sideways p-orbital overlap.
True or false: A carbon with two double bonds and no other bonds is sp hybridised. (True)
Fill in the blank: An sp² carbon has ___ unhybridised p orbital(s) available for pi bonding. (One)
True or false: A triple bond consists of three sigma bonds. (False, it is one sigma and two pi)
Fill in the blank: The bond angle around an sp³ carbon is approximately ___°. (109.5)
True or false: Lone pairs on an atom do not affect its hybridisation. (False, lone pairs count as regions of electron density)
Q: What is the hybridisation of each carbon in CH₃C≡CH?
A: The CH₃ carbon has four regions of electron density (three C–H bonds and one C–C bond), so it is sp³. The middle carbon of the triple bond has two regions (one triple bond, one single bond), so it is sp. The terminal ≡CH carbon also has two regions (triple bond and one C–H bond), so it is sp.
Q: In formaldehyde (H₂C=O), how many sigma bonds and how many pi bonds are there? Describe the orbitals involved.
A: There are three sigma bonds (two C–H and one C–O) and one pi bond (C=O). The carbon is sp². Each C–H sigma bond is sp²–1s overlap. The C–O sigma bond is sp²–sp² overlap. The C=O pi bond is p–p lateral overlap of the unhybridised p orbitals on carbon and oxygen.
Q: A carbon atom is bonded to two other carbons by single bonds and to one oxygen by a double bond. What is its hybridisation and geometry?
A: Three regions of electron density (two single bonds + one double bond) = sp² hybridisation, trigonal planar geometry, bond angles approximately 120°.
Q: Why does the oxygen in a carbonyl group (C=O) have two lone pairs in hybrid orbitals and one p orbital forming a pi bond?
A: The oxygen is sp² hybridised. Two of its sp² hybrid orbitals hold lone pairs, one sp² orbital forms the sigma bond to carbon, and the remaining unhybridised p orbital overlaps laterally with carbon's p orbital to form the pi bond.
Q: Identify the hybridisation of each carbon in vinyl cyanide (CH₂=CH–C≡N).
A: The two carbons of the C=C double bond each have three regions of electron density, so both are sp². The carbon of the C≡N triple bond has two regions of electron density, so it is sp.
Hybridisation determines molecular geometry, which feeds directly into understanding polarity and dipole moments later in the course. It also sets the stage for reaction mechanisms: sp² carbons are the ones found in carbonyl chemistry, and sp³ carbons are the targets in substitution and elimination reactions. Understanding orbital overlap here prepares you for molecular orbital theory and conjugation.
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