Difficulty: Intermediate | Prerequisites: Lewis structures, VSEPR theory, atomic orbitals (s, p, d), sigma and pi bonds.
Hybridization is the model organic chemists use to explain why carbon forms the bond angles and geometries it does. It bridges the gap between atomic orbital theory and the shapes you see via VSEPR. You need it to understand sigma/pi bonding frameworks, which in turn determine reactivity in every reaction mechanism you will meet from here on. If you are comfortable drawing Lewis structures and assigning electron-pair geometry, you are ready for this material.
Count the groups (atoms + lone pairs) around a carbon to determine its hybridization: 4 groups = sp³, 3 groups = sp², 2 groups = sp. Hybridization tells you the geometry, bond angles and which orbitals form sigma vs pi bonds. Every bond in a molecule can be described as either a sigma overlap (head-on) or a pi overlap (side-on) of specific hybrid or unhybridized p orbitals.
Hybridization
The mixing of atomic orbitals (one s and one or more p) on the same atom to produce a set of equivalent hybrid orbitals that point toward bonding partners. Think of it as: you blend orbitals together so each bond direction gets an identical orbital to work with.
sp³ hybridization
One s orbital mixes with three p orbitals to give four equivalent sp³ hybrids, arranged in a tetrahedron (109.5° bond angles). In simple terms, this is what carbon looks like when it has four single bonds and no lone pairs, as in methane or ethane.
sp² hybridization
One s orbital mixes with two p orbitals to give three equivalent sp² hybrids in a trigonal planar arrangement (120° bond angles), with one unhybridized p orbital left over perpendicular to the plane. In simple terms, this is the carbon in a double bond, a carbonyl, or a benzene ring.
sp hybridization
One s orbital mixes with one p orbital to give two sp hybrids arranged linearly (180° bond angles), with two unhybridized p orbitals remaining. Think of it as: the carbon in a triple bond or an allene.
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 simple terms, this is the strong, direct overlap that holds atoms together.
Pi (π) bond
A bond formed by side-on (lateral) overlap of unhybridized p orbitals above and below the internuclear axis. A double bond = 1σ + 1π; a triple bond = 1σ + 2π. Think of it as: the weaker, sideways overlap that sits on top of a sigma bond.
Unhybridized p orbital
A p orbital that was not mixed into the hybrid set. It participates in pi bonding or holds a lone pair that can conjugate. In simple terms, it is the leftover p orbital that sticks up and down from a flat (sp²) or linear (sp) carbon.
Draw the Lewis structure so every bond and lone pair is visible.
Count the number of electron groups (also called "steric number") around the carbon. Each single bond, double bond, triple bond, or lone pair counts as one group.
4 groups → sp³ (tetrahedral, 109.5°)
3 groups → sp² (trigonal planar, 120°)
2 groups → sp (linear, 180°)
A double bond counts as one group, not two. A triple bond also counts as one group.
Molecule (a): HC≡C–C(=O)–C(CH)(CH₃)(H₃C)
The two carbons in the C≡C triple bond each have 2 groups → sp.
The carbonyl carbon (C=O) has 3 groups (double bond to O, single bond to C on each side) → sp².
The carbon bonded to CH, CH₃ and another carbon has 4 groups → sp³.
Terminal CH₃ and CH carbons with 4 bonds each → sp³.
Molecule (b): an ester (C–C(=O)–O–C)
The carbonyl carbon (C=O with single bond to O) has 3 groups → sp².
The carbon bonded to three other atoms by single bonds only (no double bond) has 4 groups → sp³.
The carbon on the other side of the oxygen, bonded to other carbons by single bonds, has 4 groups → sp³.
Molecule (c): benzamide (benzene ring–C(=O)–NH₂)
Every carbon in the benzene ring has 3 groups (two C–C bonds in the ring + one C–H or C–C bond) → sp².
The carbonyl carbon (C=O bonded to NH₂) has 3 groups → sp².
Acetaldehyde has two carbons with different hybridizations:
Methyl carbon (CH₃): sp³ hybridized. Four sp³ orbitals point toward the three H atoms and the adjacent carbon. Each C–H bond is a σ bond formed by sp³–s overlap. The C–C bond is a σ bond formed by sp³–sp² overlap.
Carbonyl carbon (CHO): sp² hybridized. Three sp² orbitals lie in a plane, forming σ bonds to the methyl carbon (sp²–sp³), the hydrogen (sp²–s) and the oxygen (sp²–sp² or sp²–lone pair hybrid on O). One unhybridized p orbital on this carbon overlaps side-on with a p orbital on oxygen to form the π bond of C=O.
Oxygen: has two lone pairs (one in an sp² hybrid, one in the unhybridized p orbital that is perpendicular to the π system, depending on the model used) and participates in one σ bond (head-on overlap) and one π bond (lateral p–p overlap) with the carbonyl carbon.
Every bond in the molecule is either:
A σ bond (head-on overlap along the bond axis), or
A π bond (side-on p–p overlap above and below the bond axis).
The C=O double bond consists of one σ + one π. All C–H and C–C single bonds are σ only.
Steric number = (number of atoms bonded to the central atom) + (number of lone pairs on the central atom)
Steric number 4 → sp³ → tetrahedral → 109.5°
Steric number 3 → sp² → trigonal planar → 120°
Steric number 2 → sp → linear → 180°
Double bond = 1 σ + 1 π
Triple bond = 1 σ + 2 π
Single bond = 1 σ
% s-character: sp³ = 25%, sp² = 33%, sp = 50%. Higher s-character means the orbital is held closer to the nucleus, shorter bonds and greater electronegativity of that carbon.
Hybridization determines molecular shape, and molecular shape determines function. Drug molecules bind to enzyme active sites because their sp², sp³ and sp centres create exactly the right 3D geometry to fit. The rigidity of sp² carbons (flat, no rotation around double bonds) is why cis/trans isomers of fatty acids behave differently in cell membranes, and why the planar structure of graphite (all sp²) conducts electricity while diamond (all sp³) does not.
Students often count a double bond as two groups. It is one group. A C=O counts the same as a C–H for purposes of determining hybridization.
Students sometimes think lone pairs do not count as groups. They do. A nitrogen with three bonds and one lone pair has four groups and is sp³, not sp².
Students confuse bond order with group count. A triple bond is still just one group, even though it contains three bonds (1σ + 2π).
Students assume that sp² always means "double bond present." An sp² carbon can also appear in a carbocation (empty p orbital) or a radical. The key is three groups, regardless of what fills them.
⚠️ Expect questions that give you a complex molecule and ask for the hybridization of every carbon (or every non-hydrogen atom). Practice on molecules with mixed hybridizations.
⚠️ Drawing the full orbital picture of a molecule (as in the acetaldehyde problem) is a common exam question. You must label each bond as σ or π and name the overlapping orbitals (e.g., sp³–s, sp²–p).
⚠️ Understanding hybridization is a prerequisite for stereochemistry, reaction mechanisms and resonance, all of which dominate the rest of CHM 255.
True or False: A carbon with one double bond and two single bonds is sp³ hybridized. (False, it is sp².)
Fill in the blank: A triple bond consists of ___ sigma bond(s) and ___ pi bond(s). (1 sigma, 2 pi.)
True or False: The bond angles around an sp hybridized carbon are 120°. (False, they are 180°.)
Fill in the blank: An sp² carbon has ___ unhybridized p orbital(s) remaining. (1.)
True or False: A lone pair counts as a group when determining hybridization. (True.)
Q: What is the hybridization of the carbon atom in formaldehyde (H₂C=O)?
A: sp². The carbon has three groups: two C–H bonds and one C=O double bond.
Q: In acetylene (HC≡CH), describe the orbital overlap that forms the triple bond.
A: Each carbon is sp hybridized. The C–C sigma bond is formed by sp–sp head-on overlap. The two pi bonds are formed by side-on overlap of the two unhybridized p orbitals on each carbon (one pair overlapping vertically, one pair horizontally).
Q: A carbon atom is bonded to two other carbons by single bonds, one oxygen by a double bond, and has no lone pairs. What is its hybridization and geometry?
A: sp², trigonal planar with approximately 120° bond angles. Three groups total (two single bonds + one double bond).
Q: How many sigma and pi bonds are in the molecule CH₃CHO (acetaldehyde)?
A: 6 sigma bonds and 1 pi bond. Sigma bonds: 3 × C–H on the methyl group, 1 × C–C, 1 × C–H on the carbonyl carbon, 1 × C–O (the sigma component of C=O). Pi bonds: 1 × C=O (the pi component).
Q: Why does an sp² carbon form a flat arrangement while an sp³ carbon is tetrahedral?
A: An sp² carbon has three equivalent hybrid orbitals in a plane (maximising separation at 120°) and one unhybridized p orbital perpendicular to that plane. An sp³ carbon has four equivalent hybrid orbitals, which maximise their separation by pointing toward the corners of a tetrahedron at 109.5°.
This material connects directly to resonance and conjugation: a molecule can only have extended conjugation when adjacent atoms each have an unhybridized p orbital available to overlap, which requires sp² or sp hybridization. It also underpins acid-base chemistry in organic, where sp hybridized C–H bonds are more acidic because the higher s-character stabilises the resulting carbanion. Later in CHM 255, when you study E2 eliminations and addition reactions, the transition from sp³ to sp² (or vice versa) will be central to the mechanism.
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