Difficulty: Intermediate | Prerequisites: Atomic orbitals (s, p), Lewis structures, basic bonding concepts
This topic sits at the heart of organic chemistry's structural logic. Once you understand how atoms blend their orbitals to form bonds, molecular geometry, bond lengths, bond angles, and even reactivity patterns start to make sense. You should already be comfortable drawing Lewis structures and counting valence electrons. If VSEPR geometry rings a bell, you are in good shape; if not, these notes will still stand on their own.
Hybridization describes how an atom's atomic orbitals mix to form new, equivalent hybrid orbitals that point in specific directions. The type of hybridization (sp, sp², or sp³) depends on how many electron regions (bonds and lone pairs) surround an atom. More s-character in a hybrid orbital means shorter bonds, smaller orbitals, and lower energy, which is why triple bonds are shorter and stronger than double or single bonds.
Hybridization
The mixing of atomic orbitals (s and p) on a single atom to produce new, equivalent hybrid orbitals with specific geometries.
Think of it as: an atom reshuffling its orbital toolkit so each bond gets an identical, evenly spaced orbital to work with.
Sigma (σ) bond
A covalent bond formed by head-on overlap of orbitals along the internuclear axis. Every single bond is a sigma bond; double and triple bonds each contain exactly one sigma bond.
In simple terms, this is the first (and strongest) bond formed between two atoms.
Pi (π) bond
A covalent bond formed by sideways (lateral) overlap of unhybridized p orbitals above and below the bond axis. Pi bonds appear only in double bonds (one π) and triple bonds (two π).
Think of it as: the "extra" bond that sits above and below the sigma bond, locking atoms so they cannot rotate freely.
sp³ hybridization
One s orbital mixes with three p orbitals to give four equivalent sp³ hybrid orbitals, arranged in a tetrahedron (109.5° bond angle). Seen when an atom has four electron regions.
sp² hybridization
One s orbital mixes with two p orbitals to give three equivalent sp² hybrid orbitals in a trigonal planar arrangement (120° bond angle). One unhybridized p orbital remains for π bonding. Seen when an atom has three electron regions.
sp hybridization
One s orbital mixes with one p orbital to give two equivalent sp hybrid orbitals arranged linearly (180° bond angle). Two unhybridized p orbitals remain. Seen when an atom has two electron regions.
Electron regions (electron domains)
Any region of electron density around a central atom: a single bond, a double bond, a triple bond, or a lone pair each count as one region. The number of regions determines hybridization.
In simple terms, count each "thing" attached to the atom (bonds of any order, plus lone pairs). That count tells you the hybridization.
s-character
The fraction of s-orbital contribution in a hybrid orbital. sp³ = 25%, sp² = 33%, sp = 50%, pure s = 100%. Higher s-character means the orbital is held closer to the nucleus, resulting in shorter, stronger bonds.
Lone pair
A pair of valence electrons not involved in bonding but occupying a hybrid orbital. Lone pairs count as electron regions when determining hybridization.
The method is straightforward: count the number of electron regions (also called electron domains or steric number) around the atom in question.
2 regions → sp (linear, 180°)
3 regions → sp² (trigonal planar, 120°)
4 regions → sp³ (tetrahedral, 109.5°)
A "region" is any of the following: a single bond, a double bond (counts as one region, not two), a triple bond (counts as one region), or a lone pair.
Remember: a double bond = 1 σ bond + 1 π bond. A triple bond = 1 σ bond + 2 π bonds. Only the σ bonds and lone pairs sit in hybrid orbitals. The π bonds live in unhybridized p orbitals.
Carbon dioxide (CO₂): O=C=O
Carbon: 2 double bonds → 2 regions → sp
Each oxygen: 1 double bond + 2 lone pairs → 3 regions → sp²
Formaldehyde (CH₂O)
Carbon: 1 double bond (to O) + 2 single bonds (to H) → 3 regions → sp²
Peptide-bond molecule (H₃C–C(=O)–NH–CH₃ type structure)
Left carbon (bonded to 3 H and 1 C): 4 σ bonds → 4 regions → sp³
Carbonyl carbon (C=O, bonded to N and C): 1 double bond + 2 single bonds → 3 regions → sp²
Nitrogen (bonded to C, C, H, plus 1 lone pair): 3 σ bonds + 1 lone pair → 4 regions → sp³
Butadiyne (H–C≡C–C≡C–H)
Every carbon: 1 triple bond + 1 single bond → 2 regions → sp
The molecule is perfectly linear.
As you move from pure p to pure s character, orbital size decreases and electrons are held closer to the nucleus.
2p: 0% s-character (largest, highest energy)
sp³: 25% s-character
sp²: 33% s-character
sp: 50% s-character
2s: 100% s-character (smallest, lowest energy)
The key relationship: more s-character → smaller orbital → shorter bond → lower energy.
Visually, an sp³ orbital looks rounder and fatter, while an sp orbital is more elongated and compact.
The lecture compares three molecules to make this concrete:
Ethane (H₃C–CH₃), single bond
Carbon hybridization: sp³
C–C bond angle: 109.5°
C–C bond length: 1.54 Å
Ethylene (H₂C=CH₂), double bond
Carbon hybridization: sp²
C=C bond angle: 120°
C=C bond length: 1.32 Å
Acetylene (HC≡CH), triple bond
Carbon hybridization: sp
C≡C bond angle: 180°
C≡C bond length: 1.18 Å
(1 Å = 1 angstrom = 10⁻¹⁰ m)
Key observations from this comparison:
As bond angle increases, bond length decreases.
sp orbitals are shorter than sp³ orbitals because of their higher s-character.
Triple bonds store the most energy (they are the strongest C–C bonds).
Hybridization | s-Character | Bond Angle | C–C Bond Length | Geometry |
|---|---|---|---|---|
sp³ | 25% | 109.5° | 1.54 Å | Tetrahedral |
sp² | 33% | 120° | 1.32 Å | Trigonal planar |
sp | 50% | 180° | 1.18 Å | Linear |
Quick-reference relationships:
Double bond = 1 σ + 1 π
Triple bond = 1 σ + 2 π
Regions = σ bonds + lone pairs (this number sets the hybridization)
More s-character → shorter bond → stronger bond → higher bond energy
Hybridization is why diamond (all sp³ carbons in a 3D network) is incredibly hard, while graphite (all sp² carbons in flat sheets) is soft and slippery: the geometry forced by the hybridization type determines the bulk material’s properties. In drug design, knowing a nitrogen is sp³ (pyramidal, with a lone pair) versus sp² (flat) changes how a molecule fits into an enzyme’s binding pocket.
Students often count a double bond as two regions. It is one region. A double bond is two bonds sharing one region of electron density between the same two atoms.
Students often forget to count lone pairs as electron regions. A lone pair on nitrogen or oxygen occupies a hybrid orbital and must be counted when assigning hybridization.
Students sometimes assume hybridization is about the molecule as a whole. It is not. Hybridization is assigned atom by atom. Different atoms in the same molecule can have different hybridizations.
Students often confuse bond order with number of regions. A triple bond has a bond order of 3, but it counts as only 1 electron region.
⚠️ Determining hybridization from a Lewis structure is one of the most commonly tested skills in early organic chemistry exams. You will be given a molecule and asked to assign hybridization to each atom.
⚠️ Know the bond-angle and bond-length trends cold: sp³ (109.5°, longest), sp² (120°, middle), sp (180°, shortest). Expect a question asking you to rank bond lengths or predict geometry.
⚠️ The relationship between s-character and bond properties (length, strength, acidity of attached H) comes up repeatedly in later chapters on acidity and reactivity. Building this intuition now pays off.
True or False: A double bond counts as two electron regions when determining hybridization.
Answer: False. A double bond counts as one region.
Fill in the blank: An atom with three electron regions is ______ hybridized.
Answer: sp²
True or False: sp hybrid orbitals have more s-character than sp³ hybrid orbitals.
Answer: True. sp = 50% s-character; sp³ = 25%.
Fill in the blank: The C–C bond length in ethane (sp³) is ______ Å.
Answer: 1.54 Å
True or False: A lone pair does not count as an electron region.
Answer: False. Lone pairs count as electron regions.
Q: What is the hybridization of each carbon in H₂C=O (formaldehyde)?
A: The carbon has 1 double bond + 2 single bonds = 3 electron regions, so it is sp² hybridized.
Q: In CO₂, what is the hybridization of the carbon and each oxygen?
A: Carbon has 2 double bonds = 2 regions = sp. Each oxygen has 1 double bond + 2 lone pairs = 3 regions = sp².
Q: Rank the following bonds from shortest to longest: C–C (ethane), C=C (ethylene), C≡C (acetylene).
A: C≡C (1.18 Å) < C=C (1.32 Å) < C–C (1.54 Å). More s-character means shorter bonds.
Q: A nitrogen atom has three sigma bonds and one lone pair. What is its hybridization?
A: 3 bonds + 1 lone pair = 4 regions = sp³.
Q: Why does increasing s-character shorten a bond?
A: s orbitals are held closer to the nucleus than p orbitals. A hybrid orbital with more s-character keeps its electrons nearer the nucleus, making the orbital smaller and the bond shorter.
Q: What are the bond angles in a molecule where the central atom is sp hybridized?
A: 180° (linear geometry).
This connects to VSEPR theory because hybridization and VSEPR both predict molecular geometry, but hybridization explains it through orbital mixing while VSEPR explains it through electron-pair repulsion. They arrive at the same shapes.
Understanding s-character and bond strength connects directly to acidity in later chapters: a C–H bond on an sp carbon (50% s-character) is more acidic than one on an sp³ carbon (25% s-character), because the resulting anion holds its electrons in a more stable, lower-energy orbital.
Hybridization also sets the stage for understanding molecular representations (Lewis structures, line structures, 3D models), which are covered in the companion notes on Molecular Representations.
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