Hybridization & Orbitals
How atomic orbitals mix to explain molecular geometry, from sp3 to expanded octets
Why Hybridization?
Carbon has one 2s orbital and three 2p orbitals. The 2s orbital is lower in energy and spherical. The 2p orbitals are higher in energy and dumbbell-shaped, pointing along the x, y, and z axes. If carbon bonded using these pure orbitals, you would expect one bond from the s orbital (different energy) and three bonds from the p orbitals (at 90 degrees to each other).
But that is not what happens. In methane (CH₄), all four C-H bonds are identical - same length, same energy, and arranged at 109.5 degrees. Pure atomic orbitals cannot explain this.
Hybridization solves the problem. The s and p orbitals on the central atom mix (hybridize) to form a new set of equivalent orbitals that point toward the bonding partners. The number of orbitals you mix in equals the number of hybrid orbitals you get out.
Energy level diagram: sp³ hybridization in carbon
One 2s + three 2p orbitals mix to form four equivalent sp³ hybrids at intermediate energy. Each holds one electron, ready to bond.
The key idea
Hybridization is not something atoms "decide" to do. It is a mathematical model that explains observed bond angles and equivalent bonds. We look at the molecular geometry first (from VSEPR or experiment), then assign the hybridization that matches. The number of atomic orbitals mixed always equals the number of hybrid orbitals produced.
How to Predict Hybridization
Predicting hybridization requires exactly one piece of information: the steric number (also called the number of electron groups) around the atom. You get this directly from the Lewis structure.
Count every group of electrons around the atom. A single bond, a double bond, a triple bond, and a lone pair each count as one region. The total gives you the steric number, which maps directly to hybridization:
| Steric number | Orbitals mixed | Hybridization | Geometry | Angle |
|---|---|---|---|---|
| 2 | s + p | sp | Linear | 180° |
| 3 | s + p + p | sp² | Trigonal planar | 120° |
| 4 | s + p + p + p | sp³ | Tetrahedral | 109.5° |
| 5 | s + p + p + p + d | sp³d | Trigonal bipyramidal | 90°/120° |
| 6 | s + p + p + p + d + d | sp³d² | Octahedral | 90° |
Critical rule: multiple bonds count as one region
A C=O double bond occupies one hybrid orbital (the sigma bond), with the pi bond coming from an unhybridized p orbital. A C≡N triple bond also occupies just one hybrid orbital. This is why CO₂ (two double bonds, no lone pairs) has a steric number of 2, not 4.
sp³ Hybridization
When an atom has four electron groups, it mixes one s orbital with all three p orbitals to produce four equivalent sp³ hybrid orbitals. These point toward the corners of a tetrahedron, maximizing the distance between them at 109.5 degrees.
sp³ hybridization is the most common type in organic chemistry because it describes every carbon with four single bonds, every nitrogen with three bonds and one lone pair, and every oxygen with two bonds and two lone pairs.
Methane (CH₄)
4 bonds, 0 lone pairs
All four sp³ orbitals hold bonding pairs. Perfect tetrahedral geometry at 109.5°.
Ammonia (NH₃)
3 bonds, 1 lone pair
Three sp³ orbitals bond to H, one holds a lone pair. Molecular shape is trigonal pyramidal. Angles compress to 107° because the lone pair repels more strongly.
Water (H₂O)
2 bonds, 2 lone pairs
Two sp³ orbitals bond to H, two hold lone pairs. Molecular shape is bent. Angles compress further to 104.5°.
Lone pairs occupy hybrid orbitals too
A common misconception is that hybridization only applies to bonding. In water, oxygen is sp³ because it has four electron groups (2 bonds + 2 lone pairs), even though the molecular shape is bent. The electron geometry (tetrahedral) determines hybridization, not the molecular shape.
sp² Hybridization
When an atom has three electron groups, it mixes one s orbital with two p orbitals to form three sp² hybrid orbitals arranged in a trigonal plane at 120 degrees. One p orbital remains unhybridized, perpendicular to the plane.
sp² hybridization is found wherever there is a double bond. The sigma component of the double bond uses a hybrid orbital, while the pi component uses the leftover unhybridized p orbital.
Ethene (C₂H₄)
C=C double bond
Each carbon is sp². The sigma bond uses one sp² orbital from each C. The pi bond forms from the two unhybridized p orbitals overlapping above and below the plane.
Formaldehyde (CH₂O)
C=O double bond
Carbon has 3 electron groups (2 C-H bonds + 1 C=O). The double bond uses one sp² orbital (sigma) plus the unhybridized p (pi).
BF₃
Electron deficient
Boron is sp² with 3 bonds and no lone pairs. The unhybridized p orbital is empty, which makes BF₃ a strong Lewis acid.
sp Hybridization
When an atom has two electron groups, it mixes one s orbital with one p orbital to form two sp hybrid orbitals pointing in opposite directions (180 degrees). Two p orbitals remain unhybridized, perpendicular to each other and to the bond axis.
sp hybridization appears in triple bonds and in molecules with two double bonds on the same atom (like CO₂). The two unhybridized p orbitals allow up to two pi bonds.
Acetylene (C₂H₂)
C≡C triple bond
Each carbon is sp. One sp orbital bonds to H, the other to the other C (sigma bond). The two unhybridized p orbitals on each C overlap to form two pi bonds, giving the triple bond. The molecule is perfectly linear.
CO₂
Two C=O double bonds
Carbon has 2 electron groups (two double bonds). Each C=O has a sigma bond (from an sp orbital) and a pi bond (from an unhybridized p orbital). The two pi bonds are perpendicular to each other.
HCN: both sp and sp³ in one molecule
In hydrogen cyanide, the carbon is sp (2 electron groups: the C-H single bond and the C≡N triple bond). The nitrogen is also sp (2 groups: the triple bond and one lone pair). Every atom in a molecule can have a different hybridization - you determine it independently for each atom based on its own steric number.
The Unhybridized p Orbitals
The p orbitals not used in hybridization do not disappear. They remain as unhybridized p orbitals and are responsible for forming pi bonds through side-by-side overlap.
| Hybridization | Hybrid orbitals | Unhybridized p | Max pi bonds |
|---|---|---|---|
| sp³ | 4 | 0 | 0 |
| sp² | 3 | 1 | 1 |
| sp | 2 | 2 | 2 |
Why double bonds are rigid
The pi bond in a double bond requires the p orbitals to be parallel for side-by-side overlap. Rotating around the bond axis would break this overlap and destroy the pi bond. This is why double bonds prevent rotation and create geometric (cis/trans) isomers, while single bonds (pure sigma) allow free rotation.
Pi bonds are weaker than sigma bonds
Side-by-side overlap (pi) is less effective than head-on overlap (sigma). This is why the second bond of a double bond is weaker than the first. It is also why pi bonds are more reactive - they are the bonds that break first in addition reactions.
Expanded Hybridization
Elements in period 3 and below have d orbitals available for bonding. When these atoms have more than four electron groups, d orbitals participate in hybridization:
sp³d - Five electron groups
One s + three p + one d orbital mix to form 5 hybrid orbitals arranged in a trigonal bipyramidal geometry. Three orbitals are equatorial (120° apart), two are axial (90° from the equatorial plane).
Example: PCl₅ - phosphorus has 5 bonds.
sp³d² - Six electron groups
One s + three p + two d orbitals mix to form 6 hybrid orbitals arranged in an octahedral geometry. All angles are 90°.
Example: SF₆ - sulfur has 6 bonds.
Why period 2 elements cannot expand
Carbon, nitrogen, oxygen, and fluorine have no accessible d orbitals. Their valence shell (n=2) only contains 2s and 2p orbitals, limiting them to a maximum of four hybrid orbitals and eight electrons. Period 3 elements like phosphorus and sulfur have 3d orbitals that, while higher in energy, can participate in bonding when needed.
Determining Hybridization: A Practical Guide
Follow these steps for any atom in any molecule:
- 1. Draw the Lewis structure of the molecule. You need to know all bonds and lone pairs before you can count electron groups.
- 2. Pick the atom you want to find the hybridization of. You determine hybridization for each atom independently.
- 3. Count electron groups around that atom. Single bond = 1, double bond = 1, triple bond = 1, lone pair = 1.
- 4. Match the count: 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d².
Worked examples
Quick shortcut for organic molecules
In most organic molecules, you can skip the Lewis structure and just look at the bonds: atom with all single bonds = sp³, atom with one double bond = sp², atom with a triple bond or two double bonds = sp. This shortcut works because carbon, nitrogen, and oxygen almost always have zero or predictable lone pair counts.
Common Misconceptions
"Atoms choose to hybridize"
Hybridization is a model we use to explain observed geometry, not a physical process that atoms undergo. We observe that methane is tetrahedral, then we use sp³ hybridization to explain why. The geometry comes first; the hybridization label follows.
"Double bonds count as two regions"
Only the sigma component of a multiple bond occupies a hybrid orbital. The pi bond uses an unhybridized p orbital. A double bond is one sigma + one pi = one electron group for the purpose of determining hybridization.
"Lone pairs don't affect hybridization"
Lone pairs absolutely count as electron groups. Water's oxygen is sp³ (not sp) because it has 4 electron groups: 2 bonds + 2 lone pairs. Forgetting lone pairs is one of the most common errors in determining hybridization.
"Every atom in a molecule has the same hybridization"
Each atom has its own hybridization determined by its own steric number. In acetic acid (CH₃COOH), the methyl carbon is sp³ (4 single bonds), the carbonyl carbon is sp² (3 groups: C-C, C=O, C-O), and the oxygen of the O-H group is sp³ (2 bonds + 2 lone pairs).
Putting It Together
Hybridization is the bridge between Lewis structures and 3D molecular geometry:
Lewis structure tells you the steric number
Count bonds and lone pairs from the Lewis structure. This gives you the steric number, which directly determines hybridization.
Hybridization tells you the electron geometry
sp³ = tetrahedral arrangement of electron groups. sp² = trigonal planar. sp = linear. This is the arrangement of all electron groups, including lone pairs.
VSEPR refines to molecular geometry
The molecular shape only considers the positions of atoms (not lone pairs). An sp³ atom with 2 bonds and 2 lone pairs has tetrahedral electron geometry but bent molecular geometry. Hybridization gets you the electron arrangement; VSEPR gets you the visible shape.
Geometry determines bond angles and polarity
Once you know the 3D arrangement, you can predict bond angles (ideal and compressed by lone pairs) and whether bond dipoles cancel or add up to a net molecular dipole.
See hybridization in action
Watch atomic orbitals mix into hybrid orbitals in 3D, then explore how they determine molecular shape.