🟣 This is Higher Level (HL) content.
Why Do We Need Hybridization?
Look at carbon's ground-state electron configuration: 1s² 2s² 2p². There are only 2 unpaired electrons in the 2p subshell. This should mean carbon can only form 2 bonds, yet in methane (CH₄) carbon forms 4 identical bonds at equal angles. The ground-state configuration simply cannot explain this.
Step 1: Promotion
One electron from the 2s orbital is promoted to the empty 2p orbital. This requires energy, but the energy released by forming 4 bonds (rather than 2) more than compensates.
After promotion: 1s² 2s¹ 2px¹ 2py¹ 2pz¹ (4 unpaired electrons).
Step 2: Mixing
The promoted orbitals mix together to form a new set of equivalent hybrid orbitals of identical energy and shape. In CH₄, one 2s + three 2p orbitals mix to give four sp³ hybrids.
🔬 Nature of Science
Hybridization is a mathematical model, not a physical process you can observe. Orbitals do not literally "blend" before bonding occurs. The model was developed to explain why observed bond lengths and angles match symmetrical geometries that ground-state configurations cannot predict.
The Three Hybridization Types
sp³ Hybridization (4 Electron Domains)
Formation: one s orbital + three p orbitals → four equivalent sp³ hybrid orbitals. No unhybridised p-orbitals remain.
Geometry: Tetrahedral | Bond angle: 109.5°
Bonding: all four sp³ orbitals form σ bonds or hold lone pairs. Only sigma bonds are possible.
Examples: CH₄ (109.5°), NH₃ (107° due to lone pair repulsion), H₂O (105° due to two lone pairs), diamond
sp² Hybridization (3 Electron Domains)
Formation: one s orbital + two p orbitals → three equivalent sp² hybrid orbitals. One p-orbital remains unhybridised, perpendicular to the plane.
Geometry: Trigonal planar | Bond angle: 120°
Bonding: the three sp² orbitals form σ bonds. The unhybridised p-orbital overlaps laterally with an adjacent p-orbital to form one π bond.
Examples: C₂H₄ (ethene), BF₃, benzene (C₆H₆), graphite
sp Hybridization (2 Electron Domains)
Formation: one s orbital + one p orbital → two equivalent sp hybrid orbitals. Two p-orbitals remain unhybridised, perpendicular to the bond axis and to each other.
Geometry: Linear | Bond angle: 180°
Bonding: the two sp orbitals form σ bonds. The two unhybridised p-orbitals can form two π bonds (as in a triple bond or two separate double bonds).
Examples: C₂H₂ (ethyne), CO₂, HCN, BeCl₂
Summary Table
| Hybridization | Orbitals Mixed | Electron Domains | Geometry | Bond Angle | Unhybridised p | Examples |
|---|---|---|---|---|---|---|
| sp³ | 1s + 3p | 4 | Tetrahedral | 109.5° | 0 | CH₄, NH₃, H₂O |
| sp² | 1s + 2p | 3 | Trigonal planar | 120° | 1 | C₂H₄, BF₃, benzene |
| sp | 1s + 1p | 2 | Linear | 180° | 2 | C₂H₂, CO₂, HCN |
🔑 The Shortcut: Steric Number
Steric Number = Bonded Atoms + Lone Pairs = Total Electron Domains
4 domains → sp³ | 3 domains → sp² | 2 domains → sp
This is the same electron domain count you use in VSEPR, just applied to determine hybridization instead of shape.
How to Determine Hybridization
- Draw the Lewis structure of the molecule, showing all bonds and lone pairs.
- Count the electron domains around the atom in question. Each single bond, double bond, triple bond, or lone pair counts as one domain.
- Match the count: 4 domains = sp³, 3 domains = sp², 2 domains = sp.
Remember: a double bond is ONE domain, a triple bond is ONE domain. Do not count σ and π components separately.
Lone Pairs and Hybridization
Lone pairs occupy hybrid orbitals just like bonding pairs. This means the hybridization depends on the total electron domain count, but the molecular shape only describes where the atoms are.
| Molecule | Bonding Pairs | Lone Pairs | Total Domains | Hybridization | Electron Domain Geometry | Molecular Shape | Bond Angle |
|---|---|---|---|---|---|---|---|
| CH₄ | 4 | 0 | 4 | sp³ | Tetrahedral | Tetrahedral | 109.5° |
| NH₃ | 3 | 1 | 4 | sp³ | Tetrahedral | Trigonal pyramidal | 107° |
| H₂O | 2 | 2 | 4 | sp³ | Tetrahedral | Bent | 105° |
⚠️ Key Distinction
sp³ hybridized does NOT automatically mean the molecule is tetrahedral. NH₃ is sp³ but pyramidal. H₂O is sp³ but bent. Always distinguish between electron domain geometry (arrangement of all domains, including lone pairs) and molecular geometry (arrangement of atoms only).
Worked Examples
📐 Methane (CH₄) - sp³
Carbon has 4 electron domains (4 C-H bonds, 0 lone pairs).
1s + 3p → 4 sp³ hybrid orbitals, each forming a σ bond.
Shape: tetrahedral, 109.5°
Bonds: 4σ, 0π
📐 Ethene (C₂H₄) - sp²
Each carbon has 3 electron domains (2 C-H + 1 C=C).
1s + 2p → 3 sp² hybrids forming σ bonds. One unhybridised p-orbital on each C overlaps laterally to form the π bond.
Shape around each C: trigonal planar, 120°
Total bonds: 5σ, 1π
📐 Ethyne (C₂H₂) - sp
Each carbon has 2 electron domains (1 C-H + 1 C≡C).
1s + 1p → 2 sp hybrids forming σ bonds. Two unhybridised p-orbitals on each C overlap to form 2 perpendicular π bonds.
Shape: linear, 180°
Total bonds: 3σ, 2π
📐 Carbon Dioxide (CO₂) - sp
Central carbon has 2 electron domains (2 C=O double bonds).
Carbon is sp hybridised. Two unhybridised p-orbitals form π bonds with each oxygen, in perpendicular planes.
Shape: linear, 180°
Total bonds: 2σ, 2π
Mixed Hybridization in One Molecule
Different atoms within the same molecule can have different hybridization states. Always check each atom individually.
📐 Propanone (CH₃COCH₃)
CH₃ carbons: 4 electron domains (3 C-H + 1 C-C) → each is sp³ hybridised
Central C=O carbon: 3 electron domains (2 C-C + 1 C=O) → sp² hybridised
The C=O carbon has one unhybridised p-orbital that forms the π bond with the oxygen.
📐 Propanal (CH₃CH₂CHO)
CH₃ carbon: 4 domains → sp³
CH₂ carbon: 4 domains → sp³
CHO carbon: 3 domains (1 C-C + 1 C-H + 1 C=O) → sp²
Total bonds: 8σ, 1π
Connection to Sigma and Pi Bonds
Hybridization directly determines the σ-bond framework and how many unhybridised p-orbitals are available for π bonds:
- sp³: 0 unhybridised p-orbitals → only σ bonds possible → single bonds only
- sp²: 1 unhybridised p-orbital → 1 π bond possible → can form double bonds
- sp: 2 unhybridised p-orbitals → 2 π bonds possible → can form triple bonds (or two double bonds as in CO₂)
The σ bonds are always formed by the head-on overlap of hybrid orbitals. The π bonds are formed by the lateral overlap of the remaining unhybridised p-orbitals. See 2.2.15 Sigma and Pi Bonds for more detail.
Examiner Traps
⚠️ Common Errors That Lose Marks
- Counting bonds inside a multiple bond separately. A double bond = 1 domain, a triple bond = 1 domain. Do not split them into σ and π when counting domains.
- Assuming sp³ always means tetrahedral. The hybridization tells you the electron domain geometry, but lone pairs change the molecular shape (e.g. NH₃ is sp³ but pyramidal).
- Treating hybridization as a real physical event. Atoms do not "rearrange" their orbitals before bonding. It is a model used to explain observed geometries.
- Calling a double bond "two bonds of equal strength." A double bond is 1σ + 1π. The sigma component is stronger than the pi component due to greater orbital overlap.
- Ignoring hybridization on non-carbon atoms. Nitrogen and oxygen can also be hybridised. For example, the oxygen in a C=O group is sp² hybridised (1 bond + 2 lone pairs = 3 domains).