The big idea: A covalent bond is a shared pair of electrons sitting in the overlap of two atomic orbitals. How the orbitals overlap gives two different kinds of bond:
- A sigma (σ) bond forms by head-on (end-to-end) overlap, with the electron density directly along the bond axis (the line joining the two nuclei). - A pi (π) bond forms by sideways (side-on) overlap of two parallel p orbitals, with the electron density in two lobes above and below the bond axis.
The σ bond is the stronger, primary bond; a π bond is an extra bond added on top of a σ bond.
How to count σ and π bonds (the rule): Count straight off the structure:
- Every single bond is one σ bond. - Every multiple bond contains exactly ONE σ bond, plus the rest as π bonds: - a double bond = 1 σ + 1 π - a triple bond = 1 σ + 2 π
So a molecule's σ count = the total number of bonds between atoms, and its π count = (double bonds) + 2 × (triple bonds).
Why π bonds restrict rotation: A σ bond is symmetrical about the bond axis, so groups can rotate freely about a single (σ-only) bond.
A π bond needs its two p orbitals to stay parallel to keep overlapping. Twisting one end would break that overlap, so a π bond locks the two atoms in place — there is no rotation about a C=C double bond. This is exactly why alkenes show cis–trans (E/Z) isomerism.
An isolated carbon atom has one 2s and three 2p orbitals in its valence shell. Pure s and p orbitals point in fixed directions, which cannot give the symmetrical shapes molecules actually adopt. So before bonding, carbon mixes some of these orbitals into a new set of equal-energy hybrid orbitals — a process called hybridization. The number of orbitals mixed equals the number of electron domains (σ bonds + lone pairs) around the atom.
Carbon mixes its one 2s and (depending on the bonding) one, two or three 2p orbitals to make sp, sp² or sp³ hybrid orbitals. The 2p orbitals left unmixed are used for π bonds.
Interactive diagram
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Mixing the 2s with one, two or all three of the 2p orbitals gives the three hybridization states. Any 2p orbital left unmixed is free to form a π bond.
| Hybridization | Orbitals mixed | Electron domains | Geometry | Bond angle | p orbitals left for π |
|---|---|---|---|---|---|
| sp³ | 1 s + 3 p | 4 | Tetrahedral | 109.5° | 0 → no π bonds |
| sp² | 1 s + 2 p | 3 | Trigonal planar | 120° | 1 → one π bond |
| sp | 1 s + 1 p | 2 | Linear | 180° | 2 → two π bonds |
Read the superscript: The superscript on sp counts the p orbitals used in the mix, so the total = the number of electron domains:
- sp³ = 1 + 3 = 4 domains - sp² = 1 + 2 = 3 domains - sp = 1 + 1 = 2 domains
Fewer p orbitals in the hybrid means more p orbitals left over for π bonds — so the more multiple bonds an atom has, the lower its hybridization number.
Worked example — sp² carbon in ethene
Each carbon in ethene, CH2=CH2, forms two C–H σ bonds and one C=C bond. Deduce its hybridization, the geometry, and how the C=C double bond is made up.
Solution
- Count electron domains around the carbon: 2 C–H bonds + 1 C=C bond (a double bond counts as one domain) = 3 domains.
- 3 domains → mix 1 s + 2 p orbitals → sp² hybridized, trigonal planar, bond angle ≈ 120°.
- That leaves one unmixed 2p orbital on each carbon, perpendicular to the plane.
- The three sp² hybrids make σ bonds (2 C–H + 1 C–C σ); the two leftover 2p orbitals overlap sideways to make the π bond.
Final answer
Each C is sp², trigonal planar (≈120°). The C=C double bond = 1 σ (sp²–sp² head-on) + 1 π (2p–2p sideways).
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The quickest route to a carbon's hybridization is to look at its bonds: an atom with only single bonds is sp³; an atom in a double bond is sp²; an atom in a triple bond is sp. Compare the simplest two-carbon molecules.
A C–C single bond is one σ bond. Each carbon has 4 single bonds = 4 electron domains → sp³, tetrahedral.
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Worked example — sp³ carbon in ethane
Deduce the hybridization of each carbon in ethane, C2H6, and state the number of σ and π bonds in the molecule.
Solution
- Each carbon has 4 single bonds (3 C–H + 1 C–C) = 4 electron domains → sp³, tetrahedral, 109.5°.
- Count the bonds: 6 C–H bonds + 1 C–C bond = 7 σ bonds. All are single bonds.
- There are no multiple bonds, so there are 0 π bonds.
Final answer
Both carbons are sp³. The molecule has 7 σ bonds and 0 π bonds.
The C=C double bond is 1 σ + 1 π. Each carbon has 3 domains → sp², trigonal planar; the leftover 2p orbital forms the π bond.
Interactive diagram
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The C≡C triple bond is 1 σ + 2 π. Each carbon has 2 domains → sp, linear; the two leftover 2p orbitals form two π bonds.
Interactive diagram
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Worked example — sp carbon in ethyne
Deduce the hybridization of each carbon in ethyne, C2H2 (H–C≡C–H), and state the total number of σ and π bonds.
Solution
- Each carbon has 2 domains: one C–H bond + one C≡C bond (a triple bond is one domain) → sp, linear, 180°.
- σ bonds: 2 C–H σ + 1 C–C σ (the triple bond contributes exactly one σ) = 3 σ bonds.
- π bonds: the triple bond has 1 σ + 2 π, so the molecule has 2 π bonds (from the two leftover 2p orbitals on each carbon).
Final answer
Both carbons are sp. Ethyne has 3 σ bonds and 2 π bonds.
How this is tested: HL Paper 2 loves to give you an organic molecule and ask you to:
- deduce the hybridization of one or more named carbons (count the electron domains), and - count the σ and π bonds in the whole molecule.
The single most common error is treating a double or triple bond as more than one σ bond — remember, a multiple bond is always exactly one σ plus the π bonds.
A two-step routine: (1) For each atom, count electron domains (σ bonds + lone pairs): 4 → sp³, 3 → sp², 2 → sp. (2) For the whole molecule, σ = number of bonds between atoms, and π = (doubles) + 2 × (triples).
C1 (CH₃) is sp³, C2 and C3 (the C=C) are sp², C4 (the C≡N carbon) is sp.
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IB-style question — but-2-enenitrile (a) [3]
But-2-enenitrile has the structure CH3–CH=CH–C≡N. (a) Deduce the hybridization of each of the four carbon atoms, giving a reason in each case. [3]
How to score the marks
- C1 (the CH3 carbon): 4 single bonds = 4 electron domains → sp³. ✓
- C2 and C3 (the two C=C carbons): each has 3 domains (one double bond counts as one domain) → both sp². ✓
- C4 (the C≡N carbon): 2 domains (the triple bond + the C–C single bond) → sp. ✓
Final answer
C1 = sp³ (4 domains), C2 = sp² (3 domains), C3 = sp² (3 domains), C4 = sp (2 domains).
IB-style question — but-2-enenitrile (b) [3]
(b) State the total number of σ (sigma) bonds and the total number of π (pi) bonds in one molecule of but-2-enenitrile. Explain how you reached the π count. [3]
How to score the marks
- Count every bond between atoms as one σ: 5 C–H bonds (3 on C1, 1 on C2, 1 on C3), 3 C–C bonds (C1–C2, the C2=C3 σ, C3–C4) and 1 C≡N σ = 9 σ bonds. ✓
- π bonds from the C=C double bond: 1 π. ✓
- π bonds from the C≡N triple bond: 2 π, so total π = 1 + 2 = 3 π bonds (a double adds one π, a triple adds two). ✓
Final answer
9 σ bonds and 3 π bonds (1 π from the C=C, 2 π from the C≡N).