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Carbon’s electron configuration is 1s² 2s² 2p². On paper, that means it has only two unpaired electrons, in two p orbitals at 90° to each other. So why does carbon form four identical bonds in methane, pointing to the corners of a tetrahedron at 109.5°? The answer, proposed by Linus Pauling in 1931, is hybridisation: the idea that atomic orbitals mix to form new, equivalent orbitals pointing in the right directions for bonding. This article explains sp³, sp² and sp hybridisation with the classic examples.
The puzzle
Atomic orbitals have fixed shapes (see atomic orbital shapes):
- An s orbital is spherical.
- Three p orbitals are dumbbell-shaped, pointing along the x, y and z axes, at 90° to each other.
If carbon used its orbitals exactly as they are, we’d expect:
- only two bonds (from the two half-filled 2p orbitals), or
- if an electron were promoted from 2s to 2p to give four unpaired electrons, three bonds at 90° (from the p orbitals) and one different bond (from the s orbital).
Neither matches methane. Experiments show four identical C–H bonds — the same length (109 pm), the same energy — with bond angles of 109.5°.
The idea: mixing orbitals
Hybridisation says that when an atom forms bonds, its valence s and p orbitals can mix mathematically to form an equal number of new hybrid orbitals. These hybrids:
- are all identical in energy and shape within a set;
- point in directions that keep them as far apart as possible;
- each have a large lobe on one side, which overlaps strongly with orbitals on other atoms, making strong sigma (σ) bonds.
The number of hybrid orbitals always equals the number of atomic orbitals mixed. The names tell you what went in:
| Hybrid | Orbitals mixed | Number of hybrids | Arrangement | Angle | Left-over p orbitals |
|---|---|---|---|---|---|
| sp³ | 1 s + 3 p | 4 | Tetrahedral | 109.5° | 0 |
| sp² | 1 s + 2 p | 3 | Trigonal planar | 120° | 1 |
| sp | 1 s + 1 p | 2 | Linear | 180° | 2 |
Energy picture for carbon: promoting one 2s electron to the empty 2p orbital costs energy, but that cost is repaid many times over by forming four bonds instead of two.
sp³ hybridisation: methane, CH₄
Carbon mixes its 2s orbital with all three 2p orbitals to make four sp³ hybrids. Each hybrid has ¼ s character and ¾ p character. Each holds one electron.
The four hybrids point to the corners of a tetrahedron, 109.5° apart. Each overlaps head-on with a hydrogen 1s orbital to form a σ bond. The result is four identical C–H bonds in a tetrahedral molecule — exactly what’s observed.
The same picture applies to every carbon atom with four single bonds: the carbons in ethane, in the chain of any alkane, and in diamond, where each carbon forms four σ bonds to four other carbons in a 3D network.
sp³ with lone pairs: ammonia and water
Hybridisation also applies to atoms with lone pairs:
- Ammonia, NH₃: nitrogen has four electron domains — three bonding pairs and one lone pair — so it’s sp³ hybridised. Three hybrids form N–H bonds; the fourth holds the lone pair. The molecule is trigonal pyramidal with bond angles of about 107°, slightly less than 109.5°, because the lone pair repels more strongly than bonding pairs.
- Water, H₂O: oxygen has two bonding pairs and two lone pairs — also sp³. The two lone pairs squeeze the H–O–H angle to about 104.5°, giving a bent shape.
This matches the predictions of VSEPR theory (see VSEPR and molecular geometry).
sp² hybridisation: ethene, C₂H₄
In ethene, each carbon is bonded to three atoms (two H and one C) and the molecule is flat, with angles of about 120°.
Each carbon mixes its 2s orbital with two of its 2p orbitals to make three sp² hybrids, lying in a plane at 120°. One 2p orbital is left unhybridised, sticking up and down, perpendicular to the plane.
- The three sp² hybrids on each carbon form three σ bonds: two C–H and one C–C.
- The two leftover p orbitals, one on each carbon, lie parallel and overlap sideways, above and below the plane, to form a π bond.
So the C=C double bond is one σ bond + one π bond (see sigma and pi bonds). Because the π bond needs the p orbitals to stay parallel, the molecule can’t twist around the double bond — the origin of cis–trans isomerism.
Other sp² atoms include the carbon in a C=O group (such as in methanal and carboxylic acids), boron in BF₃ (which has an empty p orbital, making it an electron-pair acceptor), and the carbons in benzene and graphite. In benzene, the six unhybridised p orbitals overlap all around the ring, creating a delocalised π system (see resonance structures).
sp hybridisation: ethyne, C₂H₂
In ethyne (acetylene), H–C≡C–H, each carbon is bonded to two atoms, and the molecule is linear (180°).
Each carbon mixes its 2s orbital with one 2p orbital to make two sp hybrids, pointing in opposite directions. Two p orbitals are left unhybridised, at right angles to each other and to the molecular axis.
- The two sp hybrids on each carbon form two σ bonds: one C–H and one C–C.
- The two pairs of leftover p orbitals overlap sideways to form two π bonds, one above and below the axis and one in front and behind.
So a triple bond is one σ + two π. The two π bonds together form a cylinder of electron density around the C–C axis.
Carbon dioxide’s central carbon is also sp hybridised (two σ bonds to oxygen, two π bonds), as is the carbon in HCN.
Why s character matters
The more s character a hybrid orbital has, the closer its electrons are held to the nucleus, because s orbitals are more compact than p orbitals:
| Hybrid | s character | C–H bond length (typical) |
|---|---|---|
| sp³ | 25 % | 109 pm (ethane) |
| sp² | 33 % | 108 pm (ethene) |
| sp | 50 % | 106 pm (ethyne) |
As s character increases, bonds get shorter and carbon holds its electrons more tightly. That’s why the C–H hydrogens of ethyne are noticeably more acidic than those of ethene or ethane: when ethyne loses H⁺, the negative charge sits in an sp orbital close to the nucleus, where it’s more stable. Ethyne’s pKa is about 25, compared with roughly 44 for ethene and about 50 for ethane.
Quick way to find hybridisation
Count the electron domains around the atom: each σ bond counts as one and each lone pair counts as one (double and triple bonds count as one domain, because they contain one σ bond).
| Electron domains | Hybridisation |
|---|---|
| 4 | sp³ |
| 3 | sp² |
| 2 | sp |
For a full method with many examples, see how to determine the hybridisation of any atom.
Limits of the model
Hybridisation is a very useful model, especially in organic chemistry, but it has limitations:
- It’s a model, not a physical process: atoms don’t literally go through a “promotion then mixing” sequence.
- It doesn’t explain magnetic properties — for example, why O₂ is attracted to a magnet. That needs molecular orbital theory (see molecular orbital theory for beginners).
- For molecules like H₂S (bond angle about 92°) and PH₃ (about 93°), the angles are close to 90°, suggesting the central atom uses nearly pure p orbitals — little hybridisation at all.
- The old idea that atoms like sulfur in SF₆ use d orbitals (sp³d², etc.) to form more than four bonds isn’t supported by modern calculations; these molecules are better described in other ways.
Common misconceptions
- “Hybridisation causes the shape.” It’s a way of describing the shape we observe (or predict with VSEPR).
- “π bonds are formed by hybrid orbitals.” π bonds come from the unhybridised p orbitals.
- “Every atom is hybridised.” Terminal atoms like H use unhybridised orbitals; some heavier atoms barely hybridise.
- “sp³ means four bonds.” It means four electron domains — lone pairs count too (NH₃, H₂O).
Key takeaways
- Hybrid orbitals are mixtures of s and p orbitals that explain observed bond angles and equal bonds.
- sp³: 4 domains, tetrahedral, 109.5° (CH₄, NH₃, H₂O).
- sp²: 3 domains, trigonal planar, 120°, one leftover p orbital for a π bond (C₂H₄, BF₃, C=O).
- sp: 2 domains, linear, 180°, two leftover p orbitals for two π bonds (C₂H₂, CO₂).
- More s character gives shorter bonds and more acidic C–H hydrogens.
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