正文与公式仍可阅读；滚动同步、实验控件和答案验证需要 JavaScript。
The prose and formulas remain readable; interactive figures and checks require JavaScript.

Lesson 4 · 30 min

Hybrid orbitals

s and p orbitals blend into hybrids that point in definite directions — explaining 180°, 120° and 109.5°, and the π bonds of ethene and ethyne.

In this lesson

Say which s and p orbitals make up sp, sp² and sp³ hybrids, and the angles between them
Describe σ bonds with hybrid orbitals and π bonds with the unhybridised p orbitals
Assign a central atom’s hybridisation from its electron domains, and count σ and π bonds
Work backwards from a bond angle to s character, and explain why hybridisation is a description rather than a process

Carbon’s ground-state configuration is $1s^2\,2s^2\,2p^2$ : only two unpaired electrons, in p orbitals at 90° to each other. Taken literally, carbon should form just two bonds at 90°. Yet methane, $\ce{CH4}$ , has four identical C–H bonds at 109.5°. In 1931 Pauling proposed that when an atom bonds, its s and p orbitals can “mix” into a set of hybrid orbitals pointing in new directions.

Predict first

Suppose carbon used its 2s and three 2p orbitals unmixed to bond four H atoms. What would the H–C–H angles look like?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

All four bonds at 109.5° Three bonds at 90° to each other, the fourth different All four in one plane at 90° All four along one line

How s and p mix

Mixing one s orbital with n p orbitals gives n + 1 hybrids of equal energy, called $\mathrm{sp}^n$ :

sp: one s + one p → two hybrids at 180°, each 50 % s; two p orbitals left over.
sp²: one s + two p → three hybrids at 120°, each 33 % s; one p left over.
sp³: one s + three p → four hybrids at 109.5°, each 25 % s; no p left over.

Each hybrid has one big and one small lobe: the two lobes of a p orbital have opposite phases, and adding the s orbital reinforces one and weakens the other. The hybrid therefore leans strongly to one side, ideal for overlapping with another atom.

h = \sqrt{x}\;\mathrm{s} + \sqrt{1 - x}\;\mathrm{p},\qquad \cos\theta = -\frac{x}{1 - x}

Two equivalent hybrids with s character x are orthogonal only when their angle θ satisfies the right-hand equation (Coulson). x = ½, ⅓ and ¼ give 180°, 120° and 109.5°.

$h$: hybrid orbital
$x$: s character (the fraction of s orbital)
$\theta$: angle between two equivalent hybrids

Mixing s and p · 3D

Hybrid orbital explorer

Cross-section of the angular part (+ and − are wavefunction phases)

s character 33%

Show the s and p parts

h = \sqrt{0.33}\,\mathrm{s} + \sqrt{0.67}\,\mathrm{p} = 0.577\,\mathrm{s} + 0.816\,\mathrm{p}

Name: $\mathrm{sp^{2}}$
Angle between two equivalent hybrids: 120.0°
Back lobe / main lobe: 42%

Backwards: s character from a measured bond angle
Molecule	Angle	s share
$\ce{CH4}$	109.5°	25.0%
$\ce{NH3}$	106.7°	22.3%
$\ce{H2O}$	104.5°	20.0%
$\ce{PH3}$	93.5°	5.8%
$\ce{H2S}$	92.1°	3.5%

Loading 3D view…

Drag to rotate · scroll or pinch to zoom · arrow keys rotate too

Hybrids: 3
Ideal angles: 120°
Unhybridised p orbitals: 1

phase +
phase −
empty orbital

Drag the s-character slider: more s means a smaller back lobe and a wider angle between equivalent hybrids. Switch between sp, sp² and sp³ on the right to see them in 3D. Then use “Use this” in the table: water’s angle corresponds to about 20 % s, and H₂S bonds almost with pure p orbitals.

σ and π bonds

Hybrids overlap head-on along the line between two nuclei, making a σ bond: its electron cloud is symmetric around the bond axis, so single bonds can rotate freely. An unhybridised p orbital stands perpendicular to the σ bonds, and parallel p orbitals on neighbouring atoms can overlap side by side, making a π bond with electron density above and below the axis.

In ethene, $\ce{C2H4}$ , each carbon is sp² hybridised: its three hybrids make three σ bonds (two C–H, one C–C) and its leftover p orbital makes a π bond — a double bond is one σ plus one π bond. The π bond needs the two p orbitals to stay parallel, so all six atoms of ethene lie in one plane and the C=C bond cannot rotate — the reason cis–trans isomers exist. In ethyne, $\ce{C2H2}$ , carbon is sp hybridised and its two leftover p orbitals make two perpendicular π bonds: a triple bond is one σ plus two π bonds.

Mixing s and p · 3D

Hybrid orbital explorer

Loading 3D view…

Drag to rotate · scroll or pinch to zoom · arrow keys rotate too

Show unhybridised p orbitals

Central-atom hybridisation: $\mathrm{sp^2}$
σ bonds: 5
π bonds: 1
Empty p orbitals: 0
Measured angle: 117.4°

Each C uses three sp² hybrids for σ bonds; the parallel p orbitals on the two carbons overlap side by side to form the π bond, which locks C=C against rotation. The measured H–C–H angle is 117.4°, a little under 120° — the hybrids are not exactly equivalent.

phase +
phase −
empty orbital

Rotate ethene: the sp² hybrids lie in the molecular plane, and the p orbitals stand perpendicular to it, parallel in pairs. Switch to ethyne to see two perpendicular π bonds, then look at the empty p orbital on BF₃.

Hybridisation and electron domains
Domains	Hybrids	Angle	Examples
2	sp	180°	$\ce{BeCl2}$ , $\ce{CO2}$ , $\ce{C2H2}$
3	sp²	120°	$\ce{BF3}$ , $\ce{C2H4}$ , $\ce{SO2}$
4	sp³	109.5°	$\ce{CH4}$ , $\ce{NH3}$ , $\ce{H2O}$

Worked example

Hybridisation and bonds in acetonitrile, CH₃CN

Assign the hybridisation of both carbons and the nitrogen in acetonitrile, $\ce{CH3-C#N}$ , and count its σ and π bonds.

Concept check

What is the hybridisation of carbon in $\ce{CO2}$ ?

sp sp² sp³ unhybridised

Try the calculation

NH₃ has an H–N–H angle of 106.7°. Using $\cos\theta = -x/(1-x)$ , what is the s character of the N–H bonding hybrids (in %)?

Key takeaways

One s and n p orbitals mix into n + 1 spⁿ hybrids: sp at 180°, sp² at 120°, sp³ at 109.5°.
Hybrids overlap head-on in σ bonds; unhybridised p orbitals overlap side by side in π bonds. Double bond = σ + π, triple bond = σ + 2π.
The number of electron domains sets the hybridisation: 2 → sp, 3 → sp², 4 → sp³.
Hybridisation describes rather than causes; measured angles can be turned into s character (about 20 % for water).

Progress is stored only in this browser and never uploaded.

Data and references

L. Pauling, “The nature of the chemical bond”, J. Am. Chem. Soc. 53, 1367 (1931).
C. A. Coulson, Valence, 2nd ed. (Oxford University Press, 1961): orthogonality of equivalent hybrids.
NIST CCCBDB: experimental geometries of ethene and ethyne; photoelectron ionisation energies of water.
E. Magnusson, J. Am. Chem. Soc. 112, 7940 (1990) — d-orbital participation in hypervalent main-group molecules.