Hybridization and Sigma and Pi Bonds

What is it?

Carbon’s outer electrons are in one 2s and three 2p orbitals, which point in different directions at 90° to each other. Yet the four bonds in methane are identical and point to the corners of a tetrahedron at 109.5°. Hybridization explains this: an atom’s valence orbitals mix to form an equal number of new, identical hybrid orbitals that point where the bonds are.

Hybridization	Orbitals mixed	Hybrid orbitals	Electron domains	Geometry	Angle
sp³	one s + three p	4	4	Tetrahedral	109.5°
sp²	one s + two p	3 (+ 1 unhybridized p)	3	Trigonal planar	120°
sp	one s + one p	2 (+ 2 unhybridized p)	2	Linear	180°

Bonds come in two kinds:

A sigma (σ) bond forms when orbitals overlap head-on, along the line between the two nuclei. Every single bond is a σ bond.
A pi (π) bond forms when two parallel p orbitals overlap side-on, above and below the line between the nuclei.

A double bond is one σ + one π; a triple bond is one σ + two π.

Key idea

Count the electron domains (bonded atoms + lone pairs) around an atom, exactly as in VSEPR. 4 domains → sp³, 3 → sp², 2 → sp. The p orbitals that are left over are the ones that form π bonds.

Why does it matter?

It connects shape to orbitals. VSEPR predicts the shape; hybridization explains which orbitals make it.
Double bonds don’t rotate. Twisting a π bond would break the side-on overlap, so groups on a C=C are locked in place. This is why cis and trans isomers exist.
Reactivity. π electrons sit above and below the bond, exposed to attack, which is why alkenes react by addition (see Electrophilic Addition to Alkenes).

How does it work?

1. Finding the hybridization

Draw the Lewis structure.
Count electron domains around the atom: each bonded atom counts as one (whether the bond is single, double or triple), and each lone pair counts as one.
Match: 4 → sp³, 3 → sp², 2 → sp.

2. Three carbon compounds

Ethane, $\ce{CH3CH3}$ : each C has 4 domains, so sp³, 109.5°; 7 σ bonds, free rotation about C–C.
Ethene, $\ce{CH2=CH2}$ : each C has 3 domains, so sp², 120°, planar. The C=C is one σ (sp²–sp²) plus one π (from the leftover p orbitals).
Ethyne, $\ce{HC#CH}$ : each C has 2 domains, so sp, 180°, linear. The C≡C is one σ plus two π bonds at right angles.

3. Atoms with lone pairs

Lone pairs occupy hybrid orbitals too. In $\ce{NH3}$ , N has 3 bonds + 1 lone pair = 4 domains: sp³ (bond angle squeezed to about 107°). In $\ce{H2O}$ , O has 2 bonds + 2 lone pairs: sp³ (about 104.5°).

4. Counting σ and π bonds

Single = 1σ. Double = 1σ + 1π. Triple = 1σ + 2π. Count every bond in the full structure, including C–H bonds.

Think of it like this

Mixing orbitals is like mixing paint. One tin of white (s) and three tins of blue (p) make four identical tins of light blue (sp³). The colour is “25 % s, 75 % p”, and all four tins are the same. Mix one white with two blue and you get three tins of sp², with one blue tin left unmixed: that leftover p orbital is ready to make a π bond.

More precisely

Hybridization is a model within valence bond theory; it is a way of describing the bonding, not a step that atoms “do”. Molecular orbital theory is a more complete description that explains, for example, why $\ce{O2}$ is magnetic. A π bond is weaker than a σ bond, because side-on overlap is smaller than head-on overlap: the C=C bond (614 kJ/mol) is less than twice as strong as C–C (348 kJ/mol). Atoms with more than four electron domains, such as S in $\ce{SF6}$ , are better described without d-orbital hybrids, so they are left out here.

Visualise it

Three panels. sp3: four hybrid orbitals pointing to the corners of a tetrahedron, 109.5 degrees, as in methane. sp2: three hybrid orbitals in a plane at 120 degrees with one unhybridized p orbital above and below the plane, as in ethene. sp: two hybrid orbitals at 180 degrees with two unhybridized p orbitals at right angles, as in ethyne. Below: a sigma bond as head-on overlap and a pi bond as side-on overlap of two p orbitals. — sp³, sp² and sp hybrids, and the difference between σ (head-on) and π (side-on) overlap.

Worked example

Worked example: Hybridization of carbon

Question: Give the hybridization of carbon in (a) $\ce{CH4}$ (b) $\ce{H2C=O}$ (methanal) (c) $\ce{CO2}$ .

(a) 4 bonded atoms, no lone pairs: 4 domains, sp³.
(b) 3 bonded atoms (2 H, 1 O): 3 domains, sp², trigonal planar.
(c) 2 bonded atoms (O=C=O): 2 domains, sp, linear.

Worked example: Counting σ and π bonds

Question: How many σ and π bonds are in ethanoic acid, $\ce{CH3COOH}$ ?

Bonds: 3 × C–H, 1 × C–C, 1 × C=O, 1 × C–O, 1 × O–H.
σ bonds: every bond has one, so 3 + 1 + 1 + 1 + 1 = 7 σ.
π bonds: only the C=O has one: 1 π.

Worked example: A triple bond

Question: Describe the bonding in hydrogen cyanide, $\ce{H-C#N}$ .

Carbon has 2 domains (H and N): sp, linear, 180°.
C–H is a σ bond (sp–s). C≡N is 1 σ + 2 π.
Total: 2 σ and 2 π. Nitrogen also has 2 domains (C and its lone pair), so it is sp too.

Worked example: An atom with lone pairs

Question: What is the hybridization of the oxygen atom in water, and why is the bond angle less than 109.5°?

O has 2 bonds + 2 lone pairs = 4 domains: sp³. The lone pairs repel more strongly than bonding pairs, squeezing the H–O–H angle to about 104.5°.

Common mistake

Common mistake: Counting a double bond as two domains

A double or triple bond counts as one electron domain, because it points in one direction. $\ce{CO2}$ has two domains on carbon (sp), not four.

Common mistake: Forgetting that every bond has a σ

A double bond is σ + π, not two π bonds. The first bond between two atoms is always σ.

Common mistake: Forgetting lone pairs

Lone pairs count as domains: N in $\ce{NH3}$ is sp³, not sp², even though it has only three bonds.

Notation note

sp³ is read “s-p-three”; the superscript counts the p orbitals mixed, not electrons.
σ is the Greek letter sigma; π is pi.

Remember this

Electron domains = bonded atoms + lone pairs. 4 → sp³ (109.5°), 3 → sp² (120°), 2 → sp (180°).
σ: head-on overlap, along the bond axis. π: side-on overlap of parallel p orbitals.
Single = σ; double = σ + π; triple = σ + 2π.
Leftover (unhybridized) p orbitals make the π bonds.
π bonds prevent rotation about double bonds.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/hybridization-and-sigma-pi-bonds/