Learn · General Chemistry

sp, sp², sp³ hybridization

How to figure out the hybridization of any atom by counting groups, and what each hybridization means for bond angles and geometry.

Quick answer Count the groups on an atom — every σ bond plus every lone pair, never π bonds. 4 → sp³ (tetrahedral, 109.5°), 3 → sp² (trigonal, 120°, one leftover p), 2 → sp (linear, 180°, two leftover p).

Hybridization is bookkeeping for geometry: an atom mixes its valence orbitals into a matched set that points where the molecule needs — so you count instead of memorize.

Methane · sp³ · 109.5°
Ethene · sp² · 120°
Ethyne · sp · 180°

One carbon, three hybridizations — more bonds packed in means a straighter, tighter geometry. Drawn live.

1. Count the Groups Around an Atom to Get Its Hybridization

Add up an atom's σ bonds and lone pairs — the total is the number of orbitals mixed, and it names the hybrid: 4 → sp³, 3 → sp², 2 → sp.

Ethane · each C has 4 σ bonds → sp³
Propene · CH₃ is sp³, the C=C carbons are sp²

2. sp³ Is Tetrahedral (109.5°), sp² Is Trigonal Planar (120°), sp Is Linear (180°)

Groups spread as far apart as possible, so geometry falls straight out of the count — fewer groups means a straighter shape (lone pairs squeeze angles a little, as in water's 104.5°).

sp³ tetrahedral, 109.5°
sp² trigonal, 120°
sp central C, linear 180°

3. Hybrid Orbitals Make σ Bonds; Leftover p Orbitals Make π Bonds

σ bonds and lone pairs live in hybrid orbitals (what you count); π bonds are built from leftover p orbitals — one on sp² (one π), two on sp (two π) — which is why they never count.

Ethene · double bond = 1 σ + 1 π · sp²
Ethyne · triple bond = 1 σ + 2 π · sp

4. Heteroatoms Count the Same Way — Just Remember the Lone Pairs

Oxygen and nitrogen follow the same recipe once you include lone pairs: water's O (2 σ + 2 lp) and ammonia's N (3 σ + 1 lp) are both sp³; a nitrile C and N are both sp.

Water · O: 2 σ + 2 lp → sp³
Ammonia · N: 3 σ + 1 lp → sp³
Acetonitrile · nitrile C and N → sp

5. More s-Character Means Shorter, Stronger Bonds and More Stable Anions

s-character rises sp³ (25%) → sp² (33%) → sp (50%), pulling electrons closer to the nucleus, so bonds get shorter and stronger and anions more stable — which is why a terminal alkyne C–H is the most acidic.

sp³ · 25% s · longest, weakest C–H
sp² · 33% s
sp · 50% s · most acidic C–H

Want the full argument behind that acidity trend? See the acidity of terminal alkynes.

6. Summary

Count σ bonds + lone pairs, ignore π · 4 → sp³, tetrahedral, 109.5° · 3 → sp², trigonal, 120°, one π · 2 → sp, linear, 180°, two π · hybrids hold σ + lone pairs, leftover p holds π · s-character rises sp³→sp, so bonds shorten and anions stabilize.

Quiz yourself

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The oxygen has 2 σ bonds + 2 lone pairs = 4 groups → sp³. It is bent, and the angle compresses to ~104.5° because the two lone pairs push harder than bonding pairs, squeezing the H–O–H angle below the ideal tetrahedral 109.5°.

You count groups (σ bonds + lone pairs), not total bonds. The alkyne carbon has just 2 σ bonds (one to carbon, one to H or R) and no lone pairs → 2 groups → sp, linear at 180°. The other two bonds of the triple bond are π bonds made from its two leftover p orbitals.

One σ + one π. The σ bond is made from an sp² hybrid orbital on each atom; the π bond is made from the leftover unhybridized p orbital on each atom overlapping side-on above and below the plane.

An sp orbital has 50% s-character versus 25% for sp³. Removing the proton leaves the lone pair in an sp orbital that sits closer to the nucleus, so the resulting carbanion is more stable — and a more stable conjugate base means a stronger (more acidic) parent. Higher s-character = more stable anion = more acidic.

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