How to Determine the Hybridisation of Any Atom

Exam questions often ask “What is the hybridisation of the carbon atom marked with an asterisk?” or “State the hybridisation of each nitrogen in this molecule”. There’s a quick, reliable method that works for almost every atom you’ll meet: count the electron domains around the atom. This guide sets out the method, works through twelve examples from simple to tricky, and flags the special cases where the quick method needs care. For the theory behind hybrid orbitals, read hybridisation explained first.

The method

Step 1. Draw a complete Lewis structure, including all lone pairs (see Lewis dot structures). Lone pairs are easy to forget and they matter.

Step 2. For the atom you’re interested in, count its electron domains:

• each atom bonded to it counts as 1 (a single, double or triple bond to the same atom still counts only once, because there’s only one σ bond);
• each lone pair counts as 1.

This is also called the steric number:

Steric number = (number of atoms bonded) + (number of lone pairs)

Step 3. Match the steric number to the hybridisation:

Steric number	Hybridisation	Electron-domain geometry	Ideal angle
2	sp	Linear	180°
3	sp²	Trigonal planar	120°
4	sp³	Tetrahedral	109.5°

(For steric numbers 5 and 6, older textbooks use sp³d and sp³d². Modern bonding theory doesn’t support d-orbital involvement, so many courses no longer ask for these. Check your syllabus.)

Shortcut for carbon: four single bonds → sp³; one double bond → sp²; one triple bond or two double bonds → sp.

Worked examples

1. Carbon in methane, CH₄

• Atoms bonded: 4 (four H). Lone pairs: 0.
• Steric number 4 → sp³.

2. Nitrogen in ammonia, NH₃

• Atoms bonded: 3. Lone pairs: 1.
• Steric number 4 → sp³. (The shape is trigonal pyramidal, but the hybridisation is still sp³ — lone pairs count.)

3. Oxygen in water, H₂O

• Atoms bonded: 2. Lone pairs: 2.
• Steric number 4 → sp³.

4. Carbon in ethene, CH₂=CH₂

• Atoms bonded: 3 (two H and one C). Lone pairs: 0.
• Steric number 3 → sp². The C=C counts once.

5. Carbon in ethyne, HC≡CH

• Atoms bonded: 2. Lone pairs: 0.
• Steric number 2 → sp.

6. Carbon in carbon dioxide, O=C=O

• Atoms bonded: 2. Lone pairs: 0.
• Steric number 2 → sp. (Two double bonds, but only two σ bonds.)

7. Oxygen in carbon dioxide

• Each O is bonded to 1 atom (C) and has 2 lone pairs.
• Steric number 3 → sp².

8. Boron in BF₃

• Atoms bonded: 3. Lone pairs: 0.
• Steric number 3 → sp². Boron has an empty, unhybridised p orbital, which is why BF₃ accepts electron pairs (see dative covalent bonds).

9. Carbon and oxygen in methanal, H₂C=O

• Carbon: bonded to 3 atoms (2 H, 1 O), 0 lone pairs → sp².
• Oxygen: bonded to 1 atom, 2 lone pairs → steric number 3 → sp².

10. The atoms in ethanol, CH₃CH₂OH

• Both carbons: four single bonds each → sp³.
• Oxygen: bonded to C and H (2), with 2 lone pairs → sp³.

11. The atoms in ethanenitrile (acetonitrile), CH₃–C≡N

• CH₃ carbon: 4 atoms bonded → sp³.
• Nitrile carbon: bonded to 2 atoms (CH₃ carbon and N) → sp.
• Nitrogen: bonded to 1 atom, 1 lone pair → steric number 2 → sp.

12. The atoms in ethanoic acid, CH₃COOH

• CH₃ carbon: sp³.
• Carboxyl carbon: bonded to 3 atoms (C, =O, –OH) → sp².
• Carbonyl oxygen (=O): 1 atom + 2 lone pairs → sp².
• Hydroxyl oxygen (–OH): 2 atoms + 2 lone pairs → by counting, sp³ — but see the special case below.

Special case: lone pairs next to π bonds

The counting method assumes every lone pair sits in a hybrid orbital. But when an atom with a lone pair is next to a double bond (or part of an aromatic ring), the lone pair can be delocalised into the π system. To do that, it must occupy an unhybridised p orbital, and the atom becomes sp², even though counting gives sp³.

Common examples:

• Nitrogen in amides (e.g. in ethanamide, CH₃CONH₂, and in the peptide bonds of proteins). The nitrogen lone pair is delocalised into the C=O group, making the C–N bond partly double. The nitrogen is effectively sp² and flat. This is why peptide bonds are planar and don’t rotate freely — crucial for protein shape (see peptide bonds).
• Nitrogen in pyrrole and oxygen in furan, where the lone pair is part of an aromatic ring.
• Nitrogen in aniline (phenylamine), which is nearly flat because its lone pair interacts with the benzene ring.

The –OH oxygen in a carboxylic acid is a borderline case; many courses accept sp³ by the counting rule, while more advanced courses describe it as sp² because one lone pair is delocalised towards the C=O. Follow your course’s convention.

Rule of thumb: if an atom with a lone pair is directly attached to a C=O, C=C or an aromatic ring, and resonance can move that lone pair into a π bond, treat the atom as sp² (see resonance structures).

Checking your answer

• Geometry check: sp³ atoms have roughly tetrahedral angles (≈109°, a bit less with lone pairs); sp² atoms have ≈120°; sp atoms are linear.
• π-bond check: an sp² atom can form one π bond; an sp atom can form two; an sp³ atom forms none.
• Count check: the number of hybrid orbitals always equals the steric number.

Practice

Give the hybridisation of each marked atom:

1. The carbon in HCN.
2. The nitrogen in the ammonium ion, NH₄⁺.
3. The sulfur in H₂S.
4. Each carbon in propene, CH₂=CH–CH₃.
5. The central atom in the carbonate ion, CO₃²⁻.
6. Each carbon in benzene.

Answers:

1. Bonded to 2 atoms, no lone pairs → sp.
2. 4 atoms, 0 lone pairs → sp³.
3. 2 atoms, 2 lone pairs → steric number 4 → sp³ by counting. (Note: the measured angle of about 92° suggests sulfur uses nearly pure p orbitals; many courses still accept sp³.)
4. CH₂= carbon: sp²; =CH– carbon: sp²; CH₃ carbon: sp³.
5. Bonded to 3 oxygens, no lone pairs → sp² (trigonal planar).
6. Each is bonded to 3 atoms (2 C and 1 H) → sp²; the leftover p orbitals form the delocalised ring.

Common mistakes

• Counting double bonds as two domains. A double or triple bond to one atom counts as one domain.
• Ignoring lone pairs. Always draw them before counting.
• Confusing shape with hybridisation. Ammonia is pyramidal, water is bent, but both are sp³.
• Forgetting resonance for lone pairs next to π systems (amides, aromatic rings).
• Assigning hybridisation to hydrogen. Hydrogen uses its 1s orbital; it isn’t hybridised.

Key takeaways

• Steric number = atoms bonded + lone pairs on the atom.
• 2 → sp, 3 → sp², 4 → sp³.
• Multiple bonds count once; lone pairs always count.
• Lone pairs next to π systems often make an atom sp² because of resonance.
• Use geometry and π-bond counts to check your answer.

Next: see how σ and π bonds come from these orbitals in sigma and pi bonds.