Hybridisation questions look intimidating because the labels (sp, sp², sp³) sound like orbital theory you must derive from first principles. In practice, an examiner is almost always testing one thing: can you count the electron domains around an atom? Once you trust that count, most of these questions take seconds. This set of 15 problems starts with single atoms in small molecules and finishes with molecules where a careless count gives the wrong answer.
The rule you will use every time
For the atom you are asked about, add up:
- the number of σ (sigma) bonds it forms (every single, double or triple bond contains exactly one σ bond), and
- the number of lone pairs it carries.
That total is the steric number, or the number of electron domains. Then read off the label:
| Electron domains | Hybridisation | Electron-pair geometry | Ideal angle |
|---|---|---|---|
| 2 | sp | Linear | 180° |
| 3 | sp² | Trigonal planar | 120° |
| 4 | sp³ | Tetrahedral | 109.5° |
| 5 | “sp³d” | Trigonal bipyramidal | 90°, 120° |
| 6 | “sp³d²” | Octahedral | 90° |
The last two rows are in quotation marks for a reason. For molecules with more than eight electrons around the central atom, the sp³d and sp³d² labels are a simplified model. Modern calculations show that d orbitals play very little part in the bonding of molecules such as PCl₅ and SF₆, so some syllabuses no longer ask for these labels. Use them if your course does.
π (pi) bonds do not count as domains. They are formed from the unhybridised p orbitals left over after hybridisation: an sp² atom has one spare p orbital (room for one π bond), and an sp atom has two (room for two π bonds, or one to each of two neighbours).
If you want the background first, read hybridisation explained, and for the step-by-step method with more examples see how to determine the hybridisation of any atom.
Questions
Total: 40 marks. Draw a Lewis structure first whenever you are unsure how many lone pairs an atom has.
- State the hybridisation of carbon in methane, CH₄. (1 mark)
- State the hybridisation of nitrogen in ammonia, NH₃, and explain why the H–N–H angle is smaller than the ideal angle for that hybridisation. (2 marks)
- State the hybridisation of oxygen in water, H₂O. (1 mark)
- Give the hybridisation of boron in boron trifluoride, BF₃, and name the shape of the molecule. (2 marks)
- State the hybridisation of each carbon atom in ethene, C₂H₄, and describe how the π bond forms. (3 marks)
- In hydrogen cyanide, HCN, state the hybridisation of (a) the carbon atom and (b) the nitrogen atom. (2 marks)
- In carbon dioxide, CO₂, state the hybridisation of the carbon atom and of each oxygen atom. (2 marks)
- Give the hybridisation of each carbon atom and the oxygen atom in propanone, CH₃COCH₃. (3 marks)
- For ethanenitrile, CH₃CN, state the hybridisation of each carbon atom and of the nitrogen atom, and give the C–C–N bond angle. (3 marks)
- Propene is CH₂=CH–CH₃. (a) Give the hybridisation of each carbon, numbering from the CH₂ end. (b) Count the σ and π bonds in the molecule. (4 marks)
- State the hybridisation of the central atom in (a) NH₄⁺, (b) NO₃⁻ and (c) CO₃²⁻. (3 marks)
- Allene is H₂C=C=CH₂. (a) Give the hybridisation of each carbon. (b) Explain why the two CH₂ groups lie in planes at 90° to each other. (4 marks)
- Phosphorus pentachloride, PCl₅, has five P–Cl bonds. State the number of electron domains on phosphorus, the hybridisation label used in the simplified model, and the shape. (3 marks)
- Xenon tetrafluoride, XeF₄, has four Xe–F bonds and two lone pairs on xenon. Give the number of electron domains, the simplified-model hybridisation label and the molecular shape. (3 marks)
- In ethanamide, CH₃CONH₂, a simple count of σ bonds and lone pairs on nitrogen suggests one hybridisation, but experiments show the nitrogen and its neighbours are flat. (a) State the hybridisation a simple count predicts. (b) Explain the experimental result and give the better description. (4 marks)
Answer key
1. Methane. Carbon forms four C–H σ bonds and has no lone pairs: 4 domains, so sp³. The four hybrid orbitals point to the corners of a tetrahedron, giving 109.5°. (1)
2. Ammonia. Nitrogen has three N–H σ bonds and one lone pair: 4 domains, so sp³ (1). The lone pair is held closer to the nitrogen and repels the bonding pairs more strongly than they repel each other, squeezing the H–N–H angle to about 107° (1).
3. Water. Oxygen has two O–H σ bonds and two lone pairs: 4 domains, so sp³. The two lone pairs compress the angle further, to about 104.5°. (1)
4. Boron trifluoride. Boron makes three B–F σ bonds and has no lone pairs (only six electrons around it): 3 domains, so sp² (1). The molecule is trigonal planar, with 120° angles (1). Boron’s unhybridised p orbital is empty, which is why BF₃ behaves as a Lewis acid.
5. Ethene. Each carbon forms three σ bonds (two to H, one to the other C) and no lone pairs: 3 domains, so both carbons are sp² (1 for both). Each carbon keeps one unhybridised 2p orbital at right angles to the plane of the molecule (1). These two p orbitals overlap sideways, above and below the C–C axis, to form the π bond (1). That sideways overlap is also why there is no free rotation about the C=C bond.
6. Hydrogen cyanide. (a) Carbon forms two σ bonds (one to H, one to N) and no lone pairs: 2 domains, sp (1). (b) Nitrogen forms one σ bond to carbon and carries one lone pair: 2 domains, sp (1). The triple bond is one σ bond plus two π bonds made from the two spare p orbitals on each atom.
7. Carbon dioxide. Carbon has two σ bonds (one in each C=O) and no lone pairs: sp (1). Each oxygen has one σ bond and two lone pairs: 3 domains, sp² (1). The two π bonds lie in planes at right angles to each other, one on each side of the carbon.
8. Propanone. Each methyl carbon has four σ bonds: sp³ (1). The carbonyl carbon has three σ bonds (two to C, one to O) and no lone pairs: sp² (1). The oxygen has one σ bond and two lone pairs: sp² (1). So the central C=O group and the two carbons attached to it lie in one plane.
9. Ethanenitrile. The CH₃ carbon has four σ bonds: sp³ (1). The nitrile carbon has two σ bonds (to CH₃ and to N) and no lone pairs: sp (1). Nitrogen has one σ bond and one lone pair: sp. The C–C–N angle is 180°, because the sp carbon is linear (1).
10. Propene. (a) C1 (the CH₂=) has three σ bonds: sp². C2 (=CH–) has three σ bonds: sp². C3 (the CH₃) has four σ bonds: sp³ (2 marks for all three; 1 for two correct). (b) Count bonds: six C–H bonds (2 + 1 + 3) and two C–C connections. Every connection contains one σ bond, so there are 6 + 2 = 8 σ bonds. Only the C=C contains a π bond: 1 π bond (2). If counting is shaky, practise with sigma and pi bonds.
11. Ions. (a) In NH₄⁺, nitrogen forms four N–H σ bonds and has no lone pair: sp³ (1). (b) In NO₃⁻, nitrogen is bonded to three oxygens (one double bond in any single resonance form) with no lone pair: 3 domains, sp² (1). (c) In CO₃²⁻, carbon is bonded to three oxygens with no lone pair: sp² (1). Both NO₃⁻ and CO₃²⁻ are trigonal planar, and the π electrons are delocalised over all three oxygens.
12. Allene. (a) The two end carbons each form three σ bonds (two to H, one to the middle C): sp² (1). The middle carbon forms two σ bonds and no lone pairs: sp (1). (b) The central sp carbon has two unhybridised p orbitals at 90° to each other. One overlaps with the p orbital of the left carbon to form the first π bond; the other overlaps with the p orbital of the right carbon to form the second (1). Each end carbon’s p orbital must be perpendicular to its own CH₂ plane, so the two CH₂ planes end up twisted 90° relative to each other (1). This is a favourite “explain the shape” question, and it shows that hybridisation predicts more than just bond angles.
13. Phosphorus pentachloride. Phosphorus forms five P–Cl σ bonds and has no lone pairs: 5 domains (1). The simplified model labels this sp³d (1). The shape is trigonal bipyramidal, with three equatorial Cl at 120° and two axial Cl at 90° to them (1). Remember that the label is a convenient bookkeeping device, not evidence that d orbitals are heavily involved.
14. Xenon tetrafluoride. Four σ bonds plus two lone pairs gives 6 domains (1). The simplified-model label is sp³d² (1). The electron-pair geometry is octahedral; the two lone pairs sit opposite each other to minimise repulsion, so the atoms form a square planar shape with 90° F–Xe–F angles (1). The VSEPR practice questions cover these shapes in more depth.
15. Ethanamide. (a) Nitrogen forms three σ bonds (to C and two H) and has one lone pair: a simple count gives 4 domains, sp³ (1). (b) The nitrogen lone pair is next to the C=O π bond, so it can be delocalised onto oxygen: a resonance form with a C=N double bond and a negative oxygen contributes to the real structure (1). To take part in that π system, the lone pair must sit in a p orbital, so nitrogen is better described as sp² (1), and the nitrogen, the carbonyl carbon and their attached atoms are planar, with the C–N bond partly double and slow to rotate (1). The same effect makes peptide bonds in proteins flat.
Mark scheme notes
- Always credit a correct domain count even if the final label is wrong: examiners usually award a method mark for “3 σ bonds + 0 lone pairs”.
- In question 10(b), counting each double bond as two σ bonds is the most common error. A double bond is one σ plus one π.
- In questions 13 and 14, a correct shape with the note that sp³d/sp³d² is a simplified model earns full marks on most modern schemes.
Key takeaways
- Count σ bonds plus lone pairs on the atom in question: 2 → sp, 3 → sp², 4 → sp³.
- π bonds never count as domains; they use the leftover p orbitals, which is why sp² atoms have one π bond and sp atoms can have two.
- Hybridisation matches geometry: sp is linear, sp² trigonal planar, sp³ tetrahedral.
- Watch lone pairs next to a π system: in amides (and similar cases) resonance makes an apparently sp³ atom behave as sp².
- sp³d and sp³d² are labels from a simplified model for atoms with five or six domains. Use them only if your syllabus expects them.
- Carbon, the star of most of these questions, can adopt all three common hybridisations; see carbon for its other properties.
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