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AS · Practice questions · Sigma and pi bonds, hybridisation

Count the overlaps.

Six original Cambridge-style questions on orbital overlap: σ and π bonds, counting them in molecules, hybridisation and the consequences of the π bond.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Every bond has one σ; extras are π.

01
Explain
[3 marks]

Describe, with reference to orbitals, how the double bond in ethene is formed.

  • Each carbon is sp2 hybridised; one sp2 orbital from each carbon overlaps end-on to form a σ bond. ✓
  • Each carbon has one unhybridised p orbital, perpendicular to the plane of the molecule. ✓
  • These p orbitals overlap sideways to form a π bond, with electron density above and below the plane. ✓
02
Analysis
[2 marks]

State the number of σ bonds and π bonds in (a) ethane, C2H6, and (b) carbon dioxide, O=C=O.

  • (a) Ethane: six C–H and one C–C, all single, so 7 σ and 0 π. ✓
  • (b) CO2: two double bonds, so 2 σ and 2 π. ✓
03
Recall
[3 marks]

State the hybridisation of carbon and the bond angle around it in (a) methane, (b) ethene and (c) ethyne.

  • (a) sp3, 109.5°. ✓
  • (b) sp2, 120°. ✓
  • (c) sp, 180°. ✓
04
Explain
[2 marks]

The bond energy of C=C is about 610 kJ mol−1 and of C–C about 350 kJ mol−1. Explain why the C=C value is less than twice the C–C value.

  • A double bond is one σ bond plus one π bond. ✓
  • The π bond forms by sideways overlap of p orbitals, which is less effective than end-on overlap, so it is weaker than a σ bond (about 260 kJ mol−1 against 350). ✓
05
Explain
[2 marks]

Explain why there is no free rotation about a C=C double bond.

  • Rotation would twist the two p orbitals out of alignment. ✓
  • That would destroy the sideways overlap and break the π bond, which needs a lot of energy. ✓
06
Analysis
[2 marks]

Hydrogen cyanide has the structure H–C≡N. State the total number of σ and π bonds, the hybridisation of carbon and the H–C–N bond angle.

  • 2 σ bonds (C–H and the σ part of C≡N) and 2 π bonds. ✓
  • Carbon is sp hybridised, so the molecule is linear: 180°. ✓

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