UNIT 2 CHEMISTRY · PEARSON AAQ · E1.5
Sigma and Pi Molecular Orbitals
FREE CHEMISTRY REVISION · TOPIC 54
Sigma and Pi Molecular Orbitals
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.5
What you will learn
Explain how orbital overlap produces sigma (σ) and pi (π) bonds, describe their electron density and count them in alkanes and alkenes. An atomic orbital describes an electron wavefunction; its familiar lobe diagram indicates a region of space associated with electron probability. Bonding combines orbitals from different atoms into molecular orbitals extending over both atoms. The diagrams below focus on bonding overlap, not a full molecular-orbital energy-level treatment.
1. Sigma bonds: overlap along the axis
A σ bond forms by end-on overlap along the line joining the bonded nuclei, the internuclear axis. Its bonding electron density is concentrated between the nuclei and is cylindrically symmetrical about that axis. Appropriate s, p or hybrid orbitals can overlap end-on; sigma does not mean that only spherical s orbitals are involved.
Each ordinary C–C single bond in an alkane is one σ bond. C–H single bonds are also σ bonds. In ethane, the carbon orbitals used for bonding can be described as sp³ hybrids; their end-on overlap makes the C–C σ bond. Carbon hybrid orbitals overlap with hydrogen 1s orbitals for the C–H bonds. Each carbon has an approximately tetrahedral arrangement, with angles near 109.5°.
2. Pi bonds: sideways overlap
A π bond forms by sideways overlap of parallel p orbitals on adjacent atoms. In ethene these p orbitals point perpendicular to the plane containing the carbon atoms and their attached hydrogens. Bonding density is above and below that molecular plane. There is a nodal plane through the nuclei: π electron density is zero in that plane, so it is not concentrated on the internuclear axis.
The lobes above the plane overlap with each other; the lobes below also overlap. These are two regions of the same π bonding orbital, occupied by a bonding pair of electrons. They are not two π bonds. The colours and signs indicate the relative phases of the orbitals, not positive and negative ions.
3. Compare σ and π bonding
| Feature | Sigma, σ | Pi, π |
|---|---|---|
| Overlap | End-on along the internuclear axis | Sideways between parallel p orbitals |
| Bonding density | Includes the axis between the nuclei | On opposite sides of a nodal plane containing the axis |
| Ordinary C–C single bond | One σ | None |
| Ordinary C=C double bond | One σ | One π |
| Effect of twisting around the C–C axis | End-on overlap can be maintained | Twisting misaligns p orbitals and reduces overlap |
For a given pair of carbon atoms, sideways π overlap is generally less effective than end-on σ overlap. A C=C bond is stronger and shorter overall than C–C, but is not twice as strong: the σ and π contributions differ. Do not infer that a π bond is a weak intermolecular attraction; it is part of covalent bonding within a molecule.
4. Ethane and ethene worked examples
Worked example 1 — ethane, C₂H₆
Draw CH₃–CH₃. There are six separate C–H single bonds and one C–C single bond. Every one is σ: 6+1=7 σ bonds and 0 π bonds. Each carbon’s bond-order total is 3×1+1=4. Do not count the subscript 3 as three carbon atoms. The condensed formula groups three hydrogens on each carbon.
Worked example 2 — ethene, C₂H₄
Draw CH₂=CH₂. Four C–H bonds contribute four σ bonds. The C=C bond contributes one σ and one π. Therefore the entire molecule has 5 σ and 1 π. Each carbon has two C–H single bonds plus a C=C double bond: total bond order 1+1+2=4. There are six bonding electron pairs overall, not six sigma bonds.
Worked example 3 — explain ethene’s shape and restricted rotation
Each carbon in ethene uses three sp² hybrid orbitals for its three σ bonds: two C–H and one C–C. These directions lie approximately in a plane, with angles near 120°. One unhybridised p orbital on each carbon remains perpendicular to that plane. Parallel p orbitals overlap sideways to form the π bond. Twisting one end of the double bond would disrupt that overlap, so C=C does not rotate freely under ordinary conditions. This restriction underlies the alkene stereoisomerism studied in E1.4.
5. Extend the counting method
- Draw the connectivity and include the implicit C–H bonds.
- Count one σ contribution for every pair of atoms directly bonded, including each pair joined by a double bond.
- Add one π contribution for each C=C double bond.
- Check the atom valencies and report whether your count concerns the whole molecule or only its carbon–carbon bonds.
Worked example 4 — propane versus propene
Propane, CH₃CH₂CH₃, has eight C–H σ and two C–C σ: 10 σ, 0 π. Propene, CH₂=CHCH₃, has six C–H σ and two C–C σ: 8 σ and 1 π. Both carbon frameworks contain two C–C connections; the alkene has an additional π contribution on one connection and two fewer hydrogens.
Worked example 5 — but-2-ene
CH₃CH=CHCH₃ has eight C–H bonds and three connections between carbon atoms. All eleven give a σ contribution. The central C=C adds one π. Answer: 11 σ and 1 π. If asked for carbon–carbon bonds only, answer 3 σ and 1 π.
Worked example 6 — compare two bond descriptions
“Ethene contains two sigma bonds between its carbons” is incorrect. Its carbon–carbon bond contains one end-on σ contribution and one sideways π contribution. “Ethene has two regions of π electron density” is correct, provided both regions are understood to belong to one π bond.
Practice with answers
| Question | Answer |
|---|---|
| Count all σ/π bonds in butane, C₄H₁₀ | 10 C–H + 3 C–C = 13 σ; 0 π. |
| Count all σ/π bonds in but-1-ene, C₄H₈ | 8 C–H + 3 C–C = 11 σ; 1 π. |
| Are the two regions above/below ethene two bonds? | No: together they form one π bonding orbital. |
| Does a planar drawing make ethane planar? | No. Ethane has tetrahedral carbon environments; a connectivity drawing omits that geometry. |
Common misconceptions
- A double bond is two σ bonds: it is one σ plus one π.
- Every bond in an alkene is a π bond: its C–H bonds and the σ part of C=C are sigma bonds.
- Two π lobes mean two π bonds: count the molecular bonding orbital, not the visible regions.
- A plus/minus phase label denotes charge: orbital phase is not ionic charge.
- Pi density lies along the carbon–carbon axis: that axis is within its nodal plane.
- Breaking intermolecular forces changes σ/π counts: molecular covalent bonds normally remain intact on boiling.
Examination tips
- Use the terms end-on and sideways overlap explicitly.
- Name the internuclear axis or molecular plane when describing density.
- State one σ plus one π for each C=C.
- Show your C–H and C–C subtotals in a bond-counting answer.
- Keep overlap diagrams labelled and distinguish orbital shape from a structural bond-line drawing.
Check your understanding
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