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Organic Shapes Sigma And Pi Bonds

Learn and apply Organic Shapes Sigma And Pi Bonds in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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H1 Organic Shapes, Sigma and Pi Bonds: Orientation

Organic shape questions become predictable when you separate two decisions: determine the electron-region geometry around carbon, then identify whether each carbon–carbon bond contains only a sigma bond or both sigma and pi bonding.

H1 8873 scope
  • Apply sigma and pi bonding and molecular shape to ethane, ethene, benzene and analogous molecules.
  • Explain shape using electron-pair repulsion and the arrangement of sigma and pi bonds.
  • Hybridisation terminology and orbital-mixing models are not required in H1.

Definitions (Must Know)

  • A sigma (σ) bond forms by head-on orbital overlap, with electron density concentrated along the internuclear axis.
  • A pi (π) bond forms by sideways overlap of parallel p orbitals, with electron density on opposite sides of the internuclear axis.
  • An electron region is one bond, whether single or multiple, or one lone pair around a central atom.
  • Tetrahedral arrangement has four bonding regions around carbon and bond angles of about 109.5°.
  • Trigonal-planar arrangement has three bonding regions around carbon and bond angles of about 120°.
  • Delocalised pi electrons are shared across more than two adjacent atoms rather than confined to one pair of atoms.

Detailed Explanations

A. Sigma bonding defines the framework

Every pair of directly bonded atoms has one sigma bond. Head-on overlap places shared electron density along the line joining the nuclei, so a sigma bond provides the basic molecular framework. A C–C single bond is therefore one sigma bond.

B. Pi bonding is the extra component of a multiple bond

After the sigma framework forms, parallel p orbitals can overlap sideways. A C=C double bond contains one sigma bond and one pi bond. The two lobes of pi electron density form one bond, not two separate bonds.

C. Ethane and ethene have different local geometry

Each carbon in ethane has four bonding regions. Repulsion places them tetrahedrally, giving angles near 109.5°. Each carbon in ethene has three sigma-bonding regions. These spread into a trigonal-planar arrangement near 120°. The unhybridised p-orbital language used in deeper courses is not needed to earn the H1 shape mark.

D. The pi bond restricts rotation in ethene

Rotation about C=C would remove the parallel alignment needed for sideways overlap. It would therefore require breaking the pi component, so free rotation is restricted. This is the bonding basis of the cis–trans isomerism test.

E. Benzene has a planar sigma framework and delocalised pi system

Each benzene carbon has three sigma-bonding regions, so it is trigonal planar and the six-carbon ring is planar. The six parallel p orbitals overlap around the ring, placing delocalised pi electron density above and below the plane. Do not describe benzene as three isolated alkene double bonds.

Worked Examples

Modelled example 1

Comparing ethane and ethene around carbon

Core

Problem

Compare the shape around each carbon and the carbon–carbon bonding in ethane and ethene.
Study the worked solution

  1. Analyse ethane

    Method

    Count four sigma-bonding regions around each carbon.

    Reason

    Four regions repel into a tetrahedral arrangement.

    Working

    Each carbon is tetrahedral, with angles about 109.5°; the C–C bond is one sigma bond.

  2. Analyse ethene

    Method

    Count three sigma-bonding regions around each carbon.

    Reason

    The C=C counts as one region for shape, while its second component is a pi bond.

    Working

    Each carbon is trigonal planar, with angles about 120°; C=C contains one sigma and one pi bond.

  3. State the comparison

    Method

    Link the local geometry to the carbon–carbon bonding.

    Reason

    The extra pi component in ethene requires sideways overlap of parallel p orbitals.

    Working

    Ethane is locally tetrahedral about carbon; ethene is planar about both trigonal-planar carbon centres.

Guided practice 2

Counting sigma and pi bonds in propene

About 6 min

Problem

Propene is CH₃CH = CH₂. Determine its total number of sigma and pi bonds.

Try this before viewing the solution

Hints

Hint 1: count bonded pairs
Every directly bonded atom pair contributes one sigma bond, including the pair joined by C=C.
Hint 2: add multiple-bond components
After counting the sigma framework, add one pi bond for the extra component of C=C.
View solution step by step

  1. Count the carbon framework

    Method

    Count the two directly bonded C–C pairs.

    Reason

    Each bonded pair has one sigma component, whether drawn as a single or double bond.

    Working

    2 C–C sigma bonds.

  2. Count C–H bonds

    Method

    Add all six C–H bonds.

    Reason

    Each C–H bond is a sigma bond.

    Working

    2 + 6 = 8 sigma bonds.

  3. Count pi bonds

    Method

    Add the extra component of the one C=C bond.

    Reason

    A double bond consists of one sigma bond and one pi bond.

    Working

    Propene has 8 sigma bonds and 1 pi bond.

Common misconception 3

Correct a double-bond shape claim

Find and correct the mistake

Learner claim

A learner says, “Each carbon in ethene has two C–H bonds and a C=C double bond, so there are four electron regions and the shape is tetrahedral.” Identify the counting error and give the correct shape and angle.

Classify the C=C region

For shape prediction, C=C counts as

View solution step by step

  1. Correct the count

    Method

    Count the C=C as one electron region around each carbon.

    Reason

    Both its sigma and pi electron density join the same pair of carbon atoms and therefore point in the same overall direction for shape prediction.

    Working

    Two C–H regions + one C=C region = three regions.

  2. Deduce the geometry

    Method

    Arrange three regions as far apart as possible.

    Reason

    Three bonding regions give a trigonal-planar arrangement.

    Working

    Each ethene carbon is trigonal planar, with angles about 120°.

Examiner practice 4

Explain restricted rotation in ethene

4 marks

Problem

Explain why rotation about the C=C bond in ethene is restricted, whereas rotation about the C–C bond in ethane is possible. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Describe ethene bonding

    1 mark

    Method

    State that C=C contains one sigma bond and one pi bond.

    Reason

    The two components arise from different modes of orbital overlap.

    Working

    C=C = 1σ + 1π.

  2. Describe the pi overlap

    1 mark

    Method

    Identify sideways overlap of parallel p orbitals.

    Reason

    That alignment places pi electron density above and below the C–C axis.

    Working

    Parallel p orbitals are required for the pi bond.

  3. Explain the restriction

    1 mark

    Method

    State that rotating one carbon would destroy the sideways overlap.

    Reason

    Rotation therefore requires the pi bond to be broken.

    Working

    Breaking a bond requires energy, so free rotation is prevented.

  4. Contrast ethane

    1 mark

    Method

    State that ethane has only a C–C sigma bond.

    Reason

    Head-on overlap about the bond axis is retained during rotation.

    Working

    Rotation about ethane’s C–C sigma bond is possible.

Challenge 5

Test a folded benzene model

Minimal support

Problem

A model shows alternate carbon atoms of benzene folded above and below the ring plane. Use shape and bonding to explain why this model is inconsistent with benzene’s delocalised pi system.

Try this before viewing the solution

Hints

Hint 1: start locally
Each carbon has three sigma-bonding regions.
Hint 2: extend around the ring
Continuous sideways overlap requires the p orbital at every carbon to remain parallel to its neighbours.
View solution step by step

  1. Deduce each carbon's shape

    Method

    Count three sigma-bonding regions at every carbon.

    Reason

    Three regions give a trigonal-planar carbon centre.

    Working

    All six carbon centres favour a common planar ring.

  2. Connect planarity to overlap

    Method

    Keep the p orbitals parallel around the ring.

    Reason

    Parallel alignment permits continuous sideways overlap and delocalised pi electron density above and below the plane.

    Working

    Folding alternate carbons would disrupt p-orbital alignment, so the proposed model is inconsistent.

Mind Stretchers

Attempt each task before opening its hint.

Mind stretcher 1: Diagnosing a shape claim in propeneExtension

Question. A student says every carbon in propene has the same shape because all three atoms are carbon. Find the error and give the correct local shapes.

Show Hint

Count sigma-bonding regions around each carbon rather than identifying the element alone.

Show Answer

The methyl carbon has four sigma-bonding regions and is tetrahedral. Each carbon of C=C has three sigma-bonding regions and is trigonal planar. Shape depends on the local electron-region arrangement, not merely on the atom being carbon.

Mind stretcher 2: Connecting benzene planarity to bondingExtension

Question. Explain why a model showing alternate benzene carbons folded above and below the ring plane is inconsistent with its bonding.

Show Hint

Consider the geometry at each carbon and the alignment required for continuous sideways overlap.

Show Answer

Each benzene carbon has three sigma-bonding regions and is trigonal planar. Keeping all six carbons planar also keeps their p orbitals parallel, allowing continuous sideways overlap and a delocalised pi system above and below the ring. Folding alternate carbons would disrupt that alignment.