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Sigma and Pi Bonds (Orbital Overlap)

Learn and apply Sigma and Pi Bonds (Orbital Overlap) in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Sigma and Pi Bonds: Orbital Overlap: Orientation

This lesson explains covalent bonding through overlap of s and p orbitals, the scope specified in the syllabus. The decisive distinction is whether overlap occurs along the internuclear axis (σ) or above and below it (π).

Treat this as an extension of Atomic Structure (A Level), then use the Chemical Bonding hub to compare models across the topic.

Definitions (Must Know)

A. Sigma bond, σ

A σ bond forms by head-on overlap of orbitals along the internuclear axis.

B. Pi bond, π

A π bond forms by sideways overlap of p orbitals above and below the internuclear axis.

Four orbital-overlap diagrams: s-s, s-p and p-p head-on overlap along the internuclear axis form sigma bonds, while parallel p orbitals overlap sideways above and below the axis to form a pi bond
Within syllabus scope, sigma bonds can arise from s–s, s–p or p–p head-on overlap; a pi bond arises from sideways overlap of parallel p orbitals.

Detailed Explanations

A. Head-on overlap gives a σ bond

Head-on overlap concentrates shared electron density directly between the nuclei, along the internuclear axis. Within the specified s/p scope, this may be:

  • s–s overlap, as in the H–H bond in H₂;
  • s–p overlap, as a simplified description of H–Cl bonding;
  • p–p head-on overlap, as in the σ component of Cl₂ or N₂.

B. Sideways p–p overlap gives a π bond

Two parallel p orbitals can overlap sideways. This creates two continuous regions of shared electron density, one on each side of the internuclear axis. It is still one π bond, not two bonds.

C. Bond-counting method

  1. Every bond between two atoms contains one σ bond.
  2. Any “extra” bond is a π bond:
    • double bond = 1 extra bond → 1 π
    • triple bond = 2 extra bonds → 2 π

For example, C = C has one σ bond and one π bond, while N#N has one σ bond and two π bonds in perpendicular planes.

D. Applying overlap later

Rotation about a double bond would destroy the parallel p-orbital overlap, so rotation is restricted. The organic-chemistry consequences are developed in Isomerism and Stereochemistry and Alkenes and Electrophilic Addition.

Worked Examples

Modelled example 1

Decompose the Nitrogen Triple Bond

Core

Problem

How many σ and π bonds are in N₂?

Study the worked solution

  1. Identify the bond order

    Method

    Represent nitrogen as N#N.

    Reason

    The two nitrogen atoms are joined by a triple bond.

    Working

    Bond order = 3.

  2. Apply the overlap rule

    Method

    Assign one first bond as σ and the two additional bonds as π.

    Reason

    Every bonded atom pair has one head-on σ bond; extra bonds arise from sideways p-orbital overlap.

    Working

    N#N contains 1σ + 2π.

Guided practice 2

Count Bonds in Ethene

About 6 min

Problem

How many σ and π bonds are in CH₂ = CH₂ (ethene)?

Try this before viewing the solution

Hints

Hint 1: split the double bond
The C = C bond contributes one σ and one π bond.
Hint 2: include every single bond
Each of the four C–H bonds contributes one σ bond.
View solution step by step

  1. Count the carbon–carbon components

    Method

    Split the double bond into one σ and one π.

    Reason

    A double bond contains one first bond and one additional bond.

    Working

    C = C:1σ + 1π.

  2. Count the carbon–hydrogen bonds

    Method

    Count four C–H single bonds.

    Reason

    Every single bond is one σ bond.

    Working

    4C-H:4σ.

  3. State the totals

    Method

    Add like bond types.

    Reason

    The carbon–carbon sigma component and all C–H bonds contribute to the sigma total.

    Working

    σ = 1 + 4 = 5; π = 1.

Common misconception 3

Correct an Ethyne Bond Count

Find and correct the mistake

Learner attempt

Asked how many σ and π bonds are in HC#CH, a learner counts the C#C triple bond as three σ bonds. Identify the error and give the correct totals.

Classify the triple bond

C≡C components

View solution step by step

  1. Correct the carbon–carbon bond

    Method

    Assign one σ and two π bonds to C#C.

    Reason

    One head-on overlap lies on the internuclear axis; the two additional overlaps are sideways.

    Working

    C#C:1σ + 2π.

  2. Include the C–H bonds

    Method

    Add the two C–H single bonds to the sigma count.

    Reason

    Each C–H bond is one σ bond.

    Working

    Total σ = 1 + 2 = 3; total π = 2.

Examiner practice 4

Compare Sigma and Pi Overlap

4 marks

Problem

Compare the orbital overlap and position of shared electron density in a σ bond and a π bond. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Describe sigma overlap

    1 mark

    Method

    State that orbitals overlap head-on.

    Reason

    The overlap is aligned with the internuclear axis.

    Working

    σ: head-on overlap.

  2. Locate sigma density

    1 mark

    Method

    Place shared electron density directly between the nuclei.

    Reason

    Head-on overlap concentrates density along the internuclear axis.

    Working

    σ density lies on the axis.

  3. Describe pi overlap

    1 mark

    Method

    State that two parallel p orbitals overlap sideways.

    Reason

    Parallel alignment permits overlap on both sides of the axis.

    Working

    π: sideways p–p overlap.

  4. Locate pi density

    1 mark

    Method

    Place two connected regions of density on opposite sides of the axis.

    Reason

    Those two regions are parts of one π bond.

    Working

    π density lies above and below the axis.

Challenge 5

Explain Restricted Rotation in Ethene

Minimal support

Problem

Use orbital overlap to explain why rotation about the C = C bond in ethene is restricted, whereas rotation about a carbon–carbon single bond is much less restricted.

Try this before viewing the solution

Hints

Hint 1: track the pi orbitals
The π bond requires parallel p orbitals to maintain sideways overlap.
Hint 2: consider rotation
Rotating one carbon relative to the other destroys that parallel alignment and the π overlap.
View solution step by step

  1. Identify the alignment requirement

    Method

    State that the unhybridised p orbitals must remain parallel.

    Reason

    Only parallel p orbitals can overlap sideways to sustain the π bond.

    Working

    Parallel p orbitals → continuous π overlap.

  2. Explain the restriction

    Method

    State that rotation would break the sideways overlap.

    Reason

    Energy must be supplied to disrupt the π bond before free rotation can occur.

    Working

    Rotation destroys π overlap, so it is restricted.

  3. Contrast a single bond

    Method

    State that a C–C single bond contains only a σ bond.

    Reason

    Head-on overlap is cylindrically symmetric about the internuclear axis and is retained during rotation.

    Working

    σ-only C–C rotation is much less restricted.

Mind Stretchers

Mind stretcher 1Extension

Propene is CH₃CH = CH₂. Count the total number of σ and π bonds.

Show Answer

Mark scheme:

  • One C = C: 1 σ + 1 π.
  • One C–C single bond: 1 σ.
  • C–H bonds: CH₃ has 3, CH has 1, CH₂ has 2 → 6 C–H σ bonds.
  • Total: σ = 1(C = C) + 1(C-C) + 6(C-H) = 8, π = 1.

Mind stretcher 2Extension

A shared electron-density region lies directly between two nuclei and was formed from one s orbital and one p orbital. Identify the bond type and justify your answer.

Show Answer

Mark scheme:

  • It is a σ bond.
  • The s and p orbitals overlap head-on along the internuclear axis.
  • A π bond would require sideways overlap of two parallel p orbitals.