What are the key differences between σ and π bonds in organic molecules?

In AS-Level Organic Chemistry, σ (sigma) bonds are single covalent bonds formed by the head-on overlap of atomic orbitals, allowing free rotation around the bond axis. In contrast, π (pi) bonds are formed by the side-to-side overlap of p orbitals, which restricts rotation due to the electron cloud above and below the bond axis. Understanding these differences is crucial for predicting the shapes and reactivity of organic molecules.

How do σ and π bonds influence the shape of organic molecules?

The shape of organic molecules is significantly influenced by the presence of σ and π bonds. σ bonds allow for the free rotation of atoms, contributing to the flexibility of the molecule, while π bonds restrict rotation, leading to fixed geometries. For example, in alkenes, the presence of a π bond results in a planar structure, which is an important concept in AS-Level Organic Chemistry.

Why is the concept of hybridization important in understanding the shapes of organic molecules?

Hybridization is a key concept in AS-Level Organic Chemistry that explains the formation of σ and π bonds and the resulting molecular shapes. It involves the mixing of atomic orbitals to form new hybrid orbitals, which determine the geometry of the molecule. For instance, sp3 hybridization leads to a tetrahedral shape, sp2 to a trigonal planar shape, and sp to a linear shape, each affecting the molecule's properties and reactivity.

How do σ and π bonds affect the reactivity of organic molecules?

In AS-Level Organic Chemistry, σ bonds are generally stronger and less reactive than π bonds. π bonds, due to their electron-rich nature and weaker overlap, are more reactive and often participate in chemical reactions such as addition reactions in alkenes. Understanding the reactivity of these bonds helps predict how organic molecules will behave in different chemical environments.

What role do σ and π bonds play in the stability of organic molecules?

The stability of organic molecules is influenced by the presence of σ and π bonds. σ bonds, being stronger due to direct orbital overlap, contribute to the overall stability of the molecule. However, π bonds, while weaker, provide additional stability through resonance in conjugated systems. This balance between σ and π bonds is a fundamental concept in AS-Level Organic Chemistry, affecting molecular stability and reactivity.

Why is "Giving the wrong bond angle for the hybridisation." a common mistake in Shapes of organic molecules; σ and π bonds?

Why it happens: Not linking shape to angle. How to avoid it: sp³ 109.5°, sp² 120°, sp 180°. Source: 9701 Examiner Reports 2022-2024

Why is "Saying a C=C can rotate freely." a common mistake in Shapes of organic molecules; σ and π bonds?

Why it happens: Treating it like a single bond. How to avoid it: The π bond prevents rotation about C=C (allowing cis–trans isomers).

Why is "Counting the π bond when assigning hybridisation." a common mistake in Shapes of organic molecules; σ and π bonds?

Why it happens: Counting all bonds. How to avoid it: Hybridisation depends on σ bonds (+ lone pairs) only.

Why is "Saying the π bond is stronger than the σ bond." a common mistake in Shapes of organic molecules; σ and π bonds?

Why it happens: Assuming the 'extra' bond is the strong one. How to avoid it: The σ bond (direct overlap) is stronger; the π bond (sideways overlap) is weaker and more reactive.

An introduction to AS Level organic chemistry(AS-Level Organic Chemistry)Cambridge International A Level Chemistry 9701

Shapes of organic molecules; σ and π bonds

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Shapes of Organic Molecules; σ and π Bonds — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

How hybridisation (sp³, sp², sp) sets the shape and bond angles around carbon, the σ/π arrangement in single, double and triple bonds, and the meaning of a planar molecule such as ethene.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

13.3.1 — Describe organic molecules as either straight-chained, branched or cyclic.
13.3.2 — Describe and explain the shape of, and bond angles in, molecules containing sp, sp² and sp³ hybridised atoms.
13.3.3 — Describe the arrangement of σ and π bonds in molecules containing sp, sp² and sp³ hybridised atoms.
13.3.4 — Understand and use the term planar when describing the arrangement of atoms in organic molecules, for example ethene.

Straight-chained, branched and cyclic

Carbon chains can run straight, branch off, or close into a ring.

Carbon skeletons take three basic forms:

Straight-chained — carbons in a single continuous chain (e.g. butane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ ).
Branched — a chain with side groups (e.g. 2-methylpropane).
Cyclic — carbons joined in a ring (e.g. cyclohexane).

The shape around each carbon depends on its hybridisation, regardless of the overall chain type.

Straight-chained, branched, or cyclic.
Local shape depends on each carbon's hybridisation.

sp³, sp² and sp carbons: shapes and σ/π bonds

Count σ bonds: 4 = sp³ tetrahedral; 3 = sp² trigonal planar; 2 = sp linear.

Hybridisation (mixing s and p orbitals) sets the geometry around carbon and how σ/π bonds are arranged:

Hybridisation	σ bonds	Shape	Bond angle	π bonds	Example
sp³	4	tetrahedral	109.5°	0	alkane C, CH₄
sp²	3	trigonal planar	120°	1 (in C=C)	alkene C, ethene
sp	2	linear	180°	2 (in C≡C/C≡N)	alkyne/nitrile C

σ and π bonds (recap of 3.4):

A single bond is one σ bond (end-on overlap) — sp³ carbon.
A double bond is one σ + one π (sideways p-orbital overlap) — sp² carbon.
A triple bond is one σ + two π — sp carbon.

The leftover unhybridised p orbital(s) form the π bond(s); the hybrid orbitals form the σ framework.

sp³ 4σ tetrahedral 109.5°; sp² 3σ planar 120°; sp 2σ linear 180°.
Single = σ; double = σ+π; triple = σ+2π.
Unhybridised p orbitals make the π bonds.

Planarity — the case of ethene

The C=C and everything attached lie flat in one plane because of sp² carbons.

A molecule (or part of one) is planar if all the relevant atoms lie in the same flat plane.

Ethene, C₂H₄, is planar: both carbons are sp² hybridised (trigonal planar, 120°), so all six atoms (2 C + 4 H) lie in one plane. The π bond (sideways overlap of the two p orbitals above and below the plane) locks the geometry and prevents rotation about the C=C.

This planarity and restricted rotation are the reason alkenes can show cis–trans (geometrical) isomerism (Topic 13.4).

Planar = all atoms in one plane.
Ethene planar: sp² carbons, all 6 atoms coplanar.
π bond prevents rotation about C=C.

How it’s examined

Stating hybridisation, shape and bond angle for a named carbon, and explaining ethene's planarity, are common Paper 1/2 marks. Examiner reports flag giving 120° for an sp³ carbon and not linking restricted rotation to the π bond.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 13.3; 9701 Examiner Reports 2022-2024; 9701/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-20.

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Step-by-step worked examples — Shapes of organic molecules; σ and π bonds

Step-by-step solutions to past-paper-style questions on shapes of organic molecules; σ and π bonds, written exactly the way a tutor would explain them at the board.

1σ and π bonds in a molecule (2 marks)
Getting started• sigma pi
Question
State the number of σ and π bonds in ethyne, HC≡CH.
Step-by-step solution
1. Step 1
  Two C–H σ bonds + one C–C σ bond = 3 σ; the triple bond adds 2 π.
Answer
3 σ bonds and 2 π bonds.
Examiner tip
Every bond contributes one σ; the extra bonds of a double/triple are π.
2Shape around a saturated carbon (2 marks)
Getting started• hybridisation
Question
State the hybridisation, shape and bond angle around a carbon atom in ethane, C₂H₆.
Step-by-step solution
1. Step 1
  Each C has 4 σ bonds, no π → sp³.
2. Step 2
  4 bonding regions → tetrahedral, 109.5°.
Answer
sp³, tetrahedral, 109.5°.
3Hybridisation and shape in propene (3 marks)
Building confidence• hybridisation
Question
For propene, CH₃–CH=CH₂, state the hybridisation, shape and bond angle around the CH₃ carbon and around a C=C carbon.
Step-by-step solution
1. Step 1
  CH₃ carbon: 4 σ bonds → sp³ → tetrahedral, 109.5°.
2. Step 2
  C=C carbon: 3 σ + 1 π → sp² → trigonal planar, 120°.
Answer
CH₃: sp³, tetrahedral, 109.5°. C=C: sp², trigonal planar, 120°.
Examiner tip
Count σ bonds (+ lone pairs) for hybridisation — the π bond does NOT count.
4Why ethene is planar (3 marks)
Building confidence• planar
Question
Explain why ethene is a planar molecule.
Step-by-step solution
1. Step 1
  Both carbons are sp² hybridised (3 σ bonds each, 120°).
2. Step 2
  This trigonal planar geometry places all six atoms in one plane.
3. Step 3
  The π bond above and below the plane locks this flat shape.
Answer
sp² carbons → trigonal planar → all 6 atoms coplanar; π bond fixes the geometry.
5Restricted rotation about C=C (3 marks)
Stretch• restricted rotation
Question
Explain why there is no rotation about a C=C double bond, and state one consequence. (3)
Step-by-step solution
1. Step 1
  The π bond is sideways overlap of two p orbitals above and below the σ bond.
2. Step 2
  Rotating one carbon would break this overlap (it needs the p orbitals parallel), which costs too much energy.
3. Step 3
  Consequence: groups are fixed on each side → cis–trans (geometrical) isomerism is possible.
Answer
Rotation would break the π overlap, so the C=C is rigid → allows cis–trans isomerism.
Examiner tip
Link 'no rotation' to the π overlap, then to the possibility of cis–trans isomers.
6Orbital overlap in a C=C bond (4 marks)
Stretch• sigma pi, overlap
Question
Describe the bonding in a carbon–carbon double bond in terms of orbital overlap. (4)
Step-by-step solution
1. Step 1
  Each carbon is sp² hybridised, with one unhybridised p orbital.
2. Step 2
  The σ bond forms by direct (end-on) overlap of an sp² orbital from each carbon.
3. Step 3
  The π bond forms by sideways overlap of the two unhybridised p orbitals.
4. Step 4
  The π electron density lies above and below the σ axis and makes the C=C shorter and stronger than a C–C single bond.
Answer
C=C = one σ (sp²–sp² end-on overlap) + one π (sideways p–p overlap above/below the σ axis).
Examiner tip
Name the orbitals: sp² for the σ, unhybridised p for the π.

Key Formulae — Shapes of organic molecules; σ and π bonds

The formulae you need to memorise for shapes of organic molecules; σ and π bonds on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Hybridisation from σ bonds
$(\sigma\text{ bonds at C}) = 4\,(sp^3),\ 3\,(sp^2),\ 2\,(sp)$
$σ bonds at C$
number of σ bonds around the carbon
When to use
Assigning shape/angle: sp³ tetrahedral 109.5°, sp² trigonal planar 120°, sp linear 180°.

Key Definitions and Keywords — Shapes of organic molecules; σ and π bonds

Definitions to memorise and the exact keywords mark schemes credit for shapes of organic molecules; σ and π bonds answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

σ and π bonds
Examiner keyword
σ bond: direct end-on orbital overlap. π bond: sideways overlap of p orbitals, with density above and below the σ axis.
Hybridisation (organic)
Examiner keyword
Mixing of s and p orbitals on carbon: sp³ (tetrahedral, 109.5°), sp² (trigonal planar, 120°), sp (linear, 180°).
Planar
Examiner keyword
A molecule (or part of one) in which all the relevant atoms lie in the same flat plane (e.g. around a C=C).
Restricted rotation
Examiner keyword
The lack of rotation about a C=C bond because turning would break the π-orbital overlap; it allows cis–trans isomerism.

Common Mistakes and Misconceptions — Shapes of organic molecules; σ and π bonds

The traps other students keep falling into on shapes of organic molecules; σ and π bonds questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Giving the wrong bond angle for the hybridisation.
9701 Examiner Reports 2022-2024
Why it happens
Not linking shape to angle.
How to avoid it
sp³ 109.5°, sp² 120°, sp 180°.
✕Saying a C=C can rotate freely.
Why it happens
Treating it like a single bond.
How to avoid it
The π bond prevents rotation about C=C (allowing cis–trans isomers).
✕Counting the π bond when assigning hybridisation.
Why it happens
Counting all bonds.
How to avoid it
Hybridisation depends on σ bonds (+ lone pairs) only.
✕Saying the π bond is stronger than the σ bond.
Why it happens
Assuming the 'extra' bond is the strong one.
How to avoid it
The σ bond (direct overlap) is stronger; the π bond (sideways overlap) is weaker and more reactive.

Practice questions

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Past paper style quiz

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Shapes of organic molecules; σ and π bonds — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Hybridisation

σ bonds

Shape

Bond angle

π bonds

Example

sp³

tetrahedral

109.5°

alkane C, CH₄

sp²

trigonal planar

120°

1 (in C=C)

alkene C, ethene

linear

180°

2 (in C≡C/C≡N)

alkyne/nitrile C

Shapes of organic molecules; σ and π bonds

Short Study Notes in the form of Slides

Short Study Notes — Shapes of organic molecules; σ and π bonds

Detailed Notes

What you’ll learn

How it’s examined

1σ and π bonds in a molecule (2 marks)

2Shape around a saturated carbon (2 marks)

3Hybridisation and shape in propene (3 marks)

4Why ethene is planar (3 marks)

5Restricted rotation about C=C (3 marks)

6Orbital overlap in a C=C bond (4 marks)

Hybridisation from σ bonds

σ and π bonds

Hybridisation (organic)

Planar

Restricted rotation

✕Giving the wrong bond angle for the hybridisation.

✕Saying a C=C can rotate freely.

✕Counting the π bond when assigning hybridisation.

✕Saying the π bond is stronger than the σ bond.

Practice questions

Past paper style quiz

Assess your understanding

Shapes of organic molecules; σ and π bonds — frequently asked questions

Shapes of organic molecules; σ and π bonds

Short Study Notes in the form of Slides

Short Study Notes — Shapes of organic molecules; σ and π bonds

Detailed Notes

What you’ll learn

How it’s examined

1σ and π bonds in a molecule (2 marks)

2Shape around a saturated carbon (2 marks)

3Hybridisation and shape in propene (3 marks)

4Why ethene is planar (3 marks)

5Restricted rotation about C=C (3 marks)

6Orbital overlap in a C=C bond (4 marks)

Hybridisation from σ bonds

σ and π bonds

Hybridisation (organic)

Planar

Restricted rotation

✕Giving the wrong bond angle for the hybridisation.

✕Saying a C=C can rotate freely.

✕Counting the π bond when assigning hybridisation.

✕Saying the π bond is stronger than the σ bond.

Practice questions

Past paper style quiz

Assess your understanding

Shapes of organic molecules; σ and π bonds — frequently asked questions