# Organic molecule shapes

> CIE A-Level Chemistry · Introduction to organic chemistry
> Source: https://www.owlsprep.com/study/cie-9701-u12-organic-molecule-shapes/

This sub-topic applies VSEPR theory to predict the 3D shape and bond angles of organic molecules, based on carbon hybridization and electron domains. You will also learn to count and distinguish sigma and pi bonds across common functional groups.

**Prerequisites:** [Basic VSEPR theory](https://www.owlsprep.com/study/cie-9701-u03-shapes-of-molecules/); Carbon hybridization basics; Drawing organic structural formulae

## Learning objectives

- Predict the 3D shape and bond angles around carbon atoms in organic molecules
- Relate carbon hybridization to shape using VSEPR theory
- Count the number of sigma and pi bonds in any organic molecule
- Explain how hybridization influences molecular planarity

## Shapes around sp³ hybridized carbon

Any carbon atom that forms four single covalent bonds is sp³ hybridized. It has four bonding electron domains and no lone pairs of electrons. VSEPR theory states that electron domains repel each other to arrange as far apart as possible.

**sp³ hybridization** — Hybridization of one 2s and three 2p orbitals to form four equivalent hybrid orbitals, each forming one sigma bond.

*Notation:* sp³

*Example:* Carbon in alkanes, alcohols, and haloalkanes

**Worked example:** Predict the shape and ideal bond angle around the central carbon in 2-methylpropane, $(CH_3)_3CH$.

1. Count electron domains around the central carbon: 4 single bonds, no lone pairs = 4 electron domains.
2. All electron domains are bonding, so VSEPR repulsion produces a tetrahedral arrangement.
3. The ideal bond angle for four equal bonding domains is 109.5°.

> **exam_tip**
>
> Always assume tetrahedral geometry for any neutral carbon with four single bonds. Lone pairs on carbon are extremely rare in stable neutral organic molecules.

## Shapes around sp² hybridized carbon

A carbon atom with one double bond and two single bonds is sp² hybridized. It has three bonding electron domains, with one unhybridized p orbital that forms the pi bond of the double bond. All atoms bonded directly to the sp² carbon lie in the same plane.

**sp² hybridization** — Hybridization of one 2s and two 2p orbitals to form three equivalent hybrid orbitals, leaving one unhybridized p orbital for pi bond formation.

*Notation:* sp²

*Example:* Carbon in alkenes, carbonyl groups, and benzene

**Worked example:** Predict the shape and bond angle around the carbonyl carbon in propanone, $CH_3COCH_3$.

1. The carbonyl carbon bonds to two methyl groups via single bonds and oxygen via a double bond.
2. Count electron domains: double bonds count as one domain, so 3 bonding domains, no lone pairs.
3. Three equal bonding domains arrange to give trigonal planar geometry.
4. The ideal bond angle around the carbonyl carbon is 120°.

## Shapes around sp hybridized carbon

A carbon atom with one triple bond and one single bond, or two separate double bonds, is sp hybridized. It has two bonding electron domains, with two unhybridized p orbitals that form two pi bonds.

**sp hybridization** — Hybridization of one 2s and one 2p orbital to form two equivalent hybrid orbitals, leaving two unhybridized p orbitals for two pi bonds.

*Notation:* sp

*Example:* Carbon in alkynes and carbon dioxide

**Worked example:** Predict the shape and bond angle around each carbon in ethyne, $C_2H_2$.

1. Each carbon bonds to one hydrogen via a single bond and the other carbon via a triple bond.
2. Each carbon has two bonding electron domains, no lone pairs.
3. Two electron domains arrange to give linear geometry.
4. The bond angle around each carbon is 180°, so the entire ethyne molecule is linear.

## Identifying sigma and pi bonds

Every covalent connection between two atoms contains exactly one sigma bond. Any additional bonds between the same pair of atoms are pi bonds. The rule is: single bond = 1 sigma, double bond = 1 sigma + 1 pi, triple bond = 1 sigma + 2 pi.

**Worked example:** Count the total number of sigma and pi bonds in propene, $CH_3CHCH_2$.

1. Draw the full structural formula to show all bonds: 8 C-H single bonds, 2 C-C single bonds, 1 C=C double bond.
2. Count sigma bonds: every bond contributes one sigma, so 8 + 2 + 1 = 11 sigma bonds.
3. Count pi bonds: only the double bond has one extra pi bond, so 1 pi bond.
4. Final answer: 11 sigma bonds and 1 pi bond.

**Check your understanding**

Test your understanding

1. How many sigma and pi bonds are in a benzene molecule $C_6H_6$?

   - A: 6 sigma, 3 pi
   - B: 12 sigma, 3 pi
   - C: 12 sigma, 6 pi
   - D: 9 sigma, 3 pi

   *Why:* Benzene has 6 C-H single bonds and 6 C-C bonds in the ring, giving 12 sigma bonds total. The three delocalized double bonds each contribute one pi bond, for 3 pi total.

## Common pitfalls

- **Wrong:** Claiming the bond angle around sp³ carbon is always exactly 109.5°
  - Why it fails: Lone pairs, electronegative groups, and ring strain in cyclic alkanes distort bond angles from the ideal value
  - Correct: State 109.5° as the ideal bond angle, unless the question specifically asks for distorted values
- **Wrong:** Counting a double bond as two sigma bonds
  - Why it fails: Students forget multiple bonds only contain one sigma bond, with extra bonds being pi
  - Correct: Follow the rule: 1 sigma per covalent connection, 1 pi per double bond, 2 pi per triple bond
- **Wrong:** Predicting tetrahedral shape for a carbonyl carbon
  - Why it fails: Students incorrectly count double bonds as two electron domains in VSEPR
  - Correct: Any carbon with a double bond has 3 electron domains, so it is trigonal planar with ~120° bond angles
- **Wrong:** Claiming all atoms in propene are planar
  - Why it fails: Students forget the methyl carbon is sp³ hybridized, so its bonds are not planar
  - Correct: Only atoms bonded directly to sp² carbons are planar; sp³ carbons retain tetrahedral geometry

## Cheatsheet

| Hybridization | Electron domains | Shape | Ideal bond angle | Sigma bonds per C | Pi bonds per C |
| --- | --- | --- | --- | --- | --- |
| sp³ (4 single bonds) | 4 | Tetrahedral | 109.5° | 4 | 0 |
| sp² (1 double + 2 single) | 3 | Trigonal planar | 120° | 3 | 1 |
| sp (1 triple + 1 single) | 2 | Linear | 180° | 2 | 2 |

## What's next

Understanding the 3D shape of organic molecules is foundational for all further organic chemistry topics you will study. Molecular shape determines the reactivity of functional groups, enables different types of stereoisomerism (including cis-trans and optical isomerism), and governs how biological molecules like enzymes and drug molecules interact. You will rely on this core knowledge when learning reaction mechanisms, where the spatial arrangement of bonds directly controls how nucleophiles and electrophiles approach and react. Next, you will build on this understanding to explore stereoisomerism and apply shape concepts to aromatic compounds and their delocalized bonding.

- [Hydrocarbons](https://www.owlsprep.com/study/cie-9701-u13-overview/)
- [Alkanes](https://www.owlsprep.com/study/cie-9701-u13-alkanes/)
- [Alkenes](https://www.owlsprep.com/study/cie-9701-u13-alkenes/)

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