Molecules are not flat structures: they possess distinct three-dimensional shapes. The geometry of a molecule dictates its physical and chemical properties, such as polarity, boiling point, and how it interacts with other molecules. We predict these shapes using a simple but powerful model: Valence Shell Electron Pair Repulsion (VSEPR) theory.
🔑 Key Principle
The shape of a molecule is determined by the number of electron pairs (bonding pairs and lone pairs) surrounding the central atom. Because electrons are negatively charged, these pairs repel each other and arrange themselves to be as far apart as possible to minimize repulsion.
The VSEPR Rules
To predict the shape of any molecule or compound ion, follow these core rules:
- Identify the central atom and determine its group number to find its valence electrons.
- Add one electron for each single bond formed with surrounding atoms.
- If dealing with a charged ion, adjust the electron count: subtract 1 electron for each positive charge, and add 1 electron for each negative charge.
- Divide the total by 2 to find the number of electron pairs (or electron density regions) surrounding the central atom.
- Compare this total number of pairs to the number of bonded atoms to distinguish between bonding pairs and lone pairs.
A pair of valence electrons shared between two atoms, forming a covalent bond.
A pair of valence electrons on the central atom that is not shared with another atom.
Lone Pair Repulsion and Bond Angles
Not all electron pairs repel equally. A lone pair of electrons is held closer to the nucleus of the central atom than a bonding pair, which is stretched between two nuclei. Consequently, lone pairs exert a greater repelling force.
Relative Repulsion Strength
Lone pair : lone pair > Lone pair : bonding pair > Bonding pair : bonding pair
As a rule of thumb, each lone pair reduces the ideal bond angle by approximately 2.5°.
In written exam answers, when explaining why a molecule has a specific shape and bond angle, you must state that: (1) electron pairs repel to get as far apart as possible, (2) state the number of bonding pairs and lone pairs, and (3) explain that lone pairs repel more than bonding pairs. Missing the comparative repulsion statement is a common reason students lose marks.
Summary of Molecular Shapes
The table below summarizes the shapes and bond angles you are required to know for the AQA specification:
| Total Pairs | Bonding Pairs | Lone Pairs | Shape Name | Bond Angle | Example |
|---|---|---|---|---|---|
| 2 | 2 | 0 | Linear | 180° | \( \text{BeCl}_2 \), \( \text{CO}_2 \) |
| 3 | 3 | 0 | Trigonal Planar | 120° | \( \text{BF}_3 \), \( \text{AlCl}_3 \) |
| 4 | 4 | 0 | Tetrahedral | 109.5° | \( \text{CH}_4 \), \( \text{NH}_4^+ \) |
| 4 | 3 | 1 | Trigonal Pyramidal | 107° | \( \text{NH}_3 \), \( \text{H}_3\text{O}^+ \) |
| 4 | 2 | 2 | Bent (V-shaped) | 104.5° | \( \text{H}_2\text{O} \), \( \text{OF}_2 \) |
| 5 | 5 | 0 | Trigonal Bipyramidal | 120° (equatorial) / 90° (axial) | \( \text{PCl}_5 \) |
| 6 | 6 | 0 | Octahedral | 90° | \( \text{SF}_6 \) |
| 6 | 4 | 2 | Square Planar | 90° | \( \text{XeF}_4 \) |
How to Deduce the Shape and Bond Angle
Deducing the shapes of molecules and compound ions is a core calculation skill. Here are three step-by-step worked examples to master this method.
Step-by-step solution:
- Identify the central atom: Carbon is the central atom. Carbon is in Group 4, so it has 4 valence electrons.
- Add electrons for bonds: It forms 4 single covalent bonds with 4 hydrogen atoms, adding 4 electrons.
- Account for charge: The molecule is neutral, so no adjustment is needed. Total electrons = 8.
- Divide by 2 to find electron pairs: \( 8 / 2 = 4 \) electron pairs.
- Determine bonding and lone pairs: There are 4 surrounding hydrogen atoms. Since there are 4 electron pairs and 4 bonds, all 4 pairs are bonding pairs and there are 0 lone pairs.
- Deduce shape and angle: 4 bonding pairs and 0 lone pairs repel equally to form a regular tetrahedral shape with a bond angle of 109.5°.
Step-by-step solution:
- Identify the central atom: Nitrogen is the central atom. Nitrogen is in Group 5, so it has 5 valence electrons.
- Add electrons for bonds: It forms 3 single bonds with 3 hydrogen atoms, adding 3 electrons.
- Account for charge: The molecule is neutral. Total electrons = 8.
- Divide by 2: \( 8 / 2 = 4 \) electron pairs.
- Determine bonding and lone pairs: There are 3 surrounding hydrogen atoms. Therefore, there are 3 bonding pairs and \( 4 - 3 = 1 \) lone pair.
- Deduce shape and angle: The 4 electron pairs arrange themselves in a tetrahedral geometry. However, the presence of 1 lone pair reduces the bond angles due to greater repulsion. The molecular shape is trigonal pyramidal. The lone pair reduces the ideal tetrahedral angle of 109.5° by about 2.5° to 107°.
Step-by-step solution:
- Identify the central atom: Nitrogen is in Group 5 (5 valence electrons).
- Add electrons for bonds: 4 bonds with hydrogen atoms (add 4 electrons).
- Account for charge: The ion has a 1+ charge. A positive charge means the ion has lost 1 electron. Subtract 1 electron. Total electrons = \( 5 + 4 - 1 = 8 \).
- Divide by 2: \( 8 / 2 = 4 \) electron pairs.
- Determine bonding and lone pairs: There are 4 surrounding hydrogen atoms. Thus, there are 4 bonding pairs and 0 lone pairs.
- Deduce shape and angle: 4 bonding pairs and 0 lone pairs form a tetrahedral geometry. Because there are no lone pairs, the shape is a regular tetrahedron with a bond angle of 109.5°.
Methane (\( \text{CH}_4 \)), ammonia (\( \text{NH}_3 \)), and water (\( \text{H}_2\text{O} \)) all have a total of 4 electron pairs around the central atom. This is a common comparison question on exams. Always start by stating they all have 4 electron pairs, then list their bonding and lone pairs, and explain the differences in their bond angles (109.5° vs 107° vs 104.5°) in terms of comparative lone pair repulsion.
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