Shapes & bond angles
Shapes & bond angles
Predict shape by counting electron regions
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The shape comes from electron regions repelling one another around the central atom. They arrange as far apart as possible. Count bonding regions and lone pairs before naming the shape.
- Find the central atom's outer-shell electron number: for example, 5 for N or 6 for O.
- For a suitable species with single bonds to H or halogens, add one electron for each attached atom.
- Add electrons for a negative charge and subtract them for a positive charge.
- Divide by two to get the total pairs. Subtract the bonding pairs to find the lone pairs.
- Choose the parent electron arrangement. Then name the molecular shape using the positions of the atoms.
The shortcut has limits. With multiple bonds or a supplied electron diagram, count regions directly: a single, double or triple bond is one bonding region. Lone pairs count as regions too.
- Bonding pair
- A pair of electrons shared between two covalently bonded atoms.
- Lone pair
- A pair of outer-shell electrons not used in a covalent bond.
- Electron region
- One direction of electron density: a single, double or triple bond counts as one region.
- Molecular shape
- The arrangement of atoms around the central atom.
- Bond angle
- The angle between two bonds meeting at the central atom.
| Total regions | Parent arrangement | Ideal angles |
|---|---|---|
| 2 | Linear | 180° |
| 3 | Trigonal planar | 120° |
| 4 | Tetrahedral | 109.5° |
| 5 | Trigonal bipyramidal | 90°, 120° and 180° |
| 6 | Octahedral | 90° and 180° |
Count the pairs in ammonia
Predict the shape of NH3.
State its bond angle and explain your prediction in terms of bonding pairs and lone pairs.
Worked answer
- Nitrogen contributes five outer electrons: (5 + 3)/2 = 4 pairs.
- Three pairs form N–H bonds, leaving one lone pair.
- Four regions have a tetrahedral parent arrangement; the lone pair compresses the bonds.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) State the electron-pair repulsion principle used to predict molecular shape. [2 marks]
(b) Place these repulsions in decreasing order of strength: bond pair-bond pair, lone pair-bond pair, lone pair-lone pair. [3 marks]
(c) Explain why replacing a bonded pair with a lone pair usually decreases a bond angle. [1 mark]
Total: 6 marks
Show answer & marking points
(a) Electron pairs around a central atom repel [1] and adopt positions of maximum separation / minimum repulsion [1]. [2]
(b) lone pair-lone pair [1] > lone pair-bond pair [1] > bond pair-bond pair [1]. [3]
(c) A lone pair occupies more space and repels bonding pairs more strongly, pushing bonds closer together [1]. [1]
Linear, trigonal planar and tetrahedral shapes
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When every electron region is a bond, the arrangement of the regions gives the molecular shape directly: two gives linear, three gives trigonal planar and four gives tetrahedral.
The bond angles are 180°, 120° and 109.5° respectively. Count a multiple bond as one region when deciding the shape.
A tetrahedral diagram needs depth. Use ordinary lines for bonds in the paper, a solid wedge towards you and a hashed wedge away. Keep equivalent bond lengths consistent and finish each bond at its atom label.
| Regions | Lone pairs | Shape | Angle | Examples |
|---|---|---|---|---|
| 2 | 0 | Linear | 180° | CO2, BeCl2 |
| 3 | 0 | Trigonal planar | 120° | BF3 |
| 4 | 0 | Tetrahedral | 109.5° | CH4, NH4+ |
Why is CO2 linear?
Each C=O bond in CO₂ is a double bond. Predict the molecular shape and O–C–O bond angle, and explain them using electron-pair repulsion.
Worked answer
Each double bond occupies one direction of electron density. The two regions reach maximum separation on opposite sides of carbon.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Complete the table by giving the molecular shape and bond angle(s). Names and angles only; diagrams are not required. [12]
| Species | Bonding pairs | Lone pairs | Shape | Bond angle(s) |
|---|---|---|---|---|
| BeCl2 | 2 | 0 | … | … |
| BF3 | 3 | 0 | … | … |
| CH4 | 4 | 0 | … | … |
| NH3 | 3 | 1 | … | … |
| H2O | 2 | 2 | … | … |
| SF6 | 6 | 0 | … | … |
Total: 12 marks
Show answer & marking points
BeCl2: linear [1], 180° [1]. BF3: trigonal planar [1], 120° [1].
CH4: tetrahedral [1], 109.5° [1]. NH3: pyramidal [1], 107° [1].
H2O: non-linear [1], 104.5° [1]. SF6: octahedral [1], 90° between adjacent bonds [1]. [12]
Lone pairs compress bond angles
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A lone pair takes up more space near the central atom than a bonding pair. It is attracted to one nucleus rather than shared between two nuclei. It repels nearby bonding regions more strongly and pushes them closer together.
CH₄, NH₃ and H₂O each have four electron regions around the central atom. Replacing bonds with lone pairs changes the positions of the atoms and therefore the molecular shape.
- CH₄: four bonds, no lone pairs; tetrahedral, 109.5°.
- NH₃: three bonds, one lone pair; pyramidal, about 107°.
- H₂O: two bonds, two lone pairs; bent/non-linear, about 104.5°.
Subtracting about 2.5° for each lone pair is a memory aid for this familiar comparison. It is not a universal formula for every molecule or parent arrangement.
Terminology: pyramidal and trigonal pyramidal are two names for this required shape. NH₃ is the common example, with a bond angle of about 107°.
| Molecule | Regions | Shape | Angle |
|---|---|---|---|
| NH3 | 3 bonding + 1 lone | Pyramidal (trigonal pyramidal) | 107° |
| H2O | 2 bonding + 2 lone | Non-linear (bent) | 104.5° |
Why is water’s angle smaller?
State the H–O–H bond angle in water and the H–N–H bond angle in ammonia. Explain why the angle in water is smaller.
Worked answer
- Water has two lone pairs; ammonia has one.
- Lone pairs repel more strongly than bonding pairs.
- The O–H bonds are pushed closer together.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) State the numbers of bonding pairs and lone pairs around the central atom in NH3 and H2O. [2 marks]
(b) For each molecule, state the shape name and draw a clear 3-D diagram. Also state its bond angle. [4 marks]
(c) Explain why the H-O-H angle is smaller than the H-N-H angle. [2 marks]
Total: 8 marks
Show answer & marking points
(a) NH3: three bonding pairs and one lone pair [1]. H2O: two bonding pairs and two lone pairs [1]. [2]
(b) NH3: correctly named and drawn pyramidal shape [1], 107° [1]. H2O: correctly named and drawn non-linear shape [1], 104.5° [1]. [4]
(c) H2O has two lone pairs rather than one [1]; stronger lone-pair repulsions compress H-O-H more than H-N-H [1]. [2]
Molecular ions and up to six regions
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Treat a molecular ion with the same method as a neutral molecule. Include the charge when counting electrons, then separate the bonding regions from the lone pairs.
Four regions give NH₄⁺ a tetrahedral shape because all four are bonds. In H₃O⁺, one region is a lone pair, so the atom arrangement is pyramidal.
The named shapes include linear, non-linear, trigonal planar, pyramidal (also called trigonal pyramidal), tetrahedral and octahedral.
Trigonal bipyramidal, as in PCl₅, is a different shape and is not in that named list. You must still apply electron-pair repulsion to molecules and ions with up to six electron pairs around the central atom.
| Species | Bonding + lone regions | Shape | Angles |
|---|---|---|---|
| NH4+ | 4 + 0 | Tetrahedral | 109.5° |
| H3O+ | 3 + 1 | Pyramidal | No fixed recall angle stated here |
| SF6 | 6 + 0 | Octahedral | 90° adjacent, 180° opposite |
| PCl5 | 5 + 0 | Trigonal bipyramidal | 120° equatorial; 90° axial-equatorial; 180° axial-axial |
Account for the charge in NH4+
Predict the shape of the ammonium ion.
State its bond angle and explain your prediction in terms of electron pairs.
Worked answer
- Electron total = 5 + 4 − 1 = 8; divide by two to get four pairs.
- All four pairs are bonding pairs, so they repel equally.
- Draw the ion in square brackets with an overall + charge.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
(a) PCl5 has five bonding pairs and no lone pairs around P. SF6 has six bonding pairs and no lone pairs around S. State the name of the shape and the bond angle(s) for each molecule. [6 marks]
(b) Draw a clear 3-D diagram of each molecule to match the shape names in part (a). Use wedge-and-dash bonds where needed and keep equivalent bonds similar in length. [2 marks]
Total: 8 marks
Show answer & marking points
(a) PCl5: trigonal bipyramidal [1], with 120° between equatorial bonds [1], and 90° between axial and equatorial bonds plus 180° between the two axial bonds [1]. SF6: octahedral [1], with 90° between adjacent bonds [1] and 180° between opposite bonds [1]. [6]
(b) One chemically correct 3-D drawing for each molecule, with the five or six bonds arranged correctly [1 each]. [2]