Shapes of molecules and ions

Q: What shape and bond angle are formed by a molecule with 2 bonding pairs and 0 lone pairs?

Linear with a bond angle of $180^\circ$.

Q: What shapes and bond angles result from 3 electron pairs with 0 lone pairs versus 1 lone pair?

3 BP, 0 LP gives trigonal planar ($120^\circ$); 2 BP, 1 LP gives bent / non-linear (approx. $117.5^\circ$ to $119^\circ$).

Q: How do shapes and bond angles vary within the 4-electron-pair tetrahedral family as lone pairs increase?

4 BP, 0 LP is tetrahedral ($109.5^\circ$); 3 BP, 1 LP is pyramidal ($107^\circ$); 2 BP, 2 LP is bent ($104.5^\circ$).

Q: What are the shapes and bond angles for molecules with 5 electron pairs (0, 1, and 2 lone pairs)?

5 BP, 0 LP: trigonal bipyramidal ($90^\circ$, $120^\circ$, $180^\circ$); 4 BP, 1 LP: see-saw ($ T-shaped (\(<90^\circ$).

Q: What shapes and bond angles are observed for molecules with 6 electron pairs?

6 BP, 0 LP is octahedral ($90^\circ$, $180^\circ$); 4 BP, 2 LP is square planar ($90^\circ$, $180^\circ$).

Q: Why does the bond angle increase when ammonia ($\text{NH}_3$) reacts with $\text{H}^+$ to form the ammonium ion ($\text{NH}_4^+$)?

The lone pair in $\text{NH}_3$ forms a coordinate bond, converting 3 BP / 1 LP (pyramidal, $107^\circ$) into 4 BP / 0 LP (tetrahedral, $109.5^\circ$), because $\text{BP–BP}$ repulsion is weaker than $\text{LP–BP}$ repulsion.

Chemistry 1110 · CCEA AS Level · 9 min read · Updated: 31 Aug 2026

Welcome to Shapes of Molecules and Ions

Ever wondered why water $\text{H}_2\text{O}$ is bent like a boomerang, but carbon dioxide $\text{CO}_2$ is straight as a rod? The physical shape of a molecule influences everything from whether a substance is a liquid or gas, to how enzymes bind in our bodies. In this chapter of AS 1: Basic Concepts in Physical and Inorganic Chemistry, we will explore the rules that govern three-dimensional molecular structures. Don't worry if 3D shapes seem tricky to picture at first; once you learn the step-by-step method, predicting shapes becomes straightforward and predictable!

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1. Valence Shell Electron Pair Repulsion (VSEPR) Theory

To predict molecular shapes, chemists use a simple and powerful model called Valence Shell Electron Pair Repulsion (VSEPR) theory.

The Fundamental Idea

• Electron pairs repel: All electrons carry a negative charge. In the outer (valence) shell of a central atom, electron pairs repel each other strongly.
• Maximum separation: Because they repel, these electron pairs spread out in three dimensions to get as far apart from each other as possible, minimising mutual repulsion.

Everyday Analogy: Imagine tying identical inflated balloons together by their nozzles. They naturally push against one another and point outwards in distinct geometric patterns. Electron pairs behave in the exact same way!

The Repulsion Hierarchy: Bonding Pairs vs. Lone Pairs

Not all electron pairs repel with the same strength. We classify outer electron pairs into two types:

1. Bonding Pairs (BP): Electron pairs shared between the central atom and bonded atoms. These electrons are held between two nuclei.
2. Lone Pairs (LP): Unshared electron pairs residing entirely on the central atom. Because they are attracted to only one positive nucleus, lone pairs sit closer to the central atom's nucleus and occupy more orbital space (a greater angular volume).

Because lone pairs take up more space, they exert a stronger repulsive force on neighbouring pairs. CCEA examiners expect you to know and state this exact hierarchy:

$$\text{Lone pair–Lone pair (LP–LP)} > \text{Lone pair–Bonding pair (LP–BP)} > \text{Bonding pair–Bonding pair (BP–BP)}$$

The result: A lone pair pushes adjacent bonding pairs closer together, reducing the bond angle between them.

Special Rules to Remember

• Multiple Covalent Bonds: Double and triple bonds (such as the $C=O$ double bonds in $\text{CO}_2$) are treated as a single region/centre of negative charge when determining the basic shape.
• Octet Rule and Expansion: While many central atoms obey the octet rule (surrounded by 8 outer electrons / 4 pairs, like $\text{CH}_4$), some are electron-deficient with fewer than 8 electrons (such as $\text{BeCl}_2$ with 4, or $\text{BF}_3$ with 6), while others can expand their octet up to 10 or 12 electrons (5 or 6 electron pairs, such as $\text{PCl}_5$ or $\text{SF}_6$).

Key Takeaway: The final 3D shape of any molecule or ion depends on the total number of electron pairs around the central atom and the balance between bonding pairs and lone pairs.

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2. The Step-by-Step Method to Predict Any Shape

Follow this reliable 5-step method to deduce the shape and bond angle of any molecule or ion:

Step 1: Identify the central atom and determine its group number in the Periodic Table (this gives its number of valence electrons).
Step 2: Account for ionic charge:
• For a positive ion, subtract one electron for each $+$ charge.
• For a negative ion, add one electron for each $-$ charge.
Step 3: Count the bonded atoms (each single bond contributes 1 electron from the surrounding atom).
Step 4: Calculate total electron pairs: Add the electrons together and divide by 2.
Step 5: Determine bonding pairs (BP) and lone pairs (LP): The number of bonded atoms equals the number of BP. Any remaining pairs are LP.

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3. Standard Geometries, Bond Angles, and Benchmark Examples

A. Molecules with 2 Electron Pairs

• 2 Bonding Pairs, 0 Lone Pairs (2 BP, 0 LP)
Shape Name: Linear
Bond Angle: $180^\circ$
Examples: $\text{BeCl}_2$, $\text{CO}_2$, $\text{HCN}$, $\text{NO}_2^+$
Why? Two electron regions push as far apart as possible, sitting on opposite sides of the central atom in a straight line.

B. Molecules with 3 Electron Pairs

• 3 Bonding Pairs, 0 Lone Pairs (3 BP, 0 LP)
Shape Name: Trigonal planar
Bond Angle: $120^\circ$
Examples: $\text{BF}_3$, $\text{AlCl}_3$, $\text{NO}_3^-$, $\text{CO}_3^{2-}$
Why? Three pairs spread out flat in a single plane, pointing to the corners of an equilateral triangle.

• 2 Bonding Pairs, 1 Lone Pair (2 BP, 1 LP)
Shape Name: Bent / Non-linear / V-shaped
Bond Angle: $\approx 117.5^\circ\text{ to }119^\circ$
Examples: $\text{SO}_2$, $\text{NO}_2^-$
Why? The lone pair repels the two bonding pairs more strongly than they repel each other, squeezing the angle down slightly from $120^\circ$.

C. Molecules with 4 Electron Pairs (The Tetrahedral Family)

When there are 4 electron pairs, the pairs point to the corners of a three-dimensional tetrahedron.

• 4 Bonding Pairs, 0 Lone Pairs (4 BP, 0 LP)
Shape Name: Tetrahedral
Bond Angle: $109.5^\circ$
Examples: $\text{CH}_4$, $\text{CCl}_4$, $\text{NH}_4^+$, $\text{SO}_4^{2-}$

• 3 Bonding Pairs, 1 Lone Pair (3 BP, 1 LP)
Shape Name: Pyramidal / Trigonal pyramidal
Bond Angle: $107^\circ$
Examples: $\text{NH}_3$, $\text{H}_3\text{O}^+$, $\text{PCl}_3$, $\text{N(CH}_3)_3$

• 2 Bonding Pairs, 2 Lone Pairs (2 BP, 2 LP)
Shape Name: Bent / Non-linear / V-shaped
Bond Angle: $104.5^\circ$
Examples: $\text{H}_2\text{O}$, $\text{H}_2\text{S}$, $\text{NH}_2^-$, $\text{OCl}_2$

The $2.5^\circ$ Rule of Thumb

Notice how the bond angle decreases systematically across the tetrahedral family:
$$\text{Methane (CH}_4\text{, 0 LP)} = 109.5^\circ \xrightarrow{-2.5^\circ} \text{Ammonia (NH}_3\text{, 1 LP)} = 107^\circ \xrightarrow{-2.5^\circ} \text{Water (H}_2\text{O, 2 LP)} = 104.5^\circ$$
Every lone pair replaces a bonding pair and compresses the remaining bond angles by approximately $2.5^\circ$ due to greater $\text{LP–BP}$ repulsion.

D. Molecules with 5 Electron Pairs (Expanded Octet)

• 5 Bonding Pairs, 0 Lone Pairs (5 BP, 0 LP)
Shape Name: Trigonal bipyramidal
Bond Angles: $90^\circ$ (axial–equatorial), $120^\circ$ (equatorial–equatorial), and $180^\circ$ (axial–axial)
Examples: $\text{PCl}_5$, $\text{PF}_5$
Structure: Three bonds form a flat triangle around the equator (equatorial), while one bond points straight up and one points straight down along the vertical axis (axial).

• 4 Bonding Pairs, 1 Lone Pair (4 BP, 1 LP)
Shape Name: See-saw
Bond Angles: $< 90^\circ$ (approx. $88^\circ$) and $< 120^\circ$ (approx. $118^\circ$)
Examples: $\text{SF}_4$

• 3 Bonding Pairs, 2 Lone Pairs (3 BP, 2 LP)
Shape Name: T-shaped
Bond Angles: $\approx 86^\circ\text{ to }88^\circ$ ($< 90^\circ$)
Examples: $\text{ClF}_3$, $\text{BrF}_3$

E. Molecules with 6 Electron Pairs (Expanded Octet)

• 6 Bonding Pairs, 0 Lone Pairs (6 BP, 0 LP)
Shape Name: Octahedral
Bond Angles: $90^\circ$ (and $180^\circ$)
Examples: $\text{SF}_6$, $[\text{AlF}_6]^{3-}$, $[\text{PCl}_6]^-$
Why octahedral? Six bonds point symmetrically along the x, y, and z axes to the six vertices of an octahedron.

• 4 Bonding Pairs, 2 Lone Pairs (4 BP, 2 LP)
Shape Name: Square planar
Bond Angles: $90^\circ$ (and $180^\circ$)
Examples: $\text{XeF}_4$, $[\text{ICl}_4]^-$
Why? To minimise repulsion, the two lone pairs sit directly opposite each other at $180^\circ$ (axial positions), leaving the four bonding pairs in a flat square arrangement.

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4. Crucial Polyatomic Ions & Coordinate Bonding

Examiners frequently ask you to explain how shape and bond angles change when coordinate (dative covalent) bonds form.

1. The Ammonium Ion ($\text{NH}_4^+$)

• Formed when an ammonia molecule ($\text{NH}_3$) donates its lone pair to a hydrogen ion ($\text{H}^+$).
• In $\text{NH}_3$, there are 3 BP and 1 LP (pyramidal, $107^\circ$).
• In $\text{NH}_4^+$, the lone pair becomes a bonding pair. There are now 4 BP and 0 LP.
• The shape changes to tetrahedral and the bond angle increases from $107^\circ$ to $109.5^\circ$ because $\text{BP–BP}$ repulsion is weaker than the original $\text{LP–BP}$ repulsion.

2. The Hydroxonium Ion ($\text{H}_3\text{O}^+$)

• Formed when water ($\text{H}_2\text{O}$) acts as a base and donates a lone pair to form a coordinate bond with $\text{H}^+$.
• In $\text{H}_2\text{O}$, there are 2 BP and 2 LP (bent, $104.5^\circ$).
• In $\text{H}_3\text{O}^+$, one lone pair is converted into a bonding pair, giving 3 BP and 1 LP.
• The shape changes to pyramidal and the bond angle increases from $104.5^\circ$ to $107^\circ$.

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5. 3D Drawing Conventions (Wedge-and-Dash Notation)

When drawing molecules in exams, use standard wedge-and-dash conventions to show 3D depth:

• Solid Line (—): Represents a covalent bond lying in the plane of the paper.
• Solid Wedge: Represents a covalent bond coming forward out of the paper towards you.
• Hashed / Dashed Wedge: Represents a covalent bond pointing backwards into the paper away from you.
• Polyatomic Ions: Always draw the complete 3D skeletal shape enclosed within large square brackets, with the overall charge placed at the top-right outside the brackets (e.g. $[\text{NH}_4]^+$, $[\text{H}_3\text{O}]^+$, $[\text{AlF}_6]^{3-}$).

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6. How to Write a Perfect Exam Explanation

When a question asks: "Explain the shape and bond angle of species X," follow this 3-point template to guarantee full marks:

Point 1 (Count): State the exact number of bonding pairs AND lone pairs of electrons around the central atom.
Point 2 (VSEPR Principle): State that electron pairs repel each other to achieve maximum separation / minimum repulsion.
Point 3 (Hierarchy / Conclusion): If lone pairs are present, explicitly state that $\text{LP–LP} > \text{LP–BP} > \text{BP–BP}$, then give the final shape name and exact bond angle.

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7. Common Pitfalls & Examiner Warnings

Mistake 1: Confusing Electron Geometry with Molecular Shape
Error: Writing that $\text{H}_2\text{O}$ is "tetrahedral" because it has 4 electron pairs.
Correction: While the electron pair arrangement is tetrahedral, the observed molecular shape (the positions of the atoms) is bent ($104.5^\circ$).

Mistake 2: Forgetting to Adjust for Ionic Charges
Error: Counting 5 valence electrons for Nitrogen in $\text{NH}_2^-$.
Correction: The $-1$ charge means you must add 1 electron: $5 + 1 = 6$ electrons $\rightarrow$ 2 single bonds leave 2 lone pairs $\rightarrow$ bent shape ($104.5^\circ$).

Mistake 3: Treating Double Bonds as Multiple Repulsion Centres
Error: Treating $\text{CO}_2$ as 4 separate electron pairs and guessing tetrahedral.
Correction: Double bonds act as single electron density regions. $\text{CO}_2$ has 2 double bond regions, giving a linear shape with a $180^\circ$ bond angle.

Mistake 4: Mixing up Angles in Trigonal Bipyramidal & Octahedral Species
Error: Stating that $\text{SF}_6$ has bond angles of $109.5^\circ$ or $\text{PCl}_5$ has only $90^\circ$.
Correction: $\text{SF}_6$ is octahedral with $90^\circ$ and $180^\circ$ angles. $\text{PCl}_5$ has both equatorial ($120^\circ$) and axial ($90^\circ$) angles.

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Quick Summary Checklist

Before sitting your AS 1 exam, make sure you can:
✓ State the VSEPR rule and the repulsion hierarchy ($\text{LP–LP} > \text{LP–BP} > \text{BP–BP}$).
✓ Deduce BP and LP for any molecule or ion up to 6 pairs.
✓ Name all 10 standard shapes and quote their exact bond angles.
✓ Draw 3D structures using solid lines, wedges, dashes, and ion bracket notation.
✓ Explain the increase in bond angle when converting $\text{NH}_3 \rightarrow \text{NH}_4^+$ and $\text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^+$.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the core principle of Valence Shell Electron Pair Repulsion (VSEPR) theory?

Valence shell electron pairs repel each other due to negative charge and spread out in three dimensions to achieve maximum separation and minimum repulsion.

What is the relative order of repulsion strength between lone pairs and bonding pairs?

$\text{Lone pair–Lone pair (LP–LP)} > \text{Lone pair–Bonding pair (LP–BP)} > \text{Bonding pair–Bonding pair (BP–BP)}$.

How are multiple covalent bonds treated when predicting molecular shapes using VSEPR theory?

Double and triple covalent bonds are treated as a single region or centre of negative charge.

What is the step-by-step method to calculate the total number of electron pairs and determine bonding and lone pairs around a central atom?

Find the central atom's valence electrons, adjust for ion charge ($-1$ per $+$, $+1$ per $-$), add 1 electron per bonded atom, divide by 2 for total pairs, and subtract bonding pairs to find lone pairs.

What shape and bond angle are formed by a molecule with 2 bonding pairs and 0 lone pairs?

Linear with a bond angle of $180^\circ$.

What shapes and bond angles result from 3 electron pairs with 0 lone pairs versus 1 lone pair?

3 BP, 0 LP gives trigonal planar ($120^\circ$); 2 BP, 1 LP gives bent / non-linear (approx. $117.5^\circ$ to $119^\circ$).

How do shapes and bond angles vary within the 4-electron-pair tetrahedral family as lone pairs increase?

4 BP, 0 LP is tetrahedral ($109.5^\circ$); 3 BP, 1 LP is pyramidal ($107^\circ$); 2 BP, 2 LP is bent ($104.5^\circ$).

What are the shapes and bond angles for molecules with 5 electron pairs (0, 1, and 2 lone pairs)?

5 BP, 0 LP: trigonal bipyramidal ($90^\circ$, $120^\circ$, $180^\circ$); 4 BP, 1 LP: see-saw ($<90^\circ$, $<120^\circ$); 3 BP, 2 LP: T-shaped ($<90^\circ$).

What shapes and bond angles are observed for molecules with 6 electron pairs?

6 BP, 0 LP is octahedral ($90^\circ$, $180^\circ$); 4 BP, 2 LP is square planar ($90^\circ$, $180^\circ$).

Why does the bond angle increase when ammonia ($\text{NH}_3$) reacts with $\text{H}^+$ to form the ammonium ion ($\text{NH}_4^+$)?

The lone pair in $\text{NH}_3$ forms a coordinate bond, converting 3 BP / 1 LP (pyramidal, $107^\circ$) into 4 BP / 0 LP (tetrahedral, $109.5^\circ$), because $\text{BP–BP}$ repulsion is weaker than $\text{LP–BP}$ repulsion.

What do solid lines, solid wedges, and hashed/dashed wedges represent in 3D molecular drawings?

Solid lines lie in the plane of the paper, solid wedges point forward out of the paper, and dashed wedges point backward into the paper.

What 3-point structure should be used to answer exam questions explaining molecular shape and bond angles?

1. State the number of bonding pairs and lone pairs; 2. State that electron pairs repel to maximum separation; 3. State the repulsion hierarchy ($\text{LP–LP} > \text{LP–BP} > \text{BP–BP}$) and name the shape and bond angle.

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