Chemistry (9701) | Shapes of molecules

Understanding the Shapes of Molecules (VSEPR Theory)

Welcome to one of the most fundamental and often-tested topics in Chemical Bonding! Don't worry if three-dimensional shapes seem tricky at first—this section is built on one simple, logical principle. Once you master this principle, you can predict the shape of almost any simple molecule or ion.

Why do we care about molecular shape? Because the shape of a molecule dictates everything: its polarity, how it interacts with other molecules, its boiling point, and even how it works in biological systems (think of an enzyme needing a perfect "key" shape to fit its "lock").

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1. The Core Idea: Valence Shell Electron Pair Repulsion (VSEPR) Theory

What is VSEPR Theory?

The shape of a molecule is determined by how the electrons surrounding the central atom arrange themselves to minimise repulsion.

The key principle, or the "Golden Rule" of VSEPR, is:

Electron pairs (both bonding pairs and lone pairs) repel each other and arrange themselves as far apart as possible in space.

We refer to these areas of electron density as Electron Domains or Electron Regions.

Analogy: The Crowded Bus

Imagine you are the central atom, and the bonding pairs and lone pairs are other people trying to sit near you on an empty bus. You will naturally spread out as far away from each other as possible. The final arrangement (the shape) is the one that achieves the maximum possible distance, thus achieving the lowest energy state.

Key Terms for VSEPR

• Bonding Pair (BP): Electrons shared between two atoms (single, double, or triple bonds count as one electron domain).
• Lone Pair (LP): Electrons belonging to the central atom that are not involved in bonding.
• Electron Domain Geometry: The geometric arrangement of all electron domains (BPs + LPs) around the central atom.
• Molecular Geometry (Shape): The geometric arrangement of only the atoms (i.e., focusing only on the bonding pairs).

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2. Step-by-Step VSEPR Application

To determine the shape of a molecule (\(A\text{X}_n\)), follow these four steps:

Step 1: Identify the Central Atom (A)

This is usually the least electronegative atom, or the atom to which all others are bonded.

Step 2: Calculate the Total Number of Electron Domains (Regions)

Count how many regions of electron density are around the central atom:

• Count 1 domain for every atom bonded to the central atom (regardless of whether it's a single, double, or triple bond).
• Count 1 domain for every lone pair (non-bonding pair) on the central atom.
• Total Domains = (\(\text{Bonding Pairs}\)) + (\(\text{Lone Pairs}\))

Step 3: Determine the Electron Domain Geometry

This is the initial geometry where repulsion is minimised (the ideal shape).

Total Domains	Electron Domain Geometry	Ideal Bond Angle
2	Linear	\(180^{\circ}\)
3	Trigonal Planar	\(120^{\circ}\)
4	Tetrahedral	\(109.5^{\circ}\)
5	Trigonal Bipyramidal	\(120^{\circ}\) and \(90^{\circ}\)
6	Octahedral	\(90^{\circ}\)

Step 4: Determine the Molecular Geometry (Shape)

Use the number of lone pairs to determine the final molecular shape. Remember: lone pairs are invisible when naming the shape, but they change the arrangement of the atoms!

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3. Common Molecular Shapes (Syllabus Examples)

Let's look at the specific examples required by the syllabus, grouped by their total number of electron domains.

A. Two Electron Domains

• Example: Carbon Dioxide (\(\text{CO}_2\))
• Central Atom: C
• Bonding Pairs: 2 (The C=O double bonds count as two separate regions).
• Lone Pairs on C: 0
• Total Domains: 2
• Shape: Linear
• Bond Angle: \(180^{\circ}\)

B. Three Electron Domains

• Example: Boron Trifluoride (\(\text{BF}_3\))
• Central Atom: B
• Bonding Pairs: 3 (Three B-F single bonds).
• Lone Pairs on B: 0
• Total Domains: 3
• Shape: Trigonal Planar
• Bond Angle: \(120^{\circ}\)

Did you know? Boron is an exception to the Octet Rule, forming only three bonds and having six valence electrons in the molecule.

C. Four Electron Domains (The Tetrahedral Family)

The ideal angle is \(109.5^{\circ}\), but lone pairs compress this angle.

• Case 1: 4 BP, 0 LP
• Example: Methane (\(\text{CH}_4\))
• Total Domains: 4
• Shape: Tetrahedral
• Bond Angle: \(109.5^{\circ}\)


• Case 2: 3 BP, 1 LP
• Example: Ammonia (\(\text{NH}_3\))
• Total Domains: 4
• The lone pair strongly repels the three N-H bonding pairs.
• Shape: Pyramidal (or Trigonal Pyramidal)
• Bond Angle: \(107^{\circ}\) (Lower than \(109.5^{\circ}\) due to LP-BP repulsion).


• Case 3: 2 BP, 2 LP
• Example: Water (\(\text{H}_2\text{O}\))
• Total Domains: 4
• The two lone pairs exert very strong repulsion.
• Shape: Non-linear (or Bent)
• Bond Angle: \(104.5^{\circ}\) (Significantly lower than \(109.5^{\circ}\) due to strong LP-LP and LP-BP repulsion).

D. Five Electron Domains (Expanded Octet)

These atoms (like Phosphorus and Sulfur in Period 3 onwards) can have an expanded octet.

• Example: Phosphorus Pentachloride (\(\text{PCl}_5\))
• Central Atom: P
• Bonding Pairs: 5
• Lone Pairs on P: 0
• Total Domains: 5
• Shape: Trigonal Bipyramidal
• Bond Angles: \(120^{\circ}\) (equatorial positions) and \(90^{\circ}\) (axial positions).

E. Six Electron Domains (Expanded Octet)

• Example: Sulfur Hexafluoride (\(\text{SF}_6\))
• Central Atom: S
• Bonding Pairs: 6
• Lone Pairs on S: 0
• Total Domains: 6
• Shape: Octahedral
• Bond Angle: \(90^{\circ}\) (All positions are equivalent).

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4. Comprehensive Summary of Shapes and Angles

This table is your best friend for revision. You must know the bond angle and the specific shape name for each geometry, especially those containing lone pairs, as they are crucial for explaining bond angle deviation.

Total Electron Domains	Bonding Pairs (BP)	Lone Pairs (LP)	Molecular Shape	Bond Angle(s)	Syllabus Example
2	2	0	Linear	\(180^{\circ}\)	\(\text{CO}_2\)
3	3	0	Trigonal Planar	\(120^{\circ}\)	\(\text{BF}_3\)
4	4	0	Tetrahedral	\(109.5^{\circ}\)	\(\text{CH}_4\)
4	3	1	Pyramidal (Trigonal Pyramidal)	\(107^{\circ}\)	\(\text{NH}_3\)
4	2	2	Non-linear (Bent)	\(104.5^{\circ}\)	\(\text{H}_2\text{O}\)
5	5	0	Trigonal Bipyramidal	\(120^{\circ}, 90^{\circ}\)	\(\text{PF}_5\)
6	6	0	Octahedral	\(90^{\circ}\)	\(\text{SF}_6\)

5. Predicting Shapes of Analogous Molecules and Ions

A crucial skill in your exams is applying VSEPR to new species (ions or molecules) that are analogous to the core examples. This just means they have the same number of electron domains and lone pairs.

The key is counting the electron domains accurately!

Step-by-Step Example: Predicting the shape of the Hydronium ion, \(\text{H}_3\text{O}^+\)

1. Central Atom: Oxygen (O).

2. Bonding Pairs (BP): 3 (three single bonds to H atoms).

3. Lone Pairs (LP): Don't worry about the charge initially, just the atoms' positions. Oxygen (Group 16) has 6 valence electrons. In \(\text{H}_2\text{O}\) it uses 2 for bonding, leaving 4 (2 lone pairs). In \(\text{H}_3\text{O}^+\), it forms a dative bond using one lone pair, meaning it now only has 1 lone pair remaining.

4. Total Domains: \(3 \text{ BP} + 1 \text{ LP} = 4\). (Analogous to \(\text{NH}_3\)).

5. Electron Domain Geometry: Tetrahedral.

6. Molecular Shape: The shape is determined by the 3 bonded atoms and the 1 lone pair.

• Prediction: Pyramidal
• Bond Angle: Approximately \(107^{\circ}\) (Due to the repulsion from the single lone pair).

Key Takeaway for this Chapter

Molecular shapes are predicted using VSEPR theory, which states that electron domains repel each other to achieve maximum separation. Lone pairs repel more strongly than bonding pairs, leading to distortions and reductions in the ideal bond angle.