The main ideas of VSEPR theory
Sidgwick and Powell suggested this theory in 1940; Nyholm and Gillespie improved it in 1957. Its main ideas:
- The shape of a molecule depends on the number of valence electron pairs (bonded and lone) around the central atom.
- Electron pairs repel each other because their clouds are negative.
- They take positions that keep them as far apart as possible (least repulsion).
- A double or triple bond counts as one super pair.
- A lone pair is held by only one nucleus, so it spreads out more and repels more. Order: lp–lp > lp–bp > bp–bp.
- If a molecule has two or more resonance structures, VSEPR can be used with any of them.
Molecules with only bond pairs
| Pairs | Shape | Angle | Example |
|---|---|---|---|
| 2 | linear | 180° | BeCl₂, CO₂ |
| 3 | trigonal planar | 120° | BF₃ |
| 4 | tetrahedral | 109.5° | CH₄, NH₄⁺ |
| 5 | trigonal bipyramidal | 120° and 90° | PCl₅ |
| 6 | octahedral | 90° | SF₆ |
In PCl₅ the two axial bonds (up and down) meet three bond pairs at 90°, while the equatorial ones meet only two at 90°. So axial bonds feel more repulsion and are slightly longer (and weaker) than equatorial bonds.
Molecules with lone pairs
With lone pairs, we still place all pairs in the basic shape, but we name the shape only from the atoms. Lone pairs also squeeze the bond angles.
| Total pairs | Lone pairs | Shape | Example |
|---|---|---|---|
| 3 | 1 | bent (V) | SO₂, SnCl₂ (≈119°) |
| 4 | 1 | trigonal pyramidal | NH₃ (107°) |
| 4 | 2 | bent (V) | H₂O (104.5°) |
| 5 | 1 | see-saw | SF₄ |
| 5 | 2 | T-shaped | ClF₃ |
| 5 | 3 | linear | XeF₂, I₃⁻ |
| 6 | 1 | square pyramidal | BrF₅ |
| 6 | 2 | square planar | XeF₄ |
In a trigonal bipyramid, lone pairs always choose the equatorial places, because there they are at 90° to only two other pairs. In an octahedron, two lone pairs sit opposite each other (180°), which gives square planar XeF₄.
Counting electron pairs quickly
For a central atom A joined to other atoms by single bonds:
Pairs = ½ (V + M − c + a), where V = valence electrons of A, M = number of monovalent atoms (H, halogen) attached, c = positive charge, a = negative charge. Oxygen atoms add 0 (they use two electrons of their own).
Bond pairs = number of atoms attached. Lone pairs = total pairs − bond pairs.
Example NH₄⁺: ½ (5 + 4 − 1) = 4 pairs, 4 bonded → tetrahedral.
Try it: the balloon model
Blow up 2, 3, 4, 5 and 6 balloons and tie each set together at the knots. Look how they settle: line, triangle, tetrahedron, bipyramid, octahedron. Now make one balloon in the set of 4 bigger (a "lone pair") and see the other three crowd together, like NH₃. Predict the shape first, then check it in the free-play step.
Key formulas and definitions
- Electron pairs on central atom = ½ (V + M − c + a)
- Lone pairs = total pairs − bonded atoms
- Repulsion order: lp–lp > lp–bp > bp–bp
- 2 linear · 3 trigonal planar · 4 tetrahedral · 5 trigonal bipyramidal · 6 octahedral
- Multiple bond = one electron domain
Worked examples
1. Predict the shape of BF₃.
Step 1: B has 3 valence electrons, 3 F atoms: pairs = ½(3 + 3) = 3. Step 2: 3 bonded atoms → 0 lone pairs. Answer: trigonal planar, 120°.
2. Predict the shape and bond angle of NH₃.
Step 1: N: V = 5, 3 H: pairs = ½(5 + 3) = 4. Step 2: 3 bond pairs + 1 lone pair. Step 3: 4 pairs → tetrahedral arrangement; hide the lone pair. Step 4: lp–bp repulsion squeezes the angle below 109.5°. Answer: trigonal pyramidal, about 107°.
3. Predict the shape of H₂O and explain its angle.
Step 1: O: V = 6, 2 H: pairs = ½(6 + 2) = 4. Step 2: 2 bond pairs + 2 lone pairs. Step 3: Two lone pairs push more strongly than one (lp–lp > lp–bp). Answer: bent, 104.5° (smaller than NH₃'s 107°).
4. Predict the shape of SF₄.
Step 1: S: V = 6, 4 F: pairs = ½(6 + 4) = 5. Step 2: 4 bond pairs + 1 lone pair. Step 3: 5 pairs → trigonal bipyramid; the lone pair takes an equatorial place. Answer: see-saw.
5. Predict the shape of ClF₃.
Step 1: Cl: V = 7, 3 F: pairs = ½(7 + 3) = 5. Step 2: 3 bond pairs + 2 lone pairs, both equatorial. Step 3: The three F atoms are two axial + one equatorial. Answer: T-shaped, F–Cl–F angle slightly less than 90°.
6. Predict the shape of XeF₄.
Step 1: Xe: V = 8, 4 F: pairs = ½(8 + 4) = 6. Step 2: 4 bond pairs + 2 lone pairs. Step 3: 6 pairs → octahedron; the two lone pairs sit opposite each other. Answer: square planar, 90°.
7. Predict the shape of NH₄⁺ and of I₃⁻.
Step 1: NH₄⁺: ½(5 + 4 − 1) = 4 pairs, 4 bonded → tetrahedral. Step 2: I₃⁻: central I: ½(7 + 2 + 1) = 5 pairs, 2 bonded → 3 lone pairs, all equatorial. Answer: NH₄⁺ tetrahedral; I₃⁻ linear.
Common mistakes
- Naming the shape from all electron pairs instead of from the atoms. NH₃ has a tetrahedral arrangement of pairs but a pyramidal shape.
- Counting a double bond as two electron domains. In VSEPR a double or triple bond is one domain.
- Putting lone pairs in axial positions of a trigonal bipyramid. They always go equatorial.
- Forgetting the ion charge when counting pairs (NH₄⁺, I₃⁻, ClO₃⁻).