Question
Water (H₂O) has the formula that might suggest a linear structure. Using VSEPR theory, explain why water is actually bent (V-shaped), and give the bond angle.
Solution — Step by Step
Oxygen has 6 valence electrons. It forms 2 bonds with 2 hydrogen atoms (each using 1 electron). After bonding:
- 2 bonding pairs (O–H bonds)
- 4 remaining electrons = 2 lone pairs on oxygen
Lewis structure:
H – O – H
(2 lone pairs on O)The central atom is oxygen with a total of 4 electron domains: 2 bonding pairs + 2 lone pairs.
Steric number = bonding pairs + lone pairs = 2 + 2 = 4
With 4 electron domains, the electron geometry is tetrahedral (the four domains are as far apart as possible — at 109.5° angles in a tetrahedral arrangement).
If water had NO lone pairs (like CH₄), all four domains would be bonding pairs and the geometry would be tetrahedral with 109.5° angles.
Electron geometry = tetrahedral (considers all 4 domains including lone pairs)
Molecular geometry (shape) = only positions of atoms matter = bent/V-shaped (only 2 hydrogens are visible; the 2 lone pairs are invisible in the shape description)
The molecule’s shape is determined by where the atoms are, not where the electron pairs are. Since only 2 H atoms are bonded to O, the molecular shape is the angle formed between O–H–H — a bent shape.
In a perfect tetrahedral, bond angles are 109.5°. But in water, two of the four domains are lone pairs, not bonding pairs.
Lone pair–lone pair repulsion is stronger than bonding pair–bonding pair repulsion (because lone pairs are held closer to the nucleus and occupy more space, exerting greater electrostatic repulsion on adjacent electron clouds).
The order of repulsion strength:
- LP–LP > LP–BP > BP–BP
The two lone pairs on oxygen push the two O–H bonds closer together, compressing the bond angle below 109.5°.
Actual H–O–H bond angle = 104.5°
This is a significant compression from 109.5°: nearly 5° reduction due to the presence of two lone pairs.
| Feature | Value |
|---|---|
| Electron geometry | Tetrahedral |
| Molecular shape | Bent/V-shaped |
| Bond angle | 104.5° |
| Lone pairs on O | 2 |
| Hybridisation of O | sp³ |
Because the molecule is bent (not linear), the two O–H dipoles do NOT cancel. Each O–H bond is polar (O is more electronegative than H, with EN = 3.44 vs 2.2). In a bent molecule, the two bond dipoles add up (vector sum), giving water a net dipole moment of 1.85 D — making water a polar molecule.
This polarity is responsible for water’s exceptionally high boiling point, surface tension, and ability to dissolve ionic compounds.
Why This Works
If water were linear (like CO₂), the two O–H dipoles would point in opposite directions and cancel — water would be non-polar. But the bent structure due to lone pairs ensures the dipoles point in roughly the same direction (both directed toward the oxygen), giving a net dipole.
The lone pairs are not “invisible nothing” — they take up real space, exert real repulsive forces on bonding pairs, and compress the H–O–H angle. This is the central insight of VSEPR theory.
Alternative Method
You can compare water with related molecules to build intuition:
- BeCl₂ (steric number 2, no lone pairs): Linear, 180°
- SO₂ (steric number 3, one lone pair): Bent, ~119°
- H₂O (steric number 4, two lone pairs): Bent, 104.5°
- CH₄ (steric number 4, no lone pairs): Tetrahedral, 109.5°
As lone pairs increase, the bond angle decreases.
Common Mistake
A common mistake is saying “water has sp² hybridisation because it has a bent structure.” The bent shape does NOT mean sp² hybridisation. Bent shape with 2 lone pairs = sp³ hybridisation (4 electron domains). Bent shape with 1 lone pair (like SO₂) = sp² hybridisation (3 electron domains). Always count electron domains to determine hybridisation, not the visible atomic arrangement alone.