Question
Water (H₂O) has an oxygen atom with four electron pairs around it, which should give a tetrahedral geometry with bond angle 109.5°. But experimentally, the H–O–H bond angle is 104.5°. Why is it less than the ideal tetrahedral angle?
Solution — Step by Step
Oxygen has 6 valence electrons. It forms 2 bonds with hydrogen atoms, using 2 electrons each. The remaining 4 electrons sit as 2 lone pairs on oxygen.
Total electron pairs = 2 bond pairs + 2 lone pairs = 4 → tetrahedral electron geometry.
Not all electron pairs repel equally. The order of repulsion strength is:
This is the key rule that explains everything. Lone pairs are “fatter” — they’re held only by one nucleus and spread out more in space, creating stronger repulsion.
In H₂O, we have two lone pairs sitting on oxygen. These two lone pairs repel each other strongly, AND they each repel the O–H bond pairs.
This pushes the two O–H bonds closer together, compressing the H–O–H bond angle below 109.5°.
Each lone pair causes roughly 2.5° of compression compared to pure bond-pair repulsion. With two lone pairs, the total compression is approximately:
That’s exactly what we observe experimentally. The bond angle of H₂O is 104.5°.
Ammonia (NH₃) has one lone pair + 3 bond pairs. Only one lone pair compresses the H–N–H angle, giving 107° — less than 109.5° but more than 104.5°.
The pattern: more lone pairs → more compression → smaller bond angle.
Why This Works
VSEPR (Valence Shell Electron Pair Repulsion) theory rests on one idea — electron pairs minimize repulsion by staying as far apart as possible. When all four pairs around an atom are bond pairs (like in CH₄), they settle at the perfect tetrahedral angle of 109.5°.
Lone pairs break this symmetry. Because a lone pair is attracted only to one nucleus (unlike a bond pair shared between two), it occupies more angular space. Two lone pairs on the same atom create significant crowding that squeezes the bond pairs inward.
This is why the molecular shape (what we observe) differs from the electron geometry (what the electron pairs dictate). In H₂O: electron geometry = tetrahedral, molecular geometry = bent/V-shaped.
Alternative Method — Using Hybridisation
Oxygen in H₂O undergoes sp³ hybridisation, forming 4 hybrid orbitals at 109.5° apart. Two of these hold lone pairs, two hold bonds.
But pure sp³ orbitals would only give 109.5° if all four substituents were identical. Since lone pairs exert more repulsion than bonding pairs, they effectively “steal” angular space, contracting the H–O–H angle to 104.5°.
For quick MCQs: Bond angles follow the pattern CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°). Each lone pair drops the angle by ~2.5°. This series appears almost every year in JEE Main and NEET.
Common Mistake
Mixing up electron geometry and molecular geometry. Many students write “H₂O has tetrahedral geometry” — that’s wrong for the molecular shape. The electron pair geometry is tetrahedral (4 pairs), but the molecular geometry is bent/V-shaped because we only consider the positions of atoms, not lone pairs. In JEE Main 2024, one option trap was labeling H₂O as tetrahedral — don’t fall for it.
A second slip: saying the lone pairs “push the bonds apart.” They actually push the bonds toward each other — the lone pairs expand, compressing the space available for the O–H bonds. The bonds are the ones getting squeezed, not spreading out.