Draw Lewis Dot Structure of Water (H₂O) — Step by Step

Question

Draw the Lewis dot structure of water (H₂O). Show all bonding pairs and lone pairs. Using the structure, explain why water has a bent shape and a bond angle of 104.5°.

Solution — Step by Step

Step 1: Count Total Valence Electrons

Oxygen (Group 16): 6 valence electrons Hydrogen × 2 (Group 1): 1 × 2 = 2 valence electrons

Total = 6 + 2 = 8 valence electrons

Step 2: Identify the Central Atom

Oxygen is the central atom — it has more bonding capacity (needs 2 bonds to complete its valence requirements) and hydrogen can only ever form one bond.

Step 3: Place Single Bonds from O to Each H

Draw O in the centre, with one bond to each H:

H — O — H

Each bond uses 2 electrons. Two bonds = 4 electrons used. Remaining: 8 − 4 = 4 electrons (= 2 lone pairs)

Step 4: Distribute Remaining Electrons

Hydrogen only needs 2 electrons to fill its shell (n=1 shell holds max 2). Both H atoms already have their 2 electrons from the O–H bonds — they're done.

The remaining 4 electrons go on oxygen as 2 lone pairs.

Step 5: The Final Lewis Structure

     ·· ··
H — O — H

In standard notation:

• O has 2 bonding pairs (one to each H) and 2 lone pairs
• Each H has 1 bonding pair, no lone pairs
• Total electrons shown: 2 (bonding O–H₁) + 2 (bonding O–H₂) + 4 (lone pairs on O) = 8 ✓

Step 6: Check Octets

• Oxygen: 2 + 2 (bonding) + 4 (lone pairs) = 8 electrons → octet satisfied ✓
• Hydrogen: each has 2 electrons → duet (n=1 rule) satisfied ✓

Why This Works — From Lewis Structure to Shape

Hybridization of Oxygen

Oxygen in water has 4 electron pairs (2 bonding + 2 lone): Steric number = 4 → sp³ hybridization

Oxygen's 2s and three 2p orbitals mix to form 4 sp³ hybrid orbitals. Two orbitals form O–H σ bonds; two hold lone pairs.

VSEPR Prediction

With 4 electron pairs, the electron geometry is tetrahedral (109.5°). But molecular geometry looks only at bonded atoms — ignoring lone pairs.

Since only 2 atoms are bonded to O (the two H atoms), the molecular geometry is bent (V-shaped).

Why 104.5° and Not 109.5°?

The ideal tetrahedral angle is 109.5°. But lone pairs are not the same as bonding pairs:

• Lone pair electrons are held only by one nucleus → they spread out and occupy more angular space
• Bonding pairs are shared between two nuclei → more constrained

Repulsion order: LP–LP > LP–BP > BP–BP

The two lone pairs on O repel the two O–H bonds more strongly than the bonds would repel each other, compressing the H–O–H angle from 109.5° to 104.5°.

Why Water's Shape Matters

The bent shape of water makes it a polar molecule — the two O–H bond dipoles don't cancel (they would if water were linear). The net dipole points from the H atoms toward O (more electronegative side).

This polarity explains:

• Water's unusually high boiling point (100°C) — strong intermolecular hydrogen bonds
• Water's role as a universal solvent — dissolves ionic and polar substances
• Ice being less dense than liquid water — H-bonding in ice creates an open hexagonal lattice

Alternative Method — Using the Formula Directly

For any AXₙEₘ molecule (A = central atom, X = bonded atoms, E = lone pairs):

Water = AX₂E₂ (2 bonds, 2 lone pairs)

From the VSEPR table: AX₂E₂ → Bent, ≈ 104.5°. That's the quick answer for an exam. The step-by-step derivation above is for full-mark answers in boards or to truly understand VSEPR.

Common Mistake