Lewis structures

What a Lewis structure shows

• A Lewis structure is a diagram showing all the outer-shell electrons of a molecule.
• Bonding pairs are drawn as a line between two atoms (or as two dots).
• Lone pairs are drawn as a pair of dots on the atom they belong to.
• Every atom should end up with a full outer shell — 8 electrons for most atoms, 2 for hydrogen.

At Level 2 you only ever draw simple molecules — ones with no more than four electron pairs around any atom.

Drawing one, step by step

• Count the total outer-shell electrons — add up the group contribution of every atom.
• Choose the central atom — the one that can form the most bonds (never hydrogen, which only forms one).
• Join the central atom to each outer atom with a single bond. Each bond uses 2 electrons.
• Place the electrons that are left as lone pairs, filling the outer atoms first, then the central atom.
• Check every atom has a full shell. If the central atom is short, change a lone pair into another bonding pair — that makes a double or triple bond.

Counting outer electrons

Group	Atoms	Outer electrons
1	H	1
14	C	4
15	N, P	5
16	O, S	6
17	F, Cl, Br, I	7

• For a negative ion, add one electron per negative charge. For a positive ion, subtract one per positive charge.

Bonding pairs and lone pairs — count both

You will need both numbers on the next page to work out the shape, so record them as you go:

Molecule	Bonding pairs	Lone pairs on the central atom
$CH_4$	4	0
$NH_3$	3	1
$H_2O$	2	2
$CO_2$	2 (both double)	0

• Careful: for shape purposes a double bond counts as one region of electrons, not two.

When you need a double bond

• If the central atom does not reach 8 electrons after you place the lone pairs, it needs more sharing.
• Move a lone pair from an outer atom into the bond to make a double (or triple) bond.

Worked through for carbon dioxide:

• Total outer electrons: C gives 4, each O gives 6 → $4 + 6 + 6 = 16$.
• Single bonds O–C–O use 4, leaving 12 for lone pairs.
• Putting three lone pairs on each oxygen uses all 12 — but carbon then has only 4 electrons. Not enough.
• So move one lone pair from each oxygen into the bond, making two double bonds: $O{=}C{=}O$.
• Now carbon has 8 ✓ and each oxygen has 4 bonding + 4 lone = 8 ✓.