1.1 · Bonding & Lewis

Drawing Lewis structures

A step-by-step method for placing every valence electron.

By the end you should be able to:

Count the valence electrons in molecules and polyatomic ions
Draw Lewis structures that satisfy the duet and octet rules

Interactive

Lewis structure builder

Set bond orders and lone pairs on real molecules, check your electron count and formal charges, and compare with the best structure.

Loading…

Key idea

Counting valence electrons

Only valence electrons appear in a Lewis structure. For main-group atoms the count follows from the group number:

Group	1	2	13	14	15	16	17	18
Valence e⁻	1	2	3	4	5	6	7	8 (He: 2)

For an ion, add one electron per negative charge and remove one per positive charge:

$\ce{SO4^2-}$ : $6 + 4(6) + 2 = 32$
$\ce{CO3^2-}$ : $4 + 3(6) + 2 = 24$
$\ce{NH4+}$ : $5 + 4(1) - 1 = 8$

Key idea

The octet and duet rules

Atoms share electrons until each is surrounded by a noble-gas count: 8 (octet) for C, N, O, F and most main-group atoms, 2 (duet) for H. A bonding pair counts toward both atoms it joins; a lone pair counts only for the atom that holds it.

Typical patterns for neutral atoms with zero formal charge:

Atom	Bonds	Lone pairs
H	1	0
C	4	0
N	3	1
O	2	2
F, Cl, Br, I (terminal)	1	3

Period-2 atoms (C, N, O, F) have only four valence orbitals (2s and 2p), so they never carry more than 8 electrons.

Method

Drawing a Lewis structure, step by step

Count the total valence electrons, adjusting for the charge of an ion.
Build the skeleton. Put the least electronegative atom in the centre. H is always terminal and F is always terminal; in oxoacids such as $\ce{HNO3}$ the H atoms bond to O.
Connect every outer atom to the centre with a single bond (2 e⁻ each) and subtract these electrons from the total.
Complete the octets of the terminal atoms with lone pairs (H needs nothing beyond its bond).
Put any leftover electrons on the central atom as lone pairs.
Central atom short of an octet? Turn terminal lone pairs into double or triple bonds until it has 8 (but see octet exceptions for B and Be).
Check: the electron total matches step 1, formal charges are as small as possible, and an ion is drawn in square brackets with its charge outside.

Formula

Shortcut: how many bonds?

When every atom obeys the octet or duet rule, the number of shared electrons is

$S = N - A$

$N$ = electrons needed: 8 for each non-H atom, 2 for each H
$A$ = valence electrons available (from the count, including charge)
$S/2$ = number of bonds

For $\ce{HCN}$ : $N = 2 + 8 + 8 = 18$ , $A = 10$ , so $S = 8$ and there are 4 bonds (one single, one triple). The shortcut fails for incomplete and expanded octets and for radicals.

Common mistake

Counting and skeleton mistakes

Wrong: using 23 electrons for $\ce{NO3-}$ . Right: add one electron for the negative charge: $5 + 3(6) + 1 = 24$ .

Wrong: putting the first atom in the formula in the centre, as in H–N–C for $\ce{HCN}$ or H–Cl–O for $\ce{HOCl}$ . Right: H is never central; the skeletons are H–C–N and H–O–Cl.

Wrong: giving H or a terminal F a double bond to complete the central atom's octet. Right: H and terminal halogens always form exactly one bond.

Wrong: counting core electrons (using 7 for N). Right: only valence electrons count (5 for N).

Worked example

Worked example: HCN

Valence electrons: $1 + 4 + 5 = 10$ .
Skeleton H–C–N: C is less electronegative than N, and H is terminal.
Two single bonds use 4 e⁻, leaving 6.
Complete N's octet with 3 lone pairs, leaving 0. Carbon now has only 4 electrons.
Move two of N's lone pairs into the C–N bond to make a triple bond: $\ce{H-C#N}$ , with one lone pair left on N.

Check: H has 2 electrons, C has 8, N has 8; the structure uses exactly 10 electrons and every formal charge is 0.

Answer: H–C≡N with one lone pair on nitrogen.

Check yourself

Fresh questions every time you visit. Answers count toward your progress.

Keep practising this topic