How to draw Lewis structures: steps and examples

Key ideas

Valence electrons are the outer-shell electrons. For main group elements, the count matches the group: H 1, C 4, N 5, O 6, F and Cl 7.
Octet rule. Most atoms are most stable with 8 electrons around them, counting both shared and lone pairs. Hydrogen needs only 2.
Bonds. A single bond is 1 shared pair (2 electrons), a double bond is 2 pairs and a triple bond is 3 pairs.
Typical bonds in neutral molecules. H makes 1, O makes 2, N makes 3, C makes 4. This is a good check on any drawing.

The six steps

Count the total valence electrons. Add 1 for each negative charge on an ion, and subtract 1 for each positive charge.
Pick the central atom: usually the least electronegative atom that can make the most bonds (often C, N, S or P). Hydrogen is never central.
Connect each outer atom to the central atom with a single bond. Subtract 2 electrons for each bond.
Give each outer atom lone pairs until it has 8 electrons (hydrogen stays at 2).
Put any electrons that are left on the central atom as lone pairs.
If the central atom has fewer than 8 electrons, move a lone pair from an outer atom into a double bond. Repeat until it has 8.

Then check the formal charge of each atom. The best structure has formal charges that are zero, or as close to zero as possible:

\text{FC} = V - N - \frac{B}{2}

$V$ is the number of valence electrons of the free atom.
$N$ is the number of nonbonding electrons on the atom (2 per lone pair).
$B$ is the number of electrons in bonds to that atom (2 per bond line).

Worked examples

Example 1: water

Problem Draw the Lewis structure of $\mathrm{H_2O}$ .

Count: 2 H × 1 + 1 O × 6 = 8 valence electrons.
O is central. Two single bonds to H use 4 electrons, leaving 4.
$\mathrm{H{-}O{-}H}$
Each H already has its 2 electrons. Put the 4 leftover electrons on O as 2 lone pairs.
Check O: 2 bonds (4 electrons) + 2 lone pairs (4 electrons) = 8. Formal charge on O: $6 - 4 - \tfrac{4}{2} = 0$ .

Answer H-O-H with 2 single bonds and 2 lone pairs on O

Example 2: carbon dioxide needs double bonds

Problem Draw the Lewis structure of $\mathrm{CO_2}$ .

Count: 4 + 2 × 6 = 16 valence electrons.
C is central. Two single bonds use 4 electrons, leaving 12. Each O gets 3 lone pairs (6 electrons each), which uses all 12.
Now carbon has only 4 electrons. Move one lone pair from each O into a bond with C.
$\mathrm{O{=}C{=}O}$
Check: C has 2 double bonds = 8 electrons. Each O has 1 double bond + 2 lone pairs = 8. Every formal charge is 0.

Answer O=C=O with 2 double bonds, 2 lone pairs on each O and none on C

Example 3: a triple bond

Problem Draw the Lewis structure of hydrogen cyanide, HCN.

Count: 1 + 4 + 5 = 10 valence electrons.
H cannot be central, and C is less electronegative than N, so the order is H, C, N. Two single bonds use 4 electrons, leaving 6.
N gets 3 lone pairs (6 electrons). That uses everything, but C has only 4 electrons.
C needs 4 more, so move 2 of nitrogen's lone pairs into bonds with C.
$\mathrm{H{-}C{\equiv}N}$
Check: C has 1 single + 1 triple bond = 8 electrons. N has 1 triple bond + 1 lone pair = 8. All formal charges are 0.

Answer H-C≡N with a triple bond and 1 lone pair on N

Example 4: an ion and its formal charge

Problem Draw the ammonium ion, $\mathrm{NH_4^+}$ , and find the formal charge on N.

Count: 5 + 4 × 1 = 9, minus 1 for the positive charge = 8 valence electrons.
N is central with 4 single bonds to H. That uses all 8 electrons, so N has no lone pairs.
Formal charge on N.
$\text{FC} = 5 - 0 - \frac{8}{2} = +1$
Draw square brackets around the whole structure with the charge outside: $[\ldots]^+$ .

Answer 4 N-H single bonds, no lone pairs, formal charge on N = +1

Common mistakes and how to fix them

Forgetting the ion's charge in the electron count. Fix: $\mathrm{NO_3^-}$ has 24 electrons, not 23.
Putting hydrogen in the middle. Fix: hydrogen makes only one bond, so it is always on the outside.
Giving the outer atoms too few electrons. Fix: fill the outer atoms first, then the central atom.
Using a different number of electrons than you counted. Fix: count every dot and every line (2 each) at the end. The total must match step 1.
Forcing an octet on every atom. Some atoms break the rule: boron in $\mathrm{BF_3}$ has 6, and sulfur in $\mathrm{SF_6}$ has 12. Elements in period 3 and below can hold more than 8.

Practice questions

How many valence electrons are in formaldehyde, $\mathrm{CH_2O}$ ?
1. 10
2. 12
3. 14
4. 16
Show answer
Answer: 12
C has 4, two H have 1 each, and O has 6: $4 + 2 + 6 = 12$ .
How many valence electrons are in the nitrate ion, $\mathrm{NO_3^-}$ ?
Show answer
Answer: 24 valence electrons
N has 5, three O have 6 each (18), and the negative charge adds 1: $5 + 18 + 1 = 24$ .
How many lone pairs are on the nitrogen atom in ammonia, $\mathrm{NH_3}$ ?
1. 0
2. 1
3. 2
4. 3
Show answer
Answer: 1
8 valence electrons. Three N-H bonds use 6, so 2 electrons, or 1 lone pair, are left on N.
In the hydroxide ion, $\mathrm{OH^-}$ , oxygen has 1 bond and 3 lone pairs. What is its formal charge?
1. +1
2. 0
3. -1
4. -2
Show answer
Answer: -1
$\text{FC} = 6 - 6 - \tfrac{2}{2} = -1$ . The ion's charge sits on oxygen.
Which molecule has a triple bond?
1. $\mathrm{N_2}$
2. $\mathrm{O_2}$
3. $\mathrm{F_2}$
4. $\mathrm{H_2}$
Show answer
Answer: $\mathrm{N_2}$
$\mathrm{N_2}$ has 10 valence electrons. To give each N an octet with only 10 electrons, the atoms must share 3 pairs. $\mathrm{O_2}$ has a double bond, and $\mathrm{F_2}$ and $\mathrm{H_2}$ have single bonds.

Frequently asked questions

What is resonance?

Some molecules can be drawn more than one correct way, with a double bond in different places. Ozone and nitrate are examples. The real molecule is a blend of all the drawings, called resonance structures, and its bonds are in between single and double. Draw each one and connect them with a double-headed arrow.

How do I pick the central atom?

Pick the atom that is least electronegative and can make the most bonds, which is often the one that appears once in the formula. Carbon is almost always central. Hydrogen and fluorine are always on the outside because they make only one bond.

Why use formal charge?

When more than one structure follows the octet rule, formal charge picks the best one. The best structure has formal charges closest to zero, and any negative formal charge sits on the most electronegative atom.