Drawing the Lewis Structure of CO₂, H₂O, NH₃ and CH₄

Four small molecules turn up in nearly every beginner chemistry course: methane, ammonia, water and carbon dioxide. They are worth learning properly, not just memorising, because between them they show you almost every move you need for drawing Lewis structures: single bonds, lone pairs, and the moment when you have to switch to double bonds.

We will use the same five-step routine every time, so that by the last example it feels automatic.

The routine

1. Count the total valence electrons.
2. Draw the skeleton, with single bonds from the central atom to each outer atom.
3. Give the outer atoms full shells (8 electrons, or 2 for hydrogen).
4. Put any leftover electrons on the central atom.
5. If the central atom has fewer than 8, form multiple bonds. Finish by checking formal charges.

Valence electrons come from the group number: hydrogen 1, carbon 4, nitrogen 5, oxygen 6. If you want to see where those numbers come from, the electron configuration tool shows the outer shell of any element, and Lewis symbols for atoms shows how to draw each atom’s dots.

We start with methane, because it is the simplest, and finish with CO₂, which needs the extra step.

Example 1: methane, CH₄

Step 1: count. Carbon has 4 valence electrons. Each of the four hydrogens has 1. Total = 4 + (4 × 1) = 8 electrons, or 4 pairs.

Step 2: skeleton. Carbon goes in the middle, because hydrogen can only ever form one bond and so can never be central. Draw four C–H single bonds. Each bond uses 2 electrons, so that is 8 electrons used.

Step 3: outer atoms. Each hydrogen now has 2 electrons (its bond). That is a full shell for hydrogen. ✓

Step 4: leftovers. 8 − 8 = 0. Nothing is left over.

Step 5: check the centre. Carbon is involved in four bonds, which means 8 electrons around it. ✓ No multiple bonds are needed.

Formal charges. Carbon: 4 − 0 − ½(8) = 0. Each H: 1 − 0 − ½(2) = 0. All zero, as you would hope for a neutral, stable molecule.

The finished structure: carbon in the centre with four single bonds to hydrogen and no lone pairs anywhere.

Shape. Four bonding pairs push apart as far as possible, giving a tetrahedral molecule with H–C–H angles of about 109.5°. Methane is often drawn flat on paper as a cross, but that is only a drawing convention; the real molecule is three-dimensional.

Example 2: ammonia, NH₃

Step 1: count. Nitrogen has 5, three hydrogens have 3. Total = 5 + 3 = 8 electrons.

Step 2: skeleton. N in the centre, three N–H single bonds: 6 electrons used.

Step 3: outer atoms. Each H has its 2. ✓

Step 4: leftovers. 8 − 6 = 2 electrons. These go on nitrogen as one lone pair.

Step 5: check the centre. Nitrogen has three bonds (6 electrons) plus one lone pair (2 electrons) = 8. ✓

Formal charges. N: 5 − 2 − ½(6) = 0. Each H: 0. All zero. ✓

The finished structure: N in the centre, three single bonds to H, and one lone pair on N.

Shape. There are still four electron groups around nitrogen, as there were around carbon in methane. But one of them is a lone pair, which you cannot “see” when you describe the positions of atoms. The atoms form a trigonal pyramid, and the lone pair squeezes the bonds together slightly, so the H–N–H angle is about 107° instead of 109.5°. That lone pair is also what lets ammonia act as a base, picking up H⁺ to become NH₄⁺. The shapes of ammonia and methane article explains the angle difference in more detail.

Example 3: water, H₂O

Step 1: count. Oxygen has 6, two hydrogens have 2. Total = 6 + 2 = 8 electrons.

Step 2: skeleton. H–O–H. (Not H–H–O: hydrogen can form only one bond, so it can’t sit between two atoms.) Two bonds use 4 electrons.

Step 3: outer atoms. Each H has 2. ✓

Step 4: leftovers. 8 − 4 = 4 electrons, which become two lone pairs on oxygen.

Step 5: check the centre. O has two bonds (4) + two lone pairs (4) = 8. ✓

Formal charges. O: 6 − 4 − ½(4) = 0. Each H: 0. All zero. ✓

The finished structure: H–O–H with two lone pairs on O.

Shape. Once more there are four electron groups on the central atom, but now two of them are lone pairs. The molecule is bent (sometimes called V-shaped), with an H–O–H angle of about 104.5°, smaller again because two lone pairs push harder than one. See the shape of the water molecule for why this bend is so important for water’s properties.

A pattern to notice. Methane, ammonia and water all have 8 valence electrons and 4 electron pairs around the central atom. The only difference is how many of those pairs are bonds and how many are lone pairs:

Molecule	Bonding pairs on centre	Lone pairs on centre	Shape	Angle
CH₄	4	0	tetrahedral	~109.5°
NH₃	3	1	trigonal pyramidal	~107°
H₂O	2	2	bent	~104.5°

As you move from carbon to nitrogen to oxygen, each atom has one more valence electron and needs one fewer bond to reach eight, so one bond becomes a lone pair each time.

Example 4: carbon dioxide, CO₂

Now for the one that needs the extra step.

Step 1: count. Carbon has 4, each oxygen has 6. Total = 4 + (2 × 6) = 16 electrons, or 8 pairs.

Step 2: skeleton. Carbon in the middle (it is less electronegative than oxygen and can form more bonds): O–C–O. Two single bonds use 4 electrons, leaving 12.

Step 3: outer atoms. Each oxygen already has 2 electrons from its bond and needs 6 more, which is three lone pairs. Two oxygens × 6 = 12 electrons. That uses up everything: 12 − 12 = 0.

Step 4: leftovers. None. Carbon gets nothing extra.

Step 5: check the centre. Carbon has just two single bonds = 4 electrons. It is 4 short of an octet. ✗

The fix. Take one lone pair from each oxygen and turn it into a second bond to carbon.

• Left oxygen: one lone pair becomes a bond, so O=C.
• Right oxygen: the same, so C=O.

Now count again:

• Carbon: two double bonds = 8 electrons. ✓
• Each oxygen: one double bond (4 electrons) + two lone pairs (4 electrons) = 8. ✓
• Total electrons drawn: 8 in bonds + 8 in lone pairs = 16. ✓ (Always check the total hasn’t changed.)

Formal charges.

Atom	Lone-pair electrons	Bonds	Formal charge
C	0	4	4 − 0 − 4 = 0
each O	4	2	6 − 4 − 2 = 0

The finished structure: O=C=O, with two lone pairs on each oxygen and none on carbon.

Why not a triple bond on one side? You could also satisfy all octets with O≡C–O: the triple-bonded oxygen with one lone pair, the single-bonded one with three. Its formal charges are +1 on the triple-bonded O and −1 on the single-bonded O. That is worse than all zeros, and it would make the two C–O bonds different lengths. Experiment shows they are identical, which fits O=C=O. The formal charge guide covers this comparison in more depth.

Shape. Carbon has no lone pairs and only two electron groups (each double bond counts as one group), so the molecule is linear, 180°. Each C=O bond is polar, but they point in exactly opposite directions and cancel, so CO₂ has no overall dipole. That is why a molecule with polar bonds can still be a non-polar gas.

Bonus example: methanal (formaldehyde), CH₂O

Try this one yourself before reading the answer.

Step 1: count. C 4 + H 2 × 1 + O 6 = 12 electrons.

Step 2: skeleton. Carbon in the centre, bonded to both H atoms and to O. Three single bonds use 6, leaving 6.

Step 3: outer atoms. H atoms are done. O needs three lone pairs: 6 electrons used, 0 left.

Step 4: leftovers. None, so carbon gets no lone pair.

Step 5: check the centre. Carbon has three bonds = 6 electrons. ✗ Move one oxygen lone pair into a C=O bond. Carbon now has 8; oxygen has a double bond plus two lone pairs = 8. ✓

Formal charges: C 4 − 0 − 4 = 0; O 6 − 4 − 2 = 0; each H 0.

The finished structure: carbon with two C–H single bonds and a C=O double bond; oxygen carries two lone pairs. Three electron groups around carbon make it trigonal planar, with angles close to 120°.

Common slips to avoid

• Forgetting to check the total. After converting lone pairs into bonds, recount. The number of electrons never changes.
• Putting hydrogen in the middle. It only forms one bond.
• Adding lone pairs to carbon in CO₂ to fix its octet. You have no spare electrons; the fix is always to share existing lone pairs as extra bonds.
• Leaving lone pairs off oxygen and nitrogen. They are part of the structure and they decide the shape.

Key takeaways

• Use the same routine every time: count, skeleton, outer atoms, leftovers, fix the centre, then check formal charges.
• CH₄, NH₃ and H₂O all have 8 valence electrons and four electron pairs on the central atom; they differ only in how many pairs are lone pairs (0, 1, 2).
• Those lone pairs shrink the bond angle from about 109.5° to 107° to 104.5°.
• CO₂ has 16 electrons and needs two C=O double bonds; O=C=O with two lone pairs on each oxygen gives every atom a formal charge of zero.
• CO₂ is linear and non-polar even though each C=O bond is polar.