General chemistry · Bonding
How to draw Lewis structures, step by step
Count the valence electrons, connect the atoms, then place every electron. A Lewis structure shows bonds and lone pairs. The trick is to check both your total electron count and the electrons around each atom.
A six-step checklist
This method is for common main-group molecules and polyatomic ions. Use any atom connectivity your problem supplies; a molecular formula alone does not always identify a unique structure.
- Count the total valence electrons.
Add each atom’s contribution. Add one electron for each negative charge; subtract one for each positive charge. Write this total down before drawing.
- Choose the skeleton.
For a simple structure with one central atom, usually put the least electronegative atom in the center. Hydrogen goes on the outside and makes one bond in these structures. Fluorine is terminal. Connect neighboring atoms with single bonds.
- Subtract the bonding electrons.
Each single bond uses two electrons from your total. Do not count a shared pair twice in the overall budget.
- Fill the outside atoms, then the center.
Add lone pairs to give terminal atoms octets, except H, which needs only two electrons. Put any remaining electrons on the central atom.
- Make multiple bonds if needed.
If a central atom such as C or N is short of an octet, turn a neighboring lone pair into an extra bond. You are moving two existing electrons, not adding two new ones. Check for an octet exception before forcing a double bond.
- Audit electrons, octets, and formal charges.
Recount all bond and lone-pair electrons. Check each atom’s shell and calculate formal charges. Enclose a polyatomic ion in brackets with its overall charge outside. Include resonance contributors when needed.
| Atom(s) | Electrons per atom | Example count |
|---|---|---|
| H | 1 | 2 H contribute 2 |
| C | 4 | 1 C contributes 4 |
| N, P | 5 | 1 N contributes 5 |
| O, S | 6 | 2 O contribute 12 |
| F, Cl, Br, I | 7 | 3 F contribute 21 |
Two different counts: a C=O double bond uses four electrons in the whole drawing. When checking octets, those same four electrons count around C and around O. For formal charge, each atom is assigned half of them.
Three worked examples, one method
In the drawings below, each dot is one electron. The drawings show connections and electron placement, not necessarily the molecule’s 3D shape.
1. Water, H2O: place lone pairs
Budget: 2(1) + 6 = 8 valence electrons
- Put oxygen in the center: H–O–H.
- Two single bonds use 2 × 2 = 4 electrons. You have 8 − 4 = 4 left.
- Each H already has two electrons from its bond. Put the remaining four electrons on O as two lone pairs.
- O now has eight electrons around it: four in bonds and four in lone pairs. Each H has two. The drawing uses all eight electrons.
- Formal charges: O = 6 − 4 − 2 = 0; each H = 1 − 0 − 1 = 0.
·· H–O–H ··
2. Carbon dioxide, CO2: repair an incomplete octet
Budget: 4 + 2(6) = 16 valence electrons
- Start with O–C–O. Two single bonds use four electrons, leaving twelve.
- Give each O three lone pairs. This uses all twelve remaining electrons, but carbon has only four electrons around it.
- Move one lone pair from each oxygen into its C–O bond. The result is O=C=O, with two lone pairs on each O.
- Recount: two double bonds contain eight electrons; four lone pairs contain eight more. Total = 16. All three atoms have octets.
- Formal charges: C = 4 − 0 − 4 = 0; each O = 6 − 4 − 2 = 0.
·· ·· :O–C–O: ·· ··
·· ·· O=C=O ·· ··
3. Nitrate, NO3−: charge and resonance
Budget: 5 + 3(6) + 1 = 24 valence electrons
- Add one electron for the −1 ion charge. Put N in the center and connect it to three O atoms.
- Three single bonds use six electrons. Place the remaining eighteen as three lone pairs on each O.
- N has only six electrons around it. Move a lone pair from one O into an N–O bond to make one double bond.
- The final contributor has eight bonding electrons and sixteen nonbonding electrons: 24 total. N and all three O atoms have octets.
- N has formal charge +1. The double-bonded O has 0; each single-bonded O has −1. Their sum is +1 + 0 − 1 − 1 = −1.
- Draw brackets and the overall − charge. The double bond can be placed on any of the three O atoms, giving three equivalent resonance contributors.
⎡ ··O·· ⎤⁻ ⎢ ║ ⎥ ⎢ ·· ║ ·· ⎥ ⎢ :O⁻–N⁺–O⁻: ⎥ ⎣ ·· ·· ⎦
Resonance does not mean switching structures. The real nitrate ion has three equivalent N–O bonds. The contributors describe delocalized electrons; moving the double bond on paper does not move the atoms.
The formal-charge check
Formal charge = valence electrons of the neutral atom − nonbonding electrons − ½(bonding electrons)
For the paired-electron structures here, the shortcut is valence electrons − lone-pair electrons − bond lines. A single bond counts as one line, a double as two, a triple as three. Subtract electrons in lone pairs, not the number of pairs.
Formal charges must add to the overall charge. Among reasonable structures, prefer smaller formal charges and less charge separation; negative charge generally belongs on the more electronegative atom. These are guidelines; they never let carbon or nitrogen go past eight electrons. Formal charge is bookkeeping, not an atom’s measured partial charge.
When the octet checklist needs an exception
| Situation | What to do |
|---|---|
| Hydrogen | Stop at two electrons. An H with one single bond is complete. |
| Incomplete octets | Some compounds are electron deficient. In the usual BF3 Lewis structure, B has three single bonds and only six electrons around it. |
| Odd electron total | NO has 5 + 6 = 11 valence electrons. At least one is unpaired; you cannot give every atom an octet by pairing them all. |
| Expanded valence shells | Some period-3 and heavier central atoms are drawn with more than eight electrons, such as S in SF6 (six bonds, twelve electrons around S). Do not apply this to period-2 atoms such as C, N, O, or F. |
If your answer looks wrong: first check the ion-charge adjustment, then count bond electrons and lone-pair electrons separately. A correct total alone is not enough: the unfinished CO2 drawing used sixteen electrons but left carbon short.
Your turn: six practice questions
Draw each structure on paper, then check the key decision below. You can change an answer and check again as often as you like.
1. Adjust for an ion’s charge
How many total valence electrons are in NH4+?
Show solution
8 electrons. N supplies 5, four H supply 4, and the positive charge removes 1: 5 + 4 − 1 = 8. Draw N with four single N–H bonds and no lone pairs. N has formal charge +1; each H has 0. Bracket the ion and add + outside.
2. Turn leftover electrons into lone pairs
In H2O, how many lone pairs belong on oxygen after drawing both O–H bonds?
Show solution
2 lone pairs. The budget is 8 electrons. Two single bonds use 4, so 4 remain. Four electrons make two pairs. O has an octet and both H atoms have two electrons.
3. Fix the CO2 skeleton
Show solution
Make O=C=O by moving existing pairs. The original carbon has only four electrons around it. Two double bonds give it eight. Each O keeps two lone pairs. You still use sixteen electrons; adding new electrons would break the budget.
4. Build HCN
Show solution
H–C≡N, with one lone pair on N. The budget is 1 + 4 + 5 = 10. Start with two single bonds (4 electrons), then put the remaining 6 on N. Carbon is short, so move two of N’s lone pairs into the C–N bond. Final count: 2 in H–C, 6 in C≡N, and 2 in N’s lone pair. Formal charges: H = 1 − 0 − 1 = 0; C = 4 − 0 − 4 = 0; N = 5 − 2 − 3 = 0. Hydrogen cannot have a double bond in this structure.
5. Audit nitrate’s central atom
Show solution
+1. N’s formal charge is 5 − 0 − ½(8) = +1. The double-bonded O is 0 and each single-bonded O is −1, giving the ion’s total −1 charge. An atom’s formal charge need not equal the overall ion charge.
6. Recognize resonance
Show solution
Only electron placement changes. Keep N connected to the same three O atoms. Place the N=O double bond on a different O, adjusting lone pairs and formal charges. There are three equivalent contributors, each with 24 electrons and overall charge −1. The real ion is a resonance hybrid with three equivalent N–O bonds.
Lewis structure FAQ
How do I choose the central atom?
For simple molecules with one central atom, start with the least electronegative atom other than hydrogen. Hydrogen is terminal, and fluorine is terminal in these structures. Follow any connectivity given in the problem; a formula alone may allow more than one arrangement.
How do I know when to use a double or triple bond?
If a central atom needs an octet but all available electrons are already placed, move a lone pair from a neighboring atom into a bond. Recheck the total, octets, and formal charges after each move. Check for octet exceptions before forcing a multiple bond.
Do I add or subtract electrons for an ion?
Add one electron for each unit of negative charge and subtract one for each unit of positive charge. For example, NO₃⁻ has 24 valence electrons, while NH₄⁺ has 8. Put brackets around a polyatomic ion and write the overall charge outside.
Is a Lewis structure the same as molecular geometry?
No. A Lewis structure records atom connections, bonds, and nonbonding electrons. Use those electron groups with VSEPR to predict a three-dimensional shape. For example, water has two bonds and two lone pairs on oxygen, and its molecular shape is bent.
Can a correct Lewis structure have formal charges?
Yes. Formal charges do not all have to be zero. Their sum must equal the overall charge. In a nitrate contributor, nitrogen is +1, two oxygens are −1 each, and the double-bonded oxygen is 0, for a total of −1.