A Lewis structure is a map of a molecule's valence electrons: lines for shared pairs (bonds) and dots for unshared pairs (lone pairs). It builds on dot symbols from Valence Electrons and Dot Symbols and the bonding from Covalent Bonds, and it is the first step toward predicting a molecule's shape.
🔑 Key idea
1. Count all the valence electrons (add one for each negative charge, subtract one for each positive charge). 2. Connect the atoms with single bonds; the central atom is usually the least electronegative one, never H. 3. Give the outer atoms full octets (H gets 2) with lone pairs. 4. Put leftover electrons on the central atom. 5. If the central atom still has fewer than 8, turn lone pairs from outer atoms into double or triple bonds.
💡 Why it works
The electron count is fixed by the atoms and the charge: a structure that shows more or fewer electrons than that is describing a different particle. Each step then spends that budget so that every atom ends up surrounded by a full shell, the arrangement that is lowest in energy for these atoms. Moving a lone pair into a shared position (step 5) lets two atoms count the same electrons.
Carbon dioxide. The electron budget shows 16 valence electrons, all placed; carbon gets its octet from two double bonds.
Worked examples
✏️ Worked example: Water
Draw the Lewis structure of $\mathrm{H_2O}$.
Valence electrons: $2 \times 1 + 6 = 8$.
O in the center, bonded to each H: 2 bonds use 4 electrons.
The 4 left go on O as 2 lone pairs. O has 8, each H has 2.
Answer: H–O–H with two lone pairs on O.
✏️ Worked example: Carbon dioxide
Draw the Lewis structure of $\mathrm{CO_2}$.
Valence electrons: $4 + 2 \times 6 = 16$.
O–C–O uses 4; give each O three lone pairs (12). All 16 are placed, but C has only 4 electrons around it.
Move one lone pair from each O into a bond: O=C=O. Now every atom has 8.
Answer: O=C=O, two lone pairs on each O.
✏️ Worked example: An ion
How many valence electrons does the ammonium ion, $\mathrm{NH_4^+}$, have?
N: 5; H: $4 \times 1 = 4$. Total 9.
The + charge means one electron is missing: $9 - 1 = 8$.
Four N–H bonds use all 8.
Answer: 8 valence electrons
⚠️ Common mistake
Counting the wrong total, especially for ions (forgetting to add electrons for a negative charge or remove them for a positive one), or giving carbon five bonds. Check every structure twice: count the electrons drawn (2 per line, 1 per dot) against your total, and count the electrons around each atom (no more than 8 for C, N, O and F).
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.