Lewis Structure Chemistry Example 2

Solution

1. 1
   Count total valence electrons: C has 4, each O has 6. Total = $4 + 2(6) = 16$.
2. 2
   Place C as central atom. With single bonds to each O: $\text{O–C–O}$ uses 4 electrons, leaving 12.
3. 3
   Place remaining electrons on O atoms (3 lone pairs each = 12 electrons). But C only has 4 electrons — it needs 8.
4. 4
   Move lone pairs to form double bonds: $\text{O=C=O}$. Now C has 8 electrons (two double bonds), each O has 8 (2 bonding + 2 lone pairs).