Worked example 1: Build the Lewis picture for nitrate, NO₃^-, and explain why one drawing is not enough.

Count valence electrons.
Nitrogen contributes 5, three oxygens contribute 18, and the negative charge adds 1, for a total of 24 electrons.
Draw N in the center with three O atoms attached.
Three single bonds use 6 electrons, leaving 18 to place as lone pairs on the oxygens.
Check the center and formal charge.
With only single bonds, nitrogen lacks an octet and formal charges are large. Turning one oxygen lone pair into an N=O double bond gives nitrogen an octet and improves the charge pattern.
Recognize equivalent choices.
Any of the three oxygens can be the double-bonded oxygen in a valid drawing, so the real nitrate ion is represented by three resonance forms, not one frozen double bond.

Common mistake: drawing one nitrate structure and then speaking as if one N-O bond is permanently different. The resonance model says the electron density is delocalized across all three N-O connections.

Worked example 2: Why is BF₃ not repaired by forcing three B=F double bonds?

Count around boron in the normal Lewis structure.
Boron has three B-F single bonds, so only six electrons are assigned around B.
Check whether boron is allowed to be electron-poor.
Boron compounds commonly accept electron pairs; BF₃ behaves as a Lewis acid.
Test the forced-double-bond repair.
Multiple B=F double bonds create poor formal-charge and bonding assumptions for this early model. The exception is the point, not a mistake to hide.

Common mistake: repairing every incomplete octet with extra multiple bonds. First ask whether the central atom is a known exception.

Worked example 3: Why can SF₆ be drawn with six S-F bonds in a Lewis model?

Sulfur is in the third period.
The strict second-period octet limit is strongest for atoms such as C, N, O, and F.
Six fluorines around sulfur create six bonding domains.
The simplified Lewis structure places twelve electrons around S, an expanded-octet count.
Connect to geometry later.
Six electron domains lead naturally into an octahedral VSEPR arrangement.

Common mistake: using expanded octets for second-period atoms. Carbon, nitrogen, oxygen, and fluorine should not be casually given ten or twelve electrons.

Worked example 4: Use the formal-charge equation on one nitrate resonance form.

Use formal charge = valence electrons - nonbonding electrons - 1/2(bonding electrons).
This equation checks whether the electron bookkeeping in a Lewis drawing is plausible.
Double-bonded oxygen: 6 - 4 - 1/2(4) = 0.
That oxygen has two lone pairs and one double bond, so it carries no formal charge in that resonance form.
Single-bonded oxygen: 6 - 6 - 1/2(2) = -1.
Each singly bonded oxygen has three lone pairs and one bond, so each carries -1 formal charge.
Nitrogen: 5 - 0 - 1/2(8) = +1.
Nitrogen has four bonds and no lone pairs in this drawing. The total charge is 0 + (-1) + (-1) + (+1) = -1, matching nitrate.
Then apply resonance.
The -1 charges are not pinned to two permanent oxygens; the equivalent resonance forms spread the electron density over all three N-O connections.

Common mistake: using formal charge as a separate ritual after the structure is finished. It is an error-checking equation that tells you whether the drawing and total charge agree.

Chemical bonding