Mistake Master
Resonance and formal charge
Some molecules refuse to be captured by a single Lewis structure. Resonance draws several — but they are not snapshots the molecule flips between; they blend into one real average. Formal charge is the tool that ranks them.
§1
One hybrid, several contributors.
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When more than one valid Lewis structure can be drawn for a molecule, they are resonance structures. The true molecule is a single resonance hybrid, an average of the contributors — not any one of them, and not a molecule switching between them.
Resonance is not motion in time. The bonds do not flip back and forth; the real structure is the fixed blend, with bond character and charge spread out (delocalized) over the atoms the contributors disagree about.
Formal charge ranks the contributors. It is the charge an atom would carry if every bond were split evenly: FC = valence electrons − (lone electrons + ½ bonding electrons). The best contributors have formal charges closest to zero, with any negative charge on the more electronegative atom.
§2
Computing and using formal charge.
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Formal charge turns 'which structure is best?' into arithmetic.
- Count the atom's valence electrons. From its group — the electrons it would have as a free atom.
- Count what it 'owns' in the structure. All of its lone-pair electrons, plus half of each bonding pair it shares.
- Subtract. FC = valence − (lone + ½ bonding). A zero formal charge is ideal.
- Rank the contributors. Prefer structures with formal charges nearest zero, and with any negative formal charge on the more electronegative atom. That structure contributes most to the hybrid.
§3
The pieces you'll meet.
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Two ideas run this topic: the hybrid, and formal charge.
§4
Worked example: formal charge in nitrate, NO₃⁻.
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Question. In one resonance structure of NO₃⁻, nitrogen has one N=O double bond and two N–O single bonds. Find the formal charge on nitrogen.
Valence. Nitrogen (group 15) has 5 valence electrons as a free atom.
Owns. Nitrogen here has no lone pairs, and four bonds (one double = 2 pairs, two singles = 2 pairs) = 8 bonding electrons, half of which is 4.
Formal charge. FC = 5 − (0 + 4) = +1 on nitrogen. The three resonance structures (rotating which oxygen holds the double bond) are equivalent contributors, so the real ion is their average: the negative charge and the double-bond character are spread evenly over all three oxygens.
§5
Mistakes that cost real points.
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"The molecule rapidly flips between the resonance structures."
Resonance is not flipping in time. The molecule is one unchanging hybrid, an average of the contributors, with the shared electrons delocalized. Nitrate does not spend part of its time as each structure; it is always the blend.
Fix. Read the double-headed resonance arrow as 'averages into,' not 'flips to.' The real structure is the single, static hybrid.
"The most symmetric-looking structure is the best contributor."
Symmetry is not the selector; formal charge is. The best contributor has formal charges closest to zero, with negative charge on the more electronegative atom. A symmetric drawing with large formal charges is a minor contributor.
Fix. Compute formal charges and rank by them. Prefer near-zero charges on the right atoms, regardless of how symmetric a structure looks.
"Every molecule has one clean structure where all octets are satisfied."
Some molecules cannot be captured by a single complete-octet structure, which is exactly why resonance exists; others have odd electron counts (like NO) or electron-deficient atoms and never reach a tidy all-octet picture. Assuming one always exists leads to forcing wrong structures.
Fix. When one structure will not do, draw the resonance contributors and take the hybrid; accept that some species genuinely lack a single all-octet structure.
§6
Skill Check.
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Ten scenarios. Pick the chips that match your answer, then check. A scenario marks complete the first time every part is right. Progress saves on this device.