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Lewis structures and formal charge

Build any Lewis structure in five steps and use formal charge to pick between resonance forms — without memorization.

Quick answer Count valence electrons, draw a single-bonded skeleton, fill outer octets with lone pairs, then make multiple bonds to satisfy a short central atom. Assign each atom a formal charge = valence − lone-pair electrons − ½(bonding electrons); the best structure keeps those charges smallest, with negatives on the most electronegative atom.

Every arrow in organic chemistry starts from a lone pair or bond, and every charge comes from a formal-charge count — so resonance, acids and bases, and mechanisms all follow from these two skills.

Water
Ammonia
Carbon dioxide
Ammonium (+1 on N)

The species we will build and charge-count below — drawn live.

1. A Lewis Structure Shows Every Bonding Pair and Every Lone Pair

A Lewis structure draws a line for each bonding pair and a dot-pair for each lone pair; as bonds fall 4 → 3 → 2, lone pairs climb 0 → 1 → 2 to keep a full octet.

Methane — 4 bonds, 0 lone pairs
Ammonia — 3 bonds, 1 lone pair
Water — 2 bonds, 2 lone pairs

2. Build Any Lewis Structure in the Same Five Steps

(1) Count valence electrons (CO₂ = 4 + 2·6 = 16); (2) draw a single-bonded skeleton, least electronegative atom central; (3) fill outer octets with lone pairs; (4) make multiple bonds to satisfy a short central atom; (5) check formal charge.

CO₂ — two C=O double bonds
Formaldehyde — one C=O
Methanol — all single bonds

3. Formal Charge = Valence Electrons − Lone-Pair Electrons − Half the Bonding Electrons

Take valence electrons minus lone-pair electrons minus half the bonding electrons: water's O = 6 − 4 − 2 = 0, hydronium's O = 6 − 2 − 3 = +1, ammonium's N = 5 − 0 − 4 = +1 — the extra bond, not a missing electron, carries the charge.

Hydronium — +1 on O (3 bonds)
Ammonium — +1 on N (4 bonds)
Hydroxide — −1 on O (1 bond)

4. The Formal Charges Always Add Up to the Overall Charge

The formal charges on every atom must sum to the overall charge: nitromethane's +1 N and −1 O cancel to neutral, and methoxide's lone −1 O matches its −1 — if they don't sum, you miscounted at step 1.

Nitromethane — +1 N, −1 O, neutral overall
Methoxide — one −1 on O

5. Formal Charge Picks the Best Resonance Contributor

The major contributor has the fewest and smallest formal charges, with negatives on the most electronegative atom — though carbon monoxide's octet-satisfying triple bond forces a −1 on carbon, and ozone splits into two equivalent charged forms.

Carbon monoxide — −1 C, +1 O
Ozone — +1 center, −1 terminal

6. Where the Charge Sits: Reading Real Ions

In common ions the charge is delocalized by resonance: nitrate spreads −1 over three oxygens (N is +1), carbonate spreads −2 over three, and acetate spreads −1 over two — see resonance structures and hybridization.

Nitrate — +1 N, −1 spread over O
Carbonate — −2 over three O
Acetate — −1 over two O

7. Summary

Count valence electrons · single-bonded skeleton · fill outer octets · multiple bonds for a short center · formal charge = valence − lone-pair − ½(bonding) · use it to check (charges sum to total), locate, and judge the best structure.

Worked example

Problem. What is the formal charge on nitrogen in the nitronium ion, O=N=O?
nitronium, NO2+
  1. Formal charge = (valence e⁻) − (nonbonding e⁻) − ½(bonding e⁻).
  2. Nitrogen: 5 valence electrons; in O=N=O it has 0 lone-pair electrons and 8 bonding electrons (two double bonds).
  3. FC = 5 − 0 − ½(8) = 5 − 4 = +1.

Answer. Nitrogen is +1 (each oxygen is 0) — consistent with the overall NO2+ charge.

Quiz yourself

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+1. Oxygen's group valence = 6. It keeps one lone pair (2 nonbonding electrons) and makes three bonds (6 bonding electrons, half = 3). Formal charge = 6 − 2 − 3 = +1. The extra bond, not a missing electron, is what makes it positive.

Zero. Formal charges always sum to the overall charge: (+1) + (−1) + zeros on every other atom = 0, so the molecule is neutral even though it contains charged atoms.

The one with the −1 on oxygen. When charges are otherwise equal, the major resonance contributor places the negative charge on the more electronegative atom, and oxygen is more electronegative than nitrogen.

Because giving both atoms a full octet requires a carbon–oxygen triple bond. With three bonds and one lone pair, carbon counts 4 − 2 − 3 = −1 and oxygen counts 6 − 2 − 3 = +1. Satisfying the octet outranks the slightly awkward charge placement.

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