The big idea: Sometimes a molecule or ion can be drawn as more than one valid Lewis structure. Formal charge is a bookkeeping tool that helps you decide which drawing is most realistic.
The idea: pretend every bond is shared exactly equally, give each atom half of every bonding pair plus all of its lone-pair electrons, then compare that count with the atom's normal number of valence electrons.
If an atom 'owns' the same number it started with, its formal charge is 0. More electrons than normal → negative; fewer → positive.
What the symbols mean: For one atom in a structure:
- V — its valence electrons as a free atom (the group number: C is group 14 → 4, O is group 16 → 6). - N — its non-bonding electrons (the dots in its lone pairs). - B — its bonding electrons (count 2 per single bond, 4 per double, 6 per triple, summed over all its bonds).
Formal charge is not a real charge — it is a label that helps compare structures.
Apply the same rule to every atom. The formal charges must add up to the overall charge of the molecule or ion (0 for a neutral molecule).
- formal charge on the atom
- valence electrons of the free (neutral) atom = its group number (1–18 → 1–8)
- non-bonding electrons (electrons in lone pairs on that atom)
- bonding electrons (total electrons in the bonds to that atom)
Worked example — formal charge in the ammonium ion
Calculate the formal charge on the nitrogen atom in the ammonium ion, NH4+ (N is bonded to four H atoms, with no lone pairs on N).
Solution
- Formula first:
- Read off the three counts for nitrogen. N is in group 15, so V = 5. It has no lone pairs, so N = 0. It has four single bonds, so B = 4 × 2 = 8:
- Substitute:
- Work it out:
Final answer
The nitrogen has a formal charge of +1 — which is exactly why the ion carries an overall +1 charge.
Worked example — formal charge in the hydroxide ion
Calculate the formal charge on the oxygen atom in the hydroxide ion, OH- (O has one O–H single bond and three lone pairs).
Solution
- Formula first:
- Oxygen is in group 16 → V = 6. Three lone pairs → N = 6. One single bond → B = 2:
- Substitute and solve:
Final answer
The oxygen has a formal charge of −1, matching the −1 charge of the hydroxide ion.
Know your predicted grade
Take timed mock exams and get detailed feedback on every answer. See exactly where you're losing marks.
When two or more structures are possible, formal charge picks the most plausible one. Use two rules together:
The two selection rules: (1) The best structure has formal charges as close to zero as possible on every atom.
(2) If a structure must carry a negative formal charge, it should sit on the most electronegative atom (the one that holds electrons best).
Smaller, better-placed formal charges → the more realistic structure.
Worked example — formal charge picks the structure
Two Lewis structures can be drawn for the thiocyanate ion, SCN-: structure A is S=C=N (a double bond either side), and structure B is S–C≡N (single S–C, triple C≡N). In each, the terminal atoms complete their lone pairs. Use formal charge to decide which structure is more plausible.
Solution
- Formula first — apply FC = V − N − ½B to each terminal atom in turn.
- Structure A (S=C=N). Sulfur (V=6) has 2 lone pairs (N=4) and one double bond (B=4): FC = 6 − 4 − 2 = 0. Nitrogen (V=5) has 2 lone pairs (N=4) and one double bond (B=4): FC = 5 − 4 − 2 = −1.
- Structure B (S–C≡N). Sulfur (V=6) has 3 lone pairs (N=6) and one single bond (B=2): FC = 6 − 6 − 1 = −1. Nitrogen (V=5) has 1 lone pair (N=2) and a triple bond (B=6): FC = 5 − 2 − 3 = 0.
- Both keep the carbon at 0 and sum to −1 (the ion's charge), so rule (1) is a tie. Apply rule (2): the −1 should sit on the more electronegative atom. Nitrogen is more electronegative than sulfur, so the −1 belongs on N — that is structure B.
Final answer
Structure B (S–C≡N) is more plausible — its negative formal charge sits on the more electronegative nitrogen.
Resonance — one molecule, several drawings: Some species cannot be shown by a single Lewis structure. The carbonate ion CO32-, for example, could have its one C=O double bond on any of the three oxygens — three equally valid drawings.
The true structure is a resonance hybrid: the bonding electrons are delocalized (spread evenly) over all the positions, not stuck in one place. We draw the separate resonance structures linked by a double-headed arrow (↔) to represent this average.
One of three equivalent resonance forms: the C=O double bond can sit on any of the three oxygens.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Equal bond lengths are the evidence: Because the electrons are delocalized, the bonds become identical. In CO32- all three C–O bonds are the same length — intermediate between a true single (longer) and a true double (shorter) bond.
Nitrate NO3- behaves the same way (three equal N–O bonds), and ozone O3 has two equal O–O bonds. Equal bond lengths where you'd expect a single and a double are the experimental fingerprint of resonance.
One of three equivalent resonance forms of NO₃⁻ — all three N–O bonds are really equal.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
Ozone has two equivalent resonance forms; the real molecule has two identical O–O bonds.
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
When the octet rule breaks: The octet (eight outer electrons) is a guideline, not a law. Two HL exceptions:
- Electron-deficient species — the central atom has fewer than 8 electrons. Beryllium in BeCl2 has only 4; boron in BF3 has only 6. They are happy as they are (and act as electron-pair acceptors). - Expanded octets — a period-3-or-below central atom can hold more than 8 by using its larger valence space. Phosphorus in PCl5 has 10 outer electrons; sulfur in SF6 has 12. Period-2 atoms (C, N, O, F) can never expand — they have no room.
Boron has only 6 electrons in its outer shell — an incomplete octet (electron-deficient).
Interactive diagram
Explore the labelled diagram, charts and maps for this topic in full study mode.
| Type | Central atom | Outer electrons on centre | Examples |
|---|---|---|---|
| Obeys the octet | C, N, O, F (period 2) | 8 | CH4, NH3, H2O |
| Electron-deficient | Be, B | 4 (Be) · 6 (B) | BeCl2, BF3, BCl3 |
| Expanded octet | P, S, Cl … (period 3+) | 10, 12 (uses empty 3d-type space) | PCl5, SF6, SF4, XeF4 |