Lewis Structures & Formal Charge
- Draw valid Lewis (electron-dot) structures for molecules and polyatomic ions
- Assign formal charges to choose the best of several resonance structures
- Recognize exceptions to the octet rule (incomplete and expanded octets)
A Lewis structure is a bookkeeping map of valence electrons
A Lewis structure shows where every valence electron sits — as a shared bonding pair (a line between atoms) or a lone pair (dots on one atom). The goal is usually to give each atom a full octet (8 valence electrons), the stable noble-gas count. Hydrogen is the exception: it is happy with just 2. Getting the electron count right is the whole game.
A repeatable drawing procedure
Follow the same steps every time: (1) Sum all valence electrons — add one per unit of negative charge, subtract one per unit of positive charge. (2) Place the least electronegative atom in the center (never hydrogen). (3) Draw single bonds to each outer atom. (4) Fill outer atoms to an octet with lone pairs. (5) Put any leftover electrons on the central atom. (6) If the center is short of an octet, convert lone pairs into double or triple bonds.
Formal charge picks the best structure
When several valid structures exist, the best one puts formal charges as close to zero as possible, and places any negative formal charge on the most electronegative atom. When several equally good structures differ only in where double bonds sit, the real molecule is a blend of all of them — resonance. The true bonding is the average, drawn with a double-headed arrow (⇌ between forms).
Draw the Lewis structure of the nitrate ion, NO₃⁻, and confirm it with formal charge.
- 1.Sum valence electrons: N contributes 5, each O contributes 6 (×3 = 18), and the −1 charge adds 1 → 5 + 18 + 1 = 24 electrons.
- 2.Put N in the center with three O atoms around it; draw three N–O single bonds (uses 6 electrons, 18 remain).
- 3.Complete octets on the three oxygens with lone pairs (uses all 18 remaining electrons).
- 4.N now has only 6 electrons — short an octet — so move one O lone pair into a second N=O bond, giving one double and two single bonds.
- 5.Check formal charge: the double-bonded O = 6 − 4 − ½(4) = 0; each single-bonded O = 6 − 6 − ½(2) = −1; N = 5 − 0 − ½(8) = +1. Sum = 0 + (−1) + (−1) + (+1) = −1, matching the ion's charge.
The three resonance forms of NO₃⁻ are identical apart from which O holds the double bond, so no single form is "real." Every N–O bond is measured to be the same length — intermediate between a single and a double bond — exactly what an average predicts.
How many total valence electrons must you place when drawing the Lewis structure of the sulfate ion, SO₄²⁻?
When the octet rule breaks
Three families of exceptions appear on the AP exam. Odd-electron molecules (radicals) like NO have an unpaired electron, so one atom cannot reach eight. Electron-deficient atoms — boron and beryllium — are stable with fewer than eight (BF₃ leaves boron with only 6). Expanded octets occur for central atoms in period 3 and below (P, S, Cl, Xe), which have empty d-orbitals available and can hold 10 or 12 electrons, as in SF₆ or PCl₅.
In which molecule does the central atom have an expanded octet (more than 8 electrons around it)?
What is the formal charge on the central oxygen atom in ozone, O₃, in the resonance form where it has one single bond, one double bond, and one lone pair?
On free-response Lewis questions, always tally your electrons against your step-1 total before moving on — an off-by-two count is the single most common way to lose the point. Then verify formal charges sum to the overall charge of the species.
Answer the 3 checkpoints as you read.
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