2.5 Lewis Diagrams

Formal Charge

Click any atom on 13 Lewis diagrams to see FC = V − L − ½B worked out term by term, then step between the competing structures of BF₃, SO₂ and sulfate to see which one formal charge picks

13 SpeciesFC = V − L − ½BOctet vs ChargeCharge Balance
Topic 2.5

Lewis Diagrams

Represent a molecule with a Lewis diagram.

A Lewis diagram shows every valence electron in a molecule or polyatomic ion as either a bonding pair or a lone pair. EK 2.5.A.1 states simply that Lewis diagrams are constructed according to an established set of principles — here is that set:

  1. Count total valence electrons. Sum the valence electrons of every atom. Add one electron per unit of negative charge; subtract one per unit of positive charge.
  2. Choose the central atom. The least electronegative atom (excluding hydrogen, which is always terminal). Carbon is central whenever present.
  3. Connect with single bonds. Each bond uses 2 electrons.
  4. Complete octets on terminal atoms first using lone pairs. Hydrogen stops at 2 electrons.
  5. Place leftover electrons on the central atom.
  6. If the central atom is short of an octet, pull terminal lone pairs in as multiple bonds.

Then check: does your electron count match step 1 exactly? That check catches nearly every error.

Common departures from the octet rule that appear on the AP Exam:

  • Hydrogen — 2 electrons (a duet).
  • Boron and beryllium — often stable with fewer than 8 (BF₃ has 6 around boron).
  • Expanded valence — atoms in period 3 and beyond can hold more than 8 (PCl₅, SF₆). Note the CED's limit: when more than four electron pairs surround a central atom, you are responsible only for the resulting shape.

Key points

  • Total valence electrons is a conserved quantity — count it first, verify it last.
  • Hydrogen is never a central atom and never exceeds two electrons.
  • A central atom short of an octet is the signal to form a multiple bond, not to add more electrons.
  • Charges on polyatomic ions change the electron count and belong in brackets with the charge shown.

Common mistakes

  • Forgetting the ionic charge. SO₄²⁻ has 32 valence electrons (6 + 4×6 + 2), not 30.
  • Putting hydrogen in the middle. Water is H–O–H, never H–H–O.
  • Overfilling second-period atoms. N, O, F, and C can never exceed 8 electrons — there are no available d orbitals in n = 2.
  • Drawing bonds you did not budget for. Every line is two electrons out of your total.

Worked example

Draw the Lewis diagram for the nitrite ion, NO₂⁻, and verify the electron count.

Step 1 — count: N (5) + 2 × O (6) + 1 (charge) = 18 valence electrons.

Step 2 — skeleton: nitrogen is less electronegative, so it is central: O–N–O.

Step 3 — single bonds: 2 bonds = 4 electrons used, 14 remain.

Step 4 — octets on terminals: 3 lone pairs on each O = 12 electrons, 2 remain.

Step 5 — leftovers on N: 1 lone pair on nitrogen. Nitrogen now has 2 bonding pairs + 1 lone pair = 6 electrons, short of an octet.

Step 6 — form a double bond: move one lone pair from an oxygen into an N–O bond. Now: one N=O double bond, one N–O single bond, one lone pair on N.

Verify: 2 (single bond) + 4 (double bond) + 2 (N lone pair) + 4 (double-bonded O lone pairs) + 6 (single-bonded O lone pairs) = 18 ✓

Enclose in brackets with a −1 charge. Because the double bond could equally be on either oxygen, this ion has two resonance structures (see 2.6).

Full notes for topic 2.5 →