AP Chemistry · Lesson 2

Lewis structures, resonance and formal charge

Know Know it

To draw a Lewis structure: (1) count all valence electrons, adding one per negative charge and removing one per positive charge; (2) put the least electronegative atom in the center (hydrogen is always on the outside); (3) connect the atoms with single bonds; (4) complete the octets of the outer atoms with lone pairs; (5) put any electrons left over on the central atom; (6) if the central atom still has fewer than eight, turn outer lone pairs into double or triple bonds.

Exceptions: boron is often stable with six electrons (BF₃), a molecule with an odd number of electrons must leave one atom short (NO), and central atoms from period 3 or below can hold more than eight (PCl₅, SF₆). Period 2 atoms never exceed eight.

Formal charge = (valence electrons) − (lone-pair electrons) − ½(bonding electrons). The formal charges in a structure add up to the charge of the particle. When several structures are possible, the best one has formal charges closest to zero, with any negative formal charge on the more electronegative atom.

When a double bond could go in more than one place, the particle is a resonance hybrid. The real bonds are all the same, with a bond order equal to the total shared pairs divided by the number of bond positions they are spread over.

Worked example Easy

How many valence electrons must be shown in the Lewis structure of dichloromethane, CH₂Cl₂?

  1. 18
  2. 28
  3. 20Answer
  4. 10

How to solve it

  1. C is in group 14: 4 valence electrons. Each H brings 1: 2 in all.
  2. Cl is in group 17: 7 each, so 2 × 7 = 14.
  3. Total: 4 + 2 + 14 = 20 electrons, or 10 pairs.

Why each choice is right or wrong

  • A. Incorrect. 18 gives each chlorine only 6 valence electrons, as if it were in group 16; chlorine is in group 17 and has 7.
  • B. Incorrect. 28 counts the electrons each atom is surrounded by in the finished structure (8 for C and each Cl, 2 for each H), not the electrons the atoms bring.
  • C. Correct. Carbon brings 4, each hydrogen 1 and each chlorine 7, so 4 + 2(1) + 2(7) = 20 valence electrons.
  • D. Incorrect. 10 is the number of electron pairs (20 ÷ 2), not the number of electrons.

Trap Count the charge of an ion

For a polyatomic ion, the charge changes the electron count. The ammonium ion, NH₄⁺, has 5 + 4 − 1 = 8 valence electrons, and the hydroxide ion, OH⁻, has 6 + 1 + 1 = 8. Draw the finished structure in brackets with the charge outside.

Explain Explain it

Task verbs: determine, explain

Formal charge is bookkeeping: it assumes every bond is shared exactly evenly. It is not the real charge on the atom, but it is a good guide to which structure is closest to the real one.

To compare two structures, find every atom's formal charge in each. Prefer the one with smaller formal charges. If both have the same set of charges, prefer the one that puts the negative charge on the more electronegative atom.

Worked example Medium

A Lewis structure of the sulfate ion, SO₄²⁻, is drawn with sulfur in the center, four S–O single bonds and three lone pairs on each oxygen. What is the formal charge on the sulfur atom in this structure?

  1. 0
  2. +2Answer
  3. −2
  4. +6

How to solve it

  1. Formal charge = valence − lone-pair electrons − ½(bonding electrons).
  2. Sulfur: 6−0−12(8)=+26 - 0 - \frac{1}{2}(8) = +2.
  3. Each oxygen: 6−6−12(2)=−16 - 6 - \frac{1}{2}(2) = -1, and +2+4(−1)=−2+2 + 4(-1) = -2, the charge of the ion.

Why each choice is right or wrong

  • A. Incorrect. A formal charge of 0 on sulfur would need two S=O double bonds (6 − 0 − 6 = 0); with four single bonds and no lone pairs it is +2.
  • B. Correct. Sulfur has 6 valence electrons, no lone-pair electrons and 8 bonding electrons, so 6 − 0 − ½(8) = +2.
  • C. Incorrect. −2 is the charge of the whole ion; in this structure the −2 comes from the four oxygens at −1 each plus sulfur at +2.
  • D. Incorrect. +6 subtracts nothing for the bonding electrons; half of the 8 shared electrons count toward sulfur.

Determine Your turn

Two Lewis structures are drawn for the cyanate ion, OCN⁻, with carbon in the center. In structure I, O=C=N, oxygen has two lone pairs and nitrogen has two lone pairs. In structure II, O–C≡N, oxygen has three lone pairs and nitrogen has one lone pair. Determine the formal charge on each atom in both structures, then explain which structure better represents the ion.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the formal charges in structure I: O 0, C 0, N −1
  • Gives the formal charges in structure II: O −1, C 0, N 0
  • Chooses structure II because it puts the negative formal charge on oxygen, the more electronegative atom

Model answer

Structure I: oxygen is 6 − 4 − 2 = 0, carbon is 4 − 0 − 4 = 0 and nitrogen is 5 − 4 − 2 = −1. Structure II: oxygen is 6 − 6 − 1 = −1, carbon is 4 − 0 − 4 = 0 and nitrogen is 5 − 2 − 3 = 0. Both structures have one atom at −1 and the others at 0, so the charges alone do not decide it. Structure II is better, because it puts the negative formal charge on oxygen, which is more electronegative than nitrogen.

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