Introduction to Resonance共振结构入门

Big Idea · Structure and Properties (SAP)AP Science Practice 1 · Models and RepresentationsAP link · Unit 2~30 min

Learning goals

  • Explain what resonance structures are, and draw them for O₃, NO₃⁻, and CO₃²⁻
  • Explain that the real species is a resonance hybrid, and use average bond order to explain bond lengths and charge distribution
  • Recognize the delocalized π bonds that Chinese textbooks use to describe the same thing

Before you start

  • Drawing Lewis structures (2.4.4)

Why are ozone’s two bonds the same length?

Follow the steps of lesson 2.4.4 for ozone, O₃: 18 valence electrons, an O–O–O skeleton, and after completing the outer atoms the central oxygen is 2 electrons short, so one single bond becomes a double bond. That structure has one O=O double bond and one O–O single bond, so you’d expect one bond of about 121 pm and one of about 148 pm. But experiments show that ozone’s two bonds are the same length: both are 127.8 pm. A single Lewis structure can’t explain this.

Resonance structures

The problem is where to put the double bond: on the left or on the right, both satisfy the octet rule, and the two drawings are equally good.

OOO
↔
OOO

When a molecule or ion can be drawn as two or more valid Lewis structures with the atoms in the same places and only the electrons (double bonds and lone pairs) in different places, each one is a resonance structure共振结构. They are joined by a double-headed arrow, ↔.

The real species is none of them; it is a blend called the resonance hybrid共振杂化体. In the hybrid, the extra pair of electrons in the double bond isn’t fixed between two particular atoms but spread over all three. Electrons spread out like this are delocalized electrons离域电子.

An analogy: a mule is the offspring of a horse and a donkey, but it isn’t a horse one moment and a donkey the next; it is always a mule. A resonance hybrid likewise doesn’t flip between its resonance structures.

Average bond order

The bond order of each bond in the hybrid is the average of its bond orders in the resonance structures:

  • O₃: two structures, in which each bond is double once and single once, for an average bond order of (2 + 1) ÷ 2 = 1.5. So both bonds are the same length, between a single and a double bond.
  • NO₃⁻ and CO₃²⁻: three structures, in which each bond is double once and single twice, for an average of (2 + 1 + 1) ÷ 3 = 4/3. All three bonds are identical.

Formal charges (lesson 2.4.4) average too. In each NO₃⁻ structure, nitrogen has a formal charge of +1, the double-bonded oxygen 0, and the two single-bonded oxygens −1 each; in the hybrid, every oxygen carries an average of −⅔, so the negative charge is spread evenly over all three.

How Chinese textbooks say it: delocalized π bonds

Chinese textbooks often describe the same thing another way. In O₃, the three oxygen atoms are first joined by σ bonds; then parallel p orbitals on all three atoms combine into one π bond spread over the whole molecule. This delocalized π bond大 π 键 is written Π₃⁴ (3 atoms, 4 electrons). NO₃⁻ and CO₃²⁻ each have a Π₄⁶. “Resonance hybrid” and “delocalized π bond” both describe π electrons spread over several atoms.

Choose a species and look at its resonance structures. Use the step-through button, then compare with the hybrid below:

Try it: compare the resonance structures of O₃, NO₃⁻, and CO₃²⁻ with their resonance hybrids

Resonance structures (the double bond moves; the atoms do not):

O+O−O
OOO
↔
O+OO−
OOO

Resonance hybrid (the real structure; dashed lines are partial double bonds):

O+O−½O−½

Both O–O bonds are the same length, 127.8 pm, between an O–O single bond (148 pm) and an O=O double bond (121 pm): the average bond order is 1.5. Each end oxygen carries a formal charge of −½, and the central oxygen +1. Chinese textbooks often describe the same thing as a delocalized π bond, written Π₃⁴.

Resonance structures are joined by a double-headed arrow, ↔. The real species doesn't flip back and forth between them; it is always the hybrid.

Example 1: The nitrate ion

Draw the resonance structures of NO₃⁻ and describe its three N–O bonds.

Step-by-step solution

  1. Count the valence electrons. For an anion, add the charge:

    5+3×6+1=245 + 3 \times 6 + 1 = 24

  2. Skeleton and outer atoms. Nitrogen goes in the center; three single bonds use 6 electrons. Giving each oxygen 3 lone pairs uses the other 18.

  3. The central atom is short. Nitrogen has only 6 electrons, so move a lone pair from one oxygen to form an N=O double bond.

  4. Three equivalent drawings. The double bond can go to any of the three oxygens, so there are three resonance structures, joined by ↔.

  5. The hybrid. The real ion is the average of the three: all three N–O bonds are identical, with bond order

    2+1+13=43\frac{2 + 1 + 1}{3} = \frac{4}{3}

    Each oxygen carries an average formal charge of −⅔, so the ion’s negative charge is spread evenly over the three oxygens.

  6. A common mistake. Thinking the ion flips back and forth between the three structures. Resonance structures are just different drawings of one real structure.

Example 2: Oxygen–oxygen bonds in O₂ and O₃

Which has the shorter oxygen–oxygen bond, O₂ or O₃?

Step-by-step solution

  1. O₂’s bond order. The two oxygen atoms share a double bond: bond order 2.

  2. O₃’s bond order. O₃ has two resonance structures (one single and one double bond, swapped), so each O–O bond has an average order of

    2+12=1.5\frac{2 + 1}{2} = 1.5

  3. Compare. A higher bond order shares more electrons and makes a shorter bond, so O₂’s bond is shorter.

  4. Check with data. The bond length is 121 pm in O₂ and 127.8 pm in O₃.

Practice: how many equivalent resonance structures can be drawn for each species?
  1. O₃
  2. NO₃⁻
  3. CO₃²⁻
  4. NO₂⁻
  5. HCOO⁻ (formate)
  6. CO₂
  7. H₂O
  8. CH₄
  9. N₂

Common mistakes

Mistake 1: “Ozone flips rapidly back and forth between its two structures.” Resonance structures are only our pencil-and-paper approximations; the real molecule is always the hybrid, and its two bonds are always the same length.

Mistake 2: “You can move atoms when drawing resonance structures.” Resonance structures differ only in where the electrons are; the atoms stay put. If you move an atom, you get a different substance, not a resonance structure.

Mistake 3: “NO₃⁻ has one N=O double bond and two N–O single bonds.” That’s true of each resonance structure, but in the real ion all three N–O bonds are identical, each with a bond order of 4/3.

Quick check

Which statement about the ozone molecule, O₃, is correct?

What is the average C–O bond order in CO₃²⁻?

Summary

  • When a species can be drawn as several valid Lewis structures with the same atom positions but different electron positions, these are resonance structures, joined by ↔.
  • The real species is the resonance hybrid, with electrons delocalized over several atoms. Its bond orders and formal charges are the averages over the structures: 1.5 for O₃ and 4/3 for NO₃⁻ and CO₃²⁻.
  • Chinese textbooks often describe the same thing with delocalized π bonds: Π₃⁴ in O₃ and Π₄⁶ in NO₃⁻ and CO₃²⁻.
On to AP (for later)

AP U2.6 often asks why the three N–O bonds in NO₃⁻ are the same length. A full-credit answer says that NO₃⁻ has three equivalent resonance structures, the real structure is their hybrid with delocalized electrons, and so all three bonds are identical, with lengths between a single and a double bond. AP may also ask you to use formal charge to decide which resonance structure contributes most: the closer its formal charges are to 0, the more important it is.

Practice this chapter: 2.4 · Chemical Bonding and Molecular Structure · 10 questions

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