AP Chem 2.6: Resonance and Formal Charge

When one diagram isn’t enough

Experiments show that the three N–O bonds in $\ce{NO3-}$ are the same length and all three O–N–O angles are 120°. Yet the Lewis diagram drawn by the steps in U2.5 has one double bond and two single bonds, which seems to predict one short bond and two long ones. The problem isn’t the experiment; a single Lewis diagram just can’t describe this ion. Lesson 2.4.5 introduced resonance structures. AP topic 2.6 goes further: decide when resonance is needed, calculate average bond orders, and use formal charge to choose the most reasonable structure.

When resonance structures are needed

If, with the atoms fixed, a multiple bond (along with lone pairs) can be placed in several equivalent positions, draw all of these structures and connect them with “↔”. They are resonance structures共振结构. When drawing them:

move only electrons (lone pairs and the electrons of multiple bonds), never atoms;
every structure has the same total number of electrons, and every atom has an octet (or an expanded octet for a period 3 central atom).

Needs resonance structures	Doesn’t need them
$\ce{O3}$ , $\ce{SO2}$ , $\ce{NO2-}$	$\ce{CO2}$ (only one structure has all formal charges 0)
$\ce{NO3-}$ , $\ce{CO3^2-}$	$\ce{H2O}$ , $\ce{CH4}$ , $\ce{NH4+}$ (single bonds only; no multiple bond to move)
carboxylates (such as $\ce{CH3COO-}$ ), benzene $\ce{C6H6}$	$\ce{SF6}$ (six single bonds)

↔

The resonance hybrid and average bond order

The real species is not any one of these structures, and it doesn’t switch back and forth between them; it is their average, the resonance hybrid共振杂化体. The average bond order at each position in the hybrid is

\text{average bond order} = \frac{\text{total bonds over those positions}}{\text{number of positions}}

Species	Calculation	Average bond order
$\ce{O3}$ , $\ce{SO2}$	3 ÷ 2	1.5
$\ce{NO3-}$ , $\ce{CO3^2-}$	4 ÷ 3	4/3 ≈ 1.33
the two C–O bonds of a carboxylate	3 ÷ 2	1.5
the six C–C bonds of benzene	9 ÷ 6	1.5

The experimental evidence for resonance: bonds that would otherwise differ are the same length, between a single and a double bond; the corresponding bond angles are equal too (all 120° in $\ce{NO3-}$ ). The six C–C bonds of benzene are identical, so benzene is not “three single bonds plus three double bonds.”

Choose a species to see its resonance structures and its hybrid (dashed lines are partial double bonds):

Try it: compare the resonance structures and resonance hybrids of several species

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

↔

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

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.

Show formal charges

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.

Formal charge: scoring a structure

Sometimes a species can be drawn several ways with the same connections but different electron arrangements that are not equivalent. Then use formal charge形式电荷 to pick the most reasonable one (2.4.4):

\text{formal charge} = V - N - B

V is the number of valence electrons of the atom, N the number of electrons in its lone pairs, and B the number of bonds to it. The formal charges of all the atoms add up to the charge of the whole species. The rules for choosing:

Formal charges closest to 0 are best (the smallest sum of absolute values).
If two structures tie, the negative formal charge should be on the more electronegative atom.

Formal charge is a bookkeeping device, not the actual charge on an atom. Its job is to compare which of several structures is more reasonable.

Try it: use formal charge to pick the best Lewis structure

Each structure below has the right number of electrons, and every atom has an octet (the S in SO₄²⁻ may have more than 8). Click the one you think is best:

Show formal charges on the diagrams

V: valence electrons of the atom; N: electrons in its lone pairs; B: number of bonds to it. Rules: formal charges closest to 0 are best; if tied, the negative formal charge should be on the more electronegative atom.

Period 3 central atoms: two ways to draw them. S, P, Cl, and similar atoms can hold more than 8 electrons, so $\ce{SO2}$ and $\ce{SO4^2-}$ are often drawn two ways: keeping the octet ( $\ce{SO2}$ as two resonance structures, with +1 on S), or minimizing formal charge with an expanded octet ( $\ce{SO2}$ as O=S=O, all formal charges 0). Both models predict the same shape and the same equal bond lengths. On an exam, draw your structure completely and state your reasoning.

Example 1: The nitrate ion

Draw the resonance structures of $\ce{NO3-}$ , find the average N–O bond order, and predict the relative lengths of the three N–O bonds.

Step-by-step solution

Count the valence electrons. N has 5, each O has 6, and the 1− charge adds one more:
$5 + 3 \times 6 + 1 = 24$
Draw one Lewis diagram. N goes in the middle, bonded to three O atoms. Three single bonds use 6 electrons and the three O atoms take 6 each (18), which uses all 24, but N then has only 6 electrons; turning one O lone pair into an N=O double bond gives N an octet.
Find all the resonance structures. The double bond can be on any of the three O atoms. The atoms stay put and only the electrons move, so there are three equivalent resonance structures; the real ion is their average (the resonance hybrid).
Average bond order. Across the three positions there are 1 double bond and 2 single bonds:
$\begin{aligned} \text{bond order} &= \frac{2 + 1 + 1}{3} \\ &= \boxed{\tfrac{4}{3}} \end{aligned}$
Predict the bond lengths. The three N–O bonds are identical. A bond order of 4/3 lies between 1 and 2, so each bond is between an N–O single bond and an N=O double bond in length, and the three O–N–O angles are equal, all 120°.

Example 2: The thiocyanate ion

Three Lewis diagrams can be drawn for $\ce{SCN-}$ : [S=C=N]⁻, [S–C≡N]⁻, and [S≡C–N]⁻. Use formal charge to choose the most reasonable one.

Step-by-step solution

Recall the formula. Formal charge = valence electrons − lone-pair electrons − bonds (each bond counts 1). S has 6 valence electrons, C 4, and N 5.
Work through each structure. In the order S, C, N:
- [S=C=N]⁻: 0, 0, −1
- [S–C≡N]⁻: −1, 0, 0
- [S≡C–N]⁻: +1, 0, −2
For example, N in the first: 5 − 4 − 2 = −1; N in the third: 5 − 6 − 1 = −2.
Check the total. In every structure the formal charges add up to −1, the charge of the ion, so the arithmetic is right.
Rule 1: formal charges as close to 0 as possible. Sum of absolute values:
$\begin{aligned} \text{first} &: 0 + 0 + 1 = 1 \\ \text{second} &: 1 + 0 + 0 = 1 \\ \text{third} &: 1 + 0 + 2 = 3 \end{aligned}$
The third is out.
Rule 2: put negative formal charge on the more electronegative atom. N (3.04) is more electronegative than S (2.58), so the first structure, with the negative charge on N, is better.
Conclusion. The most reasonable structure is $\boxed{[\text{S=C=N}]^-}$ . (The second also contributes noticeably; the real structure lies between the two.)

Practice: does it need resonance structures?

$\ce{O3}$
$\ce{CO2}$
$\ce{NO3-}$
$\ce{H2O}$
$\ce{CO3^2-}$
$\ce{CH4}$
$\ce{SO2}$
$\ce{SF6}$
Benzene, $\ce{C6H6}$
Formate, $\ce{HCOO-}$
$\ce{NH4+}$
$\ce{NO2-}$

Common mistakes

Mistake 1: “The molecule switches back and forth between its resonance structures.” The real molecule is always the resonance hybrid; each resonance structure is just one incomplete way of drawing it.

Mistake 2: “Benzene has three single bonds and three double bonds.” The six C–C bonds of benzene are measured to be exactly the same length, between single and double, with an average bond order of 1.5.

Mistake 3: “Formal charge is the real charge on an atom.” Formal charge is bookkeeping. For example, N in $\ce{NH4+}$ has a formal charge of +1, but N is more electronegative than H, so the positive charge actually sits mostly on the H atoms.

Quick check

Which species must be represented with resonance structures?

What is the average bond order of each C–O bond in $\ce{CO3^2-}$ ?

Which experimental fact best shows that benzene must be represented with resonance structures?

All three O–N–O bond angles in $\ce{NO3-}$ are 120°. Which is the best explanation?

The cyanate ion, $\ce{OCN-}$ , can be drawn as [O=C=N]⁻, [O–C≡N]⁻, or [O≡C–N]⁻. Which is most reasonable?

What is the formal charge on O in $\ce{H3O+}$ ?

AP-style free response

In the formate ion, $\ce{HCOO-}$ , C is bonded to one H and two O atoms.

(a) Find the total number of valence electrons in $\ce{HCOO-}$ and draw its two resonance structures.

(b) Find the average bond order of the two C–O bonds.

(c) Methanol, $\ce{CH3OH}$ , has a C–O single bond, and formaldehyde, $\ce{CH2O}$ , has a C=O double bond. Predict how the length of the C–O bonds in $\ce{HCOO-}$ compares with these, and justify your prediction.

(d) Calculate the formal charges on the two O atoms in each resonance structure, then the average formal charge on each O atom in the resonance hybrid.

Model answer and scoring

(a) 1 + 4 + 2 × 6 + 1 = 18. C is in the middle, bonded to H and two O atoms; one O has a double bond to C (with 2 lone pairs), the other a single bond (with 3 lone pairs). The double bond can be to either O, so the two structures are connected with “↔” (the ion gets brackets and its charge).

(b) The two positions hold 1 double bond + 1 single bond = 3 bonds; 3 ÷ 2 = 1.5.

(c) Shorter than the C–O single bond in methanol and longer than the C=O double bond in formaldehyde. A bond order of 1.5 is between 1 and 2, and the higher the bond order, the more shared electrons and the shorter the bond. The two C–O bonds are the same length.

(d) Double-bonded O: 6 − 4 − 2 = 0; single-bonded O: 6 − 6 − 1 = −1. In the hybrid each O averages (0 + (−1)) ÷ 2 = −½.

Scoring: (a) 18 electrons, both structures correct and connected with “↔”; (b) 1.5; (c) “between the two,” explained with bond order; (d) 0, −1, and an average of −½.

Summary

When the atoms stay fixed and a multiple bond can sit in several equivalent positions, resonance structures are needed; move electrons only, never atoms.
The real species is the resonance hybrid: average bond order = total bonds over the positions ÷ number of positions ( $\ce{O3}$ 1.5, $\ce{NO3-}$ 4/3, benzene 1.5).
Evidence for resonance: several bonds of the same length, between single and double, and equal bond angles.
Formal charge = V − N − B, and the formal charges add up to the charge of the species; values closest to 0 are best, with any negative formal charge on the more electronegative atom.

AP tip: Three question types are common: identify which species needs resonance structures; find an average bond order from resonance structures and compare bond lengths; and choose the best of several Lewis diagrams using formal charge. In your justification, show the formal charge calculations or point out that the bonds are equivalent and have the same bond order.

Resonance and Formal Charge共振与形式电荷

Learning goals

Before you start

When one diagram isn’t enough

When resonance structures are needed

The resonance hybrid and average bond order

Formal charge: scoring a structure

Example 1: The nitrate ion

Example 2: The thiocyanate ion

Common mistakes

Quick check

AP-style free response

Summary

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