Resonance Structures Practice: Find and Repair the Error

A proposed formate structure can have the right 18 electrons and the right −1 charge while giving carbon ten electrons. Both totals pass. The drawing still fails.

Find the error, repair it, and explain the repair. Start with the worked example, then cover the answers. Every bond and lone pair is specified in text.

Keep the atoms fixed

Resonance contributors describe one species with different electron placements. Keep the same atoms connected to the same neighbors, including every hydrogen. You can rotate or rearrange a drawing on the page; you can't move a hydrogen from one atom to another. The hybrid doesn't switch between contributors, and the double-headed resonance arrow ↔ doesn't describe an equilibrium. These conventions follow OpenStax's resonance rules.

Here, – means a single bond, = a double bond, and LP a lone pair containing two electrons. Oa and Ob identify individual oxygen atoms; keep those identities throughout. Lone-pair counts are explicit below.

The exercises cover paired-electron structures containing H, C, N, and O. Hydrogen gets two surrounding electrons; second-period atoms cannot exceed eight. An incomplete octet needs attention, but isn't automatically disqualifying: a positively charged carbon with six electrons can appear in a minor contributor. Audit validity before considering importance.

A ledger that catches local mistakes

For each atom, record B, the sum of its bond orders, and L, its number of lone pairs:

• Surrounding electrons: E = 2B + 2L.
• Formal charge: FC = V − B − 2L, where V is 1 for H, 4 for C, 5 for N, and 6 for O.
• Whole-structure electron count: 2 × (sum of bond orders across unique bonds + sum of all lone pairs).

Count each bond once; summing atom E values would count bonding electrons twice. The formal-charge formula and charge-sum check follow OpenStax's electron bookkeeping. For a slower calculation refresher, use the formal-charge worked examples.

The two global checks are related: summed formal charge equals the neutral atoms' total valence electrons minus the electrons drawn. Passing both doesn't provide two independent reasons to trust a structure. The local E column catches the overfilled atom that those totals miss.

Copy this blank ledger onto paper:

Atom	V	B	L	E	FC
____	____	____	____	____	____
____	____	____	____	____	____
____	____	____	____	____	____

Target formula and charge: ____
Available valence electrons: ____
Drawn bonding electrons + lone-pair electrons: ____ + ____ = ____
Sum of formal charges: ____
Same atom identities and connections? ____
Any overfilled atom? ____
Any incomplete octet, and is it allowed here? ____
New electron placement or duplicate? ____
Decision and smallest repair: ____

Work through formate once

Formate, HCO₂⁻, has 1 + 4 + 2(6) + 1 = 18 valence electrons. Begin with H–C(=Oa)–Ob⁻: Oa has 2 LP, Ob has 3 LP, and H and C have 0 LP.

Atom	V	B	L	E	FC
H	1	1	0	2	0
C	4	4	0	8	0
Oa	6	2	2	8	0
Ob	6	1	3	8	−1

Four bond-order units contain 8 electrons; five lone pairs contain 10. Total: 18. Formal charges sum to −1.

To obtain the other equivalent contributor, use one Ob lone pair to make C=Ob and place the C=Oa π pair onto Oa. The result is H–C(–Oa⁻)=Ob, with LP counts H 0, C 0, Oa 3, Ob 2. Its ledger swaps the oxygen rows; all other entries stay unchanged.

A full-headed curved arrow tracks one electron pair from its starting bond or lone pair to its destination. It serves a different purpose from ↔ between complete contributors. OpenStax's drawing lesson develops these electron-placement patterns.

Eight diagnostic questions

Find the failed check or explain why each proposal passes. Repair drawings or labels. Parentheses mark branches on the preceding atom.

1. Missing pair: H–C(=Oa)–Ob, with LP counts H 0, C 0, Oa 2, Ob 2, is labeled “formate, −1.” Does it contain 18 electrons?
2. Carbon overload: H–C(=Oa)=Ob has LP counts H 0, C 0, Oa 2, Ob 2. Its calculated charges sum to −1. Can it represent formate?
3. Nitrogen overload: For nitrite, NO₂⁻, propose Oa=N=Ob with LP counts Oa 2, N 1, Ob 2. Both the 18-electron total and −1 charge total pass. What fails?
4. Nitrate labels: N(=Oa)(–Ob)(–Oc), with LP counts N 0, Oa 2, Ob 3, Oc 3, is labeled N⁰, Oa⁰, Ob⁻, Oc⁰. Which labels need repair?
5. Ozone labels: Oa=Ob–Oc has LP counts Oa 2, Ob 1, Oc 3. Someone labels all three oxygens neutral because O₃ is neutral. Correct the labels.
6. Hydrogen moved: Compare worked formate with H–Oa–C=Ob, with LP counts H 0, Oa 2, C 1, Ob 2. It has 18 electrons, a duet at H, octets at C and O, and charge −1. Is it another resonance contributor of formate?
7. Duplicate drawing: Compare H–C(=Oa)–Ob⁻ with Ob⁻–C(H)=Oa. Both have LP counts H 0, C 0, Oa 2, Ob 3. Have you found two distinct contributors?
8. Incomplete octet: Formaldehyde has two C–H single bonds and C=O, with LP counts C 0, H 0 each, O 2. Move only the C=O π pair onto O. Must the resulting C⁺–O⁻ contributor be rejected?

Answers and repairs

1. Only 16 electrons are drawn: 8 bonding + 8 nonbonding. E values are H 2, C 8, Oa 8, Ob 6; charges are 0, 0, 0, +1. Add a third LP to Ob and label it −1. That restores the worked formate ledger.

2. Carbon has ten electrons. Its B = 5 and L = 0 give E = 10 and FC = −1. H has E = 2, FC = 0; each oxygen has E = 8, FC = 0. The total really is 10 bonding + 8 nonbonding = 18. Convert C=Oa to a single bond and give that π pair to Oa. The result is the second worked contributor.

3. Nitrogen has ten electrons: B = 4, L = 1, so E = 10 and FC = −1. Each oxygen has E = 8 and FC = 0. Repair by converting N=Ob to N–Ob and adding one LP to Ob. Now N has B = 3, L = 1, E = 8, FC = 0; Ob has B = 1, L = 3, E = 8, FC = −1. Oa stays unchanged. Total: 6 bonding + 12 nonbonding = 18.

4. N needs +1; Oc needs −1. N has B = 4, L = 0, E = 8, FC = +1. Oa has B = 2, L = 2, E = 8, FC = 0. Each singly bonded oxygen has B = 1, L = 3, E = 8, FC = −1. The corrected sum is −1, matching nitrate; 8 bonding + 16 nonbonding = 24 electrons.

5. Ob is +1 and Oc is −1. Oa has B = 2, L = 2, E = 8, FC = 0; Ob has B = 3, L = 1, E = 8, FC = +1; Oc has B = 1, L = 3, E = 8, FC = −1. Total: 6 bonding + 12 nonbonding = 18. Neutrality constrains the sum, not each atom.

6. The connectivity changed. H is attached to Oa instead of C. This drawing has different constitutional connectivity, even though its arithmetic passes: H has E = 2, FC = 0; both oxygens have B = 2, L = 2, E = 8, FC = 0; C has B = 3, L = 1, E = 8, FC = −1. Total: 8 bonding + 10 nonbonding = 18. Electron redistribution alone cannot move H. This comparison tests connectivity; the ledger doesn't establish that the alternative arrangement is a stable species.

7. One contributor, drawn twice. C=Oa and C–Ob occur in both; every LP and charge matches the first worked ledger. Reading the chain from the opposite end adds no electron placement. Keeping atom labels prevents this counting mistake.

8. The minor contributor is valid. After the move, carbon has three single bonds: B = 3, L = 0, E = 6, FC = +1. Oxygen has B = 1, L = 3, E = 8, FC = −1. Each H has E = 2, FC = 0. Initially C and O each have E = 8, FC = 0. Both forms contain 12 electrons: initially 8 bonding + 4 nonbonding, then 6 + 6. The neutral octet form is more important; the six-electron carbon doesn't make the other form impossible. See the three-bonded carbon charge pattern and the distinction between unequal contributors.

Review the error you actually made

Turn a missed question into a repair prompt: “Nitrite totals pass. Which atom is overfilled, and where must one pair go?” Keep the bond and LP specification on the front; put the local calculation and repair on the back.

Choose the questions you missed rather than copying the whole worksheet into cards. The guides to chemistry flashcards and making better flashcards explain how to keep each review focused.

Created with substantial AI assistance for drafting, original exercise construction, and editorial revision. The atom-by-atom arithmetic was independently recalculated and checked with an automated electron-count ledger. No human expert or classroom validation was performed. The accompanying illustration is AI-generated.