ChemistryGeneral Chemistry

CH₂O₂ Lewis Structure: Formic Acid and Bond Counting

Draw the CH2O2 Lewis structure for formic acid and check all five exam options using 18 valence electrons, four sigma bonds, and one pi bond.

Question

Draw the Lewis structure for formic acid. Which of the following statements is FALSE?(a) Bonding-electron count: 10.(b) Bond multiplicities: three single; one double.(c) Nonbonding-electron pairs: four.(d) Carbon hybridization: sp².(e) σ/π bond counts: 3/1.

CH₂O₂ Lewis structure: formic acid

The false option is (e). Formic acid has four σ bonds and one π bond, not three σ bonds and one π bond.

:O:HCOH\begin{array}{ccccccc} &&\mathord{:}\mathrm{O}\mathord{:}&&&&\\ &&\Vert&&&&\\ \mathrm{H}&-&\mathrm{C}&-&\underset{\cdot\cdot}{\overset{\cdot\cdot}{\mathrm{O}}}&-&\mathrm{H} \end{array}

Each pair of dots is a lone pair. Both oxygens have two lone pairs. The upper oxygen forms C=O; the other oxygen forms the C–O and O–H single bonds. One hydrogen attaches to carbon and the other to oxygen.

Electron count and formal charges

The formula gives 4 + 2(1) + 2(6) = 18 valence electrons. Three single bonds use six electrons and one double bond uses four. The remaining eight electrons form four lone pairs. Thus 10 bonding + 8 nonbonding = 18.

Carbon has four bonding pairs and no lone pairs, so its formal charge is 4 − 0 − 4 = 0. For either oxygen, 6 − 4 − 2 = 0. Each hydrogen has formal charge 1 − 0 − 1 = 0. Carbon and both oxygens have octets; each hydrogen has a duet.

Evaluate all five options

OptionVerdictCheck against the structure
(a)TrueThree single bonds plus one double bond contain 3(2) + 4 = 10 shared electrons.
(b)TrueC–H, C–O, and O–H are single; the other C–O connection is double.
(c)TrueTwo lone pairs on each oxygen give four in total.
(d)TrueThree bonding directions around carbon correspond to sp² hybridization in the valence-bond model.
(e)FalseEach of the four connected atom pairs supplies one σ bond; the second component of C=O supplies one π bond.

Why the name matters

The question specifies formic acid, which fixes the H–C(=O)–O–H connectivity. A molecular formula alone lists atom counts, not a unique bonding arrangement. Moving a hydrogen to create a different connectivity does not produce a resonance form of formic acid. The neutral structure shown is the usual principal Lewis contributor; this task does not ask for an exhaustive set of resonance contributors.

Evidence boundary

The source is Texas A&M CHEM 101, Spring 2001, Exam 3 Form A, question 21&22. All five options are retained in equivalent compact wording with unchanged values and order. The named species is formic acid, not an unspecified CH₂O₂ isomer; the diagram is schematic, and the answer evaluates the supplied options rather than reporting measurements.

Sources

These references support the concepts and methods used in the explanation above.

CH₂O₂ Lewis Structure: Formic Acid and Bond Counting | Verla