Formal Charge Practice Problems with Full Working

Formal charge is pure arithmetic, which makes it one of the most reliable places to pick up marks, and one of the easiest places to drop them through a single miscounted lone pair. The questions below train both halves of the skill: calculating formal charges accurately, then using them to decide which Lewis structure is best. Every answer in the key is checked against the total charge of the species, and you should build the same habit.

The formula and the checks

For each atom in a Lewis structure:

Formal charge = (valence electrons) − (lone-pair electrons) − ½ (bonding electrons)

• Valence electrons are those of the free atom: 1 for H, 3 for B, 4 for C, 5 for N and P, 6 for O and S, 7 for the halogens.
• Lone-pair electrons are counted individually: a lone pair is 2 electrons.
• ½ bonding electrons equals the number of bonds to the atom: a single bond counts 1, a double bond 2, a triple bond 3.

A faster version: formal charge = valence electrons − (dots + lines) drawn on that atom.

Check 1: the formal charges in a structure must add up to the overall charge (0 for a neutral molecule, −1 for NO₃⁻, and so on). If they don’t, you have miscounted.

Check 2: when comparing structures, the best one usually has (a) formal charges as close to zero as possible, (b) no like charges on neighbouring atoms, and (c) any negative formal charge on the more electronegative atom.

Formal charge is not the same as oxidation number, and it is not the real charge on an atom. It is a bookkeeping tool that assumes every bond is shared perfectly equally. For the full method with worked examples, see how to calculate formal charge.

Questions

Total: 40 marks. Where a Lewis structure is described in words, draw it before you calculate.

1. Calculate the formal charge on nitrogen and on each hydrogen in the ammonium ion, NH₄⁺. (2 marks)
2. Calculate the formal charge on oxygen in the oxonium ion, H₃O⁺ (three O–H bonds, one lone pair on O). (1 mark)
3. Calculate the formal charge on oxygen in the hydroxide ion, OH⁻ (one O–H bond, three lone pairs on O). (1 mark)
4. In BF₄⁻, boron forms four single bonds to fluorine, and each fluorine carries three lone pairs. Find the formal charge on B and on each F. (2 marks)
5. Carbon monoxide is drawn as C≡O with one lone pair on each atom. Calculate both formal charges and comment on the result. (3 marks)
6. Calculate the formal charges in the cyanide ion, [C≡N]⁻, with one lone pair on each atom. (2 marks)
7. Compare two structures of CO₂: (A) O=C=O with two lone pairs on each O; (B) O≡C–O, with one lone pair on the triple-bonded O and three on the single-bonded O. Calculate all formal charges and choose the better structure. (3 marks)
8. In one resonance structure of the nitrate ion, NO₃⁻, nitrogen forms one double bond and two single bonds. Calculate all formal charges and check the total. (3 marks)
9. Ozone is drawn as O=O–O, with the central oxygen carrying one lone pair. Calculate all three formal charges. (2 marks)
10. Nitric acid, HNO₃, has the connectivity H–O–N(=O)–O. The central N has no lone pairs. Calculate the formal charge on every atom. (3 marks)
11. Two isomers have the formula HCN: hydrogen cyanide (H–C≡N) and hydrogen isocyanide (H–N≡C), each with one lone pair on the terminal atom. Calculate the formal charges in both and suggest which is more stable. (3 marks)
12. Dinitrogen monoxide has the atom order N–N–O. Calculate formal charges for (A) N=N=O and (B) N≡N–O, adding lone pairs so every atom has an octet. Which structure do the formal-charge rules favour? (4 marks)
13. For the cyanate ion, OCN⁻ (atom order O–C–N), draw three octet structures and calculate the formal charges. Which is best? (4 marks)
14. The fulminate ion, CNO⁻, contains the same atoms as cyanate but in the order C–N–O. Find the best octet structure by formal charge and explain why fulminate is much less stable than cyanate. (4 marks)
15. For the sulfate ion, SO₄²⁻, calculate the formal charges for (A) four S–O single bonds and (B) two S=O and two S–O bonds. State which structure formal-charge rules favour, and give one reason why the answer is debated. (3 marks)

Answer key

1. NH₄⁺. N: 5 − 0 − 4 = +1 (1). Each H: 1 − 0 − 1 = 0 (1). Total +1, matching the ion’s charge.

2. H₃O⁺. O: 6 − 2 − 3 = +1. Hydrogens are 0, so the total is +1. (1)

3. OH⁻. O: 6 − 6 − 1 = −1; H is 0. Total −1. (1)

4. BF₄⁻. B: 3 − 0 − 4 = −1 (1). Each F: 7 − 6 − 1 = 0 (1). Total −1. Note that the negative formal charge sits on boron, the least electronegative atom, even though chemists usually think of the ion’s negative charge as spread over the fluorines. Formal charge assumes equal sharing, so it can point the “wrong” way.

5. CO. C: 4 − 2 − 3 = −1 (1). O: 6 − 2 − 3 = +1 (1). Total 0. Comment: this is an unusual case where the negative formal charge is on the less electronegative atom, yet this structure is still the best one because it is the only one that gives both atoms an octet (1). The opposing formal charges partly cancel the expected bond polarity, which is why CO has a surprisingly small dipole moment.

6. CN⁻. C: 4 − 2 − 3 = −1. N: 5 − 2 − 3 = 0. Total −1. (2)

7. CO₂. Structure A: C: 4 − 0 − 4 = 0; each O: 6 − 4 − 2 = 0. All zero (1). Structure B: triple-bonded O: 6 − 2 − 3 = +1; C: 4 − 0 − 4 = 0; single-bonded O: 6 − 6 − 1 = −1. Total 0 (1). A is better: all formal charges are zero, and B places a positive charge on oxygen, a highly electronegative atom (1).

8. NO₃⁻. N (four bonds, no lone pair): 5 − 0 − 4 = +1 (1). Double-bonded O (two lone pairs): 6 − 4 − 2 = 0. Each single-bonded O (three lone pairs): 6 − 6 − 1 = −1 (1). Total: +1 + 0 − 1 − 1 = −1, matching the ion (1). Three equivalent structures exist, differing only in which O carries the double bond.

9. O₃. Double-bonded end O: 6 − 4 − 2 = 0. Central O: 6 − 2 − 3 = +1. Single-bonded end O: 6 − 6 − 1 = −1. Total 0. (2) The charge separation is part of the reason ozone has a small dipole moment.

10. HNO₃. H: 1 − 0 − 1 = 0. The O bonded to H (two bonds, two lone pairs): 6 − 4 − 2 = 0 (1). N (four bonds, no lone pairs): 5 − 0 − 4 = +1. Double-bonded O: 6 − 4 − 2 = 0 (1). Single-bonded terminal O (three lone pairs): 6 − 6 − 1 = −1 (1). Total 0. Nitric acid is closely related to nitrate: take away the H⁺ and you have one of the NO₃⁻ structures from question 8.

11. HCN vs HNC. In H–C≡N: H 0; C: 4 − 0 − 4 = 0; N: 5 − 2 − 3 = 0 (1). In H–N≡C: H 0; N: 5 − 0 − 4 = +1; C: 4 − 2 − 3 = −1 (1). HCN has no formal charges, so it is predicted to be more stable, which matches experiment: HCN is by far the more stable isomer (1).

12. N₂O. Structure A, N=N=O: terminal N (two lone pairs): 5 − 4 − 2 = −1; central N: 5 − 0 − 4 = +1; O (two lone pairs): 6 − 4 − 2 = 0 (1). Structure B, N≡N–O: terminal N (one lone pair): 5 − 2 − 3 = 0; central N: +1; O (three lone pairs): 6 − 6 − 1 = −1 (1). Both total 0 (1). Both have the same size of charges, so the tiebreaker applies: B puts the negative charge on oxygen, the more electronegative atom, so B is favoured. In reality both contribute significantly to the resonance hybrid (1).

13. Cyanate, OCN⁻. (A) O=C=N, two lone pairs on O and on N: O 0, C 0, N: 5 − 4 − 2 = −1 (1). (B) O–C≡N, three lone pairs on O, one on N: O: 6 − 6 − 1 = −1, C 0, N: 5 − 2 − 3 = 0 (1). (C) O≡C–N, one lone pair on O, three on N: O: 6 − 2 − 3 = +1, C 0, N: 5 − 6 − 1 = −2 (1). Each totals −1. C is clearly worst (large charges, positive charge on O). A and B each have a single −1, and B is favoured because the negative charge is on oxygen (3.44), which is more electronegative than nitrogen (3.04) (1). A still contributes to the real structure.

14. Fulminate, CNO⁻. The central atom is now nitrogen. (A) C=N=O: C (two lone pairs): 4 − 4 − 2 = −2; N: 5 − 0 − 4 = +1; O: 6 − 4 − 2 = 0 (1). (B) C≡N–O: C (one lone pair): 4 − 2 − 3 = −1; N: +1; O (three lone pairs): −1 (1). (C) C–N≡O: C (three lone pairs): 4 − 6 − 1 = −3; N +1; O (one lone pair): +1. All total −1. B is the best structure because its charges are the smallest (1). Even so, every fulminate structure has a positive formal charge on the central nitrogen and charges on at least two atoms, whereas cyanate has a structure with only one −1 charge on an electronegative atom. That extra charge separation is consistent with fulminate being much less stable than cyanate (1).

15. SO₄²⁻. (A) All single bonds: S: 6 − 0 − 4 = +2; each O (three lone pairs): −1; total +2 − 4 = −2 (1). (B) Two S=O, two S–O: S: 6 − 0 − 6 = 0; double-bonded O: 0; single-bonded O: −1 each; total −2 (1). Formal-charge rules favour B. The debate: B requires 12 electrons around sulfur, and modern calculations indicate that sulfur’s d orbitals contribute very little to bonding, so many chemists regard A, with strongly polar bonds, as the better description. Follow your syllabus (1). The full discussion is in the sulfate Lewis structure.

Exam tips

• Count bonds, not bonding electrons, for the ½-bonding term: it is quicker and harder to get wrong.
• Always add up your charges. One line of working (“total = −1 ✓”) catches most mistakes and often earns a mark on its own.
• Octets first, then formal charges. For second-period atoms (C, N, O), a structure that breaks the octet rule is not acceptable, however good its formal charges look.
• Don’t confuse formal charge with oxidation number. In NH₄⁺, nitrogen’s formal charge is +1 but its oxidation number is −3. The oxidation number calculator is handy for checking the second.
• More structure-drawing practice is in the Lewis structures practice set, and the resonance side of questions 8, 12 and 13 is developed in resonance structures practice.

Key takeaways

• Formal charge = valence electrons − lone-pair electrons − number of bonds.
• The formal charges in any structure must sum to the overall charge.
• Prefer structures with charges near zero, no neighbouring like charges, and negative charges on more electronegative atoms.
• Formal charge is a guide, not a law: CO and sulfate show where it needs chemical judgement alongside it.