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Lewis Structures Practice: 20 Molecules and Ions with Answers

Bonding & Molecular StructureIntermediate7 min read
On this page
  1. The method (quick reminder)
  2. Questions
  3. Answer key
  4. Scoring
  5. Frequent errors in this set
  6. Key takeaways

Drawing Lewis structures is a skill that only improves with practice. This set contains 20 molecules and ions, from straightforward to tricky: multiple bonds, charged ions, incomplete octets, expanded octets and resonance. For each one, count the valence electrons, draw the structure, check the formal charges and, as a bonus, predict the shape. Then compare with the answer key, which describes each structure in words.

The method (quick reminder)

  1. Count valence electrons: add each atom’s outer electrons (group number). Add one per negative charge; subtract one per positive charge.
  2. Choose the central atom: usually the least electronegative (never H; F is almost always terminal).
  3. Connect atoms with single bonds (2 electrons each).
  4. Complete octets on the outer atoms with lone pairs.
  5. Put leftover electrons on the central atom.
  6. If the central atom lacks an octet, move lone pairs from outer atoms into double or triple bonds.
  7. Check formal charges and total electrons (see how to calculate formal charge).

For background, see Lewis dot structures and VSEPR and molecular geometry.

Questions

Level 1: single bonds

  1. H₂O
  2. NH₃
  3. CH₄
  4. PCl₃
  5. H₂S

Level 2: multiple bonds

  1. CO₂
  2. HCN
  3. C₂H₄ (ethene)
  4. H₂CO (methanal)
  5. N₂

Level 3: ions

  1. OH⁻
  2. NH₄⁺
  3. CN⁻
  4. NO₂⁻
  5. SO₄²⁻

Level 4: exceptions and resonance

  1. BF₃
  2. PCl₅
  3. SF₆
  4. XeF₄
  5. O₃

Answer key

1. H₂O

  • Electrons: 2(1) + 6 = 8.
  • O central, two O–H single bonds, two lone pairs on O.
  • Formal charges: all 0.
  • Shape: bent, about 104.5°.

2. NH₃

  • Electrons: 5 + 3(1) = 8.
  • N central, three N–H bonds, one lone pair on N.
  • Formal charges: all 0.
  • Shape: trigonal pyramidal, about 107°.

3. CH₄

  • Electrons: 4 + 4 = 8.
  • C central, four C–H bonds, no lone pairs.
  • Shape: tetrahedral, 109.5°.

4. PCl₃

  • Electrons: 5 + 3(7) = 26.
  • P central, three P–Cl single bonds (6 electrons). Each Cl gets three lone pairs (18). Remaining 2 → one lone pair on P.
  • Formal charges: all 0.
  • Shape: trigonal pyramidal.

5. H₂S

  • Electrons: 2 + 6 = 8.
  • Like water: S central, two S–H bonds, two lone pairs on S.
  • Shape: bent (about 92° — smaller than water’s angle).

6. CO₂

  • Electrons: 4 + 2(6) = 16.
  • O=C=O, two lone pairs on each O.
  • Formal charges: all 0.
  • Shape: linear, 180°; non-polar overall.

7. HCN

  • Electrons: 1 + 4 + 5 = 10.
  • H–C≡N, one lone pair on N.
  • Formal charges: all 0.
  • Shape: linear.

8. C₂H₄

9. H₂CO (methanal)

  • Electrons: 2(1) + 4 + 6 = 12.
  • C central with two C–H bonds and a C=O double bond; two lone pairs on O.
  • Formal charges: all 0.
  • Shape: trigonal planar at C.

10. N₂

  • Electrons: 2(5) = 10.
  • N≡N, one lone pair on each N.
  • Formal charges: 0.

11. OH⁻

  • Electrons: 6 + 1 + 1 (charge) = 8.
  • O–H single bond, three lone pairs on O. Write in brackets with the charge: [O–H]⁻.
  • Formal charges: O = 6 − 6 − 1 = −1; H = 0.

12. NH₄⁺

  • Electrons: 5 + 4 − 1 (charge) = 8.
  • Four N–H bonds, no lone pairs. Brackets and + charge.
  • Formal charges: N = +1.
  • Shape: tetrahedral.

13. CN⁻

  • Electrons: 4 + 5 + 1 = 10.
  • [C≡N]⁻, one lone pair on each atom.
  • Formal charges: C = 4 − 2 − 3 = −1; N = 5 − 2 − 3 = 0.
  • Cyanide bonds to metals (such as iron in enzymes) through carbon, where the negative formal charge sits.

14. NO₂⁻

  • Electrons: 5 + 2(6) + 1 = 18.
  • N central; one N=O double bond, one N–O single bond; one lone pair on N; the double-bonded O has two lone pairs, the single-bonded O three.
  • Formal charges: N = 0; O (double) = 0; O (single) = −1.
  • Two resonance structures (the double bond on either O); both N–O bonds are equal, bond order 1½ (see resonance structures).
  • Shape: bent, about 115°.

15. SO₄²⁻

  • Electrons: 6 + 4(6) + 2 = 32.
  • S central bonded to four O.
  • Option A (octet on S): four S–O single bonds; each O has three lone pairs. Formal charges: S = +2, each O = −1.
  • Option B (minimising formal charge): two S=O and two S–O⁻; S has formal charge 0.
  • Many textbooks draw B; modern bonding theory favours A with polar bonds and resonance. Follow your syllabus.
  • Shape: tetrahedral.

16. BF₃

  • Electrons: 3 + 3(7) = 24.
  • B central, three B–F single bonds, three lone pairs on each F. B has only 6 electrons — an incomplete octet.
  • Formal charges: all 0.
  • Shape: trigonal planar, 120°.
  • The empty space on boron makes BF₃ an electron-pair acceptor — a Lewis acid that reacts with NH₃ to form a dative bond (see Lewis acids and bases and dative covalent bonds).

17. PCl₅

  • Electrons: 5 + 5(7) = 40.
  • P central with five P–Cl single bonds; each Cl has three lone pairs. P has 10 electrons around it — an expanded octet, possible for period 3 and beyond.
  • Formal charges: all 0.
  • Shape: trigonal bipyramidal (120° in the equator, 90° to the axial positions).

18. SF₆

  • Electrons: 6 + 6(7) = 48.
  • S central with six S–F bonds; three lone pairs on each F. S has 12 electrons around it.
  • Shape: octahedral, 90°; non-polar.

19. XeF₄

  • Electrons: 8 + 4(7) = 36.
  • Xe central with four Xe–F bonds (8 electrons); each F has three lone pairs (24); remaining 4 electrons → two lone pairs on Xe.
  • Formal charges: all 0.
  • Shape: square planar — the two lone pairs sit opposite each other, above and below the plane. Noble gas compounds were first made in 1962; xenon tetrafluoride followed the same year.

20. O₃

  • Electrons: 3(6) = 18.
  • O=O–O: central O has one lone pair; the double-bonded end O has two; the single-bonded end O has three.
  • Formal charges: central O = +1; single-bonded end O = −1; double-bonded O = 0.
  • Two resonance structures; both bonds are 128 pm, bond order 1½.
  • Shape: bent, about 117°.

Scoring

  • 18–20: excellent — move on to hybridisation.
  • 13–17: good; review the ions and exceptions.
  • Below 13: go back to Lewis dot structures and practise the method step by step.

Frequent errors in this set

  • Forgetting to add or subtract electrons for ions (Q11–15).
  • Giving boron an octet by adding a double bond to BF₃ — this would put a positive formal charge on fluorine, the most electronegative element, which is unfavourable.
  • Trying to expand the octet of period 2 atoms (C, N, O, F can never exceed 8).
  • Missing the lone pairs on the central atom (PCl₃, XeF₄), which changes the predicted shape.
  • Drawing only one resonance structure for NO₂⁻ and O₃.

Key takeaways

  • Always start by counting valence electrons correctly, including charges.
  • Complete outer atoms first, then the central atom; use multiple bonds if the centre lacks an octet.
  • Know the exceptions: incomplete octets (B, Be), expanded octets (period 3 and beyond), resonance (NO₂⁻, O₃, CO₃²⁻).
  • Use formal charges to check and choose structures, and lone pairs to predict shapes.

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