AP Chemistry Quiz: Lewis Diagrams

What this quiz covers

This quiz focuses on Lewis Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. H—C—N with two lone pairs on N
2. H—C=N with one lone pair on N
3. H=C—N with two lone pairs on N
4. H—C≡N with one lone pair on N (correct answer)
5. H≡C—N with one lone pair on N

Explanation: This question tests the skill of selecting correct Lewis structures for triatomic molecules with proper bond orders and lone pairs. Hydrogen cyanide, HCN, has 10 valence electrons (1 from H, 4 from C, 5 from N), and the correct structure is H—C≡N with one lone pair on N, using 2 electrons in the C-H bond, 6 in the triple bond, and 2 in nitrogen's lone pair, totaling 10. Carbon achieves an octet with the single and triple bonds, while nitrogen uses the triple bond and lone pair. Hydrogen has its duet. A tempting distractor is choice B (H—C=N), which miscounts to 8 electrons, overlooking the need for 10 total valence electrons. Start by connecting atoms with single bonds, then add triple or double bonds to match electron count and octets.

Question 2

1. N—N with three lone pairs on each N
2. N=N with two lone pairs on each N
3. N≡N with one lone pair on each N (correct answer)
4. N≡N with two lone pairs on each N
5. N—N with two lone pairs on each N

Explanation: This question tests the ability to construct Lewis structures for diatomic molecules with correct bond orders. Dinitrogen, N2, has 10 valence electrons (10 from 2N), and the correct structure is N≡N with one lone pair on each nitrogen, using 6 electrons in the triple bond and 4 in lone pairs, totaling 10. Each nitrogen achieves an octet with 2 lone electrons and 6 from the triple bond. The triple bond is necessary to satisfy octets with the electron count. A tempting distractor is N=N with two lone pairs each, which totals 12 electrons, misconstruing the valence total. Calculate total electrons and increase bond order until octets are met without exceeding the count.

Question 3

1. 0 lone pairs
2. 1 lone pair (correct answer)
3. 2 lone pairs
4. 3 lone pairs
5. 4 lone pairs

Explanation: This question tests the skill of counting central atom lone pairs in bent ion structures. The nitrite ion, NO2^-, has 18 valence electrons (5 from N, 12 from 2O, plus 1 for the charge), and in one valid structure, nitrogen has one double bond, one single bond to oxygen, and one lone pair (2 electrons), with oxygens having 4 and 6 lone electrons respectively, totaling 18. Nitrogen achieves an octet with 2 lone and 6 bonding electrons. Octets are satisfied for all. A tempting distractor is 0 lone pairs, from overusing double bonds and ignoring nitrogen's need for 8 electrons, a bond order misconception. Assign a lone pair to the central atom if bonds alone don't reach octet after electron tally.

Question 4

1. Two single bonds
2. One single bond and one triple bond
3. One double bond and one triple bond
4. Two double bonds
5. One single bond and one double bond (correct answer)

Explanation: This question tests the skill of determining bond patterns in non-linear molecules using Lewis structures without resonance. Ozone, O3, has 18 valence electrons (18 from 3O), and in one valid structure, the central oxygen has one single bond and one double bond to the terminal oxygens, using 6 electrons in bonds, 2 in the central oxygen's lone pairs, 4 in the double-bonded terminal's lone pairs, and 6 in the single-bonded terminal's lone pairs, totaling 18. This satisfies octets for all atoms. The central oxygen achieves 8 electrons from bonds and lone pairs. A tempting distractor is two double bonds, which undercounts electrons to 16, misconstruing total valence needs. Draw initial single bonds, then add multiples to central atoms for octet completion, verifying total electrons.

Question 5

1. 0 lone pairs
2. 3 lone pairs
3. 2 lone pairs
4. 1 lone pair (correct answer)
5. 4 lone pairs

Explanation: This question tests the ability to draw Lewis structures for molecules with a central atom and determine lone pair counts while accounting for total valence electrons. Nitrogen trifluoride, NF3, has 26 valence electrons (5 from N, 21 from 3F), and the correct structure features nitrogen with three single bonds to fluorine (using 6 electrons) and one lone pair (2 electrons), with each fluorine having three lone pairs (18 electrons total for F lone pairs), summing to 26. This gives nitrogen an octet: 2 electrons from the lone pair plus 6 from the bonds. Each fluorine also achieves an octet with its lone pairs and single bond. A tempting distractor is choice C (2 lone pairs), which might arise from mistakenly assigning extra electrons to nitrogen instead of bonds, a misconception of undercounting bonding pairs. When constructing Lewis structures, tally valence electrons and connect atoms with single bonds first before adding lone pairs to complete octets.

Question 6

1. 1 lone pair
2. 2 lone pairs
3. 3 lone pairs (correct answer)
4. 4 lone pairs
5. 5 lone pairs

Explanation: This question tests the skill of drawing Lewis structures for simple polyatomic ions. The hypochlorite ion ClO⁻ has 14 valence electrons: Cl contributes 7, O contributes 6, plus 1 for the negative charge. With a single Cl-O bond using 2 electrons, the remaining 12 electrons must be placed as lone pairs. Oxygen needs 3 lone pairs to complete its octet (since it has 2 electrons from the bond), using 6 electrons, which leaves 6 electrons (3 lone pairs) on chlorine. A common misconception is thinking chlorine would have only 2 lone pairs (choice B), which would account for only 10 electrons instead of the required 14. To verify Lewis structures, always count total electrons used in bonds and lone pairs to ensure it matches the total valence electrons calculated initially.

Question 7

1. Single bond
2. Double bond
3. Triple bond (correct answer)
4. Quadruple bond
5. No bond (separate ions)

Explanation: This question tests the skill of determining bond order in Lewis structures. The cyanide ion (CN⁻) has 10 total valence electrons: carbon contributes 4, nitrogen contributes 5, and the negative charge adds 1 electron. To draw the Lewis structure, we connect C and N and distribute the remaining electrons to satisfy the octet rule for both atoms. With only 10 electrons total, the only way to give both atoms complete octets is to form a triple bond between C and N (using 6 electrons), with one lone pair on carbon (2 electrons) and one lone pair on nitrogen (2 electrons). The bond order is therefore 3, corresponding to a triple bond. A common misconception is trying to form only single or double bonds, which would not allow both atoms to achieve octets with the available electrons. When determining bond order in small molecules or ions, check if multiple bonds are needed to satisfy the octet rule with the given number of valence electrons.

Question 8

1. 0 lone pairs
2. 1 lone pair (correct answer)
3. 2 lone pairs
4. 3 lone pairs
5. 4 lone pairs

Explanation: This question tests your ability to determine lone pairs on a central atom in a Lewis structure. NF₃ has 26 valence electrons (N=5, F=7×3=21). With nitrogen as the central atom, it forms three single bonds to fluorine atoms (6 electrons used). To satisfy the octet rule, each fluorine needs 3 lone pairs (18 electrons), and nitrogen needs electrons to complete its octet. Since nitrogen has used 3 electrons in bonding, it needs 5 more electrons, which means it has 1 lone pair (2 electrons) plus the 3 bonding electrons for a total of 8. Students often mistakenly think nitrogen has no lone pairs because they focus only on the bonding electrons. Remember that after drawing all bonds, distribute remaining electrons as lone pairs starting with terminal atoms, then place any remaining on the central atom.

Question 9

1. 0 lone pairs (correct answer)
2. 1 lone pair
3. 2 lone pairs
4. 3 lone pairs
5. 4 lone pairs

Explanation: This question tests the skill of drawing Lewis structures for polyatomic cations. The ammonium ion NH₄⁺ has 8 valence electrons: N contributes 5, each H contributes 1 (4 total), minus 1 for the positive charge. With nitrogen as the central atom bonded to four hydrogens, all 8 electrons are used in the four N-H single bonds (2 electrons per bond). This leaves no electrons for lone pairs on nitrogen, giving nitrogen a complete octet with its 8 bonding electrons. A common misconception is forgetting to subtract an electron for the positive charge and thinking nitrogen would have 1 lone pair (choice B), but this would give the structure 10 total electrons instead of 8. When drawing Lewis structures for cations, always remember to subtract electrons equal to the positive charge from your total valence electron count.

Question 10

1. H—C—O—H with two lone pairs on O
2. H—C(=O)—H with two lone pairs on O (correct answer)
3. H—C≡O with one lone pair on O and one lone pair on C
4. H=C—O with three lone pairs on O
5. H—C—O with one lone pair on O

Explanation: This question tests the skill of selecting accurate Lewis structures for organic molecules with heteroatoms. Formaldehyde, CH2O, has 12 valence electrons (4 from C, 2 from 2H, 6 from O), and the correct structure is H—C(=O)—H with two lone pairs on oxygen, using 4 electrons in C-H bonds, 4 in the double bond, and 4 in oxygen's lone pairs, totaling 12. Carbon achieves an octet with the double and two single bonds. Oxygen has 8 electrons from lone pairs and the double bond. A tempting distractor is H—C—O—H, which implies a single bond pattern totaling more electrons or violating octets, a connectivity misconception. Connect the central atom to all others and add double bonds to satisfy valences and octets.

Question 11

1. All atoms have complete octets and there are no unpaired electrons
2. Chlorine forms two single bonds and has two lone pairs
3. There is one unpaired electron in the structure (correct answer)
4. Each oxygen has exactly two lone pairs
5. Chlorine must have exactly one lone pair

Explanation: This question tests the understanding of odd-electron molecules in Lewis structures. Chlorine dioxide, ClO2, has 19 valence electrons (7 from Cl, 12 from 2O), an odd number, so any correct structure must include one unpaired electron, often on chlorine with one double bond, one single bond, and lone pairs adjusted accordingly. This odd count prevents all electrons from pairing, leading to a radical species. Octets may be approximated, but the unpaired electron is inevitable. A tempting distractor is all complete octets with no unpaired electrons, ignoring the odd total valence count, a misconception in electron pairing. Always check if total valence electrons are even or odd to identify potential radicals before drawing.

Question 12

1. 0
2. 1 (correct answer)
3. 2
4. 3
5. 4

Explanation: This question tests the skill of identifying lone pairs in Lewis structures of simple molecules. Ammonia (NH₃) has 8 total valence electrons: nitrogen contributes 5 and each hydrogen contributes 1 (3 total). In the Lewis structure, nitrogen is the central atom bonded to three hydrogen atoms with single bonds, using 6 electrons (2 electrons per bond). This leaves 2 electrons, which form 1 lone pair on the nitrogen atom. The nitrogen atom thus has 3 bonding pairs and 1 lone pair, giving it a complete octet. A common misconception is forgetting to account for all valence electrons or thinking nitrogen has no lone pairs because it's bonded to three atoms. To find lone pairs on central atoms, subtract the bonding electrons from the total valence electrons the atom contributes, remembering that hydrogen only needs 2 electrons total.

Question 13

1. 1 lone pair
2. 2 lone pairs (correct answer)
3. 3 lone pairs
4. 4 lone pairs
5. 0 lone pairs

Explanation: This question tests the skill of determining lone pairs on specific atoms in resonance structures. The carbonate ion CO₃²⁻ has 24 valence electrons: C contributes 4, each O contributes 6 (18 total), plus 2 for the 2- charge. In the structure with one C=O double bond and two C-O single bonds, the bonds use 8 electrons total. The remaining 16 electrons are placed as lone pairs: the double-bonded oxygen needs 2 lone pairs to complete its octet (since it already has 4 electrons from the double bond), while each single-bonded oxygen needs 3 lone pairs. A common misconception is thinking the double-bonded oxygen would have 3 lone pairs (choice C), which would give it 10 electrons total. To find lone pairs on any atom in a Lewis structure, remember that each atom needs 8 electrons total (for octet rule), so subtract the bonding electrons from 8 to find the number of lone pair electrons needed.

Question 14

1. 4 lone pairs
2. 1 lone pair
3. 2 lone pairs
4. 3 lone pairs (correct answer)
5. 0 lone pairs

Explanation: This question tests the skill of drawing simple ion Lewis structures and counting lone pairs. The hydroxide ion, OH^-, has 8 valence electrons (6 from O, 1 from H, plus 1 for the charge), and the structure is O—H with three lone pairs on oxygen (6 electrons) and the bond using 2 electrons, totaling 8. Oxygen achieves an octet with 6 lone electrons and 2 from the bond. Hydrogen has its duet. A tempting distractor is 2 lone pairs, possibly from forgetting the extra charge electron, a valence counting error. Include charge in total electrons and assign lone pairs to the more electronegative atom after bonding.

Question 15

1. 1 lone pair on each N (correct answer)
2. 2 lone pairs on each N
3. 0 lone pairs on each N
4. 4 lone pairs on each N
5. 3 lone pairs on each N

Explanation: This question tests the determination of lone pairs in molecules with multiple central atoms. Hydrazine, N2H4, has 14 valence electrons (10 from 2N, 4 from 4H), and the structure is H2N—NH2 with each nitrogen having two bonds to H, one to the other N, and one lone pair, using 10 electrons in bonds and 4 in lone pairs, totaling 14. Each nitrogen achieves an octet with 2 lone electrons and 6 from bonds. Hydrogens have duets. A tempting distractor is 2 lone pairs per N, which would under-bond the nitrogens and miscount to 12 bond electrons needed, a valence satisfaction error. Assign bonds to match typical valences, then add lone pairs to complete octets, verifying total electrons.

Question 16

1. 1 lone pair on each O
2. 2 lone pairs on each O (correct answer)
3. 3 lone pairs on each O
4. 4 lone pairs on each O
5. 0 lone pairs on each O

Explanation: This question tests the skill of identifying lone pair distributions in linear molecules from Lewis structures. Carbon dioxide, CO2, has 16 valence electrons (4 from C, 12 from 2O), and the correct structure is O=C=O with two double bonds (8 electrons) and each oxygen having two lone pairs (8 electrons total for lone pairs), summing to 16. Each oxygen thus has 4 lone electrons plus 4 from the double bond, achieving an octet. Carbon has 8 electrons from the two double bonds. A tempting distractor is 3 lone pairs on each O, which overcounts electrons to 20, reflecting a misconception of treating double bonds as single. To verify, sum bonding and lone electrons to match the total valence count while ensuring octets.

Question 17

1. 0 lone pairs
2. 1 lone pair (correct answer)
3. 2 lone pairs
4. 3 lone pairs
5. 4 lone pairs

Explanation: This question tests the ability to determine lone pairs on central atoms in trigonal pyramidal molecules. Phosphorus trichloride, PCl3, has 26 valence electrons (5 from P, 21 from 3Cl), and the structure has phosphorus with three single bonds to chlorine (6 electrons) and one lone pair (2 electrons), with each chlorine having three lone pairs (18 electrons), totaling 26. Phosphorus achieves an octet with 2 lone electrons and 6 from bonds. Each chlorine has 8 electrons. A tempting distractor is 2 lone pairs, perhaps from confusing phosphorus with nitrogen's valence, a misconception in electron counting. Tally valence electrons and assign bonds first, then lone pairs to central atom for octet completion.

Question 18

1. 0 oxygen atoms
2. 1 oxygen atom
3. 2 oxygen atoms (correct answer)
4. 3 oxygen atoms
5. 4 oxygen atoms

Explanation: This question tests the skill of analyzing lone pair distributions in resonance-stabilized ions without considering resonance. The carbonate ion, CO3^2-, has 24 valence electrons (4 from C, 18 from 3O, plus 2 for the charge), and in one valid structure, carbon is central with one double bond and two single bonds to oxygen, resulting in two oxygens (single-bonded) each with three lone pairs and one (double-bonded) with two. This uses 8 electrons in bonds and 16 in lone pairs, totaling 24, with all octets satisfied. Carbon has 8 bonding electrons. A tempting distractor is 3 oxygens with three lone pairs, which violates carbon's octet by using only single bonds, a misconception of avoiding multiples. Use double bonds to ensure central atom octet, then count lone pairs per atom type.

Question 19

1. 6 electrons (correct answer)
2. 14 electrons
3. 12 electrons
4. 10 electrons
5. 8 electrons

Explanation: This question tests the understanding of incomplete octets in Lewis structures. Boron trifluoride, BF3, has 24 valence electrons (3 from B, 21 from 3F), and the structure has boron with three single bonds (6 electrons) and each fluorine with three lone pairs (18 electrons), totaling 24, with 6 electrons surrounding boron from the bonds. Boron does not achieve a full octet, common for boron compounds. Fluorines have octets. A tempting distractor is 8 electrons, assuming boron needs an octet like carbon, a misconception in exceptions to the octet rule. Recognize elements like boron can have fewer than 8 electrons and count only bonding electrons around them in such cases.

Question 20

1. 3 lone pairs
2. 1 lone pair
3. 0 lone pairs (correct answer)
4. 4 lone pairs
5. 2 lone pairs

Explanation: This question tests the ability to construct Lewis structures for cations and count central atom lone pairs. The ammonium ion, NH4^+, has 8 valence electrons (5 from N, 4 from 4H, minus 1 for the positive charge), and the structure features nitrogen with four single bonds to hydrogen (8 electrons), leaving no lone pairs on nitrogen. This gives nitrogen an octet entirely from bonding electrons. Each hydrogen has its duet satisfied. A tempting distractor is 1 lone pair, possibly from forgetting to subtract the charge electron, a misconception in valence counting for ions. Always adjust total valence electrons for charges and distribute all as bonds or lone pairs to meet octet or duet rules.