AP Chemistry Practice Quiz: Lewis Diagrams
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) 22
B) 23
C) 24
D) 32
Correct Answer: C
To calculate the total valence electrons, sum the valence electrons for each atom and add the number of electrons corresponding to the negative charge. Carbon (Group 14) has 4, each Oxygen (Group 16) has 6. The 2- charge indicates an additional 2 electrons. Total = 4 + 3(6) + 2 = 24.
A) BF₃
B) CH₄
C) NH₃
D) SF₆
Correct Answer: D
In the Lewis diagram for SF₆, the central sulfur atom forms single bonds with six fluorine atoms. This results in the sulfur atom being surrounded by 12 valence electrons, which is an 'expanded octet'. This is possible because sulfur is in the third period and has accessible d-orbitals. BF₃ has an incomplete octet, while CH₄ and NH₃ obey the octet rule.
A) -2
B) -1
C) 0
D) +1
Correct Answer: C
The structure is [:O=C=N:]⁻. Formal Charge = (Valence Electrons) - (Non-bonding Electrons) - (1/2 * Bonding Electrons). For Oxygen (Group 16), this is 6 - 4 - (1/2 * 4) = 6 - 4 - 2 = 0. The formal charge on the oxygen atom is zero in this specific representation.
A) A diagram showing H double-bonded to C, and C single-bonded to N.
B) A diagram showing H single-bonded to N, and N triple-bonded to C.
C) A diagram showing H single-bonded to C, and C triple-bonded to N.
D) A diagram showing H single-bonded to C, and C single-bonded to N.
Correct Answer: C
The correct Lewis diagram for HCN has a single bond between H and C and a triple bond between C and N. This representation uses the correct total of 10 valence electrons (H=1, C=4, N=5) and satisfies the octet rule for both carbon and nitrogen, while hydrogen has its stable duet.
A) Places a triple bond between the two nitrogen atoms.
B) Places a triple bond between the nitrogen and oxygen atoms.
C) Minimizes the magnitude of formal charges and places the negative charge on the most electronegative atom.
D) Maximizes the number of atoms that satisfy the octet rule, regardless of formal charge.
Correct Answer: C
While multiple resonance structures for N₂O satisfy the octet rule, the most stable (and thus best) representation is the one that minimizes the magnitude of the formal charges on the atoms. Furthermore, any necessary negative formal charge should be placed on the most electronegative atom, which is oxygen in this case. The structure N≡N-O accomplishes this best.
A) F₂
B) C₂H₆
C) H₂O
D) C₂H₄
Correct Answer: D
To construct the Lewis diagram for C₂H₄ (ethene), a double bond must be placed between the two carbon atoms to satisfy the octet rule for both carbons while using the correct total of 12 valence electrons. F₂, C₂H₆, and H₂O are all correctly represented using only single bonds.
A) Nitrogen cannot form more than three bonds, so one oxygen must have a different bond order.
B) All three N-O bonds are experimentally shown to be identical in length and strength, which a single Lewis structure with one double and two single bonds cannot represent.
C) The ion has a negative charge, which must be delocalized across all three oxygen atoms equally.
D) The central nitrogen atom has an expanded octet that shifts between the three oxygen atoms.
Correct Answer: B
A single Lewis diagram for NO₃⁻ would show one N=O double bond and two N-O single bonds. However, experimental evidence shows that all three N-O bonds are identical, intermediate in length and strength between a single and a double bond. The concept of resonance, represented by drawing all three possible structures, is used to reconcile the Lewis model with this physical reality.