AP Chemistry Quiz: Resonance And Formal Charge

What this quiz covers

This quiz focuses on Resonance And Formal Charge, 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. N has $+1$; each O has $0$ (both N=O bonds double) (correct answer)
2. N has $+2$; each O has $-1$
3. N has $+1$; one O has $0$; the other O has $0$
4. N has $0$; each O has $+1$
5. N has $+1$; one O has $+1$; the other O has $-1$

Explanation: This question tests the ability to assign preferred formal charges in a resonance structure of the nitronium ion. The correct answer is E, N has +1, each O has 0 with both N=O bonds double, summing to +1. In the structure, nitrogen has two double bonds and no lone pair, leading to +1 formal charge. This is preferred for symmetry and octet satisfaction. A tempting distractor is B, N +1, one O 0, other 0, but this is incorrect as it implies inconsistent bonding, arising from the misconception of unequal bonds in symmetric ions. A transferable strategy is to use symmetric structures for ions with equivalent atoms to minimize energy.

Question 2

1. N has $+1$; the two singly bonded O atoms each have $-1$; the double-bonded O has $0$ (correct answer)
2. N has $-1$; each O atom has $0$
3. N has $+1$; one O has $-2$; the other two O atoms have $0$
4. N has $0$; one O has $-1$; the other two O atoms have $0$
5. N has $+2$; each O atom has $-1$

Explanation: This question tests the ability to determine the most stable formal charge distribution in a resonance structure of the nitrate ion by calculating formal charges and evaluating stability based on charge separation and electronegativity. The correct answer is A, where N has +1, the two singly bonded O atoms each have -1, and the double-bonded O has 0, as this distribution sums to the ion's charge of -1 and places negative charges on the more electronegative oxygen atoms. In this structure, the central nitrogen is bonded to three oxygen atoms with one double bond and two single bonds, with no lone pairs on nitrogen, leading to its +1 formal charge according to the formula valence electrons minus nonbonding electrons minus half of bonding electrons. This distribution is preferred because it minimizes the magnitude of formal charges while ensuring the positive charge is on the less electronegative nitrogen. A tempting distractor is D, where N has +2 and each O has -1, but this is incorrect because it would require all single bonds, leading to higher formal charges and less stability, stemming from the misconception that all bonds are equivalent without considering resonance. A transferable strategy is to always calculate formal charges for each atom in a Lewis structure and choose the resonance form where charges are as small as possible and negative charges are on more electronegative atoms.

Question 3

1. The structure with the negative charge on the middle carbon is most stable because it has more bonds
2. Both resonance structures are equivalent in stability because they have the same octet satisfaction and charge placement (correct answer)
3. The structure with the negative charge on an end carbon is most stable because terminal carbanions are always most stable
4. A structure with a C≡C triple bond is most stable because it reduces electron density
5. One resonance structure is invalid because carbon cannot have a negative formal charge

Explanation: This question tests the ability to assess resonance structures of the allyl anion for stability. The correct answer is B, both resonance structures are equivalent in stability because they have the same octet satisfaction and charge placement on terminal carbons. The structures CH2=CH-CH2^- and ^-CH2-CH=CH2 are identical in energy due to symmetry. This is preferred for describing delocalized charge. A tempting distractor is A, negative on middle carbon most stable, but this is incorrect because such a structure isn't a valid resonance form, stemming from the misconception that more bonds always stabilize charges centrally. A transferable strategy is to identify equivalent resonance forms by checking for symmetry in charge distribution.

Question 4

1. Two equivalent structures: one $\mathrm{C{=}O}$, one $\mathrm{C{-}O^-}$, and one $\mathrm{C{-}O(H)}$; the negative charge is on an oxygen not bonded to H (correct answer)
2. One structure dominates: the negative charge is on the oxygen bonded to H because it is stabilized by H
3. A structure with three $\mathrm{C{=}O}$ double bonds is preferred because it gives carbon an expanded octet
4. A structure with three $\mathrm{C{-}O}$ single bonds is preferred because it avoids formal charges
5. A structure with $\mathrm{C^-}$ and all three oxygens neutral is preferred because it minimizes charge separation

Explanation: This question tests your understanding of resonance in ions containing hydrogen atoms. For HCO₃⁻, there are two equivalent resonance structures where the C=O double bond can be on either oxygen that is not bonded to H, with the negative charge on the other non-hydrogen-bonded oxygen. These structures are equivalent because both involve the same types of atoms in similar environments. Option B incorrectly suggests the negative charge should be on the oxygen bonded to H, but this would place negative charge adjacent to the partially positive hydrogen, creating an unfavorable arrangement. When drawing resonance structures for ions with H atoms, avoid placing negative charges on atoms directly bonded to hydrogen.

Question 5

1. The two structures with one $\mathrm{C{=}C}$ double bond and a $+1$ charge on a terminal carbon are equivalent in stability (correct answer)
2. The structure with the $+1$ charge on the central carbon is preferred because the central carbon is more substituted
3. The structure with two $\mathrm{C{=}C}$ double bonds is preferred because it gives all carbons octets
4. The structure with all single bonds is preferred because it minimizes pi bonding
5. The structure with a $-1$ charge on one terminal carbon is preferred because opposite charges stabilize each other

Explanation: This question tests your understanding of resonance stabilization in carbocations. The allyl cation C₃H₅⁺ has two equivalent resonance structures, each with one C=C double bond and the positive charge on a terminal carbon atom. These structures are equivalent because the positive charge alternates between the two terminal carbons, which are in identical chemical environments. Option B incorrectly suggests the central carbon position is preferred for the positive charge, but resonance structures show the charge is delocalized over the terminal positions. When evaluating resonance in symmetric systems, structures that differ only in the position of charge/bonds among equivalent atoms contribute equally.

Question 6

1. The structure with one $\mathrm{C{=}O}$ and one $\mathrm{C{-}O^-}$ is preferred; the two such structures are equivalent in stability (correct answer)
2. The structure with two $\mathrm{C{=}O}$ double bonds is preferred because it minimizes formal charge
3. The structure with two $\mathrm{C{-}O}$ single bonds is preferred because it avoids double bonds
4. The structure with negative charge on carbon is preferred because carbon is less electronegative than oxygen
5. The structure with $\mathrm{O^+}$ and $\mathrm{O^-}$ on the two oxygens is preferred because it shows charge separation

Explanation: This question tests your understanding of equivalent resonance structures in formate ion. For HCO₂⁻, there are two equivalent resonance structures: one with C=O to the first oxygen and C-O⁻ to the second, and another with these bonds reversed. Both structures have carbon with 0 formal charge and one oxygen with -1 formal charge, making them equally stable and equally contributing to the resonance hybrid. Option B incorrectly suggests a structure with two C=O double bonds, which would not account for the -1 charge on the ion. When molecules have symmetrical arrangements, resonance structures that differ only in the position of bonds/charges are equivalent contributors.

Question 7

1. $^-\!$O–C≡N (negative charge on O; C≡N triple bond) (correct answer)
2. O≡C–N$^-$ (negative charge on N; O≡C triple bond)
3. O=C=N$^-$ (negative charge on N; two double bonds)
4. O=C=N (no formal charges on any atom; two double bonds)
5. O–C–N$^{2-}$ (all single bonds; $2-$ on N to complete octet)

Explanation: This question tests evaluation of resonance structures for cyanate ion using formal charge and electronegativity principles. Structure (1) ⁻O-C≡N is the most stable because it places the negative formal charge on oxygen, the most electronegative atom in the molecule. In this structure, O has -1, C has 0, and N has 0 formal charge. Structure (2) O=C=N⁻ places the negative charge on the less electronegative nitrogen, making it less stable. Structure (3) O≡C-N⁻ also places negative charge on nitrogen and creates larger formal charges. Option C incorrectly identifies structure (2) as most stable, ignoring the electronegativity consideration. The strategy is to place negative formal charges on more electronegative atoms when possible.

Question 8

1. Type I is preferred because sulfur cannot exceed an octet under any circumstances
2. Type II is preferred because it reduces formal charges while keeping negative charges on oxygen (correct answer)
3. Type II is preferred because it places the $2-$ charge entirely on sulfur
4. Type I is preferred because it gives all atoms a formal charge of $0$
5. Type I and Type II are equally preferred because resonance structures must have identical formal charges

Explanation: This question tests understanding of expanded octets and formal charge minimization in sulfate ion. Type II structures (two S=O double bonds and two S-O single bonds) are preferred because they reduce formal charges compared to Type I (all single bonds). In Type II, sulfur has formal charge +2, double-bonded oxygens have 0, and single-bonded oxygens have -1, giving smaller magnitude charges than Type I where sulfur would have +2 and all four oxygens would have -1. Sulfur, being in the third period, can accommodate more than 8 electrons through d-orbital participation. Option A incorrectly claims sulfur cannot exceed an octet, confusing it with second-period elements. The key principle is that third-period and heavier elements can exceed the octet to minimize formal charges.

Question 9

1. Both resonance structures are equivalent in stability because they have the same octet satisfaction and charge placement (correct answer)
2. One resonance structure is invalid because carbon cannot have a positive formal charge
3. The structure with the positive charge on an end carbon is most stable because terminal carbocations are always most stable
4. The structure with the positive charge on the middle carbon is most stable because it makes two double bonds
5. The structure with a triple bond is most stable because it minimizes formal charge

Explanation: This question tests the ability to evaluate resonance structures of the allyl cation based on stability and charge placement. The correct answer is C, both resonance structures are equivalent in stability because they have the same octet satisfaction and charge placement on terminal carbons. The structures CH2=CH-CH2^+ and ^+CH2-CH=CH2 are symmetric, sharing the positive charge equally. This is preferred as it reflects delocalization without favoring one form. A tempting distractor is A, positive on middle carbon most stable, but this is incorrect because it would require a different structure without terminal charges, arising from the misconception that central charges are always better. A transferable strategy is to compare symmetry and charge delocalization in resonance forms for equivalent stability.

Question 10

1. Both O atoms have $-1$; C has $+1$
2. Single-bonded O has $-2$; double-bonded O has $+1$; C has $0$
3. Carbonyl C has $-1$; both O atoms have $0$
4. Both O atoms have $0$; carbonyl C has $-1$
5. One O has $-1$ and is single-bonded; the other O has $0$ and is double-bonded; all other atoms are $0$ (correct answer)

Explanation: This question tests the ability to determine preferred formal charges in a resonance structure of the acetate ion. The correct answer is B, where one O has -1 and is single-bonded, the other O has 0 and is double-bonded, and all other atoms are 0, summing to -1 and placing the charge on oxygen. In the structure, the carbonyl carbon has a double bond to one oxygen and a single bond to the charged oxygen, with formal charges calculated accordingly. This is preferred because it ensures octets and minimal charges. A tempting distractor is A, both O -1, C +1, but this is incorrect because it ignores the resonance distinction between single and double bonds, stemming from the misconception of averaging charges in a single structure. A transferable strategy is to calculate formal charges individually for each resonance form and favor those with charges on electronegative atoms.

Question 11

1. Three Cl=O: Cl has $-1$; each O has $0$ (Cl has 12 electrons)
2. One Cl≡O and two Cl–O: Cl has $-1$; two O have $-1$; one O has $+1$
3. Two Cl=O and one Cl–O: Cl has $0$; one O has $-1$; two O have $0$ (Cl has 10 electrons) (correct answer)
4. Three Cl–O single bonds: Cl has $+2$; each O has $-1$ (Cl has an octet)
5. One Cl=O and two Cl–O: Cl has $+1$; two O have $-1$; one O has $0$ (Cl has an octet)

Explanation: This question tests the ability to select the most stable resonance structure for the chlorate ion using expanded octets and formal charge minimization. The correct answer is C, two Cl=O and one Cl–O, Cl has 0, one O has -1, two O have 0, with Cl having an expanded octet, as this minimizes formal charges to sum -1. In this structure, Cl has 12 electrons (2 nonbonding + 10 bonding), allowing FC 0 by the formal charge formula. This is preferred for third-row elements to reduce positive charge on Cl. A tempting distractor is B, Cl +1, two O -1, one O 0, but this is incorrect because it has higher absolute charges than C, arising from the misconception that octets cannot be expanded beyond 10 electrons. A transferable strategy is to allow expanded octets for elements beyond the second row to achieve lower formal charges.

Question 12

1. $\mathrm{^-O-C\!\equiv N}$ (negative charge on O; minimal charge separation) (correct answer)
2. $\mathrm{O^+\!\equiv C-N^{2-}}$ (places more negative charge on N to match its electronegativity)
3. $\mathrm{O=C=N^-}$ (negative charge on N; carbon has an expanded octet)
4. $\mathrm{O^-\!=C\!=N}$ (negative charge on O with N positive)
5. $\mathrm{O\!\equiv C-N^-}$ (triple bond to O; negative charge on N)

Explanation: This question tests the ability to select the most stable resonance structure for the cyanate ion using formal charges and electronegativity. The correct answer is A, ^-O-C≡N with negative charge on O, as oxygen is more electronegative than nitrogen, minimizing charge separation. In this structure, formal charges are O -1, C 0, N 0, summing to -1 with octets. This is preferred for placing charge on O. A tempting distractor is C, O=C=N^-, but this is incorrect because it places the charge on less electronegative N, stemming from the misconception that double bonds stabilize better than triple regardless of charge location. A transferable strategy is to prioritize structures with negative charges on the most electronegative atoms.

Question 13

1. Three Br–O single bonds: Br has $+2$; each O has $-1$ (Br has an octet)
2. One Br=O and two Br–O: Br has $+1$; two O have $-1$; one O has $0$ (Br has an octet)
3. Two Br=O and one Br–O: Br has $0$; one O has $-1$; two O have $0$ (Br has 10 electrons) (correct answer)
4. Three Br=O: Br has $-1$; each O has $0$ (Br has 12 electrons)
5. One Br≡O and two Br–O: Br has $-1$; two O have $-1$; one O has $+1$

Explanation: This question tests the ability to choose the most stable resonance structure for the bromate ion using expanded octets. The correct answer is C, two Br=O and one Br–O, Br has 0, one O has -1, two O have 0, with Br expanded octet, summing to -1. In this structure, Br has 12 electrons, achieving FC 0. This is preferred for minimizing charges. A tempting distractor is B, Br +1, two -1, one 0, but this is incorrect as it has higher charges, arising from the misconception of limiting to octet without expansion. A transferable strategy is to apply expanded octet rules to halogen anions for optimal formal charges.

Question 14

1. Central O has $+1$; the singly bonded terminal O has $-1$; the double-bonded terminal O has $0$ (correct answer)
2. Central O has $0$; both terminal O atoms have $0$
3. Central O has $-1$; one terminal O has $+1$; the other terminal O has $0$
4. Central O has $+2$; each terminal O has $-1$
5. Central O has $0$; one terminal O has $-1$; the other terminal O has $+1$

Explanation: This question tests the ability to evaluate resonance structures of ozone based on formal charge stability. The correct answer is A, where the central O has +1, the singly bonded terminal O has -1, and the double-bonded terminal O has 0, as this distribution sums to 0 for the neutral molecule and places the negative charge on a terminal oxygen. In the structure, the central oxygen has one double bond and one single bond with a lone pair, leading to its +1 formal charge. This is preferred because it achieves octet satisfaction and minimizes charge magnitude. A tempting distractor is E, central O 0, one terminal -1, other +1, but this is incorrect because it places a positive charge on a more electronegative terminal oxygen, stemming from the misconception that charges should be balanced without considering electronegativity. A transferable strategy is to prioritize resonance structures where positive charges are on less electronegative atoms and verify octet completion.

Question 15

1. $\mathrm{N^{2-}\!\!-N^+\!\!\equiv N}$ (one end $-2$, one end $0$, center $+1$)
2. $\mathrm{N\!\!\equiv N^+\!\!-N^{2-}}$ (one end $-2$, one end $0$, center $+1$)
3. $\mathrm{^-N\!\!\equiv N\!\!\equiv N}$ (center neutral, one end $-1$)
4. $\mathrm{^-N\!=N^+\!=N^-}$ (ends $-1$, center $+1$, all octets satisfied) (correct answer)
5. $\mathrm{N\!=N\!=N^-}$ (center neutral, one end $-1$, incomplete octet on center)

Explanation: This question tests the ability to identify the most stable resonance structure for the azide ion based on formal charges and octet satisfaction. The correct answer is B, ^-N=N^+=N^-, with ends -1, center +1, and all octets satisfied, as this minimizes charge magnitude and places negative charges on terminal nitrogens. In this structure, double bonds ensure all atoms have octets, and the formal charges sum to -1. This is preferred over structures with higher charges like -2 on one nitrogen. A tempting distractor is A, N≡N^+-N^{2-}, but this is incorrect because it has higher formal charge magnitudes, arising from the misconception that triple bonds always stabilize better regardless of charge increase. A transferable strategy is to select resonance forms with the lowest absolute formal charges and complete octets for all atoms.

Question 16

1. Three $\mathrm{C{=}O}$ double bonds; $\mathrm{C}$ has $-2$ formal charge and each $\mathrm{O}$ has $0$ formal charge
2. One $\mathrm{C{=}O}$ and two $\mathrm{C{-}O^-}$ bonds; $\mathrm{C}$ has $0$ formal charge and two $\mathrm{O}$ atoms each have $-1$ formal charge (correct answer)
3. Three $\mathrm{C{-}O}$ single bonds; $\mathrm{C}$ has $-1$ formal charge and each $\mathrm{O}$ has $-\tfrac{1}{3}$ formal charge
4. Two $\mathrm{C{=}O}$ and one $\mathrm{C{-}O^-}$ bond; $\mathrm{C}$ has $-1$ formal charge and one $\mathrm{O}$ has $-1$ formal charge
5. One $\mathrm{C{=}O}$ and two $\mathrm{C{-}O}$ single bonds; $\mathrm{C}$ has $-2$ formal charge and two $\mathrm{O}$ atoms each have $0$ formal charge

Explanation: This question tests your understanding of resonance structures and formal charge calculations for carbonate ion. For CO₃²⁻, the most stable resonance structure has one C=O double bond and two C-O⁻ single bonds, giving carbon a formal charge of 0 (4 - 0 - 8/2 = 0) and each singly-bonded oxygen a -1 formal charge (6 - 6 - 2/2 = -1). This arrangement satisfies the octet rule for all atoms and properly distributes the -2 charge across two oxygen atoms. Option A incorrectly suggests three double bonds, which would give carbon a -2 formal charge and fail to distribute negative charge on the more electronegative oxygen atoms. When evaluating resonance structures, prioritize placing negative formal charges on more electronegative atoms while maintaining octets.

Question 17

1. Two $\mathrm{N{=}O}$ double bonds; $\mathrm{N}$ has $-1$ formal charge and each $\mathrm{O}$ has $0$ formal charge
2. One $\mathrm{N{=}O}$ and one $\mathrm{N{-}O^-}$ bond; $\mathrm{N}$ has $0$ formal charge and one $\mathrm{O}$ has $-1$ formal charge (correct answer)
3. Two $\mathrm{N{-}O}$ single bonds; $\mathrm{N}$ has $-1$ formal charge and each $\mathrm{O}$ has $0$ formal charge
4. One $\mathrm{N\equiv O}$ and one $\mathrm{N{-}O^-}$ bond; $\mathrm{N}$ has $+1$ formal charge and one $\mathrm{O}$ has $-2$ formal charge
5. One $\mathrm{N{=}O}$ and one $\mathrm{N{-}O}$ bond; $\mathrm{N}$ has $-1$ formal charge and both $\mathrm{O}$ atoms have $0$ formal charge

Explanation: This question tests your understanding of resonance structures and formal charges for nitrite ion. For NO₂⁻, the most stable resonance structure has one N=O double bond and one N-O⁻ single bond, giving nitrogen a formal charge of 0 (5 - 2 - 6/2 = 0) and the singly-bonded oxygen a -1 formal charge (6 - 6 - 2/2 = -1). This structure satisfies the octet rule and places the negative charge on the more electronegative oxygen atom. Option A incorrectly suggests two double bonds with nitrogen having a -1 charge, which would place negative charge on the less electronegative nitrogen instead of oxygen. When evaluating resonance structures, prioritize placing negative formal charges on more electronegative atoms.

Question 18

1. Two $\mathrm{N{=}O}$ bonds and one $\mathrm{N{-}O^-}$ bond; $\mathrm{N}$ has $0$ formal charge and one $\mathrm{O}$ has $-1$ formal charge
2. Three $\mathrm{N{-}O}$ single bonds; $\mathrm{N}$ has $-2$ formal charge and each $\mathrm{O}$ has $+\tfrac{1}{3}$ formal charge
3. One $\mathrm{N{=}O}$ and two $\mathrm{N{-}O}$ single bonds; $\mathrm{N}$ has $-1$ formal charge and two $\mathrm{O}$ atoms each have $0$ formal charge
4. Three $\mathrm{N{=}O}$ double bonds; $\mathrm{N}$ has $-1$ formal charge and each $\mathrm{O}$ has $0$ formal charge
5. One $\mathrm{N{=}O}$ and two $\mathrm{N{-}O^-}$ bonds; $\mathrm{N}$ has $+1$ formal charge and two $\mathrm{O}$ atoms each have $-1$ formal charge (correct answer)

Explanation: This question tests your ability to draw resonance structures and calculate formal charges for polyatomic ions. For NO₃⁻, the most stable resonance structure has one N=O double bond and two N-O⁻ single bonds, giving nitrogen a +1 formal charge (5 valence electrons - 0 lone pair electrons - 4 bonding electrons/2 = +1) and each singly-bonded oxygen a -1 formal charge (6 - 6 - 2/2 = -1). This structure satisfies the octet rule for all atoms and minimizes formal charge magnitudes. Option B incorrectly suggests two double bonds, which would violate the overall -1 charge of the ion since it would only have one negative formal charge instead of the required net -1. When drawing resonance structures for polyatomic ions, always verify that formal charges sum to the overall charge and that the octet rule is satisfied.

Question 19

1. N≡N–N$^{2-}$ because it has a triple bond, which is always most stable
2. $^-\!$N=N$^+$=N$^-$ because it minimizes the magnitude of formal charges and keeps octets (correct answer)
3. N=N=N$^-$ with a single negative charge on the terminal N and no positive charge on the central N
4. N–N≡N$^{2-}$ because placing a $2-$ charge on nitrogen is favored over charge separation
5. N=N=N with all formal charges equal to $0$ because nitrogen can expand its octet

Explanation: This question tests evaluation of resonance structures for azide ion using formal charge rules. Structure (1) ⁻N=N⁺=N⁻ is more stable than structure (2) N≡N-N²⁻ because it minimizes the magnitude of formal charges. In structure (1), the terminal nitrogens each have -1 charge and the central nitrogen has +1, giving formal charges of ±1. In structure (2), one terminal nitrogen has -2 charge, which is a larger magnitude. All atoms maintain octets in both structures. Option E incorrectly claims nitrogen can expand its octet, but nitrogen is a second-period element limited to 8 electrons. The strategy is to prefer structures with smaller magnitude formal charges when all atoms have complete octets.

Question 20

1. $\mathrm{H-O^+=C=N^-}$ because it has a full octet on all atoms and a stronger C=O bond
2. $\mathrm{H-O-C\!\equiv N}$ because it has no formal charges and all atoms have octets (correct answer)
3. $\mathrm{H-O^-=C=N^+}$ because negative charge is best on oxygen and positive is best on nitrogen
4. $\mathrm{H-O\!\equiv C-N}$ because a triple bond to oxygen is always favored
5. $\mathrm{H-O-C\!=N}$ because carbon prefers only three bonds in stable structures

Explanation: This question tests the ability to determine the preferred resonance structure for hydrogen cyanate based on formal charges and octets. The correct answer is B, H-O-C≡N with no formal charges and all atoms having octets, as this minimizes charges entirely. In this structure, the triple bond and single bond satisfy octets without charge separation. This is preferred over charged forms for neutral molecules. A tempting distractor is A, H-O^+=C=N^-, but this is incorrect because it introduces unnecessary charges, arising from the misconception that double bonds are always better than triple for stability. A transferable strategy is to favor resonance structures with zero formal charges when possible for neutral species.