Resonance and Formal Charge - AP Chemistry

1. The carbonate ion, $\mathrm{CO_3^{2-}}$, has three resonance structures with one C=O double bond and two C–O single bonds. In any one resonance structure, which formal-charge distribution is most stable (preferred)?
   Answer: C has $0$; two O atoms have $-1$; the third O has $0$ This question tests the ability to determine the most stable formal charge distribution in a resonance structure of the carbonate ion using formal charge calculations. The correct answer is E, where C has 0, two O atoms have -1, and the third O has 0, as this sums to -2 and places negative charges on oxygen atoms. In the structure, the central carbon is bonded to three oxygens with one double bond and two single bonds, with no lone pairs on carbon, resulting in its 0 formal charge. This distribution is preferred because it minimizes formal charges and ensures octet satisfaction for all atoms. A tempting distractor is C, C +1, two singly bonded O -1, double-bonded O 0, but this is incorrect because the formal charges sum to -1 instead of -2, arising from the misconception of miscounting the ion's overall charge. A transferable strategy is to verify that formal charges sum to the molecule's total charge and prefer structures with minimal charge separation.
2. For the sulfite ion, $\mathrm{SO_3^{2-}}$, resonance structures can be drawn with one S=O double bond and two S–O single bonds (and permutations). Using the guideline that third-row S may have an expanded octet, which formal-charge distribution is most stable (preferred) for a single resonance structure?
   Answer: S has $0$; two O atoms have $-1$; one O has $0$ (one S=O, two S–O) This question tests the ability to determine the most stable formal charge distribution in a resonance structure of the sulfite ion using expanded octets. The correct answer is B, S has 0, two O atoms have -1, one O has 0, with one S=O and two S–O, summing to -2 and minimizing charges. In this structure, S has 10 electrons (2 nonbonding + 8 bonding), leading to FC 0. This is preferred as it places negative charges on oxygen. A tempting distractor is A, S +2, each O -1, but this is incorrect because it sums to -1 not -2 and doesn't expand the octet, stemming from the misconception of restricting sulfur to 8 electrons. A transferable strategy is to calculate formal charges after drawing structures with possible expanded octets for third-row elements.
3. For the formate ion, HCO$_2^-$, two resonance structures are drawn: (1) H–C(=O)–O$^-$ and (2) H–C(–O$^-$)=O. Which statement best describes the most stable resonance description of the ion?​
   Answer: Both (1) and (2) contribute equally because they are equivalent and satisfy octets This question tests understanding of equivalent resonance structures in formate ion. The two resonance structures H-C(=O)-O⁻ and H-C(-O⁻)=O are equivalent because they differ only in which oxygen bears the double bond and which bears the negative charge. Both structures have the same atoms, same connectivity, and same formal charge distribution (just on different but equivalent oxygen atoms), making them contribute equally to the resonance hybrid. Option A incorrectly suggests one structure is preferred over the other, failing to recognize their equivalence. The key principle is that resonance structures with identical energy (equivalent structures) contribute equally to the resonance hybrid.
4. The thiocyanate ion, $\mathrm{SCN^-}$, can be represented by multiple resonance structures. Which formal-charge distribution corresponds to the most stable resonance contributor?
   Answer: $\mathrm{S{=}C{=}N^-}$ with formal charges: $\mathrm{S}$ 0, $\mathrm{C}$ 0, $\mathrm{N}$ $-1$ This question tests your ability to evaluate resonance structures based on formal charge distribution for linear triatomic ions. For SCN⁻, the most stable resonance structure is S=C=N⁻ with formal charges of 0 on S (6 - 2 - 8/2 = 0), 0 on C (4 - 0 - 8/2 = 0), and -1 on N (5 - 2 - 6/2 = -1). This structure places the negative charge on nitrogen, the most electronegative atom in the ion, while maintaining zero formal charges on S and C. Option B incorrectly places the negative charge on the less electronegative sulfur atom, making it less stable despite satisfying octets. When comparing resonance structures, the most stable contributor places negative charges on the most electronegative atoms.
5. For the bicarbonate ion, $\mathrm{HCO_3^-}$, resonance occurs between the two oxygens not bonded to H (one is double-bonded to C and one is single-bonded). In a single resonance structure, which formal-charge distribution is preferred?
   Answer: The nonprotonated singly bonded O has $-1$; the other two O atoms have $0$; C has $0$ This question tests the ability to assign formal charges in a resonance structure of the bicarbonate ion considering the protonated oxygen. The correct answer is C, where the nonprotonated singly bonded O has -1, the other two O atoms have 0, and C has 0, summing to -1 and accounting for the OH group. In the structure, the carbon is double-bonded to one oxygen (0) and single-bonded to the charged oxygen (-1) and the OH oxygen (0). This is preferred because it places the charge on the nonprotonated oxygen for stability. A tempting distractor is A, the O bonded to H has -1, others 0, C 0, but this is incorrect because the protonated oxygen has two bonds, leading to 0 charge, stemming from the misconception that all single-bonded oxygens are charged. A transferable strategy is to distinguish between protonated and nonprotonated atoms when calculating formal charges in acids.
6. The formate ion, $\mathrm{HCO_2^-}$, has two major resonance structures that differ in which oxygen is double-bonded to carbon. Which statement best describes the most stable resonance contributor(s)?
   Answer: The structure with one $\mathrm{C{=}O}$ and one $\mathrm{C{-}O^-}$ is preferred; the two such structures are equivalent in stability 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.
7. The azide ion, N$_3^-$, can be drawn with resonance structures including: (1) $^-\!$N=N$^+$=N$^-$ and (2) N≡N–N$^{2-}$ (with appropriate lone pairs). Which structure is the more stable resonance contributor?​
   Answer: $^-\!$N=N$^+$=N$^-$ because it minimizes the magnitude of formal charges and keeps octets 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.
8. For the carbonate ion, $\mathrm{CO_3^{2-}}$, which resonance structure is the most stable (best single contributor) based on formal-charge placement and octet rules?
   Answer: 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 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.
9. The azide ion, $\mathrm{N_3^-}$, has resonance structures including $\mathrm{^-N\!=N^+\!=N^-}$ and $\mathrm{N\!\!\equiv N^+\!\!-N^{2-}}$ (and the reversed form). Which resonance structure is most stable (preferred)?
   Answer: $\mathrm{^-N\!=N^+\!=N^-}$ (ends $-1$, center $+1$, all octets satisfied) 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.
10. For the sulfate ion, SO$_4^{2-}$, a student proposes two types of resonance descriptions: (I) all four S–O bonds are single (with formal charges assigned), or (II) two S=O double bonds and two S–O single bonds (with formal charges assigned). Which is the best statement about the most stable resonance contributors under typical AP Chemistry formal-charge rules?​
    Answer: Type II is preferred because it reduces formal charges while keeping negative charges on oxygen 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.
11. For the formate ion, $\mathrm{HCOO^-}$, two resonance structures can be drawn by placing the C=O double bond to either oxygen. In one resonance structure, which formal-charge distribution is preferred?
    Answer: One O has $-1$ and is single-bonded; the other O has $0$ and is double-bonded; C and H are $0$ This question tests the ability to assign preferred formal charges in a resonance structure of the formate ion. The correct answer is A, where one O has -1 and is single-bonded, the other O has 0 and is double-bonded, C and H are 0, summing to -1. In the structure, the carbon has a double bond to one oxygen and a single bond to the charged oxygen. This is preferred for octet satisfaction and charge placement on oxygen. A tempting distractor is B, both O -1, C +1, but this is incorrect because it doesn't distinguish bond types in resonance, stemming from the misconception of uniform charges across atoms. A transferable strategy is to draw individual resonance structures and calculate formal charges separately for each.
12. For the acetate ion, $\mathrm{CH_3COO^-}$, two resonance structures place the negative charge on either oxygen. Considering a single resonance structure, which formal-charge distribution is preferred?
    Answer: One O has $-1$ and is single-bonded; the other O has $0$ and is double-bonded; all other atoms are $0$ 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.
13. For the nitronium ion, $\mathrm{NO_2^+}$, one can draw resonance structures placing the N=O double bond to either oxygen (the structures are equivalent). In a single resonance structure, which formal-charge distribution is preferred?
    Answer: N has $+1$; each O has $0$ (both N=O bonds double) 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.
14. For sulfur dioxide, $\mathrm{SO_2}$, two common resonance structures are drawn: (1) $\mathrm{O=S-O^-}$ with $\mathrm{S^+}$, and (2) $\mathrm{O=S=O}$ with an expanded octet on S (no formal charges). Using AP Chemistry conventions (allowing expanded octets for S), which is the more stable resonance structure?
    Answer: $\mathrm{O=S=O}$ with no formal charges because it minimizes formal charge This question tests the ability to identify the preferred resonance structure for sulfur dioxide using expanded octets and formal charges. The correct answer is B, O=S=O with no formal charges, as this minimizes charges and allows S an expanded octet. In this structure, sulfur has two double bonds and a lone pair, resulting in zero formal charges. This is preferred over charged forms for the neutral molecule. A tempting distractor is A, O=S-$O^-$ with $S^+$, but this is incorrect because it introduces charges unnecessarily, stemming from the misconception that octets cannot be expanded for zero charges. A transferable strategy is to prefer uncharged resonance structures when expanded octets permit for neutral compounds.
15. Ozone, $\mathrm{O_3}$, has 18 valence electrons. Which Lewis structure uses all 18 valence electrons, gives every atom an octet, and shows correctly assigned formal charges?
    Answer: $\mathrm{O{=}O{-}O}$ with formal charges: central $\mathrm{O}$ $+1$, single-bond terminal $\mathrm{O}$ $-1$, double-bond terminal $\mathrm{O}$ 0 This question tests your ability to draw resonance structures and calculate formal charges for ozone. The most stable resonance structure for O₃ has one O=O double bond and one O-O single bond, with the central oxygen having a +1 formal charge (6 - 2 - 6/2 = +1), the singly-bonded terminal oxygen having a -1 formal charge (6 - 6 - 2/2 = -1), and the doubly-bonded terminal oxygen having a 0 formal charge (6 - 4 - 4/2 = 0). This structure satisfies octets and minimizes formal charge magnitudes. Option E incorrectly suggests all atoms have zero formal charge, which is impossible for a bent molecule with different bond orders. When drawing resonance structures, verify that formal charges are calculated correctly using the formula: valence electrons - lone pair electrons - (bonding electrons/2).
16. In the cyanate ion, OCN$^-$, consider these resonance structures: (1) $^-\!$O–C≡N, (2) O=C=N$^-$, and (3) O≡C–N$^-$. Which resonance structure is expected to be the most stable contributor?​
    Answer: $^-\!$O–C≡N (negative charge on O; C≡N triple bond) 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.
17. For the nitrite ion, $\mathrm{NO_2^-}$, which resonance structure is the most stable (best single contributor) based on formal charges and octet considerations?
    Answer: One $\mathrm{N{=}O}$ and one $\mathrm{N{-}O^-}$ bond; $\mathrm{N}$ has $0$ formal charge and one $\mathrm{O}$ has $-1$ formal charge 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.
18. Ozone, $\mathrm{O_3}$, can be represented by two resonance structures in which one O–O bond is double and the other is single. Which resonance structure is most stable (preferred) based on formal charges?
    Answer: Central O has $+1$; the singly bonded terminal O has $-1$; the double-bonded terminal O has $0$ 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.
19. For the cyanate ion, $\mathrm{OCN^-}$, resonance structures include $\mathrm{^-O-C\!\equiv N}$ and $\mathrm{O=C=N^-}$. Which resonance structure is most stable (preferred)?
    Answer: $\mathrm{^-O-C\!\equiv N}$ (negative charge on O; minimal charge separation) 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.
20. The allyl cation, $\mathrm{C_3H_5^+}$, can be represented by two resonance structures: $\mathrm{CH_2=CH-CH_2^+}$ and $\mathrm{^+CH_2-CH=CH_2}$. Which statement best describes the most stable resonance description?
    Answer: Both resonance structures are equivalent in stability because they have the same octet satisfaction and charge placement 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.

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