Lewis Diagrams - AP Chemistry

Carbon dioxide, $\text{CO}_2$ , has a total of 16 valence electrons. In the correct Lewis structure, how many lone pairs are on each oxygen atom?
Answer: 2 lone pairs on each O 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.
Boron trifluoride, $\text{BF}_3$ , has a total of 24 valence electrons. In the correct Lewis structure with B central and three single bonds to F, how many electrons surround boron (count bonding electrons around B)?
Answer: 6 electrons 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.
Carbonate has the formula $\mathrm{CO_3^{2-}}$ and a total of 24 valence electrons. In one correct Lewis structure (no resonance required), carbon is the central atom with one C=O double bond and two C–O single bonds. How many lone pairs are on the oxygen atom that is double-bonded to carbon?
Answer: 2 lone pairs 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.
Phosphorus trichloride, $\text{PCl}_3$ , has a total of 26 valence electrons. In the correct Lewis structure (P central with single bonds), how many lone pairs are on phosphorus?
Answer: 1 lone pair 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.
Chlorine dioxide, $\text{ClO}_2$ , has a total of 19 valence electrons. Which statement must be true for any correct Lewis structure of $\text{ClO}_2$ ?
Answer: There is one unpaired electron in the structure 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.
The molecule carbon dioxide, $\mathrm{CO_2}$ , has a total of 16 valence electrons. In the correct Lewis structure, how many lone pairs are on the central carbon atom?
Answer: 0 This question tests the skill of determining lone pairs on central atoms in Lewis structures. Carbon dioxide (CO₂) has 16 total valence electrons: carbon contributes 4 and each oxygen contributes 6. In the Lewis structure, carbon is the central atom forming double bonds with each oxygen atom (O=C=O). Each double bond uses 4 electrons, so the two double bonds account for 8 electrons, leaving 8 electrons to be distributed as lone pairs on the oxygen atoms (4 electrons or 2 lone pairs on each oxygen). The central carbon atom has no lone pairs because all 4 of its valence electrons are used in the two double bonds. A common misconception is thinking that all atoms must have lone pairs, but carbon typically forms 4 bonds and has no lone pairs when it achieves an octet. To determine lone pairs on central atoms, count the bonding electrons around the atom and subtract from 8 (for octet rule) to find nonbonding electrons.
The carbonate ion, $\text{CO}_3^{2-}$ , has a total of 24 valence electrons. In one valid Lewis structure (do not include resonance) with carbon central and octets satisfied, how many oxygen atoms have three lone pairs?
Answer: 2 oxygen atoms 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.
The cyanide ion, $\text{CN}^-$ , has a total of 10 valence electrons. Which Lewis structure correctly represents $\text{CN}^-$ (showing bond order and lone pairs) while satisfying octets?
Answer: C≡N with one lone pair on C and one lone pair on N, in brackets with a $-$ charge This question tests the skill of constructing Lewis structures for polyatomic ions, ensuring the total valence electrons are accounted for and octets are satisfied. The cyanide ion, $CN^-$ , has 10 valence electrons (4 from C, 5 from N, plus 1 for the negative charge), and the correct structure is C≡N with one lone pair on C and one on N, using 6 electrons in the triple bond and 4 in lone pairs to total 10 while giving both atoms octets through shared electrons. In this arrangement, carbon achieves an octet with its lone pair (2 electrons) and three bonds (6 shared electrons), and nitrogen does the same with its lone pair and the triple bond. The negative charge is formally on carbon, but the structure satisfies the octet rule for both atoms. A tempting distractor is choice B, which shows a double bond with two lone pairs on each, but this miscounts to 12 electrons, reflecting the misconception of ignoring the total valence electron count. To approach similar problems, always calculate total valence electrons first and distribute them to satisfy octets, prioritizing central atom bonds.
Hydrogen cyanide, HCN, has a total of 10 valence electrons. Which Lewis structure correctly represents HCN with carbon as the central atom and octets satisfied?
Answer: H—C≡N with one lone pair on N 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.
Nitrogen trifluoride, $\mathrm{NF_3}$ , has 26 total valence electrons. In the correct Lewis structure with nitrogen as the central atom and all atoms satisfying the octet rule, how many lone pairs are on the central nitrogen atom?
Answer: 1 lone pair 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.
The hypochlorite ion has the formula $\mathrm{ClO^-}$ and a total of 14 valence electrons. In the correct Lewis structure with a single Cl–O bond, how many lone pairs are on the chlorine atom?
Answer: 3 lone pairs 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.
Nitrogen trifluoride, $\text{NF}_3$ , has a total of 26 valence electrons. In the correct Lewis structure (single bonds to F), how many lone pairs are on the nitrogen atom?
Answer: 1 lone pair 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.
Formaldehyde, $\text{CH}_2\text{O}$ , has a total of 12 valence electrons. Which Lewis structure correctly represents $\text{CH}_2\text{O}$ with carbon as the central atom and octets satisfied?
Answer: H—C(=O)—H with two lone pairs on O 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.
Hydrazine, $\text{N}_2\text{H}_4$ , has a total of 14 valence electrons. In the correct Lewis structure with an N—N single bond and all H atoms terminal, how many lone pairs are on each nitrogen atom?
Answer: 1 lone pair on each N 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.
The nitrite ion, $\text{NO}_2^-$ , has a total of 18 valence electrons. In one valid Lewis structure (do not include resonance) with N central and octets satisfied, how many lone pairs are on the nitrogen atom?
Answer: 1 lone pair 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.
The molecule ammonia, $\mathrm{NH_3}$ , has a total of 8 valence electrons. In the correct Lewis structure, how many lone pairs are on the nitrogen atom?
Answer: 1 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.
Ozone, $\text{O}_3$ , has a total of 18 valence electrons. In one valid Lewis structure (do not include resonance), what is the bond order pattern between the central O and the two terminal O atoms?
Answer: One single bond and one double bond 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.
The hydroxide ion, $\text{OH}^-$ , has a total of 8 valence electrons. In the correct Lewis structure, how many lone pairs are on oxygen?
Answer: 3 lone pairs 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.
Dinitrogen, $\text{N}_2$ , has a total of 10 valence electrons. Which Lewis structure correctly represents $\text{N}_2$ ?
Answer: N≡N with one lone pair on each N 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.
The cyanate ion has the formula $\mathrm{OCN^-}$ and a total of 16 valence electrons. In the Lewis structure drawn with carbon as the central atom and no resonance shown, how many lone pairs are on the oxygen atom in the structure $\mathrm{O{=}C{=}N}$ ?
Answer: 2 lone pairs This question tests the skill of determining lone pairs in Lewis structures with multiple bonds. The cyanate ion OCN⁻ has 16 valence electrons: O contributes 6, C contributes 4, N contributes 5, plus 1 for the negative charge. In the structure O=C=N, the two double bonds use 8 electrons total (4 electrons per double bond). This leaves 8 electrons to place as lone pairs: oxygen needs 2 lone pairs to complete its octet (since it already has 4 electrons from the double bond), and nitrogen needs 1 lone pair. A common misconception is thinking oxygen would have 3 lone pairs (choice C), which would give oxygen 10 electrons total, violating the octet rule. To find lone pairs on any atom, count the bonding electrons it has, then add lone pairs until it reaches 8 electrons total.

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