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.

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.

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.

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.

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.

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.

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.

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.

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.

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.