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.

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.

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 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 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 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.

This question tests the skill of determining bond order in Lewis structures. The cyanide ion (CN⁻) has 10 total valence electrons: carbon contributes 4, nitrogen contributes 5, and the negative charge adds 1 electron. To draw the Lewis structure, we connect C and N and distribute the remaining electrons to satisfy the octet rule for both atoms. With only 10 electrons total, the only way to give both atoms complete octets is to form a triple bond between C and N (using 6 electrons), with one lone pair on carbon (2 electrons) and one lone pair on nitrogen (2 electrons). The bond order is therefore 3, corresponding to a triple bond. A common misconception is trying to form only single or double bonds, which would not allow both atoms to achieve octets with the available electrons. When determining bond order in small molecules or ions, check if multiple bonds are needed to satisfy the octet rule with the given number of valence electrons.

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.

This question tests the skill of drawing Lewis structures for polyatomic cations. The ammonium ion NH₄⁺ has 8 valence electrons: N contributes 5, each H contributes 1 (4 total), minus 1 for the positive charge. With nitrogen as the central atom bonded to four hydrogens, all 8 electrons are used in the four N-H single bonds (2 electrons per bond). This leaves no electrons for lone pairs on nitrogen, giving nitrogen a complete octet with its 8 bonding electrons. A common misconception is forgetting to subtract an electron for the positive charge and thinking nitrogen would have 1 lone pair (choice B), but this would give the structure 10 total electrons instead of 8. When drawing Lewis structures for cations, always remember to subtract electrons equal to the positive charge from your total valence electron count.

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.