In the best Lewis dot structure for NO^+, how many lone pairs are present on the nitrogen atom?

• A. 1
• B. 2
• C. 3
• D. 0

Answer: (D)

Verified step by step guidance

1. Determine the total number of valence electrons for the NO\textsuperscript{+} ion. Nitrogen has 5 valence electrons, oxygen has 6, and since the species has a positive charge (+1), subtract one electron: Total electrons = 5 + 6 - 1 = 10.
2. Draw a skeletal structure with nitrogen and oxygen connected by a bond. Since NO\textsuperscript{+} is a diatomic molecule, place nitrogen and oxygen atoms side by side with a bond between them.
3. Distribute the 10 valence electrons to satisfy the octet rule for both atoms. Start by placing bonding electrons between N and O, then assign remaining electrons as lone pairs to complete octets.
4. Check formal charges to find the best Lewis structure. Calculate formal charges using the formula: \(\text{Formal charge} = \text{Valence electrons} - \text{Nonbonding electrons} - \frac{1}{2} \times \text{Bonding electrons}\). The best structure minimizes formal charges, ideally zero or close to zero on atoms.
5. Analyze the nitrogen atom in the best Lewis structure. Count the number of lone pairs on nitrogen after bonding and formal charge considerations. In NO\textsuperscript{+}, nitrogen typically has zero lone pairs because it shares electrons in multiple bonds with oxygen.