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🔗 Unit 2 · Molecular & Ionic Compound Structure & Properties 🏠 Unit Hub 🗂 Flashcards 🗺 Cheat Sheet Essentials 🎙 Podcast 🎨 Visual Review 📝 MC Practice ✍️ SAQ Practice

AP Chemistry Unit 2 SAQ Practice

Practice a College Board-style short-answer question on Molecular & Ionic Compound Structure & Properties. Write your response, then reveal the model answer to see exactly what earns each point.

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Short Answer Question · Unit 2 · Lewis Structures, VSEPR & Bonding
The nitrite ion, NO₂⁻, has a central nitrogen atom bonded to two oxygen atoms, with one N–O single bond and one N=O double bond in one possible Lewis structure, plus one lone pair on nitrogen. A second, equally valid Lewis structure can be drawn with the single and double bonds swapped between the two oxygens.
A
Explain why the existence of two equally valid Lewis structures for NO₂⁻ indicates that the molecule exhibits resonance, and describe how the actual bonding in NO₂⁻ differs from either individual structure.

✓ Model answer (earns the point)

Because both Lewis structures are equally valid (same connectivity, same formal charges, only the position of the double bond differs), NO₂⁻ exhibits resonance. The actual bonding is a hybrid of both structures — the negative charge and double-bond character are delocalized equally across both N–O bonds, rather than one bond being a true single bond and the other a true double bond.

Why it scores: Identifies resonance specifically (not just "there are two structures") AND explains that the real structure is a delocalized hybrid, not a 50/50 mixture of two discrete molecules.
B
Identify the electron-domain geometry and the molecular geometry around the central nitrogen atom in NO₂⁻, and explain why they are the same or different.

✓ Model answer (earns the point)

Nitrogen has 3 total electron domains (2 bonding domains to the oxygens, plus 1 lone pair), giving a trigonal planar electron-domain geometry. Because one of those domains is a lone pair rather than a bonded atom, the molecular geometry is bent — different from the electron-domain geometry because molecular geometry only counts bonded atoms, not lone pairs.

Why it scores: Correctly names both geometries AND explains why they differ (lone pair present), rather than just stating the two geometry names without justification.
C
Predict the hybridization of the central nitrogen atom in NO₂⁻ and justify your answer.

✓ Model answer (earns the point)

Nitrogen is sp² hybridized. Since nitrogen has 3 total electron domains (trigonal planar electron-domain geometry), it requires 3 hybrid orbitals — one s orbital mixed with two p orbitals — to accommodate them, leaving one unhybridized p orbital available for the delocalized pi bonding involved in the resonance system.

Why it scores: Names the correct hybridization AND connects it explicitly to the electron-domain count (3 domains → sp²), rather than guessing a hybridization without justification.

How to score points on SAQs