Why can one Lewis drawing be insufficient?
You will be able to: Use resonance contributors to explain equivalent bonds without implying molecular switching.
Why can one Lewis drawing be insufficient?
A nitrate ion has three equivalent N–O connections, yet one ordinary Lewis drawing puts a double bond on only one oxygen. Several drawings together represent what one cannot.
A useful starting point: How does formal charge check a Lewis diagram? →
Words and symbols before equations
- Resonance contributors
- Different valid electron placements for the same atom connectivity.
- Resonance hybrid
- One actual delocalized structure described by the contributors together.
- Delocalization
- Electron density distributed over more than one localized bond.
- Average bond order
- A useful fractional bond description for equivalent contributors.
What this picture assumes
Nitrate contributors have the same atom connectivity and 24-electron budget. Selection changes the drawing, not time. Hybrid dashed links indicate delocalized bonding, not literal fractional sticks.
Read the picture in three steps
- Read the species and labels first. A Lewis line represents two electrons; a spatial stick indicates connectivity. Use the stated quantities and units for numerical comparisons.
- One delocalized nitrate ion: all three N–O bonds equivalent, average order 4/3; net charge −1.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Keep the atom positions and connections fixed. Move electron pairs in the drawings, maintaining the electron budget and overall charge. Moving atoms would create a different structural arrangement, not another resonance contributor.
For NO₃⁻, three equivalent contributors place the double bond on three possible oxygens. Each has 24 electrons and total formal charge −1.
The real ion does not hop among one long-lived double bond and two single bonds. Its three N–O bonds are equivalent in the symmetric isolated-ion model, with average bond order (2 + 1 + 1)/3 = 4/3. Not all molecules have equivalent or equally weighted contributors.
A worked example, step by step
Compute the average bond order and formal-charge sum for symmetric nitrate.
- Each contributor has one double and two single N–O bonds.
- Total bond order across the three connections is 4.
- Equivalent connections have average order 4/3, intermediate between 1 and 2.
- N is +1; two single-bonded O atoms are −1 and one double-bonded O is 0, giving total −1 in each contributor.
Resonance drawings are not snapshots of a molecule switching bonds back and forth.
Do nitrate’s three bonds take turns becoming the short one?
Compare with an explanation
No. The symmetric ion has equivalent delocalized bonds; the separate drawings are representations.
Predict. Change one thing. Explain.
Select each nitrate contributor, then the hybrid. Predict whether the atom positions or overall charge should change. Notice that choosing a drawing is not a time animation.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
One delocalized nitrate ion: all three N–O bonds equivalent, average order 4/3; net charge −1.
Nitrate contributors have the same atom connectivity and 24-electron budget. Selection changes the drawing, not time. Hybrid dashed links indicate delocalized bonding, not literal fractional sticks.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using electron accounting, electrostatic interactions or spatial geometry. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionDraw the two usual resonance contributors of ozone, O₃. State what stays fixed, find the average bond order, and explain why a switching-bond animation would mislead.
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Compare with the answer and four-point rubric
- 1 point: Both have the same O–O–O atom connectivity.
- 1 point: The double bond and electron assignments exchange positions in the drawings, preserving 18 electrons and zero net charge.
- 1 point: The two equivalent connections have average order (2 + 1)/2 = 1.5.
- 1 point: The actual molecule has delocalized bonding, not alternating localized structures.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What moves between resonance drawings?
Electron assignments; atom connectivity stays fixed.
RECALL 2Is the hybrid an extra resonance contributor?
No. It describes the actual delocalized structure.
RECALL 3When can simple equal averaging be used?
When the contributors and compared bonds are equivalent by symmetry.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why can one Lewis drawing be insufficient?
- Equivalent NO₃⁻ N–O bonds: average order = 4/3.
- Contributors preserve atom connectivity and overall charge.
Remember: Resonance drawings are not snapshots of a molecule switching bonds back and forth.
Conditions: Nitrate contributors have the same atom connectivity and 24-electron budget. Selection changes the drawing, not time. Hybrid dashed links indicate delocalized bonding, not literal fractional sticks.
Refresh Kid · AP Chemistry Unit 2 · Objectives 2.6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.6, objectives 2.6.A. CED effective Fall 2024, current official file checked September 16, 2026, together with the published clarifications. This is Unit 2: Compound Structure and Properties, Topics 2.1–2.7. The focused lesson breakdown is Refresh Kid’s editorial sequence. Models and original practice are teaching materials, not official AP questions. Numerical potential curves, ion comparisons and orbital-alignment indices state their approximations. Five- and six-domain shapes are included; d-orbital hybridization and molecular-orbital diagrams are not required here. GitHub’s 3D website examples, including the Three.js Mars camera-control example, informed the use of rotatable scenes. Our scientific geometry and viewer code are original; no repository artwork or tutorial code was copied. The self-hosted Three.js library retains its MIT license. Camera rotation does not alter chemistry. See also the official clarifications.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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