Worked example 1
Draw the resonance structures of ozone O₃ and state the O-O bond order.
Try it first: Draw one valid structure, then ask what else is equally valid.
0 of 4 steps revealed.
What you'll be able to do: Draw a full set of resonance structures and describe the real molecule as a hybrid with delocalised electrons.
Best after: Multiple Bonds and Polyatomic Ions
Ozone has two equal bonds, yet no single Lewis structure can show that. Resonance is the fix.
These are recommended, not required. You can start this lesson at any time.
In ozone O₃ you can place the double bond on the left oxygen or the right one. Both drawings obey every rule, so neither can be preferred. Experiment shows the two O-O bonds are identical in length, intermediate between a single and a double bond, so neither drawing alone is the truth.
The real species is a single unchanging structure called the resonance hybrid, best pictured as a blend of the contributors. The double headed arrow between structures means blend, not equilibrium. Ozone does not oscillate; every molecule is the hybrid at all times.
Average bond order equals the total number of bonds between the two atoms across all contributors divided by the number of contributors. In nitrate, one N=O and two N-O give an average of 4 bonds over 3 positions, so each bond has order 1.33. Spreading electrons over more atoms lowers the energy, which is why delocalised species such as carbonate, nitrate and benzene are unusually stable.
When the contributors are not equivalent, the ones with lower formal charges and with negative formal charge on the more electronegative atom contribute more to the hybrid. That idea is developed further in the formal charge lessons.
Two or more valid Lewis structures for the same arrangement of atoms, differing only in electron placement.
The single real structure, an average of all contributors.
Only electrons move between resonance forms; moving atoms creates an isomer instead.
Three equivalent contributors give each C-O bond an average order of 1.33.
average bond order = total bonds between the pair across contributors / number of contributors
Draw the resonance structures of ozone O₃ and state the O-O bond order.
Try it first: Draw one valid structure, then ask what else is equally valid.
0 of 4 steps revealed.
How many equivalent resonance structures does the carbonate ion CO₃ 2- have, and what is each C-O bond order?
Try it first: Count electrons, then see how many positions the double bond can occupy.
0 of 4 steps revealed.
Why it's wrong: There is only one real structure, the hybrid, and it never changes.
Check instead: Describe the electrons as delocalised over the whole set of atoms.
Why it's wrong: Different atom positions mean a different compound.
Check instead: Check that every atom sits in the same place in each drawing.
Why it's wrong: Delocalisation gives fractional averages such as 1.33 or 1.5.
Check instead:
No practice questions are available for this topic yet. You can still practice the whole unit.
When more than one valid Lewis structure differs only in where the multiple bonds and lone pairs sit, the real species is a resonance hybrid, an average of the contributors. Delocalisation makes all equivalent bonds identical in length and gives fractional bond orders and fractional average formal charges.
CC BY 4.0 License