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Resonance contributors differ only in where electrons are placed; the real species is one hybrid in which charge is shared and bond orders are averaged, and moving an atom gives a different compound.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will count a species' equivalent contributors, tell resonance from a change of compound, describe the hybrid, and calculate shared charges and averaged bond orders.
You can place a formal charge on an atom and count pi bonds. Resonance is what happens when those can be placed in more than one equally good way.
A resonance contributor is one structure written for a species; contributors differ only in where electrons are placed. The resonance hybrid is the real species, an average of its contributors. Delocalized electrons are spread over more than two atoms.
Write the ethanoate ion as CH3C(=O)O−: one C=O and one C–O−. But the two C–O bonds are measured to be the same length, 127 pm, between a C=O (123 pm) and a C–O (136 pm). Move a lone pair from the negative oxygen into its bond, and the C=O pi bond onto the other oxygen, and you get CH3C(O−)=O: the same structure with the charge on the other oxygen.
Neither structure is right. The real ion is the hybrid: each C–O bond has order 1.5 and each oxygen carries half the charge. The rules for writing contributors:
Equivalent contributors contribute equally; if they differ, the one with fewer charges and charges on suitable atoms (negative on oxygen rather than carbon) contributes more.
Another way: steps
To turn a hybrid into numbers:
Spreading charge over several atoms makes a species more stable. The ethanoate ion's charge shared over two oxygens is why ethanoic acid is far more acidic than ethanol, whose conjugate base keeps its charge on one oxygen (unit 4). And a species that shares a charge over two atoms can react at either: an enolate reacts at oxygen or at carbon; the allyl cation accepts electrons at either end.
Resonance is not isomerism. CH3CHO and CH2=CHOH differ in where a hydrogen atom is, so they are two different compounds that can interconvert (tautomers), not two contributors of one.
A molecule repeatedly changes into different resonance compounds. It is one hybrid all the time.
Contributors are a mixture. They are not compounds and cannot be separated.
Moving a hydrogen gives a resonance form. Moving an atom gives a different compound.
One contributor is the real one. None is; the hybrid is.
Write N with one N=O and two N–O−; the nitrogen is +1.
One contributor.
The double bond can sit on any of the three oxygens: three equivalent contributors.
Move electrons only.
Hybrid: each N–O bond has order 4/3, and the ion's −1 shared over three oxygens gives each −1/3.
Numbers the hybrid fixes.
One contributor is O=O+–O−. Where else can the double bond sit?
On the other terminal oxygen.
So how many equivalent contributors, and what bond order?
Two; each O–O bond has order 3/2.
How many equivalent resonance contributors does the ethanoate ion, CH3COO− have?
Answer:
Each pair of structures is described below. Match each pair to what it is.
| resonance contributors of one species | different compounds: an atom has moved | |
|---|---|---|
| CH3C(=O)O− and CH3C(O−)=O | ||
| CH3CHO and CH2=CHOH | ||
| CH3CH2OH and CH3OCH3 | ||
| CH2=CH–CH2+ and +CH2–CH=CH2 |
Benzene is often written as two Kekulé structures with alternating single and double bonds. What is benzene actually like?
In benzene, C6H6, what is the order of each carbon–carbon bond in the hybrid? (A fraction is fine.)
Answer:
The enolate ion of ethanal can be written CH2=CH–O−. Which atoms share its negative charge in the hybrid?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The carbonate ion, CO3 2−, has three equivalent contributors, each with one C=O and two C–O− bonds. In the hybrid, fill in the charge on each oxygen and the order of each carbon–oxygen bond. (Fractions are fine.)
| value in the hybrid | |
|---|---|
| charge on each oxygen | |
| order of each C–O bond |
You can say where delocalized electrons are and what the hybrid looks like. Tell someone why benzene does not flip between two structures. Next: bond polarity, which says which end of a bond is short of electrons.
8. Your turn: ozone, O3, step 3