Which of the following statements regarding resonance contributors and resonance

Question

Which of the following statements regarding resonance contributors and resonance hybrids is correct?

A. The greater the predicted stability of a resonance contributor, the less it contributes to the resonance hybrid.

B. The resonance hybrid is more stable than the predicted stability of any of its resonance contributors because it is the sum of all stable resonance contributors.

C. The fewer nearly equivalent resonance contributors are in the structure, the greater the resonance energy.

D. The resonance hybrid will show all pi bonds and locations of all full charges.

I know the correct answer is B. I am having trouble understanding resonance could you please explain why A,C, and D is wrong and why B is correct.

Explanation / Answer

All resonance contributors do not necessarily contribute equally to the resonance hybrid. The degree to which each resonance contributor contributes depends on its predicted stability. Because resonance contributors are not real, their stabilities cannot be measured. Therefore, the stabilities of resonance contributors have to be predicted based on molecular features that are found in real molecules. The greater the predicted stability of the resonance contributor, the more it contributes to the resonance hybrid; and the more it contributes to the resonance hybrid, the more similar the contributor is to the real molecule.

This will explain why option a is incorrect and b is correct.

A compound with delocalized electrons is more stable than it would be if all of its electrons were localized. The extra stability a compound gains from having delocalized electrons is called delocalization energy or resonance energy. Electron delocalization gives a compound resonance, so saying that a compound is stabilized by electron delocalization is the same as saying that it is stabilized by resonance. Since the resonance energy tells us how much more stable a compound is as a result of having delocalized electrons, it is frequently called resonance stabilization.

Therefore,option c is incorrect.

Practically every time there are pi bonds in a molecule, especially if they form part of a conjugated system,
there is a possibility for having resonance structures, that is, several valid Lewis formulas for the same
compound. What resonance forms show is that there is electron delocalization, and sometimes charge
delocalization. All the examples we have seen so far show that electrons move around and are not static,
that is, they are delocalized. Charge delocalization is a stabilizing force because it spreads energy over a
larger area rather than keeping it confined to a small area. Since electrons are charges, the presence of
delocalized electrons brings extra stability to a system compared to a similar system where electrons are
localized. The stabilizing effect of charge and electron delocalization is known as resonance energy.
Since conjugation brings up electron delocalization, it follows that the more extensive the conjugated
system, the more stable the molecule (i.e. the lower its potential energy). If there are positive or negative
charges, they also spread out as a result of resonance.The corollary is that
the more resonance forms one can write for a given system, the more stable it is.
That is, the greater its resonance energy.

So,option d is incorrect.

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