I've read that the resonance structures in which there is lesser 'charge separation' contribute more to the resonance hybrid than the ones with greater charge separation.

But I am unable to understand why it is so.

I mean isn't it correct that the presence of two charges on adjacent atoms leads to "less stability" due to bond stretching or/and compression?

But apparently, the actual observation is contrary to this argument.

I guess I'm actually confused about the meaning of 'charge separation'? Does it refer to the distance between the location of the charges or the difference between their amounts?

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    $\begingroup$ Resonance structures have nothing to do with any observation, because they exist only in our mind. $\endgroup$ Apr 12, 2016 at 12:38
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    $\begingroup$ @IvanNeretin I understand that, but the contribution of the structures can be observed right? What I am asking is why the contribution of structures with less charge seperation more $\endgroup$
    – ShankRam
    Apr 12, 2016 at 12:47
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    $\begingroup$ You might want to clarify your question. I think there are two concepts being mixed up here. If I interpret properly: In a simple argument for a neutral molecule, such as hexatriene, the Couloumbic interaction between the charges would not favor charge separation. $\endgroup$
    – brose
    Apr 12, 2016 at 14:11
  • $\begingroup$ Simple classic electrostatic should help. I would say that if there must be formal charge separation, then the less unstable limiting form is that whit minimal distance between them, ie minimal "formal electrical dipole". So you have to look for what minimise separation, both for a formal physical distance as well as the overall amount of separated formal charges. $\endgroup$
    – Alchimista
    Dec 4, 2021 at 20:24

1 Answer 1


There are two ways to think of charge separation and both make resonance structures less stable:

  1. Moving electrons from a bond to a single atom to generate a cation/anion pair. For example:

    $$\ce{R2C=CH-CH=CR2 <-> R2C=CH-\overset{+}{C}H-\overset{-}{C}R2}$$

  2. Moving double bonds to separate opposite charges. For example:

    $$\ce{R2C=CH-\overset{+}{C}H-\overset{-}{C}R2 <-> R2\overset{+}{C}-CH=CH-\overset{-}{C}R2}$$

Usually, however, the first way is meant. The general idea behind required a minimal charge separation is that in general (as long as we do not form super-octet structures) double bonds are lower in energy than separate charges on adjacent atoms as you correctly mention. Therefore, if we minimise the number of formal charges the idea is that we are concurrently maximising the number of double bonds, thereby generating a better resonance structure which contributes more to the actual composition.

On the other hand, even in the second case it is better to keep the charges together (in the absence of other effects such as strongly electronegative or electropositive atoms). This is because the charges attract each other and they do this better if they are close.

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    $\begingroup$ I feel it's kinda missing point. Mesomeric structures are only lame representation of real bonding. If there's no separation of charge (or ionic character) in reality, then what would be the point of writing such structure? $\endgroup$
    – Mithoron
    Oct 14, 2017 at 23:39
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    $\begingroup$ @Mithoron I have no argument against that. With the exception that I think that is how it is taught in school (wrongly) $\endgroup$
    – Jan
    Oct 15, 2017 at 14:23
  • $\begingroup$ Exactly, they tell us Resonant structures contribute to the Resonance Hybrid based on their energy levels $\endgroup$ Jan 2, 2021 at 17:17

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