0
$\begingroup$

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?

|improve this question
$\endgroup$
  • 1
    $\begingroup$ Resonance structures have nothing to do with any observation, because they exist only in our mind. $\endgroup$ – Ivan Neretin Apr 12 '16 at 12:38
  • $\begingroup$ @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 '16 at 12:47
  • 1
    $\begingroup$ 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 '16 at 14:11
1
$\begingroup$

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.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ 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 '17 at 23:39
  • $\begingroup$ @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 '17 at 14:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.