What is the meaning of “charge separation” in resonance?

Question

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?

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.