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There are only two resonance structures of allyl anion with negative charge distributed over positions 1 and 3:

$$\ce{\overset{-}{C}H2-CH=CH2 <-> CH_2=CH-\overset{-}{C}H2}.$$

What's the criteria for electrons forming a double bond? Why isn't there a third resonance structure

$$\ce{\overset{+}{C}H2-\overset{-}{C}H-\overset{-}{C}H2}?$$

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    $\begingroup$ Without even checking the correctness of the formula, it is always the same question.... We can perhaps draw other limiting forms, but their weight would be negligible as those hypothetical structure would be highly energetic (for instance you have placed two adjacent opposite charges. This is unfavourable already in terms of coulombic repulsion, and so on. Look for standard rules, they work in most cases. $\endgroup$
    – Alchimista
    Commented Feb 18, 2021 at 9:19
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    $\begingroup$ Wouldn't the adjacent negatives be stabilized by the adjacent positive and negative charges? $\endgroup$
    – harry
    Commented Feb 18, 2021 at 9:29
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    $\begingroup$ Well, yes in a way they do. Forming a double bond. But even coulombic treatment suffices. The outermost right charge would anyway feel the adjacent more than the outermost left. $\endgroup$
    – Alchimista
    Commented Feb 18, 2021 at 9:36
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    $\begingroup$ Right, got it. It's all a matter of knowing it, and you can't always make your own predictions, right? $\endgroup$
    – harry
    Commented Feb 18, 2021 at 9:38
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    $\begingroup$ Don't put too much trust in ephemeral things. Resonance structures only exist in your head and vanish when you stop thinking about them. $\endgroup$
    – Ivan Neretin
    Commented Feb 18, 2021 at 12:09

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