Why is it that $\ce{O2NCl}$ has only two resonance structures? More specifically, why is the Lewis drawing where $\ce{N}$ is double bonded to $\ce{Cl}$ not considered a resonance structure along with the other two where $\ce{N}$ is double bonded to an oxygen?


2 Answers 2


$\ce{O2NCl}$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if weighted accordingly.

This last expression in italics is important. Not all resonance structures are equal. People teaching resonance structures usually use expressions like most probable resonance structures. To determine whether a resonance structure is more probable than another one, check the following list:

  1. does it give all main group atoms a valence electron octet?
  2. are formal charges minimised?
  3. are formal charges distributed in accordance with electronegativity (i.e. do electronegative atoms have more formal negative charge than electropositive ones)?

The list is ordered from most important to not so important although traditional teaching sometimes weighs maximised amount of bonds higher and thus arrives at hyperoctet structures.

Now let’s check the structures of $\ce{O2NCl}$ that you are talking about:

resonance strucures

We see that all structures fulfill rule 1. However, rule 2 clearly shows a difference: The two structures on the left contain $+1$ and $-1$ formal charges each while the rightmost structure has $+2$ and $-2$. We don’t need to check rule three, we can see that the rightmost structure is unfavourable and thus less close to the truth. Finally, the formal charges are distributed correctly on all three structures, because oxygen is more electronegative.

Therefore, the left two structures are equally probable while the rightmost one is not so; oftentimes in exams, only the left two will be considered ‘correct’. But as I said, the true picture (or something really close to the truth) can only be determined by quantum chemical calculations.


There's nothing wrong with your suggestion. If $\ce{N}$ was double bonded to $\ce{Cl}$, then both $\ce{N}$ and $\ce{Cl}$ would have positive charges and both $\ce{O}$ atoms would have negative charges. That structure satisfies valence bond theory and octet rules, but it looks a bit strange being doubly zwitterionic. The structure with the $\ce{N=O}$ double bond $\ce{N}$ positive and the other oxygen is negative looks the most plausible.


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