# Cis and Trans Isomerism in Resonance-Stabilized Compounds

Would it be valid to describe $N_2O_5$ isomers as cis or trans? As a matter of fact, does this molecule even have isomers, given that all its outer N-O bonds are resonance stabilized and therefore equal in length and strength?

Also for non-resonance stabilized compounds, would it be okay to describe say the N to O double bonds as cis or trans with respect to each other? I've just never heard of cis/trans being used this way.

• $\ce{N2O5}$ doesn't have cis-trans isomers. Can you add a picture to clarify what you mean by "N to O double bonds as cis or trans with respect to each other"? – ron May 15 '14 at 18:36
• They were trying to say something like this: shodor.org/Master/environmental/general/decomposition/n2o5.gif has its double bonds trans to each other. – Dissenter May 15 '14 at 19:25
• @Dissenter The "outer" $\ce{N=O}$ double bonds are delocalized within their respective $\ce{NO2}$ groups (like it is shown here). The picture you link to in your comment only shows one of the possible mesomeric structures and not a different isomere. – Philipp May 15 '14 at 19:45
• Right, that's what I thought. In a resonance stabilized compound - as I wrote in my original post - all the bonds are equal. However, my question is, if there were no resonance, could we describe molecules in this fashion? I.e. N=O is cis/trans to the other N=O? – Dissenter May 15 '14 at 19:51

Thanks for the picture, now I understand your question. All of the atoms in $\ce{N2O5}$ don't need to lie in a plane as is shown in the picture in your link. At room temperature there would be rapid rotation about the 2 bonds connecting the central oxygen to the two nitrogen atoms, so there is no "fixed" geometry from which you could use the cis and trans references. For example, imagine the conformation with one $\ce{NO2}$ group in the plane of the screen and the other perpendicular to the screen, even if each $\ce{NO2}$ group had one $\ce{N=O}$ double bond and one $\ce{N-O}$ single bond, the terms cis and trans would still have no meaning in this conformation. Further, as you noticed, all 4 bonds between nitrogen and the exterior oxygens are equivalent due to resonance, so again terms like cis and trans would have no meaning. Finally, I can't imagine a situation where an $\ce{N=O}$ double bond (e.g. where the oxygen is unsubstituted) could have cis-trans isomers because the 2 lone pairs on the oxygen are equivalent.