I have googled and found that the structure of $\ce{N4O}$ is linear with a positive and negative charge on 2 nitrogen atoms. But, the following structure seems more stable because it fulfils octet for all elements, the structure has no charge and it is resonance stabilised:

Why is this not the actual structure of $\ce{N4O}$? Are there any destabilizing effects that outweigh the stabilizing effects in this structure? The structure doesn't seem to have much angle strain. Where am I wrong?

  • 1
    $\begingroup$ The real structure, too, has a complete octet for all atoms and is resonance stabilized. As for having charge, that's not a big deal. $\endgroup$ Dec 22, 2015 at 18:42
  • $\begingroup$ en.wikipedia.org/wiki/Nitrosylazide, looks sth good about oxatetrazene is harder to find, but they both decompose easily. $\endgroup$
    – Mithoron
    Dec 22, 2015 at 22:19
  • 4
    $\begingroup$ This is the kind of molecule you’ve drawn, where I’m scared to touch the monitor because it might explode … $\endgroup$
    – Jan
    Dec 23, 2015 at 1:32
  • $\begingroup$ @Jan, could you please explain why it is so unstable? $\endgroup$
    – ShankRam
    Dec 23, 2015 at 1:33
  • 2
    $\begingroup$ You're wrong in thinking that everything can be explained using few rules of thumb. $\endgroup$
    – Mithoron
    Dec 24, 2015 at 0:52

1 Answer 1


From The Nitrosyl Azide Potential Energy Hypersurface: A High-Energy-Density Boom or Bust? J. Am. Chem. Soc. 1996, 118, 4860-4870 :

[Abstract] ...While the ring isomer is predicted to be the most stable structure on the hypersurface, the barrier to dissociation is most likely between 1 and 2 kcal mol-1 (including zero-point vibrational energy [ZPVE], the existence of any barrier becomes questionable) making isolation theoretically possible but experimentally difficult. This small barrier also detracts from the attractiveness of the N4O ring structure as a high energy-density material. The trans-chain isomer, however, lies in an energy valley with higher sides, consistent with its previous experimental observation.

[full text]...In an ideal five-membered ring with 6 $\pi$ electrons, the $\pi$ electrons would be distributed evenly among all the bonds. In the present case, the highly electronegative O atom prefers to keep electrons around itself, leading to a partial negative charge on the oxygen. Energetically, there is a certain degree of stability associated with the ring isomer, although not on the order of common aromatic systems. The ring isomer is predicted to be at most (DZP CISD) 20.9 kcal mol-1 more stable than the trans-chain isomer; however, this value decreases to 13.2 kcal mol-1 with TZ2P CCSD. The ring isomer TS to dissociation into N2 and N2O is shown in Figure 5. ... Energetically, the barrier to dissociation is at most 15.3 kcal mol-1 (DZP CISD) and drops lower with improvements in both basis set and correlation scheme. In going from a DZP to a TZ2P basis set, for example, this barrier drops by 5.2 kcal mol-1 for CISD and 4.3 kcal mol-1 for CCSD. Assuming a similar trend in moving from DZP CCSD(T) to TZ2P CCSD(T), the ring dissociation barrier is expected to drop to 1-2 kcal mol-1 with the addition of f-type functions possibly making it even lower. A barrier of this size lies below the ZPVE, throwing doubt on the existence of the N4O ring isomer.

So thermodynamically, yes, the ring form is the lowest energy isomer. However, all the isomers are unstable to decomposition and the linear isomer is in a deeper potential energy well than the ring isomer. Therefore, the linear isomer is easier to observe experimentally for kinetic reasons.

There may be additional information in Theoretical study on structures and stabilities of N4X (X = O, S, Se, Te) series International Journal of Quantum Chemistry Volume 109, pages 226–235.


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