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Here is an image showing the resonance structures of $\ce{N2O3}$ molecule,enter image description here

As the bonds numbered (2) and (3) are in resonance, they will have same bond lengths but each with bond order less than two. As the bond numbered (1) doesn't involve in resonance, it will have a bond order of two. Hence, the order of $\ce{N-O}$ bond lengths is (1)>(2)=(3).

But according to Wikipedia, all the three $\ce{N-O}$ bond lengths are of different lengths! Here is an image form Wikipedia showing the bond lengths and the bond angles in $\ce{N2O3}$,enter image description here So my question is why are the resonating bonds are of different lengths?

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    $\begingroup$ Bonds don't "resonate", and in dissymmetric molecule won't be equal $\endgroup$
    – Mithoron
    Commented Aug 27, 2022 at 15:44
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    $\begingroup$ Consider the relative energies of the two resonance structures $\endgroup$
    – Andrew
    Commented Aug 27, 2022 at 16:48
  • $\begingroup$ Could you explain what you mean by and why you think "As the bonds numbered (2) and (3) are in resonance, they will have same bond lengths"? $\endgroup$
    – Ian Bush
    Commented Aug 27, 2022 at 19:38
  • $\begingroup$ @Mithoron if the bonds aren't in resonance, then there could be a significant difference in the bond lengths of bonds (2) and (3). $\endgroup$
    – Infinite
    Commented Aug 28, 2022 at 0:49
  • $\begingroup$ @Andrew I understand what you are saying. So according to bond lengths data, second resonating structure is more contributing than the first one. Can you explain, why? $\endgroup$
    – Infinite
    Commented Aug 28, 2022 at 0:51

1 Answer 1

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The two oxygen atoms on the $\ce{NO2}$ nitrogen are not equivalent. The molecule is planar*, and one $\ce{NO2}$ oxygen (call it the "cis" oxygen) is closer to the third oxygen atom than the other $\ce{NO2}$ oxygen ("trans").

We then note that the oxygen atoms have nonbonding pairs around them, which tend to repel each other. So the cis oxygen will tend to form a stronger pi bond, pushing more nonbonding electron density onto the relatively remote trans oxygen.

*This planarity is due to hyperconjugation. See Halpern et al[1].

Reference

  1. Halpern AM, Glendening ED (2007 Apr 21). "Ab initio study of the torsional potential energy surfaces of N2O3 and N2O4: origin of the torsional barriers." J Chem Phys.; 126(15):154305. https://doi.org/10.1063/1.2713756. PMID: 17461624.
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    $\begingroup$ What is the reason the NO group orientation is locked? Does the N-N bond have partially pi-character too? As it does not seem that there is a sterical obstacle for the bond to rotate. $\endgroup$
    – Poutnik
    Commented Aug 29, 2022 at 6:59
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    $\begingroup$ Halpern et al: "N(2)O(4) and N(2)O(3) exhibit strong hyperconjugative interactions of in-plane O lone pairs with the central N-N sigma* antibond. Hyperconjugative stabilization is somewhat stronger at the planar geometries because 1,4 interactions of lone pairs on cis O atoms promote delocalization of electrons into the N-N antibond. " $\endgroup$
    – Oscar Lanzi
    Commented Aug 29, 2022 at 8:55

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