How can I compare the bond angle between $\ce{O-N-O}$ atoms in $\ce{NO2}$ and $\ce{N2O4}$?

In $\ce{NO2}$ there is lone pair-bond pair repulsion, and in $\ce{N2O4}$ there is bond pair-bond pair repulsion.

  • $\begingroup$ I think that NO2 will be a free radiacal with the radical centred on oxygen. Thus it will be a bonded pair-single electron (lone half pair) repulsion. $\endgroup$ – Nuclear Chemist May 12 '18 at 19:29

The bond angle in NO$_2$ is $\approx 134^\mathrm{o}$ and the NO bond length 120 pm, in $\ce{N2O4}$ the N-N bond is very long 175 pm and the angle $\approx 126^\mathrm{o}$.

Nitrogen dioxide is paramagnetic with the odd electron in the $\sigma$ bonding sp$^2$ orbital in the 'third' lobe as it were with the O atoms occupying the other two. Additionally there are 4 electrons in the $\pi$ system, two bonding and two antibonding. NO$_2$ also has several low lying excited states which greatly complicates matters.

Joining up two NO$_2$ naturally leads to $\ce{N2O4}$ which is planar and has a very long single $\sigma$ N-N bond. Why the molecule is planar is not clear as it might be expected to twist to remove oxygen-oxygen repulsion. Intuitively one would expect an MO to extend over the whole molecule and so prevent twisting, but the NN bond is very long and single.

Hopefully someone has more up to date data, otherwise you will have to search the literature for MO based (i.e. numerical quantum/chemistry) approaches to describe the bonding.

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