# Comparision of ONO Bond angle in NO2 and N2O4

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.

• 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. – 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.