# O-N-O bond angles in oxides of nitrogen

If we look at the bond angles of O-N-O in figure (c), (D) and (F), there not too much change in bond angle. Even though they are notable.

So my question is,

Why do we observe very small change in angles?

• This seems to be two entirely separate questions so you should ask them as so. – bon Sep 4 '15 at 8:05
• 1. A simplistic explanation would be to say, that in all relevant oxides nitrogen exists in $sp^2$ hybridization state. The reason for it is that it have to form n+1 bonds with n atoms, forcing it to form a double bond, that requires it to adopt said hybridization. || 2. When one needs to quickly estimate geometry of a molecule, quantum chemistry calculation may help. For precise experimental measurements diffraction methods are suited, say X-ray diffraction for solid state and gas electron and gas neutron diffraction for gas phase. – permeakra Sep 4 '15 at 9:09