# Question #5f272

Nitrogen dioxide, or $N {O}_{2}$, is an odd-electron molecule, i.e. the number of valence electrons the molecule has is not an even number.
Now, VSEPR Theory can be extended to accomodate odd-electron molecules. In this case, the nitrogen dioxide molecule is considered to have a $A {X}_{2} {E}_{0.5}$ geometry, which is right between $A {X}_{2} {E}_{0}$ - linear, and $A {X}_{2} {E}_{1}$ - bent.
The molecule's geometry will still be closer to bent, but the bond angle will be larger than the ideal ${120}^{\circ}$ such a geometry predicts, mostly because the oxygen atoms experience less repulsion from the single electron present on the nitrogen atom.
The actual bond angle for nitrogen dioxide is ${134}^{\circ}$. By comparison, the angle found in the nitrite ion, or $N {O}_{2}^{-}$, is ${115.4}^{\circ}$, because now nitrogen has attached a full lone pair of electrons that experience greater repulsion with the oxygen atoms.