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Some reactions between nitrogen oxides and ozone can be the following:

$$NO\cdot(g) + O_3(g) \rightarrow NO_2\cdot(g)+O_2(g)$$

$$NO_2\cdot(g) + O\cdot(g) \rightarrow NO\cdot(g)+O_2(g)$$

Where the $NO$ when transformed to $NO_2$ stays a radical. Why is that the case? Wouldn't the unbounded electron left be used to form a double bond between nitrogen and oxygen?

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Compounds of nitrogen and oxygen with the general formula $\ce{N}_x \ce{O}_y$ will always be radicals when $x$ is odd. Nitrogen atoms have five valence electrons and oxygen atoms have six valence electrons. $\ce{NO}$ will have 11 valence electrons. One will have to be unpaired. $\ce{NO2}$ has 17 valence electrons. One must be unpaired. However $x$ can be even like in $\ce{N2O}$, which has 20 valence electrons.

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