# the oxidation number of N in this equation?

the equation :

$\ce{NH4Cl + AgNO3 -> NH4NO3 + AgCl}$

I know that oxidation number of $\ce{N}$ in $\ce{NH4Cl}$ is -3 and in $\ce{AgNO3}$ is +5

but what I am confused about is the oxidation number of $\ce{N}$ in the product $\ce{NH4NO3}$

Should I consider $\ce{N}$ as $\ce{N2}$ in this compound

Or Should I get its oxidation number in $\ce{NH4}$ and in $\ce{NO3}$ and then get the sum of them ?

Or .. I really don't know :(

there are 2 different nitrogen atoms in $\ce{NH4NO3}$, one having oxidation number -3 and one with +5. They are different, having separate oxidation numbers independently of each other. It is perfectly OK to have several atoms of same element having different oxidation state in same compounds, for example in thiosulphate $\ce{SO3S^{2-}}$ one (central) sulfur atom has oxidation number +4 and another has 0, and $\ce{Fe3O4}$ is actually $\ce{Fe^{+2}Fe^{+3}2O4}$ and many more.
A reasonable trick to get the oxidation state would be to write down the ionic components of $\ce{NH4NO3}$ . They are
* $\ce{NH4+}$
* $\ce{NO3-}$