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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?

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  • $\begingroup$ First off, is this a redox reaction? $\endgroup$ – Oscar Lanzi Sep 12 at 12:30
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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 fine 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 $\ce{\overset{(II)}{Fe}\overset{(III)}{Fe2}O4}$ and many more.

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A reasonable trick to get the oxidation state of both the $\ce{N}$ atoms would be to write down the ionic components of $\ce{NH4NO3}$. They are-

  • $\ce{NH4+}$
  • $\ce{NO3-}$

Now these ions can be considered separately to get the answer which is $-3$ and $+5$ respectively.

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