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How do you determine the oxidation number of nitrogen in the nitrate ion?

1 Answer

Jun 29, 2016

The sum of the oxidation numbers equals the charge on the ion.

Explanation:

For $N O_{3}^{-}$ $NO_3^-$ , $N_{Oxidation Number} + 3 \times O_{Oxidation number} = - 1$ $N_"Oxidation Number"+3xxO_"Oxidation number"=-1$ .

Now $O$ $O$ normally takes an oxidation number of $- I I$ $-II$ and it does so here.

Thus $N_{Oxidation Number} + (3 \times - 2) = - 1$ $N_"Oxidation Number"+(3xx-2)=-1$ .

Clearly, $N_{Oxidation Number} = + V$ $N_"Oxidation Number"=+V$

What is the oxidation state/number of $N$ $N$ in $N O$ $NO$ , $N O_{2}$ $NO_2$ , and $N_{2} O_{4}$ $N_2O_4$ ?

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