What are the oxidation numbers of $F e, N$ and $O$ in the compound $F e (N O_{3})_{3}$ ?

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Solution

Clearly the oxidation number of the iron centre is $I I I +$ , as $3$ nitrate ions are bound to this centre:
$F e (N O_{3})_{3} \to F e^{3 +} + 3 N O_{3}^{-}$ ;
Now to determine the oxidation state of nitrogen, we know that
$3 \times O N_{O} + O N_{N} = - 1$ . Why? Because the sum of the oxidation numbers must equal the charge on the ion. $O N_{O}$ is normally $- I I$ , and it is here, so
$3 \times (- 2) + O N_{N} = - 1$ . Oxidation number of nitrogen is $V +$ .
I would have got the same result had I used the definition of oxidation state, i.e.
$N O_{3}^{-} \to N^{5 +} + 3 \times O^{2 -}$

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