# What are the oxidation numbers of Fe, N and O in the compound Fe(NO3)3?

Nov 1, 2015

Oxidation number is the charge left on the central atom, when all the bonding electrons are removed with the charge devolving to the most electronegative atom.

#### Explanation:

Clearly the oxidation number of the iron centre is $I I I +$, as 3 nitrate ions are bound to this centre:

$F e {\left(N {O}_{3}\right)}_{3} \rightarrow F {e}^{3 +} + 3 N {O}_{3}^{-}$;

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 $3 \times \left(- 2\right) + O {N}_{N} = - 1$. Oxidation number of nitrogen is $V +$.

I would have got the same result had I used the definition of oxidation state, i.e. $N {O}_{3}^{-} \rightarrow {N}^{5 +} + 3 \times {O}^{2 -}$.

What are the oxidation states of the metal in $F e C {l}_{4}^{-}$, $M n {O}_{4}^{-}$, and $C {r}_{2} {O}_{7}^{2 -}$?