How do I find the oxidation numbers of individual elements in an ion or molecule?

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How do I find the oxidation numbers of individual elements in an ion or molecule?

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Answer 1 · 2016-10-24T08:20:24

The sum of the oxidation numbers ALWAYS equal the charge on the ion, OR on the molecule....

Explanation:

And of course, if it is a neutral molecule, the sum of the constituent elemental oxidation numbers is ZERO.

Nitrogen oxide, $N O$ $NO$ , is a neutral molecule, and thus the sum of the oxidation numbers must be zero. The oxidation number of $O$ $O$ is normally $- I I$ $-II$ , and it is here. The oxidation number of nitrogen must of course be $+ I I$ $+II$ , because the sum of the oxidation numbers must equal the neutral charge on the molecule. For $N O_{2}$ $NO_2$ , the oxidation number of $N$ $N$ must be $+ I V$ $+IV$ . And of course, for $N_{2} O_{5}$ $N_2O_5$ the oxidation number of nitrogen is?

For nitrous oxide, $N_{2} O$ $N_2O$ , the oxidation number of $O$ $O$ is still $- I I$ $-II$ . And for this neutral molecule, the oxidation number of $N$ $N$ is $+ I$ $+I$ , i.e. $1 + 1 - 2 = 0$ $1+1-2=0$ as required.

Given all this, can you tell me the oxidation number of the metal in $C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ ?

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How do I find the oxidation numbers of individual elements in an ion or molecule?

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