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The oxidation number for 'O2' is zero(0).

However, when calculating oxidation numbers in compounds containing oxygen, then the oxygen is used as a standard at '-2'.

So as an example using potassium permanganate (KMnO4), what is the oxidation number of manganese (Mn).

In solution this dissolves producing the permanganate ion (MnO4^(-).

Note the charge on the ion is '-1'.

Since there are 4 oxygens present , then the oxidation of the oxygen moiety is 4 X -2 = -8 .

So creating a little sum

Mn + -8(oxygen moiety) = -1 (overall charge on the ion).

Mn + -8 = -1

Add '8' to both sides

Mn = (+)7 is the oxidation number of manganese.

So the formula could be written as KMn(VII)O4 . (Note the Roman numerals for oxidation number).

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lenpollock

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1y ago

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