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OpenStudy (anonymous):

Why is Fe3O4 Iron(II) diiron(III) oxide?

OpenStudy (jfraser):

same reason as the earlier question, just slightly more complicated. Oxygen has to be a -2. 4 of them make the total negative charge -8. With 3 irons, there's no way for the charge to be distributed evenly. 8/3 is not a whole #. The easiest way to split up the charge is to have 1 iron be +2, and 2 irons be +3 each 2 + 3 + 3 = 8.

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