Question

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
$bottom enclose Fe subscript 3 end enclose straight O subscript 4$

Solution

bold Fe subscript bold 3 bold O subscript bold 4 bold By bold space bold conventional bold space bold method bold comma bold space bold space bold space bold space bold space bold space bold space bold space bold x bold space bold minus bold 2 bold space bold space bold space bold space bold space bold Fe subscript bold 3 bold space bold O subscript bold 4

bold 3 bold x bold plus bold 4 bold left parenthesis bold minus bold 2 bold right parenthesis bold space bold equals bold 0 bold space bold space bold space bold space bold or bold space bold space bold space bold x bold space bold equals bold space bold 8 over bold 3 bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold plus bold 2 bold space bold minus bold 2 bold space bold plus bold 3 bold space bold minus bold space bold 2 bold By bold space bold stoichiometry bold space bold Fe subscript bold 3 bold O subscript bold 4 bold space bold equals bold space bold Fe bold space bold O bold. bold space bold Fe subscript bold 2 bold O subscript bold 3 bold therefore bold space bold space bold space bold Fe bold space bold has bold space bold oxidation bold space bold number bold space bold of bold space bold plus bold 2 bold space bold and bold space bold plus bold 3 bold.

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