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  4. Oxidation number of iron in Fe0.94 O is

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Q. Oxidation number of iron in ${Fe_{0.94} O}$ is

Redox Reactions

Solution:

O.N. of Fe in $Fe_{0.94} \, O$ is
$x \times 0.94 + 2 x (- 1) = 0$ or
$94 x = 200 $ or $x = 200/94$.