# What is the "oxidation state" of an element in a compound?

Of course, oxidation state is a formalism. Consider the water molecules, $O {H}_{2}$. We break an $O - H$ bond (i.e. 1 bond, so 2 electrons), the charge goes to the most electronegative atom; we get ${O}^{2 -}$ $+$ $2 \times {H}^{+}$. So the oxidation state of $H$ in water is $+ I$, and the oxidation state of $O$ is $- I I$ (because we have an ${O}^{2 -}$ species). Now break the $O - O$ bond in hydrogen peroxide, $H - O - O - H$; the 2 electrons are shared between each oxygen atom atom to give 2 radical species, i.e. $2 \times H - O \cdot$. This species is neutral because of course oxygen atoms have equal electronegativity, and when we break the remaining $H - O$ we get ${H}^{+}$, and ${O}^{-}$; remember this is a formalism, but this means that oxygen atom has a formal oxidation state of $- I$. This is the same when we assign oxidation states to $C$ in a chain, ${H}_{3} C - C {H}_{3}$, the $C$ atoms have a $- I I I$ oxidation state. Are we clear?
Now tell me the oxidation state of oxygen in $O {F}_{2}$? This is a real molecule. Which is the most electronegative atom that will get the charge from the $O - F$ bonds? What about the oxidation state of oxygen in $O = O \left(g\right)$?