# Why does iron lose exactly 3 electrons in oxidation?

I'm not a chemist, but I have this question. Why does iron (Fe atom) lose three electrons when exposed to oxygen for the first time? Why doesn't it lose less than or more than three electrons?

The electronic configuration of iron is $\ce{[Ar] 3d^6 4s^2 }$.
It can lose two electrons to form the ferrous $(\ce{Fe^2+})$ ion with E.C. $\ce{[Ar] 3d^6}$. This is one stable ion of iron. The second (and more stable) ferric $\ce{(Fe^3+)}$ ion is formed by losing another electron to form $\ce{[Ar] 3d^5}$. This has half-filled d subshell, and hence imparts extra stability.
Though your question is not clear. Iron can exist perfectly in both the oxidation states. Both the oxides $\ce{FeO}$ and $\ce{Fe2O3}$ are well known, although - as expected - the first one is rarer.