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?
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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.
answered May 12 '18 at 9:22
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$\begingroup$ Thank you. But why would it become more stable if it loses three electrons instead of two ? $\endgroup$ – user50321 May 12 '18 at 9:25
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2$\begingroup$ @hellothere Half filled subshells provide more stability to an atom/ion as compared to non-half-filled subshells. $\endgroup$ – Gaurang Tandon May 12 '18 at 9:26
