-1
$\begingroup$

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?

Improve this question
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy