# Why 4s lose electron to 3d? [duplicate]

In chromium and copper, the are the configuration fill up like $$\mathrm{3d^5 4s^1}$$ but not $\mathrm{4s}$ first as it has lower energy level. and why only these 2 elements.

## marked as duplicate by Ben Norris, user15489, M.A.R. ಠ_ಠ, Wildcat, jerepierreJul 27 '15 at 15:02

• Copper doesn't have $3d^5$ in its base state, but a completely filled $d$ orbital. – M.A.R. ಠ_ಠ Jul 26 '15 at 16:16
• If you've just only been recently introduced to the electron configurations, then an appropriate answer would be: "the half-filled 3d orbital in Cr and the completely filled 3d orbital in Cu are particularly stable". For a more detailed but also potentially very technical answer, look up exchange energy. – orthocresol Jul 26 '15 at 16:23
• This is a very popular question in the chemistry world, and it has been asked and answered before: chemistry.stackexchange.com/questions/151/… – Ben Norris Jul 27 '15 at 1:06

In any element after Argon [1s22s22p63s23p6], the repulsion force between the elctrons increase at a great extent and also the atom becomes very stable. So the elctrons tend to move in far 4s orbital first and after filling it completely , fill the 3d orbital. About chromium and copper, they are exceptions. Any half filled or fully-filled orbitals are more stable then rest of the configurations, so in Chromium one electron of 4s comes to 3d orbital and a stable 3d5 is obtained, similarly in Copper one electron from 4s comes to 3d orbital and forms stable 3d10.

I hope this answers your question, If you have any further doubts you can ask me :)