Why is copper #3d^10 4s^1#??

2 Answers
Jul 4, 2017

For copper? Because the #3d# orbitals are significantly lower in energy for copper (thus making a doubly-occupied #4s# orbital unfavorable enough!), AND because it can fill all the quantum states available in the #n = 3# quantum level, as seen in its electron configuration of:

#[Ar] 3d^10 4s^1#

So, it's only natural for copper to "prefer" filling its #3d# orbitals rather than its #4s#.


Consider the data here (Appendix B.9):

  • #E_(4s,Cu) = -"8.42 eV"#
  • #E_(3d,Cu) = -"13.47 eV"#

The #3d# orbitals of copper are about #"5.05 eV"# of energy lower than the #4s# orbitals of copper (which is about #"487 kJ/mol"#, almost #5# times the strength of a chemical bond!).

You can see the significant increase in favorability when you look at the big picture (graphed from the data here [Appendix B.9]), and see that the #3d# orbitals drop in energy faster than the #4s# orbitals do as the atomic number increases:

So, copper would have a more stable electron configuration by

  • filling its #3d# orbitals to occupy all possible #bb(n = 3)# quantum states.
  • taking advantage of the particularly low #bb(3d)# orbital energies compared to the #4s# orbital.
Jul 4, 2017

Answer:

You mean to ask why is the electronic configuration of #"ATOMIC COPPER"# is........

Explanation:

You mean to ask why is the electronic configuration of #"ATOMIC COPPER"# is........#[Ar]3d^(10)4s^1# rather than #[Ar]3d^(9)4s^2#?

For copper, #Z=29#; and thus in the atom there are 29 electrons to distribute in the neutral atom. Eighteen of these electrons are distributed in a core the same as that of atomic argon, i.e. #1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)#. Because these are inner core, and do not really participate in the chemistry of the proceeding elements, we can represent this configuration by #[Ar]# rather than go thru the pfaff of writing it out.

So to your question, why is the electronic configuration of #"ATOMIC COPPER"# ........#[Ar]3d^(10)4s^1# rather than #[Ar]3d^(9)4s^2#? This is probably a combination of Hund's rule, which gives a special electronic stability to HALF-FILLED SHELLS, i.e. the #4s# orbital is half-filled; and also to the special stability of FILLED ELECTRONIC shells, i.e. the #3d# shell is FULL at #3d^10#.

The electronic stability to HALF-FILLED SHELLS, is also apparent in the atomic configuration of the #Cr# atom, which has a configuration of #[Ar]4s^(1)3d^5#. I would expect a 2nd year inorganic student to know these specific, special configurations,