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

Jul 4, 2017

For copper? Because the $3 d$ orbitals are significantly lower in energy for copper (thus making a doubly-occupied $4 s$ 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:

$\left[A r\right] 3 {d}^{10} 4 {s}^{1}$

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

Consider the data here (Appendix B.9):

• ${E}_{4 s , C u} = - \text{8.42 eV}$
• ${E}_{3 d , C u} = - \text{13.47 eV}$

The $3 d$ orbitals of copper are about $\text{5.05 eV}$ of energy lower than the $4 s$ orbitals of copper (which is about $\text{487 kJ/mol}$, 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 $3 d$ orbitals drop in energy faster than the $4 s$ orbitals do as the atomic number increases: So, copper would have a more stable electron configuration by

• filling its $3 d$ orbitals to occupy all possible $\boldsymbol{n = 3}$ quantum states.
• taking advantage of the particularly low $\boldsymbol{3 d}$ orbital energies compared to the $4 s$ orbital.
Jul 4, 2017

You mean to ask why is the electronic configuration of $\text{ATOMIC COPPER}$ is........

#### Explanation:

You mean to ask why is the electronic configuration of $\text{ATOMIC COPPER}$ is........$\left[A r\right] 3 {d}^{10} 4 {s}^{1}$ rather than $\left[A r\right] 3 {d}^{9} 4 {s}^{2}$?

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. $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$. Because these are inner core, and do not really participate in the chemistry of the proceeding elements, we can represent this configuration by $\left[A r\right]$ rather than go thru the pfaff of writing it out.

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

The electronic stability to HALF-FILLED SHELLS, is also apparent in the atomic configuration of the $C r$ atom, which has a configuration of $\left[A r\right] 4 {s}^{1} 3 {d}^{5}$. I would expect a 2nd year inorganic student to know these specific, special configurations,