# Question #2e939

Apr 30, 2017

Copper has the electron configuration $\left[A r\right] 4 {s}^{1} 3 {d}^{10}$. Here is the radial density distribution for the $3 d$ vs $4 s$ orbitals, a graph of $4 \pi {r}^{2} {R}_{n l}^{2} \left(\vec{r}\right)$ vs. $\vec{r}$ ($y$ vs. $x$): (where ${R}_{n l} \left(\vec{r}\right)$ is the radial component of the wave function $\psi \left(\vec{r} , \theta , \phi\right) = {R}_{n l} \left(\vec{r}\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$.)

It tends to be that transition metals, particularly the first-period transition metals ($\text{Sc","Ti", . . . , "Zn}$) lose their $4 s$ electrons first because they are farther away from the nucleus on average.

We can tell because the most probable location of the $4 s$ electron is farther to the right-hand side of the graph than the most probable location of the $3 d$ electron.

Therefore, the $4 s$ electron comes off first, and then the second electron removed is the tenth $3 d$. This gives:

$\left[A r\right] 4 {s}^{0} 3 {d}^{9}$

The two electrons removed were therefore:

• $4 s$ $\to$ $\left(n , l , {m}_{l} , {m}_{s}\right) = \left(4 , 0 , 0 , + \frac{1}{2}\right)$
• $3 d$ $\to$ $\left(n , l , {m}_{l} , {m}_{s}\right) = \left(3 , 2 , \left\{- 2 , - 1 , 0 , + 1 , + 2\right\} , \pm \frac{1}{2}\right)$

where by our convention, we chose ${m}_{s} = + \frac{1}{2}$ arbitrarily for the one $4 s$ electron.

For the $3 d$ electron, we probably could not know whether it was ${m}_{s} = + \frac{1}{2}$ or $- \frac{1}{2}$.

(Although by Hund's rule, we maintain parallel spins to maximize the spin multiplicity [i.e. keep all the electrons for a given $l$ the same spin, usually spin-up by convention], we could just as well have removed the "wrong" [spin-up] electron instead of the spin-down electron and ended up with the remaining electron spin-flipped to be spin-up.)

We do not, however, care about which specific $3 d$ orbital it came from. It could have been anywhere in the set ${m}_{l} = \left\{- 2 , 1 , 0 , + 1 , + 2\right\}$, and we cannot be truly sure which orbital it "came from".