# What are the quantum numbers for last electron of cadmium?

May 17, 2017

$\left(n , l , {m}_{l} , {m}_{s}\right) = \left(5 , 0 , 0 , \pm \frac{1}{2}\right)$ We generally call the main transition metal block (in pink) the $d$-block, and the first two columns of the periodic table the $s$-block.

Since cadmium is in the last column of the second-row transition metals (period $5$), its valence electrons consist of the $n s$ and $\left(n - 1\right) d$ electrons.

In shorthand notation then, we write the electron configuration as:

$\left[K r\right] 4 {d}^{10} 5 {s}^{2}$

(note that I specifically said $n s$ and $\left(n - 1\right) d$. That is because $n = 5$ for the fifth period, so we have $5 s$ and $4 d$ valence orbitals, and not $5 d$.)

Since we have no way of proving how exactly the electrons fill the orbitals (we assume!), the "last" electron in cadmium is considered NOT the one that is listed last, but the one that is removed first.

That would be one of the $5 s$ electrons, since it is farther away from the nucleus on average. How do I know that? Well, its orbital potential energy is less negative by $\approx \text{8.89 eV}$ (Appendix B.9).

We do not know which electron will be removed, so we can only say:

$\textcolor{b l u e}{\left(n , l , {m}_{l} , {m}_{s}\right) = \left(5 , 0 , 0 , \pm \frac{1}{2}\right)}$