# Will electrons pair up in an orbital only when all orbitals in different sub-levels have one electron?

Aug 26, 2017

Usually, yes, but not necessarily.

For a given $n$, different sublevels have differing $l$, where $l = 0 , 1 , 2 , . . . , n - 1$.

Hund's rule states that in order to maximize spin multiplicity, an atom usually fills one orbital at a time, and pairs up after every orbital of the same energy (or neighborhood of that energy) is half-filled.

Well, a counterexample is that thorium ($\text{Th}$) has a $7 s$ orbital that is pretty much the same energy as its $6 d$ AND $5 f$ orbitals (empty circle vs. empty triangle vs. filled diamond):

And yet, its electron configuration is:

$\textcolor{b l u e}{\left[R n\right] 7 {s}^{2} 6 {d}^{2} 5 {f}^{0}}$

rather than... (as lower $n$ is usually lower in energy)

$\left[R n\right] 7 {s}^{2} 6 {d}^{0} 5 {f}^{2}$
(Aufbau)

or... (as it decreases electron pairing interactions)

$\left[R n\right] 7 {s}^{1} 6 {d}^{3} 5 {f}^{0}$
(Hund)

or even more seemingly reasonable... (as lower $n$ is usually lower in energy)

$\left[R n\right] 7 {s}^{1} 6 {d}^{0} 5 {f}^{3}$
(Aufbau + Hund)

And the reasoning behind that is the $5 f$ orbitals are particularly compact compared to the $6 d$ that $\text{Th}$ would rather have $\left[R n\right] 7 {s}^{1} 6 {d}^{2}$ rather than $\left[R n\right] 7 {s}^{1} 5 {f}^{2}$, even though (or rather, BECAUSE) they are all pretty much the same energy.

The $6 d$ orbitals are also apparently not so diffuse that a $\left[R n\right] 7 {s}^{1} 6 {d}^{3}$ ends up being viable, so that the $7 s$ being doubly-occupied leads to the lowest-energy ground-state thorium atom.