# If filling the orbitals according to the Aufbau principle, which of the following elements will have the wrong configuration?

## $\text{Fe}$, $\text{Zn}$, $\text{Cd}$, $\text{Hg}$

Jul 11, 2017

Well, first of all, that filling order has been made too specific, so this question is flawed in that $\left(i\right)$ $4 s , 3 d$ is not the ubiquitous, end-all-be-all natural filling order, and $\left(i i\right)$ even with the actual, natural filling order, there are other factors that makes it unfavorable to keep filling the $3 d$ orbitals.

Under a certain ASSUMED filling order that is blindly followed, $\boldsymbol{\text{Fe}}$ would be the odd one out here, since it uses $3 d$ orbitals AND not all orbitals are filled, allowing any too-strict interpretation of the Aufbau principle to manifest in the wrong configuration.

ORBITAL FILLING ORDERS

In general, the Aufbau principle actually says that the orbitals are USUALLY filled from $\underline{\text{lowest to highest energy}}$, from having zero electrons (not at all related to the element to the immediate left).

Some exceptions are due to relative orbital size or shape, relative closeness of certain orbital energies (e.g. $\left(n - 2\right) f$ vs. $\left(n - 1\right) d$, $n s$ vs. $\left(n - 1\right) d$, . . . ), and so on.

If we, for the moment, accept this particular filling order suggested by the question, then we would ASSUME a filling order of $3 d$ and then $4 s$ (which for the transition metals is, in principle, correct!!).

However, when one blindly fills the orbitals using this so-called "rule" without considering the orbital energy variations (i.e. by assuming no exceptions ever exist ever), and suggests filling the $3 d$ orbitals completely, the iron electron configuration would be incorrect:

$\implies {\overbrace{\textcolor{red}{\left[A r\right] 3 {d}^{8}}}}^{\text{incorrect}}$

rather than

$\implies {\overbrace{\textcolor{b l u e}{\left[A r\right] 3 {d}^{6} 4 {s}^{2}}}}^{\text{correct}}$

and this has been proven by magnetism studies wherein iron has a large magnetic moment (i.e. has many unpaired electrons). This can be found in many textbooks, such as Chemistry: A Molecular Approach (Tro).

WHY NOT THE FIRST CONFIGURATION?

The observed electron configuration (the second one) is correct, and by observing the orbital potential energies, one can make an hypothesis as to why.

As we go across the first-row transition metals, the $3 d$ orbitals decrease in energy faster (in response to increasing ${Z}_{e f f}$) than the $4 s$ orbitals do. As that occurs, it becomes more and more favorable to place valence electrons in the $3 d$ orbitals.

However, rather than continuing to add the last two electrons in the $3 d$, they are added to the $4 s$. This is likely because...

1. The $3 d$ orbitals are more compact and have no radial nodes, so for a while (for $6$ post-argon-core electrons), it is favored to fill them.
2. The $3 d$ orbitals aren't that far in energy from the $4 s$ orbitals ($\text{3.75 eV}$ difference), so eventually, together with the compactness of the $3 d$, it becomes favorable to fill the $4 s$ orbitals.

(For copper on the other hand, that's when filling the $4 s$ is too unfavorable, and we get $\left[A r\right] 3 {d}^{10} 4 {s}^{1}$. Zinc just says "fine, the $3 d$ are filled anyway.")

For perspective, the cumulative, two-electron ionization energy out of the $4 s$ orbital is about $\text{16.2 eV}$, so the $\text{3.75 eV}$ difference is rather small, making the radial nodes in the $4 s$ orbital a significant factor to the true filling order.

Why the filling of the $3 d$ orbitals stops at the 6th $3 d$ electron (rather than the 7th) and shifts to the $4 s$ is just a matter of experiment.