# Which one is the electron configuration of "Fe"^(+)? [Ar] 3d^5 4s^1, or [Ar]3d^6?

Aug 20, 2017

Neither of them are correct! The correct electron configuration is $\left[A r\right] 3 {d}^{6} 4 {s}^{2}$...

Consider the following diagram...

By the convention given in the diagram, the first electron in an orbital is taken to have spin down, ${m}_{s} = - \frac{1}{2}$. In $\left(b\right)$, note the relative energies of the $3 d$ electrons and $4 s$ electron of $\textcolor{red}{{\text{Fe}}^{+}}$:

overbrace(E_(3d(m_s = -1//2)))^("5 electrons") < overbrace(E_(4s(m_s = -1//2)))^"1 electron" < overbrace(E_(3d(m_s = +1//2)))^"1 electron" < overbrace(E_(4s(m_s = +1//2)))^"1 electron"

This means for ${\text{Fe}}^{+}$, so far, the $4 s$ orbital is such that the first electron in the $4 s$ orbital (with ${m}_{s} = - 1 / 2$) is lower in energy than the sixth $3 d$ electron.

That establishes the electron configuration $\textcolor{red}{\left[A r\right] 3 {d}^{6} 4 {s}^{1}}$ for $\textcolor{red}{{\text{Fe}}^{+}}$.

In diagram $\left(a\right)$, we see $\text{Fe}$ (not ${\text{Fe}}^{+}$), decreasing its $4 s$ orbital energy (the $4 s$ curves shifted down!), so that...

${\overbrace{{E}_{4 s \left({m}_{s} = - 1 / 2\right)}}}^{\text{1 electron" < overbrace(E_(3d(m_s = -1//2)))^("5 electrons") < overbrace(E_(4s(m_s = +1//2)))^"1 electron" < overbrace(E_(3d(m_s = +1//2)))^"1 electron}}$

This then agrees with experiment, that the electron configuration of $\text{Fe}$ is...

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

Now you may wonder,

"why is the $4 s$ ionized instead of the $3 d$ first, even though in $F e$, the sixth $3 d$ electron is highest in energy?"

The radial extent of the $4 s$ is farther out, so its electrons are more accessible for ionization:

The $3 d$ electrons are also stabilized by a greater effective nuclear charge. That rapid decrease in ${Z}_{e f f}$ across the transition metals is shown in the steeper slope of the $3 d$ curves in the above diagrams $\left(a\right)$ and $\left(b\right)$ than for the $4 s$ curves.

So, the $3 d$ is not readily ionized.