# Write the ground-state electron configurations of the following transition metals: "V"^(5+), "Au"^(3+), and "Fe"^(2+)?

Mar 16, 2016

First, let's look at those for the neutral atom, and then work our way to the cation.

$\text{V}$ is atomic number $23$, so its configuration is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{3} 4 {s}^{2}$. In shorthand it is $\left[A r\right] 3 {d}^{3} 4 {s}^{2}$. This is an expected configuration; not an oddball element.

Since the $4 s$ orbital is higher in energy, its electrons will be removed first. Not that it matters here, though, because exactly $5$ electrons are removed:

$\textcolor{b l u e}{{\text{V}}^{5 +} \to 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} \equiv \left[A r\right]}$

GOLD (III)

$\text{Au}$ is atomic number $79$, so its configuration is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6} 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$. In shorthand it is $\left[X e\right] 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$.

This is strange, not only because gold wants to acquire a filled $5 d$ subshell, but more likely also something to do with the high amount of relativistic contraction effects; the $s$ electrons move close to the speed of light and are often farther from the nucleus than closer, shrinking the radius of the $1 s$ orbital by ~22%, and other orbitals by a bit as well.

Let's examine the radial density distribution of the $6 s$, $5 d$, and $4 f$ valence orbitals to see how that turns out:

By a long shot, the $4 f$ is the most penetrating, meaning that most of its electron density is centered near the gold nucleus. Further out are the $5 d$ electrons, and then the $6 s$ electrons.

This suggests that the $6 s$ orbital, which has most electron density farther out from the nucleus, has electron(s) that are less attracted to the nucleus (smaller ${Z}_{e f f}$) than those in either the $5 d$ or $4 f$ orbitals, and thus contains the first electron that is easiest to remove during ionization.

The $5 d$ contains the next two electrons that will be removed during the second and third ionizations.

So, we would get:

$\textcolor{b l u e}{{\text{Au}}^{3 +} \to 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{10} 4 {s}^{2} 4 {p}^{6} 4 {d}^{10} 5 {s}^{2} 5 {p}^{6} 4 {f}^{14} 5 {d}^{8} \equiv \left[X e\right] 4 {f}^{14} 5 {d}^{8}}$

IRON (II)

$\text{Fe}$ is atomic number $26$, so its configuration is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$. In shorthand it is $\left[A r\right] 3 {d}^{6} 4 {s}^{2}$. This is an expected configuration; not an oddball element.

The first and second ionizations would removed the $4 s$ electrons, which are higher in energy than the $3 d$ electrons, so we get:

$\textcolor{b l u e}{{\text{Fe}}^{2 +} \to 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} \equiv \left[A r\right] 3 {d}^{6}}$