# Why is the electron configuration of chromium?

Dec 14, 2015

It is energetically favored.

#### Explanation:

Looking at the periodic table, you would expect Chromium to have this configuration:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{2} 4 {s}^{2} 3 {d}^{4}$

Or, condensed using noble gases:

$\left[\text{Ar}\right] 3 {d}^{4} 4 {s}^{2}$

However, the actual configuration of Chromium is:

$\left[\text{Ar}\right] 3 {d}^{5} 4 {s}^{1}$

So why is this? It has to do with main the arrangement of sublevels in the electron cloud. Even though $4 s$ belongs to a higher main energy level than $3 d$, the energy levels are similar. This actually allows electrons to move between the sublevels if it is energetically favorable.

This happens in Chromium, as one $4 s$ electron moves to the $3 d$ sublevel. Why? There are two main reasons:

1. The $3 d$ orbital is slightly lower in energy, and minimizing repulsions in the $4 s$ orbital by moving one of the $4 s$ electrons to a close-lying $3 d$ orbital minimizes the ground-state energy of chromium.
2. Hund's Rule: It is energetically favorable to maximize the spin state in a sublevel. Since two opposite spins result in a total spin of $0$, maximizing this tends to require as many electrons in of same spin in different orbitals as possible. So, in this case, an electron moves to $3 d$ and is unpaired, therefore maximizing the spin state.

You will see a similar situation where an $s$ electron moves to a $d$ sublevel with Molybdenum:

$\left[\text{Kr}\right] 4 {d}^{5} 5 {s}^{1}$

Copper:

$\left[\text{Ar}\right] 3 {d}^{10} 4 {s}^{1}$

Silver:

$\left[\text{Kr}\right] 4 {d}^{10} 5 {s}^{1}$

And gold:

$\left[\text{Xe}\right] 4 {f}^{14} 5 {d}^{10} 6 {s}^{1}$

In all of these scenarios, the new redistribution of the electron is energetically favorable compared to the predicted configuration.