# Why do the electron configurations of chromium and copper seem to disagree with what is expected according to the Aufbau principle?

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333
Sep 30, 2016

Some elements do not follow the Aufbau principle, there are some alternate ways that electrons can arrange themselves that give these elements better stability.

#### Explanation:

Using the Aufbau principle, you would write the following electron configurations

Cr = [Ar] $4 {s}^{2}$ $3 {d}^{4}$
Cu = [Ar] $4 {s}^{2}$ $3 {d}^{9}$

The actual electron configurations are:
Cr = [Ar] $4 {s}^{1}$ $3 {d}^{5}$
Cu = [Ar] $4 {s}^{1}$ $3 {d}^{10}$

To understand why this occurs, it is important to realize that...
1. Completely filled sublevels are more stable than partially filled sublevels.
2. A sublevel which is exactly half filled is more stable than a partially filled sublevel which is not half full.
3. Electrons are lazy and will do whatever places them in the lowest energy state = which is the most stable state

In both examples, an electron moves from the 4s sublevel to produce a 1/2 full 3d (Cr) or completely filled 3d (Cu). This gives the atom greater stability so the change is favorable.
http://ericscerri.blogspot.com/2012/07/anomalous-configuration-of-chromium.html

It is important to note that this explanation is just looking at the examples of Cr and Cu. There are other elements which will have exceptions to the rules we have come up with to predict electron configurations. For example, W will follow the Aubau principle and not display behavior like that of Cr. This is due to the fact that the atoms and electrons know nothing of the rules we (humans) come up with to try to explain and predict their behavior.

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Jun 18, 2016

Exchange energy is a quantum mechanical phenomenon that helps account for the more stable configuration of $\text{Cr}$ being $\left[A r\right] 3 {d}^{5} 4 {s}^{1}$ instead of $\left[A r\right] 3 {d}^{4} 4 {s}^{2}$.

The presence of two electrons of the same spin increases the exchange energy, which is a stabilization energy. This is straight from my textbook if you want to see it.

The more ways unpaired electrons of the same spin can exchange, the more stable the configuration overall.

My book doesn't explain enough of it in my opinion (as it makes a hand-waving argument that exchange energy is stabilizing), but maybe it has to do with the effective distribution of energy throughout the sublevel instead of:

• having some coulombic repulsion energy ${\Pi}_{c}$ from pairing electrons in the same orbital, which destabilizes the configuration due to the electrostatic repulsions.
• having two electrons of different spins in different orbitals, which doesn't allow them to exchange while still retaining a degenerate configuration (since with different spins, they are distinguishable), thus lowering the effects of the exchange energy ${\Pi}_{e}$.

Basically, exchange energy is generally calculated in units of ${\Pi}_{e}$, where ${\Pi}_{e} < 0$. The number of exchanges that can be made tell you how many ${\Pi}_{e}$ units the exchange energy is.

As a simple example, for $2$ indistinguishable electrons of parallel spin in the $3 d$ orbitals, you would normally have $2$ exchanged configurations possible:

${\uparrow}_{1} {\uparrow}_{2} \text{_ _ _}$
${\uparrow}_{2} {\uparrow}_{1} \text{_ _ _}$

So for a ${d}^{2}$ configuration, since there is one exchange possible, the exchange energy would be represented as $1 \times {\Pi}_{e}$.

For chromium, the configuration is $\left[A r\right] 3 {d}^{5} 4 {s}^{1}$.

So, we have the exchanges possible for five indistinguishable electrons:

${\uparrow}_{1} {\uparrow}_{2} {\uparrow}_{3} {\uparrow}_{4} {\uparrow}_{5}$
${\uparrow}_{2} {\uparrow}_{1} {\uparrow}_{3} {\uparrow}_{4} {\uparrow}_{5}$ ($1 \leftrightarrow 2$)
${\uparrow}_{3} {\uparrow}_{2} {\uparrow}_{1} {\uparrow}_{4} {\uparrow}_{5}$ ($1 \leftrightarrow 3$)
${\uparrow}_{4} {\uparrow}_{2} {\uparrow}_{3} {\uparrow}_{1} {\uparrow}_{5}$ ($1 \leftrightarrow 4$)
${\uparrow}_{5} {\uparrow}_{2} {\uparrow}_{3} {\uparrow}_{4} {\uparrow}_{1}$ ($1 \leftrightarrow 5$)
${\uparrow}_{1} {\uparrow}_{3} {\uparrow}_{2} {\uparrow}_{4} {\uparrow}_{5}$ ($2 \leftrightarrow 3$)
${\uparrow}_{1} {\uparrow}_{4} {\uparrow}_{3} {\uparrow}_{2} {\uparrow}_{5}$ ($2 \leftrightarrow 4$)
${\uparrow}_{1} {\uparrow}_{5} {\uparrow}_{3} {\uparrow}_{4} {\uparrow}_{2}$ ($2 \leftrightarrow 5$)
${\uparrow}_{1} {\uparrow}_{2} {\uparrow}_{4} {\uparrow}_{3} {\uparrow}_{5}$ ($3 \leftrightarrow 4$)
${\uparrow}_{1} {\uparrow}_{2} {\uparrow}_{5} {\uparrow}_{4} {\uparrow}_{3}$ ($3 \leftrightarrow 5$)
${\uparrow}_{1} {\uparrow}_{2} {\uparrow}_{3} {\uparrow}_{5} {\uparrow}_{4}$ ($4 \leftrightarrow 5$)

That gives $10$ nonredundant exchanges, so we have determined the exchange energy as $\setminus m a t h b f \left(10 \times {\Pi}_{e}\right)$.

Note: since we don't include the redundant exchanges, whereas we had $2 \leftrightarrow 3$, we didn't include $3 \leftrightarrow 2$.

Had chromium had the configuration $\left[A r\right] 3 {d}^{4} 4 {s}^{2}$, it would have an exchange energy of $6 \times {\Pi}_{e}$ instead of $10 \times {\Pi}_{e}$, and that's less stable by $4 \times {\Pi}_{e}$ units.

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