# Why is the electron configuration of chromium 1s^2 2s^2 2p^6 3s^2 3p^6 3d^color(red)(5) 4s^color(red)(1) instead of 1s^2 2s^2 2p^6 3s^2 3p^6 3d^color(red)(4) 4s^color(red)(2)?

Aug 4, 2016

It's a combination of factors:

• Less electrons paired in the same orbital
• More electrons with parallel spins in separate orbitals
• Pertinent valence orbitals NOT close enough in energy for electron pairing to be stabilized enough by large orbital size

DISCLAIMER: Long answer, but it's a complicated issue, so... :)

A lot of people want to say that it's because a "half-filled subshell" increases stability, which is a reason, but not necessarily the only reason. However, for chromium, it's the significant reason.

It's also worth mentioning that these reasons are after-the-fact; chromium doesn't know the reasons we come up with; the reasons just have to be, well, reasonable.

The reasons I can think of are:

• Minimization of coulombic repulsion energy
• Maximization of exchange energy
• Lack of significant reduction of pairing energy overall in comparison to an atom with larger occupied orbitals

COULOMBIC REPULSION ENERGY

Coulombic repulsion energy is the increased energy due to opposite-spin electron pairing, in a context where there are only two electrons of nearly-degenerate energies.

So, for example...

$\underline{\uparrow \downarrow} \text{ " ul(color(white)(uarr darr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ is higher in energy than $\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(darr color(white)(uarr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

To make it easier on us, we can crudely "measure" the repulsion energy with the symbol ${\Pi}_{c}$. We'd just say that for every electron pair in the same orbital, it adds one ${\Pi}_{c}$ unit of destabilization.

When you have something like this with parallel electron spins...

$\underline{\uparrow \downarrow} \text{ " ul(uarr color(white)(darr)) " } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$

It becomes important to incorporate the exchange energy.

EXCHANGE ENERGY

Exchange energy is the reduction in energy due to the number of parallel-spin electron pairs in different orbitals.

It's a quantum mechanical argument where the parallel-spin electrons can exchange with each other due to their indistinguishability (you can't tell for sure if it's electron 1 that's in orbital 1, or electron 2 that's in orbital 1, etc), reducing the energy of the configuration.

For example...

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ is lower in energy than $\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(darr color(white)(uarr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

To make it easier for us, a crude way to "measure" exchange energy is to say that it's equal to ${\Pi}_{e}$ for each pair that can exchange.

So for the first configuration above, it would be stabilized by ${\Pi}_{e}$ ($1 \leftrightarrow 2$), but the second configuration would have a $0 {\Pi}_{e}$ stabilization (opposite spins; can't exchange).

PAIRING ENERGY

Pairing energy is just the combination of both the repulsion and exchange energy. We call it $\Pi$, so:

$\Pi = {\Pi}_{c} + {\Pi}_{e}$

Basically, the pairing energy is:

• higher when repulsion energy is high (i.e. many electrons paired), meaning pairing is unfavorable
• lower when exchange energy is high (i.e. many electrons parallel and unpaired), meaning pairing is favorable

So, when it comes to putting it together for chromium... ($4 s$ and $3 d$ orbitals)

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$

compared to

$\underline{\uparrow \downarrow}$

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

is more stable.

For simplicity, if we assume the $4 s$ and $3 d$ electrons aren't close enough in energy to be considered "nearly-degenerate":

• The first configuration has $\setminus m a t h b f \left(\Pi = 10 {\Pi}_{e}\right)$.

(Exchanges: $1 \leftrightarrow 2 , 1 \leftrightarrow 3 , 1 \leftrightarrow 4 , 1 \leftrightarrow 5 , 2 \leftrightarrow 3 ,$
$2 \leftrightarrow 4 , 2 \leftrightarrow 5 , 3 \leftrightarrow 4 , 3 \leftrightarrow 5 , 4 \leftrightarrow 5$)

• The second configuration has $\setminus m a t h b f \left(\Pi = {\Pi}_{c} + 6 {\Pi}_{e}\right)$.

(Exchanges: $1 \leftrightarrow 2 , 1 \leftrightarrow 3 , 1 \leftrightarrow 4 , 2 \leftrightarrow 3 , 2 \leftrightarrow 4 , 3 \leftrightarrow 4$)

Technically, they are about $\text{3.29 eV}$ apart (Appendix B.9), which means it takes about $\text{3.29 V}$ to transfer a single electron from the $3 d$ up to the $4 s$.

We could also say that since the $3 d$ orbitals are lower in energy, transferring one electron to a lower-energy orbital is helpful anyways from a less quantitative perspective.

COMPLICATIONS DUE TO ORBITAL SIZE

Note that for example, $\text{W}$ has a configuration of $\left[X e\right] 5 {d}^{4} 6 {s}^{2}$, which seems to contradict the reasoning we had for $\text{Cr}$, since the pairing occurred in the higher-energy orbital.

But, we should also recognize that $5 d$ orbitals are larger than $3 d$ orbitals, which means the electron density can be more spread out for $\text{W}$ than for $\text{Cr}$, thus reducing the pairing energy $\Pi$.

That is, ${\Pi}_{\text{W" < Pi_"Cr}}$.

Since a smaller pairing energy implies easier electron pairing, that is probably how it could be that $\text{W}$ has a $\left[X e\right] 5 {d}^{4} 6 {s}^{2}$ configuration instead of $\left[X e\right] 5 {s}^{5} 6 {s}^{1}$; its $5 d$ and $6 s$ orbitals are large enough to accommodate the extra electron density.

Indeed, the energy difference in $\text{W}$ for the $5 d$ and $6 s$ orbitals is only about $\text{0.24 eV}$ (Appendix B.9), which is quite easy to overcome simply by having larger orbitals that stabilize the pairing energy.