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

Question

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

1 Answer

Impact of this question

54355 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T21:18:07

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...

$ul(uarr darr) " " ul(color(white)(uarr darr)) " " ul(color(white)(uarr darr))$ is higher in energy than $ul(uarr color(white)(darr)) " " ul(darr color(white)(uarr)) " " ul(color(white)(uarr darr))$

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...

$ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))$

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...

$ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(color(white)(uarr darr))$ is lower in energy than $ul(uarr color(white)(darr)) " " ul(darr color(white)(uarr)) " " ul(color(white)(uarrdarr))$

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 $Pi_e$ ( $1harr2$ ), but the second configuration would have a $0Pi_e$ 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:

$Pi = Pi_c + Pi_e$

Inorganic Chemistry, Miessler et al.

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... ( $4s$ and $3d$ orbitals)

$ul(uarr color(white)(darr))$

$ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))$

compared to

$ul(uarr darr)$

$ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul(color(white)(uarr darr))$

is more stable.

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

The first configuration has $\mathbf(Pi = 10Pi_e)$ .

(Exchanges: $1harr2, 1harr3, 1harr4, 1harr5, 2harr3,$
$2harr4, 2harr5, 3harr4, 3harr5, 4harr5$ )

The second configuration has $\mathbf(Pi = Pi_c + 6Pi_e)$ .

(Exchanges: $1harr2, 1harr3, 1harr4, 2harr3, 2harr4, 3harr4$ )

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

We could also say that since the $3d$ 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, $"W"$ has a configuration of $[Xe] 5d^4 6s^2$ , which seems to contradict the reasoning we had for $"Cr"$ , since the pairing occurred in the higher-energy orbital.

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

That is, $Pi_"W" < Pi_"Cr"$ .

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

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

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question