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# What is the electron configuration of chromium?

The for chromium is NOT ##1s^2 2s^2 2p^6 3s^2 3p^6 3d^4 4s^2##, but ##color(blue)(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1)##.

Interestingly enough, Tungsten is more stable with an electron arrangement of ##[Xe]4f^14 5d^4 6s^2##.

Unfortunately, there is no easy way to explain these deviations in the ideal order for each element.

To explain Chromium's electron configuration, we could introduce:

- The exchange energy ##Pi_e## (a stabilizing quantum mechanical factor that is directly proportional to the number of pairs of electrons in the same subshell or very close-energy subshells with parallel spins)
- The coulombic repulsion energy ##Pi_c## (a destabilizing factor that is inversely proportional to the number of electron pairs)
- These combine to produce an overall pairing energy ##Pi = Pi_c + Pi_e##.

The former is stabilizing and the latter is destabilizing, as shown below (suppose configuration 2 is at pairing energy ##Pi = 0##):

One explanation for Chromium, then, is that:

- The maximized exchange energy ##Pi_e## stabilizes this configuration (##3d^5 4s^1##). The maximization comes from how there are ##5## unpaired electrons, instead of just ##4## (##3d^4 4s^2##).
- The minimized coulombic repulsion energy ##Pi_c## further stabilizes this configuration. The minimization comes from having all unpaired electrons in the ##3d## and ##4s## (##3d^5 4s^1##), rather than one electron pair in the ##4s## (##3d^4 4s^2##).
- The small-enough orbital size means that the electron density is not as spread out as it could be, which makes it favorable enough for a maximum total spin to give the most stable configuration.

However, Tungsten's ##5d## and ##6s## orbitals being larger than the ##3d## and ##4s## orbitals (respectively) spreads out the electron density enough that the pairing energy (##Pi = Pi_c + Pi_e##) is small enough.

The more the electron distribution is spread out, the less electron-pair repulsion there is, and thus the lower ##Pi_c## is. Therefore, the lower ##Pi## is.

Thus, electron pairing is favorable enough for Tungsten.

There is no hard and fast rule for this, but that is an explanation that correlates with experimental data.