Exceptional electronic configuration of Chromium

I understand that the reason for the exceptional electronic configuration of Cr is the increased stability of half filled set of orbitals


But here's an excerpt from my [textbook](https://ncert.nic.in/textbook/pdf/lech108.pdf)



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> …consider the case of Cr, which has $3d^54s^1$ configuration instead of $3d^44s^2$; the energy gap between the two sets ($3d$ and $4s$) of orbitals is small enough to prevent electron entering the $3d$ orbitals.
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What do they mean by *prevent* the electron from entering $3d$ orbital? Didn't one electron from the $4s$ orbital actually go to the $3d$ orbital?


Is it likely that it's a typo to write "prevent" instead of "permit"?




Yes I believe that's a typo, it should say "permit." The reason we see these Aufbau's principle exceptions in transition metals is because the $4s$ and $3d$ orbitals are very similar in energy.


In chromium, having a $4s^2$ $3d^4$ configuration results in electron-electron repulsion due to the two electrons in the $4s$ orbital. For this reason, chromium adopts a $4s^1$ $3d^5$ configuration, in which each electron occupies its own orbital. Recall that Hund's rule essentially states that you fill each orbital once before going back with the second electron in regards to orbitals of the same energy (in the same subshell). In the case of chromium, we are dealing with orbitals that are *almost* in the energy ($3d$ and $4s$) so you can essentially view this as a special case of Hund's rule that extends to orbitals of nearly the same energy.


For what it's worth, the other notable exception is copper, which adopts a $4s^1$ $3d^9$ configuration over $4s^2$ $3d^8$. For reasons that go beyond the scope of this answer, before you hit the transition metal elements ($Z \le 21$), $3d$ is higher in energy than $4s$, so we fill $4s$ before $3d$. But beyond this point, $3d$ becomes lower in energy than $4s$, so you lose $4s$ electrons first when we're talking about ionization. Knowing this, it is energetically more favorable to have $4s^1$ $3d^9$ because the higher energy orbital only has one unpaired electrons and the lower energy orbitals have paired ones.


The important thing to note from these two examples is that *there is no special added stability form merely having a half full electron shell* (i.e., something is not magically more stable because of having a half filled subshell). Adopting this form of configuration has consequences that are perfectly explainable to students of intro chemistry—do not brush it off as mere "magic."


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