Question

Why Cr(24) has an empty 4p shell though it does not have any electrons in it?

Answer 1

One explanation for is that:
* The maximized exchange energy IIe stabilizes this configuration $\left(3d^{5}4s^1\right)$. The maximization comes from how there are 5 unpaired electrons, instead of just 4$\left(3d^{4}4s^2\right)$.
* The minimized coulombic repulsion energy $I{I}_c$ further stabilizes this configuration. The minimization comes from having all unpaired electrons in the 3d and $4s\left(3d^{5}4s^1\right)$, rather than one electron pair in the $4s\left(3d^{4}4s^2\right)$.
* 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.