Electronic Configuration of Transition Metals

The general outer electronic configuration of a first-row (3d) transition metal atom is $3d^{1-10}4s^{1-2}$, built on the argon $[\text{Ar}]$ core. Filling this pattern in order of increasing atomic number gives:

Two members, chromium and copper, depart from the pattern that a straightforward filling of $4s$ before $3d$ would predict (which would give Cr as $3d^44s^2$ and Cu as $3d^94s^2$). In both cases, one electron is promoted from the $4s$ orbital into the $3d$ subshell so that the $3d$ subshell reaches either a half-filled ($d^5$, for chromium) or a completely filled ($d^{10}$, for copper) configuration. Half-filled and completely filled subshells are unusually stable for two related reasons: the electron-electron repulsion within the subshell is minimized when electrons are spread singly across all five d orbitals (as in $d^5$) or when every orbital is doubly filled with paired spins (as in $d^{10}$), and the exchange energy (a quantum-mechanical stabilization that arises between electrons of parallel spin occupying different orbitals of the same subshell) is maximized precisely at the half-filled and completely filled configurations. The net stabilization gained by achieving $3d^5$ or $3d^{10}$ outweighs the small energy cost of promoting one electron out of the $4s$ orbital, so chromium and copper adopt $3d^54s^1$ and $3d^{10}4s^1$ respectively rather than the "expected" configurations.

A useful way to see why $4s$ fills before $3d$ (and consequently why $4s$ electrons are also the ones lost first on ionization, discussed in a later section) is that, for a neutral atom, the $4s$ orbital lies at slightly lower energy than $3d$ at the point in the periodic table where filling begins; but once several $3d$ electrons are present, their mutual shielding raises the relative energy of $4s$ compared with $3d$, which is why the singly ionized and doubly ionized transition-metal cations lose their $4s$ electrons before any $3d$ electron. …