# Question 83d74

Nov 11, 2016

Electron configurations of transition metals are in general not necessarily intuitive at first glance.

BASIC ELECTRON CONFIGURATIONS

The first-row transition metals are where we first introduce the explicit usage of the $3 d$ orbitals (instead of simply for heptavalent hybridization (EX: ${\text{SF}}_{6}$, ${\text{PF}}_{5}$, etc).

It's where the electron configuration goes ["Noble Gas"] 4s^? 3d^"?" 4p^?#, and "hiccups" down one quantum level for just the $d$ electrons, or two quantum levels for the $f$ electrons, relative to the $s$ quantum level.

IONIZATION OF TRANSITION METALS (FIRST-ROW)

Most professors will tell you that the $4 s$ electrons are lost first upon ionization, but then they go around and tell you that the $3 d$ orbitals are higher in energy.

It's good to compare to other sources (trust but verify!), and this PDF from my Inorganic Chemistry textbook says that most transition metals have orbital potential energies higher for the $4 s$ than for the $3 d$:

Then, it makes sense that the $4 s$ electrons are lost first, since they are higher in energy.
Furthermore, even if it were really the case that the $4 s$ orbital were lower in energy than the $3 d$, if you look at the radial electron density distribution of the $3 d$ orbital (which has no radial nodes) and $4 s$ orbital, the most probable position of the $4 s$ electron is farther away from the nucleus than a $3 d$ electron:
Thus, it is indeed easier to ionize a transition metal's $4 s$ electron than its $3 d$ electron.