# Why is the iron KLMN configuration 2,8,14,2 and not 2,8,8,8?

Jun 16, 2017

Because it's not going to have an electron configuration of $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 4 {p}^{6}$ in the ground state... iron has $3 d$ orbitals available... the energy ordering of the $4 s$ and $3 d$ may be lacking consistency among some textbooks, but certainly not the $4 p$ and $3 d$.

${\overbrace{2}}^{1 {s}^{2}} , {\overbrace{8}}^{2 {s}^{2} 2 {p}^{6}} , {\overbrace{14}}^{3 {s}^{2} 3 {p}^{6} 3 {d}^{6}} , {\overbrace{2}}^{4 {s}^{2} 4 {p}^{0}}$

$K , \text{ "L, " "M, " } N$

Iron cannot skip the $3 d$ orbitals and use $4 p$ orbitals. We know already that the $4 p$ orbitals are higher in energy than the $3 d$ orbitals, by virtue of the $4 p$ having a higher $l$ than the $4 s$, and the $4 s$ being at least CLOSE in energy to the $3 d$.

Hence, the $3 d$ orbitals are occupied (and NOT the $4 p$!), as expected by the Aufbau principle (iron is not some weird exception here!).

The $3 d$ orbitals are in the so-called "$M$ shell", so in "$K L M$" notation (which is not particularly clear, but is nevertheless used for X-ray experiments), one would write $\textcolor{b l u e}{2 , 8 , 14 , 2}$.

${\overbrace{2}}^{1 {s}^{2}} , {\overbrace{8}}^{2 {s}^{2} 2 {p}^{6}} , {\overbrace{14}}^{3 {s}^{2} 3 {p}^{6} 3 {d}^{6}} , {\overbrace{2}}^{4 {s}^{2} 4 {p}^{0}}$

$K , \text{ "L, " "M, " } N$