Why does iron have an abnormally high ionization energy?

Question

Along a period the ionization energy should increase because the atomic number is increasing, but there is negligible increase in shielding. However, $\mathrm{IE}_\ce{Mn} < \mathrm{IE}_\ce{Fe} > \mathrm{IE}_\ce{Co} > \mathrm{IE}_\ce{Cu}$, so why does iron have such a high ionization energy?

I've considered that iron has a $\mathrm{4s^2 3d^6}$ configuration that goes to a $\mathrm{4s^1 3d^6}$ configuration upon ionization, while $\ce{Co}$ has a $\mathrm{4s^2 3d^7}$ that goes to $\mathrm{4s^0 3d^8}$ configuration, $\ce{Ni}$ has a $\mathrm{4s^2 3d^8}$ that goes to a $\mathrm{4s^0 3d^9}$ configuration and $\ce{Cu}$ has a $\mathrm{4s^2 3d^9}$ that goes to a $\mathrm{4s^0 3d^{10}}$ configuration (Lang, J. Chem. Ed., 2003). In the Long paper, the authors explain the increase of IE at iron is due to "electronic structure" and I would like to know exactly what about the electronic structure it is that causes iron to have a high IE than expected based on a general periodic trend.

I was thinking that adding an electron to the $\mathrm{3d}$ orbitals might be energetically favorable enough to lower the ionization energy of $\ce{Ni}$, $\ce{Co}$ and $\ce{Cu}$, so rather than iron having a high ionization energy the surround metals just have a low ionization energy because the ion is stabilized by having another $\mathrm{d}$-electron added? Does that sound reasonable?

Another thing I was thinking was that iron is the first transition metal to have a paired $\mathrm{d}$-electron, could that effect the shielding and increase the ionization energy?

Answer 1

Why does Iron have an abnormally high ionization energy?

First, I'm assuming your question is essentially why doesn't iron follow the general trend where ionization energy increases when moving across the periodic table? So why doesn’t theory match observations? An over simplistic answer would be our understanding of the quantum realm is still quite limited. When we’re considering the behavior of electrons, we’re most certainly in the quantum realm. You can’t see an electron because of Heisenberg’s Uncertainty Principle (essentially you can know the position or momentum of an electron with some certainty, but not both at the same time). The best we can do is ‘map’ the orbital using surface topology like TEM. What you’re taught in general chemistry about orbital shapes is an oversimplification that doesn’t entirely equate to reality. As you add different orbitals (d & f particularly) the idea that you have distinct orbital shapes where the electron is orbiting the nucleus breaks down. Like you stated Iron has a 4s^2 3d^6 configuration that goes to 4s^1 3d^6 upon ionization while Cobalt has a 4s^2 3d^7 that goes to 4s^0 3d^8 configuration. I think your thought process is correct that adding an electron to the 3d orbital is energetically favorable and will lower the ionization energy of Cobalt. Think about the Aufbau principle, subshells of the lowest available energy are filled first. This is what’s being observed here.