# What happens to the $3d$ and $4s$ levels during the ionisation of a transition metal?

I'm just wondering about the transition state of a transitional metal before (ex: $$[\mathrm{Ar}]\, 3d^3 \,4s^2$$) and after ionisation (which means $$[\mathrm{Ar}]\, 3d^3 \,4s^1$$) because before ionisation the $$4s$$ level is closer to the radius (the $$4s$$ orbital is filled before the $$3d$$ orbital). However, during ionisation, a $$4s$$ electron is knocked off which means that $$4s$$ has to be the outermost shell of the atom.

• Scenario 1: We blast the atom with photon of different energy and record the energy of the photon that causes the atom to be ionised. We found out that the energy of this photon is equal to that of a $$4s$$ orbital. (The atom somehow can tell that we are going to ionised it and therefore changes its electron configuration)?

• Scenario 2: The energy of the specific photon is equal to that of a $$3d$$ orbital but then due to increase of effective nuclear charge and the decrease of repulsive term, $$E_\mathrm{rep}(3d,3d) - E_\mathrm{rep}(3d,4s),$$ in total energy of an ionised atom the electron configuration is modified by having a $$4s$$ electron drops to $$3d$$ level.

(See, in particular, this chemguide page regarding why the assumption that $$4s$$ electron levels are lower than of $$3d$$ is wrong.)

This depends on the energy of the photon.

• If the photon has less energy than the difference $$E_{g, \rm V^+} - E_{g, \rm V}$$ between the ground state $$[\mathrm{Ar}]\, 3d^3 \,4s^2$$ (assuming we're doing vanadium) of the neutral atom and the ground state of the singly-ionized cation $$\rm V^+$$, then no ionization will take place. (Instead, the photon will only be able to excite the atom to higher eigenstates of the neutral, assuming that it is resonant with any relevant transitions.)
• If the photon has enough energy $$\hbar\omega$$ to go from the neutral's ground state $$E_{g, \rm V}$$ to the ground state $$E_{g, \rm V^+}$$ of the cation, then ionization will take place, with any excess energy $$E_{\mathrm{kin},g} = \hbar \omega - (E_{g, \rm V^+} - E_{g, \rm V})$$ going into the kinetic energy of the electron.
• If the photon has enough energy $$\hbar\omega$$ to go from the ground state to the neutral to an excited state of the cation, $$E_{e,\rm V^+}$$, then you will also have ionization onto that channel, where the ion is left in that excited state and the photoelectron carries away the remaining energy, $$E_{\mathrm{kin},e} = \hbar \omega - (E_{e, \rm V^+} - E_{g, \rm V}),$$ as kinetic energy. This will be less than the kinetic energy carried away on the ground-state channel $$E_{\mathrm{kin},g}$$ by $$E_{e, \rm V^+} - E_{g, \rm V^+}$$, so if you look at the photoelectron energy spectrum, you will see two distinct peaks separated by that energy.

Here it is important to note that it is the energies of the eigenstates of the ion that matter, and those are equally subject to the vicissitudes of transition-metal electron-configuration rules as the neutrals are.

In particular, for vanadium, while it would indeed be reasonable to think that the ground-state configuration of the $$\rm V^+$$ ion would be $$[\mathrm{Ar}]\, 3d^3 \,4s^1$$, it appears (as explained here) that the added bonus of getting closer to a half-filled $$3d$$ shell makes it more beneficial to go to a $$[\mathrm{Ar}]\, 3d^4$$ configuration as the ground state.

I'm unsure what the electron configurations would be for the first few excited states - for an atom that's this complex, there's no substitute for actually going and doing the experiment, and my go-to resource for this (the NIST Atomic Spectra Database) is currently closed.

Ultimately, you're not that wrong in your second scenario.

The energy of the specific photon is equal to that of a $$3d$$ orbital

This is wrong - this single-electron picture doesn't work. It's the global energies of the entire system that matter, and trying to split them up into single-electron contributions is only going to lead you up all sorts of wrong alleys.

but then due to increase of effective nuclear charge and the decrease of repulsive term ($$E_\mathrm{rep}(3d,3d) - E_\mathrm{rep}(3d,4s)$$) in total energy of an ionised atom the electron configuration is modified by having a $$4s$$ electron drops to $$3d$$ level.

This isn't that wrong. Within the Hartree-Fock picture, it is indeed correct to note that the transition in the ionization to the $$\rm V^+$$ ground state, $$[\mathrm{Ar}]\, 3d^3 \,4s^1 \longrightarrow [\mathrm{Ar}]\, 3d^4 + e^-,$$ is a two-particle transition, and those are in principle forbidden in the initial naive theory. Your interpretation ("$$4s$$ electron drops to $$3d$$ level") is ultimately wrong, though it does represent the kind of hand-waving language that's used at professional levels to explain what happens in the process.

However, the fact that the process is forbidden in the naive theory (Hartree-Fock with single-photon single-particle processes) means that a full understanding of this ionization pathway is going to be tricky to get, and it won't fit within any theoretical framework that you're likely to be taught before graduate school. There are plenty of post-Hartree-Fock theory frameworks that are able to handle this (generally by allowing states to contain superpositions of multiple configurations) but they just don't fit into simple hand-wavy language - they're just a bunch of technical quantum chemistry, and there's ultimately no way around that.

• Thanks a lot for the answer. Is there the possibility of having a photon excite inner electrons to higher eigenstates ? $$\begin{array}{l} {E}_{ph}= a + b +c\\ a={E}_{g,{V}^{+}}-{E}_{g,V}\\ b={E}_{e,{V}^{+}}-{E}_{g,{V}^{+}}\\\end{array}$$ Why does b not just simply go into the increase of $$\begin{array}{l} \\ {E}_{kin}\end{array}$$ of photoelectron ? – Jung Jan 29 at 20:24
• c is just a small extra amount of energy. Yes, I do think the various assumptions in HF-theory make it insufficient or invalid for large multi-electron systems. But like you said, I was only introduced to HF and perturbation theory in the introductory course. Though I believe that considering the electron-electron repulsion term as interaction with an average charge cloud becomes less true as the no of e- increases. – Jung Jan 29 at 20:38
• @Jung The concept of "repulsion energy" is a distinctly HF type of concept, and it's not sufficient to explain this ionization process. You're on the boundaries of some rather nontrivial beyond-HF effect, and indeed I suspect that you could mount a pretty serious cutting-edge research project on some of the questions you've raised, requiring the sharpest tools we have available on both theory and experiment but also potentially producing some very fresh new insights. The road from what you're asking to "nobody knows right now" isn't all that long. – Emilio Pisanty Jan 30 at 1:51
• As for your first comment - the breakup as you've written it down does not represent what your words say, but as for the text - yes, it's indeed possible for a photon to excite inner electrons to higher shells. However, one needs to keep a very sharp eye on how much energy that takes: in a large majority of that class of processes, it takes more energy to excite an inner-shell electron to a higher shell than it takes to fully remove a valence electron. – Emilio Pisanty Jan 30 at 1:56
• When that happens, the state is called an auto-ionizing state, and the pure excitation is impossible - it just becomes a so-called Fano resonance in the photoionization cross-section. (Which, by the way, is also the focus of a lot of current attention, particularly as regards the time-domain picture of those ionization processes. Try searching, in particular for attosecond transient absorption spectroscopy. Though that's generally in helium and neon and the like, which are easier to describe theoretically, but also easier to deliver as gas-phase atoms within a vacuum chamber.) – Emilio Pisanty Jan 30 at 1:57