I recently came across a question Why is the vanadium(3+) ion paramagnetic?, where the asker is wondering how $\ce{V^{3+}}$ is paramagnetic (he used Aufbau in reverse to remove the electrons), while the correct answerer to that question remarked that removing electrons must follow the order from outer shell to inner shell. For example, in the case of $\ce{V^{3+}}$ the electronic configuration of $\ce{V}$ is $\ce{[Ar] 3d^3 4s^2}$. The asker used reverse-Aufbau, which is $\ce{[Ar] 3d^0 4s^2}$; while the actuality was $\ce{[Ar] 3d^2 4s^0}$.

My question: Why does filling electrons follow a certain rule, while removing them follows a different rule?

  • $\begingroup$ Well in our textbook it is written that "while removing electrons always remove electrons from the last shell". AFAIK, The exceptions to this rule are few and dont occour till 4d series. $\endgroup$
    – Kartik
    Commented Apr 25, 2017 at 14:46
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    $\begingroup$ Aufbau works for the wrong reasons. The 4s orbital is not below 3d in energy. pubs.acs.org/doi/abs/10.1021/ed055p2 Many people have written on this already. chemistry.stackexchange.com/questions/8357/… $\endgroup$ Commented Apr 25, 2017 at 16:34
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    $\begingroup$ The whole idea of "adding" or "subtracting" an electron from the electron configurations of different atoms, or even the same atom in different ionization states, is fundamentally wrong. The overall energy for every possible electronic configuration must be determined separately for every combination of the number of electrons and number of protons in an atom, and whichever has the lowest energy is the ground state, irrelevant what the result is for the element or ion next to it. $\endgroup$ Commented Apr 27, 2017 at 12:36

1 Answer 1


Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom.

When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with different interactions between nucleus and electrons, which affects the energy of the orbitals.

Also keep in mind, that the orbital picture is just a simplification. Strictly speaking (and from a purely theoretical point of view) orbitals do not exist, they are merely a mathematical crutch to solve the Schrödinger Equation. Although in practice this picture works surprisingly well, hence people use it a lot.

A proper quantum chemical calculation will give you the correct ground state, but you might have problems identifying it as something like $\ce{[Ar] 4s^2 3d^0}$. It will be a mixture (linear combination) of many electron configurations, although $\ce{[Ar] 4s^2 3d^0}$ might be the most important one for $\ce{V^{3+}}$.


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