When Vanadium is ionised it loses the 4s electron first, meaning that it's 3+ ion has a different electron configuration to Calcium despite it being isoelectronic. Can it be explained in terms of radial distribution functions? Maybe the maxima of the 3d and 4s become further apart as the effective nuclear charge increases?


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I think @NicolauSakerNeto mentioned a number of excellent points in his linked answer. Frankly, I think he deserves a bounty.

There are several effects occurring, which give different electron configurations to the transition metals. A very nice review article is available: "The Full Story of the Electron Configurations of the Transition Elements" J. Chem. Educ., 2010, 87 (4), pp 444–448

  • "d orbital collapse": Because of the difference in angular-momentum "centrifugal force" in quantum mechanics, incomplete shielding of core electrons, and nuclear attraction, the $Z_{eff}$ - effective nuclear charge - for 3d and 4s electrons are different. As a rule of thumb, electrons are typically lost from the 4s first,

  • electrostatic repulsion in 4s vs. 3d. It's interesting you mention V, since $\ce{V^0}$ is $[3d^3 4s^2]$ and $\ce{V+}$ is $[3d^4]$. The usual explanation is that while 3d is closer to the nucleus and energetically favorable, once you start filling the 3d electrons, the electrostatic repulsion is high, and for the neutral atoms, it's favorable to have one or two 4s electrons, which are more diffuse.

There are other things going on, particularly later in the periodic table, where relativistic and spin-orbit coupling effects occur. Also, many times the electron configuration can change in a free atom or ion compared to a complex.


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