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I am learning about electron configurations and I came across the "exceptions" of $\ce{Cr}$ and $\ce{Cu}$ where electrons actually half fill the $\mathrm d$ orbitals instead of the expected $\mathrm s$ orbitals. I thought maybe vanadium would do the same, but my textbook gives the answer of an electron configuration exactly following the Aufbau principle.

I've attached a picture of my drawing for the textbook answer and what I had thought might happen — could someone explain why vanadium is not an "exception" like chromium or copper?

Orbital Diagram


marked as duplicate by Jan, ron, pentavalentcarbon, Jon Custer, airhuff Nov 23 '17 at 18:26

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Electrons always fill in the lowest energy configuration possible.

Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order. In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion. In their case, the energy to promote an s electron to the d orbitals is compensated for by the effects of exchange energy and no repulsion. This effect is called "the stability of half filled subshells", or something to that effect, in most textbooks.

The energy cost to promote 2 electrons from the s subshell, as would be the case for Vanadium, is too much to be compensated for by the exchange energy gained and the repulsion lost, so the lower energy configuration is to have two electrons in the s subshell and 3 in the d.

Here's a nice website that shows the electron configuration for all elements: with http://www.ptable.com/#Orbital


Its very simple. You cant skip orpitals. Since 4s is smaller than 3d you will have to fill 4s first and than 3d. But u cant excite both the electrons from 4s cause it would take a very high amount of energy and even if u did the 4s becomes vacant half filling the 3d orbital and thats would be highly unstable. So the only possible stable configuration would be 4s2 3d3. And in Cr and Cu the excites 1 electrons because by doing that both 4s and 3d orbital gains half filled i.e the stable configuration. So in vanadium if u excite 1 electron the d orbital will still not be half filled. But 4s orbital will be half filled. Since fully filled orbital are more stable than half filled the vanadium will have 2 electrons in 4s orbital.


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