Why are the electronic configuration of V and Nb different?
Submitted by satishburnwal on 7 December 2004 - 3:44pm.
Just worried: why are the electronic configuration of V(23 - [Ar]3d(3)4s(2)) and Nb(41 - [Kr]4d(4)5s(1)) different for the last 5 electrons (these 2 elements are in the same group VB) ?
Can anybody tell me the reason..?

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These 'ground state configurations' aren't really that important,
as they are for isolated gas phase atoms.
But anyways, it just works out that those are the lowest energy configurations. (At least, that's what we observe in the spectra)
If you have that amount of positive electric field surrounding a nuleus, and you feed in that number of electrons to dance around it, that's what you get.
For the "why", you could look at the wavefunction equations for the orbitals involved - but they are very complicated mathematical expressions, and moreover, the notion of having separate individual orbitals is only an approximation - in reality, you need one big wavefunction for the nucleus and all the electrons together, taking into account all the nuclear-electron attractions and all the electron-electron repulsions...
..which would be a huge mathematical mess. :lol: