![]() ![]() Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells. However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. Thus, its electronic configuration can be written as, Cu 291s 2,2s 2,2p 6,3s 2,3p 6,3d 10,4s 1 (Since, fully filled orbitals are more stable, so one electron from 4s orbital transfers into 3d orbital. When you populate an $\mathrm$ subshells. Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.Īs I understand this, there are basically two effects at work here. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. What I'm really looking for is why the d orbital is more stable this way. When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.
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