Electron orbitals of iron and copper

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The discussion centers on the electronic configurations of iron and copper, highlighting the differences in their d and s orbitals. Iron has an incomplete d orbital (3d^6) and a full s orbital (4s^2), while copper features a full d orbital (3d^10) and a partially filled s orbital (4s^1). The order of orbital filling is influenced by energy levels, with stability being a key factor. The stability of configurations is affected by the pairing of electrons and the effective nuclear charge experienced by d-electrons, which increases with filling. This leads to greater stabilization for fully filled (d10) and half-filled (d5) d orbitals. The discussion references empirical rules, specifically Slater's rules, to explain the effective nuclear charge and the energetic advantages of certain electron configurations.
Johnleprekan
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I was reading an article about the electron orbitals of iron and copper:

The electronic configuration of iron is 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 2, 3d 6

And that of copper is 1s 2, 2s 2, 2p 6, 3s 2, 3p 6, 4s 1, 3d 10


Iron has an incompletely filled d orbital while copper has a full d orbital. The s orbital of iron is full while copper is not. Why is the d orbital of copper full when the s (of copper) one isn't? (s orbitals being able to hold two electrons that is) Why doesn't the s orbital fill up and leave the d orbital incomplete?
 
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Order in which orbitals are filled depends on the energy - the lower the energy of the atom, the more stable the configuration. Energies of these orbitals are almost identical, and some very tiny effects (like pairing or not pairing spins on a particular orbital) make one of the configurations slightly more stable. These differences are typically so small they require quite elaborate models to be reproduced.

You can also see an explanation saying that there is an additional energy gain in the case of half filled or full filled d orbital which makes d5 and d10 particularly stable. As far as I am aware it is not entirely true, even if we do observe d5 and d10 quite often. But I am not going into details, as it is not a subject I feel comfortable with.
 
Because it is more favorable (in terms of energy) to have a filled 3d and a partially filled 4s than the other way around. Why that is is much more complicated to answer. Check the following thread for more information: https://www.physicsforums.com/showthread.php?t=712360
 
There is an empirical set of rules, called the Slater rules
http://en.wikipedia.org/wiki/Slater's_rules
, to estimate the effective nuclear charge seen by electrons in the different sub-shells.
The effective nuclear charge seen by the d-electrons increases with increasing filling of the d-shell resulting in the d shell becoming more and more stabilized when the number of d-electrons increases.
This is understandable, as electrons in the same sub shell are not very efficient in screening each other from the nuclear charge.
Hence the energetic stabilization of the d-shell has two maxima at d5 and d10 respectively, as at going from d5 to d6 an electron has to be put into an orbital which contains already an electron, which is not as favourable as putting an electron into an empty orbital.
 
Delete please.
 
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