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    Why does copper have such a weird electron configuration?
    I don't understand why it's [Ar]4s1,3d10 instead of [Ar] 4s2, 3d9
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    I think it's because having 9 electrons in the 3d sub-shell would be too unstable, and it's more stable for it to have a full 3d shell than a full 4s shell.
    Hope this helps
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    to make it more stable.
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    you fill up 3d before 4s because they're so close together
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    [QUOTE=z33;60031967]you fill up 3d before 4s because they're so close together[/QUOTE

    why isn't this the case with scandium, titanium, vanadium etc?
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    In a subshell, every pair of electrons with parallel spins has what's known as an exchange energy contribution which lowers the overall energy of the system. So with Scandium (d1) there is no exchange energy contribution, but with Titanium (d2) there are are now two parallel electrons (remember, electrons added to orbitals go into empty orbitals before they pair up with other electrons- this is the reason for that) so the two electrons interact and lower the overall energy. At d3, the energy is lowered even more as all three electrons have the same spin and interact favourably. At d5 this favourable interaction between the electrons is at a maximum and the energy is lowered by a very large amount.

    However at Iron (d6) the electrons start pairing up with the other electrons- one electron is added with opposite spin. There are no other electrons with the same spin as this new electron so there is no decrease in the overall energy. But when you start adding the electrons with d6, d7, d8 etc the energy begins to decrease once again. Basically all the "up" electrons interact favourably with each other and all the "down" electrons interact favourable with each other (but "ups" do not interact favourably with "downs")

    Because there are so many electrons with parallel spins, the d5 and d10 configurations lower the overall energy of the atom by a large amount. This makes the d5 and d10 configurations exceptionally stable. So stable that in Cr and Cu, one electron will drop down from 4s to 3d.
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    (Original post by PurpleCherry)
    I think it's because having 9 electrons in the 3d sub-shell would be too unstable, and it's more stable for it to have a full 3d shell than a full 4s shell.
    Hope this helps
    This is incorrect. The preference for 4s13d10 over 4s23d9 comes from the added exchange energy that you get from sticking an extra spin-down electron in the 3d subshell, not from relieving some sort of instability arising in the 3d9 configuration.

    (Original post by honeylooloo)
    why isn't this the case with scandium, titanium, vanadium etc?
    It's certainly true that the 3d and 4s subshells are close in energy for first row transition metal atoms. You fill the 4s subshell first because 4s is lower in energy than 3d. The energy difference is enough to make adding electrons to 4s twice (with some electron repulsion to go with it as a result) better than one in 4s and one in 3d, except in rare circumstances. These rare circumstances, the reason for Cu and Cr having an extra electron in 3d instead of 4s, are explained well by TheWiseSalmon.

    Similarly, I suppose, vanadium is 4s13d3 instead of 4s03d5 because the extra exchange energy you'd get from putting those two electrons into 3d wouldn't make up for the higher energy (i.e. less favourable) of the 3d orbital in and of itself.
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    (Original post by BJack)
    Similarly, I suppose, vanadium is 4s13d3 instead of 4s03d5 because the extra exchange energy you'd get from putting those two electrons into 3d wouldn't make up for the higher energy (i.e. less favourable) of the 3d orbital in and of itself.
    Actually, by Scandium, the 3d orbital has dropped below the 4s so is actually at a slightly lower energy level (because the 3d orbital drops in energy much more rapidly than 4s as nuclear charge increases). The reason why the electrons go into 4s before 3d is not because 4s is lower in energy, but because the electron-electron repulsion is much higher in d-orbitals than in s-orbitals (look at the shapes of the orbitals- d orbitals have 2 nodes so there are more regions of space where electrons have 0 probability of existing; they are more restricted).

    The favourable exchange energy contribution for the d5 and d10 configurations makes it "worthwhile" for one electron to drop from 4s to 3d because the stability gained due to additional exchange energy overcomes the increased electron-electron repulsion in the d-orbitals.
 
 
 
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