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Why do Copper and Chromium not follow the normal electron shell trend? watch

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    Why do copper and chromium only have 1 electron in their 4s orbital as oppose to 2 electrons, like all the other transition metals in that period?
    Please don't just say "It's more stable for them to have half filled or fully filled D sub-shells, and that there isn't much energy difference between the S and the D sub-shell so it becomes more energetically efficient to do so."
    I've read that countless times, but that just RESTATES the question
    of course if it happens naturally its more stable
    but WHY IS IT MORE STABLE? I can't seem to find that answer anywhere on t'interweb

    Thanks in advance TSR
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    It is more stable BECAUSE the 4s and 3d subshells are close in energy. Therefore, the increase in energy due to filling a 4s subshell (ie. having 2 electrons of opposite spin occupying one orbital) is greater than the increase in energy due to putting the electron in the slightly higher energy unfilled 3d orbital.

    This explains Cr at least. Cu must have a similar explanation but I'm a bit rusty on Hund's Rules etc.
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    Its basically the same as Cr. But the d shell is full instead of half full.

    Atoms and molecules look for stability. A half full subshell is more stable than an empty one and one with 1 set of opposing spins.
    Also as the 4s subshell requires less energy to excite electrons from than the 3d so it is more energetically favourable.
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    I was thinking though... when you go to put the last electron in Cu, you can put it into an already half-filled 4s subshell (which will then be full), or you can put it into an already half-filled 3d orbital (which is part of an almost-filled 3d subshell). So why, if 4s is a bit lower in energy, does it go into the 3d subshell?

    It might be something to do with the radial density function? ie. 3d has a lot of density closer to the nucleus than most of the 4s density, so as the 3d fills, the 4s is raised in energy. (I think this is the explanation for why 4s electrons are lost before 3d)
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    I think it is purely to do with stability. I would look into why but, lab reports :pinch:

    I might have a go later.

    Or
    (Original post by EierVonSatan)
    .
    (Original post by charco)
    .
    (Original post by Plato's Trousers)
    .
    (Original post by shengoc)
    .
    Any ideas?
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    (Original post by mdr)
    It is more stable BECAUSE the 4s and 3d subshells are close in energy. Therefore, the increase in energy due to filling a 4s subshell (ie. having 2 electrons of opposite spin occupying one orbital) is greater than the increase in energy due to putting the electron in the slightly higher energy unfilled 3d orbital.

    This explains Cr at least. Cu must have a similar explanation but I'm a bit rusty on Hund's Rules etc.
    So why is the increase in energy needed of filling the 4s subshell greater than the energy needed to occupy the higher 3d orbital?
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    (Original post by Loz17)
    I think it is purely to do with stability. I would look into why but, lab reports :pinch:

    I might have a go later.

    Or





    Any ideas?
    in atomic forms, these are the so called half - shell stability; which qualitatively comes out from hund's rules;
    1) spin correlation => maximise no of unpaired spins
    2) orbital correlation => maximise orbital angular momentum

    in ionic form, ie Cu2+ and Cr2+; be aware of Jahn Teller distortion conferring extra stability to their complexes formed(in solutions).
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    (Original post by guitarmike456)
    Why do copper and chromium only have 1 electron in their 4s orbital as oppose to 2 electrons, like all the other transition metals in that period?
    Please don't just say "It's more stable for them to have half filled or fully filled D sub-shells, and that there isn't much energy difference between the S and the D sub-shell so it becomes more energetically efficient to do so."
    I've read that countless times, but that just RESTATES the question
    of course if it happens naturally its more stable
    but WHY IS IT MORE STABLE? I can't seem to find that answer anywhere on t'interweb

    Thanks in advance TSR
    Look up 'Exchange energy', but I warn you it involves hard sums
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    (Original post by guitarmike456)
    So why is the increase in energy needed of filling the 4s subshell greater than the energy needed to occupy the higher 3d orbital?
    there is a crossover point in orbital energies(effect of shielding/penetration) from main group elements in period 4 going into transition metal block.

    subtle point; 4s is lower in energy than 3d for K and Ca; but 3d is lower in energy than 4s from Sc onwards.

    you fill the lower energy orbital first, but in terms of ionisation, you take electrons away from higher energy orbital.
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    (Original post by guitarmike456)
    So why is the increase in energy needed of filling the 4s subshell greater than the energy needed to occupy the higher 3d orbital?
    I think because of the fact the electrons have opposite spins. Or you could think of it as having 2 electrons in the same orbital, which repel eachother.
 
 
 
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