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    In a Cu^2+ iron, all five d-orbitals have the same energy. However, when the octahedral complex ion [Cu(H20)v6]^2+ is formed, the d orbitals split into different energy levels

    a) Draw a diagram to describe the energy levels of the split d-orbitals

    So I know that there are two orbitals which are at a higher level than three orbitals but which electrons go into which orbitals.

    Say an orbital is defined by [] and ^ and v are used to describe individual electrons.

    So we start with [^v] [^v] [^v] [^v] [^]

    Does it just become
    [^v] [^v]
    [^v] [^v] [^]


    Or does the singular electron have to be in the higher energy orbital because with d orbital splitting, that is the electron which gets promoted?
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    The single electron is in the higher energy level.

    Lower energy orbitals filled first. Aufbau principle.
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    (Original post by gilbo)
    The single electron is in the higher energy level.

    Lower energy orbitals filled first. Aufbau principle.
    So it becomes

    [^v] [^]
    [^v] [^v] [^v]

    right? does the order matter, is that different from

    [^] [^v]
    [^v] [^v] [^v]
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    The top configuration is correct
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    Both the orbitals at the top are degenerates, so pairing with any of them shouldn't make a difference, but i think it is just convention(normal) for people to pair up the electrons on the furthest left(ie where they start filling up electrons)
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    (Original post by shengoc)
    Both the orbitals at the top are degenerates, so pairing with any of them shouldn't make a difference, but i think it is just convention(normal) for people to pair up the electrons on the furthest left(ie where they start filling up electrons)
    This is true but it makes a difference once you come to look at bonding and symmetry as the top left is dz^2 and the right dx^2 - y^2
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    (Original post by gilbo)
    This is true but it makes a difference once you come to look at bonding and symmetry as the top left is dz^2 and the right dx^2 - y^2
    I think that's just by convention, rather than anything else. They both transform the same way until you change the point group, at which point all this octahedral field stuff doesn't apply.
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    (Original post by gilbo)
    This is true but it makes a difference once you come to look at bonding and symmetry as the top left is dz^2 and the right dx^2 - y^2
    Why does the unpaired electron go with dx^2 - y^2 rather than dz^2? thanks also for the help !
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    (Original post by Kyle_S-C)
    I think that's just by convention, rather than anything else. They both transform the same way until you change the point group, at which point all this octahedral field stuff doesn't apply.
    That, indeed sir, makes perfect sense. Perhaps why you are at Cambridge.
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    (Original post by Kyle_S-C)
    I think that's just by convention, rather than anything else. They both transform the same way until you change the point group, at which point all this octahedral field stuff doesn't apply.
    One thought maybe you can clear this up.

    Doesn't Cu2+ generally Jahn Teller distort along the axial which would mean 2e- in the dz^2. Would that be correct or not?
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    (Original post by gilbo)
    One thought maybe you can clear this up.

    Doesn't Cu2+ generally Jahn Teller distort along the axial which would mean 2e- in the d. Would that be correct or not?
    Sorry, I've been horrendously busy this week. You make a very good point, I forgot all about Jahn-Teller. However, with J-T it would be incorrect to label the energy levels as degenerate; that's the whole point of the J-T effect, it breaks degeneracy to allow the complex to have a lower energy. I can't remember how you tell whether the distortion is going to be an axial compression or elongation though.

    As for the Cambridge thing, that probably makes me less qualified to answer a chemistry question than a proper chemist because of all the other nonsense we do. :p:
 
 
 
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