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    Came across this one while working out electronic configs...

    Transition metals have unfilled d-orbitals, yes? So the progression goes (through Period 4):
    Scandium (Sc)- 2,8,9,2
    Titanium: (Ti)- 2,8,10,2
    Vanadium (V)- 2,8,11,2
    etc etc ad nauseam UNTIL:
    Nickel (Ni)- 2,8,16,2
    Copper (Cu)- 2,8,18,1
    Zinc (Zn) 2,8,18,2

    Now, given the above electron configs, one can assume the most stable ion by referring to the fully or half filled orbital rules. Therefore, logically, the most stable ion of copper would be 1+ (giving 2,8,18,0- a fully filled (and stable) d-orbital).

    SO WHY THEN... does copper most commonly form the MORE stable Cu2+ ion?
    Cu2+ has a config of 2,8,17- the d-orbital requires another electron to be filled, making it more reactive.

    This is annoying me, if any ideas please reply!

    Ta

    UDLM
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    (Original post by UpsidedownLandMan)
    Came across this one while working out electronic configs...

    Transition metals have unfilled d-orbitals, yes? So the progression goes (through Period 4):
    Scandium (Sc)- 2,8,9,2
    Titanium: (Ti)- 2,8,10,2
    Vanadium (V)- 2,8,11,2
    etc etc ad nauseam UNTIL:
    Nickel (Ni)- 2,8,16,2
    Copper (Cu)- 2,8,18,1
    Zinc (Zn) 2,8,18,2

    Now, given the above electron configs, one can assume the most stable ion by referring to the fully or half filled orbital rules. Therefore, logically, the most stable ion of copper would be 1+ (giving 2,8,18,0- a fully filled (and stable) d-orbital).

    SO WHY THEN... does copper most commonly form the MORE stable Cu2+ ion?
    Cu2+ has a config of 2,8,17- the d-orbital requires another electron to be filled, making it more reactive.

    This is annoying me, if any ideas please reply!

    Ta

    UDLM
    This is an answer I found:
    http://uk.answers.yahoo.com/question...5123158AAQFjcO
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    Chromium does a similar thing

    It says in the textbook that 'it is suggested that electron repulsions between the outer electrons are minimised, resulting in an increased stability of the chromium and copper atoms'

    So my idea is that if you have the same number of electrons in each orbital in the 3d subshell, it's more stable. So copper upgrades a 4s electron (because it has less energy) to fill in the last orbital. Same goes for chromium I guess.

    Hope this vaguely helped
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    (Original post by UpsidedownLandMan)
    Came across this one while working out electronic configs...

    Transition metals have unfilled d-orbitals, yes? So the progression goes (through Period 4):
    Scandium (Sc)- 2,8,9,2
    Titanium: (Ti)- 2,8,10,2
    Vanadium (V)- 2,8,11,2
    etc etc ad nauseam UNTIL:
    Nickel (Ni)- 2,8,16,2
    Copper (Cu)- 2,8,18,1
    Zinc (Zn) 2,8,18,2

    Now, given the above electron configs, one can assume the most stable ion by referring to the fully or half filled orbital rules. Therefore, logically, the most stable ion of copper would be 1+ (giving 2,8,18,0- a fully filled (and stable) d-orbital).

    SO WHY THEN... does copper most commonly form the MORE stable Cu2+ ion?
    Cu2+ has a config of 2,8,17- the d-orbital requires another electron to be filled, making it more reactive.

    This is annoying me, if any ideas please reply!

    Ta

    UDLM
    You are quite right that on the basis of electronic configuration alone, you would expect Cu (I) to be the most stable.

    However, in aqueous media Cu (II) is more stable as it has a much higher hydration enthaply (Cu (I) has a very low hydration enthalpy meaning it is only really found in the solid state).


    (Original post by student777)
    Chromium does a similar thing

    It says in the textbook that 'it is suggested that electron repulsions between the outer electrons are minimised, resulting in an increased stability of the chromium and copper atoms'

    So my idea is that if you have the same number of electrons in each orbital in the 3d subshell, it's more stable. So copper upgrades a 4s electron (because it has less energy) to fill in the last orbital. Same goes for chromium I guess.

    Hope this vaguely helped
    You are talking about the copper atom which promotes an s-electron to gain the extra stability of a full d-orbital. But the question was about the stability of Cu(I) vs Cu(II)
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    Jahn Teller distortion? ie the d9 configuration has larger LFSE for the octahedral complexes or a +2 ion just gives the best lattice energy/ coulombic interactions in the compounds/complexes in compensation for the ionisation energy needed to remove electrons
 
 
 
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