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    Could someone try and explain this to me in very basic terms so i can understand it? I need to understand why lower oxidation states become more stable with heavier elements.

    any help would be greatly appreciated

    thanks
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    I think it's always best to describe this using examples :yes:

    Take carbon and consider it's electronic configuration:



    We know that carbon forms 4 bonds and to do this it needs 4 half filled orbitals, so in order to bond (in an sp3 fashion anyway) it needs to spend some energy to get to an excited state:



    The energy spent getting to this state is counter balanced by the greater stability of forming a compound, so it's energetically worthwhile :yep:

    Go further down group 14 and this excitation state becomes harder to reach as the energy gap increases and the counter energy is reduced as the bonds become weaker.

    So the two electrons in the s subshell become increasingly inert down the group and hence the name 'inert pair effect'
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    (Original post by Tori1607)
    Could someone try and explain this to me in very basic terms so i can understand it? I need to understand why lower oxidation states become more stable with heavier elements.

    any help would be greatly appreciated

    thanks
    Much of chemistry is moulding theory to suit observation.

    In the heavier elements (around lead) the oxidation state +II becomes stable. This would leave lead with two 's' electrons and a full 'd' subshell.

    So, it seems that this configuration is relatively stable compared to the +IV state (the group valency), which would require loss of the two 's' electrons as well.

    Hence, the idea that these two 's' electrons are relatively stable in their orbital... an inert pair.

    Extending this idea it makes the +I state become more stable descending group III and the +III state descending group V.

    But, usually at A' level it's the change from IV to II stability descending group IV that is put down to the 'inert pair effect'
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    [QUOTE=EierVonSatan;28813319]I think it's always best to describe this using examples :yes:

    Take carbon and consider it's electronic configuration:

    [img]http://image.wistatutor.com/

    thank you, that was very helpful! just one thing: when we go down a group why does the excitation stage become harder to reach?
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    (Original post by Tori1607)
    thank you, that was very helpful! just one thing: when we go down a group why does the excitation stage become harder to reach?
    That's a bit more involved, do you know much about orbital penetration or relativistic contraction? Basically the heavier the element the greater the difference between s and p orbitals increase as 's electrons' are better able to penetrate through the core electron shielding of the atom than electrons in p orbitals - thus the s orbitals are more stable in comparison, being closer to the nucleus.
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    (Original post by EierVonSatan)
    That's a bit more involved, do you know much about orbital penetration or relativistic contraction? Basically the heavier the element the greater the difference between s and p orbitals increase as 's electrons' are better able to penetrate through the core electron shielding of the atom than electrons in p orbitals - thus the s orbitals are more stable in comparison, being closer to the nucleus.
    Hey, i'm not sure if we have learnt that yet, or if we have I missed the lecture! you have been a lot of help! can i ask are you at uni studying chemistry? and if so what year are you in?
 
 
 
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