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Why do transition metals have multiple oxidation states?

  1. Feb 27, 2016 #1

    TT0

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    I have googled it and cannot find anything. Apparently the rule that transition metals want full or half-full orbitals is false. I understand why the 4s orbital would be lost but I don't understand why some d electrons would be lost. For example in Mn6+, the electron configuration would be [Ar] 3d1. I don't understand why the Mn atom would lose 4 d electrons but keep 1. Can someone try to explain this?

    Thanks
     
  2. jcsd
  3. Feb 28, 2016 #2

    Borek

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    Staff: Mentor

    In short: "rule" about full or half orbitals is oversimplified, and predicts (if anything) only ground states. Once you come to compounds, you can no longer talk about just the metal. It becomes part of a molecule (even in simple salts it is rarely just a bare ion, typically it is at least hydrated, so it is a complex molecule) and things get more complicated, as it is molecules as a whole that needs to be taken into account.
     
  4. Feb 28, 2016 #3

    TT0

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    I see so there is no high school level explanation as to why there are multiple oxidation states?
     
  5. Feb 28, 2016 #4
    Do you mind if I explain this in terms of potential energy?

    As @Borek said, the metal is rarely an individual entity. It is mostly a part of a complex or a molecule. In such a situation, the energies of the orbitals which surround the metal ion are altered and can no longer be said to be as atomic orbitals. One has to invoke the MOT or (to an extent) the Crystal Field Theory to understand complex salts.

    Now coming to the various oxidation states a metal can possess, this is easily explained by potential energy. @TT0 .. when you were wondering about the oxidation states, did you also wonder why only some of these states are stable? In case you notice harder, you will observe that the so called "half filled" and "fully filled" oxidation states are the ones which are stable. What I am driving at is that the other oxidation states are formed in situ i.e, as part of the reaction. They may be partly stable, but eventually the metal will reconfigure to achieve a more stable oxidation state provided the necessary conditions are present.

    PS: I have not mentioned how potential energy explains these oxidation states. I believe you can figure it out. Think in terms of collison theory of reactions.
     
  6. Feb 28, 2016 #5

    DrDu

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    Just want to remark that any oxidation state is only partly stable. If not, we would not observe any chemistry at all.
     
  7. Feb 29, 2016 #6

    TT0

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    Thanks, I don't really know the answer to
    :(

    I am presuming that potential energy is the bonds. But I am not too sure about the rest and how it explains it. Cheers!
     
  8. Feb 29, 2016 #7

    DrDu

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    I think much can be explained by simple stochiometry. The electrons from the transition metal have to be taken up by some other atom. Hence the oxidation state will depend on the number of electron acceptors. This reasoning can be extended to a thermodynamic reasoning. I.e. the oxidation state will depend on the chemical potential of both electron donors and acceptors in the reaction mixture. They will depend crucially on concentration. Finally, also take in mind that different oxidation states are not peculiar to transition metals. For example for nitrogen, every oxidation state ranging from -3 to +5 has been observed in simple compounds made up of only N, H and O. Same for Sulphur or Phosphorus.
     
  9. Feb 29, 2016 #8
    Yes, by potential energy I do mean the bonds. When two reactant species collide, the product formed will be such that it has the least possible PE configuration possible for the given conditions. I specify conditions here because otherwise all we would observe was chargeless species all the time, or just one single oxidation state. For example, consider the Fe(IV) ion. It exists only in highly acidic medium (pH=0). As soon as the medium turns alkaline, it reconfigures to Fe(III). There are other conditions of temperature and solvent involved as well (both are highly important to determine PE because PE is partially dependant on the KE with which the reactants collide).

    Thus, the conditions which I have specified include, but are not limited to, reactants, temperature and solvent. I believe @DrDu could a bit more on this for the benefit of us all.
     
  10. Feb 29, 2016 #9

    DrDu

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    Yes, I take your example of Fe(IV) and Fe(III). In an acidic solution there are many competing electron acceptors, namely ##\mathrm{H_3O^+}## and few potential electron donors, namely ##\mathrm{OH^-}##. Hence Fe(IV) is stable because there are few reducing species as ##\mathrm{OH^-}##. This can be made quantitative looking at the redox potentials of the relevant species. The redox potential is proportional to the chemical potential I mentioned earlier. Have a look here where the stability regions of different compounds containing elements in different oxidation states is discussed as a function of pH:
    http://chemwiki.ucdavis.edu/Textboo...4:_Electrochemistry/24.4:_The_Nernst_Equation
     
  11. Feb 29, 2016 #10
    The Nernst Equation is very convenient for judging the feasibility of a reaction as it takes into consideration all the aspects I mentioned earlier - temperature, solvent and reactants. Cheers, DrDu...! Hope @TT0 understood the discussion.
     
  12. Mar 1, 2016 #11

    TT0

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    I see thanks guys, I think I am getting it a bit :P
     
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