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Electron transfer and oxidation state

  1. Dec 25, 2016 #1
    When beryllium donates a pair of electrons to oxygen, that's oxidation, and the metal is Be(II).

    When carbon and oxygen share 6 electrons, 2 from carbon and 4 from oxygen, this is also oxidation. Of carbon. Even though it got more electrons, and even, I heard, has a negative charge.

    When carbonyl donates a pair of electrons to nickel, that is called adduction (or complexation?), and nickel stays at 0.

    I heard that oxidation states are a bit of voodoo, and can accept the CO thing. But what is so special about nickel carbonyl? Why is this reaction not considered a redox?

    As I understand, the only difference between this bond (dative covalent) and the covalent bonds in, say, water is that nickel doesn't share electrons of its own. Is that the critical piece?
  2. jcsd
  3. Dec 25, 2016 #2


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    You heard wrong. Oxygen pulls electron density from carbon, leaving a partial positive charge on carbon. Formally, oxygen almost always has an oxidation number of -2, especially when bonded to carbon. So the reaction of oxygen with carbon-containing molecules will generally result in oxidation of carbon.
    Basically. It's kind of like adding F- to BF3. The oxidation states don't change. Instead the fluoride anion donates electrons into the empty p orbital of the boron. These types of interactions are best described as Lewis acid/base interactions instead of redox reactions.
  4. Dec 25, 2016 #3
    What do you mean by empty p-orbital? Free boron is 1s2 2s2 2p1, while oxidized boron is also 1s2. So if an electron goes to boron, would it not be on 2s? Is 2s skipped? Why?

    Also, here's what Wikipedia says about CO:

    "This causes a C ← O polarization of the molecule, with a small negative charge on carbon and a small positive charge on oxygen."
  5. Dec 25, 2016 #4


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    Boron in BF3 is sp2 hybridized, and the 3 sp2 orbitals are filled, leaving the higher energy p orbital empty. This makes BF3 a good electron pair acceptor. This property is known as Lewis basicity.

    CO is an oddball. It has a tiny dipole moment. In fact, until recently there was quite a debate in the literature about the direction of this dipole moment, for the very reason that every other carbon-oxygen bond ever observed is polarized with a (usually substantial) positive charge on carbon.
  6. Dec 26, 2016 #5
    So I had two problems, one being that I didn't know what a hybrid orbital is, and the other is that I forgot that the bonds in BF3 are covalent.

    So it actually goes up to 2p4, although it sees the electrons less frequently thanfluorine does. And then the 2p4 mix with 2s2 to form 2 sp2 orbitals, but this still leaves a free p orbital.

    Is any of this wrong?
  7. Dec 26, 2016 #6


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    I don't understand what this means. What goes up to 2p4? Boron? The interaction of boron with three fluorines is such that mixing (hybridizing) two of boron's three 2p orbitals with the 2s orbital results in a lower-energy compound than not hybridizing them. The leftover p orbital is empty and can accept an electron pair.
  8. Dec 26, 2016 #7
    I meant, that's what it would do if hybrid orbitals didn't exist. It just helped me picture what's going on when I wrote it that way.
  9. Dec 27, 2016 #8
    Look up LCAO theory. It will explain a lot. Don't confuse atomic orbitals with molecular orbitals. Metal complexes belong to ligand field theory.
  10. Dec 31, 2016 #9
    Why is it that this empty p-orbital in BF3 has to be filled by a base, and not a reducing agent like sodium?
  11. Dec 31, 2016 #10


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    Why do you think it wouldn't react with sodium?
  12. Dec 31, 2016 #11


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    So I got to thinking about it and this question is quite a bit deeper than I had originally anticipated. First off, Na does react with BF3, but it gives you NaF and elemental boron. This is a redox reaction (more on that in a sec). But if you compare the interaction of Na with the empty p orbital of BF3 with, e.g., the interaction of NH3 with BF3, you see that in the former case, you would have a one electron bond, whereas in the latter case you'd have a two electron bond. So you'd expect the latter to be the stronger interaction. That said, I'm not sure exactly what the mechanism is for the reaction of sodium with BF3, that is, whether it forms a one electron bond briefly before abstracting a fluoride ion. And I don't have my home computer set up currently to run ab initio calculations on the system. But it could be an interesting computational project.

    Anyways, all that aside, to answer your real question, it's not always obvious what's a redox reaction and what's a Lewis acid-base interaction. The easy answer is that oxidation states change in a redox reaction but not necessarily in a Lewis acid-base reaction. So for example, when Na reacts with F, the Na atom loses an electron to become Na+ and the F picks one up. In the process, their oxidation numbers change. Whereas an NH3 electron pair interacts with an empty orbital on BF3 and there's quite a bit of sharing of electrons that goes on, without a change in oxidation states. But even this is kind of a cop out because organic oxidations (like alcohol to ketone) also feature lots of electron sharing, but with changes in oxidation states. The basic answer is that redox reactions involve electron transfer from one group to another, while Lewis acid-base reactions involve the sharing of electrons between species (not necessarily with transfer).
  13. Jan 3, 2017 #12


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    That explains it all. Oxidation numbers were invented as an accounting device for balancing redox reactions. There is no measurable property of an atom that we can check to see what its oxidation state is. Making predictions based on some randomly assigned numbers* is asking for troubles :wink:

    *yes, there are some arbitrary rules used for assigning these numbers, but they can be changed and the system will still work, it just has to be consistent with the accounting.
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