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Paramagnetic and diamagnetic species

Discussion in 'MCAT Study Question Q&A' started by growingpains, 04.14.12.

  1. growingpains

    growingpains

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    I'm having trouble understanding why molecular fluorine and hydrogen are diamagnetic. In F2, after the fluorine atoms form a single bond with each other, each atom has 6 valence electrons left over, Are 2 of these electrons in the 2s orbital and the other 4 in the 3 2p orbitals? If so, wouldn't the fluorine have an unpaired electron in the 2p orbitals?
  2. pfaction

    pfaction

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    Diamagnetic: everything is paired.
    Paramagnetic: one electron THINK RADICAL is left out.

    So F2 = complete 16
    BrO is paramagnetic: 7+6 = 13
    Fluorine by itself is paramagnetic.
  3. growingpains

    growingpains

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    Ah, so the bonding electrons are included in both fluorines. Thanks.
  4. folktale

    folktale

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    Can you explain how you got 13 for BrO?
  5. growingpains

    growingpains

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    Br has 7 valence e-, O has 6. 7+6=13
  6. folktale

    folktale

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    Then why would F2 be 16, if both atoms have 7 valence e-
  7. chiddler

    chiddler

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    should be 14 i think. fluorine by itself has 5 electrons in the p shell so it is paramagnetic. then you make it F2 so that electron #5 is shared with the other F and the other F shares electron #5 with the first F. Now they both have 6 electrons in their p shell and there is no unfilled orbitals. They are diamagnetic.
  8. folktale

    folktale

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    I think of it as F2 having a total of 18 electrons b/c each F atom will have 9 e- each. Then if you add that up, that's 18. There are 3 atomic orbitals in the p-orbital and 18e- are completely paired up w/ each other.

    EDIT: and even if you just count valence e-, it'd be 7 + 7 = 14. And an F2 molecule will have all its valence e- paired F-F. So it is diamagnetic.
    Last edited: 04.14.12
  9. chiddler

    chiddler

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    yeah that sounds right.

    oops @ my numbers.
  10. folktale

    folktale

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    This is a great topic. Let's keep going with other examples.

    How about CN-
  11. growingpains

    growingpains

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    Alright let me take another stab at this. CN- has a triple bond and both C and N have 8 valence e- by sharing the triple-bond e-. Thus, with 8 valence e- each, the compound is diamagnetic?
  12. folktale

    folktale

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    See this is what I'm confused about.

    I'm fine with single atoms. Like Li would be paramagnetic b/c it has one unpaired electron in the 2s orbital, and Be would be a diamagnetic because it has the 2s orbital completely filled. But I am somewhat still confused on molecules like CN- or any other ones. Anyone wanna take a stab at this?
  13. hellocubed

    hellocubed

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    I'm not quite sure how you came onto the figure 8

    C has 4 valence electrons, nitrogen has 5. They should have 9 valence electrons together. If they had 8 valence electron each they would both be noble gases, and the bond would not be possible.

    9e-, and hence should be diamagnetic...
  14. folktale

    folktale

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    But there's an extra e- because CN- is negatively charged, so overall it would be 10e- correct? and how is 9e- going to result to all electrons paired? Isn't a 10e- going to make CN- diamagnetic? 9e- seems like it would have one unpaired electron, only filling two of the 3 2d orbitals, leaving the last one with an unpaired e-.
  15. hellocubed

    hellocubed

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    Oh wow I screwed up everything on that response. haha

    10 e-, diamagnetic
  16. growingpains

    growingpains

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    When we're looking at paramagnetism/diamagnetism of compounds, aren't we looking at the paramagnetism/diamagnetism of each of the atoms in the compound first? (e.g. If a compound contains all diamagnetic atoms, the compound is diamagnetic, but if the compound contains at least one paramagnetic atom, the compound is paramagnetic).

    In CN-, both C and N have full octets (8 e-) via the triple bond. So would they be both diamagnetic, making CN- diamagnetic? Is this the right approach to paramagnetism/diamagnetism of compounds?
  17. EnginrTheFuture

    EnginrTheFuture

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    http://scramlinged.com/resources/Notes on Molecular Orbital Theory.pdf

    You have to count up the total electrons available (not just valence)... then you have to throw it in the MO diagram shown in the link above. I believe this includes antibonding orbital filling which is why people are getting thrown off attempting the individual fill out method. I do not think we are responsible for knowing how to predict paramagnetism/diamagnetism of molecules, just atoms. Although who knows.

    I do not think you can just take "half the bonding electrons" or all the bonding electrons and analyze each atom individually. (not sure if thats what people were doing up there, I skimmed). One thing is for certain, if you add up all the electrons in the system and it's not an even number --> definitely paramagnetic. Other than that you have to fill out the orbital diagram or have the numbers of electrons needed memorized. This isn't like doing the usual 1s 2s 2p 3s filling!!! These are bonding orbitals, not the usual individual atomic orbitals. Not only that but there are two forms of bonding theory, the molecular orbital theory and valence bonding theory. This is all explained fairly well in the above link.

    If total electrons add up to 2,4,6,8,12, 16, 20, or 22... it's diamagnetic BUT if it's a bond between oxygens or fluorines the orbital filling changes. This gets real nasty and this is why I think your better off sticking to individual atoms and ions.

    I could be wrong and please let me know if this is the case... but from what I remember of g-chem years ago and some searching around, this is what I came up with.
    Last edited: 04.14.12
  18. chiddler

    chiddler

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    for mcat, I don't think we have to know molecular orbital theory, do we?

    after all, for something like cyanide, it would take too long to draw out the orbitals and see if it is indeed para or diamagnetic. plus i couldn't find it on the list.
  19. johnwandering

    johnwandering

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    You are definitely right
    It's not quite as simple as we are making it

    http://www.youtube.com/watch?v=KcGEev8qulA

    O2 and N2 both have even number of valence electrons (Now that I think about it, all real compounds apparently do, or else they would be free radicals. BrO has a free radical, it is supposed to be BrO2)

    But O2 is paramagnetic, even though the Oxygen atoms are DIAMAGNETIC.

    weird~
  20. folktale

    folktale

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    I still don't see how an Oxygen atom by itself is diamagnetic. it would have 4 e- in its 2p orbital, with one of the 2p orbitals filled and 2 2p orbitals with an unpaired e- each.

    And to clarify, do we use the orbitals s p d f for elements/atoms like N and the LUMO/HOMO bonding orbitals for molecules like N2?
  21. EnginrTheFuture

    EnginrTheFuture

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    Check out the link I posted a few up... it should explain everything you are curious about in a fairly easy-to-read 2 page pdf

    cheers!
  22. johnwandering

    johnwandering

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    Ah, so... It has really not much to do with being "even or odd."
  23. EnginrTheFuture

    EnginrTheFuture

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    well if the total # of electrons in the molecule is odd... Boom, you know there is no way of pairing --> paramagnetic. If it's even number though, things get complicated.
  24. folktale

    folktale

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    Alright I get it now. So my statement holds true. If an atom is in question, you can use the regular 1s2s2p orbitals, but if we're talking about a molecule, then MO diagrams are necessary. I'm just going to remembre the 1s, 1s*, 2s, 2s*, 2py-2pz, 2px, etc. Hopefully, the MCAT shouldn't be this complicated.
  25. EnginrTheFuture

    EnginrTheFuture

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    That should do it for most cases, yup. If anything more complicated comes up... everyone is in the same boat to wtf-town
  26. folktale

    folktale

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    Yep. If something comes up that starts dealing with the d-orbital comes up, I'm gonna mark that question and move on lol. Well, I'd guess first at least. But I'm thinking if they're gonna ask something in this topic, it would be like an atom instead of a molecule. That'd be a great discrete.
  27. Dasypus

    Dasypus

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    Do we really have to know any of this stuff? It doesn't seem to be on the AAMC outline....
  28. MrNeuro

    MrNeuro Gold Donor

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    we do need to know how to determine if an atom is diamagnetc/paramagnetic

    we don't need to know mo theory...although it could show up in a passage where it is explained....(TPR had a passage on this)
  29. pfaction

    pfaction

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    Wait, how is O2 paramagnetic?
  30. Dasypus

    Dasypus

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