What Can Be Magnetic?

Discussion in 'Chemistry' started by Orleander, Sep 24, 2009.

  1. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    No problem. I am please to help, teaching and correcting errors when I can is why I post.
     
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  3. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Plasma can be very good electrical conductors (better than copper). All good conductors resist the application of a magnetic field by creating near surface current whose magnetic field is approximately "equal and opposite." In this sense you should call a plasma diamagnetic, not magnetic, if you want to be more accurate. Most people understand "magnetic" to mean ferromagnetic, like iron and plasma certainly is not that.

    If you form the plasma in a region of space that contains a magnetic field already then ions and electrons will gryrate about the field lines (with their "Lamor Frequency" at least until a colision give them an cente of gyration). If you now trun off the current that was making the initial field, perhaps you could call the plasma "paramagnetic" (but I don't think anyone does) as it will induce currents in itself to try to preserve the field internally but it will decay due to the collisons.

    If you drive an electric current thru the plasma, then that current will make a magnetic field like any cuurent does, however, unlike a current in a wire of rigid geometry, the shape of the current path can change - many plasma instalbilites are related to this.
     
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  5. Nasor Valued Senior Member

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    My main source of confusion, I think, is that I was equating "outer shell" with "highest energy" rather than "furthest from the nucleus". However, I still think that the reason the rare-earths make such strong magnets is the large orbital angular momentum contribution toward the magnetic moment. In the famous Nd2Fe14B, for example, almost half of the Nd's magnetic moment comes from orbital angular momentum.
     
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  7. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    All of Nd's would be orbital as all the electrons are paired - net total spin is zero.
     
  8. Nasor Valued Senior Member

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    Nd has four unpaired 4f electrons.
     
  9. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I think that must be wrong. Each added electron goes into the lowest eergy level it can (as you conceptually build up the periodic table of elements) Electron follow the Pauli exclusion principle inthe electric field of the nucleus quantum mechanics first fills the n = 1 shell of He with two electrons of exactly the same electric potenital energy (one being spin up and the other spin down, to keep Pauli happy)

    That process continues for quite a few elements filling the smallest n before going to the next and using s (zero angular moment) electrons before p or d electrons (1 or 2 units of angulat maomentum) but at some point the next lowest unoccupied level is an s electron of a higher n shell than a d electon in the more inner shell. (am not botherning to look up as just illustrating)

    For example a 5s electon may have lower energy level than a 4d electron.
    Now to come more directly to may point: The second 5s electron does too. I.e. it can be spin up and the first spin down in the very same n shell and the very same orbital qunatum nuber (s, p, d, f, ... )

    These two in the same nuclear potential well have the same electrostatic potential energy so they alway fill as a pair. (If there is a nucelar spin, there can be a difficult to even detect energy difference between the spin up and spin down pair - the hyperfine splitng of energy levels by electron spin / nuclear spin interaction. (assuming I am recalling correctly)

    So to summarize: The first and second of Nd's four 4f electrons will both go into the very same electrostatic potential energy state. I.e. are identiacal in their principle quantum numbers, and with angular momentum projection number m = 0 (almost sure of that). The third and fourth also form a pair with the magnitude of m = 1 (As I stated before, I am not sure if that is both m = 1 or both m = -1, or one with m = 1 and the other with m = -1.) If no magnetic field is applied all the m projections have essentially the same energy - See Zeeman splitting.

    In a few cases the energy levels are so close that while the hyper fine splitting still does not matter, the interactions with other electons does. I.e. the second electron in high A atoms may not always find pairing with the first is the lowest next available unoccupied state Pauli allows.

    This is becasue the 4s, & 3d levels have almost idential energy as do the 5s, & 4d states and also the two tirads (6s,4f, 5d) & (7s, 5f, 6d). Note the larger n goes with the s. This is because as is the case with all the rare Earths, with two 6s electrons, the 6s electron spends a lot of its time inside all lessor n shell (inside shells n = 5,4,3,2 &1) as it has zero angular momentum it does NOT orbit the nuclues (to speak in classial terms) but goes straight thru the nucleus, despite being the most outter shell electron - It is sort of like an coment spending a few days near the sun and 100 years far from the sun.

    I don't know if Leonards I Schiff's book Quantum Mechanics is still available but all this is discussed in chapter XI (especailly pages 279 & 278 of the 2nd edition) IMHO, Schiff's book was the best on QM back then.

    WHY WOULD YOU THINK THAT THE SECOND OF TWO ADDED ELECTRONS DID NOT GO INTO THE SAME ENERGY LEVEL AS THE FIRST AND BE ITS PAIR? Are you assuming it will go into an higher energy level and leave the other spin orientation state of the lower level un occupied? Water does not run up hill, nor do electrons - they ALWAYS take the lowest energy level Pauli will allow, not a higher one. I.e. except in a few very rear cases the second takes the same n, the same (s, p, d, or f) and the least m magnitude avilable that the first did.
     
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  10. Nasor Valued Senior Member

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    Are you proposing that the 4 electrons in Nd's 4f orbital will pair up so that two electrons are sharing the same magnetic quantum number when other possible magnetic quantum numbers are unused? Because my understanding is that electrons will always occupy vacant orbitals with different magnetic quantum numbers (but the same principle and orbital angular quantum number) before pairing. The seven different 4f orbitals are all equal in energy, and it's energetically favorable for electrons in the same shell to have the same spin due to exchange interactions, and slightly energetically unfavorable to pair them in the same orbital.

    In other words, in the 4f orbital it's energetically favorable to have two electrons that are m=0, s=1/2 and m=1, s=1/2 than to have two electrons that are m=0, s=1/2 and m=0, s=-1/2.
    On the contrary, I think that

    1) all the possible values of the magnetic quantum number (m) are equal in energy for a given principle quantum number (n) and orbital angular momentum quantum number (l)

    2) it is energetically unfavorable to pair electrons of opposite spin in the same value of m, l, and n

    3) it is energetically favorable to have electrons of the same spin with the same values of l and n but different values of m.

    Edit: My textbook seems to agree with me on this...
     
    Last edited: Oct 6, 2009
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I think your (2) IS FALSE. Whenever there is spin there is both angular momentum and energy associated with it. By paring you can make the net spin zero and REDUCE the energy.

    I believe this is what happens in superconductors - even the bulk free electrons pair into a lower energy state and at low temperatures cannot be thermally busted apart. (One cannot loses any energy in a collison with a rare phonon as the other "pulls it back into sync" (but as I recall they do not need to be physically close - they are "joined at the waist" in "spin space", not our 3D world.)

    I am not so sure about your (3) but especially in the Rare Earths or other with an intrinisc magnet moment I think it false too. I think the various possible m values, normally "degenerate" (same enery) are slightly split as if an external field were applied (sort of an internal Zeeman effect)

    If you can provide some reputable source support for putting only one electron (of the two possible spin orientation ) into each m state then I will assume I have forgotten my facts*, as I am a little uncertain of my memory that tells me the first two of the 4f electrons both have the same m = 0 projection.

    My copy of Schiff got badly water damaged and I only saved some of the most valuable reference pages so cannot now check up with old familiar turf. Wiki must have something on this.
    ----------
    *It has been ~50 years, so I am not doing too badly.
     
    Last edited by a moderator: Oct 6, 2009
  12. Nasor Valued Senior Member

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    I don't have any references handy, but if you google "electron pairing energy" plenty of things pop up indicating that there's an energy penalty for putting two electrons in the same orbital.

    Here's a random one talking about what determines the ground spin state of a molecule (second paragraph from the bottom):
    http://books.google.com/books?id=WL...3#v=onepage&q=electron pairing energy&f=false
     
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Thanks for the link, but they are speaking of molecultes, not atoms. Then the meaning of "orbital" is entirely different from the s, p, d , f orbitals of an atom.

    Molecular orbitals have the axis of the molecule as a spatial reference, atoms do not as they have spherical symmetry. I never knew much about molecular orbitals and there energy relaionships, but can certainly agree that the second electron put into a different molecular orbit (different spatial geometer wrt the molecular axis) will be on average be further from the first and thus at a lower energy than if both are rammed into the same orbital.

    In the atomic case, however, especially in atoms with relatively large magnetic moments, I would think that the "internal Zeeman effect" would remove the "m degeneracy" and putting two (spin up & down) into the same m value would be the lower energy choice.

    I am however will to drop the subject as I suspect we are discussing fine detail of little interest to most, but if you can find evidence that this "internal Zeeman splitting" argument is wrong at least PM me with a link, but to atomic, not molecular states.
     
  14. Forceman May the force be with you Registered Senior Member

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    My input is that the reason solids are ideal for being magnets (excluding the fact that it's much more cheaper and convenient to have a solid bar magnet than a liquid or plasma) is that domains are created and properly aligned to allow for magnetic lines to properly propagate and return to the proper source for reiteration and reiteration. In liquids and gases, magnetic properties are present, but the atoms aren't aligned and therefore the strongest potential for magnetic attraction isn't there.
     
  15. Agonaces Ecbatana Banned Banned

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    Everything is magnetic including water.
     

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