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Why is an atom with a full shell more stable?

  1. Oct 20, 2005 #1
    Hello,
    I am trying to understand why atoms have a tendancy to completely fill their valence shell. What force attracts the electrons in the first place and keeps them from flying off. There is very little explianation for this that I know of besides the 'happy' atom.
    Thanks,
    Scott
     
  2. jcsd
  3. Oct 21, 2005 #2
    The Coulomb force.
     
  4. Oct 21, 2005 #3

    Astronuc

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    As inha mentioned, electrons (- charge) are attracted to the nucleus which contains positive protons (+ charge) and neutrons, which are neutral.

    Electrons occupy specific energy levels ('orbit') in the atom, which is described by quantum mechanics of the atomic structure.

    See - http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qnenergy.html#c1

    http://en.wikipedia.org/wiki/Electron_configuration

    http://www.chemicalelements.com/show/electronconfig.html

    The elements may be arranged in a periodic table, which is organized by the mass, atomic number and electron configuration.
     
  5. Oct 21, 2005 #4

    Gokul43201

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    This is not true. Most atoms certainly do nothing of the sort. It's only some of the smaller atoms (from He to Al) that try to fill their valence shell. Give me an example of an element after Al that tries to fill its valence shell.

    <I'll get back to this when I find more time.>
     
  6. Oct 21, 2005 #5
    I understand that the electrons are attracted to protons, I am just trying to understand why atoms, such as elements in the second row, have a tendancy ,up to carbon, to lose electrons, while at and above carbon until Neon they have a tendancy to gain electrons. Either way the atom is perfectly balanced magnetically, at least in an ideal element, with an equal number of protons and electrons, why should atoms react at all? What stability do the noble gases have that prevents them from reacting, atleast under normal circumstances.
    -Scott
     
    Last edited: Oct 21, 2005
  7. Oct 21, 2005 #6

    ZapperZ

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    It's due to shielding of the nucleus by the inner-shell electrons. When you're just filling up the p and d orbitals, for example, the inner s orbitals, which are spherically symmetric around the nucleus, can blanket the total nuclear charge from this p and d electrons. Since p and d orbitals tend to extend further way from the nucleus, this results in those electrons being very weakly bound and in many cases, these will result in ionic bonds with other atoms with a higher electron affinity.

    However, as you add more electrons to fill up the p and d orbitals, you also add more protons to the nucleus. The nucles has increased in charge, even if the inner shell electrons are still shielding it by a constant amount. So these p and d electrons are seeing a larger attractive coulomb force than before. They now tend to be bound better to the nucleus.

    Zz.
     
    Last edited: Oct 21, 2005
  8. Oct 21, 2005 #7
    I apprecite that, since it improves my understanding, but when using that same logic how is it possible that atoms can be stable with additional electrons. I see that the shielding of orbitals can allow electrons to be lost easily, but with that same concept, how can an atom have a tendancy to completly fill it valence shell.
    -Scott
     
  9. Oct 21, 2005 #8
    I don't think so.
    First, how does that shielding reduce or eliminate the reactivity of full shell atoms?
    Second, let's consider the following:
    Chlorine 1s(2)2s(2)2p(6)3s(2)3p(5)
    Argon 1s(2)2s(2)2p(6)3s(2)3p(6)
    Chlorine forms a zillion compounds,both inorganic and organic.Argon reacts with nothing. Zero known compounds,a perfectly inert atom.
    How come?
     
  10. Oct 21, 2005 #9

    ZapperZ

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    Let's take the filling of the d-orbital. If you start with 1, 2, or 3 electrons in the valence shell, these electrons are highly shielded from the nucleus by the inner shell electrons. In fact, in the transition elements, it is more likely that the 4s orbital begins to fill up first before the 3d for exactly this reason.

    However, as you fill up the 4d orbital, you are also, at the same time, adding more positive charge to the nucleus without significant increase in the radius of the atom, since you're still filling the same orbital. So while there is still shielding, the nuclear charge is getting larger.

    Now there are ALWAYS exception because I've told you the simplest explanation for the GENERAL TREND. There are special things happening as you get towards half-filling, full-filling, etc. The rules are also modified when you go away from isolated atoms and start to allow them to form molecules. So how much more complicated do you want this to get?

    Zz.
     
  11. Oct 21, 2005 #10
    I think I understand you so far. I would like to get deeper, but lets just clarify on your past point. So you are essentially saying that even though, the filling up of the 4D orbital increases the negative charge, because it is within the range of the postive charge of the proton, it can still exsist as an stable atom. It is the photon that communicates this force, correct?
    -Scott
     
  12. Oct 22, 2005 #11

    ZapperZ

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    Not sure why we need to invoke the "photon" here. Let's not make this more complicated than it should, because you definitely do not want to bring QFT into this (people in Chemistry certainly didn't).

    Keep in mind that as you go through the periodic table, you are no only increasing the number of electrons, you are also increasing the number of protons in the nucleus by the same amount, or else the atom will not be neutral. As you are filling an orbital shell, the size of the atom doesn't increase much, since you are filling the same shell. At the same time, both the electron and proton number increases by the same amount to keep the atom neutral.

    This is quite different than fillilng across an orbital, i.e. going from the last filling of 4s and into the 3d. There's a substantial increase in "size" when that happens.

    Zz.
     
  13. Oct 22, 2005 #12

    Gokul43201

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    Actually, during the first process (filling the same orbital) there is a decrease in the size of the atom, because each extra proton added to the nucleus increases the effective nuclear charge.

    Further, there isn't an increase even when you go from filling the 4s and starting to fill the 3d. The 3d is not sufficiently bigger than 4s that it can offset the increase in effective nuclear charge. For instance Calcium (=[Ar] 4s2) is about 15% bigger than Scandium (= [Ar] 4s2 3d1), and similarly Sr is bigger than Y.

    The ONLY time that going to the next atomic number increases the size of the atom is when you go from a configuration of Ns2 Np6 (the Noble Gas config) to (N+1)s1 (the alkali metal config). This is the only time that the new electron gets added to the next shell, which is big enough to offset the increasing effective nuclear charge.
     
  14. Oct 22, 2005 #13

    ZapperZ

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    The average position of the 3d is larger than 4s. It is why the 4s fills first than the 3d. The 3d orbitals almost have no density near the nucleus, while the 4s does. They also extend significantly further away at their maximum location than teh 4s. If we go by just the principle quantum number alone, one would expect the 3's to start filling first before the 4's. Clearly this isn't true for the transition elements.

    Zz.
     
  15. Oct 22, 2005 #14

    Gokul43201

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    Zz, you misunderstood me. I said "the 3d is not sufficiently bigger than the 4s that it can...". I didn't say it wasn't bigger.
     
  16. Oct 22, 2005 #15
    So you are saying that for an atom like Lithium, because the next electron begins in the next shell, it is sufficently far enough from the nucleus of the atom, that it is not recieving a very strong postive charge, which could result in the forming of an postive ion. However, as the atom increases in its atomic number, the expansion of the shell, is minimal and allows the postive charge to be recieved at a higher level, allowing such atoms as Flourine to gain 1 electron. Now does this same priciple have to do with the strange orbital fillings in transition elements? Does it have to do something with the configeration of the d orbital?
     
    Last edited: Oct 23, 2005
  17. Oct 22, 2005 #16

    Gokul43201

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    I've answered this and similar questions repeatedly in chemistry. You can do a search there. Here's the latest related thread : https://www.physicsforums.com/showthread.php?t=91938
     
  18. Oct 22, 2005 #17
    Just to confirm before I am done with this question, the point at which Fe can be most stable is essentailly the half full shell, since the atom would become to unstable if it were to take on a noble gas-like configeration. When one starts dealing with the d orbitals, the locations of stability vary since, these orbitals overlap quite a bit. The fact of the filling up of the half orbital with 3d with 5 electrons and 4p with 1 is a result of negative charge repelling? Since the energy levels are not that defined, within these two energy levels wouldn't be easy for the force of repulsion to push the electrons away from each other creating this organization? Thanks for all of the help you have given me so far.
    Scott
     
  19. Oct 24, 2005 #18
    I think I understand now anyways and I appreciate your guys' help greatly (finally my first thread ending without me being massacred.)
    -Scott
     
  20. Oct 24, 2005 #19

    Gokul43201

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    I'll respond to the one question in this post :

    1. I think you wanted to say "4s" instead of "4p"
    2. Yes, the reason is Coulomb repulsion. The half-filled orbital configuration is relatively stable because adding an extra electron forces pairing to happen - and that's something that comes with a large Coulomb energy cost.
     
  21. Oct 25, 2005 #20
    Thanks once more Gukel43201 and ZapperZ.
    -Scott
     
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