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Final Valence Question, I hope.

  1. Dec 10, 2004 #1
    Besides, the fact that I don't understand why atoms with three shells can be stable even without filling their D orbitals, I have finally come to one final question.
    Why do atoms gain a special stabilization, when the entire valence shell is filled?
  2. jcsd
  3. Dec 11, 2004 #2


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    With a full p-subshell - all the electrons are paired. For the details, one will have to immerse oneself in the Quantum (Wave) Mechanics of atomic (electron) orbitals.
  4. Dec 11, 2004 #3


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    In case you didn't get this before: Only some atoms (the smaller ones, as explained in the other thread) gain additional stability by fully filling the valence shell. The bigger atoms attain highest stability by having 8 (not 2N^2) electrons in the valence shell. In these atoms, it is IMPOSSIBLE to fully fill the valence shell, because it is energetically advantageous to start filling a higher shell before completely filling the next inner one. This is because, the energy of an electron goes roughly like the square of (n+l).

    So, the question should then read :
    This is a very sensible question to ask, but unfortunately, is a little hard to satisfactorily (and classically) answer. I'll need some time to think about it, before I make an attempt. Perhaps Zz, Astronuc or someone else might find a good way to resolve this first.
    Last edited: Dec 11, 2004
  5. Dec 11, 2004 #4


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    There are two rules that you will get used to with time (and I'll attempt to explain their origin, further down) :

    Rule #1 : The energy of an electron increases with increasing n+l, where n is the shell number, and l is the sub-shell (for s, p, d, f, l = 0, 1, 2, 3 respectively) number. (Note : there can be conditions when this rule gets slightly altered)

    Rule #2 : When two subshells have the same n+l value (like 2p and 3s), the one with the lower n has a lower energy.

    From these above rules (which are outcomes - albeit approximate - of QM calculations), one can determine the order in which sub-shells are chosen for occupancy.

    This is the order (verify using above rules, and correct me, if I've made a mistake) :

    1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < ...

    Argument for the Origin of the above Rules :

    Let's look at what happens to the energy of the atom/ion when you add electrons. There are two effects that dictate the changes in energy.

    (i) Initially, adding more electrons to a given shell reduces the energy, because it increases the attractive force on the nucleus (which has a negative sign). Now, at the same time, this addition will also increase the repulsion (positive sign) between the electrons themselves, but this increase is found to be smaller than that with the nucleus. Since the reduction in energy (due to increased attraction from the nucleus) dominates the increase in energy (due to increased repulsion between electrons), the net effect is a reduction of energy - or an improvement of stability. However, as more electrons are added to the shell, the shell starts to get more crowded, so after a point, the repulsion between the electrons becomes more dominant. As a result, further addition of electrons beyond a particular number causes the energy to go up. Typically, this starts to happen whenever you've just filled (or sometimes, exactly half-filled) a sub-shell. Sometimes, this increase can be small, but other times it can be large.

    (ii) Adding an electron into a new shell increases the energy too. Since this new (bigger) shell is farther away from the nucleus, the increase in attraction to the nucleus is smaller (remember, the electrostatic force is inversely proportional to the square of the distance between the charges) and so the repulsion is more dominant, causing the energy to go up (and usually by a significant amount).

    So at certain points in the filling process (usually when a sub-shell has just been filled), you have to make a decision between filling the next sub-shell in the same shell (which increases the energy) and going to the next shell itself (this too increases the energy). The decision depends on which route causes a smaller increase in energy. Sometimes, it's better to go to the next subshell and at other times it's preferable to go to the next shell. Stated differently, it is possible for a lower subshell in a higher shell to have a lower energy than a higher subshell in the same shell. The energies of these sub-shells have been calculated using QM and the essence of the results of this calculation is conveyed by the 2 rules above.

    (While this explains why the filling order follows some complex pattern, it does not explain why this pattern can be described (approximately) by the above rules. To understand this, you must know how to do the calculations.)

    Let's now apply the rules to the elements to better understand how they work.

    Start with no electrons, and keep adding one at a time, putting the first electron in 1s. The addition of the second electron to 1s reduces the energy as explained in (i), so He (=1s^2) is more stable than H(=1s^1). Now the next electron must go to 2s, causing the energy to increase (by quite a large number actually). Thus, Li (=1s^2 2s^1) is more unstable than He. So, He sits at something like a stability peak (Duet/Fully Filled Shell Rule). Adding a second electron to 2s makes Be (=1s^2 2s^2) more stable than Li. But adding the first electron to 2p causes the energy to increase, though by a small enough amount that this is still preferable to 3s. So B (=1s^2 2s^2 2p^1) is marginally more unstable than Be. Adding more electrons to 2p makes C (=1s^2 2s^2 2p^2), N (=1s^2 2s^2 2p^3),..., Ne (=1s^2 2s^2 2p^6) more and more stable. The only deviation from this trend is that O (=1s^2 2s^2 2p^4) is actually a little bit more unstable than N, because N has an exactly half-filled 3p sub-shell (the reason for this increase in energy beyond half-filling has to do with the spins of the electrons). Now, as expected, adding an electron to Ne involves starting a new shell, making Na much more unstable than Ne. So, Ne occupies the second major stability peak. But keep in mind that, along the way, Be and N also occupied little peaks themselves.

    (Check the bar graph at this link to confirm that the above analysis is indeed correct. The ionization energy is a measure of stability)

    Now finally, we can show why the Octet Rule works, based upon the ordering of sub-shells as determined by the above two rules. Let's list the order again :
    First 1s is filled (Duet/Fully Filled Shell Rule) .

    Then 2s and 2p are filled completing the second shell ((Octet/ Fully Filled Shell Rule)

    Next, 3s and 3p are filled. But the next electron goes to 4s (a new shell), making a large drop in stability. So, Ar (=1s^2 2s^2 2p^6 3s^2 3p^6) sits at a stability peak with 8 electrons in the valence (n=3) shell (Octet Rule).

    Similarly, the filling of 4s, 3d (which is more energetic than 4s, but better than 4p) and 4p gives us Kr (=1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6). But since the next electron goes to 5s (a new shell), Kr sits at another peak with 8 electrons in the valence (n=4) shell (Octet Rule).

    Checking the others in the list shows that the Octet Rule applies to all atoms beyond He.

    Further Insights :

    This pattern (which is a result of the two rules) suggests that the d sub-shells are energetically expensive, because whenever some {N}p (n+l =N+1) sub-shell gets filled, the next electron goes to {N+1}s (n+l = N+1) rather than {N}d (n+l = N+2), since clearly N+1 < N+2. This is, in fact true. The shape of the d sub-shells makes them relatively unfavorable.

    Conclusion :

    So, in short, the "lower n" rule (#2) ensures that the valence shell gets an Octet (since {N+1}s will not be filled before {N}p, as long as {N}p exists), and the "n+l" rule (#1) ensures that you won't get more than an Octet (since {N+1}s will get filled before {N}d).
    Last edited: Dec 11, 2004
  6. Dec 11, 2004 #5


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    Thanks,Gokul,for providing this great insight to Hund's rules and how they help building the periodic table.(I my school,in the atom and molecule phyiscs course,they were both missing,though i can remember discussing the periodic table... :grumpy: )
    I would have done it myself,had i been able to come up with an english text,but the only one i could find among my documents was in German. :biggrin: And translating it to English would have taken me a couple of days. :yuck:

    Last edited: Dec 11, 2004
  7. Dec 11, 2004 #6


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    It took me all of the two hours following the first post to think and write up the second one. <tired>
  8. Dec 11, 2004 #7
    Well, I greatly appreciate it. I finally understand, this also explains the other post on titanium. That was an EXTREMELY useful and descriptive post, I thank you very much. This clears up all my questions. You must be a physics teacher!
    I do have one question, though, how does adding electrons to the atom increase the attractive force to the nucleus?
    Last edited: Dec 12, 2004
  9. Dec 14, 2004 #8
    Anyone here?
  10. Dec 14, 2004 #9


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    Dual, I think I clarified in my PM to you, that adding electrons does not increase the net force, but each new attractive force vector is associated with a new (negative) energy term (due to the attractive interaction between the nucleus and the electron).

    If you've done some basic electrostatics (usually covered late in high school), you'll understand what I mean by this. If not I can find you some links to look up.
  11. Dec 14, 2004 #10
    I'm not following you.
  12. Dec 15, 2004 #11
    But no good arguments for the origins of these rules were given...

    That Rule #1 is not really correct. In hydrogen or hydrogen-like atoms, the orbital energy is independent of [tex]l[/tex].

    As to the origin of Rule #2: The outer electrons do not feel a pure [tex]1/r^2[/tex] force, but the nuclear charge is partly screened by the inner electrons. Orbitals with high angular momentum are like classical circular orbitals. They do not penetrate this screening charge cloud, and they do not feel as much of the nuclear attraction as the the elliptical orbitals with lower [tex]l[/tex].

    In particular, s-orbitals have a non-zero electron density near the nucleus. That is why the 4s-orbital is lower in energy than the 3d. (At least in the Aufbau; when you look at core binding energies of say silver, the 3d binding energy is higher than that of 4s.)
    Last edited: Dec 15, 2004
  13. Dec 15, 2004 #12


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    Yes, I should have called them "arguments for the existence of the rules". I thought going any deeper would have been befyond the scope of the OP.

    Thanks for the critique. Yes, for Hydrogen-like atoms, E(n) ~ 1/n^2. That includes H, He+, Li++..., where there are no electron-electron interactions. Among these only H is a stable atom.

    I tried to avoid raising screening (or shielding), at this point, because I couldn't think of a good way to explain it (quickly) to a 9/10 grade student (who, I think, has not yet had instruction in electrostatics, and probably has never heard the term orbital angular momentum). In retrospect, I realize this may be an omission on my part to not mention it.

    On the other hand, you don't really have to go to Ag or Au (where the screening effect is pronounced) to see screening. The reason that 2p is higher than 2s (in non H-like atoms) is itself partially due to a screening effect.
  14. Dec 15, 2004 #13
    I just mentioned silver to show that the quantitative details cannot be explained by simple arguments. Screening always results in 3d being higher than 3p, but to know whether 3d is below 4s (as in silver) or not (as in potassium, calcium, iron...) requires numerical computer calculations.

    My impression of th OP's questions is that he wants to "see" a physical explanation of why all these rather arbitrary-looking rules are what they are. But yes, it would be easier to explain things sitting beside him, or making sketches on the blackboard.
  15. Dec 15, 2004 #14


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    My original intention was to only point out the fallacy in the OP's understanding that a fully filled valence shell created stability. But what started out as a 5-line post, went through successive edits to try and convey more explanation, without invoking too many different effects, as well as trying to avoid mentioning words like 'probability', 'charge density' or others such, that would only result in an explosion of more questions.
  16. Dec 15, 2004 #15
    Gokul, I understand the basics, it's just the way you're wording that last part. Here this is an E-mail I wrote up for a friend, I've been searching the answer to covelant bonding for some time now.
    So, it's not going to result in an explosion of questions.
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  17. Dec 15, 2004 #16
    Okay, I think I know what you're saying, but correct me if I'm wrong.

    The protons are attracted to the other atom's electrons, and visa versa. The electrons are attracted to the other atom's protons, and visa versa.
    As the atom gets closer, the repullsiveness of the electrons would grow
    just as much as the attractiveness, cancelling out the force. But, the
    electrons may spin in opposite direction, reducing the repullsive force.
    Meaning, that the attractiveness is larger than the repullsiveness, but
    eventually, the repullsiveness catches up.
    The reason the electrons are in orbitals, is because that's the least
    repullsive form.
    Even in the ball park?
  18. Dec 16, 2004 #17
    Am I right?
  19. Dec 16, 2004 #18


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    Actually, in the discussion till now we only talk about a single atom, and not the effect of bringing atoms together. This gives rise to behavior that really can not be explained classically. Band formation and ligand field effects are quite complex (not to mention, Quantum Mechanical effects), and we have not involved them in the picture so far.

    Electrons are in orbitals
    (i) to reduce the repulsion, and
    (ii) because of certain quantum mechanical rules that they are forced to obey (Exclusion)
  20. Dec 16, 2004 #19
    Geeze, why is this so hard?
    Do you have any links?
    Okay, I understand it all. Only one part I don't understand.
    One second.
    Last edited: Dec 16, 2004
  21. Dec 16, 2004 #20


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    So hard ? Hardly !

    The real problem is way harder than any of us have described here. It involves very complicated math, and a radically counter-intuitive framework of physics that defies visualization. In fact, calculating the energies of the various levels in a single atoms becomes incredibly difficult if you have more than just a couple of electrons. It takes powerful computers to solve for such simple atoms as Li, Be, and B.

    To address your second question, one needs to know a little about your current level of physics education. Have you been tought electrostatics ? Do you know the expressions for force (coulomb's law), and work, and how to use them ?
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