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A deeper understanding of the octet rule.

  1. Aug 9, 2012 #1
    I believe that many chemistry undergrads have misconceptions about basic concepts such as the octet rule. I wanted to make sure this wasn't the case for me.

    When we are taught the octet rule, we are taught it in simplistic terms of "this atom wants this electron or it wants to get rid of this electron."

    So for example, if you have lithium (1 valence electron) and fluorine (7 valence electrons) lithium can ionically bond with fluorine by donating an electron to fluorine. In this way an electron in a higher energy level is placed in a low energy level and the result is more stable.

    But really, you have to look at the thermodynamics of the process right? For example, what if an atom with a single electron in the valence shell is combined with an atom which has an almost full valence shell that is at a higher energy? Wouldn't the transfer not happen then? Even though a full shell could be achieved, if it must be placed in a higher energy level then it wouldn't happen right?

    So does the octet rule not hold when the principle quantum number n is very different between the two atoms? It should't be about the magic number of a full shell, it should be about what energy levels the electron ends up in compared to where it started. And it just happens to hold for low quantum number atoms (also taking into account lattice energy, etc)?

    Like I said, I'm trying to move away from the simplistic understanding of bonding offered in general chemistry courses.
     
  2. jcsd
  3. Aug 9, 2012 #2
    The octet rule is a simple rules that yields good qualitative predictions.

    Why not? The result would still yield a compound that is more stable than the two isolated atoms. For example, lithium chloride is a compound in which an electron goes from an orbital with a lower n to one with a larger n. The fact that Li+ and Cl- are much more stable than Li and Cl makes up for the energy difference between energy levels. When talking about non-ionic compounds it becomes even harder to rationalize this, since no electrons are "donated".
     
  4. Aug 9, 2012 #3
    But how are Li+ and Cl- more stable then? You are taking a low energy electron and putting it in a higher energy orbital. This is unfavorable, yes? Is the compensating factor the lattice energy then?
     
  5. Aug 9, 2012 #4
    I have no idea. I don't know if you can explain this in simple terms. Chemical bonding is a quantum mechanical phenomenon. Unfortunately the SE is not exactly solvable for molecules with more than one electron, but it turns out that, as I said, these molecules are more stable than their component atoms taken in isolation.
     
    Last edited: Aug 9, 2012
  6. Aug 9, 2012 #5

    DrDu

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  7. Aug 9, 2012 #6
    You are thinking in terms of *atomic* orbitals of *isolated* atoms. That's what the textbook Schrodinger's Equation in 3-D spherical coordinates gives you. However you don't have isolated atoms in a molecule. There's 2 or more nuclei next to each other and their combined potential will no longer yield things corresponding to atomic orbitals. The V(x) is now very different, and we know that the V(x) determines the problem in wave mechanics.

    That means when we talk about molecules, we need to talk about *molecular* orbitals. Qualitatively, it turns out that when molecular orbitals form, their energy is usually in between the higher and lower atomic orbitals. Quantitatively, you need a computer to do it.
     
  8. Aug 9, 2012 #7
    Hello,
    I think electronegativity, molecule structure and a whole bunch of things add up to the energy before and after. Just taking the orbital in account is not enough to conclude that it will bond or not bond. The reaction will happen if the gibbs free energy drops. Your argument is that for example a Li to bond with Astatine the electron will have to go up in energy because Astatine's shell that has room is very high up there. I think the lattice structure will answer this if not some other energy lowering phenomenon that happens when you bond them. Which is what you said in a previous post. SO I THINK YOUR RIGHT !!!=)


    Here is how i see bonding. The octet is just a more convenient and quicker way to see if things will bond. Works most of the time but doesnt have to especially when you get into grey areas like element P, B and C etc.I think even in a NaCl salt the Na are bonded to more than Cl at once or the Cl is bonded to more than one Na at once at least some part of the time. Breaking the octet rule. kinda like resonance but not rlly. The fact that there is a bond strength scale to classify covalent and ionic and mere intermolecular forces introduces a grey area of bonding, meaning there is never a bond and no bond thing like the octet rule implies but instead of how much or how little of a bond it is.


    This is my understanding of it plz correct me if I am wrong, I have never taken quantum mechanic but had part and glimspe of it but this is my impression of bonding from all the schooling I had.
    Thank you=)
     
  9. Aug 9, 2012 #8
    As was said before, you have to look at the energy of a molecular orbital formed from the atomic orbitals of the separate atoms (provided that symmetry allows the formation of the bond). This could lead to a molecular orbital where one or more electrons could occupy a lower energy state.

    Just to be 100% clear though, "orbitals" are just mathematical constructs meant to demarcate basis functions in a quantum chemical calculation all that the above REALLY means is that there is a chance that when you calculate the total energy of the system, it could be lower in the case of the molecule than the separate case. All of THAT is also just a theoretical construct and is meant to explain and rationalize and sometimes predict experimental observations.

    Of course, quantum chemistry is on much MUCH firmer footing than something like the octet rule, which is only uncontroversial and without special cases for the first couple of rows.

    To know if two things will react, the best way is to see if they react.
     
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