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Lone pairs and dipole moments

  1. Jun 13, 2004 #1
    Do lone pairs in a molecule have dipole moments? and does net dipole always go in the direction of the lone pairs?
     
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  3. Jun 13, 2004 #2
    you can't have an electric dipole without a net positive pole and a net negative pole, so to talk about a lone pair having a dipole moment doesn't entirely make sense.

    but, a lone pair will most definitely contribute to a dipole moment as long as there isn't another lone pair to "cancel it out" symmetrically.

    as for the second part, a lone pair would give net negative charge to that part of the molecule, so the dipole has a good chance to point in the general direction of that lone pair, the actual net direction of the dipole moment would be perturbed, though, by the effect of the whole charge surface of the molecule.
     
  4. Jun 13, 2004 #3

    GCT

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    Lone pairs are not accounted for when determining dipole moments, dipole moments pertain to electronegativity, to the bonds involved since polarity pertains to when one element (locality) is electron deficient as a result of a stronger electronegative-neighbor element Relating to polarity. etc...


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  5. Jul 2, 2004 #4
    Do you mean to say that the lone pair has no role in defining the dipole moment vector? None at all?

    Cheers
    Vivek
     
  6. Jul 2, 2004 #5

    GCT

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    Remember that dipole moments always pertains to the relative electronegativity relationship between elements in a molecule and thus lone pairs are usually not pertinent in determining dipole moments unless you bring up more complicated situations; the existence of lone pairs usually influence the charge or oxidation state of an element however this is probably not significantly important for you at this point.

    hope this helps.


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  7. Jul 4, 2004 #6
    An Important Exception: ammonia and nitrogen trifluoride

    Okay here is something I'd like to say:

    Lone pairs decide the geometry and depending on which structure has an optimum distribution of minimum lone-pair repulsions at bond angles less then or equal to 90 degrees, classifications are made (based on VSEPR).

    If water were linear and not bent for instance, its dipole moment would be zero. Since it is bent (two sigma bonds and two lone pairs on oxygen, which has a sp3 hybridization state) its dipole moment is nonzero.

    Also [tex]NH_{3}[/tex] and [tex]NF_{3}[/tex] are important exceptions to the non-participation of lone pairs in dipole moments. You might think that ammonia has smaller dipole moment than nitrogen trifluoride but actually its the opposite way. Why?

    Well, firstly, the lone pair in the sp3 hybridized orbital of the nitrogen atom which is directed towards one end of the tetrahedral electron pair geometry of these molecules has a contribution to the dipole moment of the molecules. Its own "bond" dipole moment is directed away from the central atom, Nitrogen in this case. Now in case of ammonia, the bond dipoles all point towards nitrogen as between N and H, N is more electronegative. In case of nitrogen trifluoride, F is more electronegative than N and so the bond dipoles all point towards fluorine (there are 3 bond dipoles in each case).

    So if you draw a vector diagram and actually calculate the net resultant dipole moment in either case (or reason intuitively) you will find that the lone pair dipole moment in ammonia adds to the 3 bond dipole moment vectors corresponding to the 3 N-H bobnds whereas in case of nitrogen trifluoride,
    this lone pair dipole moment is in the direction opposite to the 3 bond dipole moment vectors corresponding to the 3 N-F bonds and therefore (based on actual calculations) ammonia has greater dipole moment than nitrogen trifluoride.

    So this is an exception to the "rule" or "fact" that lone pairs have no role in deciding dipole moments. In case you are aware of more such exceptions, please post them here...I would be interested to know.

    Cheers
    Vivek
     
  8. Jul 5, 2004 #7
    To add to my last post, my chemistry teacher tells me that the net dipole moment has a magnitude equal to the summation (over i),

    [tex]
    \sum q_{i}r_{i}
    [/tex]
    (r is the distance)

    Simply put, you have to take into account lone electron pairs as well, even though practically their contributions might be small enough to distort geometry/affect dipole moment.
     
  9. Jul 5, 2004 #8
    yeah, i can't see why you wouldn't take into account the entire charge surface

    i mean it's all just vector addition
     
  10. Jul 5, 2004 #9

    GCT

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    Why? The difference electronegativity between N and F are not as great as N and H. Again, dipole moment is a result of net transfer of electrons based on differences of electronegativity, it is similar to oxidation states in ionic compounds in that the latter has an almost complete charge separation, a molecule possesses a so called dipole due to partial charge separation, unlike ionic compounds...dipole moments pertain to oxidation states.

    Saying that lone pairs contribute to the net dipole moment in that it contributes to the shape of the molecule is a bit of a stretch. There's no need to ask if a molecule has lone pairs to determine dipole moments and this was my point from the beginning.

    "does this molecule have lone pairs?" or should we rather ask "what is the shape of this molecule."

    Your argument pertains to a insignificant technicality and not chemistry.

    I hope I did not sound condenscending, I intention is not to discourage, however it is important that when learning chemistry one thinks chemistry and not technicality; such students always seem confused.

    Again, what you want to consider is differences in electronegativity, and not whether the molecule has lone pairs. This should be the conclusion of this discussion.


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    Last edited: Jul 5, 2004
  11. Jul 6, 2004 #10
    Now, from your last post, I do not infer anything technical. And I am damn sure (having read sources and participated in lectures of two different chemistry teachers within the past 7 days) that ammonia has a greater dipole moment than nitrogen trifluroide and this is due to the contribution of the lone pair.

    I too do not wish to argue with you...its just that this case came up in a class discussion and I thought I would bring it to your notice since this subject was already on.

    Cheers
    Vivek
     
  12. Jul 6, 2004 #11
    The dipole moment of a BOND depends on differences in e-negativity between the 2 atoms in the BOND.

    The dipole moment of a MOLECULE is then the resultant vector of all the BOND dipoles. The shape of the molecule determines where all the vectors point and hence the direction of the resultant vector. The shape is a result of orbital hybridization (and VSEPR). Lone pairs contribute to the shape.
     
  13. Jul 6, 2004 #12

    GCT

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    Obviously you have not even vaguely grasped what I said in the last post. Forget it than, just forget it.

    Also, I'm pretty sure that the greater dipole moment of NH3 is due to the greater difference in electronegativity btw N and H...when you add up the vectors, this results in a greater magnitude in the final composite vector.

    Dipole moments = electron distribution? Think again. You may have encounted applications of electron distribution in organic chemistry and thus you may know that it takes into account van der walls attractions also. Electron distributions are NOT dipole moments (it is a much broader category), in general chemistry discussions of dipole moments pertain to chemical bonds.

    If you are to take into account the lone pairs why don't you take into account the electrons in other orbitals; and no they are not in different dimensions.

    The conclusion of this discussion should not be "lone pairs are important in determining dipole moments" but rather "dipole moments are relevant to electronegativity."

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  14. Jul 6, 2004 #13
    Summarizing our ideas,

    Okay, shrumeo has put my ideas into words exactly. This is what I said earlier about lone pairs contributing to the geometry (distorting regular geometry in certain cases).
     
  15. Jul 7, 2004 #14
    Ok,

    This quote is from Atkins second addition "Physical Chemistry"

    "When two charges q and q- are separated by a distance R they constitute an electric dipole of magnitude qR. The dipole has a direction as well as a magnitude, and it is conventional to regard it as being directed from the negatively charges end toward the positive."

    i would say that there is a del-negative charge around a lone pair

    from another page:
    "Once the electronegativities have been established they can be usedto estimate the polarity of the bond, dipole moments, and even the bond strengths themselves."

    e-negativities are used to determine BOND dipole moments

    from another:
    "Dipole moments can be measured in different ways..."
    I'll summarize about 2 pages here:
    You can do capacitance measurements. You must know the capacitance between the two plates of your instrument. When the charges have been established there is a potential difference between the plates. The potential difference can be written in terms of the charge density and the separation of the plates. You measure the capacitance with a vacuum and compare it to the capacitance with your sample in there. This will give you the relative electric permittivity. The next step is to relate that to a molecular property. This is acheived by introducing the polarization of the dialectric. This polarization is the charge per unit area of the sample. It is also the average dipole moment per unit volume. The individual dipoles of the medium (sample) align with the applied field and give rise to a charge on the surface of the dialectric but cancel in the bulk. The charge surface .....

    crap this too long. Long story short, it's the charge surface of the molecule that is measured to determine the permanent dipole moment of the molecule.

    Atkins, Physical Chemistry, 2nd ed. pg. 768-770
     
  16. Jul 7, 2004 #15

    GCT

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    Its good that you guys are so persistent this however, sometimes you have to simplify. Realize the difference between dipole moments and electron distribution and whenever you decide to pursue an argument I suggest that you don't do it for technical reasons but rather for chemical (chemistry) sense.

    "e-negativities are used to determine BOND---dipole moments"

    My point exactly, there is no other "dipole moments." There is a broader subject, it's called electron distribution both of you have the former confused with the latter term.


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    Last edited: Jul 7, 2004
  17. Jul 7, 2004 #16
    Wha??

    Look here:
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diph2o.html#c2

    They are calling the little arrow the "dipole moment" of the water molecule. It is not pointing along any bonds. It is the molecular dipole moment.

    The dipole moment of a molecule is the result of the electron distribution.

    But, perhaps you are right. "Di"-pole means 2 poles. Using the term to describe the sum of many di-poles is inaccurate. Unfortunately, it's the convention. I've never seen it any other way.
     
  18. Jul 9, 2004 #17

    GCT

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  19. Jul 13, 2004 #18
    You're right about the difference between dipole moments and electron distributions. They are indeed different and as you consider (precisely) the cases in complex molecules, it is not possible to give a plausible and correct statement using oversimplified ideas of dipole moments, but I don't think we've been doing that. Nevertheless, I believe that if you analyze things from a physics-centered perspective, you will reach conclusions that shrumeo and I have.

    Cheers
    Vivek
     
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