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Problems with understanding polarity

  1. Oct 31, 2003 #1
    I am having some problems with understanding polarity.
    I will start the problems one by one, whenever one problem is solved i will go to the next one.
    Ok, here is the first one.
    I am unable to understand why we say that BeF2 is not polar. When i try to analyze it, i think that we should consider it as a molecule with two negative poles, and some positive charge at the middle (and therefore, polar, right ?).
    Any help for me to understand this would be appreciated.
    Last edited by a moderator: Feb 7, 2013
  2. jcsd
  3. Oct 31, 2003 #2
    Be has 2 outermost shell electrons. Each of them combines with one fluorine atom to form BeF2.

    Since the electronegativity between Be and F is different, so a dipole moment is created in each Be-F bond.

    As you can see from the picture, the molecular shape of BeF2 is linear (F-Be-F). Since the molecule is symmetric, so all the dipole moments are cancelled out. That's why it is non-polar.

    dot and cross diagram of BeF2

    PS. You can treat dipole moments as vectors.
  4. Oct 31, 2003 #3
    This is my problem actually.
    Why can we deal with dipole moments as vectors ?
    This is what my teacher said in the class, he said we deal with the dipole moments as vectors, and the two bonds are symetric in BeF2, so the dipole moments cancel out.
    But i don't see WHY do we do this. It would be more logical to actually consider BeF2 like a double poled molecule (with 2 negative poles). Why don't we do that ?
  5. Oct 31, 2003 #4
    BeF2 is electrically neutral.

    Well, in fact dipole mement is a vector quantity, similar to moment(distance x force) in physics. Dipole moment = net charge x distance between the two molecules. The dipole moment of the molecule is given by the vector sum of the dipole moments of all the bonds. If the vector sum is zero, the dipole moment of the molecule is zero and the molecue is described as non-polar. Non-polar molecules are molecules without polar bonds or those with polar bonds symmetrically arranged so that all the dipole moments are cancelled out (just like BeF2 since it is symmetric).

    Electronegativity is defined as the relative tendency of an atom to attract a bonding pair of electorns towrads itself.

    The bonding electron cloud will be displaced towards the more electronegative atom and the result is polar covalent bond.

    F is a very electronegative atom (in fact it is the most electronegative atom among all), thus it likes pulling electrons towards itself. (I think we may treat the pull as a force) The electronegativity of Be is less than that of F. So in the Be-F bond, F tends to pull the bond pair electrons towards itself, so the bonding electrons are not shared equally. Is it ok up till now ?

    Due to the unequal share of electrons, F seems to be more negative than Be in F-Be bond, right? So Be is slightly positive and F is slightly negative and creates a dipole moment between them. (I think we may treat the dipole moment as a force though the unit of dipole moment is Debye(D), not Newton)

    Since BeF2 is linear, so a "force" is pulling to the left and another "force" with equal magnitude is pulling to the right . The resultant dipole moment (force) is zero. Thus it is non-polar.


    edit: typo
    Last edited: Oct 31, 2003
  6. Oct 31, 2003 #5
    I don't mean to be offensive, but what u have said is what i have heard from my teacher, which is what is causing me confusion.
    I think you can't deal with the dipole moment as a force, at least taking into account the fact that the two 'forces' are not acting on the same electrons.
    The first force is acting on the electrons between Be and the first F, while the second force is acting on the electrons between Be and the second F, so how will you actually take the resultant force ? (remember that in physics, to take the resultant force on a body, all forces should be affecting on this body).
    As u have said, F will pull the electrons towards it, and since the two F's on the sides of the molecule will both pull the electrons towards the sides, i think the final molecule should have 2 negative poles (as i have said before).
    Let me rephrase my question, why we don't consider BeF2 to actually have 2 negative poles (and therefore, be polar) ?
  7. Oct 31, 2003 #6
    Yes, you're right that dipole moment does not equal to force (the unit of dipole moment is Debye(D), not Newton) but I just used it as an analogy.

    Well, I understand your question now. Whenever I deal with questions about whether a molecule is polar or non-polar, I only consider its resultant dipole moment. I think your question is a good one, sorry that I can't help.
  8. Oct 31, 2003 #7
    You are right in that if you look at an electron plot of BeF2 you will see to electron rich lobes at the end with an electron deficient center. So there are indeed to dipoles. But the effect that these dipoles have on neighboring molecules is negliegible, since the sum of the dipoles is zero. When we talk about polarity we talk in terms of bulk properties, and BeF2 is no more polar than something without any dipoles at all. There is no greater intermolecular force between BeF2 molecules greater than Van der Waals forces, and as such BeF2 is written off as nonpolar.
  9. Nov 1, 2003 #8


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    Re: Polarity

    I'm going to paraphrase your question just a little bit --- "If we know that there is a non-uniform charge distribution in BeF2, why do we throw the information away by saying it's "non-polar" when discussing dipole moments?" Hopefully, I haven't mis-interpreted you too much.

    There ARE circumstances in which BeF2 is considered as a quadrupole (two dipoles) --- it is symmetric, exhibits no net dipole moment, but if the charge distribution is of interest, it is mapped and analyzed as a quadrupole, or octapole, or hexadecapole (16), ----. As CSF has said, for most chemical purposes, details of the charge distribution are more clutter and distraction from problems at hand than they are illuminating/informative.

    The theoretical types who play games with quadrupoles, 8s, 16s, and so on are looking for fairly subtle effects in intermolecular interaction potential functions, and have never been successful at communicating to me any real insights/understandings --- that could be their problem, or my level of understanding; the other application for multipolar analyses is in spectroscopy --- obviously one gets a better idea of emission and absorbtion spectra from looking at the individual dipoles within a molecule rather than a net zero dipole.
  10. Nov 1, 2003 #9
    OK, this is lot of progress. Thanks.
    Let's continue.
    I conclude from what u have said before (correct me if i am wrong), that a molecule can have dipoles even if its dipole moment resultant is zero, but this will not affect its properties (therefore we say, it is not not a dipole molecule).
    Or in other words :
    The dipole moment resultant gives us an (accurate) idea about whether the molecule has dipole properties in bulks or not.
    If the last statement is right, how did the scientists reach the idea of taking the resultant of dipole moments, why did they choose the moment and not only the charge on the sides of the each bond (ie, on what basis did they chose to take the distance into consideration).
    If i was a scientist, i would have introduced a more complicated term, that would take the distance into consideration accoding to coloumb's law (not simply charge*bond_length )
  11. Nov 1, 2003 #10


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    The coulombic energy of the dipole itself is of less interest in most cases than the energy of interaction of the dipole with an external electric field (dipole moment x field, or charge distance product x potential per distance, is coulomb.volts).

    The coulombic energy of the dipole could be of more interest if QM were more quantitative in describing molecular systems.
  12. Nov 2, 2003 #11
    I still can't understand, why is it that when the molecule has a zero dipole moment resultant, it acts as not being dipole ?
    (i know the question seem meaningless, i am not really able to express my question .. )
    I will try to make it clearer ...
    The way i see it, scientists introduced a certain term, which was the dipole moment for some reason.
    Then, they started to add it as vector, and said that when the resultant is 0, the molecule doesn't have dipole properties. But on what did they base this conclusion ? Why does the molecule that doesn't has a dipole moment resultant of 0 have no dipole properties ? (althought we have seen that it might have 2 negative poles).

    Please if you see the question to be unclear, ask me to (try to) make it clearer.
  13. Nov 2, 2003 #12


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    Actually, STAii, you've got a very interesting point.

    Gedanken experiment: CO2, or trans dichloroethylene, CS2, or some other molecule having a zero "net" moment is placed in an electric field having a non-zero gradient --- such molecules HAVE to move in such a field. Somebody else needs to either back me up on this, or call me a complete idiot before you take it too seriously.

    The "net zero" case MAY (this is NOT an area I've played in a whole lot) be an artifact of methods used to measure dipole moment --- tuck a sample between the plates of a capacitor in an AC bridge, and start looking at reactance vs. bridge excitation frequency; the measurement method is simply too crude to respond to component dipoles of the "net zero."

    Sorry I've turned your question into an even bigger question.
  14. Nov 3, 2003 #13
    Can u please explain this more ? (my main problem is with the words, you see, english isn't my native language).
    For example, what do u mean by the "AC Bridge" ? "Bridge excitation frequency" ?
    Actually, i like to know more and more (but this doesn't always work, unfortunately).

    Thank you.
  15. Nov 4, 2003 #14


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    Are you familiar with the Wheatstone bridge? Rather than me wave my hands, try http://www.dwiarda.com/scientific/Bridge.html

    AC bridges employ the same configuration to compare reactances, complex impedances, and are excited with alternating current.

    Measurement of dipole moments is (was when the dinosaurs ruled the earth, anyway) accomplished by placing samples in a cell containing a set of cylindrical plates, radii 1-3 cm, and spaced 3-10 mm (this is an approximation of a parallel plate capacitor); excitation frequency is varied over one and half to two orders of magnitude, and capacitance for the cell plus sample is determined as a function of frequency.

    As far as the cell is concerned, we're looking at dielectric constant between the plates rather than dipole moment; however, at some frequency, the changing field polarity between the plates is "in tune" with the rotational frequency (plus some hand-waving arguments about the Boltzmann distribution) of the dipoles --- this augments the dielectric effect. Please don't ask me to recall and derive in detail the steps in the data reduction (that's a looonnnngggg time ago for me), but what eventually drops out is a dipole moment --- the method does not actually measure net charge separations and distances, but only the charge-distance product --- that's where THAT came from.

    Given X-ray data, or some other information on bond lengths, people can make the assertion that the charge separation producing the dipoles is localized on a specific bond, divide by bond length, and assign charge values, electron densities, to specific nuclei in a molecular structure --- this is on the ragged edge of OVERinterpreting data.

    Keep at it, STAii --- you DO have the knack for coming up with great questions.
    Last edited: Nov 4, 2003
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