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Chemistry Question.

lil_killer129lil_killer129 Posts: 4,706Registered User Senior Member
edited August 2007 in AP Tests Preparation
9.10
Which of the following molecule has the largest dipole moment?
A. XeF2
B. XeF4
C. PF5
D. SF4
E. SF6


How do you do this question?
Post edited by lil_killer129 on
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Replies to: Chemistry Question.

  • chris07chris07 Posts: 530Registered User Member
    I don't have the periodic table in front of me, but dipole moment is created by:
    -electronegativity
    -molecule shape
    The compound with two elements furthest away from each other will be the most polar as long as it doesn't have a symmetrical structure (this causes the polarity to cancel out). All of those are symmetrical except A. XeF2, but I may be wrong.
  • PTynanPTynan Posts: 3Registered User New Member
    D. is also not symmetrical.

    And S gives up electrons easier than Xe.
  • lil_killer129lil_killer129 Posts: 4,706Registered User Senior Member
    LOL. You made an account.
  • chris07chris07 Posts: 530Registered User Member
    Ok, it's D then.
  • afruff23afruff23 Posts: 1,966Registered User Senior Member
    Ok, I haven't done molecular geometry in a while, but I am 100% sure about this:

    When I say symmetrical, I mean that each dipole moment is counterbalanced by another equal dipole moment. This can best be explained by drawing a diagram of PF5, the molecule is not symmetric but has balanced dipoles.

    As far as this definition, they are all symmetrical, except for D. SF4. The central S atom has a lone pair of electrons that isn't countered by another lone pair of electrons pulling in the opposite direction.

    If you can't visualize lewis dot diagrams in 3D, then just type in the molecule name in Google images (i.e. PF5 or XeF4).
  • afruff23afruff23 Posts: 1,966Registered User Senior Member
    Just so we are clear: XeF2 is symmetrical (according to my above definition).
  • GoldShadowGoldShadow Posts: 6,160Registered User Senior Member
    That's correct. D is the only one that is asymmetrical.
  • fhqwgads2005fhqwgads2005 Posts: 944Registered User Member
    draw the lewis dot structures. if it is "lopsided" or asymetrical, it has a dipole moment because the electrons are being pulled towards the more electronegative atom.
  • lil_killer129lil_killer129 Posts: 4,706Registered User Senior Member
    I have a question about intermolecular forces.

    1.Why is this statement FALSE? "Van der Waals forces account for the cohesive energy of an ionic liquid.

    2. What intermolucular forces exist between Ne?
  • jenksterjenkster Posts: 842Registered User Member
    Ionic liquids exhibit ion-dipole and ion-ion interactions, both of which are stronger than Van der Waals, so no, it does not account for the "cohesive energy" (just wondering, what exactly is that? never heard of it before).

    2. Van der Waals
  • afruff23afruff23 Posts: 1,966Registered User Senior Member
    Yeah, "ionic liquid" sounds weird. I looked it up:
    http://en.wikipedia.org/wiki/Ionic_liquid

    This doesn't sound like something you go over in a hgih school chemsitry or even AP chemistry course. But if this is what you meant, then the answer is "Ion-ion forces account for the cohesive energy of an ionic liquid".

    My source: http://acs.confex.com/acs/glrm06/techprogram/P26544.HTM

    "the dipole moment of an ion is not a well-defined quantity, and cannot be applied to ionic liquids"

    If you actually meant an ionic compound dissolved in a polar solvent (e.g., water), then I am not completely sure due to the wording. Cohesiveness is the sticking together of similar species, so does that mean the answer is dipole-dipole? If it is dipole-dipole, then why mention "ionic"?

    I believe it meant ionic liquid as defined by the wikipedia link.

    FYI, Van der Waals force is the same thing as London Dispersion force, so don't be surprised if the answer key says LD force. I'm not sure, but I think the more common answer is LD force to play it safe.
  • lil_killer129lil_killer129 Posts: 4,706Registered User Senior Member
    I have this flow chart in my textbook for comparing intermolecular forces. You may have seen something similar in yours.

    One of the questions asks:
    Are hydrogen atoms bonded to N, O, or F atoms?

    If NO, then dipole-dipole forces are present.
    If YES, then hydrogen bonding is present.


    With that in mind, here is the question:
    Which intermolecular forces are important in the carboxylic acid?
    1. LDF only
    2. LDF and dipole-dipole
    3. LDF, dipole-dipole, H-bonding
    4. only dipole-dipole
    5. only H-bonding

    In carboxylic acid H is bonded to O, therefore, I would answer YES to the question in the flowchart. Hence H-bonding is present. LDF is present in between ALL molecules. So I would come the conclusion that carboxylic acid has LDF and H-bonding. But the correct answer is actually 3 which includes "dipole-dipole." Why are there both dipole-dipole AND h-bonding present? I thought I can go through the flow chart once answering EACH question ONCE.
  • wxmannwxmann Posts: 855Registered User Member
    The flowchart is wrong that H-bonding involves H w/ a N, O, or F atom WITH A LONE PAIR.

    O is much more electronegative than H, thus there is a dipole. The electrons are pulled toward the oxygen in the OH group at the end, AND there's a double-bonded O to the C, also pulling on the electrons. Thus the H has a partial positive charge, thus the affinity for the H to break off and become a H+ ion, thus the nomenclature of "carboxylic acid".

    Also, I'll add that H-bonding is a special case of dipole-dipole. So that makes 3. the automatic answer.
  • GoldShadowGoldShadow Posts: 6,160Registered User Senior Member
    Actually, the flowchart is right in that regard. Any H bonded to F, O, or N will have hydrogen bond interactions, even if the molecule has no net dipole (ie, even if there are no lone pairs/unbalanced lone pairs).
  • lil_killer129lil_killer129 Posts: 4,706Registered User Senior Member
    So if there is H-bonding, there automatical dipole-dipole interactions as well?
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