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Figuring molecule shapes and polarity

  1. Mar 16, 2006 #1
    I am having a lot of trouble understanding the following:

    1) How do I determine if the shape is a trogonal planar? Is it because it would have more bonds than the tetrahedral?

    2)Is figuring the polarity of a molecule only based on the electronegativity?

    Please help! I have a big test tomarrow and I need ot understand this material!
     
  2. jcsd
  3. Mar 16, 2006 #2
    Seriously, it would be amazing if someone could help!!!!
     
  4. Mar 16, 2006 #3

    mrjeffy321

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    1) How do I determine if the shape is a trogonal planar? Is it because it would have more bonds than the tetrahedral?
    Trigonal planar structures follow the form of AX3, they have 3 bonding pairs of electrons and no non-bondin pairs. An example would be BF3.

    2)Is figuring the polarity of a molecule only based on the electronegativity?
    No, it is also based on geometry. There, are instances when the atoms in molecules have large enough electronegativity differences to make them polar, but due to their geometry, it all cancels out and it is non polar. A good example of this would be Carbon Monoxide and Carbon Dioxide. CO is polar due to the electronegativity differece between C and O, but in CO2, since the Oxygen atoms are located opposite eachother, the polarities cancel out and it is nonpolar.
     
  5. Mar 16, 2006 #4
    Great!! That was all I needed! Thanks!
     
  6. Mar 16, 2006 #5
    Just to add on, non-bonding or lone pairs can also affect the geometry of the molecule. Take for instance XeF4. While one would initially think it would be tetrahedral, but when one counts up the electrons (36 in this case) and places them on Xenon (it can expand its octet, since it is period 3 or beyond). In actuality, this molecule is octahedral. The Fluorine atoms take on equatorial positions and the two lone pairs take up the axial positions. The bond angles for this one, since it is octahedral, are 90 degrees instead of 109.5 for tetrahedral.

    In terms of AXn notation, XeF4 is AX4, but the hybridization is sp3d2 due to the lone pairs.

    Also, just in case you didn't know, cis- molecules tend to be polar compared to trans- molecules. If you draw out a molecule that has a cis- or trans- distinction, you'll notice that there is a net dipole moment on the cis- molecule, as the placement of the atoms in the trans- molecule cancel any charge out.

    Good luck for your exam. Hope that helps :)
     
    Last edited: Mar 16, 2006
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