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Sulfur trioxide hybridization?

  1. May 20, 2008 #1
    I'm just curious about how the sulfer's molecular orbitals would be described in sulfer trioxide. I can see that the molecule is trigonal planar, but how is it involved in 3 pi bonds? Also I'm sure that a d orbital or two is involved... but can't sp3d, sp3d2 orbitals only do sigma bonds? Thanks.
     
  2. jcsd
  3. May 20, 2008 #2
    In sulfur trioxide the central sulfur is bound in a trigonal planar geometry with three oxygens. Once you have the fundamental geometry you can determine the orbital hybridization. Since there are three "things" attached to the sulfur and it has no lone pair, it must be sp2 hybridized creating three equal energy orbitals protruding at 120 degree angles from the central sulfur in the same plane. Any pi bonding takes place in unhybridized orbitals...in this case an unhybridized p orbital. If you do the Lewis dot structure you will quickly learn that there is resonance in the three S-O bonds, making each approximately a 1 and 1/3 bond and creating a pi bonding system. My knowledge is limited in the area of molecular orbitals so my answer stops here. Hope this helps.
     
  4. May 21, 2008 #3

    GCT

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    The following link should answer your questions - if you have additional ones post them here.

    http://www.colby.edu/chemistry/webmo/SO3-2.html
     
  5. May 21, 2008 #4
    That link is about the sulfite ion though (SO3 2-) and not the neutral molecular sulfur trioxide. To which is the question referring?
     
  6. May 21, 2008 #5

    GCT

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    You're right - sulfur trioxide is one of those molecules counted as an octet rule anomaly.
     
  7. May 21, 2008 #6
    Ah, I also forgot the formal charge rule, making the central sulfur double bonded to each of the three oxygen atoms to give a formal charge of 0 (6 valence e- - 6 bonds=0), rather than the 1 and 1/3 bond I mentioned in my earlier post. This then gets more into molecular orbital theory, with which I am vaguely acquainted.
     
  8. May 22, 2008 #7

    GCT

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    If you draw out the plane structure without considering the resonance for the sulfite the sulfur has a lone pair thus the basic approximation should be

    (6 valence e - - 2 lone pair electrons - 4 bonds ) = 0

    This was exactly my point - sulfite has a lone pair on the sulfur which is not true for the sulfur trioxide - if my recollection serves me correctly.
     
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