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Geometric / Optical Isomers

  1. May 6, 2007 #1
    For the molecule [tex]\textrm{Ni}\left(\textrm{OH}\right)_{2}\textrm{Cl}\left(\textrm{NH}_{3}\right)_{3}[/tex], how do you determine the number of geometric and optical isomers? I first drew the molecule with a central [tex]\textrm{Cl}[/tex] bonded to three [tex]\textrm{NH}_{3}[/tex] molecules and a [tex]\textrm{Ni}[/tex]; the [tex]\textrm{Ni}[/tex] was in turn bonded to two [tex]\textrm{OH}[/tex] molecules. From this, it appeared that no matter how I moved the atoms surrounding the central atom, the configuration would always be the same; it would just be rotated counterclockwise or clockwise.
  2. jcsd
  3. May 7, 2007 #2


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    This is a nickel (III) octahedral complex. Try drawing nickel with a square planar bipyramid around it. There will be four groups in the square plane and two other groups in the axial positions. Switch out the groups to determine how many isomers you have.
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