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VSEPR t-shaped geometry vs trig. planar

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data


    i'm confused as to why a molecule with 3 bonding pairs and 2 lone pairs takes on a t-shape rather than a trigonal planar shape.


    My notes say that this is because in a t-shape, there are less 90 degree angles between the lone pairs and the bonding pairs than in a trigonal planar shape. However, i thought that the repulsion between lone pairs was greater than repulsion between lone-bonding pairs, so i would have expected the lone pair electrons would want to get as far away from each other more so than they would want to get away from the bonding electrons, and in trigonal planar they are as far away as possible from each other.




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2013 #2
    This is a really good question that I can't answer but I hope someone can. Maybe it has to do with hybridization?

    By trigonal planar I'm assuming you mean that the molecule has trigonal bipyramidal geometry but places the lone pairs at 180 from each other?
     
  4. Apr 23, 2013 #3
    yes that is what i meant. I basically want to know why the lone pairs go to the equatorial positions in a trigonal bipyramidal shaped molecule instead of going to the axial positions. And then, whatever the reason, why doesn't this happen if the molecule is octahedral? When the molecule is octahedral the lone pairs go to the axial position, not the equatorial positions. Seems inconsistent to me! :)
     
  5. Jul 8, 2014 #4
  6. Jan 7, 2016 #5
    Hmm... I think I get it. In the video, he says it's because we are trying to minimise the number of 90 degree angles between the lone pairs and the bonding pairs. With axial lone pairs, he says, there are many more 90 degree angles.

    At first I thought, surely it is six either way? With axial lone pairs, there is one 90 degree angle per lone pair / bond pair. With equatorial lone pairs, it's still one 90 degree angle per interaction.

    But I think I understand that the point is in the second case, the bond angles are not 90 degrees: they are bent slightly wider. Is that it?

    But that leaves me wondering why the same thing doesn't apply to square planar.
     
    Last edited: Jan 7, 2016
  7. Jan 8, 2016 #6

    DrDu

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    In the t-shape you have 1 lone - lone interactions with 120 deg, 4 lone - bound interactions with 90 deg, 2 lone - bound interactions with 120 deg, and 2 bound bound interactions with 90 deg.
    In the trigonal shape you have 6 lone - bound interactions with 90 degs, and 3 bound-bound interactions at 120 deg.

    But the repulsion between two bond pairs at 120 degrees will be much smaller than at 90 degrees. So you can concentrate on the interactions of pairs at a 90 degree angle.
    For the t-shape you have 4 lone-bound and 2 bound- bound interactions compared to 6 lone-bound interactions in the trigonal shape.
    Hence the t-shape is cleary energetically more stable.
     
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