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Why would a carbanion have a tetrahedral shape?

  1. Apr 18, 2014 #1
    In my book they explained that a trigonal planar shape is not good because of the 3 lone pair bond-pair repulsions between the unhybridized p orbitals and the 3 trigonal planar bonds. So it would be more favourable to have the orbitals hybridized into a tetrahedral shape to minimize those 90 degrees lone pair-bond pair repulsions.

    However, when drawing out the speculative trigonal planar shape of the molecule i thought that the repulsions between the dumbbell shaped lone pair and the 3 bonds should actually cancel out. Because the dumbbell shaped p orbital would protrude out above and beneath the carbon atom like this: http://imgur.com/Sr5h2iR so shouldn't the repulsions between the 3 bond pair and the lone pair be cancelled out? And so wouldn't the trigonal planar structure be more favourable for the carbanion?

    But i know this is not true but i can't think of a reason for this. Thanks in advance for the help :)
  2. jcsd
  3. Apr 18, 2014 #2


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    The planar structure is an energetic maximum between the two pyramidal structures with the lone pair above or below the plane of the bonds. At a maximum the first derivative (=force) vanishes.
  4. Apr 18, 2014 #3
    Hi can you emphasize on this I don't quite understand what you're trying to say. Also to avoid the vagueness from the earlier post this is what i mean.

    In the carbanion molecule if it were to have a sp2 hybridized orbitals with the lone pair in a unhybridized p orbital (Scenario 1), why would there be a greater inter-electronic repulsion than a sp3 hybrized molecule (Scenario 2)?

    Because my reasoning is that the p orbital is going to be above and beneath the carbon atom. And so i thought the repulsions should cancel out. And since now the repulsion between the lone pair-bond pair is cancelled out, the repulsions between the bond pair-bond pair are also cancelled out due to the 120 degrees angle. So with all the repulsions cancelled out, shouldn't this shape (sp2 hybridized orbitals with the lone pair in the p orbital) be stable?
  5. Apr 18, 2014 #4


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    You have to distinguish between the energy of a given configuration and the forces acting on the atoms. The energy is the integral over the forces starting from some reference position. It should be clear that in a planar configuration, the mean distance between the lone pair electron and the bonds is smaller than in a tetrahedral configuration although there is no net force acting on the atoms in the planar configuration. Nevertheless, the repulsive potential energy is maximal there.
  6. Apr 19, 2014 #5
    Hi thanks for the reply I think I kinda understand it now. An analogue for this would be like having a rock crushing an apple. If the rock is heavy then it would get crushed and if the rock is light the apple won't be crushed. But still in both cases the apple won't have any net force acting on it but still in one case the apple isn't crushed.

    Is this a similar scenario? Like in the planar scenario there is no net repulsive forces but still the magnitude of the repulsions added together is greater in the planar scenario than the tetrahedral making it more unstable?
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