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Tetrahedron; Sum of Bond Angles

  1. May 7, 2014 #1

    Qube

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    1. The problem statement, all variables and given/known data

    Prove that if bonding-pair repulsions were maximized in CH3X, then the sum of the bond angles would be 450°.

    2. Relevant equations

    In a perfect tetrahedral molecule (e.g. methane), the sum of the bond angles is about 438 degrees (109.5° times 4).

    3. The attempt at a solution

    Well, if the tetrahedron were flattened as to give us a trigonal planar base and one attachment sticking off perpendicular to the base, we would have three 90 degree bond angles. That's not very helpful in achieving the 450 degree sum.

    So I'm guessing that the base of the tetrahedron has been very nearly flattened. In addition, a sum of 450 degrees implies an average bond angle of 112.5 degrees.

    How do we go about proving this though? There are bent molecules with bond angles of approximately 112.5 degrees, and these approximate this tetrahedron - bent molecules have two lone pairs and lone pair/lone pair repulsion is rather great.
     
  2. jcsd
  3. May 10, 2014 #2

    And what about the bond angles between those bonds forming the base?
     
  4. May 10, 2014 #3

    AGNuke

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    There are actually 6 angles which can be formed between the 4 bonds. (4-1)!
     
  5. May 10, 2014 #4

    Qube

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    Well, we were given that the sum of bond angles in methane is 438 degrees. So we seem to be ignoring the bonds formed below. Otherwise the sum of the bond angles would be 180 degrees times 3.
     
  6. May 14, 2014 #5

    Qube

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    So in other words we are talking about the internal angles. So let's consider the in-plane angles of a tetrahedron - the z-axis hydrogen and the x-axis hydrogen. Bond angle is 109.5 degrees; bond length is 1.09 Å. Sum of angles in a triangle is 180 degrees; two legs are the same length here, so the other two angles must be identical - i.e. [180 - 109.5]/2 degrees. 35.25 degrees.

    If we want a sum of 450 degrees we need each internal bond angle to be about 112.5 degrees. One of legs here stays the same but the other is different; we now have a C-X bond. I believe that X stands for a halogen substituent (as X commonly does in organic chemistry). Let's use Br as a starting guess because Br is a huge substituent and would likely push down dramatically on the whole tetrahedral, thereby maximizing repulsion. C-Br bond length is 1.91 Å. Now, we just have to do some geometry and play around with bond radii and diameter I think.
     
    Last edited: May 14, 2014
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