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Number of identical bond angles in PBr2F3

  1. May 4, 2014 #1

    Qube

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

    Consider the geometric isomer of PBr2F3 with a Br-P-Br bond angle somewhat larger than 120 degrees. It is also true that for this isomer ...

    There are two sets of identical bond angles.

    2. Relevant equations

    PBr2F3 is trigonal bipyramidal (although distorted).

    3. The attempt at a solution

    Er, two identical sets of bond angles? I can come up with more than that.

    First, we consider the equatorial trigonal plane. Since the Br-P-Br bond angle is larger than 120 degrees, the Br-P-F bond angles should each be 1) less than 120 degrees and 2) still identical. If you think about it, we started with an equatorial triangle and we simply changed the length of one of the sides (the side with the two bromine atoms). This in turn changes only one of the angles of the triangle and should leave two identical bond angles in its wake.

    So that's one set of identical bond angles.

    Now we consider more steric effects and interactions between the axial and equatorial atoms.

    We expect the bromine to push down on the fluorines because bromine is huge compared to fluorine. This distorts the perfect 180 degree bond angle in a perfect trigonal bipyramidal molecule and similarly the 90 degree bond angles in an ideal trigonal bipyramidal molecule.

    This in turn creates two sets of identical bond angles for a total of three. We have:

    1) F-axial, P, and Br-equatorial times 2 (there are two bromine atoms and two axial fluorines). This is somewhat greater than 90 degrees.

    2) The somewhat smaller than 90 degrees F-axial, P, and F-equatorial bond. Diagram attached.

    So is my teacher wrong?

    uta9etej.jpg

    Picture on left is an ideal representation with the perfect 90 degree bond angles explicitly shown; picture to the right below shows the repulsions.

    arara5e4.jpg
     
  2. jcsd
  3. May 5, 2014 #2
    "First, we consider the equatorial trigonal plane. Since the Br-P-Br bond angle is larger than 120 degrees, the Br-P-F bond angles should each be 1) less than 120 degrees and 2) still identical. If you think about it, we started with an equatorial triangle and we simply changed the length of one of the sides (the side with the two bromine atoms). This in turn changes only one of the angles of the triangle and should leave two identical bond angles in its wake."

    I agree with you.


    "equatorial triangle"

    Maybe you mean equilateral?


    I think that everything you said is correct. Is your teacher wrong? Well, it depends of what it claims.


    BTW, what did you use to draw those pics?
     
  4. May 5, 2014 #3

    Qube

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    I mean equatorial. Equatorial triangle is a reference to the triangle formed by the elements in the equatorial position. If you think about it, a trigonal bipyramidal shape is simply a trigonal planar center with a linear axes transversing the trigonal plane. I do not mean equilateral because the trigonal planar center of the molecule I referenced is distinctly not equilateral.

    Here's the original question:

    8e3ety4a.jpg

    I used Penultimate to draw these pictures.
     
  5. May 5, 2014 #4
    I understand what you mean by equatorial triangle, but when you said that "we started with an equatorial triangle and we simply changed the length of one of the sides" I thought it made a lot of sense to talk about an equilateral triangle. I know that at the end we will NOT have an equilateral triangle, but we "start" with one in our "model construction".

    Thank you for the answer.
     
  6. May 5, 2014 #5

    Qube

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    Okay, the center of the trigonal bipyramidal molecule is both an equatorial triangle and an equilateral triangle (in an ideal trigonal bipyramidal molecule).
     
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