Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric Interpretation of VSEPR Theory

  1. Apr 8, 2014 #1
    I am making a Geometric model for VESPR theory, which states that valence electron pairs are mutually repulsive, and therefore adopt a position which minimizes this, which is the position at which they are farthest apart, still in their orbitals.

    For example, the 2 electron pairs on either side of the central atom of Carbon Dioxide both repel each other equally (Dot Diagram Below), giving the molecule a linear arrangement (Shown Below):

    Dot Diagram of Carbon Dioxide:

    170px-Carbon-dioxide-2D-dimensions.svg.png

    Linear Arrangement:

    100px-Linear-3D-balls.png

    Other molecules like water molecules have 4 pairs (2 bonded pairs, 2 unbonded pairs), which all repel each other, giving it a bent configuration.

    Dot Diagram of Water:

    lewis%2Bdot%2Bdiagram%2B2.png

    Bent Arrangement:

    100px-Bent-3D-balls.png

    I figured it'd be pretty simple and straightforward to create a geometric interpretation of this. This has probably already been done, but just doing it for fun.

    I figured I could accomplish this by creating a 3-D space, plotting the pairs as points, and then using the 3-D distance formula, find the arrangement at which all the pairs are at the greatest distance (Equal distance) from each other. I could then also calculate the angle of the bonds by calculating the angle between those lines.

    All the points would have to be an equal distance from the nucleus (origin), which can be set to an arbitrary distance. Let's just say 1.

    Here is what I have so far (For a simple 2 pair configuration):
    __________________________________________________________________________________
    E1 = Electron Pair 1

    E2 = Electron Pair 2

    Origin = Nucleus

    Distance from E1 to Nucleus = Distance from E2 to Nucleus:

    √(E1x-0)^2 + (E1y-0)^2 + (E1z-0)^2 = √(E2x-0)^2 + (E2y-0)^2 + (E2z-0)^2


    Distance from E1 to E2 (Needs to be maximized):

    √(E1x-E2x)^2 + (E1y-E2y)^2 + (E1z-E2z)^2

    __________________________________________________________________________________

    Now how could I calculate the coordinates of E1x, E1y, E1z, E2x, E2y, and E2z such that the distance between them is a maximum, yet they are the same distance from the origin? I figured i'd need differential geometry for that bit.

    Once I find the coordinates for each pair, I figured I could then find the angle between them using this:
    __________________________________________________________________________________

    Θ = tan^-1(m1-m2/1+m1m2)

    __________________________________________________________________________________

    Where m1 and m2 are the slopes of the lines from each point to the origin.

    Any help or comments would be greatly appreciated.

    Thanks in advance.
     
  2. jcsd
  3. May 4, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. May 10, 2014 #3
    Unfortunately not... I thought this would be pretty a simple problem for someone who knows Differential Geometry. Isn't maximizing distances something that is done all the time?
     
  5. Jun 8, 2014 #4
    With your current approach, there is no way to determine the orientation of your coordinate system. To remove this issue you need to define one full set of coordinates and one element of the other. I would place E1 on the positive x axis (1,0,0) and eliminate the z-coordinate from the problem by setting E2z = 0. Now you should be able to solve as a system to get two solutions, one for a positive bend and one for a negative one (or a double root if they are opposite). That should also make it easier to calculate slopes as you are already working in an effectively 2d plane. It gets trickier though if you add additional pairs.

    You might consider approaching this problem with vectors instead of just coordinates. If the pairs are spread evenly about the nucleus, unit vectors drawn to each pair should all sum to 0, which gives you some nice equations to work with:
    E1x+E2x+...+Ekx=0
    E1y+E2y+...+Eky=0
    E1z+E2z+...+Ekz=0
    I haven't checked if this helps at all but I thought it was worth mentioning.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook