intothevoidx Posted March 20, 2007 Share Posted March 20, 2007 The angles in a tetrahedral molecular shape are 109.5 degrees according to my chemistry book. If I have only taken a geometry course is it possible for me to figure out why this is so? Could anyone give me a simplified explanation of how this is done? Thanks Link to comment Share on other sites More sharing options...

The Thing Posted April 22, 2007 Share Posted April 22, 2007 Yes you can. A couple of things to keep in mind: 1. The central atom of the tetrahedral molecule is the geometric center of the tetrahedron. 2. Assume the bond lengths are the same. (Like in CH4) 3. The tetrahedron is a regular tetrahedron. (Again, like in CH4.) So each face is an equilateral triangle. Sorry. I couldn't draw a tetrahedron on paint (died horribly:D ). But visualize a tetrahedron ABCD, with its center at O. Then the "angles" you're trying to find is the angle between any 2 vertices and O (such as Angle AOB, or Angle BOC, etc...) All of these angles are the same. So let's find, say, angle AOB. Try the following: 1. Give the tetrahedron a side length, like 2 or 3 or 4, any number will do. 2. Now, the center of the tetrahedron will be directly above the center of any of the faces. So it will be directly above the center of triangle BCD (let's use this as the base). 3. So knowing that, find the distance between B and the center of BCD. Try it! 4. Now, let's name the center of BCD, err, P. Consider the triangle ABP. 5. You have just found the distance BP in step 3. You know AB, so use Pythagoras to find AP. 6. In ABP, remember that the bond lengths are the same. So AO=BO. Now, you have to find out the distance of AO (or BO). Try this. Hint: Set up an equation that uses Pythagoras! 7. After you have the distance of AO and BO, you can use the cosine law on Angle AOB. 8. And you're done. The result should be [math]cos^{-1}(-\frac{1}{3})[/math] Good luck and try it! Tell me if you need any steps clarified. Link to comment Share on other sites More sharing options...

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