Why the tetrahedron is equal to the angle of 109.5degrees but it says in an article that if you divide the number of bonds thats the angle but when i divide 4 to 360 the answer is 90 they did not match pls answer my question tnx?

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Why the tetrahedron is equal to the angle of 109.5degrees but it says in an article that if you divide the number of bonds thats the angle but when i divide 4 to 360 the answer is 90 they did not match pls answer my question tnx?

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Answer 1 · 2017-09-10T12:53:12

A tetrahedron is not flat...

Explanation:

If you have a flat symmetrical molecule such as Borane $B H_{3}$ $BH_3$ then dividing $360^{\circ}$ $360^@$ by the number of bonds makes sense, but in the case of a three dimensional tetrahedral molecule such as Methane $C H_{4}$ $CH_4$ it does not.

You could model the shape by imagining a carbon atom at $(0, 0, 0)$ $(0, 0, 0)$ and hydrogen atoms at: $(1, 1, 1)$ $(1, 1, 1)$ , $(1, - 1, - 1)$ $(1, -1, -1)$ , $(- 1, 1, - 1)$ $(-1, 1, -1)$ and $(- 1, - 1, 1)$ $(-1, -1, 1)$ (alternate vertices of a cube form the vertices of a tetrahedron).

Then the angle between the line joining $(0, 0, 0)$ $(0, 0, 0)$ and $(1, 1, 1)$ $(1, 1, 1)$ and the line joining $(0, 0, 0)$ $(0, 0, 0)$ and $(1, - 1, - 1)$ $(1, -1, -1)$ is:

$arccos (\frac{< 1, 1, 1 > . < 1, - 1, - 1 >}{| < 1, 1, 1 > | \cdot | < 1, - 1, - 1 > |}) = arccos (- \frac{1}{3}) \approx {109.5}^{\circ}$ $arccos((<1,1,1> . <1,-1,-1>)/(abs(<1,1,1>)*abs(<1,-1,-1>))) = arccos(-1/3) ~~ 109.5^@$

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Why the tetrahedron is equal to the angle of 109.5degrees but it says in an article that if you divide the number of bonds thats the angle but when i divide 4 to 360 the answer is 90 they did not match pls answer my question tnx?

1 Answer

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