- Orbitales hybrides sp3 et liaisons sigma
- Orbitales hybrides sp2 et liaisons pi
- Hybridation sp³
- Nombre stérique et hybridation sp3
- Hybridation sp²
- Hybridation sp
- Exercices sur l'hybridation
- Tetrahedral bond angle proof (Vidéo Non Traduite)
Calcul mathématique de l'angle de liaison dans le méthane (une molécule tétraédrique). Créé par Jay.
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Transcription de la vidéo
On the left, we have the dot structure for methane. And we've seen in an earlier video that this carbon is sp3 hybridized, which means that the atoms around that central carbon atom are arranged in a tetrahedral geometry. It's very difficult to see tetrahedral geometry on a two-dimensional Lewis dot structure. So it's much easier to see it over here on the right with the three-dimensional representation of the methane molecule. So if I'm trying to see the four sides of the tetrahedron, I could find my first sides by connecting these hydrogen atoms like that. So there's the first side of my tetrahedron. And if I'm going to find the second side, I could connect these hydrogen atoms like that. And there's my second side. And to find my last two sides, if I connect this hydrogen atom to this one down here, I can now see the four sides of my tetrahedron. We're also concerned with the bond angle. So what is the bond angle? What is the angle between that top hydrogen, the essential carbon, and this hydrogen over here on the left? It turns out that bond angle is 109.5 degrees. And it's the same all the way around. So you could say that this angle is 109.5 degrees, or the angle back here. It's all the same. And so an sp3 bond angle is 109.5. And the proof for this was shown to me by two of my students. So Anthony Grebe and Andrew Foster came up with a very nice proof to show that the bond angle of an sp3 hybridized carbon is 109.5 degrees. And what they did was they said let's go ahead and take that tetrahedron, and let's go ahead and put it on the xyz axes. And let's put carbon at the center here. And we can choose any four points to represent the four hydrogen atoms of our tetrahedron, if we satisfy two conditions. Each point that we choose for our hydrogens is equidistant from the other three points, and also each point that we choose for our hydrogens is equidistant from the central carbon atom itself. And if you fulfill those two criteria, you guarantee that the points that you choose form a tetrahedron. And so, here we have the tetrahedron on our axes. And let's go ahead and look at the first point, so this point right here. And they chose this point to be at square root of 2, 1, and 0, meaning positive square root of 2 on the x-axis, positive 1 on the y-axis, and 0 on the z-axis. And then this point over here on the left, they were very clever and said this point is going to be in the same the plane. So this point on the left is in the same plane as the point we just talked about, the xy plane. And therefore, the coordinates for that point would be negative square root of 2, 1, and 0. We go to the hydrogen down here. So this point of our tetrahedron is located at 0, negative 1, and square root of 2. And then finally, this point going away from us right here would be at 0, negative 1, and negative square root of 2. So once again, you could choose any points that you want as long as you meet that criteria. And orienting the molecule in this way allows us to find this bond angle. So this is the bond angle that we are going for. And we don't know that bond angle yet, but we can figure out this angle right here. So I'm going to call this theta for this triangle that's formed. And I know that this x distance down here is positive square root of 2. And then we go up 1 on the y-axis and then 0 on the z-axis. So I can find out what theta is, because I know that tan of theta is equal to opposite over adjacent. So for this triangle I have here, the opposite side would be 1 and the adjacent side would be square root of 2. So to find theta, all I have to do is take inverse tan. So I take inverse tan of 1 over square root of 2 on my calculator, and I get 35.26 degrees. So I know that theta, this angle right in here, is 35.26 degrees. And therefore, this angle is also 35.26 degrees. So this is also going to be theta in here. And if I want to find my bond angle in here, I know that those three angles have to add up to equal 180 degrees since they're all in the same plane here. So to find my bond angle, all I have to do is take 180 degrees, and from that, we're going to subtract 2 times 35.26 degrees. And we, of course, come out with a bond angle of 109.5 degrees. So again, special thanks to my two students for showing me this proof.