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## Chimie organique

# 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.