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Topic: CCl4 dipole moment calculation  (Read 8922 times)

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Offline Rutherford

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CCl4 dipole moment calculation
« on: January 13, 2014, 08:07:49 AM »
Does anyone have a link of the calculation that proves CCl4 has zero dipole moment?

I tried myself, but I failed. In the picture a tetrahedron is produced from four vectors representing the dipole moments of C-Cl bonds whose intensity I will mark as unknown x, α=109.5°. Any two vectors are in the same plane, so using law of cosine I calculated the sum of two vectors which I mark as y: y=1.154x. y is in a plane which makes the degree alpha with a third vector, and their sum is: z=1.25x. This vector should be in the same line but oppositely directed as the fourth (last vector), and their sum is μ=0.25x≠0! Where am I wrong at?

Offline sjb

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Re: CCl4 dipole moment calculation
« Reply #1 on: January 13, 2014, 09:05:26 AM »
Does anyone have a link of the calculation that proves CCl4 has zero dipole moment?

I tried myself, but I failed. In the picture a tetrahedron is produced from four vectors representing the dipole moments of C-Cl bonds whose intensity I will mark as unknown x, α=109.5°. Any two vectors are in the same plane, so using law of cosine I calculated the sum of two vectors which I mark as y: y=1.154x. y is in a plane which makes the degree alpha with a third vector, and their sum is: z=1.25x. This vector should be in the same line but oppositely directed as the fourth (last vector), and their sum is μ=0.25x≠0! Where am I wrong at?

Were you working in radians or degrees?

Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #2 on: January 13, 2014, 09:13:10 AM »
Degrees.

Offline Corribus

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Re: CCl4 dipole moment calculation
« Reply #3 on: January 13, 2014, 12:08:39 PM »
Basically you need to show that the amount of "chlorine" is effectively zero in each cardinal direction. Put the origin at carbon and the top chlorine along the z-axis. Let one of the chlorines rest on the xz plane. Let the bond length be 1 in arbitrary units for convenience.

Do the easiest one first. Show that the chlorine above the xy plane cancels out the three chlorines below the xy plane.

Now think of how to do the same for the xz and yz planes.
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Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #4 on: January 13, 2014, 12:34:56 PM »
Forgot to put coordinates in my drawing.

How to show that they cancel out?

Offline Corribus

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Re: CCl4 dipole moment calculation
« Reply #5 on: January 13, 2014, 12:43:20 PM »
Magnitude of dipole moment is basically charge times distance, right? So if the center of the molecule is placed at the cartesian origin, then the sum of all the vectors (charge times displacement) must be zero for a molecule with zero dipole moment. To make things easy, then, you can separate this into three problems, one for each unit direction (x,y,z). So, for the z-direction you need to show that the charge times distance for each atom along the z direction equals zero. Here you have things easy because each chlorine has the same charge, which we can assign as "1".  Therefore you just need to show that Σdzi = 0, where dzi is the displacement of the ith chlorine along the z axis. For example, dzi for the chlorine just above the carbon center (along the z-axis, in the system I described in the previous post), would be exactly 1. So you need to find the analogous values for the other three chlorines, and show that they equal -1 in total, so that the overall total z-displacement of charge is equal to zero. Then do the same for x and y.

(The z direction is fairly easy. The x and y direction requires some more complicated geometry and trigonometry, but still the approach is generally the same.)
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Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #6 on: January 14, 2014, 06:10:17 AM »
I see now, it's basically the same I previously did.

So for z axis, I drew everything in the image attached. I marked all four vectors differently. The displacement may be 1, and the angle between each two 109,5°.
On the left part of the image I drew c+b, I calculated its intensity using law of cosine:
(c+b)=sqrt(1+1-2cos70.5°)=1.154.
On the right side I did the sum of c+b and d:
(c+b+d)=sqrt(1.1542+1-2·1.154cos70.5°)=1.25.
The sum of c+b and d is 1.25 ???. It is different from 1 and the vector can't cancel out with a :(. The final displacement on z axis would be -0.25. Where is the mistake?

Offline Borek

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Re: CCl4 dipole moment calculation
« Reply #7 on: January 14, 2014, 07:56:08 AM »
Are you sure angle between d and (c+b) is identical to angle between c and b? I am not.
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Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #8 on: January 14, 2014, 07:58:56 AM »
c and b are in the same plane, so is c+b. As the angle between d and either c or b is 109.5°, then the angle between d and c+b must be the same, right? The angle between d and the plane where c, b, and c+b are situated is 109.5°.
« Last Edit: January 14, 2014, 08:12:02 AM by Raderford »

Offline Borek

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Re: CCl4 dipole moment calculation
« Reply #9 on: January 14, 2014, 08:31:46 AM »
c and b are in the same plane, so is c+b. As the angle between d and either c or b is 109.5°, then the angle between d and c+b must be the same, right? The angle between d and the plane where c, b, and c+b are situated is 109.5°.

With this logic angle between d and -(c+b) would be 109.5° too, which is clearly impossible.
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Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #10 on: January 14, 2014, 08:55:51 AM »
Why again 109.5°? Shouldn't it be 70.5°?

Offline Corribus

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Re: CCl4 dipole moment calculation
« Reply #11 on: January 14, 2014, 09:58:33 AM »
Raderford, you are overcomplicating it.

For z direction, because of symmetry all the chlorines below the xy plane have the same z component to their displacement vector, so the overall z component of the dipole moment vector is proportional to z1 + 3*z2, where the constant of proportionality is the chlorine electronic charge. (Doesn't matter what it is, as we'll see.) We can use real bond lengths if we want to but it's easy just to normalize the bond length to aribrary units of one. Because chlorine 1 lies on the z axis and the carbon is at the origin, z1 = 1.  Chlorine 2 lies on the xz plane. The angle formed between Chlorine 1, the carbon and Chlorine 2 is 109.5.  Subtract 90 degrees to get the angle between Chlorine 2, the carbon and the x axis to get 19.5 degrees. The z2 component is therefore the sine of 19.5 degrees, or 0.3338. Because the chlorine 2 is below the xy plane, the sign on this z component is negative, as it is for the other two chlorine atoms.

Therefore the total z component of the dipole moment vector is 1 - 3*sin(19.5) ≈ 0.

Now you need to do x and y components. 

If it is helpful to see this as mathematical formalism, I can do it.
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Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #12 on: January 14, 2014, 10:08:34 AM »
Understood, thanks, I will try for x and y.

Offline Rutherford

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Re: CCl4 dipole moment calculation
« Reply #13 on: January 14, 2014, 10:17:24 AM »
But I could rotate the coordinate system (or just imagine that I watch it from a different angle) so that y becomes z, z becomes x, x becomes y, and it is the same calculation you wrote. Still don't see where was my way wrong ???.

An interesting implication of this is that CH3X and CHX3 (where X is a halide) would have the same dipole moment if C-H bond is regarded as absolutely non-polar.

Offline Corribus

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Re: CCl4 dipole moment calculation
« Reply #14 on: January 14, 2014, 10:24:15 AM »
I didn't look at your way. All you really need to do is calculate the cartesian coordinates of each atoms and then sum them up. This is what a computational package basically does.

As to your second point, no they would not be the same. For one thing, CH and CCl bonds are different lengths. For another, the charge density on a chlorine is quite a bit different than that of H. So the displacement vector will not be the same, and nor will the summed charges on each nucleus.
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