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I was studying chemical bonding when I noticed something odd.

We say compounds like $\ce{CCl4}$ and $\ce{CH4}$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.

But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.

But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?

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    $\begingroup$ Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-) $\endgroup$
    – Karl
    Commented Jul 18, 2019 at 21:40
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    $\begingroup$ @Karl Based on your very helpful answer to another question by uhoh, I will make a wild guess: dielectric relaxation spectroscopy? ;-) $\endgroup$
    – Ed V
    Commented Jul 19, 2019 at 13:35
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    $\begingroup$ @EdV Well, some substances have a finite dipole moment in the terminal (--> zero frequency) regime. Another word for frequency dependent dipole moment is complex permittivity. ;-) Your guess can tell the answer ... $\endgroup$
    – Karl
    Commented Jul 19, 2019 at 22:39
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    $\begingroup$ Actually, I'm not 100% sure if the answer to my question is "of course not, silly, it cancels out", or "obviously, stupid". If someone would explain that, I'd throw a 100pt bounty into the audience. $\endgroup$
    – Karl
    Commented Jul 19, 2019 at 22:43

3 Answers 3

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Karsten's answer is excellent, but here is a figure that shows the mathematics involved:

Tetrahedral angle

The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.

Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - \theta$ ($\approx 70.52877940 ^\circ$) away from being directly "under" the central atom, so each contributes $\cos(1 - \theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.

But $\cos(1 - \theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.

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    $\begingroup$ Nice figure and geometry treatment. I'm so happy someone actually wants to know the math! $\endgroup$
    – Karsten ♦
    Commented Jul 18, 2019 at 15:45
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    $\begingroup$ If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer). $\endgroup$
    – Karsten ♦
    Commented Jul 18, 2019 at 15:47
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    $\begingroup$ Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components. $\endgroup$
    – Ed V
    Commented Jul 18, 2019 at 15:55
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    $\begingroup$ @andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up. I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?) $\endgroup$
    – Karsten ♦
    Commented Jul 18, 2019 at 18:01
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    $\begingroup$ @andselisk Simply stating that "centrosymmetric molecules have zero dipole moment" would probably just produce the response "but why?" from the OP. If he/she could "see" that was true without any help, the question about a particular geometry would never have been asked. In fact asking "what exactly does centrosymmetric mean" is not a silly question in its own right. $\endgroup$
    – alephzero
    Commented Jul 18, 2019 at 23:48
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Orientation 1

Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.

For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.

enter image description here

If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.

If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.

Orientation 2

Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.

The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.

enter image description here

Orientation 3

In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.

enter image description here

Orientation 4

Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.

enter image description here

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"In the beginning we saw that [methane molecules] are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero."

So what happens if you turn the molecule around so that one of the other H atoms is now at the top?

Remember that a tetrahedron is symmetric, so that it looks the same when viewed from any of its four corners. So if the tetrahedral methane molecule had a non-zero dipole moment pointing towards one of the hydrogens, by symmetry it would also have to have an equal dipole moment pointing towards all of them. But then it wouldn't be a dipole, but an octupole — a dipole can, by definition, only point in one direction.

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    $\begingroup$ Nice symmetry argument! This could be generalized to "any molecule with centrosymmetry has a dipole moment of zero", but the proof did not fit into this margin (@andselisk) $\endgroup$
    – Karsten ♦
    Commented Jul 19, 2019 at 12:55
  • $\begingroup$ But is the symmetry perfect? What if we account for tiny aberrations (thinking of spins)? $\endgroup$
    – Hagen von Eitzen
    Commented Jul 20, 2019 at 14:39

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