# Why does dipole must lie parallel to a rotation symmetry axis?

While studying group theory, I've got confused by the following statement:

For a molecule to have a permanent dipole moment, it must have an asymmetric charge distribution. The point group of the molecule not only determines whether the molecule may have a dipole moment but also in which direction(s) it may point. If a molecule has a $$C_n$$ axis with $$n > 1,$$ it cannot have a dipole moment perpendicular to the axis of rotation (for example, a $$C_2$$ rotation would interchange the ends of such a dipole moment and reverse the polarity, which is not allowed — rotations with higher values of $$n$$ would also change the direction in which the dipole points). Any dipole must lie parallel to a $$C_n$$ axis.

why does the dipole need to be parallel to the $$C_n$$ axis? If I put a fluorine atom in just one of the light gray atoms of the following molecule, wouldn't it have a molecular dipole that isn't parallel to the main axis? • In that case you would no longer have a main axis at all. – Ivan Neretin Nov 13 '19 at 19:20
• Perhaps you are thinking about this in the wrong way. Changing to different types of atoms produces the dipole and at the same time changes the point group, for example octahedral MX$_6$ to MX$_5$Y. – porphyrin Nov 13 '19 at 21:55

## No, it would not

A dipole means that there is a difference (in charge) from one "side" to another "side".

If we have determined that something is symmetric, it means the "sides" are equal to each other — and that as you rotate $$\left(\frac{1}{n}\right)$$ around the symmetry axis, everything stays the same. If there is anything that shifts out of position, that symmetry is not there (it is broken).

Now, your image is a bit hard for me to read, is it supposed to be linear or is it on purpose not? I am going to assume it is not linear, but that the light grey are symmetrically placed… So in your image, there is only a $$C_2$$ rotational axis, in the approximate "up/down" direction. By placing a fluorine in one end (and something else at the other end) you would destroy the $$C_2$$ symmetry. If you rotate it $$180$$ degrees around (if you could see it through the backside of your monitor…) the fluorine would be on the other side, clearly not identical to $$E$$, the original. No symmetry.

If it was linear, the light grey and dark gray arranged as a line, then it would still retain a $$C_\infty$$ rotational axis along the dark grey atoms. Rotating around this in any fraction or direction would still keep everything looking the same.

That the dipole moment must be parallel to (and aligned with) the axis (all axes) of rotational symmetry of a molecule is a direct consequence of the observed symmetry.

Essentially, you can assume each bond between two atoms to carry an individual bond polarity moment. This is a vector that will be aligned with and parallel to the bond axis. To arrive at an overall dipole moment for the entire molecule, these individual bond polarity moments must be added, but because they are vectors they follow the rules of vector addition.

If there is an axis of rotational symmetry in the molecule, then one $$\ce{A-B}$$ bond will necessarily be transformed onto another $$\ce{A^*-B^*}$$ bond in the molecule (either A and A* or B and B* can be identical). The underlying mathematics show that for all sets of $$\ce{A-B}$$ bonds that are symmetry equivalent, the ultimate resulting vecor will be parallel to the axis of rotation. Thus, you end up with a number of intermediate result vectors all of which are parallel to the axis of rotation; the resulting dipole moment must also be. (They can, however, cancel out; this is typically the case if there are orthogonal axes of symmetry that must be considered. The resulting dipole moment vector turns out to be $$\vec 0$$.)

If, however, you change your molecule (as you suggest in your example replacing one hydrogen with a fluorine atom) you likely have just destroyed the axis of rotation as suddenly an $$\ce{A-B}$$ bond will be transformed onto an $$\ce{A^*-C}$$ bond. Since $$\ce B \ne \ce C$$, the axis of symmetry no longer exists and the dipole moment can be anywhere.