3
$\begingroup$

What is the relationship between the three concepts in the title?

I know all centrosymmetric molecules are achiral, but not all achiral molecules are centrosymmetric. Thus, there are no molecules that are centrosymmetric and chiral, but all other combinations exist.

What about having a dipole moment vs. centrosymmetry? And what about having a dipole moment vs. chirality? Is there a relationship between them, i.e. are some combinations possible and others not?

And if there is a link between these concepts, what is the proof? For example, for a centrosymmetric molecule, I could apply the inversion operation (which does not change the molecule as it has centrosymmetry) and then rotate it by 180$^\circ$, showing that there is a mirror symmetry. Thus, the molecule is achiral.

$\endgroup$
  • 1
    $\begingroup$ A dipole has a dipole moment, which is a vector. Arrows are not centrosymmetric. Thats a bit too obvious, so I have to ask: What is your actual question? $\endgroup$ – Karl Oct 29 at 22:38
  • $\begingroup$ @Karl Looking at the answer by Tyberius, I think my actual question is molecules of what point group symmetry are chiral, and molecules of what point group symmetry have (or can have?) a non-zero dipole moment. That would allow statements like X is necessary but not sufficient for Y, where X and Y could be any of chirality, achirality, certain symmetry elements, non-zero dipole moment, a vanishing dipole moment. $\endgroup$ – Karsten Theis Oct 30 at 11:06
  • $\begingroup$ So, in short, do all chiral molecules have a dipole moment? That`s an interesting question. ;) $\endgroup$ – Karl Oct 30 at 19:49
8
$\begingroup$

A molecule can't be chiral if it has an $S_n$ improper rotation (note that $S_1$ is equivalent to a mirror plane and $S_2$ is equivalent to inversion). Chirality is only possible in the $C_n$ and $D_n$, as well the less common $T$, $O$, and $I$, point groups, which only have rotation axes as symmetry elements.

A dipole moment can only occur along the primary $C_n$ rotation axis of a molecule (if it has one) and if there are any perpendicular mirror planes or rotation axes, it will have to be zero by symmetry. A $S_n$ improper rotation with $n>1$ also precludes a dipole. This limits the point groups that can have dipoles to $C_n$, $C_{nv}$, and $C_s$.

Centrosymmetry is equivalent to having the $S_2$ symmetry operation. As mentioned above, this eliminates the possibility of chirality or a dipole. However, centrosymmetry is just a special case and molecule won't be chiral and/or won't have a dipole under other conditions as well. That is to say, centrosymmetry is sufficient, but not necessary for eliminating chirality and/or a dipole.

To summarize: $$\begin{array}{|c|c|c|} \hline \text{Condition} & \text{Exact Point Group} \\ \hline \text{Chiral/Nonpolar} & I,O,T,D_{n} \\ \text{Achiral/Polar} & C_{nv}, C_s \\ \text{Chiral/Polar} & C_n \\ \hline \end{array}$$

Where I'm taking "exact point group" to mean the highest symmetry group the molecule is contained in. This delineation has some minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion.

$\endgroup$
  • 1
    $\begingroup$ @BuckThorn I was basically trying to get across that centrosymmetry is sufficient for there to be no dipole or chirality, but not necessary. $\endgroup$ – Tyberius Oct 30 at 12:23
  • $\begingroup$ Thanks for giving the entire picture! At symmetry.jacobs-university.de, they list point groups O, T and I as chiral as well. To summarize your answer with that addition, is it correct to say: "Molecules of $C_n$ symmetry are chiral and polar, $C_{nv}$ and $C_s$ are achiral and polar, and $D_n$ as well as $T, O, I$ are chiral and not polar. All others are neither chiral nor polar."? $\endgroup$ – Karsten Theis Oct 30 at 13:05
  • 1
    $\begingroup$ @KarstenTheis Careful. You'd need to specify that it is the maximal point group, since for example, $C_{n}\subset D_{n}$. $\endgroup$ – Zhe Oct 30 at 13:42
  • 1
    $\begingroup$ @KarstenTheis I had missed those cases, but they do fit the criteria of not having a $S_n$. I will add those in. Your summary looks correct, though there are minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion. $\endgroup$ – Tyberius Oct 30 at 13:45
5
$\begingroup$

Concerning chirality, the best link is to remember that any molecule that has an improper rotation symmetry (an $S_n$ axis in the Schoenflies notation or $\overline n$ in Hermann-Mauguin notation) is achiral. $S_1$ or $\overline 2$ happens to be equivalent to a plane of symmetry while $S_2$ or $\overline 1$ happen to be identical to a centre of symmetry.

The reason behind this is that an $S_n$ axis is essentially rotation followed by mirroring; however, the order of operations is irrelevant (they are commutative) so you can also generate the mirror image first and then rotate the mirrored molecule to make it identical to the original one. The Hermann-Mauguin formalism is based on centrosymmetry rather than a plane of symmetry, which means that the angle of rotation is different but otherwise the two are equivalent. (Each Schoenflies improper rotation can be transformed into a Hermann-Mauguin one and vice-versa.)

As for dipole moment, this is essentially a vector. It doesn’t matter whether your vector points from the positive to the negative pole or vice-versa (physics uses one definition, chemistry the other); the important bit is that it has a value and a direction. As all symmetry operations a molecule has must transform all atoms onto an atom of the same kind, these symmetry operations must also transform the dipole moment and any other vector-type properties onto itself.

Thus, a dipole moment must always be fully contained within all planes of symmetry and if there is an axis of rotation the dipole moment must align with it. It follows that in all the following cases a molecule cannot have a dipole moment:

  • if it is centrosymmetric (the only vector transformed onto itself under inversion is the null vector)
  • if it has an axis of rotation that is not contained in a plane of symmetry (the axis and the plane only intersect in a single point which again confines the dipole moment to the null vector)
  • if it has more than two planes of symmetry that cannot be transformed into each other by consecutive rotations
  • if it has more than two axes of symmetry (which by definition can only and must intersect in a point)

These descriptions are mathematically very similar as they all boil down to a small set of common elements.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.