The Oh character table has ten symmetry classes. I am trying to identify them, but my stumbling block is in identifying the order of each class.

I will use this diagram as a reference, from Sydney Kettle's Symmetry and Structure: Readable Group Theory for Chemists pg 191 (Wiley, 3rd edition).

The classes (with their respective orders) for the Oh table are:

$$\begin{array}{cccccccccc}E & 8C_3 & 6C'_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d}\end{array}$$

  • $E$ and $i$ are obvious.

  • $6C'_2$ and $3C_2$ are shown in the diagram.

  • $8C_3$ is shown in the diagram above; however, it only displays 4 axes. The character table indicates there are 8 equivalent rotations for this symmetry operation. Where do the other 4 axes for the $C_3$ class come from?

  • $6C_4$ is the same issue. It's only shown as 3 axes. Where are the other 3 $C_4$ axes from?

The improper rotations are given in another diagram, this time on page 194 from the same book.

  • The mirror and dihedral planes, $\sigma_\mathrm{h}$ and $\sigma_\mathrm{d}$, are both shown to match the order in the character table.

  • Similar to above for $C_3$ and $C_4$, the $S_4$ and $S_6$ rotations are only half the order supplied in the character table.

In summary, from 10 total classes in the octahedral table, I can identify all the corresponding axes/planes for 6 of those 10 classes. For the remaining 4 classes, $C_3$, $C_4$, $S_4$, and $S_6$, I can only identify half the axes. At first I thought the character table was counting both positive and negative rotations, but then the order for the $C'_2$ and $C_2$ rotations should be doubled if that were the case ($12C'_2$ and $6C_2$).

Can anyone please advise on exactly what the $C_3$, $C_4$, $S_4$, and $S_6$ operations correspond to in an octahedral complex?

  • 1
    $\begingroup$ Many rotatable 3D complexes are shown with their symmetry operations in the web site molecule-viewer.com . You can add/remove each of the elements at will. $\endgroup$
    – porphyrin
    Dec 17, 2016 at 19:08
  • 2
    $\begingroup$ As porphyrin mentioned, there are pages that show the operations, for example: symmetry.otterbein.edu/gallery/index.html $\endgroup$
    – Tyberius
    Sep 29, 2017 at 13:17

2 Answers 2


At first I thought the character table was counting both positive and negative rotations, but then the order for the $C_2'$ and $C_2$ rotations should be doubled if that were the case ($12C_2'$ and $6C_2$).

You were very close. The missing symmetry operations are indeed ‘positive and negative’ but only those that lead to non-identical transformations. In a $C_2$ axis, it does not matter whether you turn clockwise or counter-clockwise. Consider the following mock-up of the central planar square of an octahedron with the ligands labelled A, B, C and D:

BMD                      (1)

If you apply $C_2$, you will arrive at the same orientation (2), regardless of which direction you rotate in:

DMB                      (2)

However, when rotating by $C_4$ clockwise and counterclockwise actually lead to two different results (3) and (4):

CMA                      (3)

AMC                      (4)

These two are not identical transformations. In fact, they have distinct symbols; depending on your direction of rotation one is $C_4$ and the other is $C_4^3$. You can also interpret them as rotation by $90^\circ$ and rotation by $3 \times 90^\circ = 270^\circ$. $C_2$ and $C_4^2$ correspond to the same transformation as well.

The same case can be made for $C_3$, the reverse rotation is $C_3^2$.

The $S_n$ cases are not quite as simple. Well, actually $S_4$ is ($S_4$ and $S_4^3$) but that’s a coincidence. $S_6$ shows this more clearly. Remember that $$S_6 = C_6 \oplus \sigma_\mathrm{h}$$

Therefore, performing $S_6$ twice is equivalent to $C_3$:

$$S_6 \oplus S_6 = S_6^2 = C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_3 \oplus E = C_3$$

Thus, $S_6$ applied twice and $S_6^5$ applied twice give $C_3$ and $C_3^2$, respectively, so $S_6^2$ and $S_6^4$ are already taken care of. A last case needs to be made for $S_6^3$: This is identical to $i$ as can be seen below:

$$S_6^3 = S_6 + S_6 + S_6 = C_6 \oplus C_6 \oplus C_6 \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} \oplus \sigma_\mathrm{h} = C_2 \oplus \sigma_\mathrm{h} = i$$

Thus, while you could imagine up to six different $S_6$ transformations, only two of those ($S_6$ and $S_6^5$) give new transformations not covered elsewhere.


Keep in mind that a class doesn't indicate equivalent symmetry elements, merely equivalent operations.

For example, given $\ce{O_h}$, the ${C_3}$ and ${C_3^2}$ operations are class-equivalent. You're correct that there are only 4 axes, so the other 4 operations are ${C_3^2 = C_3^{-1}}$. Similarly, the 4 ${C_4}$ axes generate the ${C_4}$ operations and their inverse - which are in the same class: ${C_4^3 = C_4^{-1}}$.

It's similar with ${S_4}$ and ${S_6}$.


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