Identifying the C3, C4, S4, and S6 symmetry operations in the Oh point group

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).


![](https://i.stack.imgur.com/31zBl.png)


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.


![](https://i.stack.imgur.com/aQFuk.png)


* 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?





> 
> 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`:



```
 A
BMD                      (1)
 C

```

If you apply $C\_2$, you will arrive at the same orientation `(2)`, regardless of which direction you rotate in:



```
 C
DMB                      (2)
 A

```

However, when rotating by $C\_4$ clockwise and counterclockwise actually lead to two *different* results `(3)` and `(4)`:



```
 B
CMA                      (3)
 D

 D
AMC                      (4)
 B

```

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.


 8

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 ${S\_6}$.


 5