# How does one recognized Td/Oh symmetry in molecules?

The identification of point groups of a molecule is usually done following a strict scheme, either manually or algorithmically. In all textbooks I could find, however, the first step of the scheme is actually not explicited: in this example scheme drawn from Housecroft and Sharpe,

you can see that there is a very unhelpful step “Does this molecule have $I_h$, $O_h$ or $T_d$ symmetry?”. It is implied that one would recognize e.g. an icosahedral molecular compound on sight. However, I wonder: how can one strictly identify these “special” point groups? What set of rules is to be followed (again, either manually or algorithmically)?

• This has been address before, see here: scicomp.stackexchange.com/q/135 – Chris May 10 '12 at 16:56
• @Chris thanks for the link! It contains very useful information (such as a link to an generic algorithm and code implementing it). It doesn't provide the specific solution to this question, however (and it couldn't be a duplicate anyway, because it's on another site). – F'x May 10 '12 at 17:21
• In fact most people see this symmetries "at once". The main problem was the lack of apropriate sketches in books formerly. A org prof of me used to say: progress of sterechemisty was mostly due to progress in printing technology/cost. Another problem is that some persons lack the ability to see/think 3D more or less. – Georg Jul 12 '12 at 9:00

You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups.

According to this untitled document, which I presume1 is by W.C. Trogler, the elements are as follows:

$T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$

$O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$, $3\sigma{_h}$, $6\sigma{_d}$

$I_{h}$: $E$, $6C_{5}$, $10C_{3}$, $15C_{2}$, $i$, $6S_{10}$, $10S_6$, $15\sigma$

Obviously, you don't need to check for the identity symmetry $E$.

If you're a visual sort of person a symmetry spot check is to mentally superimpose a tetrahedron, cube or dodecahedron over the molecule and see whether the view down the surface normal of each face is identical. Cubes and octahedra are duals of each other, as are dodecahedra and icosahedra. The tetrahedron is self-dual.

Interestingly, H&S does not list chiral forms of these point groups, probably because they are so rarely encountered, however researchers have come up with molecules that obligately satisfy $T$, $I$ and $O$ symmetry2 (I have not yet read the paper).

(1) I would be indebted to anyone who can furnish me with a full citation to this work, and confirm the authorship.

(2) Narasimhan, S. K., Lu, X. and Luk, Y.-Y. (2008), Chiral molecules with polyhedral T, O, or I symmetry: Theoretical solution to a difficult problem in stereochemistry. Chirality, 20: 878–884.

• Does one really have to check for all the symmetries? Isn’t there a shorter way? (some invariant that these molecules would satisfy, or something like that) – F'x May 10 '12 at 10:22
• The syllabus for the course makes it clear that the instructor is Bill Trogler – F'x May 10 '12 at 10:26
• @F'x - Mentally superimposing the corresponding Pythagorean solid is the best heuristic that I know of. However on reflection (see what I did there?) it will fail to distinguish the chiral and achiral forms of the point group. In the end it's only a heuristic, and if you go and check every face and edge rigorously you will be implicitly checking that each and every symmetry element is satisfied. However, hopefully someone savvier will chime in with the easy method. – Richard Terrett May 10 '12 at 10:46
• Adamantane is point group $T_d$, but you won't easily figure that unless you already have a molecular model in front of you. $C_{60}$ is $I_h$, but you'll only realize that if you know the relationships between the dodecahedron, icosahedron, and truncated icosahedron... huh. – user95 May 13 '12 at 14:52

The quick way of doing that first step in the flowchart is to look for too many principal $C_n$ axes. (Since the next step is to look for a principal axis anyway, this is a natural step to take.) In particular, you are looking for more than one axis of order > 2. A threefold axis with three twofold axes perpendicular to it? Just $D_{3*}$ (where $* = d$, $h$, or nothing is yet to be determined, by using the flowsheet). But more than one threefold (or fourfold, or fivefold) axis? Has to be one of these special point groups.

The next question, of course, is which one. Again the key is these "excess" principal axes. Lots of fivefold axes is the giveaway for icosahedral symmetry (to be sure, count six of them); lots of fourfold for octahedral (look for three); and neither of these means tetrahedral (you can look for four threefold axes to be sure, but you need to eliminate the other two since both of these also have threefold axes).

In practice you can assume that icosahedral means $I_h$, octahedral $O_h$, and tetrahedral $T_d$. To be complete, however: for octahedral and icosahedral you need to find a centre of symmetry, otherwise it's just $I$ or $O$. For tetrahedral, if you have a centre of symmetry it's $T_h$; if you have mirror planes but no centre of symmetry it's $T_d$, and with neither it's $T$.

Essentially this is just the same answer as Richard's, but my point is just that there are heuristics you can carry around in your head without having to check all 120 (or however many) symmetry elements. To take J.M.'s examples: adamantane has several (you don't really even have to count four) threefold axes but nothing higher: must be $T_d$. Buckyballs have a few fivefold axes: surely $I_h$.

• I'm not sure what you mean by the question being "unhelpful," especially if you thought it was worth answering. I suggested an edit. – Colin McFaul Jul 11 '12 at 22:00
• Thanks for the edit - I was quoting the OP's referring to a "very unhelpful step" in the flowchart rather than suggesting that the question asked was unhelpful! – Aant Jul 12 '12 at 7:25
• I understand now. I did not see the word "unhelpful" in the question, so I was confused. – Colin McFaul Jul 12 '12 at 13:22