Edited by
textbook forum
RALPHK.
BIRDWHISTELL University 01 West Florida Pensacola, FL 32504
Why Is Ammonia Not Optically Active? S. F. A. Kettle University of East Anglia, Norwich, NR4 7TJ, U.K. In compilations of character tables it is customary to list a t the right-hand side some of those functions that serve as bases for individual irreducible representations. This inclusion is easily justified by the utility of basis functions, but it has always to be borne in mind that these entries are not subject to the rigor of the rest of the table. This means that, on occasions, problems arise; we have recently discussed one such problem1, and the present article is devoted to another. The point group of the ammonia molecule is Ca,and it is the E irreducible representation of this particular point group that we shall consider. Almost all listings of basis sets for this irreducible representation contain, as separate pairs of basis functions, the entries (x,y) and (R,, R,). Remembering that in all point groups the translations T,, T,, and T, have the same symmetries as their axial counterparts x, y, and z (so that (T,, T,) are a perfectly good basis set for the E irreducible representation of CaJ, we are led to the problem posed by the title of this article. When a translation and rotation transform under the same irreducible representation of a point group, we must expect molecules of that symmetry to be optically active. Physically, because basis functions are not all-or-nothing quantities, in real life they must he expected to mix. So, in the ammonia molecule a displacement of electron density (such as happens when light, electromagnetic radiation, falls on the molecule) of E symmetry will not behave as either (T,, T,) or (R,, R,) but, rather, as a bit of both. Electron density displacements which behave like a mixture of T,and R, bbhaverather like screw threads (such threads combine rotation about the axis with translation along it) and so cannot be superimposed on their mirror image. This, of course, leads to the phenomenon of optical activity. Why, then, is ammonia not optically active? We all know that ammonia is not optically active, so something must he wrong. What is it? The answer is "the basis functions in the character table". Rather than being written (I, y), (RI, R,), as it almost always done, they should be printed (x, y ) (R,, R,). To see the implications of this interchange, consider yet another basis set for the E irreducible representation of the C3"point group, the pair (xy, x2 - y2). If we wished to set this alongside (x, y), which would be the correct pattern (x, y) (xy, x2 - y2) or (x, y) (x2 - y2, xy)? Restated, this could be put "if we had a real problem in which one of the objects under consideration-perhaps a wave function-had a functional form close to the function x, which of x2 - y2 and xy is likely to also contribute? Because x and its partner appear in a single object, and this
Flgwe I. DBRnHlon 01 the x a d y axes ot the f ~ n a i o n sw a n d Z operatbn discussed In Ih-3 ted.
- P.and the
single object, clearly, can only transform in a single way under any symmetry operation, it follows that the partner of x is that one of x2 - y2 and xy that transforms in the same way as x (the one remaining would then transform in the same way as y). As Figure 1shows, the matrix representing the transformation of (x, y ) under the C3+operation, which we take as an example, with C3+, x and y defined as shown in Figure 1,is cosl20 sinlZO)(x) ~ a + ( x ) = (-sin120 coal20
Similarly, for the pair (x2 - y2,xy), we have
' Kettle, S. F. A. J. Cham. Educ., in press. Volume 66
Number 10 October 1989
841
T~ Flgre 2. Comwlpon d the m o w s represantlng R, and T,(best viewed down themreet0ldaxis)showsmatmesetwotransform isomhically. Similarly lw
&and );.
The matrices to the left on the right-hand side in eqs 1 and 2 become identical if we interchange x and y in eq 1. We conclude, therefore, that 1: is partnered by xy and y by x2 y2. Returning to the nonoptical activity of ammonia, this clearly means that, in fact, T , must be partnered by R, and T , by R,. It is almost trivial to show diagramatically that this is so. Figure 2 shows T,, T,, R, and R, for the ammonia molecule viewed down the threefold axis. Simnlv lookine at the directions in which the (projected) arrowspoint is &fficient to demonstrate the isomorohism of T. with R. and of T , with R,. There is an alternative way of proceeding that, although not without its own difficulties, avoids the need to be able to produce a diagram analogous to Figure 2 for every problem (basis functions need not be simple either mathematically or pictorially). This is by a descent-in-symmetry method. Consider the following problem.
842
Journal of Chemical Education
"It is clear that methane cannot be optically active because in the Td point group the translations (T,, T,, T , ) transform as T 2while the rotations (R,,R,, R,) transform as T I . However, if the methane molecule is slightly flattened along an Saaxis so that the symmetry becomes D2d,the pairs (T,, T,) and (R,, R,) both form bases for the E irreducible representation of D2d. IS this flattened methane optically active?'' Of course, the answer is "no", for the same reason as in the ammonia case. We can show this by distorting the methane molecule still further by destroying the Sq axis, giving a methane molecule with C2"symmetry. The basis functions in the C2. character table show that x transforms under the same irreducible representation as R,. Similarly, y and R, "go together". Now, a reduction in symmetry can only allow previously distinct functions to mix, never unmix functions. I t therefore follows from the fact that T , and R, (and T , and R,) are distinct in Cz, that they must also be so in D2d. We formally conclude that tetragonally distorted methane is not optically active. Notice two things. Firstly, in chosing a subgroup of D2d we chose to retain as many improper rotation operations as possible. The alternative, of retaining as many proper rotations as possible, would have led us to the D2 point group and to the erroneous conclusion that tetragonally distorted methane is optically active! Secondly, while we could have used a similar argument for the ammonia case, by dropping in symmetry to C,, this has the disadvantage that axis labels conventionally change in moving from Cg to C3.Of course, this in no way changes the final conclusions-it just makes it a hit more difficult to work out what happens in Cs starting from C,.
