13-4 The Concept of Class

Chapter 13 Group Theory distinguished from the full group by virtue of the fact that none of them exhausts all the possible physically achievable, nonredundant, indistinguishable conﬁgurations for the molecule.

EXAMPLE 13-1 BH3, like ammonia, has a three-fold symmetry axis. However, BH3 is planar. As a result, there is an extra symmetry operation—reﬂection through the molecular plane—that does not move any nuclei. Thus, both the identity operation and this reﬂection leave all nuclei unmoved. Is this an example of redundant operations?

SOLUTION These operations are not redundant. Even though all the nuclei stay ﬁxed for both operations, reﬂection interchanges the spaces above and below the plane, whereas the identity operation changes nothing. Two operations are redundant only if they move all points in space in the same way.

13-4 The Concept of Class Imagine that we are subjecting our ammonia model to the various symmetry operations and that someone in a parallel universe is subjecting his ammonia model to symmetry operations too. Suppose that the parallel universe differs from ours in that everything is reﬂected through the xz plane. We shall refer to this as “the mirror A universe.” We now pose the following problem. Suppose we initially have a particular conﬁguration and our “mirror A” man has the corresponding conﬁguration (achieved by reﬂecting through plane A) as shown in Fig. 13-6. If the mirror A man now performs some symmetry

Figure 13-6 Operation X in this universe parallels operation F in the mirror A universe.

Section 13-4 The Concept of Class operation on his model, say, counterclockwise rotation by 120◦, F , he ends up with the ﬁnal conﬁguration shown at the lower right of the ﬁgure. (We continue to deﬁne “clockwise” from clocks in our universe.) For us to arrive at the corresponding ﬁnal orientation, what operation must we perform? We symbolize this unknown operation

X. We can ﬁnd out what our ﬁnal conﬁguration must be by taking the reciprocal of reﬂection A on the mirror man’s ﬁnal conﬁguration. (A takes us from here to there; A−1 takes us from there to here.) But A−1 = A, so we obtain the result at the lower left of the ﬁgure. It is evident that X must be D, clockwise rotation by 120◦. This is not surprising. It is related to the fact that, when you turn counterclockwise before a mirror, your image appears to turn clockwise.

We can repeat the solution to this problem in a more formal way by requiring that the operation X followed by A (into the mirror A universe) bring us to the same conﬁguration as operation A followed by F . That is, we seek X such that AX = FA. Multiplying both sides from the left by A−1 gives X = A−1FA. This is merely a mathematical statement of our above discussion. It says, “What operation is equivalent to the process of reﬂection through plane A followed by counterclockwise rotation by 120◦ followed by the inverse of reﬂection through plane A?” Using our multiplication table, we see that A−1 = A, and so X = AFA = AB = D. Therefore, D = A−1FA.

We can ﬁnd the mirror A universe equivalents to all the symmetry operations in the group in a similar way. Thus, A−1EA = E, A−1AA = A, A−1BA = C, A−1CA = B, A−1DA = F, A−1F A = D

There is no need to stop here. We can imagine other parallel universes corresponding to reﬂections B and C and rotations D and F . (The E universe is the one we inhabit.)

We can use the same sort of technique to accumulate corresponding operations for all these universes. The results are given in Table 13-3.

If we examine this table, we note certain patterns. The operation E in our universe corresponds to E in all the universes. The reﬂections (A, B, and C) always correspond to reﬂections, and the rotations (D and F ) always correspond to rotations. This makes physical sense. If we “do nothing” (E) in our universe, we expect the people in the other universes to do nothing also. If we rotate by 120◦, we expect the people to perform TABLE 13-3 X = Y −1ZY as a Function of Y and Z

Z

E

A

B

C

D

F

E

E

A

B

C

D

F

A

E

A

C

B

F

D

Y

B

E

C

B

A

F

D

C

E

B

A

C

F

D

D

E

C

A

B

D

F

F

E

B

C

A

D

F

Chapter 13 Group Theory rotations by 120◦ too, although not necessarily always in the same direction. The argument for reﬂection is the same. These are examples of three classes of operation.

The formal deﬁnition of class is as follows: If P and Q in a group have the property that X−1P X = P or Q and X−1QX = P or Q for all members X in the group, then P and Q belong to the same class. Physically, operations in the same class are of the “same kind”—all reﬂections, all rotations, etc.

EXAMPLE 13-2 BH3, like ammonia, has three planes of reﬂection symmetry that contain the three-fold rotational axis. However, BH3, being planar, also has a reﬂection operation through the molecular plane. Is this reﬂection operation in the same class as the other three?

SOLUTION No. In both molecules, the three planes each interchange two hydrogens and keep one ﬁxed in space. The extra reﬂection plane in BH3 keeps all three hydrogens ﬁxed. If we reﬂect so as to keep all hydrogens ﬁxed, there is no way a person in a symmetry-related universe could make a corresponding reﬂection by using a reﬂection that interchanges two hydrogens and keeps one ﬁxed.

We shall see that classes are important subdivisions of groups. Note, however, that a class need not be a subgroup. For instance, D and F constitute a class, but, as the class does not include E, it is not a subgroup.

In discussing the concept of class, it is unnecessary to postulate parallel universes, and the reader should not be disturbed by this pedagogical device. The people in the “other universes” are merely working with ammonia models that have been reﬂected or rotated with respect to the model orientation that we chose in Fig. 13-3. Operations in the same class are simply operations that become interchanged if our coordinate system is subjected to one of the symmetry operations of the group.

13-5 Symmetry Elements and Their Notation

There are ﬁve kinds of symmetry operations that one can utilize to move an object through a maximum number of indistinguishable conﬁgurations. One is the trivial identity operation E. Each of the other kinds of symmetry operation has an associated symmetry element1 in the object. For example, our ammonia model has three reﬂection operations, each of which has an associated reﬂection plane as its symmetry element.

It also has two rotation operations and these are associated with a common rotation
axis as symmetry element. The axis is said to be three-fold in this case because the associated rotations are each one-third of a complete cycle. In general, rotation by 2π/n radians is said to occur about an n-fold axis. Another kind of operation—one we have encountered before is inversion, and it has a point of inversion as its symmetry element. Finally, there is an operation known as improper rotation. In this operation, we ﬁrst rotate the object by some fraction of a cycle about an axis, and then reﬂect it through a plane perpendicular to the rotation axis. The axis is the symmetry element and is called an improper axis.

The following two examples should help clarify the nature of these symmetry ele ments. Consider ﬁrst the ethane molecule in its eclipsed conformation (I).

1This term is not to be confused with a group element.

Section 13-5 Symmetry Elements and Their Notation

The following symmetry elements can be identiﬁed. (It is important that you satisfy yourself that you see these elements in the sketch.)

1. One three-fold axis coincident with the C–C bond.

2. Three two-fold axes perpendicular to the C–C bond and intersecting its midpoint.

3. Three reﬂection planes, each containing the C–C bond and a pair of C–H bonds.

4. One reﬂection plane perpendicular to the C–C bond and bisecting it.

5. No point of inversion.

6. One three-fold improper axis coincident with the C–C bond.

[Note that these operations are to be applied to the rigid molecule. Movement of only one methyl group and not the other (i.e., internal rotation) is not allowed.]

Now let us consider staggered ethane (II) (i.e., one methyl group rotated 60◦ from its eclipsed position). Its symmetry elements are 1. One three-fold axis coincident with the C–C bond.

2. Three two-fold axes perpendicular to the C–C bond and intersecting its midpoint.

3. Three reﬂection planes, each containing the C–C bond and a pair of C–H bonds.

4. No reﬂection plane perpendicular to the C–C bond.

5. One point of inversion at the midpoint of the C–C bond.

6. One six-fold improper axis coincident with the C–C bond.

In going from the eclipsed to the staggered conformation, we have lost a reﬂection plane perpendicular to the C–C bond, gained a point of inversion, and changed the order of the improper axis.

Usually it is not difﬁcult to “see” most elements of symmetry in molecules, the exception being improper axes, which tend to be a little tricky. The six-fold improper rotation in staggered ethane is not too hard to envision when it is applied once. If we rotate clockwise by 60◦ and reﬂect, H1 replaces H5, H5 replaces H2, etc. It is a little Chapter 13 Group Theory more difﬁcult when we try to imagine applying this operation twice in succession. The reader should recognize that two 60◦ rotations and two reﬂections result in H1 replacing H2, H2 replacing H3, H4 replacing H5, etc.

EXAMPLE 13-3 BH3 has a three-fold axis and a reﬂection plane in the molecular plane. Does BH3 have a three-fold improper rotation axis as a necessary member of its group?

SOLUTION  A three-fold improper axis coincident with the “regular” three-fold rotational axis appears to have the correct symmetry characteristics: It rotates each hydrogen to the next position and then interchanges the spaces above and below the plane. Do we need it in the group? Only if there is no other single operation that accomplishes the same thing. There is no such operation, so this improper axis is indeed a nonredundant symmetry element.

A notation for these symmetry elements (and for the operations related to them) has come to be generally accepted in chemistry (exclusive of crystallography, which uses a different notation). This notation is summarized in Table 13-4. The symmetry elements of eclipsed ethane would be, by these conventions, C3, 3C2, σ (perpendicular to C–C), 3σ (containing C–C), S3. It often occurs that several classes of reﬂection planes and axes for proper or improper rotations are present in a system. In eclipsed ethane, we can imagine that the three reﬂection planes containing the C–C bond might all be in the same class, that is, might be interchanged by a symmetry operation. In fact, it is easy to see that they are interchanged by rotations about the threefold axis. However, none of these could ever be equivalent to the reﬂection perpendicular to the C–C bond because this reﬂection interchanges the carbons, whereas the others do not. For this reason, we keep separate tally of the different classes of reﬂection in eclipsed ethane, rather than simply writing “4σ .”

There is a geometric convention that aids discussion of molecules having axes of rotation. One looks for a unique axis Cn (usually the axis of highest order) and imagines this axis to be vertical (coincident with the z axis). Then a reﬂection plane perpendicular to this axis is horizontal and is labeled σh. Planes containing this axis are necessarily vertical and are subdivided into dihedral and vertical planes. Dihedral planes must contain the unique reference axis, Cn, and must also bisect the angles between two-fold TABLE 13-4  Symbols for Symmetry Elements and Operations

Symmetry operation

Symmetry element

Symbol “Do nothing” (identity)

None

E

Rotation by 2π/n radians

An n-fold (proper) axis

Cn

Reﬂection through plane

A plane

σ

Inversion through a point

A point

i

Rotation through 2π/n radians

An n-fold (improper) axis

Sn

followed by reﬂection through a plane perpendicular to the rotation axis axes perpendicular to Cn. Such planes are labeled σd. Vertical planes that do not bisect two-fold axes2 are labeled σv. In eclipsed ethane, the C3 axis is the principal axis, so the C–C bond is oriented vertically. Therefore, our symmetry elements are labeled C3, 3C2, σh, 3σv, S3. (The vertical planes are labeled σv because they contain, rather than bisect, the two-fold axes that are perpendicular to the principal axis.) For staggered ethane, we have C3, 3C2, i, 3σd, S6. (Here the vertical planes do not contain the two-fold axes, but do bisect them. Hence, we label them σd. Making a simple sketch will aid in clarifying this distinction.)

Once one has recognized the set of symmetry elements associated with a given object, it is a straightforward matter to list the symmetry operations associated with the set. Simplest are the operations associated with elements σ and i, because each such element gives rise to only one operation. Proper and improper axes are somewhat more complicated. Let us return to our ammonia molecule for illustration of this. There we had a threefold axis C3 and we noted that we could rotate by 2π/3 (C+) to get one con 3

ﬁguration, and 4π/3 [C+C+ = (C+)2] to get another. Alternatively we could choose

3

3

3

to rotate by 2π/3 and −2π/3 (C−), the latter easily being shown to be equivalent to

3

(C+)2. But if we rotate by (C+)3, we return to our original conﬁguration. That is,

3

3

(C+)3 = E. Therefore, a C

3

3 axis produces two unique operations. The reader can easily generalize this to the statement that a Cn axis yields n − 1 unique symmetry operations Cn, Cn2, Cn3, . . . , Cn−1

n

, where we assume rotation to be in the clockwise direction in all cases. Benzene has a C6 axis with ﬁve associated nonredundant symmetry operations (C6, C2, C3, C4, C5). Now C3 is equivalent to a rotation by π radians, an operation

6

6

6

6

6

we normally write as C2. Similarly, C2 = C

6

3. The result of such reductions is the set of operations C6, C3, C2, C2, C5 for the C

3

6

6 axis of benzene. There are several classes of rotation here. If we use a compressed notation, we can write this set as 2C6, 2C3, C2.

This indicates that the presence of a sixfold axis implies the existence of coincident twofold and threefold axes. However, these implied elements are not listed for such a system since their operations are all contained in the set of operations of the C6 axis.

Improper axes can also be associated with several symmetry operations. We noted earlier that S6, applied twice in succession, results in a simple 2π/3 rotation about the S6 axis. In other words, we can write the set S6, S2, S3, S4, S5 as S .

6

6

6

6

6, C3, S2, C2

3

We stop at S5 because S6 = E due to the combination of C6 and an even number of

6

6

6

reﬂections. S3 is equivalent to S

6

2 because it contains three rotations by 2π/6 and an odd number of reﬂections, and S2 means one rotation by π and one reﬂection. The operation S2, however, is easily shown to be equivalent to an inversion, and so we have, using a compressed notation, 2S6, 2C3, i associated with the S6 axis. Since we have already explicitly listed the elements C3 and i in our set of elements for staggered ethane (or any other system containing an S6 axis) only the 2S6 operations are unique to the S6 axis. The generalization of this case is that an S2n axis with odd n implies that elements Cn and i are also present. Of the 2n − 1 operations associated with S2n, n − 1 are preempted by the Cn axis and 1 by the element i leaving n − 1 operations to be attributed to the S2n axis. [If n = 1, we have S2, C1, and i as elements. But C1 = E, so we ignore it. There is only one operation here (2 · 1 − 1 = 1) and it is preempted by i.

Therefore, S2 has no unique operations and it is not listed as a symmetry element.] For S2n with n even, i is not implied and there are n unique operations.

2In certain cases there will be two geometrically nonequivalent sets of vertical planes that cannot be distinguished on the basis of bisecting two-fold axes (e.g., in square-planar XeF4). In such cases the convention is to take the σv planes to be those containing the larger number of atoms.

Chapter 13 Group TheoryEXAMPLE 13-4 How many operations must be associated with S4 in the square planar molecule XeCl4?

SOLUTION  This molecule possesses a C2 and a C4 axis coincident with S4. C2 usurps S24 since the latter has two reﬂections that cancel, making it just like C2. (C2 also usurps C2.) S4 = E.

4

4

Only two operations, S1 and S3, survive as unique symmetry operations

4

4

In eclipsed ethane, we have the element S3, which generates operations S3, S2, S3, S4, S5, S6, . . . In deciding where to stop here, we note that S3 = E because we

3

3

3

3

3

3

have here an odd number of reﬂections. Therefore, S3 = σ , where we understand this

3

to be reﬂection through the plane perpendicular to the S3 axis. For eclipsed ethane, this is σh. The element S6 has an even number of reﬂections and is equal to E. How

3

about S4? This has an even number of reﬂections, so it is identical to C4 which is the

3

3

same as C3C

3

3 = EC3 = C3. Thus, our S3 element has operations that are consistent with the presence of C3 and σh elements. We note that eclipsed ethane was indeed found to have these elements. Once again, we allow these elements to preempt their operations from S3. This leaves only two unique operations, S3 and S5, corresponding

3

to clockwise and counterclockwise improper rotations by 2π/3 radians. In general, S2n+1 will occur only when C2n+1 and σ (perpendicular to C2n+1) are also present.

These preempt a total of 2n + 1 operations from the total of 2(2n + 1) − 1 = 4n + 1 we can achieve with S2n+1 alone; and we are left with 2n operations for S2n+1. The above conclusions are summarized in Table 13-5.

The set of symmetry operations is contrasted to the set of symmetry elements for eclipsed and staggered ethane in Table 13-6. Note that there are 12 symmetry operations in each case. By setting up the multiplication table for either of these sets of 12 operations, we can show that the mathematical requirements for a group are satisﬁed.

Thus, each of these sets of symmetry operations constitutes a separate group of order 12.

The multiplication table for the ammonia model is given in Table 13-7 in terms of the symmetry symbols just described. (A, B, C are taken to be σ1, σ2, σ3, respectively.)

It would be possible, using what has been described up to this point, to construct a list of symmetry operations for any object we please. Then we could test the set to see if it satisﬁed the various requirements for a group. In practice, this is not done.

It turns out that (1) only a rather limited number of distinct kinds of symmetry are possible, and (2) in each case, the symmetry operations for the object do form a group.

The practical question, then, is, given an object, to which of the known groups does it TABLE 13-5  Number of Operations Associated with Symmetry Elements

Element (Other elements present) No. operations

i

—

1

σ

—

1

Cn

—

n − 1

n = odd : Cn and i present n − 1

S2n

n = even : Cn present

n

S2n+1

C2n+1, σ

2n