GERALD KOKOSZKA
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CHEMISTRY 430
GERALD KOKOSZKA |
In this section a set of simple mechanical rules will be developed to allow a student to calculate a reducible representation for a molecule. We will first concentrate on the five symmetry operations as they apply to the xyz coordinate system.
First let us consider the identity operation which is to be represented by a simple 3x3 matrix. The rules of matrix multiplication are reviewed in an appendix.
The question to be answered is: What matrix is needed to convert the column matrix (x,y,z) into itself. Clearly (the appropriate large brackets will be discussed in class.)
1 0 0 x x
0 1 0 y = y
0 0 1 3 z
will provide the necessary result. In other words a matrix with all diagonal elements equal to one and all off-diagonal elements equal to zero leaves the coordinate system in its original position. The character of such a matrix is 3, the sum of the diagonal elements.
For the operation of inversion we ask: What matrix is needed to convert (x,y,z) into (-x, -y, -z). The answer is:
-1 0 0 x -y
0 -1 0 y = -y
0 0 -1 z -z
The trace of this matrix is -3.
The reflection operation requires that a particular plane be selected. Any plane will do but let us choose the xy plane. Then z should go into -z upon the performance of reflection in the xy plane while the x and y axes remain unchanged. The appropriate matrix operator is:
1 0 0 x x
0 1 0 y = y
0 0 -1 z -z
The sum of the diagonal element is +1.
For rotation the problem is a little more complicated. Let us consider the rotation of the vector rl. with components xl and yl in the xy plane through an angle 8. For convenience let rl lie in the first quadrant and the vector into which it is transformed. r2 lies in the second quadrant.
Now the component xl is transformed according to
Xl ->- Xl cos6i + Xl sin6j after rotation by an angle 8. The symbols i and j are unit vectors along the x and y axes respectively.
Similarly the component of rl originally in the y direction is transformed according to
yl = -yl sin 0i + yl cos Oj
therefore the components along the x and y axes of the new vector r2 are
x2 = xl cos8 -yl sin8
y2 = xl sin8 + yl cos6
Had we explicitly considered a z component it would have remained unchanged.
zn - zl
Therefore the appropriate matrix is
Cos -sin 8 0 xi x2
sin Cos 8 0 Yl ' Y2
0 0 1 zl z2
The character of this matrix is 1 + 2 cos8 .
For a rotation about the z axis followed by a reflection in the xy plane the matrix is
cos8 -sin8 0 xi x2
sinO cos8 0 Yl = Y2
0 0 -1 zl z2
The character for the Sn operation is 2cos8 -1.
We have considered the calculation of trances for the xyz coordinate system. It turns out if a molecule is subjected to a symmetry, operation then any atom (or orbital or whatever physical aspect is the object of interest) which moves does not contribute to the character:: That is if the origin of the coordinate system centered on a given atom physically moves under the operation, then there is no contribution to the character under that operation. For those atoms that do not move the results listed above apply. These are summarized as follows:
Let us consider an xyz coordinate system on each atom. Then
1. Under the operation E, each atom in the molecule contributes 3 to the trace. For example, for H20 the character would be 9.
2. Under the operation i, each atom not at the center of symmetry contributes zero and the atom at the center of symmetry contributes -3. For benzene the character is zero while for square planar Cu(NH3)4+2 the character is -3.
3. Under reflection, each atom in the plane of reflection contributes +1 to the total character. So, for example, in water there are two planes of reflection. The one which includes the two hydrogens have a character of +3, the one which does not has a character of +1.
4. Under the operation Cn, the contribution is 2 cos +1 for each atom on the axis of rotation. For water the character is 2(-1)+l = -1. For a 3-fold rotation in PC15 the result is 3 times 2(0)+1 which is 3. The reason for multiplying by 3 is that a phosphorus and 2 chlorinated atoms lie on the axis of rotation. Also note that cos 120 is zero which is quite convenient.
5. Under the operation Sn the contribution is 2 cos8-1 but this can only contribute a non-zero if an atom lies at the center point of the molecule. The S4 axis in CH4 contributes 2(0)-1 = -1 to the character and this is the total character.
Molecular Vibrations
A convenient way to illustrate some of the utility of group theoretical methods is to apply them to problems of molecular vibrations in small molecules. There are several methods for analyzing the so-called "normal modes" of vibrations but it is important at the outset to emphasize that the number of different ways that a molecule can vibrate is fixed and that usually most atoms participate in the basic or "normal" modes. Of course in some biological molecules or larger organic molecules function group and bonding properties often effectively "localize" a mode which is why the IR and Raman techniques may be useful although in very large molecules the spectra become so complex that they are often uninterpretable. In this section we will not attempt a full vibrational analysis for even the relatively small molecules that are to be discussed. Rather we shall concentrate on which bands are IR active, which are Raman active and which are either, neither or both. The real purpose here is to gain some familiarity with the methodology.
In analyzing molecular vibrations, the molecule may be treated classically and thus 3N velocity components are needed to specify the motion in the x, y or z directions of each of the N atoms in a molecule. The number of unique vibrational modes may be determined by subtracting from this number, 3 coordinates corresponding to the translation of the center of mass of the molecule and 3 coordinates corresponding to the three independent rotational modes for a non-linear molecule. Thus there are 3N-6 normal modes of modes in a non-linear molecule and any arbitrary vibrational. motion can be decomposed in one or more of these modes. Furthermore there can be no more than 3N-6 fundamental bands to be found by analyzing the vibrational spectrum. If the molecule is linear there are only two rotational bands and the number of independent vibrational bands is 3N-5.
The distinction between the number of normal modes in linear molecules (3N-5) and those in non-linear molecules (3N-6) is important but is not as discontinuous as one concludes without reflection. Consider the linear OCO molecule. An analysis of its normal modes (3x3-5=4) produces one symmetrical stretch, one antisymmetrical stretch, an in-plane angular deformation and an out-of-plane angular deformation. If the molecule were to become bent in the plane (see figure) then the out-of-plane angular stretch becomes a rotational band. So a vibrational mode "disappears" and is replaced by a rotational mode. Of course the process may be viewed as a continuous one.
Let us now consider the bent molecule H20 which belongs to the point group C2v. Let the Z axis by the two-fold axis and the YZ plane be the plane of the molecule. Now a coordinate system should be placed at the center of each of the three atoms. Thus there are 9 coordinates in all and their time derivatives (which are velocity vectors along the three axes and therefore have the same symmetry properties as the axes themselves) form the basis for constructing the character of the reducible representation corresponding to the full molecular motion of water.
Under operation E the character of the reducible representation is 9 because there are three coordinates for each of the three atoms and none of them are transformed by the identity operation.
The operation C2, the hydrogen atoms are interchanged and therefore both of their origins physically move. Therefore, they contribute nothing to the character. The single atom on the axis of rotation is the oxygen atom and it is rotated by 180°. The character is therefore 2 cos 180+1 = 2(-1)+l=-1.
For reflection in the XZ plane which corresponds to the three individual XZ planes located on each of the atoms, the two hydrogen atoms again are interchanged and therefore contribute nothing to the trace under this operation. The rule listed above that each atom on the plane of reflection contributes +1 to the total trace leads to a character of +1 for the reducible representation corresponding to the full molecular motion of H20.
Finally for reflection in the YZ plane, the total character is +3 because
each of the three atoms in the molecule lies in this plane. The result is
E C2 a(XZ) 6(YZ)
rred 9 -1 +1 +3
The next step is to decompose the reducible representation into a sum of the appropriate irreducible representations. There are four irreducible representations in the point group C2v: The number of times Ni that the with irreducible representation occurs in the reducible representation r red is given by
Ni = 1 E WRXR(irr. rep.) XR(1)
h R
where h is the order of the group and is equal to the number of actual operations, R are the operations in the group. WR is the coefficient in front of the operation which really indicates how many times that operation or another operation in the same class (and therefore with the same character) appears in the group. XR (irr. rep.) is the character of the irreducible representation under the operation R and XR(i) is the character of the with irreducible representation under the operation R.
In the group C2v there are four operations and the weighting factor for each of them is one. Thus h=4 and w=1 in all cases. The number of times that Al occurs in red is given by
NA, = _1 [(9xlxl) + (lx-lxl) + (lxlxl) + (3xlxl)] = 3
4
The final result must be whole number and must be less than 9, the dimensionality of the reducible representation. In a similar way:
NA2 = _1 [9xlxl) + (-lxlx1) + (lx-lx-1) + (lx3x-1)]= 1
4
NBl - _1 [(9xlx1) + (x-lx-1) + (lxlxl) + 3x-lx-1)] = 2
4
N B2 = 1 [(9xlxl) + (-lxlx-1) + (lxlx-1) + (3xlx1)l = 3
4
Therefore the reducible representation is equal to
3A1 + A2 + 2B1 + 3B2
Note that the dimensionality of the reducible representation is 9 and that the four different irreducible representations are all of dimensionality 1: Thus there are nine dimensions present after the decomposition as there should be.
Since water is a non-linear molecule there are 3 translational and 3 rotational components to the total molecular motion. Thus there should be 3 normal modes of vibration making a total of 9 "degrees of freedom" or 9 independent kinds of motion. The three translational degrees of freedom transform in the same way as the coordinates. Thus, from the character table, we can see the Z transform as A1, X and B1 and Y as B2. The rotation degree of freedom can also be specified as to their symmetry type by direct reference to the character table: RZ belongs to A2, RX to 82 and RY to Bl. The 6 reducible representations for translational and rotational motion are:
Al + A2 + 281 + 2B2
When these are subtracted from the sum of the decomposition for the total motion, the vibrational motion transforms as
2A1 + B2
We now wish to determine which bands are infrared active and which are Raman active. Before carrying out the details of this operation it should be noted that the three species associated with vibrational motion are actually the symmetry of the first excited state in each case as the ground state for all modes is the total symmetric representation Al or Alg. The justification of this statement appears in a later chapter.
In order to determine if a band is infrared active in Z polarized electromagnetic radiation of the proper frequency, we must calculate the reducible representations of the product of the symmetry type of the (1) ground state (A1), (2) the appropriate electric dipole Hamiltonian (Z) corresponds to Al and the excited state (Al or Bl).
For Al as an excited state the character under all the operations is 1. (The product of lxlxl for each of the four operations) and the reducible representation is, in fact, an irreducible representation A1. In fact, the product of any three one-dimensional irreducible representations will always be an irreducible representation. It may also be noted that multiplying any irreducible representation by A1 (or Alg) will leave it unchanged. Here A1 x A1 x Al = Al.
Since the product representation contains A1 (in this case is equal to A1) the transition is allowed in Z polarized light. Further since there are two Al vibrational modes, two bands would appear in this hypothetical experiment. If X polarized light of the same frequency were directed at the molecule then the appropriate direct product would be A1 x B1 x A1 = Bl (ground state) (Hamiltonian)(excited state) = product and the transitions would not be allowed for y polarized light:
A1 x B2 x A1 = B2
and again the transition would be symmetry forbidden even-though the energy of the incoming radiation would match the separation between the energy levels. The other vibrational band is of type B2. The product:
Al x Al x B2 = B2
produces the result that this transition is forbidden in Z polarized light. For X polarized light the product:
A1 x Bi x B2
must be calculated. The character under E is lxlxl, under C2 lx-lx-1, under
c5(XZ) l*lx-l and under (;(YZ) lx-1x1. Thus it is
E C2 a(XZ) 6(YZ)
1 1 -1 -1
and the product representation is identical with A2 and therefore the transition is forbidden. Finally, for Y polarized light the product is:
Al x 82 x B2
The character under E is lxlxl = 1, under C2 it is lx--lx-1 = 1, under C(XZ) it is lx-lx-1 = 1 and under 6(YZ) it is lxlxl = 1. Thus all the four characters are 1 and the representation must be Al. Therefore the B2 vibration band would be observed with Y polarized light. A general result may be stated here that the product of an irreducible representation with itself will always contain or be equal to Al (or Alg).
For water we have the result that two of the bands are allowed in Z polarized light and a third in Y polarized light. If a large number of water molecules are present with random orientation and with unpolarized light then all three possible bands will be observed in the infrared spectrum.
In Raman spectroscopy, the term for the electromagnetic radiation corresponds to the irreducible representation for any product of coordinates. Since the ground state always transforms as Al, the product of the radiation term and the fundamental must be examined to see if the Al bands are Raman active we ask if there is a r that satisfies.
rAl = Al
in which 1' is an irreducible representation which has a product of
coordinates in it. The answer is yes if it is Al itself. Also can we find a r
such that
TB2 = Al
Again, the answer is affirmative with r= B2
Thus all three bands are IR and Raman active. Water is a simple molecule and we might imagine that the normal modes could correspond to simple angular or radial displacements. Let Rl be one HO vector, r2 be the second HO vector and be the HOH angle. Then the representations for 'these coordinates are:
E C3 C7(Xz) cF(YZ)
fa 1 1 1 1
ra 2 0 0 2
Thus
ra= Al
rr = A1 + B2
As it turns out there is a mixing of the radial and angular coordinates of type Al. In S02 and 0120 this mixing can be quite complicated. But in water one of the AI's is mostly an OH stretch and the other an angular deformation. The B2 mode is antisymmetrical stretch. This kind of detailed analysis is beyond the scope of the present text.
Another example is NH3, which belongs to the group C2v„ There are 4 atoms in the molecule and the total number of coordinates for. a complete specification of the molecular motion is 12. The character of the reducible representation is
E 2C3 3Crv
12 0 2
Note the order of the group is 6. because there are 6 operations: one identity, 2 rotation and 3 reflections. In order to decompose the reducible representation we use the formula listed earlier:
NA, = _1 (12+0+61 = 3
6
NA2 = _1 [12+0-61 = 1
6
NE = _1 [24+0+01 = 4
6
Therefore
rred=3A1+1A2+4E
The dimensionality of the reducible representation is 12 and it is decomposed into 4 one-dimensional representations and 4 two-dimensional representations so that the total dimensionality is preserved. There are three dimensions associated with the total translational motion (Al + E) and three with the rotational motion (A2 + E) leaving 2A + 2E as the irreducible representations for vibrational motion. Note tat there are 6 possible normal modes [4x3-6j but the two fold degeneracy of two modes mans that a maximum of four bands can be observed. All four are both infrared and Raman active.
Cis N2F2 is in the group C 2v, the same as water and can be analyzed by the student as an exercise while trans N2F2 is a centrosymmetric molecule in the group C2h. The reducible representation is
E C2 i O'h
12 0 0 4
which can be decomposed into
4A g + 2B g + 2Au + 4Bu
After the translational (Au + 2Bu) and rotational (A g + 2Bg) modes have been eliminated, the vibrational possibilities are
3A g + Au + 2Bu
There are 6 possible fundamental absorptions which might be observed but the exclusion rate associated with the center of symmetry is operative here. Three bands, Au + 3Bu are infrared active and the 3A g bands are Raman active.
Optical Spectra of Metal Complexes
As another example of electric dipole induced transitions the simple crystal field spectra of ion with a single d electron (say V4+ or Ti+3) in a low symmetry environment may be considered. Let us consider a distortion which reduces the Td symmetry to C2v. The energy level diagram places the X2-Y2 (Al) level lowest followed by Z2(Al), XY(A2), XZ(Bl) and YZ(B2). The direct product is to be evaluated to see if it is equal to or contains Al where GS and ES stand for ground state and excited state and EM stands for the polarization of the electric field portion of the incident radiation. The latter is assumed to have the correct energy. The table summarizes the results;
Ground Excited Allowed Forbidden
State State Polarization Polarization
X2y2 Z2 Z X,Y
X2y2 XY - X,Y,Z
X2y2 XZ X Y,Z
XY2 YZ Y X,Z
Of course in high symmetry environments, especially when a center of symmetry is present, d to d transitions are electric dipolar forbidden. Historically this is referred to as Laporte forbidden bends. The symmetry of the two d orbitals must be even and the electric dipole transition operator (which is proportional to T') must be odd. Therefore the integral over a symmetrical region is zero. The same agreement applies to p ; p or f ~ f transitions.