Appendices

Appendices

APPENDICES APPENDIX 1 Projection Operators In Chapter 6 (Section 6.3) the limitations of a pictorial approach to generating MOs was touched on. A SA...

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APPENDICES

APPENDIX 1 Projection Operators In Chapter 6 (Section 6.3) the limitations of a pictorial approach to generating MOs was touched on. A SALC is a linear combination of atomic orbitals: SALC 

= N[c11 + c22 + c33………]

(eqn. A.1)

where N is a normalisation constant and i are the wavefunctions for individual AOs. In the qualitative approach to describing MOs adopted throughout this book the coefficients (ci) associated with each AO contribution are ignored (i.e. assumed to be equal). Although this does not affect significantly the pictorial description of the MO, it lacks a level of detail which can, at times, lead to problems. It has already been commented on that these coefficients are not equal when the AO energies are different, but, in addition, symmetry also dictates that they need not necessarily be equal even when the AO energies are the same. This Appendix will set out the rigorous approach that is required to generate exact forms of the SALCs using a technique called projection operators. This is a difficult topic and only the basic methodology will be described. In addition to its application to MO theory, the same technique can be used to generate the forms of the vibrational modes modes discussed in Part 2 of this book (see Section 5.1), which are themselves SALCs, but now linear combinations of individual vibrational modes. A.1 THE BASICS – NON-DEGENERATE SALCs In Section 7.1 we saw that the SALCs for the two hydrogen atoms in water (C2v) could be described by the irreducible representations a1 + b2. Since there are only two ways in which the two hydrogen AOs can combine it was trivial to draw these possibilities: z y

How can we arrive at these in a more formal way, so that more difficult situations can be analysed? We do this using a projection operator technique, which is essentially a mechanism for automatically generating algebraic functions such as in equation A.1. As with the reduction formula, the appearance of the projection operator formula looks intimidating but is relatively easy to apply: Pij =  (  ( IR ) R j ) sum over all classes of operation R R

(eqn. A.2)

Pi is the projection operator for a given irreducible representation j is the chosen basis (or generating) function (IR) is the character in the irreducible representation for operation R Rj is the effect of performing operation R on the basis function j

174

[App. 1

Projection Operators

This rather daunting equation says "the result of applying the projection operator Pi to a chosen basis function j is arrived at by summing, over all operations R in the point group, the product of the character in the irreducible representation for the operation (IR) and the effect of the operation on the basis function"! An example will illustrate this is not as difficult as it sounds. Let us take one of the H 1s orbitals as our basis function, 1: z y

1

2

We need to find the SALCs associated with the a1 and b2 representations, so for a1: C2v 1  (IR) a1 (IR)R1

E 1 1 1

C2 2 1 2

(xz) 2 1 2

(yz) 1 1 1

= 21 + 22

Since we are only interested in the relative contributions of the two AOs to the MO, this can be simplified to: a1 = 1 + 2 which is what is expected for the in-phase combination of AOs that make up the bonding MO. The SALC needs finally to be normalised to be completely accurate, so that the total electron density sums to one electron. Since the probability of finding an electron at any point is given by 2, we need a normalisation constant N which reduces the sum of the squares of the coefficients (ci in equation A.1) to unity. Since, for the a1 SALC, c1 = c2 = 1, N = 1/2. i.e. a1 = (1/2)1 + (1/2)2 = 1/2(1 + 2), so that (1/2)2 + (1/2)2 = 1 In general, N = 1/n, where n = (ci)2. SAQ A.1 : Using the projection operator method with 1 as basis function, show that the MO of b2 symmetry in H2O is of the form 1/2(1 - 2). Answers to all SAQs are given in Appendix 3. A final check to make sure the SALCs that have been generated are correct is that they must be orthogonal to each other. The mathematics of this need not concern us, we simply note how it is done:  functions are orthogonal if the sum of the products of corresponding coefficients is zero i.e. product of coefficients of 1 + product of coefficients of 2 ..etc.

App.1]

175

Projection Operators

For the two MO SALCs of water this is simply (1  1) + (1  1) = 0 A.2 DEGENERATE SALCs Problem 1 at the end of Chapter 8 looked at the MO diagram for the -bonds in BF3 (D3h). 3

z

C2 (1), v (1) 1 2

The pz orbital on boron has a2'' symmetry, while the SALCs associated with the three pz orbitals on fluorine transform as a2'' + e'' (see Appendix 4, Chapter 8, Problem 1). The form of these SALCs can easily be drawn using the methodology of Chapter 8 and SAQ 8.2:

a2''

e''

The match beween the fluorine SALC and boron AO each of a2'' symmetry is clear, but close consideration of the mismatch between this AO on boron and the e'' fluorine SALCs raises a problem: the pz on boron has equal amounts of bonding and anti-bonding overlap with one SALC making it a non-bonding interaction overall, but with the other SALC there appears more bonding than anti-bonding, which cannot be correct. A consideration of the exact form of the SALC will rationalise this apparent anomaly. We begin by applying the projection operator method using 1 as the generating function. In seeing how this transforms under the operations of the point group, we must consider all the operations of the point group individually. For the a2'' SALC: D3h E 1  1 1 (IR) a2'' (IR)R1 1 …continued D3h S3 1 1  2 (IR) a2'' 1 (IR)R1 2

C3 1 2 1 2

C32 3 1 3

C2(1) 1 1 1

S3 5 3 1 3

v(1) 1 1 1

v(2) 3 1 3

C2(2) 3 1 3

v(3) 2 1 2

C2(3) 2 1 2

h 1 1 1

176

[App. 1

Projection Operators

Note that the "2S3" operations in the shortened form of the character table refer to S31 and S35; S32  C32, S33  h, S34  C31 and S36  E. Thus, a2'' = 41 + 42 + 43, or a2'' = 1 + 2 + 3 considering only the relative contributions of each AO. Including the normalisation coefficient, this becomes 13(1 + 2 + 3). By the same method, for the e'' SALC: E 1 2 21

C31 2 1 2

C32 3 1 3

C2(1) 1 0

C2(2) 3 0

…continued D3h S3 1 1  2 1 (IR) e'' (IR)R1 2

S3 5 3 1 3

v(1) 1 0

v(2) 3 0

v(3) 2 0

D3h 1  (IR) e'' (IR)R1

C2(3) 2 0

h 1 2 21

e'' = 41 - 22 - 23 or e'' = 21 - 2 - 3 [normalised : 1/6(21 - 2 - 3)] Pictorially, we should redraw the second e'' SALC showing the double contribution from 1, and which now has equal amounts of bonding and antibonding overlap with the pz on boron making the interaction, correctly, non-bonding:

SAQ A.2 : Show that the two SALCs derived above of a2'' and e'' symmetry are orthogonal. While this is one problem resolved, another has been raised. Using 1 as generating function has only yielded the form of one of the two SALCs of e'' symmetry. To generate the form of the other half of this degenerate pair we need a new generating function and this is where the use of projection operators becomes less intuitive. It might seem reasonable to try either 2 or 3 as an alternative generating function, but neither of these gives acceptable answers. Using 2 as generating function, and ignoring those operations for which (IR) is zero for e'': D3h 2  (IR) e'' (IR)R1

E 2 2 22

C31 3 1 3

C32 1 1 1

h 2 2 22

S3 1 3 1 3

S3 5 1 1 1

App.1]

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Projection Operators

This gives e'' = 42 - 23 - 21 or e'' = 22 - 3 - 1 (ignoring normalisation). However, this is not orthogonal to the other e'' function, as multiplying common coefficients for 1, 2 and 3 shows : (2  1) + (1  2) + (1  1) = 3, not 0 as required for othogonality. SAQ A.3 : Determine the e'' function that is generated using 3 as generator and show that this is not orthogonal to the function generated for e'' using 1. The remaining e'' function can be arrived at using 2 - 3 as the generating function. It is not at all intuitive that this is the generator to chose, and this is one of the limitations of the projection operator method at a basic level of group theory. Rational ways of determining the choice of basis functions for degenerate MOs (and vibrational modes) i.e.those of e or t symmetry, can be found in more advanced texts on the subject.† Using 2 - 3 as generator and the same methodology as before gives: D3h 2 - 3  (IR) e'' (IR)R1

E 2 - 3 2 22 -23

C31 3 - 1 1 -3+ 1

C32 1 - 2 1 -1+ 2

h -2 + 3 2 22- 23

S3 1 -3+ 1 1 -3+ 1

S3 5 -1 + 2 1 -1+ 2

From which: e'' = 62 - 63 or e'' = 2 - 3 [1/2(2 - 3) including normalisation] This is both orthogonal to the other e'' SALC [(2  0) + (1  1) + (1  1) = 0] and consistent with the pictorial representation of the SALC shown at the beginning of this Section. A.3 VIBRATIONAL MODES We can also use projection operators to generate the functions which represent modes in a vibrational spectrum, as these can be considered as linear combinations of individual stretching or bending modes, in a manner dictated by symmetry, just as SALCs of AOs have been derived above. Using the methodology described in Chapters 3 and 5, the vibrational modes of NH3 (C3v) can be shown to be: vib = 2A1 + 2E N-H = A1 + E bend = A1 + E

For example: Molecular Symmetry and Group Theory, R L Carter, John Wiley and Sons, 1998. Chemical Applications of Group Theory 3rd Edition, F A Cotton, John Wiley and Sons, 1990. †

178

[App. 1

Projection Operators

We can use the vectors v1, v2, v3 to generate the forms of the stretching modes and the angles α1, α2, α3 for the bends:

v3

H

N

H

H

H



v1, v (1)

N H

v2



v (1)

H 

For the stretching modes, using v1 as generator gives: C3v v1  (IR) A1 (IR)Rv1 (IR) E (IR)Rv1

C31 v2 1 v2 1 v2

E v1 1 v1 2 2v1

C32 v3 1 v3 1 v3

σv(1) v1 1 v1 0

σv(2) v3 1 v3 0

σv(3) v2 1 v2 0

→ 1/3(v1 + v2 + v3) → 1/6(2v1 - v2 - v3)

The second component of the E mode requires v2 - v3 as generating function: C3v v2 - v3 (IR) E (IR)Rv1

E v2 - v3 2 2v2 - 2v3

C31 v3 - v1 1 -v3 + v1

C32 v1 - v2 1 -v1 + v2

→ 1/2(v2 - v3)

H

H

Pictorially, these appear as: N

H H

H

N

H H

N H

H

A1 E For completeness, in each case the central nitrogen will also move to keep the molecule stationary, but this has been omitted for clarity of the stretching motions, both above and in further examples in this Appendix.

SAQ A.4 : Use 1 and 2 - 3 as generators to determine the functions which describe the bending modes for NH3. There are clear analogies between the forms of these vibrational SALCs and the SALCs for cyclic combinations of AOs (Chapter 8). The A1 bending mode (1 + 2 + 3) requires all angles to behave the same (the "+" in Figure A.1 shows them all expanding at the same time) which occurs by each hydrogen moving up out of the H3 plane (the molecule becomes less pyramidal). The first of the E bends has one angle expanding and two contracting, analogous to the AO SALC in which two AOs are

App.1]

179

Projection Operators

in-phase but out-of-phase with the third. The second E bend has one angle expanding at the expense of one other, the third angle remaining static. AO SALCs N

stretching modes

H

H

H

H N

+

H

H

_ _

_

H

H H

H

+ +

H

H

H

bending modes

N

N

H

N H

H

N

+

H

+

H

Figure A.1. Hydrogen SALCs and vibrational modes for NH3.

Thus, without recourse to projection operators the method for generating cyclic SALCs from Chapter 8 also offers a simplistic way of depicting the vibrational modes. A final example, referring back to our analysis of the vibrational spectrum of [PtCl4]2 (Chapter 5), will reinforce this, but also highlight that this pictorial approach too has its limitations. From a knowledge of how the four SALCs for a cyclic array of AOs are arranged (SAQ 8.2), we would predict the following for the vibrational modes of [PtCl4]2: SALCs

Stretching modes Cl

Cl

Cl

Cl

Cl

Cl

Cl

Pt

Pt

Cl

Cl

Cl

Pt Cl

Cl

Cl

Cl

Cl

Pt

Cl

In-plane bending modes Cl

+

Cl

Pt

Cl

+

+ Cl

+

+ Cl

Cl

_

+ Pt

_

Cl _

Cl

Cl

_

Cl

_

+ Pt

Cl

+ Cl

Cl

Cl

+ Pt

Cl _

+ Cl

180

[App. 1

Projection Operators Out-of-plane bending modes

Cl

Pt

Pt

Cl

Cl Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Pt

Pt

Cl

Cl

Cl

Cl

Figure A.2. Predicted vibrational modes for [PtCl4]2.

From Chapter 5 (including SAQ 5.1), the vibrational modes for [PtCl4]2 were shown to be (redundancies in parentheses): vib

= A1g + B1g + B2g + A2u + B2u + 2Eu

Pt-Cl

= A1g + B1g + Eu

in-plane

= (A1g) + B2g + Eu

out-of-plane = (Eg) + A2u + B2u The functions corresponding to the stretching modes can be determined simply from the projection operator method. v4 Cl

Cl

4 v3

Cl

Cl

Pt

v1

C2 (1)', v (1)

 Cl

Pt

Cl



Pt

Cl

Cl

 Cl



Cl

C2(1)'', d(1)

v2

Cl 

3

Cl

1

SAQ A.5 : Use v1 (for A1g, B1g, Eu) and v2 (for Eu) as generators to determine the functions which describe the stretching modes for [PtCl4]2. The A1g (v1 + v2 + v3 + v4) and B1g (v1 - v2 + v3 - v4) modes are as in Figure A.2, while the Eu modes (v1 - v3; v2 - v4) are simply the symmetry-equivalent sum and difference pairs from those drawn (see Section 9.1 for a related case):

Cl

Cl

Cl

Cl

Cl

Cl Cl

=

Pt Cl

Cl Pt

Cl

Cl

Cl

= Cl

Cl

Cl

Cl

Cl

-

Pt Cl

Pt

+

Pt

Cl

Cl

Cl

Cl

Pt Cl

Cl

App.1]

Projection Operators

181

The out-of plane bending modes are arrived at in an analogous manner, using angles 1 and 2 as generating functions. SAQ A.6 : Use 1 (for A2u, B2u, Eg) and 2 (for Eg) as generators to show the functions which describe the out-of-plane bending modes for [PtCl4]2 are: A2u = 1/2(1 + 2 + 3 + 4) B2u = 1/2(1 - 2 + 3 - 4) Eg = 1/2(1 - 3), 1/2(2 - 4) Note that the Eg redundancy among the out-of-plane modes is because these are, in fact, rotations about x and y (Fig. A.2). Deriving functions for the in-plane bending modes has the recurrent difficulty for this basic description of projection operator methods of choosing the correct generating function. While α1 and α2 would seem logical based on the methodology of SAQ A.5 and SAQ A.6, these are not completely acceptable. They can be used as generators for the singly-degenerate A1g and B2g modes, but not for the degenerate Eu in-plane bends. This is because it is a requirement of the generating functions for degenerate systems that they all have the same symmetry, and in this respect α1 and α2 differ from v1 and v2 (or 1, 2) used to derive the functions for the degenerate Eu stretching and E out-of-plane bending modes. For example, under C2'(1), v1  v1 (i.e. itself) but α1  α4 (not itself). Functions (α1 + α2) and (α1 + α4) do meet these requirements [e.g. C2'(1), (α1 + α4)  (α1 + α4)] and the interested reader may wish to show that using this combination of functions generates the following, consistent with Figure A.2: A1g = 1/2(α1 + α2 + α3 + α4) B2g = 1/2(α1 - α2 + α3 - α4) Eu = 1/2(α - α3), 1/2(α2 - α4) This appendix concludes with a note of caution. Depicting molecular vibrations using combined qualitative approach in parallel with the use of projection operators is, in itself, not without limitations. We have already seen in the platinum example that apparently different, but symmetry-equivalent, SALCs can be derived from the two methods. This is usually a problem only with doubly- and triply-degenerate species (E, T labels), and it is up to the user to make use of the most appropriate form for the problem being addressed.

APPENDIX 2 Microstates and Term Symbols This brief overview sets out a stepwise strategy for determining the microstates for a given electronic configuration and how these are divided up into groups, each described by a term symbol. The microstates and term symbols for the d2 configuration will be used as an example. Step 1 : Determine the degeneracy of the configuration This is done using eqn.12. 1: Dt = (N)! / (Ne)!(Nh)! Dt = (2  5)! / 2!  8! = 10  9 / 2 = 45 Step 2 : Determine the maximum L and S for the configuration For each d-electron, ml = 2, 1, 0, 1, 2, so maximum L = 4 (2 + 2). The two electrons can both have ms = ½ so maximum S = 1. Step 3 : Determine the range of ML and MS values associated with maximum L, S L = 4 has associated ML values of 4, 3, 2, 1, 0, 1, 2, 3, 4; S = 1 has associated MS values of 1, 0, 1. The range of possible microstates must lie within a grid which spans these ML / MS ranges (Table A.1). Step 4 : Generate and complete a grid of microstates The maximum ML (= 4) arises from ml = 2 + 2. As the two electrons are in the same d-orbital their spins must be paired given the table entry (2+,2-). The next highest possible ML = 3, and arises from ml = 2 + 1. As the two electrons are in different orbitals they can take either spin, giving rise to four microstates (2+,1+), (2+,1-), (2-,1+), (2-,1-). This process can be followed upto and including the row for ML = 0, after which point the entries for ML = 1, 2, 3, 4 mirror those for ML = 1, 2, 3, 4. The complete grid of microstates is given in Table A.1. Step 5 : Identify the terms and their associated microstates from the grid At this point it is worth copying Table A.1 so that the microstates can be crossed off as each term is identified. The first entry in the Table is (2+,2-) corresponding to M = 4 and MS = 0. This is one microstate that is part of the 1G term (L = 4  G; S = 0  singlet) with an overall degeneracy of (2L+1)(2S+1) = 9. The remaining eight microstates all have S = 0 (this is the only value associated with MS = 0) and ML = 3, 2, 1, 0, 1, 2, 3, 4 and are shown shaded in the partial grid of Table A.2 e.g. (2+,1-), though it doesn't matter which entry in any given ML / MS square is removed. These microstates should now be crossed off from Table A.1 Of the microstates remaining the one with highest L and S is (2+,1+), with L = 3 and S = 1; this is part of a 3F term (L = 3  F; S = 1  triplet). 3F has a degeneracy

App.2]

183

Microstates and Term Symbols

Table A.1 Microstate table for a d2 configuration.

ML 4 3

(2+,1+)

2

(2+,0+)

1

(2+,-1+) (1+,0+)

0

(2+,-2+) (1+,-1+)

-1

(-2+,1+) (1+,0+)

-2

(-2+,0+)

-3

(-2+,-1+)

1

-4

MS 0 (2+,2-) (2+,1-) (2-,1+) (2+,0-) (2-,0+) (1+,1-) (2+,-1-) (2-,-1+) (1+,0-) (1-,0+) (2+,-2-) (2-,-2+) (1+,-1-) (1-,-1+) (0+,0-) (-2+,1-) (-2-,1+) (-1+,0-) (-1-,0+) (-2+,0-) (-2-,0+) (-1+,-1-) (-2+,-1-) (-2-,-1+) (-2+,-2-)

-1 (2-,1-) (2-,0-) (2-,-1-) (1-,0-) (2-,-2-) (1-,-1-)

(-2-,1-) (-1-,0-) (-2-,0-) (-2-,-1-)

Table A.2 Partial microstate table for a d2 configuration highlighting some of the 1G microstates.

ML 4 3

(2+,1+)

2

(2+,0+)

1

(2+,-1+) (1+,0+)

0

(2+,-2+) (1+,-1+)

1

MS 0 (2+,2-) (2+,1-) (2-,1+) (2+,0-) (2-,0+) (1+,1-) (2+,-1-) (2-,-1+) (1+,0-) (1-,0+) (2+,-2-) (2-,-2+) (1+,-1-) (1-,-1+) (0+,0-)

-1 (2-,1-) (2-,0-) (2-,-1-) (1-,0-) (2-,-2-) (1-,-1-)

184

[App. 2

Microstates and Term Symbols

of 21 (7  3), and the remaining twenty microstates have combinations of ML = 3, 2, 1, 0, 1, 2, 3 and MS = 1, 0, 1 (excluding 2+,1+ already identified). For example, for ML = 3, MS = 1, 0, -1 so a microstate from each of the three boxes in the ML = 3 row need to be removed; the process is then repeated removing a microstate from each of the three boxes in the ML = 2 row etc. These, and the other microstates of the 3 F term, are shown in Table A.3 in bold e.g. (2+,1+) and should be crossed off the table. Table A.3 Microstate table for a d2 configuration grouping microstates of the same term a

ML 4 3

(2+,1+)

2

(2+,0+)

1

(2+,-1+) (1+,0+)

0

(2+,-2+) (1+,-1+)

-1

(-2+,1+) (1+,0+)

-2

(-2+,0+)

-3

(-2+,-1+)

1

-4

MS 0 + (2 ,2 ) (2+,1-) (2-,1+) (2+,0-) (2-,0+) (1+,1-) (2+,-1-) (2-,-1+) (1+,0-) (1-,0+) (2+,-2-) (2-,-2+) (1+,-1-) (1-,-1+) (0+,0-) (-2+,1-) (-2-,1+) (-1+,0-) (-1-,0+) (-2+,0-) (-2-,0+) (-1+,-1-) (-2+,-1-) (-2-,-1+) (-2+,-2-)

-1 (2-,1-) (2-,0-) (2-,-1-) (1-,0-) (2-,-2-) (1-,-1-)

(-2-,1-) (-1-,0-) (-2-,0-) (-2-,-1-)

G shaded e.g. (2+,2-); 3F in bold e.g. (2+,1+); 1D in italic e.g. (1+,1-); 3P in underscore e.g. (1+,0+); 1S in standard font e.g. (0+,0-).

a1

The process is now continued, starting with (1+,1-) which is the microstate with the highest ML /MS remaining. This is the start of five microstates belonging to a 1D term (remove these), then nine microstates forming a 3P term (remove these), leaving, finally, a single 1S microstate. This is left as an exercise for the reader.

APPENDIX 3 Answers to SAQs Chapter 1 SAQ 1.1 There are C4 and C2 axes which are coincident and pass through the centre of the ring, perpendicular to the ring plane. Additional C2 axes are indicated in the figure. The principal axis is C4, the axis with highest n. C2 H

H C . C C C

H

H

SAQ 1.2 h contains molybdenum and the four CO groups; there are two v planes which contain Mo and pairs of CO groups mutually trans to each other, and there are two d which contain molybdenum and bisect the angles C-Mo-C. SAQ 1.3 In all questions of this type, it is imperative to first determine the correct molecular shape, using VSEPR if the compound has one of the s- or p-block elements as the central atom. In this case, I = 7e, 7 fluorines contribute a total of 7e, making 14e i.e. 7 e pairs in total. As there are 7 bonds, there are no residual lone pairs and the molecule has a pentagonal bipyramidal shape. The S5 axis is coincident with the axial F1-I-F2 fragment. S51 moves each of the equatorial fluorines around by one position (a rotation of 72o) but leaves F1 and F2 fixed. The reflection part of the operation, which takes place with respect to the equatorial IF5 plane, then exchanges F1 and F2. After S55, the five “rotation through 72o then reflect” operations have brought all the equatorial fluorines back to their original positions, but the odd number of reflections required by S55 means that F1 and F2 are still swapped around. It is only when ten “rotation-reflection” operations have been carried out i.e. S510 that the original position is reverted to. Thus, in general, when n is odd, Sn2n  E. SAQ 1.4 After five S5 operations all the equatorial fluorines have returned to their original locations, but the odd number of reflections means that the axial fluorines have swapped positions. S55 is thus equivalent to h.

186

[App. 3

Answers to SAQs

SAQ 1.5 S2  i. The S2 point group arises when a molecule possesses only an S2 axis i.e. S2 in combination with C1. Since an S2 improper axis is equivalent to i (C2 + h), then the S2 point group is equivalent to the Ci point group (only a inversion centre in addition to C1). SAQ 1.6 The shape of PF5 is trigonal bipyramidal. It is not one of the high symmetry linear or cubic point groups though there is a C3 principal axis (along Fax-P-Fax). There are three C2 axis perpendicular to C3, (along each P-Feq bond) and the PF3 equatorial plane is a h. The point group is therefore D3h. SAQ 1.7 (a) Cs ; not chiral but polar. (b) C2 ; chiral and polar. (c) D2d ; neither chiral nor polar. The symmetry elements in (c) and (d) are most easily seen in Newman projections:

d C2

Me

Me C

C2

H

C2

H H

C

H

d

C2

H

H

(c)

(d)

Chapter 2 SAQ 2.1 E C2 (xz) (yz)

E E, (yz) C2, (xz) C2, (xz) E, (yz)

C2 C2, (xz) E, (yz) E, (yz) C2, (xz)

(xz) C2, (xz) E, (yz) E, (yz) C2, (xz)

(yz) E, (yz) C2, (xz) C2, (xz) E, (yz)

SAQ 2.2 The inverse of C31 (rotate through 120o) is C32 (rotate through 240o) i.e. C31  C32 = E. The inverse of S53 is S57, as S510  E. Although S52 would complete a 360o rotation, the odd number of reflections requires a second series of five improper rotations (see SAQ 1.3), so S52 is not the inverse of S53.

App.3]

187

Answers to SAQs

SAQ 2.3 The product matrix has 3 rows and 1 column.  2  3    2

4 5 7

 1 0  3

 2  3     2

18 = 21 e.g. z21 = (3  2) + (5  3) + (0  –2) = 21 11

1 0  0  0 0

0 0 0 0 0 0 0 1  0 0 1 0  1 0 0 0 0 1 0 0

SAQ 2.4

SAQ 2.5

(E ) Tx (C2) Tx ((xz)) Tx ((yz)) Tx

C  C  H  H   1  4 H 2  = H 3      H 3  H1   H 4   H 2 

= (1) (Tx) = (–1) (Tx) = (1) (Tx) = (–1) (Tx)

SAQ 2.6 [(xz)  (yz)]  C2 = [(1)  (–1)]  (–1) = 1 (xz)  [(yz)  C2] = (1)  [(–1)  (–1)] = 1 SAQ 2.7 For example: (xz)  (yz) = C2, but –1  –1  –1 SAQ 2.8

C31  C32 = E For the representations based on Tz and Rz, this becomes 1  1 which = 1, the character for E in each case. For the representations based on (Tx, Tz) and (Rx, Rz): 3   1  3    1 2  2 2 =  2  3 2  1 2    3 2  1 2   1 x  1    3 x  3  12 x 2 2 2 2  3 x  3 2 x  1 2   1 2 x  3 2 2



 

  

3

2

3

   

2

3



x  12   1 x 1  2 2  2

=

1 0   0 1  

188

[App. 3

Answers to SAQs

SAQ 2.9  1 0   0 1    1  3  2  2  3 2  1 2 

 1 2  3   2

3





2 1  2

 1 0    0 1   12  3 2   3  1  2 2  

1+1=2 –½ + (–½) = –1

 1  2  3 2

–½ + (–½) = –1

3

  1  2  2

1 + (–1) = 0 –½ + ½ = 0

–½ + ½ = 0

Chapter 3 SAQ 3.1  1 0  0  0 0  0 0  0 0 

0 0 1 0

0 0

0 0  S x    S x      0 0  S y   S y   0 0 0 0  S z   S z  1  1   0 0 0 0  O x    O x  1 1 0 0 0 0  O y  =  O y    1   1 0 0 0 O z   O1z  0  1 0 0 O 2x    O 2x      0 0 1 0 O 2y   O 2y      0 0 0 1 O 2z   O 2z  while vectors along x on all three atoms are

0 0 0 0

0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0

0 0

0 0

0 0 0 0

0 0

0 0

All y, z vectors are unmoved reversed.

0 0

SAQ 3.2 The six vectors lying in the yz plane are unmoved (6  1) while all the vectors lying along x are reversed on reflection in the yz mirror plane (3  –1). Thus,  = 6 – 3 = 3. SAQ 3.3 Total number of operations (g) = 6 (E, 2C3, 3v). Remember, the “2C3” refers to the two operations C31 and C32 and not two C3 axes. C33 is already in the character table as E. The number of operations in each class (nR) is 1 (E), 2 (C3) and 3 (3v). SAQ 3.4

C4v 3N

E 21

2C4 3

C2 –3

2v 5

2d 3

App.3]

189

Answers to SAQs

E : 21  1, as all vectors unmoved. C4 : Only the vectors on W and the halogens on the z-axis are unmoved, all other vectors move. For the three atoms on z, their vector on z is unmoved (3  1) while their vectors along x and y rotate by 90o to new positions (6  0). C2 : Only vectors associated with the atoms lying on the z-axis contribute (others: 12  0); the vectors along z for these atoms are unmoved (3  1), though their x, y vectors rotate through 1800 to their reverse (6  –1). v : Only the five atoms lying on the xz mirror plane contribute. For each atom, two vectors lie in the xz mirror plane (10  1), while the y vector on each atom is reversed (5  –1). d : Only the vectors on the three atoms on z contribute; the x, y vectors swap places (6  0) while their z vectors are unmoved (3  1). A1 A2 B1 B2 E

= = = = =

1/8 [(1  21  1) + (2  3  1) + (1  –3  1) + (2  5  1) + (2  3  1)] = 1/8 [(1  21  1) + (2  3  1) + (1  –3  1) + (2  5  –1) + (2  3  –1)] = 1/8 [(1  21  1) + (2  3  –1) + (1  –3  1) + (2  5  1) + (2  3  –1)] = 1/8 [(1  21  1) + (2  3  –1) + (1  –3  1) + (2  5  –1) + (2  3  1)] = 1/8 [(1  21  2) + (2  3  0) + (1  –3  –2) + (2  5  0) + (2  3  0)] =

3N = 5A1 + A2 + 2B1 + B2 + 6E

5 1 2 1 6

(3N = 21)

trans + rot = A1 + A2 + 2E We expect three translations and three rotations, and we have labels describing these six movements since each E describes a doubly degenerate movement (Tx, Ty; Rx, Ry) (Section 2.5). vib = 3N – trans + rot = 4A1 + 2B1 + B2 + 4E This totals 15 vibrational modes consistent with 3N-6 (note: again each E is a doubly degenerate mode). SAQ 3.5 For S4, u.a. = –1 + 2cos = –1 + 2cos(90) = –1 + 2(0) = –1. SAQ 3.6 u.a. = 1 + 2cos For C2 : u.a. = 1 + 2 cos (180) = 1 + 2(–1) = –1 For C4 : u.a. = 1 + 2 cos (90) = 1 + 2(0) = 1 Other u.a. are given in Section 3.4: E = 3,  = 1 E

2C4

C2

2v

2d

7

3

3

5

3

 u.a.

3

1

 –1

1

1

3N

21

–3

5

3

C4v unshifted atoms

3 (as in SAQ 3.4)

190

[App. 3

Answers to SAQs

Chapter 4 SAQ 4.1 Only A2u (same symmetry as Tz) and Eu (same symmetry as Tx, Ty) are infrared active; all the other modes are infrared inactive. SAQ 4.2 A1g (x2 + y2, z2), B1g (x2 - y2) and B2g (xy) are all Raman active, while the remaining modes are Raman inactive. SAQ 4.3 FN=NF can exist as either cis- or trans- isomers: F F

F N

N

N

N

F

As there are no coincidences in the infrared and Raman spectra, the molecule must possess an inversion centre and is thus the trans-isomer; the inversion centre is at the mid-point of the N-N bond.

Chapter 5 SAQ 5.1 D4h un.atoms  u.a. 3N

E 5 3 15

2C4 1 1 1

C2 1 –1 –1

2C2' 3 –1 –3

i 1 –3 –3

2C2'' 1 –1 –1

2S4 1 –1 –1

h 5 1 5

2v 3 1 3

2d 1 1 1

Using the reduction formula: 3N = A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu trans + rot = A2g + Eg + A2u + Eu vib

=

3N – trans + rot

SAQ 5.2 O

N O

O

The nitrate ion belongs to the D3h point group, assuming complete delocalisation of the negative charge, making all three N-O bonds equivalent. D3h N-O

E 3

2C3 0

3C2 1

h 3

2S3 0

3v 1

= A1' + E'

App.3]

191

Answers to SAQs A1' : E' :

Raman-only (pol) active. Infrared and Raman (depol) active.

SAQ 5.3 O

N O

D3h in-plane

E 3

2C3 0

3C2 1

O

2S3 0

h 3

3v 1

= A1' + E', as above

Note that under both C2 and v one double-headed arrow swaps ends but remains indistinguishable from the original (count 1) while two double-headed arrows move to new locations (count 0, 0). The A1' mode is a redundancy, as it has already been accounted for by an N-O stretching mode (see SAQ 5.2). As with the example of [PtCl4]2, this corresponds to all bond angles opening or closing simultaneously. SAQ 5.4 O

N

O

O

D3h out-of-plane

E 3

2C3 0

3C2 1

Given: vib N-O in-plane you should predict: out-of-plane

h 3

2S3 0

3v 1

= A2'' + E''

= A1' + 2E' + A2'' = A1' + E' = E' = A2''

The E'' is a redundancy and corresponds to rotation of the whole molecule about x, y (Rx, Ry). SAQ 5.5 vib = A1' + 2E' + A2''

192

[App. 3

Answers to SAQs Infrared (solid) 1383 825 720

Raman (solution) 1385 (depol) 1048 (pol) 718 (depol)

E' A1' A2'' E'

N-O stretch N-O stretch out-of-plane bend in-plane bend

The A1' band is the Raman-only polarised band and is an N-O stretch. The A2'' band is the infrared-only band and is an out-of-plane bend. The two E' modes are active in both the infrared and Raman (depol); they can be distinguished by the N-O stretch coming at higher energy than the in-plane bend.

Chapter 6 SAQ 6.1 energy

SAQ 6.2 C........C

[C3]-

C-

.... SALC

2pz

....

The three pz orbitals combine to form three -MOs (assuming the molecule lies in the xy plane). There are two ways in which the terminal atom pz AOs can combine (left) and only the in-phase combination can combine with the pz on the central carbon, to generate bonding and anti-bonding MOs. The out-of-phase combination of terminal atom pz AOs is non-bonding. Each carbon has four electrons, three of which are in sp2 hybrids, leaving pz with 1 electron for -bonding. Along with the electron for the negative charge on the

App.3]

193

Answers to SAQs

anion, this gives four -electrons which fill the bonding and non-bonding MOs. The -bond order is 1 i.e. 0.5 per C-C bond. This correlates with the following resonance forms of the allyl anion: H

H H

H H

H

H H

H

H

Chapter 7 SAQ 7.1 s : a1g (always transforms as the perfectly symmetrical representation) px, py : eu (same as Tx, Ty) pz : a2u (same as Tz)

Chapter 8 SAQ 8.1 E 3

C3v H 1s

2C3 0

3v 1

= a1 + e

This is reasonable as three combinations are expected; one SALC is unique (a1) while the other two form a degenerate pair (e). Symmetries of AOs on nitrogen: 2s : 2px, 2py : 2pz :

a1 (all s-orbitals are perfectly symmetrical) e (Tx, Ty) a1 (Tz)

SALC

AO

Label

2s

a1

2px

e

2py

e

2pz

a1

MOs

194

[App. 3

Answers to SAQs

SAQ 8.2

energy

Chapter 9 SAQ 9.1 Oh E 5 d-orbitals a =C42

8C3 –1

6C2 1

6C4 –1

3C2a 1

i 5

6S4 –1

8S6 –1

3h 1

6d 1

This can be converted to eg and t2g using the reduction formula, or: Oh Eg T2g d-orbitals

E 2 3 5

8C3 –1 0 –1

SAQ 9.2

6C2 0 1 1

6C4 0 –1 –1

i 2 3 5

3C2 2 –1 1

6S4 0 –1 –1

Oh dxy, dxz, dyz : t2g

D4h dxy : b2g dxz, dyz : eg

dz2 , dx2-y2

dx2-y2 dz 2

: eg

8S6 –1 0 –1

3h 2 –1 1

6d 0 1 1

: b1g : a1g

As the ligand along z is removed the eg degeneracy is lost. Similarly, the t2g trio are separated, with dxy differing from the two d-orbitals with a z component. SAQ 9.3 Td 4 L

E 4

8C3 1

3C2 0

6S4 0

6d 2

= a1 + t2

App.3]

195

Answers to SAQs

The AO symmetries are : s px, py, pz dz2 , dx2-y2 dxy, dxz, dyz

: : : :

a1 t2 e t2

Remembering that the AO energies are d < s
M

ML4

L4 3a1*

(t2) p

3t2

(a1) s 2t2* t 1e

(t2 + e) d

4L  SALCs (a1 + t2) 1t2 1a1

The 8 ligand electrons (4  2) fill the 1a1 and 1t2 MOs, while the metal-based Mn+ electrons fill the 1e and 2t2*. The 1e - 2t2* energy gap corresponds to t in crystal field theory, though here it is described as an e - t2 separation. Note that the t2 AOs px, py, pz could interact with the other orbitals of t2 symmetry (particularly the SALCs) if the energy match is correct. This would lower 1t2 and 2t2* while raising the energy of 3t2. 2t2* would become more non-bonding in character while 3t2 would become anti-bonding. However, the overall result of 1a1 and 1t2 accommodating the ligand electrons and 1e, 2t2* the metal electrons remains unchanged.

Chapter 11 SAQ 11.1 C2V B1 B1 B1  B1

E 1 1 1

C2 1 1 1

(xz) 1 1 1

(yz) 1 1 1

= A1

196

[App. 3

Answers to SAQs

SAQ 11.2 E 1 1 1

C2v B1 A1 A1  A1  B1  A1

C2 1 1 1

(xz) 1 1 1

(yz) 1 1 1

= B1

SAQ 11.3  μx    A1   μy   A2  μz   

 Tx    = A1   Ty   A2 = A1   Tz   

 B1     B2   A2  A1   

The direct products are: A1  B1  A2 = B2 A1  B2  A2 = B1 A1  A1  A2 = A2 As none of these direct products are A1 the transition is forbidden. SAQ 11.4 The allowed transition (a) is 1A1  1B1, while (b) is 1A1  3B1 and spin forbidden. SAQ 11.5 C6v E1 E2 E1  E2

E 2 2 4

2C6 1 1 1

2C3 1 1 1

C2 2 2 4

3v 0 0 0

3d 0 0 0

= B1 + B2 + E1

SAQ 11.6 C2h : (Ag  Au)  Bu = Au  Bu = Bg D3h : (A1''  E'')  A2' = E'  A2' = E' Td : (E  T1)  T2 = (T1 + T2)  T2 = (T1  T2) + (T2  T2) = A1 + A2 + 2E + 2T1 + 2T2 Oh : (Eg  A2g)  T1u = Eg  T1u = T1u + T2u SAQ 11.7 Under Oh symmetry,  has T1u symmetry. A1u  T2g : A1u  T2g = T2u. This does not match T1u so the transition is forbidden.

App.3]

197

Answers to SAQs

Eu  T1g : Eu  T1g = T1u + T2u. As this contains T1u so the transition is allowed.

Chapter 12 SAQ 12.1 Dt = (2  5)! / 2!  8!

= 45

SAQ 12.2 E 3

C2v p-orbitals

C2 –1

(xz) 1

(yz) 1

= A1 + B1 + B2

These correspond to Tz, Tx, Ty, repectively, as expected. SAQ 12.3 Eg  Eg = A1g + A2g + Eg For (eg)2: Dt = (2  2)! / 2!  2! = 6 SAQ 12.4 4

F has a maximum ML = 3 (ml = 2 + 1 + 0). Thus, ML = 3, 2, 1, 0, -1, -2, -3 and S = (3 x 1/2) so MS = 3/2, 1/2, -1/2, -3/2. The total mumber of microstates is 7  4 = 28. This can also be arrived at by (2S + 1)(2L + 1). In a weak octahedral field, 4F spilts into 4A2g + 4T1g + 4T2g each of which corresponds to 4, 12 and 12 microstates, respectively.

3

/2

SAQ 12.5 As the 1G term arises from a combination of d-orbitals the terms must have g symmetry and "+" should be used in the relevant equations. E 9

Oh G

8C3 0

6C2 1

6C4 1

3C2a 1

i 9

6S4 1

8S6 0

3h 1

6d 1

= A1g + Eg + T1g + T2g The singlet nature of the 1G term carries through to the ligand-field terms, so these become 1A1g + 1Eg + 1T1g + 1T2g. They total 1 + 2 + 3 + 3 = 9 microstates, as required (Section 12.1). SAQ 12.6 High-spin d7 = (t2g)5(eg)2 = 2T2g  3A2g = 4T1g Low-spin d7 = (t2g)6(eg)1 = 1A1g  2Eg = 2Eg

198

[App. 3

Answers to SAQs

Chapter 13 SAQ 13.1 Oh A1g T1u A1g  T1u =

E 1 3 3

8C3 1 0 0

6C2 1 1 1

6C4 1 1 1

3C2a 1 1 1

i 1 3 3

6S4 1 1 1

8S6 1 0 0

3h 1 0 0

6d 1 1 1

= T1u

SAQ 13. 2 Γvib = A1g + Eg + 2T1u + T2g + T2u For 1T1g  1A1g: T1g  T1u  A1g = (A1u + Eu + T1u + T2u)  A1g = A1u + Eu + T1u + T2u As this spans Γvib (T1u, T2u are common to both), the transition is vibronically allowed. For 1T2g  1A1g: T2g  T1u  A1g = (A2u + Eu + T1u + T2u)  A1g = A2u + Eu + T1u + T2u which is also vibronically-allowed. SAQ 13.3 High-spin Co3+ is d6, and the spin allowed transition is 5Eg  5T2g SAQ 13.4 [V(H2O)6]3+ is V3+ and d2 so we expect 3T2g  3T1g(F), 3T1g(P)  3T1g(F) and 3A2g  3T1g(F) spin-allowed transitions. SAQ 13.5 [FeCl4]2- is d6 ( octahedral d4) so a single 5T2 ← 5E transition would be expected, and one is seen at ca. 4,000 cm-1. SAQ 13.6 [NiCl4]2- is d8 ( octahedral d2) so the three spin-allowed transitions are : 3

T2  3T1(F), 3T1(P)  3T1(F) and 3A2  3T1(F)

SAQ 13.7 The three possible transitions are 3T2  3A2, 3T1(F)  3A2 and 3T1(P)  3A2

App.3]

199

Answers to SAQs

The symmetry of the dipole moment is T2 under Td symmetry, so the transition integrals are: A2  T2  T2 = T1  T2 = A2 + E + T1 + T2 A2  T2  T1 = T1  T1 = A1 + E + T1 + T2 (for two transitions) The 3T2  3A2 excitation is symmetry-forbidden while the two 3T1  3A2 transitions are allowed as they span the totally symmetric representation A1. SAQ 13.8 (t2g)6 has 1A1g symmetry, therefore singlet states associated with (t2g)5(eg)1 need to be identified. (t2g)5(eg)1  (t2g)1(eg)1 (by the hole formalism) = 2T2g  2Eg 2

T2g  2Eg = 1T1g + 1T2g + 3T1g + 3T2g

Spin-allowed transitions: 1

T1g 1A1g and 1T2g 1A1g

SAQ 13.9 D4h B2u Eu B2u + Eu a = C42

E E 1 2 3

6C4 2C4 1 0 1

C2 1 2 1

3C2a 2C2' 1 0 1

6C2 2C2'' 1 0 1

i i 1 2 3

6S4 2S4 1 0 1

h 1 2 1

3h 2v 1 0 1

6d 2d 1 0 1

Oh

T2u

SAQ 13.10 [Cr(H2O)6]2+ is Cr(II) d4 [(t2g)3(eg)1] and is subject to a Jahn-Teller distortion. The 5Eg ground term splits into 5B1g and 5A1g, while the 5T2g exhibits a much smaller splitting into 5B2g and 5Eg, analogous to d9. The spectrum can be interpreted in terms of 5A1g  5B1g and unresolved 5B2g  5B1g / 5Eg  5B1g transitions.

Appendix 1 SAQ A.1 C2v 1  (IR) b2 (IR)R1

E 1 1 1

C2 2 1 2

(xz) 2 1 2

(yz) 1 1 1

= 21 - 22

200

[App. 3

Answers to SAQs

Since we are only interested in the relative contributions of the AOs, this is the same as 1 - 2. Including the normalisation coefficient b2 = 1/2(1 - 2). SAQ A.2 Multiplying pairs of coefficients (a2''  e'') for 1, 2 and 3: (1  2) + (1  1) + (1  1) = 0, so orthogonal SAQ A.3 C31 1 1 1

E 3 2 23

D3h 3  (IR) e'' (IR)R1

C32 2 1 2

h 3 2 23

S3 1 1 1 1

S3 5 2 1 2

e'' = 43 - 21 - 22 or e'' = 23 - 1 - 2 (2  1) + (1  1) + (1  2) = 3, not 0 as required for orthogonality. SAQ A.4 C3v 1  (IR) A1 (IR)Rv1 (IR) E (IR)Rv1 C3v 2 - 3 (IR) E (IR)Rv1

C31 2 1 2 1 2

E 1 1 1 2 21 E 2 - 3 2 22 - 23

C32 3 1 3 1 3

C31 3 - 1 1 -3 + 1

σv(1) 3 1 3 0

σv(2) 2 1 2 0

σv(3) 1 1 1 0

→ 1/3(1 + 2 + 3) → 1/6(21 - 2 - 3)

C32 1 - 2 1 -1 + 2

→1/2(2 - 3)

SAQ A.5 For the singly-degenerate A1g and B1g modes: D4h v1 (IR) A1g (IR)Rv1 (IR) B1g (IR)Rv1

E v1 1 v1 1 v1

C41 v2 1 v2 1 v2

C43 v4 1 v4 1 v4

C2 v3 1 v3 1 v3

C2'(1) v1 1 v1 1 v1

C2'(2) v3 1 v3 1 v3

C2''(1) v2 1 v2 1 v2

C2''(2) v4 1 v4 1 v4

i v3 1 v3 1 v3

S4 1 v2 1 v2 1 v2

App.3]

201

Answers to SAQs

continued D4h S43 v4 v1 1 (IR) A1g v4 (IR)Rv1 (IR) B1g 1 (IR)Rv1 v4

σv(1) v1 1 v1 1 v1

σh v1 1 v1 1 v1

σv(2) v3 1 v3 1 v3

σd(1) v2 1 v2 1 v2

σd(2) v4 1 v4 1 v4

= 4v1 + 4v2 + 4v3 + 4v4 = 4v1 - 4v2 + 4v3 - 4v4

For the doubly-degenerate Eu mode: D4h v1 (IR)Eu (IR)Rv1 v2 (IR)Eu (IR)Rv1

E v1 2 2v1 v2 2 2v2

C2 v3 2 2v3 v4 2 2v4

i v3 2 2v3 v4 2 2v4

σh v1 2 2v1 v2 2 2v2

= 4v1 - 4v3 = 4v2 - 4v4

From which: A1g = 1/2(v1 + v2 + v3 + v4) B1g = 1/2(v1 - v2 + v3 - v4) Eu = 1/2(v1 - v3), 1/2(v2 - v4) SAQ A.6 For the singly-degenerate A2u and B2u modes: D4h  1 (IR) A2u (IR)Rv1 (IR) B2u (IR)Rv1

E 1 1 1 1 1

continued D4h S43  1 4 (IR) A2u 1 (IR)Rv1 4 1 (IR) B2u (IR)Rv1 4

C41 2 1 2 1 2

σh 1 1 1 1 1

C43 4 1 4 1 4

C2 3 1 3 1 3

σv(1) 1 1 1 1 1

C2'(1) 1 1 1 1 1 σv(2) 3 1 3 1 3

C2'(2) 3 1 3 1 3

σd(1) 2 1 2 1 2

C2''(1) 2 1 2 1 2

σd(2) 4 1 4 1 4

C2''(2) 4 1 4 1 4

i 3 1 3 1 3

S41 2 1 2 1 2

= 41 + 42 + 43 + 44 = 41 - 42 + 43 - 44

202

Answers to SAQs

For the doubly-degenerate Eg mode: D4h  1 (IR) Eg (IR)Rv1  2 (IR) Eg (IR)Rv1

E 1 2 21 2 2 22

C2 3 2 23 4 2 24

i 3 2 23 4 2 24

From which: A2u = 1/2(1 + 2 + 3 + 4) B2u = 1/2(1 - 2 + 3 - 4) Eg = 1/2(1 - 3), 1/2(2 - 4)

σh 1 2 21 2 2 22

= 41 - 43 = 42 - 44

[App. 3

APPENDIX 4 Answers to Problems Chapter 1 1.

NH3: C3 (passing through N and the mid-point of the H3 triangle), three v (each containing one N-H bond). AsH5: C3 and S3 coincident (along the axial H-As-H unit), three C2 (along each equatorial As-H bond), h (containing the equatorial AsH3 unit), three v (each containing the axial H-As-H unit and one equatorial As-H bond). cyclo-B3N3H6: C3, S3 coincident (perpendicular to the molecular plane and passing through the mid-point of the ring), three C2 (containing pairs of B, N atoms on opposite sides of the ring), h (molecular plane), three v (perpendicular to the molecular plane and each containing a pair of B,N atoms). B(H)(F)(Br): the molecule is trigonal planar and only has a mirror plane of symmetry (containing all four atoms). Since there is no main axis the mirror plane is given the symbol  (without qualification) as the subscripts h and v have no significance. C4H4F2Cl2: i (at mid-point of the four-carbon ring), three S2 (e.g. passing through i and perpendicular to the four-carbon plane, plus two others orthogonal to this). i  any one of three orthogonal S2 axes since (x, y, z)→(-x, -y, -z) can be brought about in any one of three ways. SiH4: C2, S4 d H H Si H

d

H

H

Si H

H

H

C3

There are four C3 axes (lying along each Si-H bond), three C2 axes (one through the centre of each face of the cube), three S4 coincident with C2 and six d (the two shown are replicated across each face of the cube). 2.

The pairs of molecules have been chosen to highlight the importance of getting the correct molecular shape as a pre-requisite for assigning symmetry. CO2 is linear, Dh while SO2 is angular and C2v.

204

[App. 4

Answers to Problems

Ferrocene (staggered) D5d and ruthenocene (eclipsed) D5h. Both have a C5 axis and five C2 axes, but only ruthenocene has a h; ferrocene does have five d. cis- and trans-Mo(CO)4Cl2 are both octahedral in shape, and have C2v and D4h symmetry, respectively. [IF6]+ is perfectly octahedral Oh, while [IF6]- is pentagonal pyramidal by VSEPR (like IF7 but a lone pair in an axial site) and is C5v. SnCl(F) is angular and Cs, while XeCl(F) is linear and Cv. mer- and fac- WCl3F3 are both octahedral in shape, of C2v, C3v symmetry, respectively. 4.

The cis-isomer of CrCl2(acac)2 belongs to the C2 point group and is chiral and polar, while the trans-isomer belongs to the D2h point groups and is neither chiral nor polar. cyclo-(Cl2PN)4 belongs to the S4 point group and is therefore not chiral but does have a dipole.

Chapter 2 1.

A:=1+1+ 2=4

B :  = 2 + (–2) + 3 = 3

1 2 1 2 0 2 4 1 7  C = 3 1 1 1  2 1 = 7 1 10  = 4 + 1 + 8 = 13   1 0 2 0 3 3 2 6 8   

2.       

1 0 0 1

0 0

0 0

0 0

0 1 0 0

0 0

0

0

1

0

0  0 0  1 0

d z 2    d x 2  y 2  d xy    d yz   d  xz 

 d z2     d x 2  y 2  =  d xy     d xz   d   yz 

dz2 transforms onto itself as it lies along the z axis. dx2-y2 and dxy both turn through 90o, so they remain in their same positions but in each case the + lobes are transformed onto – lobes and vice versa. The orbitals thus transform onto the reverse of themselves. dxz turns through 90o and maps onto dyz while dyz similarly maps onto dxz.

Chapter 3 1.

C2v 

E 6

C2 4

(xz) –2

(yz) 0

= 2A1 + 3A2 + B2

App.4]

205

Answers to Problems

A1 = 1/4[(1  6  1) + (1  4  1) + (1  –2  1) + (1  0  1)] = 2 A2 = 1/4[(1  6  1) + (1  4  1) + (1  –2  –1) + (1  0  –1)] = 3 B1 = 1/4[(1  6  1) + (1  4  –1) + (1  –2  1) + (1  0  –1)] = 0 B2 = 1/4[(1  6  1) + (1  4  –1) + (1  –2  –1) + (1  0  1)] = 1 E 9

Td 

8C3 3

3C2 1

6S4 3

6d 3

= 3A1 + T1 + T2

A1 = 1/24[(1  9  1) + (8  3  1) + (3  1  1) + (6  3  1) + (6  3  1)] = 3 A2 = 1/24[(1  9  1) + (8  3  1) + (3  1  1) + (6  3  –1) + (6  3  –1)] = 0 E = 1/24[(1  9  2) + (8  3  –1) + (3  1  2) + (6  3  0) + (6  3  0)] = 0 T1 = 1/24[(1  9  3) + (8  3  0) + (3  1  –1) + (6  3  1) + (6  3  –1)] = 1 T2 = 1/24[(1  9  3) + (8  3  0) + (3  1  –1) + (6  3  –1) + (6  3  1)] = 1 E 15

C3v 

2C3 0

3v 3

= 4A1 + A2 + 5E

A1 = 1/6[(1  15  1) + (2  0  1) + (3  3  1)] = 4 A2 = 1/6[(1  15  1) + (2  0  1) + (3  3  –1)] = 1 E = 1/6[(1  15  2) + (2  0  –1) + (3  3  0)] = 5 2.

NH3 (belongs to the C3v point group) C3v unshifted atoms  u.a. 3N

E 4 3 12

2C3 1 0 0

3v 2 1 2

Using the reduction formula: = 3A + A2 + 4E (= 3N i.e. 12, which is correct remembering E means doubly degenerate)

3N trans + vib

rot

= A1 + A2 + 2E (= 6) = 2A1 + 2E

(= 3N – 6 i.e. 6, again remembering E means doubly degenerate)

cis-N2H2 (belongs to the C2v point group; let yz be the molecular plane) C2v unshifted atoms  u.a. 3N

E 4 3 12

C2 0 –1 0

(xz) 0 1 0

(yz) 4 1 4

206

[App. 4

Answers to Problems Which reduces to: 3N = 4A1 + 2A2 + 2B1 + 4B2 (= 3N i.e. 12) trans + rot = A1 + A2 + 2B1 + 2B2 (= 6) vib = 3A1 + A2 + 2B2 (= 3N – 6 i.e. 6) SO3 (belongs to the D3h point group) D3h unshifted atoms  u.a. 3N

E 4 3 12

2C3 1 0 0

3C2 2 –1 –2

h 4 1 4

2S3 1 –2 –2

3v 2 1 2

Which reduces to: 3N = A1' + A2' + 3E' + 2A2'' + E'' (= 3N i.e. 12 ) trans + rot = A2' + E' + A2'' + E'' (= 6) vib = A1' + 2E' + A2'' (= 3N – 6 i.e. 6) 3.

C2h unshifted atoms  u.a. 3N

E 10 3 30

C2 0 –1 0

i 0 –3 0

Which reduces to: 3N = 10Ag + 5Bg + 5Au + 10Bu trans + rot = Ag + 2Bg + Au + 2Bu vib = 9Ag + 3Bg + 4Au + 8Bu

h 10 1 10 (= 3N i.e. 30) (= 6) (= 3N – 6 i.e. 24)

Chapter 4 1.

[ClO4]- belongs to the Td point group. A1 : same symmetry as x2 + y2 + z2, thus Raman-only active; as this representation is perfectly symmetrical the Raman band will be polarised. E : has the symmetry of 2z2 – x2 – y2, x2 – y2 and is also Raman-only active; this is an unsymmetrical modes and is Raman depolarised. T2 : this has the symmetry of Tx, Ty, Tz and xy, yz, zx and is thus both infrared and Raman active; the Raman band is depolarised as it is an unsymmetrical mode.

2.

[BrF2] : VSEPR predicts this is linear and thus belongs to the Dh point group. The anion has an inversion centre on the bromine and thus its vibrational spectrum should have no coincidences between the infrared and Raman spectra. It is spectrum A.

App.4]

207

Answers to Problems

[BrF2]+ : VSEPR predicts an angular structure i.e. C2v point group. This has no inversion centre and thus coincidences between the infrared and Raman spectra are to be expected. This is spectrum B.

Chapter 5 2. E 15

C3v 3N

2C3 0

3v 3

= 4A1 + A2 + 5E

trans + rot = A1 + A2 + 2E = 3A1 + 3E

vib

Both A1 and E are infrared and Raman active; A1 is Raman pol, E is Raman depol. 1 3 6

P-O P-Cl Cl-P-Cl/Cl-P-O

1 0 0

1 1 2

= A1 = A1 + E = 2A1 + 2E

Infrared (liquid; cm-1)

Raman (liquid ; cm-1)

1292 580 487 340 267 not accessible

1290 (pol) 581 (depol) 486 (pol) 337 (depol) 267 (pol) 193 (depol)

A1 E A1 E A1 E

One A1 is redundant.

P=O stretch P-Cl stretch P-Cl stretch deformation deformation deformation

3. E 12

C2h 3N

C2 0

i 0

v 4

= 4Ag + 2Bg + 2Au + 4B2u

trans + rot = Ag + 2Bg + Au + 2Bu vib

= 3Ag + Au + 2Bu

Ag is Raman-only active and pol; Au, Bu both infrared-only active. N-N N-F F-N-N out-of-plane

1 2 2 2

1 0 0 0

1 0 0 0

1 2 2 2

= = = =

Ag Ag + Bu Ag + Bu Au + Bg Bg is redundant (corresponds to Rx, Ry)

208

[App. 4

Answers to Problems Infrared (gas, cm-1)

Raman (gas, cm-1) 1636 (pol) 1010 (pol)

989

592 (pol)

412 360

Ag Ag Bu Ag Au or Bu Au or Bu

N=N stretch N-F stretch N-F stretch F-N-N deformation out-of-plane deformation out-of-plane deformation

Decreasing energy : N=N stretch > N-F stretch > bending modes

Chapter 6 1. Xe+

[XeF]+

F

Xe

XeF2

F2

In [XeF]+, Xe has one electron in pz (allowing for the loss of an electron to generate the cation), while fluorine also has one. The Xe-F bond order is 1. In XeF2, the total of four electrons (Xe: 2; F: 2  1) are distributed equally over the bonding and non-bonding MOs. There is a total bond order of 1 i.e. 0.5 per Xe-F bond. 3. Li

LiH

H

-5.4 eV 2s

1s -13.6 eV

The two electrons in the bonding MO are localised on hydrogen, consistent with a bond polarity of Li+-H-.

App.4]

209

Answers to Problems

4. C

CO2

O2

 2pz



2p z

2s





C

CO2

O2

 2px , 2py

2px , 2p y



For the -bonds, one of the two combinations for the pz orbitals on oxygen combines with the s-orbital on carbon, while the other combines with the carbon pz; both bonding and anti-bonding combinations are formed. For the -bonds, the in-phase combinations formed by both px and py on oxygen combine with their equivalent orbital on carbon to give a pair of bonding and a pair of * anti-bonding MOs. The out-of-phase pair of oxygen combinations have no match with an AO on carbon and are non-bonding. There are a total of twelve bonding electrons (C: 2s2 2p2; O 2p4) which fill the two  bonding MOs, the two -bonding MOs and the two non-bonding MOs. The total bond order is 4 i.e. a bond order of 2 per C-O unit.

Chapter 7 1.

C2v O px N 2px

E 2 1

C2 0 –1

(xz) 0 1

(yz) –2 –1

= a2 + b1 = b1

The symmetry of the nitrogen px could have been read directly from the character table : same symmetry as Tx.

210

[App. 4

Answers to Problems O........O

NO2-

N2b1*

.... 2px b1)

-13.1 eV

1a2

-15.9 eV 2pz

.... (a2 + b1) SALCs 1b1

Each atom contributes one -electron, with an additional -electron for the negative charge; this has been arbitrarily assigned to nitrogen in the MO diagram. The -bond order is 1 i.e. 0.5 per N-O -bond.

Chapter 8 1.

BF3 can be considered as sp2 hybridised at boron, leaving an empty pz orbital free for -bonding with the filled pz lone pair orbitals on fluorine. z

D3h F pz

E 3

2C3 0

3C2 –1

h –3

2S3 0

3v 1

a2'' + e''

The pz AO on boron has a2'' symmetry (symmetry of Tz). BF3

B

F3

2a2'' * -8.3 eV

2pz (a2'')

1e''

1a2''

2pz

-18.7 eV

SALCs (a2'' + e'')

The pz on boron is vacant while each pz on fluorine contains two electrons. -Bond order = 1 i.e. 0.333 per B-F bond.

App.4]

211

Answers to Problems

Boron in BF3 should be a strong Lewis acid as (i) it has a vacant p-orbital capable of accepting an electron pair and (ii) it should be strongly B+ by virtue of the three very electronegative fluorines. However, intramolecular bond formation means that the pz on boron is not vacant and is thus only a weak Lewis acid. 6. D6h  C pz

E 6

2C6 0

2C3 0

C2 0

3C2' –2

3C2'' 0

i 0

2S3 0

2S6 0

h –6

3d 0

3v 2

= b2g + e1g + a2u + e2u (this is reasonable as we expect six MOs) The e1g and e2u MOs each comprise a pair of SALCs with either one node or two nodes, which can be distinguished by the subscripts g and u. Of the two singly-degenerate MOs, a2u is the one with no nodes (u) while the most antibonding MO, with three nodes, is b2g (g).

b2g

e2u

e2g

a2u

Symmetry labels for the a2u and b2g SALCs can be checked by treating each of them as a complete unit and counting 1, 0, -1 if they are unmoved, moved or reversed under each of the D6h symmetry operations; the pairs of SALCs described by the e labels cannot be analysed in this way.

Chapter 9 1.

C6v 6 C pz

E 6

2C6 0

2C3 0

C2 0

3v 2

3d 0

= a1 + b1 + e1 + e2

212

[App. 4

Answers to Problems

x y

a1

e1

e2

b1

The e1 and e2 pairs can be distinguished by their matches with the d-orbitals on iron. AOs on iron:

dz 2 dx2-y2 dxy dxz dyz

: : : : :

a1 e2 e2 e1 e1

5. D4h  Fp

E 4

2C4 0

C2 0

This reduces to:

2C2' 2

2C2'' 0

i 0

2S4 0

h 4

2v 2

2d 0

a1g + b1g + eu (correct, as we expect four SALCs)

The symmetry labels for the valence orbitals on xenon, read from the character table are: s : a1g dxz, dyz : eg px, py : eu dx2-y2 : b1g pz : a2u dz 2 : a1g dxy : b2g MO combinations:

Note: In the eu SALCs two p orbitals in each case make non-bonding contributions and can be ignored.

App.4]

213

Answers to Problems

Chapter 10 2.

C2h 6 C pz

E 6

C2 0

i 0

h 2

= 2ag + bg + au + 2bu

The three SALCs associated with each allyl group are:

Combining identical pairs, both in-phase and out-of-phase, gives the following, exemplified by SALCs associated with combinations of the inphase set of three pz AOs:

Ni

Ni

1

2

3 - 6 are shown in the Table, below. C2h 1 2 3 4 5 6

E 1 1 1 1 1 1

C2 1 –1 1 –1 –1 1

i 1 –1 –1 1 –1 1

h 1 1 –1 –1 1 1

= ag = bu = au = bg = bu = ag

The AO column (below) shows all symmetry-allowed matches between nickel AOs and ligand SALCs; the best matches are shown without parentheses. This analysis is a simplification in that further mixing, e.g. of all AOs / SALCs of ag or bu symmetry, will take place to some extent. SALC

AO

Label

Ni

s, dx2-y2 (dz2, dxy)

ag

Ni

px (py)

bu

x

1 y

2

214

[App. 4

Answers to Problems Question 2, continued.

Ni

pz

au

Ni

dxz, dyz

bg

5

Ni

py (px)

bu

6

Ni

dz2, dxy (s, dx2-y2)

ag

3

4

Chapter 11 1.

The ground state (a1)2(e)4(a1)2 has 1A1 symmetry. The excited states and their symmetries are: (a1)2(e)4(a1)1(a1*)1 = A1  A1  A1  A1 = 1A1 or 3A1 (a1)2(e)4(a1)1(e*)1 = A1  A1  A1  E = 1E or 3E Only singlet excited states will afford spin-allowed transitions. The symmetry of  is A1 + E, so the transition integrals are: (a1)2(e)4(a1)2  (a1)2(e)4(a1)1(a1*)1 : A1  A1  A1 (=A1) or A1  E  A1 (a1)2(e)4(a1)2  (a1)2(e)4(a1)1(e*)1 : A1  A1  E or A1  E  E (contains A1) So, both 1A1  1A1 and 1A1  1E transitions are symmetry allowed.

3.

The ground state (au)2(bg)2 has 1Ag symmetry. The excited states and their symmetries are: (au)2(bg)1(au*)1 = Ag  Bg  Au = 1Bu or 3Bu (au)2(bg)1(bg*)1 = Ag  Bg  Bg = 1Ag or 3Ag Only singlet excited states will afford spin-allowed transitions. The symmetry of  is Au + Bu, so the transition integrals are:

App.4]

215

Answers to Problems

(au)2(bg)2  (au)2(bg)1(au*)1 = Ag  Au  Bu (= Bg) or Ag  Bu  Bu (= Ag) (au)2(bg)2  (au)2(bg)1(bg*)1 = Ag  Au  Ag (= Au) or Ag  Bu  Ag (= Bu) Thus, only the 1Ag  1Bu (bg  au*) transition is symmetry-allowed as the transition integral contains Ag. 4.

The ground state (a2u)2(eg)4 has 1A1g symmetry. The excited states and their symmetries are: (a2u)2(eg)3(b2u*)1 = A1g  Eg  B2u = Eu ; (eg)3  (eg)1 by the hole formalism (a2u)1(eg)4(b2u*)1 = A2u  A1g  B2u = B1g Only singlet excited states will afford spin-allowed transitions. The symmetry of  is A2u + Eu, so the transition integrals are: (a2u)2(eg)4  (a2u)2(eg)3(b2u*)1 = A1g  A2u  Eu (= Eg) or = A1g  Eu  Eu (contains A1g) (a2u)2(eg)4  (a2u)1(eg)4(b2u*)1 = A1g  A2u  B1g (= B2u) or A1g  Eu  B1g (= Eu) Thus, only the 1A1g  1Eu [(a2u)2(eg)4  (a2u)2(eg)3(b2u*)1] transition is symmetry-allowed as the transition integral contains A1g.

Chapter 12 1.

The degeneracy of the configuration is given by the product of the degeneracies of its components: (s)1 = 2 and (p)1 = 6 (using eqn 12.1), so the total degeneracy for (s)1(p)1 = 12 For the (s)1, ml = 0 and ms = ½ and for (p)1 ml = 1, 0 or -1 and ms = ½ Maximum L = 1 (ml = 0 for s and 1 for p) and maximum S = ½ + ½ = 1. The grid for the microstates is as follows: ML 1

1 (1+,0+)

0

(0+,0+)

-1

(-1+,0+)

MS 0 (1+,0-) (1-,0+) (0+,0-) (0-,0+) (-1+,0-) (-1-,0+)

-1 (1-,0-) (0-,0-) (-1-,0-)

The ground term corresponds to maximum L, S and is 3P (ML= 1, 0, -1; MS = 1, 0, -1); this corresponds to the nine microstates in bold. Of the microstates

216

[App. 4

Answers to Problems

remaining, the one with maximum L, S is (1-,0+) which arises from a 1P term and accounts for the remaining three microstates. 4.

From Table 12.6 (Section 12.5): (t2g)4(eg)1 = 3T1g  2Eg = T1g + T2g ignoring spin multiplicities For 3T1g S = 1 (2S + 1 = 3) and for 2Eg S = ½, so maximum S = 3/2 and 2S + 1 = 4, thus, (t2g)4(eg)1 = 3T1g  2Eg = 4T1g + 4T2g Similarly, (t2g)3(eg)2 = 4A2g  3A2g = 6A1g The ground term has maximum S (= ½ + ½) and maximum L. L = 6 (3 + 3) is incompatible with maximum S, so the maximum L is 5 (3 + 2). The ground term is 3H. Using Eqn. 12.2 – 12.6 with "+" in the formulae (f-electrons are u, so f2 = u  u = g) gives:

5.

Oh S

E 11

8C3 -1 3

6C2 -1

6C4 1

3C2a -1

i 11

6S4 1

8S6 -1

3h -1

6d -1

= Eg + 2T1g + T2g

H splits into 3Eg + 2 3T1g + 3T2g ; 33 microstates, as given by (2S + 1)(2L + 1).

Chapter 13 1.

The symmetry of the dipole moment is T1u (Oh) or T2 (Td), so the transition integrals are: Oh : 2Eg  2T2g : T2g  T1u  Eg = (A2u + Eu + T1u + T2u)  Eg = Eu + A1u + A2u + Eu + T1u + T2u + T1u + T2u Td : 2T2  2E : E  T2  T2 = (T1 + T2)  T2 = A2 + E + T1 + T2 + A1 + E + T1 + T2 Only 2T2  2E spans the totally symmetry irreducible representation for its point group (A1) so is allowed while 2Eg  2T2g, which does not span A1g, is forbidden.

3.

(t2g)6 has 1A1g symmetry as all the t2g orbitals are filled. (t2g)5(eg)1  (t2g)1(eg)1 by the hole formalism = T2g  Eg = T1g + T2g For (t2g)5(eg)1 both singlet and triplet states are possible but only the singlet excited states will afford spin-allowed transitions, which therefore are: 1

T1g  1A1g and

1

T2g  1A1g

App.4] 4.

217

Answers to Problems

(eg)1 = 2Eg, while from Table 12.6 in Section 12.5: (t2g)3 = 4A2g + 2Eg + 2T1g + 2T2g (t2g)3(eg)1 = (4A2g + 2Eg + 2T1g + 2T2g)  2Eg

Considering each binary direct product in turn: 4

A 2g  2Eg = 5Eg + 3Eg (see Table 13.2 for combinations of spin multiplicities)

2

Eg  2Eg = 3A1g + 3A2g + 3Eg + 1A1g + 1A2g + 1Eg

2

T1g  2Eg = 3T1g + 3T2g + 1T1g + 1T2g ;

2

T2g  2Eg gives the same result.

Overall: (t2g)3(eg)1 = 5Eg + 23Eg + Eg + 3A1g + 3A2g + 1A1g + 1A2g+ 23T1g + 23T2g + 21T1g + 21T2g Degeneracy: (t2g)3 = 6!/(3!  3!) = 20; (eg)1 = 4 so (t2g)3(eg)1 = 80 (t2g)3(eg)1 = 10 + 12 + 2 +3 + 3 + 1 + 1 + 18 + 18 + 6 + 6 = 80 Spin allowed transitions from 3T1g ground state are: 3 3

Eg  3T1g T1g  3T1g

3 3

A1g  3T1g T2g  3T1g

3

A2g  3T1g

APPENDIX 5 Selected Character Tables Cs A´ A´´

E 1 1

C2v A1 A2 B1 B2

E 1 1 1 1

C2 1 1 -1 -1

(xz) 1 -1 1 -1

C3v A1 A2 E

E 1 1 2

2C3 1 1 -1

C4v A1 A2 B1 B2 E

E 1 1 1 1 2

2C4 1 1 -1 -1 0

C6v A1 A2 B1 B2 E1 E2

E 1 1 1 1 2 2

h 1 -1

2C6 1 1 -1 -1 1 -1

2C3 1 1 1 1 -1 -1

x2, y2, z2, xy yz, xz

Tx, Ty, Rz Tz, Rx, Ry

(yz) 1 -1 -1 1

Tz Rz Tx, Ry Ty, Rx

x2, y2, z2 xy xz yz

3v 1 -1 0

Tz Rz (Tx, Ty), (Rx, Ry)

x2 + y2, z2

C2 1 1 1 1 -2

2v 1 -1 1 -1 0

C2 1 1 -1 -1 -2 2

3v 1 -1 1 -1 0 0

2d 1 -1 -1 1 0

3d 1 -1 -1 1 0 0

(x2 - y2, xy), (yz, xz)

Tz Rz (Tx, Ty), (Rx, Ry)

x2 + y2, z2 x2 - y2 xy (yz, xz)

Tz Rz

x2+y2,z2

(Tx,Ty), (Rx,Ry)

(xz, yz) (x2 – y2, xy)

App.5] C2h Ag Bg Au Bu

E 1 1 1 1

C2 1 -1 1 -1

i 1 1 -1 -1

h 1 -1 -1 1

D2h Ag B1g B2g B3g Au B1u B2u B3u

E 1 1 1 1 1 1 1 1

C2(z) 1 1 -1 -1 1 1 -1 -1

C2(y) 1 -1 1 -1 1 -1 1 -1

C2(x) 1 -1 -1 1 1 -1 -1 1

i 1 1 1 1 -1 -1 -1 -1

(xy) 1 1 -1 -1 -1 -1 1 1

D3h A1´ A2´ E´ A1´´ A2´´ E´´

E 1 1 2 1 1 2

2C3 1 1 -1 1 1 -1

3C2 1 -1 0 1 -1 0

h 1 1 2 -1 -1 -2

2S3 1 1 -1 -1 -1 1

3v 1 -1 0 -1 1 0

2C2' 1 -1 1 -1 0 1 -1 1 -1 0

2C2'' 1 -1 -1 1 0 1 -1 -1 1 0

i 1 1 1 1 2 -1 -1 -1 -1 -2

D4h A1g A2g B1g B2g Eg A1u A2u B1u B2u Eu

E 1 1 1 1 2 1 1 1 1 2

219

Selected Character Tables

2C4 1 1 -1 -1 0 1 1 -1 -1 0

C2 1 1 1 1 -2 1 1 1 1 -2

x2, y2, z2, xy yz, xz

Rz Rx, Ry Tz Tx, Ty

2S4 1 1 -1 -1 0 -1 -1 1 1 0

(xz) 1 -1 1 -1 -1 1 -1 1

(yz) 1 -1 -1 1 -1 1 1 -1

Rz Ry Rx

x2, y2, z2 xy xz yz

Tz Ty Tx

x2 + y2, z2

h 1 1 1 1 -2 -1 -1 -1 -1 2

Rz (Tx, Ty)

(x2 – y2, xy)

Tz (Rx, Ry)

(xz, yz)

2v 1 -1 1 -1 0 -1 1 -1 1 0

2d 1 -1 -1 1 0 -1 1 1 -1 0

x2+y2,z2 Rz (Rx,Ry) Tz (Tx,Ty)

x2 – y2 xy (xz, yz)

220

Selected Character Tables

[App. 5

App.5]

Selected Character Tables

221

222

Td A1 A2 E T1 T2

Dh g+ gg g u + u u u

[App. 5

Selected Character Tables

E 1 1 2 3 3

E 1 1 2 2 1 1 2 2

8C3 1 1 -1 0 0

2C 1 1 2cos 2cos2 1 1 2cos 2cos2

3C2 1 1 2 -1 -1

…. … …. …. …. …. …. …. ….

6S4 1 -1 0 1 -1

v 1 -1 0 0 1 -1 0 0

6d 1 -1 0 -1 1

i 1 1 2 2 -1 -1 -2 -2

x2+y2+z2 (Rx, Ry, Rz) (Tx, Ty, Tz)

2S 1 1 -2cos 2cos2 -1 -1 2cos 2cos2

…. …. …. …. …. …. …. …. ….

(2z2- x2–y2, x2–y2) (xy, xz, yz)

C2 1 -1 0 0 -1 1 0 0

x2+y2, z2 Rz (Rx,Ry)

Tz (Tx, Ty)

(xz, yz ) (x2–y2, xy)