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[c11 + c22 + c33………]
(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 – NONDEGENERATE 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: Pij = ( ( 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 Rj 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)R1
E 1 1 1
C2 2 1 2
(xz) 2 1 2
(yz) 1 1 1
= 21 + 22
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 inphase 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 antibonding overlap with one SALC making it a nonbonding interaction overall, but with the other SALC there appears more bonding than antibonding, 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)R1 1 …continued D3h S3 1 1 2 (IR) a2'' 1 (IR)R1 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'' = 41 + 42 + 43, or a2'' = 1 + 2 + 3 considering only the relative contributions of each AO. Including the normalisation coefficient, this becomes 13(1 + 2 + 3). By the same method, for the e'' SALC: E 1 2 21
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)R1 2
S3 5 3 1 3
v(1) 1 0
v(2) 3 0
v(3) 2 0
D3h 1 (IR) e'' (IR)R1
C2(3) 2 0
h 1 2 21
e'' = 41  22  23 or e'' = 21  2  3 [normalised : 1/6(21  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, nonbonding:
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)R1
E 2 2 22
C31 3 1 3
C32 1 1 1
h 2 2 22
S3 1 3 1 3
S3 5 1 1 1
App.1]
177
Projection Operators
This gives e'' = 42  23  21 or e'' = 22  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)R1
E 2  3 2 22 23
C31 3  1 1 3+ 1
C32 1  2 1 1+ 2
h 2 + 3 2 22 23
S3 1 3+ 1 1 3+ 1
S3 5 1 + 2 1 1+ 2
From which: e'' = 62  63 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 NH = 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
inphase but outofphase 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
Inplane 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 Outofplane 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
PtCl
= A1g + B1g + Eu
inplane
= (A1g) + B2g + Eu
outofplane = (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 symmetryequivalent 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 outof 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 outofplane 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 outofplane modes is because these are, in fact, rotations about x and y (Fig. A.2). Deriving functions for the inplane 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 singlydegenerate A1g and B2g modes, but not for the degenerate Eu inplane 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 outofplane 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 symmetryequivalent, SALCs can be derived from the two methods. This is usually a problem only with doubly and triplydegenerate 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 delectron, 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 dorbital 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 CMoC. 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 pblock 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 F1IF2 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 “rotationreflection” 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 FaxPFax). There are three C2 axis perpendicular to C3, (along each PFeq 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, 3v). 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 (3v). SAQ 3.4
C4v 3N
E 21
2C4 3
C2 –3
2v 5
2d 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 zaxis 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 zaxis 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 3N6 (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
2v
2d
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 transisomer; the inversion centre is at the midpoint of the NN 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
2v 3 1 3
2d 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 NO bonds equivalent. D3h NO
E 3
2C3 0
3C2 1
h 3
2S3 0
3v 1
= A1' + E'
App.3]
191
Answers to SAQs A1' : E' :
Ramanonly (pol) active. Infrared and Raman (depol) active.
SAQ 5.3 O
N O
D3h inplane
E 3
2C3 0
3C2 1
O
2S3 0
h 3
3v 1
= A1' + E', as above
Note that under both C2 and v one doubleheaded arrow swaps ends but remains indistinguishable from the original (count 1) while two doubleheaded arrows move to new locations (count 0, 0). The A1' mode is a redundancy, as it has already been accounted for by an NO 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 outofplane
E 3
2C3 0
3C2 1
Given: vib NO inplane you should predict: outofplane
h 3
2S3 0
3v 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'
NO stretch NO stretch outofplane bend inplane bend
The A1' band is the Ramanonly polarised band and is an NO stretch. The A2'' band is the infraredonly band and is an outofplane bend. The two E' modes are active in both the infrared and Raman (depol); they can be distinguished by the NO stretch coming at higher energy than the inplane 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 inphase combination can combine with the pz on the central carbon, to generate bonding and antibonding MOs. The outofphase combination of terminal atom pz AOs is nonbonding. 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 nonbonding MOs. The bond order is 1 i.e. 0.5 per CC 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
3v 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 sorbitals 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 dorbitals a =C42
8C3 –1
6C2 1
6C4 –1
3C2a 1
i 5
6S4 –1
8S6 –1
3h 1
6d 1
This can be converted to eg and t2g using the reduction formula, or: Oh Eg T2g dorbitals
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 , dx2y2
dx2y2 dz 2
: eg
8S6 –1 0 –1
3h 2 –1 1
6d 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 dorbitals with a z component. SAQ 9.3 Td 4 L
E 4
8C3 1
3C2 0
6S4 0
6d 2
= a1 + t2
App.3]
195
Answers to SAQs
The AO symmetries are : s px, py, pz dz2 , dx2y2 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 metalbased 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 nonbonding in character while 3t2 would become antibonding. 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
3v 0 0 0
3d 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 porbitals
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 dorbitals 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
3h 1
6d 1
= A1g + Eg + T1g + T2g The singlet nature of the 1G term carries through to the ligandfield terms, so these become 1A1g + 1Eg + 1T1g + 1T2g. They total 1 + 2 + 3 + 3 = 9 microstates, as required (Section 12.1). SAQ 12.6 Highspin d7 = (t2g)5(eg)2 = 2T2g 3A2g = 4T1g Lowspin 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
3h 1 0 0
6d 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 vibronicallyallowed. SAQ 13.3 Highspin 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) spinallowed 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 cm1. SAQ 13.6 [NiCl4]2 is d8 ( octahedral d2) so the three spinallowed 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 symmetryforbidden 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
Spinallowed 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
3h 2v 1 0 1
6d 2d 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 JahnTeller 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)R1
E 1 1 1
C2 2 1 2
(xz) 2 1 2
(yz) 1 1 1
= 21  22
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 23
D3h 3 (IR) e'' (IR)R1
C32 2 1 2
h 3 2 23
S3 1 1 1 1
S3 5 2 1 2
e'' = 43  21  22 or e'' = 23  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 21 E 2  3 2 22  23
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(21  2  3)
C32 1  2 1 1 + 2
→1/2(2  3)
SAQ A.5 For the singlydegenerate 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 doublydegenerate 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 singlydegenerate 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
= 41 + 42 + 43 + 44 = 41  42 + 43  44
202
Answers to SAQs
For the doublydegenerate Eg mode: D4h 1 (IR) Eg (IR)Rv1 2 (IR) Eg (IR)Rv1
E 1 2 21 2 2 22
C2 3 2 23 4 2 24
i 3 2 23 4 2 24
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 21 2 2 22
= 41  43 = 42  44
[App. 3
APPENDIX 4 Answers to Problems Chapter 1 1.
NH3: C3 (passing through N and the midpoint of the H3 triangle), three v (each containing one NH bond). AsH5: C3 and S3 coincident (along the axial HAsH unit), three C2 (along each equatorial AsH bond), h (containing the equatorial AsH3 unit), three v (each containing the axial HAsH unit and one equatorial AsH bond). cycloB3N3H6: C3, S3 coincident (perpendicular to the molecular plane and passing through the midpoint 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 midpoint of the fourcarbon ring), three S2 (e.g. passing through i and perpendicular to the fourcarbon 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 SiH 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 prerequisite for assigning symmetry. CO2 is linear, Dh 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 transMo(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 Cv. mer and fac WCl3F3 are both octahedral in shape, of C2v, C3v symmetry, respectively. 4.
The cisisomer of CrCl2(acac)2 belongs to the C2 point group and is chiral and polar, while the transisomer 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. dx2y2 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
6d 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
3v 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
3v 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)
cisN2H2 (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
3v 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 Ramanonly 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 Ramanonly 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 Dh 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
3v 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
PO PCl ClPCl/ClPO
1 0 0
1 1 2
= A1 = A1 + E = 2A1 + 2E
Infrared (liquid; cm1)
Raman (liquid ; cm1)
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 PCl stretch PCl 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 Ramanonly active and pol; Au, Bu both infraredonly active. NN NF FNN outofplane
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, cm1)
Raman (gas, cm1) 1636 (pol) 1010 (pol)
989
592 (pol)
412 360
Ag Ag Bu Ag Au or Bu Au or Bu
N=N stretch NF stretch NF stretch FNN deformation outofplane deformation outofplane deformation
Decreasing energy : N=N stretch > NF 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 XeF bond order is 1. In XeF2, the total of four electrons (Xe: 2; F: 2 1) are distributed equally over the bonding and nonbonding MOs. There is a total bond order of 1 i.e. 0.5 per XeF 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 sorbital on carbon, while the other combines with the carbon pz; both bonding and antibonding combinations are formed. For the bonds, the inphase 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 * antibonding MOs. The outofphase pair of oxygen combinations have no match with an AO on carbon and are nonbonding. 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 nonbonding MOs. The total bond order is 4 i.e. a bond order of 2 per CO 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 NO 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
3v 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 BF bond.
App.4]
211
Answers to Problems
Boron in BF3 should be a strong Lewis acid as (i) it has a vacant porbital 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
3d 0
3v 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 singlydegenerate 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
3v 2
3d 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 dorbitals on iron. AOs on iron:
dz 2 dx2y2 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
2v 2
2d 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 dx2y2 : b1g pz : a2u dz 2 : a1g dxy : b2g MO combinations:
Note: In the eu SALCs two p orbitals in each case make nonbonding 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 inphase and outofphase, 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 symmetryallowed 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, dx2y2 (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, dx2y2)
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 spinallowed 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 spinallowed 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 symmetryallowed 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 spinallowed 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 symmetryallowed 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 (felectrons 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
3h 1
6d 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 spinallowed 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
3v 1 1 0
Tz Rz (Tx, Ty), (Rx, Ry)
x2 + y2, z2
C2 1 1 1 1 2
2v 1 1 1 1 0
C2 1 1 1 1 2 2
3v 1 1 1 1 0 0
2d 1 1 1 1 0
3d 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
3v 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)
2v 1 1 1 1 0 1 1 1 1 0
2d 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
Dh g+ gg 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
6d 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)