UFDC Home  Search all Groups  UF Institutional Repository  UF Institutional Repository  UF Theses & Dissertations  Vendor Digitized Files   Help 
Material Information
Subjects
Notes
Record Information

Full Text 
A NEW WEAK FIELD COUPLING SCHEME FOR COMPLEXES AND APPLICATION OF COMPLETE d3 ENERGY LEVEL CALCULATIONS TO SOME TETRAGONAL CHROMIUM (III) COMPLEXES BY JOHN ANTHONY COLLINS A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1982 I dedicate this work to Jesus of Nazareth. ACKNOWLEDGMENTS With sincere gratitude I acknowledge my advisor, Professor Jayarama R. Perumareddi, whose guidance, patience and encouragement made this work possible. I owe a special debt of gratitude to those who typed this manuscript: Mrs. Rita Levy, Mrs. Bobbi Rice and, especially, Mrs. Joyce Phillips. I would like to thank Dr. Timothy Cotton for his help with the computer programming. Most importantly, I would like to thank my Mother and Father for their encouragement, prayers, love and support which sustained me throughout this endeavor. TABLE OF CONTENTS ACKNOWLEDGMENTS . . . ABSTRACT . . . CHAPTER I. INTRODUCTION . . . II. THE WEAK FIELD COUPLING SCHEMES . The Interelectronic RepulsionsFree Ion Terms The {L,S,XC,xXQ,r} Coupline Scheme . The {L,S,XCTC,IQ} Coupling Scheme . Checking of the Weak Field Perturbation Matrices . . . III. THE STRONG FIELD COUPLING SCHEME . The Cubic Ligand Field .. .. ... The Interelectronic Repulsions . The SpinOrbit Interaction . . The Axial Ligand Field ........... Checking of the Strong Field Perturbation Matrices . . . IV. THE PERCENT PURITIES OF THE EIGENFUNCTIONS IN THE VARIOUS COUPLING SCHEMES . . The Percent Purities of the Cubic Levels . The Percent Purities of the Tetragonal Levels V. APPLICATIONS TO SOME TETRAGONAL CHROMIUM (III) COMPLEXES . . . Trans[Cr(PDC) . . fCr(NH3)5(NCO)jI2 ICr(NH3)5(NCO)]+2 . . Summary and Discussion . . BIBLIOGRAPHY . . . BIOGRAPHICAL SKETCH . . . . iii V S 1 . 17 17 . 47 S. 146 S. 188 204 . 204 . 207 . 223 . 247 . 247 256 257 269 . 282 S 292 S 295 297 300 302 . . Abstract of Dissertation Presented to the Graduate Council of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy A NEW WEAK FIELD COUPLING SCHEME FOR COMPLEXES AND APPLICATION OF COMPLETE d3 ENERGY LEVEL CALCULATIONS TO SOME TETRAGONAL CHROMIUM (III) COMPLEXES By John A. Collins December, 1982 Chairman: Jayarama R. Perumareddi Major Department: Chemistry In the wellknown weak field sequence of perturbations, which is called {L,S,J} coupling scheme, one considers the spinorbit interaction immediately after the interelectronic repulsions but prior to the ligand field potential. For 3dn systems the spinorbit interaction is not large relative to the ligand field potential. Therefore, it is of value to obtain energy levels in a coupling scheme, {L,S,X}, which considers the ligand field potential prior to the spinorbit interaction. We have devised this new {L,S,X} sequence, for both cubic and tetragonal symmetry, for the case of d3 (and d7) electronic configuration. We have found that the {L,S,X} sequence yields purer eigenfunctions than does the {L,S,J} sequence. We have also compared the {L,S,X} weak field sequence with the wellknown strong field sequence. To expand this comparison we have devised an additional tetragonal strong field coupling scheme. We find in the cubic limit, for both d3 and d7, that the strong field sequence yields purer eigenfunctions than the comparable weak field sequence. We find in the tetragonal limit that the strong field sequence yields purer eigenfunctions for d3 but for d7 in a strong tetragonal field it is the {L,S,X} sequence which yields the purest eigenfunctions. In general, we find that considering the spinorbit interaction as the last perturbation in the sequence yields purer eigenfunctions than would be obtained if the spinorbit interaction were considered earlier in the sequence. Lastly, we have, using complete energy level calcula tions, fitted the spectra of some tetragonal chromium (III) complexes in which the five lowest energy intraconfigurational doublets have been identified. CHAPTER I INTRODUCTION Crystal field theory attempts to rationalize the spectral and magnetic properties of transition metal compounds with a simple electrostatic model. In this model the ligands, i.e., the entities directly bound to the central ion, are considered to be point charges or point dipoles fixed in space about the central transition metal ion. The electric field generated by the ligands (henceforth "the ligand field") gives rise to a splitting of the dorbitals of the metal ion. This splitting of the electronic energy levels of the free ion upon complex formation can be determined qualitatively by symmetry arguments alone, as has been shown by Bethe.1 An important application of crystal field theory is in the interpretation of the near infrared, visible and near ultraviolet absorption spectra of coordination compounds. These low intensity bands are now generally recognized as arising due to electronic trans itions between the no longer degenerate dorbitals. When quantitative calculations of the electronic energy levels are undertaken with the ligands approximated as point charges or point dipoles very poor agreement with experimental data is obtained.2 1H. Bethe, Ann. Physik., [5], 3, 807(1935) 2S. Sugano, Y. Tanabe and H. Kamimura, Multiplets of Transition Metal Ions In Crystals, Academic Press, New York, 1970. 1 This is not surprising since the simple electrostatic theory neglects all covalent bonding and attempts to describe the many electron and, often, the many atom ligands as just point charges or dipoles. Molecular orbital theory, which takes account of ligandmetal orbital overlap, gives probably the best description of bonding in transition metal compounds. Unfortunately, quantitative molecular orbital calculations have to be carried out for virtually every single system and it is difficult to generalize the results for a series of systems. The hybrid of the best points of crystal field theory and molecu 2 lar orbital theory is found in modern ligand field theory. Qualita tive aspects of the crystal field theory and the molecular orbital theory can be obtained on the basis of symmetry arguments alone. Ligand field theory incorporates the symmetry aspects of both the theories but rather than attempting to calculate the energy differences rigorously, they are taken as adjustable parameters. These ligand field parameters are then obtained by fitting the absorption spectra of each complex with the calculated parametric energy levels. Ligand field theory is thus the parametric form of crystal field theory or of molecular orbital theory. The Hamiltonian for a free ion, manyelectron system is given below J. P. Dahl and C. J. Ballhausen, Advan. Quantum Chem., 4, 170 (1968) and references cited therein. C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 1962. h2 v2 h 2 ze2 8T 2M 8T2m i r. (1) 2 + I T + 7 (ri s + e + (ri)i'*si, i where h is Planck's constant, M is the mass of the nucleus, 2 2 2 2 d 3 9 2 is the Laplacian, i.e., 2 + 2 3x y y 3z for the nucleus, m is the mass of an electron, V. is the Laplacian for the i electron, ze is the effective nuclear charge, e is the charge of a single electron, ri is the distance between the nucleus and the i electron, rij is the distance between the ih and th j electrons and + > ((ri).is.i is the term in the Hamiltonian due to the inter action of the magnetic fields generated by the spin and orbital angular moment of the electron, this is called the "spinorbit interaction." The Hamiltonian for a free ion, manyelectron system can be rewritten as 2 H = Ho + e + E(ri).*s (2) i where Ho is a sum of one electron and free ion Hamiltonians and the terms which arise due to the interelectronic repulsions and the spinorbit interaction are considered to be perturbations. When a free ion forms a complex, a fourth term is added to the Hamiltonian: the ligand field potential (VLF). This potential is due to the coulombic repulsions between the central ion's d electron and the ligands. The ligand field potential has the form qke VLF(Xi) r (3) k ik th where xi is the coordinate of the it electron, qk is the magnitude of the charge on the kth ligand, and r.ik is the distance between the ith electron and the kth ligand. There are now three perturbations to be added on to the single electron Hamiltonian: the interelectronic repulsions, the spinorbit interaction, and the ligand field potential. The order in which the perturbations are considered will follow the decreasing order of the magnitude credited to each effect. This is reflected in the secular determinant in the extent to which each perturbation is diagonalized. In the strong field approach we consider the ligand field potential first. In (j,j) coupling we carry out the spinorbit interaction as the first perturbation. Lastly, in the weak field procedure we consider the interelectronic repulsions to be the first perturbation. For transition metal complexes, especially for first row tran sition metal complexes, the spinorbit interaction is always a minor perturbation relative to the interelectronic repulsions and the ligand field potential.1 Therefore, the (j,j) coupling approach, which implies that the spinorbit interaction is the major perturba tion, is inappropriate for transition metal complexes (especially first row transition metal complexes). For this reason, we will be concerned only with the weak field and the strong field approaches. After specifying which perturbation will be taken first,we must still decide the order of the remaining perturbations. Clearly, there are two possibilities for each of the three approaches we have just described. Each particular sequence of perturbations is known as a coupling scheme or a representation. When complete configuration interaction is included all coupling schemes for a given electronic configuration and symmetry will yield the same set of eigenvalues. However, how apt the description of the eigenfunctions is, in terms of the basis eigenvectors of the coupling scheme, will be dependent on the sequence of perturbations chosen. Thus, the choice of an appropriate coupling scheme is important C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 962. because only then can reasonable parentages be ascribed to the various levels and a reasonable approximation to the true eigenfunction be gained from the basis eigenvectors. Also, if configuration interac tion is neglected, the various coupling schemes for a given electronic configuration and symmetry will no longer yield identical sets of eigenvalues. With neglect of configuration interaction only an appropriate coupling scheme (i.e., a coupling scheme in which the sequence of perturbations reflects the relative magnitudes of the perturbations) will yield a reasonable approximation to the true set of eigenvalues. Thus, for the octahedral (Oh) case there are two possible weak field and two possible strong field coupling schemes. These coupling schemes can be represented as shown in Table 1. Henceforth, we will refer to the octahedral case as the cubic case. To date only the {L,S,J} sequence of perturbations has been used in complete weak field calculations.1 The {L,S,J} sequence 1A. D. Liehr, J. Phys. Chem., 64, 43(1960), (dl and d9); A D. Liehr and C. J. Ballhausen, Ann. Phys., (N.Y.), 6, 134(1959), (d and d8); H. A. Weakliem, J. Chem. Phys., 36, 2117(1962), (d3 and d/); J. P. Jesson, J. Chem. Phys., 48, 161(1968), (d3 and d7); J. Ferguson, Aust. J. Chem., 23, 635(1970), 7d3 and d7); T. M. Dunn and WaiKee Li, J. Chem. Phys., 47, 3783(1967); Erratum, Ibid., 53, 2132(1970), (d4 and db); E. Konig and S. Kremer, Z. Naturforschung, 29a 31(1974), (d4 and d6); WaiKee Li, Spectrochim. Acta, 24A, 1573(1968); Erratum, Ibid., 2289(1971), (d5); E. Konig, R. Schnakig and S. Kremer, Z. Naturforschung, 29a, 419(1974, (d5); W. Low and G. Rosengasten, J. Mol. Spec., 12, 319(1964), (dO). TABLE 1 Weak and Strong Field Coupling Schemes in Cubic Symmetry Title of the Coupling Sequence of Perturbations Scheme 2 e 2/r, then ((r)as, then VLF {L,S,J} e /rij, then VLF, then ((r).s {L,S,X} + + 2 C C C VLF, then ((r)'s, then e2/rij {x ,y , VLF, then e2/rij, then s(r)ts {xC ,XC,FC has been found convenient because it follows the free ion sequence of perturbations. That is, when carrying out an {L,S,J} coupling scheme one need not repeat the calculations for the first two perturbations (the interelectronic repulsions and the spinorbit interaction). This duplication of the free ion case for the first two perturbations greatly reduces the computational effort required in an {L,S,JI coupling scheme relative to the other cubic coupling schemes. As we have mentioned, for complexes of the 3d transition metals, in particular for those members with low atomic numbers, the spinorbit interaction's effect is small and can often be neglected. Therefore, obtaining energy levels in the {L,S,X} weak field coupling scheme, in which the spinorbit interaction can be con sidered to be a minor perturbation relative to the ligand field potential, could be of value. The first and the main purpose of this investigation is to devise the new {L,S,X) coupling scheme for the various d" elec tronic configurations and to demonstrate the superiority of the {L,S,X} coupling scheme over the {L,S,J} coupling scheme for the first row transition metal ions. We have previously carried out the {L,S,X} coupling scheme for the case of d2 (and d8) electronic configuration.1 Here we extend the {L,S,X} coupling scheme to the case of d3 (and d7) electronic configuration. The {x ,y ,r } cubic strong field coupling scheme considers the spinorbit interaction as the second perturbation. Hence, the {x ,y ,C} coupling scheme is the strong field analog to the {L,S,J} weak field coupling scheme. The {x ,X ,C } cubic strong field coupling scheme considers the spinorbit interaction as the last perturbation. It is likely that the {xC,X C,r} scheme, in which the spinorbit interaction is the last perturbation considered, is more appropriate for first row transition metal complexes than is the J. A. Collins and J. R. Perumareddi, Theoret. Chim. Acta, 54, 325 (1980). {x ,yC,r } scheme, in which the spinorbit interaction is the second perturbation considered. The {x ,X ,FC} cubic strong field coupling scheme for the case of d3 (and d7) electronic configura tion has been carried out by Eisenstein. We have repeated the SC C 3 7 calculation of the {x ,X ,C } scheme for the case of d (and d7) electronic configuration in order to have available to us the requi site wave functions so as to be able to extend this strong field coupling scheme to a lower symmetry. When the symmetry of the complex is lowered from Oh to D4h, either by substitution or by distortion, then the ligand field changes from cubic to tetragonal. We denote the ligand field poten tial of tetragonal (or quadrate) symmetry by VQ. It is also possible to express the tetragonal potential as a sum of cubic (VC) and axial (Va) potentials,2 as is illustrated in Fig. 1. In this latter case, tetragonal energy levels are obtained by going through the cubic levels and finding how they are perturbed by the axial field potential. We have seen that two weak field schemes are feasible for cubic symmetry: the {L,S,J} and the {L,S,X} coupling schemes. The {L,S,J} sequence of perturbations gives rise to two coupling schemes in tetragonal symmetry. In the first coupling scheme the cubic and the axial fields can be considered separately, always taking the cubic field first since the added axial field acts as a small perturbation 1J. C. Eisenstein, J. Chem. Phys., 34, 1528(1961). 2j. R. Perumareddi, Phys. Stat. Sol. (b), 59, K127(1973). C C U  " Cr"  LLn r r* c o o 0 E *r C (a 4a) Iv a *r ..J e ., I a r to the cubic field, both after spinorbit interaction. In the second coupling scheme the cubic and the axial fields may be taken together, after the spinorbit interaction. The {L,S,X} sequence of perturbations gives rise to three coupling schemes in tetragonal symmetry. When the cubic and the axial fields are taken separately,the axial field may be considered either before or after the spinorbit interaction. The third scheme results when the tetragonal potential is applied as a whole before the spinorbit interaction. The two cubic and the five tetragonal weak field coupling schemes are listed in Table 2 with special emphasis being placed on the cubic parentage of the tetragonal coupling schemes. We have carried out the {L,S,XC,C,F } and the {LSXC,XQ,FQ tetragonal weak field coupling schemes for d3 (and d7) electronic configuration. The {L,S,XC,rC, Q} tetragonal scheme is an extension of the {L,S,X} cubic scheme. The {L,S,X ,r } tetragonal weak field coupling scheme differs only slightly from the {L,S,XC,XQ, } tetragonal weak field scheme. Therefore, we have not carried out the {L,S,X ,r } scheme for the case of d (and d7) electronic configuration. The d3 wave functions and the Hamiltonian (energy) matrices for the {L,S,XC,C, Q} and the {L,S,XC,XQ,FQ} coupling schemes will be presented along with the details of their derivation. J. A. Collins, Master's Thesis, Florida Atlantic University, 1976. (I E an u a) cn 4' 0 3 0 u... 4 a o O S r) cu f i "3 cr 0 cn (U *r 3 *r 0 3 1^1 r_1 oC cr 0 0 0 S Cy cY Cr X D O U CY X X X a f> a 1 1 1 O CY o c m A > * ca u c > > ! C .C +3 + c m c =  L n 4 I *rS *13 C' C'~ Nc N me 0} *r" E I 4 rc o e 0 E re ar I, L >  c r en * u a 10 e + * =! 4 .. O *r** Sv u Lc i= CO a, c 0) c, 4> C 01 .0; r iLJ tt) fv, ti L C cD s, rl cu * i luvP v r: +inu cu * .c s e > c, C C( n CY U > >= + cu > C cc CU = .2 +J 4+> *0 *r3 *p N N Co' M u U C c 4< . 42 v LLP ft +*? 0 *r e c. n *'r1 C\J The strong field approach yields six coupling schemes when the symmetry is lowered from cubic to tetragonal. The two cubic and the six tetragonal strong field coupling schemes are listed in Table 3. A special emphasis has been placed, in Table 3, on the cubic parentage of the tetragonal coupling schemes. The {xC,yC C ,FQ} and {x ,C ,Y ,rQ} strong field coupling schemes consider the spinorbit interaction as the second perturba tion, prior to the interelectronic repulsions. The {xQ,y Q,Q} strong field coupling scheme also considers the spinorbit interaction prior to the interelectronic repulsions. It is unlikely, especially for complexes of first row transition metal ions, that the spinorbit interaction would be more important than the interelectronic repul sions. Consequently, the {xCyC, ,Q}, {xCy Q} and {xQ,yQ,rQ} strong field schemes are not expected to be important for complexes of the transition metal ions (especially those of the first row). The {x ,XQ,r } strong field coupling scheme considers the interelectronic repulsions prior to the spinorbit interaction but after the axial ligand field perturbation. This sequence of perturba tions is almost certainly an improvement over those of the three previously mentioned strong field coupling schemes in which the spinorbit interaction is taken prior to the interelectronic repulsions. Nonetheless, the axial ligand field is often due to a relatively small distortion in the cubic environment and, as such, is not expected to be of greater importance than the interelectronic repulsions. The {xQ,XQ,r } scheme might well find application in the study of 0) E s C) ul *r Ec: 0 0 4) o S.. C Ve LJ. a0 0.2 (I) E r 0 C ( S CD i r CI a) L O LC t) , C *r a, 0) UT r CA O v3 CD C7 C) C. . X X >< 0 0 X X X 4 .4) a (v t ( S. S C C 4C a, c ci, c *f 5 S. 5 ' >"S CC"C C0 I CY 4 ' 5 cu 55 .> 4 4 C 0) c ~> C 0 c r , SS S 4) .4 4) L) u *(0 *r *i ci (U CM 0 0) C) > tC C C S. C C 4' C 0 CY > > > *r L C. 0 .2 fto) ~> square planar complexes. A square planar complex is, of course, an extreme case of tetragonal distortion in which two transligands are completely removed. The {x xC,XC, FQ} and {xC,XC, C,FQ} strong field coupling schemes both consider the interelectronic repulsions as the second perturbation. It is likely that of the six possible strong field schemes the {xCXC,XQ,FrQ and {xC,XrCrCQ} schemes are the most appropriate for transdisubstituted and moderately distorted octa hedral complexes. The {xC,XC,XQ,FQ} coupling scheme has been carried out recently by Perumareddi for d3 (and d7) electronic configuration.1 As the second purpose of this investigation we have carried out the {xC,C,rC,rQ} strong field coupling scheme for d3 (and d7) electronic configuration, in order to compare this tetragonal coupling scheme with the {xCxC,XQ,X Q} strong field coupling scheme and with the two tetragonal weak field coupling schemes which we have carried out. The {xC,XC, C ,Q} tetragonal scheme is an extension of the {x ,X C} cubic scheme. The d3 wave functions and the Hamiltonian (energy) matrices for the {xC,XC,rC ,Q} coupling scheme will be presented along with the details of their derivation. The detailed comparison of the various coupling schemes for d3 (and d7) electronic configuration and both cubic and tetragonal J. R. Perumareddi, to be published. symmetries will be presented. This comparison will be based on the percent purities of the eigenfunctions yielded by each scheme. The spinorbit interaction usually need not be included in the interpretation of the spinallowed dd spectra of transition metal complexes. This is so because the spinorbital components of the spinallowed bands are seldom resolved. The interpretation of the spinforbidden transitions, however, requires the inclusion of the spinorbit interaction. As the third purpose of this investigation we will attempt the fitting of the spectra of some tetragonal chromium (III) complexes, taken from the literature. The complexes that we have chosen are those in the spectra of which all the ex pected low energy intraconfigurational spinforbidden doublets have been identified. CHAPTER II THE WEAK FIELD COUPLING SCHEMES The Interelectronic RepulsionsFree Ion Termsi In the weak field coupling schemes the interelectronic repulsions constitute the first perturbation. Hence, we need calculate only once the interelectronic repulsion energies for all of the weak field schemes. Furthermore, the wave functions derived in order to calculate the free ion interelectronic repulsion energies will serve as the basis functions for the other perturbations in the weak field coupling schemes. The three electron wave functions are normalized Slater determi 2 nants; that is for y1',2',3 wave functions for three isolated electrons, with total generalized coordinates X1,X2,X3 respectively, we define the threeelectron wave function (Y) by 1(X1) 1(X2) ~1(X3) '(X1,X2,X3) = '2(X1) 2(X2) 2(X3) / v3 3(X1) 3(X2) 3(X3) (4) 1E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, University Press, Cambridge, 1951. C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 1962. = (1(X1)P2(X23(X3) (X1)2(X33(2) 1(X2)2(X1)3(X3) + 1(X2)2(X3)3(X1) + (X3)2(1)(X2) 1(X3)2(X2)3(X1))//6. These wave functions are antisymmetric in accordance with the Pauli exclusion principle. Throughout, we will use a shorthand nota tion in which the fact that the wave functions are Slater determinants will be understood. Thus, we will write the above expansion of Y(X1,X2,X3) as simply ('1(1)Y2(2)i3(3)) or even more briefly as For a 3d electron the four principal quantum numbers are n = 3 (5a) S= 2 (5b) m = 2, 1, 0 (5c) and ms = 1/2 (5d) As usual, we have assumed the single electron hydrogenlike orbitals for the many electron case. For the many electron case we have the quantum numbers L,S,ML and MS, where ML = m(m.) and (6a) 1 MS = (msi) (6b) *1 with the summation being over all the electrons. The L and S are analogous to the single electron quantum numbers Z and s. Thus, we have ML = L, Ll,..., L and (7a) MS = S, S1,..., S (7b) The Pauli principle states that no two electrons can have the same wave function. Therefore, 3d electrons in the same system must differ in either mR or ms. Furthermore, due to interelectronic repulsions, not all the L and S combinations allowed by the Pauli principle will have the same energy. For each electronic configura tion, we need to pick out the allowed combinations of L and S, known as "terms," and, in order to calculate the energies, determine the wave functions associated with each term. We shall use the RussellSaunders (LS coupling) procedures for finding the energies corresponding to the terms. This is a parametric method the details of which we will now describe for the case of d3 outer electronic configuration. For a 3d3 ion we can construct the microstate table as shown in Table 4, obeying at all times the Pauli principle. In writing the microstates the fact that n=3 and k=2 is understood. Thus, the microstate (221) means that electron one has 1C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 1962. 20 TABLE 4 The Weak Field Microstate Table for d +3/2 +1/2 1/2 3/2 ML +5 (211) (2T1) (270) (270) +4 (21T) (71T) (210) (210) (2TO0) (210) (210) (210) +3 (710) (271) (2T10) (27T) (21T) (2T1) (2T1) (21T) +2 (211) (711) (272) (2TT) (227) (21T) (20U) (ITO) (200) (110) (217) (2T2) (2T2) (717) (212) (712) (20T) (212) (701) (717) +1 (201) (2U1) (701) (0T) (2T1) (7T) (10) (11) () 10) (1TT) (207) (202) (202) (202) (202) (702) (T01)2) (2) (TO1) (202) 0 (101) (1O2) (10T) (TOT) (10T) (101) (IT2) (21T) (1T7) (71T) The rest are obvious m equal to +2 equal to +2 and equal to +1 and or (221) is not We can now which we denote and ms equal to +1/2, that electron two has mk ms equal to 1/2, and that electron three has m. ms equal to +1/2. Obviously, the microstate (221) possible. derive the RussellSaunders terms (LS coupling terms) by where symbols are used for L as shown in Table 5. TABLE 5 Term Symbols L Symbol The spin multiplicity is given by (2S+1) and the tiplicity by (2L+1). Thus, the total multiplicity of by the product (2S+1)(2L+1). The microstate (221) yields ML = 5 which, bein ML possible, belongs to a term of L=5. Since for tl orbital mul a term is given g the maximum his term MS can only be +1/2 or 1/2, then S is 1/2. Therefore, we have a 2H term whose total multiplicty is 22. The 2H(22) term requires a microstate in each box from ML = +5 to ML = 5 with M = +1/2 and 1/2, 22 in all. Turning to the microstate (220) we have ML = 4. Now ML = 4 is maximum since the microstate (221) has already been claimed by 2H. Thus, (220) belongs to a term with L = 4. Since for this term MS can be only +1/2 or 1/2 then, S = 1/2. Therefore, we have a 2G term whose total multiplicity is 18. The 2G(18) term requires a microstate in each box from ML = 4 to ML = 4 with MS = +1/2 and 1/2, 18 in all. The microstate (210) has ML = 3, now the maximum possible, and MS = 3/2, also the maximum. Therefore, we have a 4F term whose total multiplicty is 28. The 4F(28) term requires a microstate in each box from ML = 3 to ML = 3 with MS = 3/2, 1/2, 1/2, and 3/2, 28 in all. Continuing we find a 2F(14) term, two 2D(10) terms, a 4p(12) term and a 2P(6) term. In summary, for d3 we have the following terms: 2H(22), 2G(18), 4F(28), 2F(14), 2Da(10), 2Db(10), 4p(12) and 2P(6). Now we must decide which microstates or linear combinations of A A microstates are eigenfunctions of the L and S terms (the circumflex indicates an operator). The appropriate combinations of the microstates will then serve as the wave functions for the terms and will be known as the IL,S;ML,Ms> functions. When only one microstate has a particular ML and MS combination, this is straightforward. Thus, since only the microstate (221) has ML = 5 and MS = 1/2, the function 15,1/2;5,1/2> is L (221) and is one of the 22 eigenfunctions of 2H. There are two microstates with ML = 4 and MS = 1/2; these are (220) and (211). Finding the linear combination of these two microstates that is the eigenfunction 15,1/2;4,1/2> of 2H is our problem. Application of 1 ^ ^ the angular momentum raising and lowering operators, L. and S., will allow us to form the proper IL,S;ML,MS> functions for 2H. The use of these raising and lowering operators is illustrated below. L+IL,ML> =.f[(L+ML+1)(LML)]1/2IL,ML+1> (9a) L_IL,ML> =,K (LML+1)(L+ML)]1/2 L,ML1> (9b) S+IS,Ms> = [(S+Ms+1)(SMS)] 1/21S,Ms+1> (9c) S_IS,MS> = [(SMS+1)(S+Ms)] 1/2S,MS1> (9d) The operators are to be applied to each electron separately; that is, L = Z(1) + Z(2) + ...(+(n) (o1a) and S+ = s(1) + s.(2) + ...s(n) (10b) when there are n electrons. C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 1962. Thus, L_5,1/2;5,1/2> = 4[(55+1)(5+5)]1/2 5,1/2;4,1/2> = [_(1) + Z (2) + Z (3)](221) =4[(22+1)(2+2)]1/2(121) +,6[(22+1)(2+2)]1/2(211) +~[(21+1)(2+1)]1/2(220) so 15,1/2;4,1/2> = [/3(220) /2(211)]/J5. By repeated application of these operators we can derive the complete set of IL,S;ML,MS> functions for the 2H term. For the 2G term we do not have a unique microstate for any ML and MS combination to begin the analysis with. For the box cor responding to ML = 4 and MS = 1/2 we have the linear combination of (220) and (211) which is the function 15,1/2;4,1/2> of 2H. We know that the function 14,1/2;4,1/2> of 2G must be orthogonal to 15,1/2;4,1/2> of 2H. Also, we have the normalization condition. Using these two conditions we can write down 14,1/2;4,1/2> of 2G. This function is 14,1/2;4,1/2> = [/2(220) + /3(211)]//5 Note that the choice of phase for 4,1/2;4,1/2> of 2G is initially arbitrary but once made must be adhered to throughout. Then by application of the appropriate raising and lowering operators we can derive the complete set of IL,S;ML,MS> functions for the 2G term. For 4F we have a unique microstate with ML = 3 and MS = 3/2. Thus, the function 13,3/2;3,3/2> is (210). Applying S_ we obtain S_13,3/2;3,3/2> = [(3/23/2+1)(3/2+3/2)]1/2 3,3/2;3,1/2> S[S_(1) + S_(2) + S_(3)](210) =,[(1/21/2+1)(1/2+1/2)]1/2(210) + ,[(1/21/2+1)(1/2+1/2)]1/2(210) +K[(1/21/2+1)(/1/2+1/2)] (210) and 3,3/2;3,1/2) = [(210) + (210) + (210)]//3 Continued application of the appropriate raising and lowering operators allows us to derive the complete set of IL,S;ML,MS> functions for the F term. The use of orthonormality and the raising and lowering operators allow us to determine the IL,S;ML,MS> functions for the other terms (2F, 0a, 2b, 4P and 2P) as well. The case of the two D terms 2 2 (2Da and Db) is particularly difficult since two independent functions must be found simultaneously for the same ML and MS combinations.1 E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, University Press, Cambridge, 1951. The complete set of the L,S;ML,MS> functions for the d3 terms is given in Table 6. In the perturbation theory of degenerate systems one constructs a secular determinant which is then diagonalized to obtain the perturbation energies. In the case of the interelectronic repulsion perturbation the microstates are the unperturbed basis functions and what RussellSaunders coupling does is to obtain the IL,S;ML,MS> eigenfunctions as the linear combinations of these basis functions that diagonalize the perturbation matrix. Also, for a term of definite L and S the set of (2S+1)(2L+1) states will have the same energy since the energy is independent of MS and ML* The energy of a term then is calculated by the expression E = /L H TdT (11) where Y is the IL,S;ML,MS> function of that term and H is the interelectronic repulsion perturbation ( I e2 /r). For 2D a 2 by 2 i We can expand 1/ri. in terms of the wellknown spherical 1 mR harmonics, YZ (86,): i * CO +i) r m m 1 41n r< m . S(2+1 .1+1 Y (i' i)'y (oj',ij) (12) ij Z=0 mz=z r where r< is the lesser and r> the greater of the two distances ri and rj and Y (96,) are the spherical harmonics. 1H. Eyring, J. Walter and G. E. Kimball, Quantum Chemistry, John Wiley and Sons, Inc., New York, 1944. TABLE 6 The IL,S;ML,Ms> Functions for 1,1/2;1,1/2> = (4/3(212)+2/3(2I2)+2/3(212)+4/2(20I)+/2(201) 5/2(201)3/3(111)3/3(100))//210 1,1/2;0,1/2> = (4(202)3/6(1i2)+8(202)4(202)(101)+3/6(21 ) +2(101)(101))//210 1,1/2;1,1/2>= (2/3(212)5/2(102)+4/2(152)+/2(I02)+2/3(212) 4/3(212)+3/3(111)+3/3(001))//210 1,1/2;1,1/2>= (2/3(212)2/3(212)+4/3(212)+5/2(20i)4/2(201) 2(20)3333 111 /3 (100))/V210 1,1/2;0,1/2>= (8(202)3/6(li2)+4(202)+4(262)2(101) +3/6(21i)+(0I)+(101) )//210 I1,1/2;11/2>= (5/2(i2)/2(102)4/2(102)+4/3(212)2v3(212) 2/3(212)+3/3(i1i)+3/3(001))//210 Da 12,1/2;2,1/2> = ((211)+(200)(222)(211))/2 12,1/2;1,1/2> = ((111)+(100)+(212)(212))/2 12,1/2;0,1/2> = ((10i)+(202)(i01)(202))/2 2,1/2;1,1/2>= ((111)+(212)+(001)(212))/2 TABLE 6Continued 12,1/2;2,1/2> 12,1/2;2,1/2> 12,1/2;1,1/2> 2,1/2;0,1/2> = ((i12)+(oO2)(222)(1I2))/2 = ((1iI)+(I0O)+(zI2)(2I2))/2 =~l)(b 2I 12)/ 12,1/2;1,1/2> = ((I1I)+(2i2)+(00f)(2I2))/2 12,1/2;2,1/2> = ((12)+(002)(22)(Ii2))/2 2b 12,1/2;2,1/2> = (3(211)5(222)+(211)4(211)3(200) 2/6(110))//84 = (3(212)+(212)4(212)(111)+3(100) 2,1/2;0,1/2> = (3(202)2/6(112)3(202)3(101) +3(10I)2/6(21I))//84 2,1/2;1,1/2> 12,1/2;2,1/2> = (2/6(102)(212)2/6(102)3(212) +3(001)(11I)+4(2i2))//84 = (4(112)5(222)3(002)(112) 3(I12)2/6(01I))//84 12,12;211> =(4(2TT)5(222)3(211)(21 T) 12,1/2;1,1/2> 2/6(201)+2/6(2D1))/84 2,1/2;2,1/2> 3(200)2/6(110))//84 TABLE 6Continued = (4(212)3(212)(212)(11i)+3(iOO) 12,1/2;0,1/2> = (3(202)2/6(112)3(202)3(101) +3(101)2/6(211))//84 12,1/2;1,1/2> = (2v6(102)2/6(102)4(212)+3(001) (I11)+(212)+3(212))//84 12,1/2;21/2> 13,1/2;3,1/2> 13,1/2;2,1/2> 13,1/2;1,1/2> 13,1/2;0,1/2> =((I12)+3(1I2)5(222)3(o62) = (/6(221)+2(210)(210)(210))/2/3 = ((211)+2(222)V6(110)(211))/2/3 = (2/2(111)+/3(201)+3/2(212)2v2(212) +3/2(100)/3(201)V2(212))//60 = (2(iO1)4(101)+V6(12)/6(21I)+4(202) 2(202)+2(10I)2(202))//60 3,1/2;1,1/2> 13,1/2;2,1/2> 13,1/2;3,1/2> = (2/2(11I)3/2(001)/3(I02)+/3(102) +3/2(2I2)2V2(212)/2(212))//60 = (/6(011)+(1I2)2(222)(I12))/2/3 = (2(012)(012)76(122)(012))/2/3 2,1/2;1,1/2> 2/6(201)+2/6(20i))/184 4(112)2/6(011))//84 13,1/2;3,1/2> 13,1/2;2,1/2> 13,1/2;1,1/2> 13,1/2;0,1/2> 13,1/2;1,1/2> 3,1/2;2,1/2> 3,1/2;3,1/2> TABLE 6Continued = (/6(221)+(210)+(210)2(2ii))/2/3 = ((2i1)+2(222)/6(1i0)(21i))/2/3 = (2/2(11i)+V3(201)+/2(212)+2/2(212) 3/2(212)+3/2(100)V3(201))//60 = (4(i)2)2(1)2(10)+/6(1I2)/6(21i) (20+2(22)(202)4(202))//60 = (2/2(i1i)3/2(001)V3(i02)+/3(102) +2/2(2i2)+2(2i2)3/2(212) )//60 = (/6(01I)+(1I2)2(222)(112))/2/3 = ((012)+(0i2)2(0i2)/6(I22))/2/3 2G 14,1/2;4,1/2> = (/2(220)+/3(211))//5 14,1/2;3,1/2> = (3(210)2(210)(210)+v6(221))/2/5 14,1/2;2,1/2> = (4/3(200)+3/3(21I)+/2(110)+4/3(211) +/3(211)+2/3(222))/2/35 14,1/2;1,1/2> = (/6(100)3(201)+7(20I)+V66(12)+2/6(111) 4(201)2V6(212)+v6(212))/2/35 14,1/2;0,1/2> = (J2(10I)+/3(21I)+/2(202)V2(101) +/3(1I2)V2(202))//14 TABLE 6Continued 14,1/2;2,1/2> 14,1/2;3,1/2> 14,1/2;4,1/2> 14,1/2;4,1/2> 14,1/2;3,1/2> 14,1/2;2,1/2> 14,1/2;0,1/2> = (2/6(111)+4(102)/6(2I2)+2V6(212) +/6(001)7(102)(2)+3(1/2)6( 2))/2/35 = (V2(011)/3(1I2)+4/3(112)+2/3(222) = ((012)+2(012)+/6(122)3(012))/2/5 = (/3(112)+/2(022))//5 = (/2(220)+/3(211))//5 = ((210)+2(210)3(210)+/6(22I))/2/5 = (4/3(200)+4/3(21I)+/2(110)3V3(2I1) /3(21I)+2/3(222))/2/35 = (/6(00)7(201)+3(201)+4(201)+2/6(212) +2/6(11I)/6(212)/6(212))/2/35 = (/2(10I)+/3(21I)+/2(202)2(101) +/3(112)/2(202))//14 4,1/2;1,1/2> = (2/6(111)+7(102)2/6(212)+/6(212) +/6(212)+/6(00I)4(102)3(IO2))/2/35 14,1/2;21/2> = (/2(01T)+v'3(I12)+3/13(12)+2/3(222) +4V3(002)4/3(112))/2/35 4,1/2;1,1/2> +4/3(002)3/3(112))/2/35 4,1/2;1,1/2> TABLE 6Continued 14,1/2;3,1/2> = (3(012)+/6(122)2(012)(012))/2/5 14,1/2;4,1/2> = (3(112)+/2(022))//5 2H 15,1/2;5,1/2> 15,1/2;4,1/2> 15,1/2;3,1/2> 15,1/2;2,1/2> 15,1/2;1,1/2> 15,1/2;0,1/2> 15, 1/2 ;14/2> 15,1/2;2,1/2> = (221) = (v3(220)/2(211))//5 = (v3(221)/2(210)+/8(210)/2(210))//15 = (/6(110)3(200)(21I)+3(21(211)2( )+(222))//30 = (8(1i1)6(100)+3/6(201)2/6(20I)(212) +4(212)/6(201)3(1)212)210 = (4(101)2(I01)+/6(1i2)/6(21i)(202) = (6(001)8(11I)+3V6(102)2/6(102) +4(212)3(212)/6(102)(212))//210 = (/6(011)+3(002)2(112)+3(112) (I12)(222))//30 15,1/2;3,1/2> 15,1/2;4,1/2> 15,1/2;5,1/2> = (v8(012)/2(012)2(012)3(122))/15 = (V3(022)+/2(112))//15 = (122) +2(202)(202)2(101))//42 TABLE 6Continued 15,1/2;5,1/2> 15,1/2;4,1/2> = (221) = (/3(220)/2(211))//5 15,1/2;,3,1/2> = (/3(22I)/8(210)+/2(210)+/2(210))//15 5,1/2;2,1/2> = (/6(1i0)3(200)3(21I)+(211) = (8(1II)6(Io1)+2,/6(2O1)+/6(20I)3/6(20) 4(212)+(212)+3(212))//210 15,1/2;0,1/2> =(2(Ib1)+2(16I)4(rIOI)+/6(1T2)/6( 21f 2(202)+(202)+(202))//42 5,1/2;1,1/2> = (6(00i)8(11I)+/6(i02)+2/6(102)3/6(i02) +3(2I2)+(2I2)4(212))//210 i5,1/2;2,1/2> = +(1I2)(222))//30 5,1/2;3,1/2> = (/2(0i2)+/2(0I2)/8(012)/3(I22))//15 5,1/2;4,1/2> = (/3(022)+/2(12))//5 5,1/2;5,1/2> = (122) 1,3/2;1,3/2> 11,3/2;0,3/2> = (/2(201)/3(212))//5 = (2(101)(202))//5 +2(2I1)+(222))//30 15,1/2;1,1/2> (/6(011)+3(002)3(112)+2(112) 11,3/2;1,3/2> 11,3/2;1,1/2> 11,3/2;0,1/2> 11,3/2;1,1/2> 11,3/2;1,1/2> 1,3/2;0,1/2> 1,3/2;1,1/2> 1,3/2;1,3/2> I1,3/2;0,3/2> 11,3/2;1,3/2> 13,3/2;3,3/2> 13,3/2;2,3/2> TABLE 6Continued = (/2(102)/3(212))//5 = (/2(201)+/2(201)+/2(20I)/3(212) /3(212)V3(212))//15 = (2(101)+2(101)+2(10I)(202) (202)(202))//15 = (/2(IO2)+/2102(12)+2(102)3(212) J3(212)/3(212))//15 = (/2(201)+/2(0I)+/2(20I)/3(2i2) /3(212)/3(2I2))//15 = (2(101)+2(101)+2(101)(202) (20(2)(202))//15 = (V2(102)+/2(102)+/2(102)/3(2i2) /3(212)/3(2I2))//15 = (/2(20I)/3(212))//5 =(2 (10I)(202))//5 =(2(102)/3(212))//5 4F = (210) = (211)  13,3/2;1,3/2> 13,3/2;0,3/2> 13,3/2;1,3/2> 13,3/2;2,3/2> 13,3/2;3,3/2> 13,3/2;3,1/2> 13,3/2;2,1/2> 13,3/2;1,1/2> 13,3/2;0,1/2> 13,3/2;1,1/2> 13,3/2;2,1/2> 13,3/2;3,1/2> 3,3/2;3,1/2> 3,3/2;2,1/2> TABLE 6Continued = (/3(201)+/2(212))//5 = ((101)+2(202))//5 = (/3(102)+/2(212))//5 = (112) = (012) = ((210)+(210)+(210))//3 = ((211)+(211)+(211))//3 = (/3(201)+/3(201)+/3(20i)+/2(212) +/2(212)+v2(212))//15 = ((i01)+(161+((101)+2(202) +2(201)+2(202))//15 = (/3(102)+/3(102)+/3(102)+/2(212) +/2(2I2)+/2(212))//15 = ((112)+(1T2)+(112))//3 = ((012)+(012)+(012))//3 = ((210)+(210)+(210))//3 = ((i1)+(21)+(21I))//3 ~I~ TABLE 6Continued 13,3/2;1,1/2> = (/3(21)+/3(201)+/3(2(1)+/2(2i2) +/2(212)+/2(212))//15 13,3/2;0,1/2> = ((1I1)+(10I)+(10i)+2(202)+(202) +(202))//15 13,3/2;1,1/2> = (/3(162)+/3(102)+/3(102)+/2(212) +/2(212)+/1 212))//15 13,3/2;2,1/2> = ((112)+(112)+(112))//3 13,3/2;3,1/2> = ((012)+(012)+(012))//3 13,3/2;3,3/2> = (210) 13,3/2;2,3/2> = (211) 13,3/2;1,3/2> = (/3(201)+/2(212))//5 13,3/2;0,3/2> = ((101)+2(262))//5 3,3/2;1,3/2> = (/3(i02)+/2(2l2))//5 13,3/2;2,3/2> = (112) 13,3/2;3,3/2> = (012) The wave functions associated with the microstates are the single electron wave functions that are the solutions of Schrodinger's equation for the hydrogen atom case. These have the form m, o = R,(r), y (,) (13) with Rn, (r) being the radially dependent factor. For a 3d ion we will write R ,(r) as R3d(r). As an example let us consider the calculation of the energy of the 2H term. We could use any of the 22 degenerate IL,S;ML,MS> functions which belong to 2H. For the sake of simplicity we choose 5,1/2;5,1/2> = (221). We must use the expanded Slater determinant for the calculation. Thus, 15,1/2;5,1/2> = [(221)(212)(221) + (122)+(212)(122)]//6 The energy of 2H is given by E'(2H) = f/ / [(221)(212)(221)+(122) 1 2 3 123 + (212)(122)1*(e2/r12 + e2/r13 + e2/r23)[(221) (212)(221)+(122)+(212)(122)]dT1 dT2 di3/6 (14) To illustrate the method of calculation let us take a fragment of the above expression for the energy of 2H. Let us consider only Sf f (221) (e /r12 + e2/r13 + e2/r23)(221)dTr dT2 dT3 123 Expanding we have l a(1)*a(1)ds1 f2 8(2)*B(2)ds2 f3 a(3)~(3)ds3 r f f f R3d(1)R3d(2)R 3d (3)e 2 y <+lJR3d(1)R3d(2) 1 2 3 =0 r S2 2* R3d(3)r drIr2 dr2r3 dr3 f f f Y2 (1)Y2 (2) 123 j=3 Y21 (3)[ i =00 ZO +z Q=0 m =t 4TT (2Z+1) * m m. Y (i, i)Y' (ej, j) Y22(1)Y22(2)Y21(3)sine1deldy1sine2de2dp2sine3de3d 3 The spin functions are orthornormal, that is a ads ds = i a ads = / 8 ds = 1 I 8 ads = / a Bds = 0 Continuing, we have o R3d (1)R3d(1) R3d (2) (2)R(3)R3d(3) 0 0 0 2 < 2 2 2 e [ <+ I]r 2drlr2 dr2r3 dr3 I=0 1 1 2 ),=O r and (15a) (15b) 2* 2 1" Sf f [Y22 (1)Y22(1)Y2 (2)Y 22(2)Y 2 (3)Y21(3)] * 123 S 2 I I R=0 m =R 4 m7 m , *(2 2 [V (1' r (e2,2) + m m m m Y (61,'l)Y9 (13,3) + Y ( 2'2)Y (e3,23)] sineldEd1d 1sin92d62d 2sin93de3d3 . Further, 00 m 3) f f f [R3d (1)R3d(1)R3d (2)R3d(2)R3d (3)R3d(3)]. 123 2 0o e [Z Z=0o r< 2 2 2 + +1]rl dr r2 r dr dr3 dr r> =0 mt=O 47T (2Z+1) f[ / / Y22 (1)Y22(1)Y2 (2)Y22(2)Y21(3)Y21(3) 1 2 3 123 Yz (al1)Yz (2,0 2)sine9de ld1sin2 d2 d92sin93dO3d3 + / y Y22 (1)2(Y2Y2 (2)Y22(2)Y21 (3)Y21(3)Y z (e1'1)Yz (63'3) 1 2 3 r% sine91d61d sine2de2d 2sin93de3dq3 2* 2 2* 2 1* S Y22 (1)y22(1)Y22 (2)Y22(2)Y21*(3)Y21(3)Yn d (02',2) 123 Y am .(03,3)sinldold~lsinld2dsin83Gd3] 3 Which further reduces to f f R3d (1)R3d(1)R3d (2)R3d(2)R3d(3)R3d(3) 123 e co r Ir r 2 dr 00 e [ r+ 1 drlr2 dr2r3 dr3 Z=0 r> t=O mR=z ^> )v 47T (2k+1) m m [ f Y2 (1)Y22(1)Y22 (2)Y22(2)Y9 (el')Y9 Z(e2'2) 12 sinOedeold 1sine2de2d02 + / Y22 ()Y22(1) sin3d3d3 + f f Y2 (2)Y22(2)Y2 (3)Y21(3) 2 3 * m r Y (e2',2)Y due to the normality of The product of two of spherical harmonics. Y22*Y2 = '2 2 1* 1 Y Y 1 = *2 2 ((03',3)sin92dO2d42sin93dO3d#3] the spherical harmonics. spherical harmonics can be expressed as a sum Thus, we find that Y40/14/T /5 Y20/7/7 + Y00/2/ 2Y 0/7/7 + /5 Y20/14/T + Y 0/2/ . 4 2 0 A list of other relevant products written as sums is given in Table 7. 41 TABLE 7 Products of Spherical Harmonics Expressed as Linear Combinations of Spherical Harmonics * = 0 /14/,T  4= = 5Y+1 /14/Tr 4 * /v5Y /7/Tv 2 * + Y0 /2/u 0 *1 + /30Y+1 /14/VT 2 = /5Y1 /14/v /30YT1 /14,/T 4 2 Y2 y2 2 Y2 Y2 Y+1 2 2 2 2 y2 y0 2 2 Y2* YTl Y2 Y2 Y01y2 2 2 * Y Y 2 2 Y2 Y2 * Y1 Y 2 2 Y+1 +1 2 2 * 1 0 Y Y2 2 2 * = /5Y3 /2/7$ 4 = /5Y /2+/7T 4 * = /5Y4 //147 4 * = 2Y4 /17/T + = v/30Y1 /14/Tv 4 * /15Y2 /14/Tv 4 5 /5Y /14/T 2 *  /5Y2 /7/v 2 * + Y /2/T 0 * + /5Y /14/Tr 2 1* * = /30Y+1 /14/Tr /5Y+1 /14/5 4 2 S 0 2  2 2 42 TABLE 7Continued * = /10Yo2 /7V /15Y2 /7 4 2 0 + 3Y /7/ + J/Y /7/T + 4 2 v/2T Y /20 r 0 y+1 Y+1 2 2 S0* Y Y2 2 2 2* 2 1* 1 Since the products Y2 2 and Y2 2 are functions of 0Y0 0 Y 4,Y20 and Y0 we need only consider in the summation those terms with A = 0,2,4 and m, = 0. Thus, the integrals we have been considering reduce to e2 ( ) R3d 3d(l)R3d (2)R3d(2)R3d (3)R3d(3) 4 r<2 2 2 ( 5)r dr r2 dr2r3 dr3 + r> 5 1 1 2 2 3 3 e2( 2 R3d(1)R3d(1)R3d (2)R3d(2)3d(3)R3d(3) 2 r< 2 2 2 (3 )r 2drlr2 dr2r3 dr3 23 d * e (1) I / / R3d (1)R3d(1)R3d (2)R3d(2)R3d (3)R3d(3) ^1 2 2 (r)r 2drlr2 dr2r3 dr3 In ligand field calculations the radial integrals are not solved directly but rather defined as parameters. For the interelectronic repulsions perturbation the radial integrals become the wellknown Condon and Shortly parameters. These CondonShortley parameters are defined as 2 mco *o Fk = () ) / R3d (1)R3d (2)R3d (3)R3d(1)R3d(2)R3d(3) r when k = 0,2 or 4 then Dk = 1,49 or 441, respecitvely. There fore, the fragment of 2H which we have been considering reduces to 3FO 7F4. To obtain the complete energy of 2H we must return to equation (14) and proceed as we have for the (221) fragment. Equa tion (14) reduces to E'(2H) = [6J(22) + 12J(21) 6K(21)]/6 In the above expression for E'(2H) we have used the following simplified notation f / (ab)(e2/r2)(cd)dT1dT2 = <(ab)(cd)> (17a) 12 <(ab) (cd)> = <(cd)l(ab)> (17b) <(ab)I(ab)> = J(ab) (17c) <(ab)l(ba)> = K(ab) (17d) These integrals (17ad) are the electron correlation integrals for d electrons. The J(ab) and K(ab) are known, respectively, as the Coulomb and Exchange integrals. The nonzero electron correlation integrals are listed in Table 8. The energy of 2H is thus E'(2H) = (Fo+4F2+F4) + 2(FO2F24F4) (6F2+5F4) E'(2H) = 3F0 6F2 12F4 TABLE 8 The NonZero Electron Correlation Integrals of the d Electrons a) J(ab) J(+2+2)=J(+2;2)=Fo+4F2+F4=A+4B+2C J(+2+1)=J(+2t1)=F02F24F4=A2B+C J(+20)=FO4F2+6F4=A4B+C J(t)=(+1+l)=+11)=F +F2+16F4=A+B+2C J(10)=FO+2F224F4=A+2B+C J(00)=Fo+4F2+36F4=A+4B+3C b) K(ab) K(+2+1)=6F2+5F4=6B+C K(+20)=4F2+15F4=4B+C K(+211)=35F4=C K(+272)=70F4=2C K(+10)=F2+30F4=B+C K(+11)=6F2+40F4=6B+2C c) <(ab)l(cd)> <(+2;2)I (t~l)>=/6(F25F4)=/6B <(+2T1) (+10)>=V6(F25F4)=V6B <(+2;1) (0+1)>=2J6(F25F4)=2/6B TABLE 8Continued <(+2;2) (t+1i )>=6F25F4=6BC <(+2;2)1 (1+1)>=35F4=C <(+2;2) (00)>=4F2+15F4=4B+C <(tl1)i (00)>=F230F4=BC <(+1+2) (10)>=2/6(F25F4)=2/6B <(+1;2) (01)>=V6(F25F4)=/6B <(0O2) I (;1l1)>=,6(F25F4)=,/6B We can also give the energy of 2H with the Racah parameters A, B and C by making use of the following relationships. A = F0 49F4 (18a) B = F2 5F4 and (18b) C = 35F4 (18c) So that we have, alternatively, E'(2H) = 3A 6B + 3C The energies of the other terms are calculated in the same manner. In the case of 2D a two by two secular determinant must be 2 2 solved since the cross term between Da and 2Db is nonzero. These interelectronic repulsions energies for the d3 terms are given in Table 9. In Fig. 2 the RussellSaunders splitting is illustrated. We have omitted in this figure the A values since they are the same for each level and we are interested in energy differences. Note that 2H and P are accidentally degenerate. The {L,S,XC,X Q} Coupling Scheme This new {L,S,X} coupling scheme for d3 considers the cubic and axial potentials individually, both before the spinorbit inter action. We must now determine how the terms are affected when we lower the symmetry of the central ion by introducing the ligand field potential. 48 TABLE 9 The Electronic Repulsion Energies for the d3 Terms = 3A 15B = 3A = 3A 6B + 3C = 3A 11B + 3C = 3A + 9B + 3C Da 2Db Da 3A+7B+7CE1 Db 3/21B 3A+3B+3CE2 = 0 2 3A 6B + 3C P = 3A 6B + 3C 3/21B FIGURE 2. Free Ion Energy Levels for d3 The Cubic Ligand Field The cubic and quadrate wave functions First we take the cubic case, that is, the octahedral arrangement of six equivalent ligands around the central ion (cf. Fig. 1). The point group for this symmetry is Oh. For Oh we have possible the following symmetry operations, grouped in classes: 3C4, 3C2(=C42), 6C2 8C3, E, i, and i combined with each of the previous operations. Since the d orbitals are even to inversion only the rotation opera tions will bring us new information. We will assume here that the spin, is, and space, %0, parts of the wave function can be separated, thus S= 0 s (19) and that the spin function has no angular dependence. Therefore, we need consider only the spatial part of the wave function. For the single electron case we have the orbitals, functions of Z and mR, while for a many electron system we have the terms, functions of L and ML. The angular form of the orbitals is given by m __ im g Y = 0 (O)1/27 e (20) where Z,m (e) are normalized, Legendre polynomials and e and } are the polar angles. Analogously, the angular dependence of the terms is ML iML YL L,ML 1/21 e (21) The relationship of the polar coordinates r, 8 and Q with the cartesian axis x, y and z is shown in Fig. 3. Obviously, no physical quantity can depend on how we choose to locate our coordinate axis. Thus, for each of the rotation operations of Oh we can take as the axis of rotation the z axis. Clearly, 6 and r are invariant to rotations about the z axis (cf. Fig. 3). So we need find only how iML the 0 dependent part of the wave function, e is affected by the various rotations. Let us take the general case of rotation by a degrees about the z axis, R(a), and assume that this rotation returns the molecule to an equivalent configuration. Hence, we have iMLP iML(4+a) R(a)e = e (22) For a P term, L equals one and ML can have the values +1, 0 and 1. Therefore, rotation by a degrees on a P term produces the following change. ei I ei[+] R(a) e0 = e (23) Sei\ ei[4+a]/ The same result can be obtained by multiplying the P functions by the matrix (eia 0 0 0 e0 0 (24) 0 0 eim Figure 3. The Polar Coordinates r, e and 6 . This matrix is the transformation matrix of the symmetry operation in question, that is, rotation by a degrees about the z axis. The set of these matrices for different rotations constitute the represen tation of L with the dimension (2L+1). The above result is not dependent on the value of ML. That is, since ML can taken all integral values between L and L, we can state that the general transformation matrix which corresponds to the rotation operations of Oh is i Lc e 0 000 0 0 0 ei(L1)a 000 0 0 0 0 000 ooo ei(1L)a 0 0 0 000 ooo 0 i Lc e (25) The sum of the diagonal elements, the character of the matrix, is in the form of a geometric progression and is given by X(c) = sin[(L+1/2)a] sin(a/2) (26) By use of (26) we can easily derive the characters corresponding to the rotation operations for different L values. It is well known that the character of each class of a reducible representation is equal to the sum of the characters of the irreducible representa tions which are contained in that particular representation. For d3 we see that the P term yields an irreducible represen tation in Oh, namely T1g. The D term yields a reducible representation in Oh composed of E and T2g. The F term yields a reducible representation in Oh composed of A2g, T1g and T2g. The G term yields a reducible representation in Oh com posed of Alg, Eg, T1g and T2g. Lastly, the H term yields a reducible representation in Oh composed of Eg, two T1g's and T 2g These results for d3 are summarized in Tables 10 and 11. If we use directly the freeion basis functions, i.e., the IL,S;ML,MS> functions, to calculate the perturbation energy due to the cubic ligand field potential, we will need to solve two secular determinants, a 40x40 (for spin quartets) and an 80x80 (for spin doublets). If the IL,S;ML,MS> functions are linearly combined such that they are the basis functions for the irreducible representations of the cubic group then the secular determinants can be block diagonal ized because of the result that the matrix elements between two different irreducible representations are zero. The proof of the theorem on which this latter result is based follows. Consider the following integral I = I /A B dT (27) where WA and tB are functions which serve as basis functions for two irreducible representations of a group. When we take the product of two representations of a group and, thus, obtain a third representation,the characters of the new representation will be equal to the product of the characters of the TABLE 10 a) The Character Table of the Terms of for the Rotation Operations of O0 L (Term) x(E) X(C4) X(C2) x(C3) X(C2) 1 (P) 3 1 1 0 1 2 (D) 5 1 1 1 1 3 (F) 7 1 1 1 1 4 (G) 9 1 1 0 1 5 (H) 11 1 1 1 1 b) The Character Table of the Irreducible Representations for the Rotation Operations of Oh Irreducible Representation X(E) x(C4) X(C2) X(C3) (C2) Alg 1 1 1 1 1 A2g 1 1 1 1 1 E 2 0 0 1 2 T1g 3 1 1 0 1 T2g 3 1 1 0 1 TABLE 11 The Splitting of of d3 in Free Ion Terms Symmetry Term State(s) in Oh 2D(a and b) 2E +2T2 g 2g 2A2g +' A1g + E ig g T1g 2T2g + 2 Tg + 2T2g 2E + 22T1g + 2T2 4T1g 4A2g 41g 4T2g two representations that were multiplied together to give the new representation. That is, if the product of representations A and B is C, then X X X C A f (28) for each class. This is known as the direct product of the representa tions A and B. We have assumed that iA and rB are bases for irreducible representations of a group. The product of two irreducible representa tions may or may not be irreducible. If A transforms as rA and iB as FB where FA and FB are irreducible representations of our group, then A B = r = C.ir A 'B C i i (29) If C is reducible,then EC.P. is the sum of irreducible representa i 1 tions which compose FC. Let us operate on I with one of the operations, R, of the group to which FA and FB belong. The value of any integral will be just a number which will not be changed by a symmetry operation. That is further RI = IR = I I = f/RAyBdT = RFABdT , (30) I = f/RC FidT i i i (31) Since the value of the integral does not change when we apply a sym metry operation to it, i.e., RI = I, then the form of the integral, fCC.r dT, must also be invariant under all the symmetry operations of ill the group. This will be so only when one of the ri is the totally symmetric representation A1. This is the case because A, is the sole representation that is invariant under all the symmetry opera tions of the group. For the product FA iB to contain the totally symmetric represen tation, FA and r must be identical. That is AFB . Therefore, we can conclude that integrals of the form /f A B dT can be nonzero only when the representations rA and FB, for which A and B serve as basis functions, are identical. The matrix elements have the familiar form of Hij = f i H' i. dT = The Hamiltonian, of which H' is part, is an expression, in quantum mechanical operator form, of the energy of a system. Application of a symmetry operation of a molecule cannot change the energy of a molecule and all symmetry operations will leave the Hamiltonian in variant. The Hamiltonian operator thus belongs to the totally symmetric representation A1. The integral f i H' j dT can therefore be nonzero only if the direct product rirj contains Al and, as we have observed, this only occurs when ri = rj. Hence, the 1 matrix elements can be nonzero only if the wavefunctions 'i and 4j belong to the same irreducible representation. For the d3 cubic case the secular determinant is thus block diagonalized as shown in Fig. 4. Our aim then is to find the linear combinations of the freeion functions which serve as the basis functions of the various irre ducible representations. The IL,S;ML,MS> functions have the same angular dependence as the spherical harmonics and these [L,ML> functions that serve as the bases for the irreducible representations of the cubic group are the cubic oriented spherical harmonics. To illustrate the formation of the cubic oriented spherical harmonics we shall take the P term as an example. From Fig. 3 it is easy to see that the cartesian coordinates relate to the polar coordinates as follows: z = r cos6, (33a) y = r sine sinp and (33b) x = r sine cos~ (33c) Thus, we can write the spherical harmonics for a P term as a func tion of x, y and z. We obtain 1 Y1 1 = / (xiy)/r (34a) Y10 = v3/4 z/r and (34b) Y11 = v/8 (x+iy)/r (34c) 2C g 4x4 2TC 1g 5x5 2TC 2g 5x5 4A C 2g 1x1 The cubic secular determinant for d3 coupling scheme. ~ . 2AC 1g 1x1 2AC 2g 1x1 4T C 1g 2x2 Figure 4. {L,s,xC,X ,x } 4TC 2g 1x1 for the Our phases will agree with those of Condon and Shortley, i.e., the spherical harmonics with mR odd and positive are given a minus sign. The real forms of these spherical harmonics correspond to the following combinations: z Y1 (35a) y i(Y1 +Y )/I 2 and (35b) x ~ (Y1Y1 1)/12 (35c) These are known as the "porbitals" in the case of a single electron and have the same symmetry properties as z,y and x. Let us examine how these cartesian functions z,y and x will be affected by the rotation operations of Oh symmetry. We will take the z axis as the principal rotation axis, i.e., C4(z). The other rotations, each representing a class, are C2 = C42(z), a C2 colinear with the y axis which we shall call C2'(y), the threefold body diagonal rotation C3(z') and, of course, the identity operation E. Using counter clockwise rotations and "replaced by" operations, the transformation properties of z,y and x are given below. z z y C4(z) x (36a) x y E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, University Press, Cambridge, 1951. C 2( C2 = C42(z) C2' (y) C3(z') E iz y x z X (36b) (36c) (36d) (36e) The matrices which will accomplish the above transformations when applied to y are listed in Table 12 underneath the appro priate symmetry operations. These matrices constitute a representation of the basis functions z,y and x. The characters of each matrix are also given in Table 12. By referring back to Table 10(b) we see that z,y and x form a basis for the irreducible representation T1g of Oh symmetry. Since z,y and x form a basis for the irreducible representation T1g, we can use the real forms of the spherical harmonics which cor respond to z,y and x, i.e., (28), as the wave functions for the triply degenerate T1g representation that arises from P. These are the cubic oriented spherical haimonics for L = 1. TABLE 12 Matrices and Characters of the Representation Spanned by (z,y,x) in Oh Symmetry E C4(z) C2EC42(z) C21(y) C3(z') 1 0 0\ 1 0 0 1 0 0 1 0 0 0/ 1 0 (z,y,x) 0 1 0 0 1 0 1 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 X(R) 3 1 1 1 0 Immediately, the question arises as to which is the a, b and c component of T C(P). Though the choice is completely arbitrary it is important that the convention chosen is adhered to strictly in all the following calculations. First, let us see how the T1g representation decomposes when the symmetry is lowered to tetragonal. It is necessary to consider this now because it will play a role in our choosing of the a, b and c components of Tlg, Here, we need to consider only those rotations that are common to Oh and D4h. These are the C4(z), C2 C42(z), C2 (y) and E. It is obvious that the transformation properties of z,y and x will not differ in going from Oh to D4h for these operations. But the representation spanned by them may not be irreducible in D4h. The characters for the basis (z,y,x) and for the irreducible represen tations of D4h for the operations we are dealing with are compared in Table 13. TABLE 13 Character Table for the Representation for Which (z,y,x) is a Basis in D4h and for the Irreducible Representations of D4h X(E) X(C4) X(C2) X(C21) (z,y,x) 3 1 1 1 Alg 1 1 1 1 A2g 1 1 1 1 B1g 1 1 1 1 B2g 1 1 1 1 E 2 0 2 0 The cartesian axes, (z,y,x), do not form a basis for an ir reducible representation in D4h. This is obvious since D4h has no representation of dimension greater than two. Decomposing we find (z,y,x) is composed of A2g and E . Reviewing the transformation properties of z in D4h (cf. equns. 36ac), which are the same as those in Oh for the rotations common to the two point groups, we see that z always transforms as plus or minus of itself while y and x mix. This indicates that z alone is the basis for the singly degenerate A2g representation and (y,x) together form the basis for the doubly degenerate E representation. This is easily confirmed by comparing the characters representation. This is easily confirmed by comparing the characters of z and (y,x) with those of A2g and Eg under the rotations of D4h. Now we can state our convention for choosing the a, b and c components of T1g. First we choose z as the basis for the a com ponent of T1g. Thus, when the symmetry is lowered to D4h we have TC  Tga 2g Next we decide that under C3(z') of Oh a will be replaced by b, b by c, and c by a. From (29d) we see that C3(z') replaces z by y, y by x and x by z. Hence, y and x become the b and c components, respectively, of T1g. Further, we specify that the b component of T1g becomes the a component of E and the c component of T1g becomes the b component of E when the symmetry is lowered to tetragonal. That is TC E and Igb ga C Q T E Igc b " Finally, in Table 14 below we have summarized the wave functions, the real forms of the spherical harmonics, for T1g from P and the decomposition properties of T1g when the symmetry is lowered to tetragonal. TABLE 14 Wave Functions and Decomposition Properties of TC (P) Cartesian Oh D4h Wave Function Representation Representation Representation Y z TC A 1 Iga 2g (YV" 1+Y1 )//2 y TC E 1 1gb ga (Y 1"Y 1)/V2 x TC EQ 1 1gc gb The wave functions and decomposition of the other terms are found in exactly the same manner. A complete list of the cubic oriented spherical harmonics corresponding to the cubic irreducible representations of all the d3 terms is given in Table 15. We have used the shorthand notation IL,ML> for the spherical harmonics in Table 15. In Table 16 the decomposition of the cubic levels ingoing to D4h is given. To obtain the cubic oriented spherical harmonics in microstate form one inserts the IL,S;ML,MS> functions (cf. Table 6) for each corresponding IL,ML> spherical harmonic. We choose the maximum MS for each IL,S;ML,MS> function which is to be substituted for IL,ML>'s in writing down the cubic oriented spherical harmonics in microstate form. The cubic oriented spherical harmonics in microstate form are listed in Table 17. it C 0 * L a) S. a) C. 1p S. Q1 c, C: S A + A I AA I A O r *I S t Cn r4 o LO E I  L.0 .J  CL 0 U  0r a3) I < U 0 u C. ro r" 1 ar (31 N C> 0 S o E 5 L 0 a) Cci A C A A I NCM C\j A A (,J I  +  A I A + N A r A I r I N I C O N "    C*M * CY) (N (N S. L L 0 0 N (N T~i cM C\D SN X X .0) 0') 0' :j N C\j N N u 0') (NJ (NI A SA i A A a e CY CD n A n oo n +  C: A I + I A + 0 A A A *r~ C C) CV (v +N A Y) UJ  a CD S D C V A CA) Sj A +  + SI A N. A n a A S A n La CD a +( n o CV+ rA A n C"  rQ CJ + C SA L ) L  a N i C) C) N 10 C, C) CN X N nC N \ >  e  m N S. C) S S S  S S. S.. S.. M.. S. M.. M . > 0 0 C) w1) j CJl Cn C" C CC C i < N I t U S. S 0 l CM 1 1c A A A A 11 A A O O a . ^  I N 1 N N O N CD C CT) N Lii Cj M A A c. n ^ A IA + ,a + 3 ?   A  + A a I A A A + + A A  A a  Id Cd +~ C  I C a A N A A A1 + A A  A A I Cd I   N N  r r^ " h X N I I I v N N SN N A I A A A I A CM A t In Ln CM C I 1d I N CM S7 N i _r .0 .U S> N C CM C (U L  I I I LU UJ _ N CM r\ CM NJ CM CM CN CM aS CM 70 A A Ln I A A A + an " A L * m I LO S A + A I LO I n I A LAO C M o LO 'I . 0 LA .0 O 0 rI  *r + 0 I C A + C LO A A A C A A A SU LA O A  LU LO *c a LA a L Ln n L .0 LA L L a +  A L A > A I+ + + I C j 0 Ln A A I A A I A I I LO r LA A  a jI LA LA I LA LO LA LO + LA LO LO + S.  O A V) >. C A >LL I j A M .J LA LALA LO LO LO 0 C. j N N 5 & N N N I C V C C N N N1 A N N N N~ N N S N dxX X x 0 N N CI C) CV) C i I I I S*r  LA X X N l N x S Cco C L LA L m CM ON CM (M M cm U) N N N S1 I 1 1 + X ) N NC CN CN CN N 1 N^ C1 CM CM1 CM C\J CM 71 A A (? (A t A n 0 LO SI  A C 0) Sco co 4 ( S. I 'h + 4 a 4 ) U A Or C 10 A A A c =3 n  LL L..I. 0 * Ua ) +  LA S  > A +  0 t3 LOn A I A 4 Sn n LO LO n Ca i 4)  .r I  + 3 cO.) c) C ( S *r .( F 0  S0 a0 ) 4) I Ir0 S N ' o 4 SM 0 1^ C: 4 =n p S . 0 ci) 53 I I 1  ft  c i) (4 0.0 a s.. (A Y  ( a ) O) 0 f) (2 4' Q o o I * I .' TABLE 16 Decomposition of the Cubic Representations in Going to Tetragonal Symmetry Cubic Representation Tetragonal Representation Alg Alg A2g Blg Ega Alg Egb B1g Tlga A2g Tlgb Ega Tlgc Egb T2ga B2g T2gb Ega T2gc Egb TABLE 17 The Cubic Oriented Spherical Harmonics in Microstate Form for the d3 Terms = (4(202)3/6(112)+8(202)4(202)  (101)+3/6(211)+2(11)(101))/v'210 = i(4/3(212)+2/3(2I2)+2/3(212) + 4/2(201)+/2(201)5/2(201)3/3(111)3/3(100) + 2 /3(212)5/2(102)+4v2(102)+/2(102)+2/3(212)  4/3(212)+3/3(111+33/3(001))//420 2TC = 2E Igc gb = (4v'3(212)2/3(2i2)2V3(212)  4/2(201)/2(201)+5/2(201)+3/3(111)+3/3(100) + 2V3(212)5/2(102)+4/2(102)+/2(102)+2/3(212)  4/3(212)+3v3(111)+3/3(O01))//420 2 C ga 2 C gb 2 T2ga = 2Ag ig S2 Q S lg 2 Q S 2g Da = ((10I)+(202)(101)(202))/2 = ((211)+(200)(222)(211)+(T12) + (002)(222)(112))//8 = i((I12)+(002)(222)(112)  (211)(200)+(222)+(211))//8 2 C iga 2TC 1gb = A2g 2g 2Q ga TABLE 17Continued 2g = ((1)+(212)+(00)(212)(1 1) 2gb = Ega (100)(212)+(212))//8 2 TC 2 Q + (00)+(212)(212))/ 8 2Db 2 AQ = ig 2 BQ = 28g g = (3(202)2V6(112)3(201)3(101) + 3(10i)2/6(21i))//84 = (4(112)5(222)3(002)(112)3(i12)  2v6(01)+3(211)5(222)+(211)4(211)3(200)  2J6(110))/2/42 2TC 2 Q 2ga 2g 2T C 2gb 2 C Tgc 2gc 2Q ga 2Q = Egb gb = i(4(112)5(222)3(002)(112)  3(I12)2/6(01I)3(21i)+5(222)(2l1)+4(211) + 3(200)+2/6(110))/2/42 = (2V6(102)(212)2/6(102)3(212) + 3(001)(11)+4(2i2)3(212)(212)+4(212) + (111)3(100)+2/6(201)2v/6(21))/2/42 = i(2V6(102)(212)2/6(102)  3(212)+3(001)(111)+4(212)+3(212)+(212)  4(2i2)(1I1)+3(100)2/6(201)+2/6(201))/2/42 2C Ega ga 2 C gb TABLE 17Continued 2F 2AC = 2 Q 2g = Ig 2TC Iga 2TC lgb = 2A 2g  2 Q ga 2TC = 2EQ Igc gb = i(/6(01i)+(112)2(222)(112)  (2i1)2(222)+/6(1i0)+(21i))/2/6 = (2(101)4(101)+/6(1l2)/6(21I) + 4(202)2(202)+2(10i)2(202))//60 = i(2/6(111)+3(201)+3/6(2I2)2/6(212) + 3/6(100)3(201)V6(212)+2/6(11i)3/6(001)3(102) + 3(102)+3(2)6(212)2/6(212)+10(12)  5(012)5/6(12 512)+5/6(221)+10(210)  5(210)5(210))/8V15 = (2V6(111)+3(201)+3/6(212)2/6(212) + 3/6(100)3(201)/6(212)2/6(111)+3/6(001) + 3(102)3(1l2)3/6(22)6(21)+6(212)+6(12) + 10(012)5(0I2)5/6(1222)512)5/6(221)  10(210)+5(210)+5(210))/8/15 2TC = 28Q 2ga 2g 2TC 2E 2gb ga ((211)+2(222)I6(1I0)(21i)+/6(01i) (112)2(222)(I12))/2/6 = i(2/2(111)+/3(201)+3/2(2i2)2/2(212) + 3/2(100)/3(201)J2(212)+2/2(11I)3/2(001)  13(102)+/3(102)+ 3/2(2(2)+2(212)  V2(212)223(1 3(2)++3/2(122)+v/3(12)  3/2(221)2/3(210)+/3(210)+/3(210))/4/12 TABLE 17Continued 2Tg = 2EQ = (2v2(li1)+/3(201)+3v2(212) 2gc gb 2/2(212)+3/2(100)v3(20I)/2(212)2/2(111) + 3/2(001)+v'3(102) 3(102)3/2(212) + 2/2(212)+/2(212)+2/3(012)/3(012)3/2(122) 13(O12)32(21)23(210)+3(210)+/3(210))/412 2G 2A lg 2 AQ = g "ig 2EC 2 Q ga lg 2EC gb 2TC T Iga 2 C Igb  2A 2g 2Q ga = (v2(10i)+/3(21i)+/2(202)v2(101) + 33(112)V2(201)+V2(2I2)+/3(211)+V3(112) + /2(022))/2/6 = (10(101)+5/6(21I)+10(202)10(101) + 5/6(112)10(202)14(220)7/6(211)7v6(1I2)  14(022))/4/105 = (4/3(200)3/3(21I)72(1i0) + 4/3(211)323(21)23(222)2(01)+3(1I2)  4V3(112)2/3(222)4/3(002)+ 3/3(112))/2/70 = i(/3(1!2)+/2(022)/2(220)  v3(211))//10 = (2/6(111)+4(102)/6(2i2) + 2/6(212)+/6(001)7(102)+3(102)V6(212)  /6(100)+3(201)7(201)v6(212)2/6(111) + 4(201)+2/6(212)/6(212)+3(210)2(210) TABLE 17Continued  (210)+/6(221)(012)2(012),6(122) + 3(012))/8/5 2TC T gc 1gc 2= gb 2TC = 2Q 2ga 2g 2TC 2gb 2 Q ga 2TC = 2Q 2gc gb = i(2/6(111)+4(102)/6(2I2) + 2/6(212)+/6(001)7(102)+3(102)/6(212) + 6(100)3(201)+7(201)+/6(212)+2/6(li1)  4(201)2/6(212)+/6(2I2)+3(210)2(210)(2I0) + J6(221)+(0I2)+2(012)+/6122)3(012))/8/5 = i(/2(011)/3(112)+4/3(112) + 2/3(222)+4J3(002)3/3(112)4/3(200)  3/3(211)/2(1i0)+4/3(211)/3(211)2/3(222))/2v70 = (/6(100)3(201)+7(20i)+/6(212) + 2/6(111)4(201)2/6(2122)+6(22)2/6(11)  4(102)+/6(21))2/6(212)6(001)+7(02)  3(102)+/6(212)+21(210)14(210)7(210)+7/6(221)  7(012)14(012)7/6(122)+21(012))/8/35 = i(/6(100)3(201)+7(201)+/6(212) + 2/6(111)4(201)2/6(212)+/6(212)+2v6(11I)+4(102)  /6(2I2)+2/6(212)+/6(001)7(102)+3(102)  /6(212)21(210)+14(210)+7(210)7/6(221)  7(0i2)14(012)7/6(122)+21(012))/8/35 TABLE 17Continued 2EC ga 2C Egb = 2Q ig 2 C 1=2A 2Tga 2Ag 2TC i2.0 1gb J2ca 2TC 1=2EQ Igc gb =i (3(220)J2(211)+/3(022)  /2(112))//10 = i(/6(110)3(200)(211)+3(2i1)  2(211)+(222)+/6(011)3(002)+2(112)  3(112)+(112)(222))//60 = (4(101)2(101)+/6(12) /6(21)(202)+2(202)(202)2(10i))//42 = i(21/3(122)+21/3(221)  7/2(012)72(12)7v3(122)+7/3(221)+14v2(012)  7/2(210)+14/2(210)7/2(210)+6/3(001)8/3(111) + 9/2(102)6J2(102)+4/3(2i2)3/3(212)  3J2(102)J3(212)+8/3(11 6V1)63(10)+9/2(201)  6/2(201)/3(212)+4J3(212)3/2(201)  3/3(212))/16J21 = (21J3(122)21J3(221)14V2(0I2) + 7/2(012)+7/2(012)+7/3(122)+7/3(221)  7J2(210)+14/2(210)7/2(210)+6/3(001)8/3(111) + 9/2(102)6i/2(102)+4/3(212)3/3(212)3/2(102)  /3(212)8/3(1T1)+6/3(100)9/2(201)+6/2(201) + /3(212)4/3(212)+3/2(201)+3/3(212))/16/21 2 H TABLE 17Continued 2TC 2=2A iga 2g 2TC 2=2EQ Igb ga 2TC 2=2E Igc gb 2TC 2Bg 2ga 2g 2TC 2Q 2gb ga = (/3(220)/2(211)/3(022) + /2(1I2))//10 = i(5/3(122)+5(221)18/2(012)+9/2(012) + 9/2(012)+9/3(122)9/3(221)+9/2(210)18/2(210) + 9/2(210)+6/3(001)8/3(111)+9/2(102)6/2(i02) + 4/3(212)3/3(212)3/2(102)v3(212)+8/3(1i1)  6/3(100)+9V2(201)6/2(201)/3(212)+4/3(212)  3/2(201)3/3(212))/16/15 = (5/3(122)5/3(221)+18/2(0I2)  9/2(012)9/2(012)9/3(122)9/3(221)+9/2(210)  18/2(210)+9/2(210)+6/3(001)8/3(111)+9/2(102)  6/2(102)+4/3(212)3/3(212)3/2(102)/3(212)  8/3(111)+6/3(100)9/2(201)+6/2(201)+/3(212)  4/3(212)+3/2(201)+3/3(212))/16/15 = (/6(1i0)3(200)(21i)+3(211)2(211) + (222)/6(01I)+3(002)2(112)+3(112)(I12)  (222))//60 = i(30(122)30(221)+4/6(0I2)  2/6(012)2/6(012)6(122)+6(221)2/6(210) + 4/6(210)2/6(210)+12(001)16(11i)+6/6(102)  4/6(102)+8(212)6(212)2V6(102)2(212) + 16(111)12(100)+6/6(201)4/6(20I)2(212)+8(212)  2/6(201)6(212))/16/15 TABLE 17Continued = (30(122)30(221)+4/6(0i2)2/6(012)  2/6(O12)6(122)6(221)+2V6(210)4v6(210)+2/6(210)  12(001)+16(111)6V6(102)+4/6(102)8(212)+6(212) + 2/6(102)+2(212)+16(111)12(100)+6v6(201)4V6(20I)  2(212)+8(212)2/6(201)6(212))/16/15 = (2(101)(201))//5 = i(v2(102)/3(212)+/2(201)  /3(212))//10 = (v2(102)/3(212)/2(201)+/3(212))//10 4Ag 2g 4TCa 4AQ igb 2g Igb ga 4TC = 4EQb Igc gb = i((112)(211)//2 = ((101)+2(202))//5 = i(3(201)+v6(212)+3(102)+/6(212) + 5(012)+5(210))/4/5 = (3(201)+/v6(212)3(102)/6(212) + 5(012)5(210))/4/5 2TC 2gc =2E gb 4TCg 4 ga 4TC Igb 4TC Igc =4A 42g = 4Q ga S4E gb 4F 4B The cubic ligand field potential The cubic potential, Vc E VOh, and the added axial potential, Va, that constitute the quadrate field arise due to the coulombic interactions between the central ion's d electrons and the charged ligands. Thus, the general form of the ligand field potential is given by (3). In this coupling scheme, {L,S,XC,XQ,rQ}, we treat V and V c a as individual perturbations. Let us first consider the form of the ligand field potential for an octahedral arrangement of the ligands. As always, the ligands are assumed to be fixed in space at equal distances, a, along the three cartesian axis. The charge of the octahedral ligands is taken as qk. Utilizing the expansion of 1/rik we have = +R 4 r Vc = qe[ ( ) 4 c k Z =0 m = ( 1 r Yt ( i)Y (ek' k)] (37) If we consider the electron to be somewhere between the central ion and the ligand (cf. Fig. 5), then r< is equal to ri, the distance between the central ion and the electron, and r> is equal to a, the central ionligand separation. In Fig. 6 we have numbered the ligands and show their locations along the cartesian axis. Because the ligands are taken as fixed in space,the angles ek and hk are constant for each ligand. Therefore, we can calculate d electron r. Sa central ligand ion Position of a d Electron in a Complex '3 1 v6 I Figure 6. The Numbering of the Ligands for an Octahedral Complex Figure 5. X4 1 x m the spherical harmonics for the ligands, i.e., Y, (ek, k). The octahedral potential is thus given by o + m V0 = qke[ X DkY (6i',i)] (38) h k Z=0 m =z where * 4 r m (39) D = __ 1 1( a+ ) Y (9kk ) (39) "k (2Z+1) a+1 ; a with mz (2k+1) (Z m Zl). 1/2 m im O Y (k'k) I 2 (t +m P (cose),/1/2T e (40) The Legendre polynomials have the form m m 2 I dImZI 0 P (cose) Pz (z) = (1z2) 2 dm P. (z) and (41a) dz p0(z) 1 d 2_ P (z) d (z2 1)~ (41b) 2 ( 2l dz( For both cubic and tetragonal symmetries the potential must be the same at the points p() and p(,+T/2). We know that the ( dependent part of the spherical harmonics is given by im . ~1/Zr~e {42) Therefore, im 0 im ( +T/2) im imZ(r/2) e =e =e *e im (Tr/2) S= e and m ZT m 7r 1 = cos( 2 ) + i sin( 2 ) .(43) To satisfy (43) m, is restricted to values of 0, 4, 8, etc. The product of the spherical harmonics for two d electrons, i.e., the spherical harmonics with Z = 2, when written as a sum of spherical harmonics can contain only terms with Z = 0, 2 and 4. Thus, in the expansion of the ligand field potential, (38), we need only consider those spherical harmonics with a equal to 0, 2 and 4 and mI equal to 0 and 4. Since, as we have noted, the ligands are fixed in space we can calculate Dk using the constant values of the polar angles 6k and hk and the sine and cosine of the two polar angles for the six cubic ligands as given in Table 18. Hence, the final form of the potential will contain only those spherical harmonics of the electron for which Dk is not zero. Evaluating (38) we obtain for the octahedral po tential 0 + (0 5 4] (44 VOh q ke/[(12/a)YO + (7r/3a )Y4 + ( r/6a )(Y4+Y4 )] (44) h i TABLE 18 The 6 And p Values For The Ligands Arranged in an Octahedron Ligand a 4sine cosO sinp cost 1 w/2 0 1 0 0 1 2 T/2 r/2 1 0 1 0 3 T/2 T 1 0 0 1 4 i/2 3T/2 1 0 1 0 5 0 0 0 1 0 1 6 T 0 0 1 0 1 0 0 0 The Y term is a constant, i.e., Y = /4,. This Y term in 00 0 the potential, though of great importance in calculating thermodynamic quantities,l can be neglected in deriving the spectral transition energies since it makes a constant contribution to all the levels. Thus, the final form of our cubic potential is V qke(r /a)[(7/3)Y +(4 7/6)(YY4)] (45) h i The wave functions are linear combinations of the microstates. The microstates are product functions of the single electron orbitals that obey the Pauli principle and are Slater determinants written in shorthand notation as i1(1).2(2)iP3(3) for d3 electronic C. J. Ballhausen, Introduction to Ligand Field Theory, McGraw Hill Book Co., New York, 1962. configuration. The cubic potential is summed over electrons one, two and three. That is VOh = V0h(1) + VOh(2) + VOh(3) Voh + Vh Vh (46) Therefore, the general form of the integrals with which we need concern ourselves here is / / / $1(1)2(2)i3(3)iV0 (1) + V0 (2) + V0 (3)1 1 2 3 h h h p4(1)p5(2)'6(3)dT2dT2dT3 (47a) This becomes 11 '1(1)VOh(1) 94(1)dT1 2 02(2)05(2)dT2 " f3 13(3)Y6(3)dT3 + f2 i2(2)VOh (2) 15(2)dT2 fl 41(1)4(1)dT1 f3 13(3) 6(3)dT3 + /3 3(3) IVOh(3) 16(3)dT3 f1 0(l)4(l)dTI f2 02(2)5(2)dr2 . (47b) Due to the orthonormality (47) will be zero unless i3 = 06 and/or '1 4 = 4 of the single electron functions we see that 2 = i5 and '3 = i6 and/or p = '4 and and 2 = r5' Thus, the matrix elements reduce to single electron integral. Since the radial part, which we call R3d, and Z, which is equal to 2 for the delectrons, are constant for the microstates these integrals will be distinguishable only by the different mZ values. Therefore, the single d electron integrals we must calculate are <21Vhl 2>, <2 Vhl 1>, <21Vh 0>, <21VOh 1>, <2 VOh 12>, <1 VOh 1>, <1 V0 0>, <11 V0 11> and <0 V0 0>. h h h 2 2 1 +1 Only the spherical harmonics products Y2 Y2 Y Y2 0* 0 2* v2 0 4 Y2 Y 2 and Y2 Y2 contain Y4 or Y 4 (cf. Table 7), which are the spherical harmonics present in the cubic potential (45). Therefore, only the integrals containing these products can be non zero. These integrals are <21V0h 2>, <1V0h 1>, <0VO h0> and <21VOhl 2>. As an illustration of the method of 'solving' the four remaining integrals let us take as an example <21Vh 12>. Thus, <21VOh2> = / R3d 2 V R3dY2d (48a) From Table 7 we obtain, neglecting the constant Y 0 term, * Y Y2 = Y4 /14/ ~ Y2 /7/v . 2 2 4 2 Expanding (48a) we obtain 4 <* r. <21V0 2> = / R3d R3d(qke)( ) h o a 2x O* * r2dr f / [Y /14/T /5 Y2 /7/] * oo 4  [7A/ Y40/3 + V/70(Y44+Y 44)/6sineded) (48b) Because of the orthonormality of the spherical harmonics (48b) reduces to <2VOh 2> = (qke/6a5) R3d Rd(ri4)r2dr (48c) <21V Oh2> = (qke/6a5) As is the usual procedure in ligand field calculations, we will 4 make no effort to solve for the average value of r4. Rather, we define the single electron matrix element to be the wellknown cubic parameter Dq, that is, Dq <2 VOh 2> = (qke/6a5) Because the octahedral potential contains only fourth order harmonics, i.e., Z = 4, we need only one parameter, Dq, for the cubic case. The other nonzero integrals are calculated in the same fashion and they are given below. <21VOh 2> = Dq (50a) <1 V hl1> = 4Dq (50b) <01VOh 0> = 6Dq (50c) <21VOhl 2> = 5Dq (50d) Using the single electron integrals tabulated above we can easily find the values of the d3 matrix elements in terms of the cubic field parameter Dq. The energy matrices thus obtained from the block diagonalized secular determinant, (cf. Fig. 4) are listed in Table 19. The Axial Ligand Field In order to describe the tetragonal potential as the sum of a cubic and an axial field we must define the charge on the axial ligands, which we will call q', to be q' = qa e (51) where qa is the charge on the two axial ligands and qe is the charge on the four equatorial ligands of the tetragonal complex, (cf. Fig. 1). The form of the axial potential is analogous to that of the cubic potential and is given by m +Z mt Va = q'e[ I Y k Y (i,i)] (52) k l=0 m =S For d electrons, as we have noted, only those terms in the summation TABLE 19 The Elements of the Cubic Energy Matrices (the Hik's) for d Electronic Configuration in the {L,S,XC,XQ,rQl Coupling Scheme 2AC 1g j k Hjk 2G 2G 11B+3C2Dq 2AC 2EC g j k Hjk j k Hjk 2Da 2Da 7B+7C+3Dq 2Db 2G 24/5Dq/7 20a 2Db 3/21B15Dq//21 2Db 2H ilODq//70 2Da 2G 60Dq//105 2G 2G 11B+3C2Dq/7 2Da 2H i/30Dq G 2H i36Dq//14 2Db 2Db 3B+3C19Dq/7 2H 2H 6B+3C+2Dq 2TC 1g 6B+3C il8Dq//42 2p 2G 2P 1(2H) 22Dq//14 2/5Dq TABLE 19Continued S k Hjk j k Hjk 2P 2(2H) 6Dq//7 2G 1(2H) i/210Dq/7 2F 2F 9B+3C+Dq 2G 2(2H) i3/6Dq 2F 2G i/3q 1(2H) 1(2H) 6B+3C2Dq 2F 1(2H) 10Dq//70 1(2H) 2(2H) 2v35Dq/7 2F 2(2H) /2Dq 2(2H) 2(2H) 6B+3C+2Dq 2G 2G 11B+3CDq 2TC 2g 2Da 2Da 7B+7C2Dq 2Db 2H i40Dq/3/70 2Da 2Db 3/21B+10Dq//21 2F 2F 9B+3CDq/3 2Da 2F ilODq//3 2F 2G il5Dq//35 2Da 2G 30Dq//105 2F 2H VlODq/3 2Da 2H i400q//30 2G 2G 11B+3C+13Dq/7 2Db 2Db 3B+3C+38Dq/21 2G 2H i6Dq//14 2Db 2F i20Dq/3/7 2H 2H 6B+3C4Dq/3 2Db 2G 12/5Dq/7 4AC 2g j k Hjk 4F 4F 15B12Dq TABLE 19Continued 4TC 1g j k Hjk 4p 4p 0 4p 4F 4Dq 4F 4F 15B+6Dq 4T C S1B2 4F 4F 15B2Dq with Z = 0, 2 and 4 need be considered. Also, since for axial sym metry the principal rotation axis is C. the potential must be the same for all values of p. That is, p(p) = p(p+a) with a taking all real values. Thus, imz) im.,((+a) e = e (53) which can hold true for all values of a only if m = 0. Hence, the axial potential can be written as Va = I q'e[ I Dk'.Y (i,i) (54a) k Z<4 with z D I4 r ) Y (kRk) (54b) k (2k+1) a+1 Z k'Y k a The axial field potential for the two ligands turns out to be Va = q'e7[(4/a)Y00 + (42/5a3)Y20 + (4ri4/3a5)Y40] (55a) or, leaving out the constant Y,0 term, Va = q'ev/7[4ri2/a3 )20 + (4ri4/3a5)Y40] (55b) Y2*2 1* 1 Only the spherical harmonics products Y22 Y22, Y 21 Y2 0* 0 c o and Y20*Y20 contain Y20 or Y4, (cf. Table 7), which are the spherical harmonics present in the axial potential (55b). Therefore, only the integrals containing these products can be nonzero. These integrals are <21Val2>, As an illustration of the method of 'solving' these three remain ing integrals let us take as an example <21Va12>. Thus, 2* 2 <21Vaj2> = a R3d 2 VaIR3dY22 d. (56a) Expanding (56a) we obtain 2 <21Va 2> = I R3d*R3d(q'e)( 3)r2dr 0a 2r7r 07 4 0 f f (y/ Y2 /7vr)(4/r Y20//5)sinededp + 0 0 4 r 2x 2 * f R3d R(q'e)( )r2dr f (Y4 /14V/) (r,/ Y40/3)sinededp (56b) Because of the orthonormality of the spherical harmonics (56b) reduces to <21Va12> = (4q'e/7a3) As in the cubic case we will make no effort to solve for the 2 4 average values of ri and r. Rather, we define two new parameters, Dt and Ds, as follows 