1016 S. POPULATION ANALYSIS N = A r∈A B s∈B P rs S rs Afterwards we may choose the following partitionings: Atomic partitioning: N = A q A q A = r∈A B s∈B P rs S rs where q are called the Mulliken charges. They are often calculated in practical ap- Mulliken charges plications and serve to provide information on how much of the electronic den- sity ρ is concentrated on atom A. Such a quantity is of interest because it may be directly linked to the reactivity of atom A, often identified with its ability to be attacked by nucleophilic or electrophilic agents. 2 Also, if we measure the di- pole moment, we would like to know why this moment is large in a molecule. By performingMullikenanalysis,wemightbeabletoidentifythoseatomsthatare responsible for this. This might be of value when interpreting experimental data. Atomic and bond partitioning: The summation may also be performed in a slightly different way N = A rs∈A P rs S rs + A<B 2 r∈A s∈B P rs S rs = A ¯ q A + A<B ¯ q AB The first term represents the contributions ¯ q A of the atoms, the second pertains to the atomic pairs ¯ q AB . The latter populations are large and positive for those pairs of atoms for which chemists assign chemical bonds. The bond population ¯ q AB may be treated as a measure of whether in the A −B atomic interaction, bonding or antibonding character prevails. 3 If, for two atoms, ¯ q AB < 0, we may say that they are not bound by any chemical bond, if ¯ q AB is large, then we may treat it as an indication that these two atoms are bound by a chemical bond or bonds. 2 We have to remember that, besides electrons, this atom has a nucleus. This has to be taken into account when calculating the atomic net charge. 3 P rs is the sum (over the occupied orbitals) of the products of the LCAO coefficients of two atoms in each of the occupied molecular orbitals. The equal signs of these coefficients (with S rs > 0) means a bonding interaction (recall Chapter 8 and Appendix R on p. 1009) and such a contribution increases P rs . The opposite signs of the coefficients (with S rs > 0) corresponds to the antibonding interactions and in such a case the corresponding contribution decreases P rs .IfS rs < 0, then the words “bonding” and “antibonding” above have to be exchanged, but the effect remains the same. This means that the product P rs S rs in all cases correctly controls the bonding (P rs S rs > 0) or antibonding (P rs S rs < 0) effects. S. POPULATION ANALYSIS 1017 Example 1. The hydrogen molecule. Let us take the simplest example. First, let us consider the electronic ground-state in the simplest molecular orbital approxima- tion, i.e. two electrons are described by the normalized orbital in the form (ab denote the 1s atomic orbitals centred on the corresponding nuclei; note that this is the famous bonding orbital) ϕ 1 =N 1 (a +b) where N 1 =(2 +2S) − 1 2 ,andS ≡(a|b). Then, P sr = i 2c ∗ ri c si =2c ∗ r1 c s1 =(1 +S) −1 independent of the indices r and s.Ofcourse, S = 1 S S 1 and therefore PS= 11 11 Thus, Tr ( PS ) = 2 = the number of electrons =P 11 S 11 +P 22 S 22 +2P 12 S 12 =q A + q B +q AB ,withq A =q B =(1+S) −1 ,andq AB = 2S 1+S > 0. Thus we immediately see that the HH bond has an electronic population greater than zero, i.e. the atom–atom interaction is bonding. LetusnowconsiderH 2 with two electrons occupying the normalized orbital of a different character 4 ϕ 2 =N 2 (a −b),withN 2 =(2 −2S) − 1 2 ,then P sr = i 2c ∗ ri c si =2c ∗ r2 c s2 =(1 −S) −1 for (r s) =(11) and (r s) =(22) while P rs =−(1 −S) −1 for (r s) =(12) and (r s) =(21). Now, let us calculate PS= 1 −1 −11 and Tr(PS) = 2 = the number of electrons =P 11 S 11 + P 22 S 22 + 2P 12 S 12 = q A + q B +q AB ,butnowq A = q B = (1 −S) −1 and q AB =− 2S 1−S < 0. Thus, a glance at q AB tells us that this time the atoms are interacting in an antibonding way. A similar analysis for polyatomic molecules gives more subtle and more inter- esting results. Other population analyses Partitioning of the electron cloud of N electrons according to Mulliken population analysis represents only one of possible choices. For a positively definite matrix 5 S (and the overlap matrix is always positively definite) we may introduce the powers 4 We do not want to suggest anything, but this orbital is notorious for antibonding character. 5 I.e.alltheeigenvaluespositive. 1018 S. POPULATION ANALYSIS of the matrix 6 S x ,wherex is an arbitrary real number (in a way shown in Appen- dix J on p. 977), and we have S 1−x S x =S.Thenwemaywrite 7 N =Tr (PS) =Tr S x PS 1−x (S.4) Now, we may take any x and for this value construct the corresponding partition of N electronic charges into atoms. If x =0or1,thenwehaveaMullikenpopula- tion analysis, if x = 1 2 then we have what is called the Löwdin population analysis,Löwdin population analysis etc. Multipole representation Imagine a charge distribution ρ(r). Let us choose a Cartesian coordinate system. We may calculate the Cartesian moments of the distribution: ρ(r)dV ,i.e.the total charge, then xρ(r)dV , yρ(r) dV , zρ(r) dV ,i.e.thecomponentsofthe dipole moment, x 2 ρ(r)dV , y 2 ρ(r)dV , z 2 ρ(r)dV , xyρ(r)dV , xzρ(r) dV , yzρ(r) dV –the components of the quadrupole moment, etc. The moments mean a complete description of ρ(r) as concerns its interaction with another (distant) charge distribution. The higher the powers of x y z (i.e. the higher the moment) the more important distant parts of ρ(r) are. If ρ(r) extends to infinity (and for atoms and molecules it does), higher order moments tend to infinity. This means trouble when the consecutive interactions of the multipole moments are calcu- lated (multipole expansion, Appendix X) and indeed, the multipole expansion “explodes”, i.e. diverges. 8 This would not happen if the interacting charge distrib- utions could be enclosed in two spheres. There is also another problem: where to locate the origin of the coordinate sys- tem, with respect to which the moments are calculated? The answer is: anywhere. Wherever such an origin is located it is OK from the point of view of mathematics. However, such choices may differ enormously from the practical point of view. For example, let us imagine a spherically symmetric charge distribution. If the origin is located at its centre (as “most people would do”), then we have a quite simple de- scription of ρ(r) by using the moments, namely, the only non-zero moment is the charge, i.e. ρ(r)dV . If, however, the origin is located off centre, all the moments would be non-zero. They are all needed to calculate accurately the interaction of the charge distribution (with anything). As we can see, it is definitely better to lo- cate the origin at the centre of ρ(r). Well, and what if the charge distribution ρ(r) were divided into segments and each segment represented by a set of multipoles? It would be all right, although, in view of the above example, it would be better to locate the corresponding origins at the centre of the segments. It is clear that, in particular, it would be OK if the segments were very small, e.g., the cloud was cut into tiny cubes and we consider 6 They are symmetric matrices as well. 7 We easily check that Tr(ABC) =Tr (CAB). Indeed, Tr(ABC) = ikl A ik B kl C li ,whileTr(CAB) = ikl C ik A kl B li . Changing summation indices k → i, l → k, i → l in the last formula, we obtain Tr (ABC). 8 Although the first terms (i.e. before the “explosion”) may give accurate results. S. POPULATION ANALYSIS 1019 every cube’s content as a separate cloud. 9 But, well , what are the multipoles for? Indeed, it would be sufficient to take only the charges of the cubes, because they approximate the original charge distribution. In this situation higher multipoles wouldcertainlybeirrelevant!Thuswehavetwoextremecases: • a single origin and an infinite number of multipoles, • or an infinite number of centres and monopoles (charges) only. We see that when the origins are located on atoms, we have an intermediary situation and it might be sufficient to have a few multipoles per atom. 10 This is what the concept of what is called the cumulative multipole moments is all about cumulative multipole moments (CAMM 11 ). Besides the isotropic atomic charges q a =M (000) a calculated in an ar- bitrary population analysis, we have, in addition, higher multipoles M (klm) a (atomic dipoles, quadrupoles, octupoles, etc.) representing the anisotropy of the atomic charge distribution (i.e. they describe the deviations of the atomic charge distribu- tions from the spherical) M (klm) a = Z a x k a y l a z m a − r∈a s D sr r x k y l z m s − k k l l m m (k l m )=(klm) k k l l m m ×x k−k a y l−l a z m−m a ·M k l m a where M (klm) a is the multipole moment of the “klm” order with respect to the Cartesian coordinates x y z located on atom a, M (000) a stands for the atomic charge, e.g., from the Mulliken population analysis, Z a is the nuclear charge of the atom a, (r|x k y l z m |s) stands for the one-electron integral of the corresponding multipole moment, and D sr χ ∗ r χ s represents the electronic density contribution re- lated to AO’s: χ s and χ r and calculated by any method (LCAO MO SCF, CI, MP2, DFT, etc.). We may also use multipole moments expressed by spherical harmonic functions as proposed by Stone. 12 9 The clouds might eventually overlap. 10 If the clouds overlapped, the description of each centre by an infinite number of multipoles would lead to a redundancy (“overcompleteness”). I do not know of any trouble of that kind, but in my opinion trouble would come if the number of origins were large. This is in full analogy with the overcomplete- ness of the LCAO expansion. These two examples differ by a secondary feature: in the LCAO, instead ofmoments,wehavethes,p,d, orbitals,i.e.somemomentsmultipliedbyexponentialfunctions. 11 W.A. Sokalski and R. Poirier, Chem. Phys. Lett. 98 (1983) 86; W.A. Sokalski, A. Sawaryn, J. Chem. Phys. 87 (1987) 526. 12 A.J. Stone, Chem. Phys. Lett. 83 (1981) 233; A.J. Stone, M. Alderton, Mol. Phys. 56 (1985) 1047. T. THEDIPOLEMOMENTOFALONE ELECTRON PAIR The electronic lone pairs play an important role in intermolecular interactions. In particular, a lone pair protruding in space towards its partner has a large dipole moment, 1 which may interact electrostatically with itspartner’s multipole moments (see Appendix X, p. 1038). Let us see how the dipole moment depends on the atom to which it belongs and on the type of hybridization. Suppose the electronic lone pair is described by the normalized hybrid h = 1 1 +λ 2 (2s) +λ(2p x ) with the normalized 2s and 2p x atomic orbitals. The coefficient λ may change from −∞ to +∞ giving a different degree of hybridization. Fig. T.1 shows for comparison two series of the hybrids: for carbon and fluorine atoms. If λ = 0, we have the pure 2s orbital, if λ =±∞we obtain the pure ±2p x orbital. The dipole moment of a single electron described by h is calculated 2 as (N = 1 √ 1+λ 2 ): μ x =h|−x|h=−N 2 2s|x|2s+λ 2 2p x |x|2p x +2λ2s|x|2p x μ y = μ z =0 where x stands for the x coordinate of the electron. The first two integrals equal zero, because the integrand represents an odd func- tion 3 with respect to the reflection in the plane x =0. As a result μ x =−N 2 2λ2s|x|2p x We will limit ourselves to λ 0, which means we are considering hybrids pro- truding to the right-hand side 4 as in Fig. T.1, and since 2s|x|2p x > 0 then μ x 0. The negative sign stresses the fact that a negative electron is displaced to the right- hand side (positive x). 1 Calculated with respect to the nucleus; a large dipole moment means here a large length of the dipole moment vector. 2 Atomic units have been used throughout, and therefore μ is expressed in a.u. 3 Please recall that the orbital 2p x represents a spherically symmetric factor multiplied by x. 4 The hybrids with λ<0 differ only by protruding to the left-hand side. 1020 T. THE DIPOLE MOMENT OF A LONE ELECTRON PAIR 1021 Fig. T.1. The length of the dipole moment vector μ lone (in a.u.) as a function of the mixing parameter λ for carbon (upper curve) and fluorine (lower curve) atoms. The figure shows the shape of different hybrids h = 1 √ 1+λ 2 [(2s)+λ(2p x )]which correspond to various mixing of the 2s and 2p x carbon Slater orbitals (with exponential factor ζ =1625) and fluorine orbitals (ζ =260); from the left: λ =0, λ =1 (sp), λ =141 (sp 2 ), λ =173 (sp 3 ), λ =1000. All the hybrids are shown in square windows of 10 a.u. The fluorine orbitals are more compact due to the larger charge of the nucleus. A hybrid orbital which corresponds to λ<0 looks exactly like that with λ =−λ, except it is reflected with respect to the yz plane. The maximum dipole moment corresponds to the sp hybridization. To calculate 2s|x|2p x we need to specify the atomic orbitals 2s and 2p.Asthe 2s and 2p atomic orbitals, let us take Slater type orbitals: 2s =N r exp(−ζr) 2p x = N x exp(−ζr) where the exponential factor ζ (the same for both orbitals) is calculated using simple rules for building the Slater orbitals, see p. 355. Using the integral ∞ 0 x n exp(−αx) dx =n!α −(n+1) we obtain the normalization constants N =ζ 2 ζ 3π and N =ζ 2 ζ π . The contri- bution of two electrons (“lone electron pair”) to the dipole moment is, therefore, equal to 1022 T. THE DIPOLE MOMENT OF A LONE ELECTRON PAIR μ lone =2μ x =−N 2 |λ|(2s|xp x ) =−2N 2 N N (2λ) rx 2 exp(−2ζr)dv =−2N 2 N N 2λ r 3 x 2 exp(−2ζr)sinθ dr dθ dφ =−2N 2 N N 2λ ∞ 0 drr 5 exp(−2ζr) π 0 sin 3 θ dθ 2π 0 cos 2 φ dφ =−2N 2 N N 2λ 5! (2ζ) 6 4 3 π =− 4λ (1 +λ 2 ) ζ 2 ζ 3π ζ 2 ζ π 5! (2ζ) 6 4 3 π =− λ 1 +λ 2 10 ζ √ 3 THE DIPOLE MOMENT OF A LONE PAIR μ lone =− λ 1+λ 2 10 ζ √ 3 . The dipole moment at λ =0, i.e. for the pure 2s orbital, is equal to 0, for λ =∞, i.e. for the pure 2p x orbital it is also equal 0. It is interesting to see for which hybridization the length of dipole moment has a maximum. We easily find ∂|μ lone | ∂λ = 10 ζ √ 3 (1 +λ 2 ) −2λ 2 (1 +λ 2 ) =0 which gives λ =±1, independently of ζ. Thus the maximum dipole moment is at the 1 : 1 mixing of 2s and 2p,i.e.for digonal hybridization (for any element), Fig. T.1. From Table T.1 it is seen that the dipole moment of a lone pair strongly depends on the chemical element, 5 and to a lesser extent on hybridization. Table T.1. The length of the dipole moments μ lone (a.u.) corresponding to doubly occu- pied hybrid atomic orbitals. The orbital exponents of 2s and 2p STO’s are identical and calculated using the rules given by Slater: ζ C =1625, ζ N =195, ζ O =2275, ζ F =260 Atom Digonal λ =1Trigonalλ = √ 2Tetrahedralλ = √ 3 C1776 1675 1538 N1480 1396 1282 O1269 1196 1099 F1110 1047 0962 5 From the practical point of view, it is probably most important to compare nitrogen and oxygen lone pairs. Thus, coordination of a cation by amines should correspond to a stronger interaction than that by hydroxyl groups. U. SECOND QUANTIZATION When we work with a basis set composed of Slater determinants we are usually confronted with a large number of matrix elements involving one- and two-electron operators. The Slater–Condon rules (Appendix M) are doing the job to express these matrix elements by the one-electron and two-electron integrals. However, we may introduce an even easier tool called second quantization, which is equivalent to the Slater–Condon rules. The vacuum state In the second quantization formalism we introduce a reference state for the system under study, which is a Slater determinant (usually the Hartree–Fock wave func- tion) composed of N orthonormal spinorbitals, where N is the number of elec- trons. This function will be denoted in short by 0 or in a more detailed way by N (n 1 n 2 n ∞ ). The latter notation means that we have to do with a normal- ized N electron Slater determinant, and in parenthesis we give the occupancy list (n i = 0 1) for the infinite number of orthonormal spinorbitals considered in the basis set and listed one by one in the parenthesis. This simply means that some spinorbitals are present in the determinant (they have n i = 1), while others are absent 1 (n i = 0). Hence, i n i = N. The reference state is often called the vac- uum state. The subscript 0 in 0 means that we are going to consider a single- determinant approximation to the ground state. Besides the reference state, some normalized Slater determinants of the excited states will be considered, with other occupancies, including those corresponding to the number of electrons which differs from N. The creation and annihilation of electrons Let us make quite a strange move, and consider operators that change the number of electrons in the system. To this end, let us define the creation operator 2 ˆ k † of the electron going to occupy spinorbital k and the annihilation operator ˆ k of an electron leaving spinorbital k: 1 For example, the symbol 2 (001000100000) means a normalized Slater determinant of dimen- sion 2, containing the spinorbitals 3 and 7. The symbol 2 (001000) is a nonsense, because the num- ber of “ones” has to be equal to 2, etc. 2 The domain of the operators represents the space spanned by the Slater determinants built of spinor- bitals. Richard Feynman, in one of his books, says jokingly that he could not understand the very sense of the operators. If we annihilate or create an electron, then what about the system’s electroneutrality? Happily enough, these operators will always act in creator–annihilator pairs. 1023 1024 U. SECOND QUANTIZATION CREATION AND ANNIHILATION OPERATORS ˆ k † N (n k )=θ k (1 −n k ) N+1 (1 k ) ˆ k N (n k )=θ k n k N−1 (0 k ) where θ k =(−1) j<k n j . The symbol 1 k means that the spinorbital k is present in the Slater determi- nant, while 0 k means that this spinorbital is empty, i.e. is not present in the Slater determinant. The factors (1 −n k ) and n k ensure an important property of these operators, namely that any attempt at creating an electron on an already occupied spinorbital gives zero, similarly any attempt at annihilating an empty spinorbital also gives zero. It can be easily shown, 3 that (as the symbol suggests) ˆ k † is simply the adjoint operator with respect to ˆ k. The above operators have the following properties that make them equivalent to the Slater–Condon rules: ANTICOMMUTATION RULES ˆ k ˆ l + = 0 ˆ k † ˆ l † + = 0 ˆ k † ˆ l + = δ kl where the symbol [ ˆ A ˆ B] + = ˆ A ˆ B + ˆ B ˆ A is called the anticommutator. 4 It is simpleranticommutator than the Slater–Condon rules, isn’t it? Let us check the rule [ ˆ k † ˆ l] + = δ kl .We have to check how it works for all possible occupancies of the spinorbitals k and l, (n k n l ): (0 0), (0 1), (1 0) and (1 1). Case: (n k n l ) =(0 0) ˆ k † ˆ l + N (0 k 0 l )= ˆ k † ˆ l + ˆ l ˆ k † N (0 k 0 l ) = ˆ k † ˆ l N (0 k 0 l )+ ˆ l ˆ k † N (0 k 0 l ) 3 Proof. Let us take two Slater determinants a = N+1 (1 k )and b = N (0 k ),inboth of them the occupancies of all other spinorbitals are identical. Let us write the normalization condition for b in the following way: 1 = b |θ k ˆ k a =θ k b | ˆ k a =θ k ˆ k # b | a ,whereas ˆ k # has been denoted the operator adjoint to ˆ k, θ k appeared in order to compensate for (θ 2 k =1) the θ k produced by the annihilator. On the other hand, from the normalization condition of a we see that 1 = a | a = θ k ˆ k † b | a .Hence,θ k ˆ k # b | a =θ k ˆ k † b | a or ˆ k # = ˆ k † , This is what we wanted to show. 4 The above formulae are valid under the (common) assumption that the spinorbitals are orthonor- mal. If this assumption is not true, only the last anticommutator changes to the form [ ˆ k † ˆ l] + = S kl , where S kl stands for the overlap integral of spinorbitals k and l. U. SECOND QUANTIZATION 1025 = 0 + ˆ lθ k N+1 (1 k 0 l ) = θ k ˆ l N+1 (1 k 0 l ) = θ k δ kl θ k N (0 k ) = δ kl N (0 k ) So far so good. Case: (n k n l ) =(0 1) ˆ k † ˆ l + N (0 k 1 l )= ˆ k † ˆ l + ˆ l ˆ k † N (0 k 1 l ) = ˆ k † ˆ l N (0 k 1 l )+ ˆ l ˆ k † N (0 k 1 l ) = θ k θ l N (1 k 0 l )−θ k θ l N (1 k 0 l ) = δ kl N (0 k 1 l ) This is what we expected. 5 Case: (n k n l ) =(1 0) ˆ k † ˆ l + N (1 k 0 l )= ˆ k † ˆ l + ˆ l ˆ k † N (1 k 0 l ) = ˆ k † ˆ l N (1 k 0 l )+ ˆ l ˆ k † N (1 k 0 l ) = (0 +0) N (1 k 0 l ) = δ kl N (1 k 0 l ) This is OK. Case: (n k n l ) =(1 1) ˆ k † ˆ l + N (1 k 1 l )= ˆ k † ˆ l + ˆ l ˆ k † N (1 k 1 l ) = ˆ k † ˆ l N (1 k 1 l )+ ˆ l ˆ k † N (1 k 1 l ) = ˆ k † ˆ l N (1 k 1 l )+0 = θ 2 k δ kl N (1 k 1 l ) = δ kl N (1 k ) Theformulahasbeenproved. Operators in the second quantization Creation and annihilation operators may be used to represent one- and two- electron operators. 6 The resulting matrix elements with Slater determinants cor- respond exactly to the Slater–Condon rules (see Appendix M, p. 986). 5 What decided is the change of sign (due to θ k ) when the order of the operators has changed. 6 The original operator and its representation in the language of the second quantization are not identical in practical applications. The second ones can act only on the Slater determinants or their combinations. Since we are going to work with the creation and annihilation operators in only those methods which use Slater determinants (CI, MC SCF, etc.), the difference is irrelevant. . charge. 3 P rs is the sum (over the occupied orbitals) of the products of the LCAO coefficients of two atoms in each of the occupied molecular orbitals. The equal signs of these coefficients (with S rs > 0). clouds overlapped, the description of each centre by an infinite number of multipoles would lead to a redundancy (“overcompleteness”). I do not know of any trouble of that kind, but in my opinion trouble. Then, P sr = i 2c ∗ ri c si =2c ∗ r1 c s1 =(1 +S) −1 independent of the indices r and s.Ofcourse, S = 1 S S 1 and therefore PS= 11 11 Thus, Tr ( PS ) = 2 = the number of electrons =P 11 S 11 +P 22 S 22 +2P 12 S 12 =q A + q B +q AB ,withq A =q B =(1+S) −1 ,andq AB = 2S 1+S >