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1016 S. POPULATION ANALYSIS XXXX N = PrsSrs: A r∈A B s∈B Afterwards we may choose the following partitionings: Atomic partitioning: X N = qA; XµXX ¶ qA = PrsSrs ; A r∈A B s∈B Mulliken charges where q are called the Mulliken charges. They are often calculated in practical ap-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 performing Mulliken analysis, we might be able to identify those atoms that are 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 X X X XX X X N = PrsSrs + 2 PrsSrs = qA + qAB: A r;s∈A A0) means a bondinginteraction(recallChapter8andAppendixRonp.1009)andsuchacontributionincreases Prs. The opposite signs of the coefficients (with Srs > 0) corresponds to the antibonding interactions and in such a case the corresponding contribution decreases Prs. If Srs < 0, then the words “bonding” and “antibonding” above have to be exchanged, but the effect remains the same. This means that the product PrsSrs in all cases correctly controls the bonding (PrsSrs > 0) or antibonding (PrsSrs < 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 (a;b denote the 1s atomic orbitals centred on the corresponding nuclei; note that this is the famous bonding orbital) ϕ1 =N1(a+b); where N1 =(2+2S)−1 , and S ≡(a|b). Then, Psr =X2c∗ csi =2c∗ cs1 =(1+S)−1; i independent of the indices r and s. Of course, µ S = S S¶ 1 µ and therefore PS = 1 1¶ 1 Thus, Tr(PS)= 2 = the number of electrons =P11S11 +P22S22 +2P12S12 =qA + qB +qAB, with qA =qB =(1+S) , and qAB = >0. Thus we immediately see that the HH bond has an electronic population greater than zero, i.e. the atom–atom interaction is bonding. Let us now consider H2 with two electrons occupying the normalized orbital of a different character4 ϕ2 =N2(a−b), with N2 =(2−2S)−2 , then Psr =X2c∗ csi =2c∗ cs2 =(1−S)−1 i for (r;s) = (1;1) and (r;s) = (2;2) while Prs = −(1 −S)−1 for (r;s) = (1;2) and (r;s)=(2;1). Now, let us calculate µ ¶ PS = −1 1 and Tr(PS) = 2 = the number of electrons = P11S11 + P22S22 + 2P12S12 = qA + qB +qAB, but now qA = qB = (1 −S) and qAB = − < 0. Thus, a glance at qAB 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 matrix5 S (and the overlap matrix is always positively definite) we may introduce the powers 4We do not want to suggest anything, but this orbital is notorious for antibonding character. 5I.e. all the eigenvalues positive. 1018 Löwdin population analysis S. POPULATION ANALYSIS of the matrix6 Sx, where x is an arbitrary real number (in a way shown in Appen-dix J on p. 977), and we have S1−xSx =S. Then we may write7 N =Tr(PS)=Tr¡SxPS1−x¢: (S.4) Now, we may take any xand for this value construct the corresponding partition of N electronic charges into atoms. If x=0 or 1, then we have a Mulliken popula-tion analysis, if x = 1 then we have what is called the 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. the components of the dipole moment, x2ρ(r)dV , y2ρ(r)dV , z2ρ(r)dV , xyρ(r)dV , xzρ(r)dV , yzρ(r)dV –thecomponentsofthequadrupolemoment,etc.Themomentsmean 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 6They are symmetric matrices as well. 7We easily check that Tr(ABC)=Tr(CAB). Indeed, Tr(ABC)= i;k;l AikBklCli, while Tr(CAB)= i;k;l CikAklBli. Changing summation indices k → i, l → k, i → l in the last formula, we obtain Tr(ABC). 8Although the first terms (i.e. before the “explosion”) may give accurate results. S. POPULATION ANALYSIS everycube’scontentasaseparatecloud.9 But,well... ,whatarethemultipolesfor? 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 would certainly be irrelevant! Thus we have two extreme cases: • 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 (CAMM11). Besides the isotropic atomic charges qa =M(000) calculated in an ar-bitrary population analysis, we have, in addition, higher multipoles M(klm) (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) = Zaxkylzm −XXDsr¡r¯xkylzm¯s¢ X X r∈a µ k ¶µ l ¶µ m ¶ k06k l06l m06m; k0 l0 m0 (k0;l0;m0)=(k;l;m) ×xk−k0yl−l0zm−m0 ·Mk0l0m0; 1019 cumulative multipole moments where M(klm) is the multipole moment of the “klm” order with respect to the Cartesian coordinates x;y;z located on atom a, M(000) stands for the atomic charge, e.g., from the Mulliken population analysis, Za is the nuclear charge of the atom a, (r|xkylzm|s) stands for the one-electron integral of the corresponding multipole moment, and Dsrχ∗χs represents the electronic density contribution re-lated to AO’s: χs and χr 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 9The clouds might eventually overlap. 10If 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 of moments, we have the s,p,d,...orbitals, i.e. some moments multiplied by exponential functions. 11W.A. Sokalski and R. Poirier, Chem. Phys. Lett. 98 (1983) 86; W.A. Sokalski, A. Sawaryn, J. Chem. Phys. 87 (1987) 526. 12A.J. Stone, Chem. Phys. Lett. 83 (1981) 233; A.J. Stone, M. Alderton, Mol. Phys. 56 (1985) 1047. T. THE DIPOLE MOMENT OF A LONE 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 whichmayinteractelectrostaticallywithitspartner’smultipolemoments (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= p 1 £(2s)+λ(2p )¤; 1+λ2 with the normalized 2s and 2px 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 ±2px orbital. The dipole moment of a single electron described by h is calculated2 as (N = √ 1 ): 1+λ μx = hh|−x|hi=−N2£h2s|x|2si+λ2h2px|x|2pxi+2λh2s|x|2pxi¤; μy = μz =0; where x stands for the x coordinate of the electron. The first two integrals equal zero, because the integrand represents an odd func-tion3 with respect to the reflection in the plane x=0. As a result μx =−N22λh2s|x|2pxi: We will limit ourselves to λ > 0, which means we are considering hybrids pro-truding to the right-hand side4 as in Fig. T.1, and since h2s|x|2pxi>0;then μx 60. The negative sign stresses the fact that a negative electron is displaced to the right-hand side (positive x). 1Calculated with respect to the nucleus; a large dipole moment means here a large length of the dipole moment vector. 2Atomic units have been used throughout, and therefore μ is expressed in a.u. 3Please recall that the orbital 2px represents a spherically symmetric factor multiplied by x. 4The hybrids with λ<0 differ only by protruding to the left-hand side. 1020 ... - tailieumienphi.vn
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