理论化学导论（英文版）：Solutions

分类：化学格式：pdf 日期：2006年01月20日

1 Solutions 1. a. First determine the eigenvalues: det ?? ? ?? ?-1 - λ 2 2 2 - λ = 0 (-1 - λ)(2 - λ) - 22 = 0 -2 + λ - 2λ + λ2 - 4 = 0 λ2 - λ - 6 = 0 (λ - 3)(λ + 2) = 0 λ = 3 or λ = -2. Next, determine the eigenvectors. First, the eigenvector associated with eigenvalue -2: ?? ? ?? ?-1 2 2 2 ?? ? ?? ?C11 C21 = -2 ?? ? ?? ?C11 C21 -C11 + 2C21 = -2C11 C11 = -2C21 (Note: The second row offers no new information, e.g. 2C11 + 2C21 = -2C21) C112 + C212 = 1 (from normalization) (-2C21)2 + C212 = 1 4C212 + C212 = 1 5C212 = 1 C212 = 0.2 C21 = 0.2 , and therefore C11 = -2 0.2 . 2 For the eigenvector associated with eigenvalue 3: ?? ? ?? ?-1 2 2 2 ?? ? ?? ?C12 C22 = 3 ?? ? ?? ?C12 C22 -C12 + 2C22 = 3C12 -4C12 = -2C22 C12 = 0.5C22 (again the second row offers no new information) C122 + C222 = 1 (from normalization) (0.5C22)2 + C222 = 1 0.25C222 + C222 = 1 1.25C222 = 1 C222 = 0.8 C22 = 0.8 = 2 0.2 , and therefore C12 = 0.2 . Therefore the eigenvector matrix becomes: ?? ? ?? ?-2 0.2 0.2 0.2 2 0.2 b. First determine the eigenvalues: det ?? ?? ?? ??-2 - λ 0 0 0 -1 - λ 2 0 2 2 - λ = 0 det [ ]-2 - λ det ?? ? ?? ?-1 - λ 2 2 2 - λ = 0 From 1a, the solutions then become -2, -2, and 3. Next, determine the eigenvectors. First the eigenvector associated with eigenvalue 3 (the third root): 3 ?? ?? ?? ??-2 0 0 0 -1 2 0 2 2 ?? ?? ?? ??C11C21 C31 = 3 ?? ?? ?? ??C11C21 C31 -2 C13 = 3C13 (row one) C13 = 0 -C23 + 2C33 = 3C23 (row two) 2C33 = 4C23 C33 = 2C23 (again the third row offers no new information) C132 + C232 + C332 = 1 (from normalization) 0 + C232 + (2C23)2 = 1 5C232 = 1 C23 = 0.2 , and therefore C33 = 2 0.2 . Next, find the pair of eigenvectors associated with the degenerate eigenvalue of -2. First, root one eigenvector one: -2C11 = -2C11 (no new information from row one) -C21 + 2C31 = -2C21 (row two) C21 = -2C31 (again the third row offers no new information) C112 + C212 + C312 = 1 (from normalization) C112 + (-2C31)2 + C312 = 1 C112 + 5C312 = 1 C11 = 1 - 5C312 (Note: There are now two equations with three unknowns.) Second, root two eigenvector two: 4 -2C12 = -2C12 (no new information from row one) -C22 + 2C32 = -2C22 (row two) C22 = -2C32 (again the third row offers no new information) C122 + C222 + C322 = 1 (from normalization) C122 + (-2C32)2 + C322 = 1 C122 + 5C322 = 1 C12 = (1- 5C322)1/2 (Note: again, two equations in three unknowns) C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) Now there are five equations with six unknowns. Arbitrarily choose C11 = 0 (whenever there are degenerate eigenvalues, there are not unique eigenvectors because the degenerate eigenvectors span a 2- or more- dimensional space, not two unique directions. One always is then forced to choose one of the coefficients and then determine all the rest; different choices lead to different final eigenvectors but to identical spaces spanned by these eigenvectors). C11 = 0 = 1 - 5C312 5C312 = 1 C31 = 0.2 C21 = -2 0.2 C11C12 + C21C22 + C31C32 = 0 (from orthogonalization) 0 + -2 0.2(-2C32) + 0.2 C32 = 0 5C32 = 0 5 C32 = 0, C22 = 0, and C12 = 1 Therefore the eigenvector matrix becomes: ?? ?? ?? ??0 1 0-2 0.2 0 0.2 0.2 0 2 0.2 2. a. K.E. = mv 2 2 = ?? ? ?? ?m m mv2 2 = (mv)2 2m = p2 2m K.E. = 12m(px2 + py2 + pz2) K.E. = 12m?? ? ?? ? ?? ? ?? ?h? i ? ?x 2 + ?? ? ?? ?h? i ? ?y 2 + ?? ? ?? ?h? i ? ?z 2 K.E. = -h ?2 2m?? ? ?? ??2 ?x2 + ?2 ?y2 + ?2 ?z2 b. p = mv = ipx + jpy + kpz p = ?? ? ?? ? i?? ? ?? ?h? i ? ?x + j?? ? ?? ?h? i ? ?y + k?? ? ?? ?h? i ? ?z where i, j, and k are unit vectors along the x, y, and z axes. c. Ly = zpx - xpz Ly = z ?? ? ?? ?h? i ? ?x - x ?? ? ?? ?h? i ? ?z 6 3. First derive the general formulas for ??x , ??y , ??z in terms of r,θ, and φ, and ??r , ??θ , and ??φ in terms of x,y, and z. The general relationships are as follows: x = r Sinθ Cosφ r2 = x2 + y2 + z2 y = r Sinθ Sinφ sinθ = x 2 + y2 x2 + y2 + z2 z = r Cosθ cosθ = z x2 + y2 + z2 tanφ = yx First ??x , ??y , and ??z from the chain rule: ? ?x = ?? ? ?? ??r ?x y,z ? ?r + ?? ? ?? ??θ ?x y,z ? ?θ + ?? ? ?? ??φ ?x y,z ? ?φ , ? ?y = ?? ? ?? ??r ?y x,z ? ?r + ?? ? ?? ??θ ?y x,z ? ?θ + ?? ? ?? ??φ ?y x,z ? ?φ , ? ?z = ?? ? ?? ??r ?z x,y ? ?r + ?? ? ?? ??θ ?z x,y ? ?θ + ?? ? ?? ??φ ?z x,y ? ?φ . Evaluation of the many "coefficients" gives the following: ?? ? ?? ??r ?x y,z = Sinθ Cosφ , ?? ? ?? ??θ ?x y,z = Cosθ Cosφ r , ?? ? ?? ??φ ?x y,z = - Sinφ r Sinθ , ?? ? ?? ??r ?y x,z = Sinθ Sinφ , ?? ? ?? ??θ ?y x,z = Cosθ Sinφ r , ?? ? ?? ??φ ?y x,z = Cosφ r Sinθ , ?? ? ?? ??r ?z x,y = Cosθ , ?? ? ?? ??θ ?z x,y = - Sinθ r , and ?? ? ?? ??φ ?z x,y = 0 . 7 Upon substitution of these "coefficients": ? ?x = Sinθ Cosφ ? ?r + Cosθ Cosφ r ? ?θ - Sinφ r Sinθ ? ?φ , ? ?y = Sinθ Sinφ ? ?r + Cosθ Sinφ r ? ?θ + Cosφ r Sinθ ? ?φ , and ? ?z = Cosθ ? ?r - Sinθ r ? ?θ + 0 ? ?φ . Next ??r , ??θ , and ??φ from the chain rule: ? ?r = ?? ? ?? ??x ?r θ,φ ? ?x + ?? ? ?? ??y ?r θ,φ ? ?y + ?? ? ?? ??z ?r θ,φ ? ?z , ? ?θ = ?? ? ?? ??x ?θ r,φ ? ?x + ?? ? ?? ??y ?θ r,φ ? ?y + ?? ? ?? ??z ?θ r,φ ? ?z , and ? ?φ = ?? ? ?? ??x ?φ r,θ ? ?x + ?? ? ?? ??y ?φ r,θ ? ?y + ?? ? ?? ??z ?φ r,θ ? ?z . Again evaluation of the the many "coefficients" results in: ?? ? ?? ??x ?r θ,φ = x x2 + y2 + z2 , ?? ? ?? ??y ?r θ,φ = y x2 + y2 + z2 , ?? ? ?? ??z ?r θ,φ = z x2 + y2 + z2 , ?? ? ?? ??x ?θ r,φ = x z x2 + y2 , ?? ? ?? ??y ?θ r,φ = y z x2 + y2 , ?? ? ?? ??z ?θ r,φ = - x2 + y2 , ?? ? ?? ??x ?φ r,θ = -y , ?? ? ?? ??y ?φ r,θ = x , and ?? ? ?? ??z ?φ r,θ = 0 Upon substitution of these "coefficients": ? ?r = x x2 + y2 + z2 ??x + y x2 + y2 + z2 ??y + z x2 + y2 + z2 ??z 8 ? ?θ = x z x2 + y2 ??x + y z x2 + y2 ??y - x2 + y2 ??z ? ?φ = -y ? ?x + x ? ?y + 0 ? ?z . Note, these many "coefficients" are the elements which make up the Jacobian matrix used whenever one wishes to transform a function from one coordinate representation to another. One very familiar result should be in transforming the volume element dxdydz to r2Sinθdrdθdφ. For example: ??f(x,y,z)dxdydz = ? � � ? f(x(r,θ,φ),y(r,θ,φ),z(r,θ,φ)) ? ? ? ? ? ? ? ???? ????x?r θφ ? ?? ????x?θ rφ ? ?? ????x?φ rθ ?? ? ?? ??y ?r θφ ?? ? ?? ??y ?θ rφ ?? ? ?? ??y ?φ rθ ?? ? ?? ??z ?r θφ ?? ? ?? ??z ?θ rφ ?? ? ?? ??z ?φ rθ drdθdφ a. Lx = h ? i ??? ?? ?? ?? ? y ??z - z ??y Lx = h ? i ?? ? ?? ? rSinθSinφ ?? ? ?? ? Cosθ ??r - Sinθr ??θ -h ? i ?? ? ?? ? rCosθ ?? ? ?? ? SinθSinφ ??r + CosθSinφr ??θ + CosφrSinθ ??φ Lx = - h ? i ?? ? ?? ? Sinφ ??θ + CotθCosφ ??φ b. Lz = h ? i ? ?φ = - ih? ? ?φ Lz = h ? i ?? ? ?? ? -y ? ?x + x ? ?y 9 4. B dB/dx d2B/dx2 i. 4x4 - 12x2 + 3 16x3 - 24x 48x2 - 24 ii. 5x4 20x3 60x2 iii. e3x + e-3x 3(e3x - e-3x) 9(e3x + e-3x) iv. x2 - 4x + 2 2x - 4 2 v. 4x3 - 3x 12x2 - 3 24x B(v.) is an eigenfunction of A(i.): (1-x2) d 2 dx2 - x d dx B(v.) = (1-x2) (24x) - x (12x2 - 3) 24x - 24x3 - 12x3 + 3x -36x3 + 27x -9(4x3 -3x) (eigenvalue is -9) B(iii.) is an eigenfunction of A(ii.): d2 dx2 B(iii.) = 9(e3x + e-3x) (eigenvalue is 9) B(ii.) is an eigenfunction of A(iii.): x ddx B(ii.) = x (20x3) 10 20x4 4(5x4) (eigenvalue is 4) B(i.) is an eigenfunction of A(vi.): d2 dx2 - 2x d dx B(i) = (48x2 - 24) - 2x (16x3 - 24x) 48x2 - 24 - 32x4 + 48x2 -32x4 + 96x2 - 24 -8(4x4 - 12x2 + 3) (eigenvalue is -8) B(iv.) is an eigenfunction of A(v.): x d 2 dx2 + (1-x) d dx B(iv.) = x (2) + (1-x) (2x - 4) 2x + 2x - 4 - 2x2 + 4x -2x2 + 8x - 4 -2(x2 - 4x +2) (eigenvalue is -2) 5. 11 z x y z x y 12 x y x z y z 6. 13 0 10 20 30 40 -0.05 0.00 0.05 0.10 0.15 0.20 0.25 Hydrogen 4s Radial Function r (bohr) R4s(r) 0 10 20 30 40 -0.02 0.00 0.02 0.04 0.06 Hydrogen 4p Radial Function r (bohr) R4p(r) 14 0 10 20 30 40 -0.02 -0.01 0.00 0.01 0.02 0.03 Hydrogen 4d Radial Function r (bohr) R4d(r) 0 10 20 30 40 0.00 0.01 0.02 Hydrogen 4f Radial Function r (bohr) R4f(r) 15 7. 0.0 0.1 0.2 0.3 0.4 0 20 40 60 80 100 120 Si 1s r (bohr) Radial Function R(r) Z=14 16 0.0 0.2 0.4 0.6 0.8 1.0 -10 0 10 20 30 40 Si 2s r (bohr) Radial Function R(r) Z=14 Z=12 0.0 0.5 1.0 1.5 0 2 4 6 8 Si 2p r (bohr) Radial Function R(r) Z=12 Z=14 17 0 1 2 3 -3 0 3 6 9 12 15 Si 3s r (bohr) Radial Function R(r) Z=14 Z=4 0 1 2 3 4 5 -2 -1 0 1 2 3 4 5 Si 3p r (bohr) Radial Function R(r) Z=14 Z=4 8. 18 i. In ammonia, the only "core" orbital is the N 1s and this becomes an a1 orbital in C3v symmetry. The N 2s orbitals and 3 H 1s orbitals become 2 a1 and an e set of orbitals. The remaining N 2p orbitals also become 1 a1 and a set of e orbitals. The total valence orbitals in C3v symmetry are 3a1 and 2e orbitals. ii. In water, the only core orbital is the O 1s and this becomes an a1 orbital in C2v symmetry. Placing the molecule in the yz plane allows us to further analyze the remaining valence orbitals as: O 2pz = a1, O 2py as b2, and O 2px as b1. The (H 1s + H 1s) combination is an a1 whereas the (H 1s - H 1s) combination is a b2. iii. Placing the oxygens of H2O2 in the yz plane (z bisecting the oxygens) and the (cis) hydrogens distorted slightly in +x and -x directions allows us to analyze the orbitals as follows. The core O 1s + O 1s combination is an a orbital whereas the O 1s - O 1s combination is a b orbital. The valence orbitals are: O 2s + O 2s = a, O 2s - O 2s = b, O 2px + O 2px = b, O 2px - O 2px = a, O 2py + O 2py = a, O 2py - O 2py = b, O 2pz + O 2pz = b, O 2pz - O 2pz = a, H 1s + H 1s = a, and finally the H 1s - H 1s = b. iv. For the next two problems we will use the convention of choosing the z axis as principal axis for the D∞h, D2h, and C2v point groups and the xy plane as the horizontal reflection plane in Cs symmetry. D∞h D2h C2v Cs N 1s σg ag a1 a' N 2s σg ag a1 a' N 2px pixu b3u b1 a' 19 N 2py piyu b2u b2 a' N 2pz σu b1u a1 a'' 9. a. Ψn(x) = ?? ? ?? ?2 L 1 2 Sinnpix L Pn(x)dx = | |Ψn 2(x) dx The probability that the particle lies in the interval 0 ≤ x ≤ L4 is given by: Pn = ?? 0 L 4 Pn(x)dx = ?? ? ?? ?2 L ?� ? 0 L 4 Sin2?? ? ?? ?npix L dx This integral can be integrated to give : Pn = ??? ???Lnpi ?? ? ?? ?2 L ?� ? 0 npi 4 Sin2?? ? ?? ?npix L d?? ? ?? ?npix L Pn = ??? ???Lnpi ?? ? ?? ?2 L ??0 npi 4 Sin2θdθ Pn = 2npi?? ?? ?? ?? - 14Sin2θ + θ2 ?? ??npi4 0 = 2npi?? ? ?? ? - 14Sin2npi4 + npi(2)(4) 20 = 14 - 12pin Sin?? ? ?? ?npi 2 b. If n is even, Sin?? ? ?? ?npi 2 = 0 and Pn = 1 4 . If n is odd and n = 1,5,9,13, ... Sin?? ? ?? ?npi 2 = 1 and Pn = 14 - 12pin If n is odd and n = 3,7,11,15, ... Sin?? ? ?? ?npi 2 = -1 and Pn = 14 + 12pin The higher Pn is when n = 3. Then Pn = 14 + 12pi3 Pn = 14 + 16pi = 0.303 c. Ψ(t) = e -iHt h? [ ]aΨn + bΨm = aΨne -iEnt h? + bΨme -iEmt h? HΨ = aΨnEne -iEnt h? + bΨmEme iEmt h? < >Ψ|H|Ψ = |a|2En + |b|2Em + a*be i(En-Em)t h? < >Ψn|H|Ψm + b*ae -i(Em-En)t h? < >Ψm|H|Ψn Since < >Ψn|H|Ψm and < >Ψm|H|Ψn are zero, < >Ψ|H|Ψ = |a|2En + |b|2Em (note the time independence) 21 d. The fraction of systems observed in Ψn is |a|2. The possible energies measured are En and Em. The probabilities of measuring each of these energies is |a|2 and |b|2. e. Once the system is observed in Ψn, it stays in Ψn. f. P(En) = ?? ??< >Ψn|Ψ 2 = |cn|2 cn = ?� ? 0 L 2 LSin?? ? ?? ?npix L 30 L5 x(L-x)dx = 60L6?� ? 0 L x(L-x)Sin?? ? ?? ?npix L dx = 60L6?? ?? ?? ?? L?� ? 0 L xSin?? ? ?? ?npix L dx - ?� ? 0 L x2Sin?? ? ?? ?npix L dx These integrals can be evaluated to give: cn = 60L6?? ? ?? ? L?? ? ?? ?L2 n2pi2Sin?? ? ?? ?npix L - Lx npiCos?? ? ?? ?npix L ?? ?L 0 - 60L6??? ? ?? ?? ?? ? ?? ?2xL2 n2pi2Sin?? ? ?? ?npix L - ?? ? ?? ?n2pi2x2 L2 - 2 L3 n3pi3Cos?? ? ?? ?npix L ?? ?L 0 cn = 60L6 { L3n2pi2( )Sin(npi) - Sin(0) - L 2 npi( )LCos(npi) - 0Cos0 ) - (2L2n2pi2( )LSin(npi) - 0Sin(0) 22 - ( )n2pi2 - 2 L 3 n3pi3 Cos(npi) + ?? ? ?? ?n2pi2(0) L2 - 2 L3 n3pi3 Cos(0))} cn = L-3 60 {- L 3 npi Cos(npi) + ( )n2pi2 - 2 L3 n3pi3 Cos(npi) + 2L 3 n3pi3 } cn = 60??? ???- 1npi(-1)n + ( )n2pi2 - 2 1n3pi3(-1)n + 2n3pi3 cn = 60??? ?????? ???-1npi + 1npi - 2n3pi3 (-1)n + 2n3pi3 cn = 2 60n3pi3 )( )-(-1)n + 1 |cn|2 = 4(60)n6pi6 )( )-(-1)n + 1 2 If n is even then cn = 0 If n is odd then cn = (4)(60)(4)n6pi6 = 960n6pi6 The probability of making a measurement of the energy and obtaining one of the eigenvalues, given by: En = n 2pi2h?2 2mL2 is: P(En) = 0 if n is even P(En) = 960n6pi6 if n is odd 23 g. < >Ψ|H|Ψ = ?� �? 0 L ?? ? ?? ?30 L5 1 2x(L-x) ?? ? ?? ?-h?2 2m d2 dx2 ?? ? ?? ?30 L5 1 2x(L-x)dx = ??? ???30L5 ?? ? ?? ?-h?2 2m ?� ? 0 L x(L-x)?? ? ?? ? d 2 dx2 ( )xL-x2 dx = ?? ? ?? ?-15h?2 mL5 ??0 L x(L-x)(-2)dx = ?? ? ?? ?30h?2 mL5 ? ? 0 L xL-x2dx = ?? ? ?? ?30h?2 mL5 ?? ? ?? ? Lx 2 2 - x3 3 ?? ?L 0 = ?? ? ?? ?30h?2 mL5 ?? ? ?? ?L3 2 - L3 3 = ?? ? ?? ?30h?2 mL2 ?? ? ?? ?1 2- 1 3 = 30h ?2 6mL2 = 5h?2 mL2 10. < >Ψ|H|Ψ = ∑ ij Ci*e iEit h? < >Ψi|H|Ψj e -iEjt h? Cj 24 Since < >Ψi|H|Ψj = Ejδij < >Ψ|H|Ψ = ∑ j Cj*CjEje i(Ej-Ej)t h? < >Ψ|H|Ψ = ∑ j Cj*CjEj (not time dependent) For other properties: < >Ψ|A|Ψ = ∑ ij Ci*e iEit h? < >Ψi|A|Ψj e -iEjt h? Cj but, < >Ψi|A|Ψj does not necessarily = ajδij because the Ψj are not eigenfunctions of A unless [A,H] = 0. < >Ψ|A|Ψ = ∑ ij Ci*Cje i(Ei-Ej)t h? < >Ψi|A|Ψj Therefore, in general, other properties are time dependent. 11. a. The lowest energy level for a particle in a 3-dimensional box is when n1 = 1, n2 = 1, and n3 = 1. The total energy (with L1 = L2 = L3) will be: Etotal = h 2 8mL2( )n12 + n22 + n32 = 3h2 8mL2 25 Note that n = 0 is not possible. The next lowest energy level is when one of the three quantum numbers equals 2 and the other two equal 1: n1 = 1, n2 = 1, n3 = 2 n1 = 1, n2 = 2, n3 = 1 n1 = 2, n2 = 1, n3 = 1. Each of these three states have the same energy: Etotal = h 2 8mL2( )n12 + n22 + n32 = 6h2 8mL2 Note that these three states are only degenerate if L1 = L2 = L3. b. ?↑ ? ?? ?? distortion????→ ?? ?? ?↑ ? ?↑?↓ ?↑?↓ L1 = L2 = L3 L3 ≠ L1 = L2 For L1 = L2 = L3, V = L1L2L3 = L13, Etotal(L1) = 2ε1 + ε2 = 2h 2 8m?? ? ?? ?12 L12 + 12 L22 + 12 L32 + 1h2 8m?? ? ?? ?12 L12 + 12 L22 + 22 L32 = 2h 2 8m?? ? ?? ?3 L12 + 1h2 8m?? ? ?? ?6 L12 = h2 8m?? ? ?? ?12 L12 For L3 ≠ L1 = L2, V = L1L2L3 = L12L3, L3 = V/L12 Etotal(L1) = 2ε1 + ε2 26 = 2h 2 8m?? ? ?? ?12 L12 + 12 L22 + 12 L32 + 1h2 8m?? ? ?? ?12 L12 + 12 L22 + 22 L32 = 2h 2 8m?? ? ?? ?2 L12 + 1 L32 + 1h2 8m?? ? ?? ?2 L12 + 4 L32 = 2h 2 8m?? ? ?? ?2 L12 + 1 L32 + 1 L12 + 2 L32 = 2h 2 8m?? ? ?? ?3 L12 + 3 L32 = h2 8m?? ? ?? ?6 L12 + 6 L32 In comparing the total energy at constant volume of the undistorted box (L1 = L2 = L3) versus the distorted box (L3 ≠ L1 = L2) it can be seen that: h2 8m?? ? ?? ?6 L12 + 6 L32 ≤ h2 8m?? ? ?? ?12 L12 as long as L3 ≥ L1. c. In order to minimize the total energy expression, take the derivative of the energy with respect to L1 and set it equal to zero. ?Etotal?L1 = 0 ? ?L1 ?? ? ?? ?h2 8m?? ? ?? ?6 L12 + 6 L32 = 0 But since V = L1L2L3 = L12L3, then L3 = V/L12. This substitution gives: ? ?L1 ?? ? ?? ?h2 8m?? ? ?? ?6 L12 + 6L14 V2 = 0 ?? ? ?? ?h2 8m?? ? ?? ?(-2)6 L13 + (4)6L13 V2 = 0 ?? ? ?? ? - 12L13 + 24L1 3 V2 = 0 27 ?? ? ?? ?24L13 V2 = ?? ? ?? ?12 L13 24L16 = 12V2 L16 = 12 V2 = 12( )L12L3 2 = 12 L14L32 L12 = 12 L32 L3 = 2 L1 d. Calculate energy upon distortion: cube: V = L13, L1 = L2 = L3 = (V) 1 3 distorted: V = L12L3 = L12 2 L1 = 2 L13 L3 = 2??? ???V2 1 3 ≠ L 1 = L2 = ?? ? ?? ?V 2 1 3 ?E = Etotal(L1 = L2 = L3) - Etotal(L3 ≠ L1 = L2) = h 2 8m?? ? ?? ?12 L12 - h2 8m?? ? ?? ?6 L12 + 6 L32 = h 2 8m?? ? ?? ?12 V2/3 - 6(2)1/3 V2/3 + 6(2)1/3 2V2/3 = h 2 8m?? ? ?? ?12 - 9(2)1/3 V2/3 Since V = 8?3, V2/3 = 4?2 = 4 x 10-16 cm2 , and h 2 8m = 6.01 x 10-27 erg cm2: ?E = 6.01 x 10-27 erg cm2?? ? ?? ?12 - 9(2)1/3 4 x 10-16 cm2 28 ?E = 6.01 x 10-27 erg cm2??? ???0.664 x 10-16 cm2 ?E = 0.99 x 10-11 erg ?E = 0.99 x 10-11 erg ??? ???1 eV1.6 x 10-12 erg ?E = 6.19 eV 12. a. H = -h ?2 2m ?? ? ?? ??2 ?x2 + ?2 ?y2 (Cartesian coordinates) Finding ??x and ??y from the chain rule gives: ? ?x = ?? ? ?? ??r ?x y ? ?r + ?? ? ?? ??φ ?x y ? ?φ , ? ?y = ?? ? ?? ??r ?y x ? ?r + ?? ? ?? ??φ ?y x ? ?φ , Evaluation of the "coefficients" gives the following: ?? ? ?? ??r ?x y = Cosφ , ?? ? ?? ??φ ?x y = - Sinφ r , ?? ? ?? ??r ?y x = Sinφ , and ?? ? ?? ??φ ?y x = Cosφ r , Upon substitution of these "coefficients": ? ?x = Cosφ ? ?r - Sinφ r ? ?φ = - Sinφ r ? ?φ ; at fixed r. ? ?y = Sinφ ? ?r + Cosφ r ? ?φ = Cosφ r ? ?φ ; at fixed r. 29 ?2 ?x2 = ?? ? ?? ? - Sinφr ??φ ?? ? ?? ? - Sinφr ??φ = Sin 2φ r2 ?2 ?φ2 + SinφCosφ r2 ? ?φ ; at fixed r. ?2 ?y2 = ?? ? ?? ?Cosφ r ? ?φ ?? ? ?? ?Cosφ r ? ?φ = Cos 2φ r2 ?2 ?φ2 - CosφSinφ r2 ? ?φ ; at fixed r. ?2 ?x2 + ?2 ?y2 = Sin2φ r2 ?2 ?φ2 + SinφCosφ r2 ? ?φ + Cos2φ r2 ?2 ?φ2 - CosφSinφ r2 ? ?φ = 1r2 ? 2 ?φ2 ; at fixed r. So, H = -h ?2 2mr2 ?2 ?φ2 (cylindrical coordinates, fixed r) = -h ?2 2I ?2 ?φ2 The Schr?dinger equation for a particle on a ring then becomes: HΨ = EΨ -h?2 2I ?2Φ ?φ2 = EΦ ?2Φ ?φ2 = ?? ? ?? ?-2IE h?2 Φ The general solution to this equation is the now familiar expression: Φ(φ) = C1e-imφ + C2eimφ , where m = ??? ???2IEh?2 1 2 30 Application of the cyclic boundary condition, Φ(φ) = Φ(φ+2pi), results in the quantization of the energy expression: E = m 2h?2 2I where m = 0, ±1, ±2, ±3, ... It can be seen that the ±m values correspond to angular momentum of the same magnitude but opposite directions. Normalization of the wavefunction (over the region 0 to 2pi) corresponding to + or - m will result in a value of ??? ???12pi 1 2 for the normalization constant. ∴ Φ(φ) = ??? ???12pi 1 2 eimφ ?? ?? (±4)2h?2 2I ?? ?? (±3)2h?2 2I ?? ?? (±2)2h?2 2I ?↑?↓ ?↑?↓ (±1) 2h?2 2I ?↑?↓ (0) 2h?2 2I b. h ?2 2m = 6.06 x 10-28 erg cm2 h?2 2mr2 = 6.06 x 10-28 erg cm2 (1.4 x 10-8 cm)2 = 3.09 x 10-12 erg ?E = (22 - 12) 3.09 x 10-12 erg = 9.27 x 10-12 erg 31 but ?E = hν = hc/λ So λ = hc/?E λ = (6.63 x 10 -27 erg sec)(3.00 x 1010 cm sec-1) 9.27 x 10-12 erg = 2.14 x 10-5 cm = 2.14 x 103 ? Sources of error in this calculation include: i. The attractive force of the carbon nuclei is not included in the Hamiltonian. ii. The repulsive force of the other pi-electrons is not included in the Hamiltonian. iii. Benzene is not a ring. iv. Electrons move in three dimensions not one. 13. Ψ(φ,0) = 43pi Cos2φ. This wavefunction needs to be expanded in terms of the eigenfunctions of the angular momentum operator, ?? ? ?? ? -ih? ??φ . This is most easily accomplished by an exponential expansion of the Cos function. Ψ(φ,0) = 43pi?? ? ?? ?eiφ + e-iφ 2 ?? ? ?? ?eiφ + e-iφ 2 = ?? ? ?? ?1 4 4 3pi( )e2iφ + e-2iφ + 2e(0)iφ 32 The wavefunction is now written in terms of the eigenfunctions of the angular momentum operator, ?? ? ?? ? -ih? ??φ , but they need to include their normalization constant, 12pi . Ψ(φ,0) = ?? ? ?? ?1 4 4 3pi 2pi ?? ? ?? ?1 2pi e 2iφ + 1 2pi e -2iφ + 2 1 2pi e (0)iφ = ?? ? ?? ?1 6 ?? ? ?? ?1 2pi e 2iφ + 1 2pi e -2iφ + 2 1 2pi e (0)iφ Once the wavefunction is written in this form (in terms of the normalized eigenfunctions of the angular momentum operator having mh? as eigenvalues) the probabilities for observing angular momentums of 0h? , 2h? , and -2h? can be easily identified as the squares of the coefficients of the corresponding eigenfunctions. P2h? = ?? ? ?? ?1 6 2 = 1 6 P-2h? = ?? ? ?? ?1 6 2 = 1 6 P0h? = ?? ? ?? ? 2 16 2 = 46 14. a. 12 mv2 = 100 eV ?? ? ?? ?1.602 x 10-12 erg 1 eV v2 = ?? ? ?? ?(2)1.602 x 10-10 erg 9.109 x 10-28g 33 v = 0.593 x 109 cm/sec The length of the N2 molecule is 2? = 2 x 10-8 cm. v = dt t = dv = 2 x 10 -8 cm 0.593 x 109 cm/sec = 3.37 x 10-17 sec b. The normalized ground state harmonic oscillator can be written as: Ψ0 = ?? ? ?? ?α pi 1/4e-αx2/2, where α = ?? ? ?? ?kμ h?2 1 2 and x = r - r e Calculating constants; αN2 = ?? ? ?? ?(2.294 x 106 g sec-2)(1.1624 x 10-23 g) (1.0546 x 10-27 erg sec)2 1 2 = 0.48966 x 1019 cm-2 = 489.66 ?-2 For N2: Ψ0(r) = 3.53333?- 1 2 e-(244.83?-2)(r-1.09769?)2 αN2+ = ?? ? ?? ?(2.009 x 106 g sec-2)(1.1624 x 10-23 g) (1.0546 x 10-27 erg sec)2 1 2 = 0.45823 x 1019 cm-2 = 458.23 ?-2 For N2+: Ψ0(r) = 3.47522?- 1 2 e-(229.113?-2)(r-1.11642?)2 c. P(v=0) = ?? ??< >Ψv=0(N2+)?Ψv=0(N2) 2 Let P(v=0) = I2 where I = integral: I= ? �? -∞ +∞ (3.47522?- 1 2e-(229.113?-2)(r-1.11642?)2) . 34 (3.53333?- 1 2 e-(244.830?-2)(r-1.09769?)2)dr Let C1 = 3.47522?- 1 2 , C 2 = 3.53333? -12 , A1 = 229.113?-2, A2 = 244.830?-2, r1 = 1.11642?, r2 = 1.09769?, I = C1C2 ? ? -∞ +∞ e-A1(r-r1)2e-A2(r-r2)2 dr . Focusing on the exponential: -A1(r-r1)2-A2(r-r2)2 = -A1(r2 - 2r1r + r12) - A2(r2 - 2r2r + r22) = -(A1 + A2)r2 + (2A1r1 + 2A2r2)r - A1r12 - A2r22 Let A = A1 + A2, B = 2A1r1 + 2A2r2, C = C1C2, and D = A1r12 + A2r22 . I = C ? ? -∞ +∞ e-Ar2 + Br - D dr = C ? ? -∞ +∞ e-A(r-r0)2 + D' dr where -A(r-r0)2 + D' = -Ar2 + Br - D -A(r2 - 2rr0 + r02) + D' = -Ar2 + Br - D such that, 2Ar0 = B 35 -Ar02 + D' = -D and, r0 = B2A D' = Ar02 - D = A B 2 4A2 - D = B2 4A - D . I = C ? ? -∞ +∞ e-A(r-r0)2 + D' dr = CeD' ? ? -∞ +∞ e-Ay2 dy = CeD' piA Now back substituting all of these constants: I = C1C2 piA1 + A2 exp?? ? ?? ?(2A1r1 + 2A2r2)2 4(A1 + A2) - A1r12 - A2r22 I = (3.47522)(3.53333) pi(229.113) + (244.830) . exp??? ???(2(229.113)(1.11642) + 2(244.830)(1.09769))24((229.113) + (244.830)) . exp( ) - (229.113)(1.11642)2 - (244.830)(1.09769)2 I = 0.959 P(v=0) = I2 = 0.92, so there is a 92% probability. 15. 36 a. Eν = ?? ? ?? ?h?2k μ 1 2??? ???ν + 1 2 ?E = Eν+1 - Eν = ?? ? ?? ?h?2k μ 1 2 ?? ?? ? ?? ?? ? ν +1 + 12 - ν - 12 = ?? ? ?? ?h?2k μ = ?? ? ?? ?(1.0546 x 10-27 erg sec)2(1.87 x 106 g sec-2) 6.857 g / 6.02 x 1023 1 2 = 4.27 x 10-13 erg ?E = hcλ λ = hc?E = (6.626 x 10 -27 erg sec)(3.00 x 1010 cm sec-1) 4.27 x 10-13 erg = 4.66 x 10-4 cm 1 λ = 2150 cm-1 b. Ψ0 = ?? ? ?? ?α pi 1/4e-αx2/2 < >x = < >Ψv=0?x?Ψv=0 = ?? -∞ +∞ Ψ0*xΨ0dx = ?� ? -∞ +∞ ?? ? ?? ?α pi 1/2xe-αx2dx = ?� ? -∞ +∞ ?? ? ?? ?α -α2pi 1/2e-αx2d(-αx2) 37 = ?? ? ?? ?-1 αpi 1/2e-αx2 ?+∞ ?∞ = 0 < >x2 = < >Ψv=0?x2?Ψv=0 = ?? -∞ +∞ Ψ0*x2Ψ0dx = ?� ? -∞ +∞ ?? ? ?? ?α pi 1/2x2e-αx2dx = 2?? ? ?? ?α pi 1/2 ?? 0 +∞ x2e-αx2dx = 2?? ? ?? ?α pi 1/2 ?? ? ?? ?1 21+1α ?? ? ?? ?pi α 1/2 = ??? ???12α ?x = (<x2> - <x>2)1/2.= ??? ???12α = ?? ? ?? ?h? 2 kμ 1 2 = ?? ? ?? ?(1.0546 x 10-27 erg sec)2 4(1.87 x 106 g sec-2)(6.857 g / 6.02 x 1023) 1 4 = 3.38 x 10-10 cm = 0.0338? c. ?x = ?? ? ?? ?h? 2 kμ 1 2 The smaller k and μ become, the larger the uncertainty in the internuclear distance becomes. Helium has a small μ and small attractive force between atoms. This results in 38 a very large ?x. This implies that it is extremely difficult for He atoms to "vibrate" with small displacement as a solid, even as absolute zero is approached. 16. a. W = ?? -∞ ∞ φ*Hφdx W = ?? ? ?? ?2b pi 1 2 ?�? -∞ ∞ e-bx 2 ?? ? ?? ? - h ?2 2m d2 dx2 + a|x| e -bx2dx d2 dx2 e -bx2 = d dx? ?? ???-2bx e-bx2 = ( )-2bx ?? ? ?? ? -2bx e-bx 2 + ?? ? ?? ? e-bx 2 ( )-2b = ?? ? ?? ? 4b2x2 e-bx 2 + ?? ? ?? ? -2b e-bx 2 Making this substitution results in the following three integrals: W = ?? ? ?? ?2b pi 1 2 ??? ???- h?2 2m ? �? -∞ ∞ e-bx 2 4b2x2 e-bx 2 dx + ?? ? ?? ?2b pi 1 2 ??? ???- h?2 2m ? �? -∞ ∞ e-bx 2 -2b e-bx 2 dx + 39 ?? ? ?? ?2b pi 1 2 ?� ? -∞ ∞ e-bx 2 a|x|e-bx 2 dx = ?? ? ?? ?2b pi 1 2 ??? ???-2b2h?2 m ? �? -∞ ∞ x2e-2bx 2 dx + ?? ? ?? ?2b pi 1 2 ??? ???bh?2 m ? �? -∞ ∞ e-2bx 2 dx + ?? ? ?? ?2b pi 1 2 a ?� ? -∞ ∞ |x|e-2bx 2 dx = ?? ? ?? ?2b pi 1 2 ??? ???-2b2h?2 m 2 ?? ? ?? ?1 222b pi 2b + ?? ? ?? ?2b pi 1 2 ??? ???bh?2 m 2 ?? ? ?? ?1 2 pi 2b + ?? ? ?? ?2b pi 1 2 a ??? ???0! 2b = ?? ? ?? ? -bh ?2 m ?? ? ?? ?1 2 + ?? ? ?? ?bh?2 m + ?? ? ?? ?2b pi 1 2??? ???a 2b W = ?? ? ?? ?bh?2 2m + a ?? ? ?? ?1 2bpi 1 2 b. Optimize b by evaluating dWdb = 0 dW db = d db?? ?? ?? ?? ?? ? ?? ?bh?2 2m + a ?? ? ?? ?1 2bpi 1 2 = ?? ? ?? ?h?2 2m - a 2 ?? ? ?? ?1 2pi 1 2 b- 3 2 So, a2 ??? ???12pi 1 2 b- 3 2 = ??? ???h?2 2m or, b -32 = ?? ? ?? ?h?2 2m 2 a ?? ? ?? ?1 2pi -12 = ?? ? ?? ?h?2 ma 2pi , 40 and, b = ??? ???ma2pi h?2 2 3 . Substituting this value of b into the expression for W gives: W = ?? ? ?? ?h?2 2m ?? ? ?? ?ma 2pi h?2 2 3 + a ??? ???1 2pi 1 2 ?? ? ?? ?ma 2pi h?2 -13 = ?? ? ?? ?h?2 2m ?? ? ?? ?ma 2pi h?2 2 3 + a ??? ???1 2pi 1 2 ?? ? ?? ?ma 2pi h?2 -13 = 2- 4 3 pi- 1 3h? 2 3 a 2 3 m- 1 3 + 2- 1 3 pi- 1 3h? 2 3 a 2 3 m- 1 3 = ?? ? ?? ? 2- 4 3pi- 1 3 + 2- 1 3pi- 1 3 h? 2 3 a 2 3 m- 1 3 = 3 2 ( )2pi -13h?23 a23 m-13 = 0.812889106h? 2 3 a 2 3 m-1/3 which is in error by only 0.5284% !!!!! 17. a. H = - h ?2 2m d2 dx2 + 1 2 kx2 φ = 1516 a- 5 2 (a2 - x2) for -a < x < a φ = 0 for |x| ≥ a ?? -∞ +∞ φ*Hφdx 41 = ?� �? -a +a 15 16 a -52 (a2 - x2) ?? ? ?? ? - h ?2 2m d2 dx2 + 1 2kx2 15 16 a -52 (a2 - x2) dx = ?? ??1516 a-5 ?� ? -a +a (a2 - x2)?? ? ?? ? - h ?2 2m d2 dx2 + 1 2kx2 (a2 - x2) dx = ?? ??1516 a-5 ?� ? -a +a (a2 - x2)?? ? ?? ? - h ?2 2m d2 dx2(a2 - x2) dx + ?? ??1516 a-5 ?� ? -a +a (a2 - x2)12kx2(a2 - x2) dx = ?? ??1516 a-5 ?� ? -a +a (a2 - x2)?? ? ?? ? - h ?2 2m (-2) dx + ?? ??1532 a-5 ?? -a +a (kx2)(a4 -2a2x2 + x4) dx = ?? ? ?? ?15h?2 16m a-5 ??-a +a (a2 - x2) dx + ?? ??1532 a-5 ?? -a +a a4kx2 -2a2kx4 + kx6 dx = ?? ? ?? ?15h?2 16m a-5?? ? ?? ? a2x?? ? a -a - 1 3 x3?? ? a -a + ?? ??1532 a-5?? ? ?? ?a4k 3 x3?? ? a -a - 2a2k 5 x5?? ? a -a + k 7 x7?? ? a -a = ?? ? ?? ?15h?2 16m a-5?? ? ?? ? 2a3 - 23 a3 + ?? ??1532 a-5?? ? ?? ?2a7k 3 - 4a7k 5 + 2k 7 a7 42 = ?? ??1516 a-5?? ? ?? ?4h?2 3m a3 + a7k 3 - 2a7k 5 + k 7 a7 = ?? ??1516 a-5?? ? ?? ?4h?2 3m a3 + ?? ? ?? ?k 3 - 2k 5 + k 7 a7 = ?? ??1516 a-5?? ? ?? ?4h?2 3m a3 + ?? ? ?? ?35k 105 - 42k 105 + 15k 105 a7 = ?? ??1516 a-5?? ? ?? ?4h?2 3m a3 + ?? ? ?? ?8k 105 a7 = 5h?2 4ma2 + ka2 14 b. Substituting a = b?? ? ?? ?h?2 km 1 4 into the above expression for E we obtain: E = 5h ?2 4b2m?? ? ?? ?km h?2 1 2 + kb2 14 ?? ? ?? ?h?2 km 1 2 = h? k 1 2 m- 1 2 ??? ???5 4 b-2 + 1 14 b2 c. E = 5h ?2 4ma2 + ka2 14 dE da = - 10h?2 4ma3 + 2ka 14 = - 5h?2 2ma3 + ka 7 = 0 5h?2 2ma3 = ka 7 and 35h? 2 = 2mka4 So, a4 = 35h ?2 2mk , or a = ?? ? ?? ?35h?2 2mk 1 4 Therefore φbest = 1516 ?? ? ?? ?35h?2 2mk -58 ?? ? ?? ? ?? ? ?? ?35h?2 2mk 1 2 - x2 , 43 and Ebest = 5h ?2 4m?? ? ?? ?2mk 35h?2 1 2 + k 14?? ? ?? ?35h?2 2mk 1 2 = h? k 1 2 m- 1 2 ??? ???5 14 1 2 . d. Ebest - EtrueEtrue = h? k 1 2 m- 1 2 ?? ? ?? ? ?? ??514 1 2 - 0.5 h? k 1 2 m- 1 2 0.5 = ?? ??514 1 2 - 0.5 0.5 = 0.0976 0.5 = 0.1952 = 19.52% 18. a. H0 ψ(0)lm = L 2 2mer02 ψ (0) lm = L2 2mer02 Yl,m(θ,φ) = 12mer02 h? 2 l(l+1) Yl,m(θ,φ) E(0)lm = h ?2 2mer02 l(l+1) b. V = -eεz = -eεr0Cosθ E(1)00 = < >Y00|V|Y00 = < >Y00|-eεr0Cosθ|Y00 = -eεr0< >Y00|Cosθ|Y00 Using the given identity this becomes: E(1)00 = -eεr0< >Y00|Y10 (0+0+1)(0-0+1)(2(0)+1)(2(0)+3) + -eεr0< >Y00|Y-10 (0+0)(0-0)(2(0)+1)(2(0)-1) 44 The spherical harmonics are orthonormal, thus < >Y00|Y10 = < >Y00|Y-10 = 0, and E(1)00 = 0. E(2)00 = ∑ lm≠00 ?? ??< >Ylm|V|Y00 2 E(0)00 - E(0)lm < >Ylm|V|Y00 = -eεr0< >Ylm|Cosθ|Y00 Using the given identity this becomes: < >Ylm|V|Y00 = -eεr0< >Ylm|Y10 (0+0+1)(0-0+1)(2(0)+1)(2(0)+3) + -eεr0< >Ylm|Y-10 (0+0)(0-0)(2(0)+1)(2(0)-1) < >Ylm|V|Y00 = -eεr03 < >Ylm|Y10 This indicates that the only term contributing to the sum in the expression for E(2)00 is when l=1, and m=), otherwise < >Ylm|V|Y00 vanishes (from orthonormality). In quantum chemistry when using orthonormal functions it is typical to write the term < >Ylm|Y10 as a delta function, for example δlm,10 , which only has values of 1 or 0; δij = 1 when i = j and 0 when i ≠ j. This delta function when inserted into the sum then eliminates the sum by "picking out" the non-zero component. For example, < >Ylm|V|Y00 = -eεr03 δlm,10 , so E(2)00 = ∑ lm≠00 e 2ε2r02 3 δlm'102 E(0)00 - E(0)lm = e 2ε2r02 3 1 E(0)00 - E(0)10 45 E(0)00 = h ?2 2mer02 0(0+1) = 0 and E (0) 10 = h?2 2mer02 1(1+1) = h?2 mer02 Inserting these energy expressions above yields: E(2)00 = -e 2ε2r02 3 mer02 h?2 = - mee2ε2r04 3h?2 c. E 00 = E(0)00 + E(1)00 + E(2)00 + ... = 0 + 0 - mee 2ε2r04 3h?2 = -mee 2ε2r04 3h?2 α = -? 2E ?2ε = ?2 ?2ε ?? ? ?? ?mee2ε2r04 3h?2 = 2mee 2r04 3h?2 d. α = 2(9.1095x10 -28g)(4.80324x10-10g 1 2cm 3 2s-1)2r 04 3(1.05459x10-27 g cm2 s-1)2 α = r04 12598x106cm-1 = r04 1.2598?-1 αH = 0.0987 ?3 αCs = 57.57 ?3 19. 46 1pig1pig 3σu 3σg 1piu1piu 2σu 2σg 1σu 1σg 2pz 2py 2px2px 2py 2pz 2s 2s 1s1s N2 NN The above diagram indicates how the SALC-AOs are formed from the 1s,2s, and 2p N atomic orbitals. It can be seen that there are 3σg, 3σu, 1piux, 1piuy, 1pigx, and 1pigy SALC- AOs. The Hamiltonian matrices (Fock matrices) are given. Each of these can be diagonalized to give the following MO energies: 3σg; -15.52, -1.45, and -0.54 (hartrees) 3σu; -15.52, -0.72, and 1.13 1piux; -0.58 1piuy; -0.58 1pigx; 0.28 1pigy; 0.28 47 It can be seen that the 3σg orbitals are bonding, the 3σu orbitals are antibonding, the 1piux and 1piuy orbitals are bonding, and the 1pigx and 1pigy orbitals are antibonding. 20. Using these approximate energies we can draw the following MO diagram: H C H z y x 2b2 4a1 1b1 3a1 1b2 2a1 1a1 H2C 1b2 1a1 3a11b21b1 2a1 1a1 2py 1σu 1σg 2pz2px 2s 1s This MO diagram is not an orbital correlation diagram but can be used to help generate 48 one. The energy levels on each side (C and H2) can be "superimposed" to generate the reactant side of the orbital correlation diagram and the center CH2 levels can be used to form the product side. Ignoring the core levels this generates the following orbital correlation diagram. Orbital-correlation diagram for the reaction C + H2 -----> CH2 (bent) a1(bonding) b2(antibonding) a1(antibonding) b1(2ppi) a1(non-bonding) b2(bonding) CH2 (bent)C + H2 σg(a1) 2s(a1) σu(b2) 2px(b1) 2py(b2) 2pz(a1) 21. 49 z y x P F F F F F 9 68 7 5 4 3 2 1 a. The two F p orbitals (top and bottom) generate the following reducible representation: D3h E 2C3 3C2 σh 2S3 3σv Γp 2 2 0 0 0 2 This reducible representation reduces to 1A1' and 1A2'' irreducible representations. Projectors may be used to find the symmetry-adapted AOs for these irreducible representations. φa1' = 12(f1 - f2) φa2'' = 12(f1 + f2) b. The three trigonal F p orbitals generate the following reducible representation: D3h E 2C3 3C2 σh 2S3 3σv Γp 3 0 1 3 0 1 This reducible representation reduces to 1A1' and 1E' irreducible representations. 50 Projectors may be used to find the symmetry-adapted -AOs for these irreducible representations (but they are exactly analogous to the previous few problems): φa1' = 13(f3 + f4 + f5) φe' = (1/6)-1/2 (2 f3 – f4 –f5) φe' = 12(f4 - f5) . c. The 3 P sp2 orbitals generate the following reducible representation: D3h E 2C3 3C2 σh 2S3 3σv Γsp2 3 0 1 3 0 1 This reducible representation reduces to 1A1' and 1E' irreducible representations. Again, projectors may be used to find the symmetry-adapted -AOs for these irreducible representations: φa1' = 13(f6 + f7 + f8) φe' = 16(2f6 - f7 - f8) φe' = 12(f7 - f8) . The leftover P pz orbital generate the following irreducible representation: D3h E 2C3 3C2 σh 2S3 3σv Γpz 1 1 -1 -1 -1 1 This irreducible representation is A2'' 51 φa2'' = f9. Drawing an energy level diagram using these SALC-AOs would result in the following: | | | || | | | | | a'1 e'* e' a''2 a''2* a'1* a'1 22. a. For non-degenerate point groups, one can simply multiply the representations (since only one representation will be obtained): a1 ? b1 = b1 Constructing a "box" in this case is unnecessary since it would only contain a single row. Two unpaired electrons will result in a singlet (S=0, MS=0), and three triplets (S=1, MS=1; S=1, MS=0; S=1, MS=-1). The states will be: 3B1(MS=1), 3B1(MS=0), 3B1(MS=- 1), and 1B1(MS=0). b. Remember that when coupling non-equivalent linear molecule angular momenta, one simple adds the individual Lz values and vector couples the electron spin. So, in this case 52 (1piu12piu1), we have ML values of 1+1, 1-1, -1+1, and -1-1 (2, 0, 0, and -2). The term symbol ? is used to denote the spatially doubly degenerate level (ML=±2) and there are two distinct spatially non-degenerate levels denoted by the term symbol Σ (ML=0) Again, two unpaired electrons will result in a singlet (S=0, MS=0), and three triplets (S=1, MS=1;S=1, MS=0;S=1, MS=-1). The states generated are then: 1? (ML=2); one state (MS=0), 1? (ML=-2); one state (MS=0), 3? (ML=2); three states (MS=1,0, and -1), 3? (ML=-2); three states (MS=1,0, and -1), 1Σ (ML=0); one state (MS=0), 1Σ (ML=0); one state (MS=0), 3Σ (ML=0); three states (MS=1,0, and -1), and 3Σ (ML=0); three states (MS=1,0, and -1). c. Constructing the "box" for two equivalent pi electrons one obtains: ML MS 2 1 0 1 |pi1αpi-1α| 0 |pi1αpi1β| |pi1αpi-1β|, |pi-1αpi1β| From this "box" one obtains six states: 53 1? (ML=2); one state (MS=0), 1? (ML=-2); one state (MS=0), 1Σ (ML=0); one state (MS=0), 3Σ (ML=0); three states (MS=1,0, and -1). d. It is not necessary to construct a "box" when coupling non-equivalent angular momenta since vector coupling results in a range from the sum of the two individual angular momenta to the absolute value of their difference. In this case, 3d14d1, L=4, 3, 2, 1, 0, and S=1,0. The term symbols are: 3G, 1G, 3F, 1F, 3D, 1D, 3P, 1P, 3S, and 1S. The L and S angular momenta can be vector coupled to produce further splitting into levels: J = L + S ... |L - S|. Denoting J as a term symbol subscript one can identify all the levels and subsequent (2J + 1) states: 3G5 (11 states), 3G4 (9 states), 3G3 (7 states), 1G4 (9 states), 3F4 (9 states), 3F3 (7 states), 3F2 (5 states), 1F3 (7 states), 3D3 (7 states), 54 3D2 (5 states), 3D1 (3 states), 1D2 (5 states), 3P2 (5 states), 3P1 (3 states), 3P0 (1 state), 1P1 (3 states), 3S1 (3 states), and 1S0 (1 state). e. Construction of a "box" for the two equivalent d electrons generates (note the "box" has been turned side ways for convenience): MS ML 1 0 4 |d2αd2β| 3 |d2αd1α| |d2αd1β|, |d2βd1α| 2 |d2αd0α| |d2αd0β|, |d2βd0α|, |d1αd1β| 1 |d1αd0α|, |d2αd-1α| |d1αd0β|, |d1βd0α|, |d2αd-1β|, |d2βd-1α| 55 0 |d2αd-2α|, |d1αd-1α| |d2αd-2β|, |d2βd-2α|, |d1αd-1β|, |d1βd-1α|, |d0αd0β| The term symbols are: 1G, 3F, 1D, 3P, and 1S. The L and S angular momenta can be vector coupled to produce further splitting into levels: 1G4 (9 states), 3F4 (9 states), 3F3 (7 states), 3F2 (5 states), 1D2 (5 states), 3P2 (5 states), 3P1 (3 states), 3P0 (1 state), and 1S0 (1 state). 23. a. Once the spatial symmetry has been determined by multiplication of the irreducible representations, the spin coupling gives the result: 1 2( )|3a1α1b1β| - |3a1β1b1α| b. There are three states here : 56 1.) |3a1α1b1α|, 2.) 12( )|3a1α1b1β| + |3a1β1b1α| , and 3.) |3a1β1b1β| c. |3a1α3a1β| 24. a. All the Slater determinants have in common the |1sα1sβ2sα2sβ| "core" and hence this component will not be written out explicitly for each case. 3P(ML=1,MS=1) = |p1αp0α| = | 12(px + ipy) α(pz)α| = 12( )|pxαpzα| + i|pyαpzα| 3P(ML=0,MS=1) = |p1αp-1α| = | 12(px + ipy) α 12(px - ipy) α| = 12( )|pxαpxα| - i|pxαpyα| + i|pyαpxα| + |pyαpyα| = 12( )0 - i|pxαpyα| - i|pxαpyα| + 0 = 12( )-2i|pxαpyα| = -i|pxαpyα| 57 3P(ML=-1,MS=1) = |p-1αp0α| = | 12(px - ipy) α(pz)α| = 12( )|pxαpzα| - i|pyαpzα| As you can see, the symmetries of each of these states cannot be labeled with a single irreducible representation of the C2v point group. For example, |pxαpzα| is xz (B1) and |pyαpzα| is yz (B2) and hence the 3P(ML=1,MS=1) state is a combination of B1 and B2 symmetries. But, the three 3P(ML,MS=1) functions are degenerate for the C atom and any combination of these three functions would also be degenerate. Therefore, we can choose new combinations that can be labeled with "pure" C2v point group labels. 3P(xz,MS=1) = |pxαpzα| = 12( )3P(ML=1,MS=1) + 3P(ML=-1,MS=1) = 3B1 3P(yx,MS=1) = |pyαpxα| = 1i( )3P(ML=0,MS=1) = 3A2 3P(yz,MS=1) = |pyαpzα| = 1i 2( )3P(ML=1,MS=1) - 3P(ML=-1,MS=1) = 3B2 Now, we can do likewise for the five degenerate 1D states: 1D(ML=2,MS=0) = |p1αp1β| = | 12(px + ipy) α 12(px + ipy) β| 58 = 12( )|pxαpxβ| + i|pxαpyβ| + i|pyαpxβ| - |pyαpyβ| 1D(ML=-2,MS=0) = |p-1αp-1β| = | 12(px - ipy) α 12(px - ipy) β| = 12( )|pxαpxβ| - i|pxαpyβ| - i|pyαpxβ| - |pyαpyβ| 1D(ML=1,MS=0) = 1 2( )|p0αp1β| - |p0βp1α| = 12??? ???|(pz)α 12(px + ipy)β| - |(pz)β 12(px + ipy)α| = 12( )|pzαpxβ| + i|pzαpyβ| - |pzβpxα| - i|pzβpyα| 1D(ML=-1,MS=0) = 1 2( )|p0αp-1β| - |p0βp-1α| = 12??? ???|(pz)α 12(px - ipy)β| - |(pz)β 12(px - ipy)α| = 12( )|pzαpxβ| - i|pzαpyβ| - |pzβpxα| + i|pzβpyα| 1D(ML=0,MS=0) = 1 6( )2|p0αp0β| + |p1αp-1β| + |p-1αp1β| = 16(2|pzαpzβ| + | 12(px + ipy)α 12(px - ipy)β| + | 12(px - ipy) α 12(px + ipy) β|) = 16(2|pzαpzβ| 59 + 12( )|pxαpxβ| - i|pxαpyβ| + i|pyαpxβ| + |pyαpyβ| + 12( )|pxαpxβ| + i|pxαpyβ| - i|pyαpxβ| + |pyαpyβ| ) = 16( )2|pzαpzβ| + |pxαpxβ| + |pyαpyβ| ) Analogous to the three 3P states, we can also choose combinations of the five degenerate 1D states which can be labeled with "pure" C2v point group labels: 1D(xx-yy,MS=0) = |pxαpxβ| - |pyαpyβ| = ( )1D(ML=2,MS=0) + 1D(ML=-2,MS=0) = 1A1 1D(yx,MS=0) = |pxαpyβ| + |pyαpxβ| = 1i( )1D(ML=2,MS=0) - 1D(ML=-2,MS=0) = 1A2 1D(zx,MS=0) = |pzαpxβ| - |pzβpxα| = ( )1D(ML=1,MS=0) + 1D(ML=-1,MS=0) = 1B1 1D(zy,MS=0) = |pzαpyβ| - |pzβpyα| = 1i( )1D(ML=1,MS=0) - 1D(ML=-1,MS=0) = 1B2 1D(2zz+xx+yy,MS=0) = 1 6( )2|pzαpzβ| + |pxαpxβ| + |pyαpyβ| ) = 1D(ML=0,MS=0) = 1A1 The only state left is the 1S: 1S(ML=0,MS=0) = 1 3( )|p0αp0β| - |p1αp-1β| - |p-1αp1β| 60 = 13(|pzαpzβ| - | 12(px + ipy)α 12(px - ipy)β| - | 12(px - ipy) α 12(px + ipy) β|) = 13(|pzαpzβ| - 12( )|pxαpxβ| - i|pxαpyβ| + i|pyαpxβ| + |pyαpyβ| - 12( )|pxαpxβ| + i|pxαpyβ| - i|pyαpxβ| + |pyαpyβ| ) = 13( )|pzαpzβ| - |pxαpxβ| - |pyαpyβ| ) Each of the components of this state are A1 and hence this state has A1 symmetry. b. Forming symmetry-adapted AOs from the C and H atomic orbitals would generate the following: 61 H H C H H C H H C H H C H H C H H C H1s + H1s = σg = a1 H1s - H1s = σu = b2 C2s = a1 C2p = a1 C2p = b2 C2p = b1 z xy The bonding, nonbonding, and antibonding orbitals of CH2 can be illustrated in the following manner: H H C H H C H H C H H C H H C H H C σ = a1 σ = b2 n = a1 ppi = b1 σ* = a1 σ* = b2 c. 62 Orbital-correlation diagram for the reaction C + H2 -----> CH2 (bent) a1(bonding) b2(antibonding) a1(antibonding) b1(2ppi) a1(non-bonding) b2(bonding) CH2 (bent)C + H2 σg(a1) 2s(a1) σu(b2) 2px(b1) 2py(b2) 2pz(a1) d. - e. It is necessary to determine how the wavefunctions found in part a. correlate with states of the CH2 molecule: 3P(xz,MS=1); 3B1 = σg2s2pxpz ??→ σ2n2ppiσ* 3P(yx,MS=1); 3A2 = σg2s2pxpy ??→ σ2n2ppiσ 3P(yz,MS=1); 3B2 = σg2s2pypz ??→ σ2n2σσ* 1D(xx-yy,MS=0); 1A1 ??→ σ2n2ppi2 - σ2n2σ2 1D(yx,MS=0); 1A2 ??→ σ2n2σppi 1D(zx,MS=0); 1B1 ??→ σ2n2σ*ppi 1D(zy,MS=0); 1B2 ??→ σ2n2σ*σ 1D(2zz+xx+yy,MS=0); 1A1 ??→ 2σ2n2σ*2 + σ2n2ppi2 + σ2n2σ2 Note, the C + H2 state to which the lowest 1A1 (σ2n2σ2) CH2 state decomposes would be σg2s2py2. This state (σg2s2py2) cannot be obtained by a simple combination of the 1D states. In order to obtain pure σg2s2py2 it is necessary to combine 1S with 1D. For example, 63 σg2s2py2 = 16( )6 1D(0,0) - 2 3 1S(0,0) - 12( )1D(2,0) + 1D(-2,0) . This indicates that a configuration correlation diagram must be drawn with a barrier near the 1D asymptote to represent the fact that 1A1 CH2 correlates with a mixture of 1D and 1S carbon plus hydrogen. The C + H2 state to which the lowest 3B1 (σ2nσ2ppi) CH2 state decomposes would be σg2spy2px. 64 (3B13B23A2) (1B21A11A11A21B1) C(1D) + H2 29.2 Kcal/mole 1A1(σ2σ2n2) 3B1(σ2σ2nppi) 3A2(σ2σn2ppi) 3B2(σ2σn2σ?) 3B1(σ2n2σ?ppi) 3B1 C(3P) + H2 σg2spy2px 3B1 3B1 3B2 3A2 1A1 78.8 Kcal/mole 97.0 Kcal/mole f. If you follow the 3B1 component of the C(3P) + H2 (since it leads to the ground-state products) to 3B1 CH2 you must go over an approximately 20 Kcal/mole barrier. Of course this path produces 3B1 CH2 product. Distortions away from C2v symmetry, for example to Cs symmetry, would make the a1 and b2 orbitals identical in 65 symmetry (a'). The b1 orbitals would maintain their different symmetry going to a'' symmetry. Thus 3B1 and 3A2 (both 3A'' in Cs symmetry and odd under reflection through the molecular plane) can mix. The system could thus follow the 3A2 component of the C(3P) + H2 surface to the place (marked with a circle on the CCD) where it crosses the 3B1 surface upon which it then moves and continues to products. As a result, the barrier would be lowered. You can estimate when the barrier occurs (late or early) using thermodynamic information for the reaction (i.e. slopes and asymptotic energies). For example, an early barrier would be obtained for a reaction with the characteristics: Progress of Reaction Energy and a late barrier would be obtained for a reaction with the characteristics: 66 Progress of Reaction Energy This relation between reaction endothermicity or exothermicity and the character of the transition state is known as the Hammond postulate. Note that the C(3P1) + H2 --> CH2 reaction of interest here has an early barrier. g. The reaction C(1D) + H2 ---> CH2 (1A1) should have no symmetry barrier (this can be recognized by following the 1A1 (C(1D) + H2) reactants down to the 1A1 (CH2) products). 25. This problem in many respects is analogous to problem 24. The 3B1 surface certainly requires a two configuration CI wavefunction; the σ2σ2npx (pi2py2spx) and the σ2n2pxσ* (pi2s2pxpz). The 1A1 surface could use the σ2σ2n2 (pi2s2py2) only but once again there is no combination of 1D determinants which gives purely this configuration (pi2s2py2). Thus mixing of both 1D and 1S determinants are 67 necessary to yield the required pi2s2py2 configuration. Hence even the 1A1 surface would require a multiconfigurational wavefunction for adequate description. C: H H H C C H x z y+ C n σ?CC σCC σ?CC σCC 2px(b1) 2py(b2) 2pz(a1) pi*(b2) 2s(a1) pi(a1) C2H2 + C C3H2 b2(bonding) a1(non-bonding) b1(2ppi) a1(antibonding) b2(antibonding) a1(bonding) Orbital-correlation diagram for the reaction C2H2 + C -----> C3H2 Configuration correlation diagram for the reaction C2H2 + C ---> C3H2. 68 Ea 3B1 ?E 3B1 Ea > ?E (for 3B1) Ea = ?E (for 1A1) pi2s2pypz3B2 pi2s2pxpy3A2 pi2s2pxpz3B1 pi2s2py2 1A1 C(1D) + C2H2 1A1(σ2σ2n2) 3B1(σ2σ2nppi) 3A2(σ2σn2ppi) 3B2(σ2σn2σ?) 3B1(σ2n2σ?ppi)3B 1 C(3P) + C2H2 pi2spy2px 26. a. CCl4 is tetrahedral and therefore is a spherical top. CHCl3 has C3v symmetry and therefore is a symmetric top. CH2Cl2 has C2v symmetry and therefore is an asymmetric top. b. CCl4 has such high symmetry that it will not exhibit pure rotational spectra because it has no permanent dipole moment. CHCl3 and CH2Cl2 will both exhibit pure rotation spectra. 69 27. NH3 is a symmetric top (oblate). Use the given energy expression, E = (A - B) K2 + B J(J + 1), A = 6.20 cm-1, B = 9.44 cm-1, selection rules ?J = ±1, and the fact that μ0→ lies along the figure axis such that ?K = 0, to give: ?E = 2B (J + 1) = 2B, 4B, and 6B (J = 0, 1, and 2). So, lines are at 18.88 cm-1, 37.76 cm-1, and 56.64 cm-1. 28. To convert between cm-1 and energy, multiply by hc = (6.62618x10-34J sec)(2.997925x1010cm sec-1) = 1.9865x1023 J cm. Let all quantities in cm-1 be designated with a bar, e.g. Be? = 1.78 cm-1. a. hcBe? = h _2 2μRe2 Re = h _ 2μhcBe? , μ = mBmOmB + mO = (11)(16)(11 + 16) x 1.66056x10-27 kg = 1.0824x10-26 kg. 70 hcBe? = hc(1.78 cm-1) = 3.5359x10-23 J Re = 1.05459x10 -34 J sec (2)1.0824x10-26 kg.3.5359x10-23 J Re = 1.205x10-10 m = 1.205 ? De = 4Be 3 h_ωe2 , De ? = 4Be ?3 ωe? 2 = (4)(1.78 cm-1)3 (1885 cm-1)2 = 6.35x10-6 cm-1 ωexe = h _ω e2 4D0e , ωexe ? = ωe ? 2 4D0e? = (1885 cm-1)2 (4)(66782.2 cm-1) = 13.30 cm-1. D00 = D0e - h _ω e 2 + h_ωexe 4 , D00 ? = D0 e ? - ωe ? 2 + ωexe? 4 = 66782.2 - 18852 + 13.34 = 65843.0 cm-1 = 8.16 eV. αe = -6Be 2 h_ωe + 6 Be3h_ωexe h_ωe αe? = -6Be ?2 ωe? + 6 Be?3ωexe? ωe? αe? = (-6)(1.78) 2 (1885) + 6 (1.78)3(13.3) (1885) = 0.0175 cm-1. B0 = Be - αe(1/2) , B0? = Be? - αe?(1/2) = 1.78 - 0.0175/2 = 1.77 cm-1 B1 = Be - αe(3/2) , B1? = Be? - αe?(3/2) = 1.78 - 0.0175(1.5) = 1.75 cm-1 71 b. The molecule has a dipole moment and so it should have a pure rotational spectrum. In addition, the dipole moment should change with R and so it should have a vibration-rotation spectrum. The first three lines correspond to J = 1 → 0, J = 2 → 1, J = 3 → 2 E = h_ ωe(v + 1/2) - h_ ωexe(v + 1/2)2 + BvJ(J + 1) - DeJ2(J + 1)2 ?E = h_ ωe - 2h_ ωexe - B0J(J + 1) + B1J(J - 1) - 4DeJ3 ?E? = ωe? - 2ωexe? - B0? J(J + 1) + B1? J(J - 1) - 4De? J3 ?E? = 1885 - 2(13.3) - 1.77J(J + 1) + 1.75J(J - 1) - 4(6.35x10-6)J3 = 1858.4 - 1.77J(J + 1) + 1.75J(J - 1) - 2.54x10-5J3 ?E? (J = 1) = 1854.9 cm-1 ?E? (J = 2) = 1851.3 cm-1 ?E? (J = 3) = 1847.7 cm-1 29. The C2H2Cl2 molecule has a σh plane of symmetry (plane of molecule), a C2 axis (⊥ to the molecular plane), and inversion symmetry, this results in C2h symmetry. Using C2h symmetry, the modes can be labeled as follows: ν1, ν2, ν3, ν4, and ν5 are ag, ν6 and ν7 are au, ν8 is bg, and ν9, ν10, ν11, and ν12 are bu. 72 30. R θ HH y z C Molecule I Molecule II RCH = 1.121 ? RCH = 1.076 ? ∠HCH = 104° ∠HCH = 136° yH = R Sin (θ/2) = ±0.8834 yH = ±0.9976 zH = R Cos (θ/2) = -0.6902 zH = -0.4031 Center of Mass(COM): clearly, X = Y = 0, Z = 12(0) - 2RCos(θ/2)14 = -0.0986 Z = -0.0576 a. Ixx = ∑ j mj(yj2 + zj2) - M(Y2 + Z2) Ixy = -∑ j mjxjyj - MXY Ixx = 2(1.121)2 - 14(-0.0986)2 Ixx = 2(1.076)2 - 14(-0.0576)2 = 2.377 = 2.269 73 Iyy = 2(0.6902)2 - 14(-0.0986)2 Iyy = 2(0.4031)2 - 14(-0.0576)2 = 0.8167 = 0.2786 Izz = 2(0.8834)2 Izz = 2(0.9976)2 = 1.561 = 1.990 Ixz = Iyz = Ixy = 0 b. Since the moment of inertia tensor is already diagonal, the principal moments of inertia have already been determined to be (Ia < Ib < Ic): Iyy < Izz < Ixx Iyy < Izz < Ixx 0.8167 < 1.561 < 2.377 0.2786 < 1.990 < 2.269 Using the formula: A = h8pi2cIa = 6.626x10 -27 8pi2(3x1010)Ia X 6.02x1023 (1x10-8)2 A = 16.84Ia cm-1 similarly, B = 16.84Ib cm-1, and C = 16.84Ic cm-1. So, Molecule I Molecule II y ? A = 20.62 y ? A = 60.45 z ? B = 10.79 z ? B = 8.46 x ? C = 7.08 x ? C = 7.42 c. Averaging B + C: B = (B + C)/2 = 8.94 B = (B + C)/2 = 7.94 A - B = 11.68 A - B = 52.51 74 Using the prolate top formula: E = (A - B) K2 + B J(J + 1), Molecule I Molecule II E = 11.68K2 + 8.94J(J + 1) E = 52.51K2 + 7.94J(J + 1) Levels: J = 0,1,2,... and K = 0,1, ... J For a given level defined by J and K, there are MJ degeneracies given by: (2J + 1) x ?? ? ?? ? 2 for K ≠ 0 1 for K = 0 d. Molecule I Molecule II HH C z => Ib y => Ia HH y => Ia z => Ib C e. Assume molecule I is CH2- and molecule II is CH2. Then, ?E = EJj(CH2) - EJi(CH2-), where: E(CH2) = 52.51K2 + 7.94J(J + 1), and E(CH2-) = 11.68K2 + 8.94J(J + 1) For R-branches: Jj = Ji + 1, ?K = 0: ?ER = EJj(CH2) - EJi(CH2-) 75 = 7.94(Ji + 1)(Ji + 1 + 1) - 8.94Ji(Ji + 1) = (Ji + 1){7.94(Ji + 1 + 1) - 8.94Ji} = (Ji + 1){(7.94- 8.94)Ji + 2(7.94)} = (Ji + 1){-Ji + 15.88} For P-branches: Jj = Ji - 1, ?K = 0: ?EP = EJj(CH2) - EJi(CH2-) = 7.94(Ji - 1)(Ji - 1 + 1) - 8.94Ji(Ji + 1) = Ji{7.94(Ji - 1) - 8.94(Ji + 1)} = Ji{(7.94- 8.94)Ji - 7.94 - 8.94} = Ji{-Ji - 16.88} This indicates that the R branch lines occur at energies which grow closer and closer together as J increases (since the 15.88 - Ji term will cancel). The P branch lines occur at energies which lie more and more negative (i.e. to the left of the origin). So, you can predict that if molecule I is CH2- and molecule II is CH2 then the R-branch has a band head and the P-branch does not. This is observed, therefore our assumption was correct: molecule I is CH2- and molecule II is CH2. f. The band head occurs when d(?ER)dJ = 0. d(?ER) dJ = d dJ [(Ji + 1){-Ji + 15.88}] = 0 = ddJ(-Ji2 - Ji + 15.88Ji + 15.88) = 0 = -2Ji + 14.88 = 0 76 ∴ Ji = 7.44, so J = 7 or 8. At J = 7.44: ?ER = (J + 1){-J + 15.88} ?ER = (7.44 + 1){-7.44 + 15.88} = (8.44)(8.44) = 71.2 cm-1 above the origin. 31. a. D6h E 2C6 2C3 C2 3C2 ' 3C2 " i 2S3 2S6 σh 3σd 3σv A1g 1 1 1 1 1 1 1 1 1 1 1 1 x2+y2,z2 A2g 1 1 1 1 -1 -1 1 1 1 1 -1 -1 Rz B1g 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 B2g 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 E1g 2 1 -1 -2 0 0 2 1 -1 -2 0 0 Rx,Ry (xz,yz) E2g 2 -1 -1 2 0 0 2 -1 -1 2 0 0 (x2-y2,xy) A1u 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 77 A2u 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 z B1u 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 B2u 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 E1u 2 1 -1 -2 0 0 -2 -1 1 2 0 0 (x,y) E2u 2 -1 -1 2 0 0 -2 1 1 -2 0 0 ΓC-H 6 0 0 0 0 2 0 0 0 6 2 0 b. The number of irreducible representations may be found by using the following formula: nirrep = 1g∑ R χred(R)χirrep(R) , where g = the order of the point group (24 for D6h). nA1g = 124∑ R ΓC-H(R).A1g(R) = 124 {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1) +(3)(0)(1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(1) +(2)(0)(1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(1)} = 1 nA2g = 124 {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1) +(3)(0)(-1)+(3)(2)(-1)+(1)(0)(1)+(2)(0)(1) +(2)(0)(1)+(1)(6)(1)+(3)(2)(-1)+(3)(0)(-1)} 78 = 0 nB1g = 124 {(1)(6)(1)+(2)(0)(-1)+(2)(0)(1)+(1)(0)(-1) +(3)(0)(1)+(3)(2)(-1)+(1)(0)(1)+(2)(0)(-1) +(2)(0)(1)+(1)(6)(-1)+(3)(2)(1)+(3)(0)(-1)} = 0 nB2g = 124 {(1)(6)(1)+(2)(0)(-1)+(2)(0)(1)+(1)(0)(-1) +(3)(0)(-1)+(3)(2)(1)+(1)(0)(1)+(2)(0)(-1) +(2)(0)(1)+(1)(6)(-1)+(3)(2)(-1)+(3)(0)(1)} = 0 nE1g = 124 {(1)(6)(2)+(2)(0)(1)+(2)(0)(-1)+(1)(0)(-2) +(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(1) +(2)(0)(-1)+(1)(6)(-2)+(3)(2)(0)+(3)(0)(0)} = 0 nE2g = 124 {(1)(6)(2)+(2)(0)(-1)+(2)(0)(-1)+(1)(0)(2) +(3)(0)(0)+(3)(2)(0)+(1)(0)(2)+(2)(0)(-1) +(2)(0)(-1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)} = 1 nA1u = 124 {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1) +(3)(0)(1)+(3)(2)(1)+(1)(0)(-1)+(2)(0)(-1) +(2)(0)(-1)+(1)(6)(-1)+(3)(2)(-1)+(3)(0)(-1)} = 0 79 nA2u = 124 {(1)(6)(1)+(2)(0)(1)+(2)(0)(1)+(1)(0)(1) +(3)(0)(-1)+(3)(2)(-1)+(1)(0)(-1)+(2)(0)(-1) +(2)(0)(-1)+(1)(6)(-1)+(3)(2)(1)+(3)(0)(1)} = 0 nB1u = 124 {(1)(6)(1)+(2)(0)(-1)+(2)(0)(1)+(1)(0)(-1) +(3)(0)(1)+(3)(2)(-1)+(1)(0)(-1)+(2)(0)(1) +(2)(0)(-1)+(1)(6)(1)+(3)(2)(-1)+(3)(0)(1)} = 0 nB2u = 124 {(1)(6)(1)+(2)(0)(-1)+(2)(0)(1)+(1)(0)(-1) +(3)(0)(-1)+(3)(2)(1)+(1)(0)(-1)+(2)(0)(1) +(2)(0)(-1)+(1)(6)(1)+(3)(2)(1)+(3)(0)(-1)} = 1 nE1u = 124 {(1)(6)(2)+(2)(0)(1)+(2)(0)(-1)+(1)(0)(-2) +(3)(0)(0)+(3)(2)(0)+(1)(0)(-2)+(2)(0)(-1) +(2)(0)(1)+(1)(6)(2)+(3)(2)(0)+(3)(0)(0)} = 1 nE2u = 124 {(1)(6)(2)+(2)(0)(-1)+(2)(0)(-1)+(1)(0)(2) +(3)(0)(0)+(3)(2)(0)+(1)(0)(-2)+(2)(0)(1) +(2)(0)(1)+(1)(6)(-2)+(3)(2)(0)+(3)(0)(0)} = 0 We see that ΓC-H = A1g⊕E2g⊕B2u⊕E1u 80 c. x and y ? E1u , z ? A2u , so, the ground state A1g level can be excited to the degenerate E1u level by coupling through the x or y transition dipoles. Therefore E1u is infrared active and ⊥ polarized. d. (x2 + y2, z2) ? A1g, (xz, yz) ? E1g, (x2 - y2, xy) ? E2g ,so, the ground state A1g level can be excited to the degenerate E2g level by coupling through the x2 - y2 or xy transitions or be excited to the degenerate A1g level by coupling through the xz or yz transitions. Therefore A1g and E2g are Raman active.. e. The B2u mode is not IR or Raman active. 32. a. Evaluate the z-component of μfi: μfi = <2pz|e r Cosθ|1s>, where ψ1s = 1pi ??? ???Za0 3 2 e -Zr a0 , and ψ2p z = 1 4 2pi ? ?? ???Za 0 5 2 r Cosθ e -Zr 2a0 . μfi = 14 2pi ??? ???Za0 5 2 1 pi ? ?? ???Za 0 3 2 <r Cosθ e -Zr 2a0 |e r Cosθ|e -Zr a0 > = 14pi 2 ??? ???Za0 4 <r Cosθ e -Zr 2a0 |e r Cosθ|e -Zr a0 > 81 = e4pi 2 ??? ???Za0 4 ?? 0 ∞ r2dr?? 0 pi Sinθdθ ?? 0 2pi d?? ?? ??? r2 e -Zr 2a0 e -Zr a0 Cos2θ = e4pi 2 2pi ??? ???Za0 4 ? �? 0 ∞ ?? ? ?? ? r4 e -3Zr 2a0 dr ?? 0 pi SinθCos2θdθ = e4pi 2 2pi ??? ???Za0 4 4! ?? ? ?? ?3Z 2a0 5 ?? ? ?? ?-1 3 Cos3θ?? ??pi 0 = e4pi 2 2pi ??? ???Za0 4 2 5a054! 35Z5 ?? ? ?? ?-1 3 ( )(-1)3 - (1)3 = e2 2 8a0 35Z = ea0 Z 28 235 = 0.7449 ea0 Z b. Examine the symmetry of the integrands for <2pz| e x |1s> and <2pz| e y |1s>. Consider reflection in the xy plane: Function Symmetry 2pz -1 x +1 82 1s +1 y +1 Under this operation, the integrand of <2pz| e x |1s> is (-1)(1)(1) = -1 (it is antisymmetric) and hence <2pz| e x |1s> = 0. Similarly, under this operation the integrand of <2pz| e y |1s> is (-1)(1)(1) = -1 (it is also antisymmetric) and hence <2pz| e y |1s> = 0. c. τR = 3h -4c3 4(Ei - Ef)3|μfi|2 , Ei = E2pz = -14 Z2 ?? ? ?? ?e2 2a0 Ef = E1s = -Z2 ?? ? ?? ?e2 2a0 Ei - Ef = 38 ?? ? ?? ?e2 a0 Z2 Making the substitutions for Ei - Ef and |μfi| in the expression for τR we obtain: τR = 3h -4c3 4?? ? ?? ?3 8 ?? ? ?? ?e2 a0 Z2 3 ?? ? ?? ? ?? ? ?? ?ea0 Z 28 235 2 , = 3h -4c3 4 3 3 83 ?? ? ?? ?e6 a03 Z6 ?? ? ?? ?e2a02 Z2 216 (2)310 , = h -4c3 38 a0 e8 Z4 28 , 83 Inserting e2 = h -2 mea0 we obtain: τR = h -4c3 38 a0 me4a04 h-8 Z4 28 = 38 28 c3 a05 me4 h-4 Z4 = 25.6289 c 3 a05 me4 h-4 Z4 = 25,6289 ??? ???1Z4 x (2.998x1010 cm sec-1)3(0.529177x10-8 cm)5(9.109x10-28 g)4 (1.0546x10-27 g cm2 sec-1)4 = 1.595x10-9 sec x ??? ???1Z4 So, for example: Atom τR H 1.595 ns He+ 99.7 ps Li+2 19.7 ps Be+3 6.23 ps Ne+9 159 fs 33. 84 a. H = H0 + λH'(t), H'(t) = Vθ(t), H0?k = Ek?k, ωk = Ek/h- ih-?ψ?t = Hψ let ψ(r,t) = ih-∑ j cj(t)?je-iωjt and insert into the Schr?dinger equation: ih-∑ j ?? ??c?j - iωjcj e-iωjt?j = ih-∑ j cj(t)e-iωjt(H0 + λH'(t)) ?j ∑ j ?? ??ih-c?j + Ejcj - cjEj - cjλH' e-iωjt?j = 0 ∑ j ?? ??ih-c?j<m|j> - cjλ<m|H'|j> e-iωjt = 0 ih-c? m e-iωmt = ∑ j cjλH'mj e-iωjt So, c? m = 1ih-∑ j cjλH'mj e-i(ωjm)t Going back a few equations and multiplying from the left by ?k instead of ?m we obtain: ∑ j ?? ??ih-c?j<k|j> - cjλ<k|H'|j> e-iωjt = 0 ih-c? k e-iωkt = ∑ j cjλH'kj e-iωjt So, 85 c? k = 1ih-∑ j cjλH'kj e-i(ωjk)t Now, let: cm = cm(0) + cm(1)λ + cm(2)λ2 + ... ck = ck(0) + ck(1)λ + ck(2)λ2 + ... and substituting into above we obtain: c? m(0) + c? m(1)λ + c? m(2)λ2 + ... = 1ih-∑ j [cj(0) + cj(1)λ + cj(2)λ2 + ...] λH'mj e-i(ωjm)t first order: c? m(0) = 0 ? cm(0) = 1 second order: c? m(1) = 1ih-∑ j cj(0) H'mj e-i(ωjm)t (n+1)st order: c? m(n) = 1ih-∑ j cj(n-1) H'mj e-i(ωjm)t Similarly: first order: c? k(0) = 0 ? ck≠m(0) = 0 second order: c? k(1) = 1ih-∑ j cj(0) H'kj e-i(ωjk)t 86 (n+1)st order: c? k(n) = 1ih-∑ j cj(n-1) H'kj e-i(ωjk)t So, c? m(1) = 1ih- cm(0) H'mm e-i(ωmm)t = 1ih- H'mm cm(1)(t) = 1ih- ?? 0 t dt' Vmm = Vmmtih- and similarly, c? k(1) = 1ih- cm(0) H'km e-i(ωmk)t = 1ih- H'km e-i(ωmk)t ck(1)(t) = 1ih- Vkm ?? 0 t dt' e-i(ωmk)t' = Vkmh-ω mk[ ] e-i(ωmk)t - 1 c? m(2) = 1ih-∑ j cj(1) H'mj e-i(ωjm)t c? m(2) = ∑ j≠m 1ih- Vjmh-ω mj[ ] e-i(ωmj)t - 1 H'mj e-i(ωjm)t + 1ih- Vmmtih- H'mm cm(2) = ∑ j≠m 1ih- VjmVmjh-ω mj ?? 0 t dt' e-i(ωjm)t' [ ]e-i(ωmj)t' - 1 - VmmVmmh-2 ?? 0 t t'dt' = ∑ j≠m VjmVmjih-2ω mj ?? 0 t dt'[ ]1 - e-i(ωjm)t' - |Vmm| 2 h-2 t2 2 87 = ∑ j≠m VjmVmjih-2ω mj ?? ? ?? ? t - e -i(ωjm)t - 1 -iωjm - |Vmm|2 h-2 t2 2 = ∑ j≠m 'VjmVmjh-2ω mj2 ( )e-i(ωjm)t - 1 + ∑ j≠m ' VjmVmjih-2ω mj t - |Vmm| 2 t2 2h-2 Similarly, c? k(2) = 1ih-∑ j cj(1) H'kj e-i(ωjk)t = ∑ j≠m 1ih- Vjmh-ω mj[ ] e-i(ωmj)t - 1 H'kj e-i(ωjk)t + 1ih- Vmmtih- H'km e-i(ωmk)t ck(2)(t) = ∑ j≠m ' VjmVkjih-2ω mj ?? 0 t dt' e-i(ωjk)t' [ ]e-i(ωmj)t' - 1 - VmmVkmh-2 ?? 0 t t'dt' e-i(ωmk)t' = ∑ j≠m 'VjmVkjih-2ω mj ?? ? ?? ?e-i(ωmj+ωjm)t - 1 -iωmk - e-i(ωjk)t - 1 -iωjk - VmmVkmh-2 ?? ? ?? ? e-i(ωmk)t'?? ? ?? ?t' -iωmk - 1 -(iωmk)2 t 0 = ∑ j≠m 'VjmVkjh-2ω mj ?? ? ?? ?e-i(ωmk)t - 1 ωmk - e-i(ωjk)t - 1 ωjk 88 + VmmVkmh-2ω mk ??? ???e-i(ωmk)t'??? ???t'i - 1ωmk t0 = ∑ j≠m 'VjmVkjEm - Ej ?? ? ?? ?e-i(ωmk)t - 1 Em - Ek - e-i(ωjk)t - 1 Ej - Ek + VmmVkmh-(E m - Ek) ??? ???e-i(ωmk)t??? ???ti - 1ωmk + 1ωmk So, the overall amplitudes cm, and ck, to second order are: cm(t) = 1 + Vmmtih- + ∑ j≠m ' VjmVmjih-(E m - Ej) t + ∑ j≠m ' VjmVmjh-2(E m - Ej)2 ( )e-i(ωjm)t - 1 - |Vmm| 2 t2 2h-2 ck(t) = Vkm(Em - Ek)[ ]e-i(ωmk)t - 1 + VmmVkm(Em - Ek)2 [ ]1 - e-i(ωmk)t + VmmVkm(Em - Ek) th-i e-i(ωmk)t + ∑ j≠m 'VjmVkjEm - Ej ?? ? ?? ?e-i(ωmk)t - 1 Em - Ek - e-i(ωjk)t - 1 Ej - Ek b. The perturbation equations still hold: c? m(n) = 1ih-∑ j cj(n-1) H'mj e-i(ωjm)t ; c? k(n) = 1ih-∑ j cj(n-1) H'kj e-i(ωjk)t So, cm(0) = 1 and ck(0) = 0 c? m(1) = 1ih- H'mm 89 cm(1) = 1ih- Vmm ?? -∞ t dt' eηt = Vmme ηt ih-η c? k(1) = 1ih- H'km e-i(ωmk)t ck(1) = 1ih- Vkm ?? -∞ t dt' e-i(ωmk+η)t' = Vkmih-(-iω mk+η)[ ] e-i(ωmk+η)t = VkmE m - Ek + ih-η[ ] e-i(ωmk+η)t c? m(2) = ∑ j≠m '1ih- VjmE m - Ej + ih-η e-i(ωmj+η)t Vmj eηt e-i(ωjm)t + 1 ih- Vmm eηt ih-η Vmm e ηt cm(2) = ∑ j≠m '1ih- VjmVmjE m - Ej + ih-η ?? -∞ t e2ηt'dt' - |Vmm| 2 h-2η ?? -∞ t e2ηt'dt' = ∑ j≠m ' VjmVmjih-2η(E m - Ej + ih-η) e2ηt - |Vmm| 2 2h-2η2 e 2ηt c? k(2) = ∑ j≠m '1ih- VjmE m - Ej + ih-η e-i(ωmj+η)t H'kj e-i(ωjk)t + 1ih- Vmm e ηt ih-η H'km e -i(ωmk)t ck(2) = ∑ j≠m '1ih- VjmVkjE m - Ej + ih-η ?? -∞ t e-i(ωmk+2η)t'dt' - 90 VmmVkm h-2η ?? -∞ t e-i(ωmk+2η)t'dt' = ∑ j≠m ' VjmVkj e -i(ωmk+2η)t (Em - Ej + ih-η)(Em - Ek + 2ih-η) - VmmVkm e-i(ωmk+2η)t ih-η(Em - Ek + 2ih-η) Therefore, to second order: cm(t) = 1 + Vmme ηt ih-η + ∑j VjmVmj ih-2η(Em - Ej + ih-η) e 2ηt ck(t) = Vkmih-(-iω mk+η)[ ] e-i(ωmk+η)t + ∑ j VjmVkj e -i(ωmk+2η)t (Em - Ej + ih-η)(Em - Ek + 2ih-η) c. In part a. the c(2)(t) grow linearly with time (for Vmm = 0) while in part b. they remain finite for η > 0. The result in part a. is due to the sudden turning on of the field. d. |ck(t)|2 = ?? ?? ?? ??∑ j VjmVkj e -i(ωmk+2η)t (Em - Ej + ih-η)(Em - Ek + 2ih-η) 2 = ∑ jj' VkjVkj'VjmVj'm e -i(ωmk+2η)tei(ωmk+2η)t (Em-Ej+ih-η)(Em-Ej'-ih-η)(Em-Ek+2ih-η)(Em-Ek-2ih-η) = ∑ jj' VkjVkj'VjmVj'm e 4ηt [(Em-Ej)(Em-Ej')+ih-η(Ej-Ej')+h-2η2]((Em-Ek)2+4h-2η2) d dt |ck(t)|2 = ∑ jj' 4η VkjVkj'VjmVj'm[(E m-Ej)(Em-Ej')+ih-η(Ej-Ej')+h-2η2]((Em-Ek)2+4h-2η2) Now, look at the limit as η → 0+: 91 d dt |ck(t)|2 ≠ 0 when Em = Ek limη→0+ 4η ((Em-Ek)2+4h-2η2) α δ(Em-Ek) So, the final result is the 2nd order golden rule expression: d dt |ck(t)|2 = 2pi h- δ(Em-Ek) limη→0+ ?? ?? ?? ??∑ j VjmVkj(E j - Em - ih-η) 2 34. a. Tnm ≈ |<n|V|m>| 2 h-2ωnm2 evaluating <1s|V|2s> (using only the radial portions of the 1s and 2s wavefunctions since the spherical harmonics will integrate to unity) where V = (e2/r), the change in Coulomb potential when tritium becomes He: <1s|V|2s> = ?� ? 2??? ???Za0 3 2 e -Zr a0 1r 1 2 ? ?? ???Za 0 3 2 ??? ???1 - Zr2a 0 e -Zr 2a0 r2dr <1s|V|2s> = 22 ??? ???Za0 3 ?? ?? ?? ?? ?� ? re -3Zr 2a0 dr - ?�? Zr22a 0 e -3Zr 2a0 dr = 22 ??? ???Za0 3 ?? ?? ?? ??1 ?? ? ?? ?3Z 2a0 2 - ?? ? ?? ?Z 2a0 2 ?? ? ?? ?3Z 2a0 3 92 <1s|V|2s> = 22 ??? ???Za0 3 ?? ? ?? ?22a02 32Z2 - 23a02 33Z2 <1s|V|2s> = 22 ??? ???Za0 3 ?? ? ?? ?(3)22a02 - 23a02 33Z2 = 8Z 227a0 Now, En = - Z 2e2 n22a0 , E1s = - Z2e2 2a0 , E2s = - Z2e2 8a0 , E2s - E1s = 3Z2e2 8a0 So, Tnm = ?? ? ?? ?8Z 227a0 2 ?? ? ?? ?3Z2 8a0 2 = 26Z226a02 (2)38a02Z4 = 211 38Z2 = 0.312 (for Z = 1) b. ?m(r) = ?1s = 2??? ???Za0 3 2 e -Zr a0 Y00 The orthogonality of the spherical harmonics results in only s-states having non-zero values for Anm. We can then drop the Y00 (integrating this term will only result in unity) in determining the value of A1s,2s. ψn(r) = ψ2s = 12 ??? ???Za0 3 2 ??? ???1 - Zr2a 0 e -Zr 2a0 Remember for ?1s Z = 1 and for ψ2s Z = 2 Anm = ?� ? 2??? ???Za0 3 2 e -Zr a0 1 2 ? ?? ???Z+1a 0 3 2 ??? ???1 - (Z+1)r2a 0 e -(Z+1)r 2a0 r2dr Anm = 22??? ???Za0 3 2??? ???Z+1a 0 3 2?�? e -(3Z+1)r 2a0 ??? ???1 - (Z+1)r2a 0 r2dr 93 Anm = 22??? ???Za0 3 2??? ???Z+1a 0 3 2 ?? ?? ?? ?? ?� ? r2 e -(3Z+1)r 2a0 dr - ?�?(Z+1)r32a 0 e -(3Z+1)r 2a0 dr We obtain: Anm = 22??? ???Za0 3 2??? ???Z+1a 0 3 2 ?? ?? ?? ??2 ?? ? ?? ?3Z+1 2a0 3 - ?? ? ?? ?Z+1 2a0 (3)(2) ?? ? ?? ?3Z+1 2a0 4 Anm = 22??? ???Za0 3 2??? ???Z+1a 0 3 2 ??? ???24a03 (3Z+1)3 - (Z+1) (3)24a03 (3Z+1)4 Anm = 22??? ???Za0 3 2??? ???Z+1a 0 3 2 ??? ???-25a03 (3Z+1)4 Anm = -2 [ ] 23Z(Z+1) 3 2 (3Z+1)4 The transition probability is the square of this amplitude: Tnm = ? ?? ??? -2 [ ] 23Z(Z+1) 3 2 (3Z+1)4 2 = 2 11Z3(Z+1)3 (3Z+1)8 = 0.25 (for Z = 1). The difference in these two results (parts a. and b.) will become negligible at large values of Z when the perturbation becomes less significant than in the case of Z = 1. 35. 94 ε→ is along Z (lab fixed), and μ→ is along z (the C-I molecule fixed bond). The angle between Z and z is β: ε→ . μ→ = εμCosβ = εμD001* (αβγ) So, I = <DM'K'J' | ε→ . μ→ |DMKJ > = ?� ? DM'K'J' ε→. μ→DMKJ Sinβdβdγdα = εμ??DM'K'J' D001* DMKJ Sinβdβdγdα. Now use: DM'n'J'* D001* = ∑ jmn <J'M'10|jm>*Dmnj* <jn|J'K'10> *, to obtain: I = εμ ∑ jmn <J'M'10|jm>*<jn|J'K'10> *??Dmnj* DMKJ Sinβdβdγdα. Now use: ??Dmnj* DMKJ Sinβdβdγdα = 8pi 2 2J+1 δJjδMmδΚn, to obtain: I = εμ 8pi 2 2J+1 ∑ jmn <J'M'10|jm>*<jn|J'K'10> *δJjδMmδΚn = εμ 8pi 2 2J+1 <J'M'10|JM><JK|J'K'10>. We use: 95 <JK|J'K'10> = 2J+1(-i)(J'-1+K) ?? ??K' 0 KJ' 1 J and, <J'M'10|JM> = 2J+1(-i)(J'-1+M)?? ??M' 0 MJ' 1 J to give: I = εμ 8pi 2 2J+1 2J+1(-i)(J'-1+M)?? ??M' 0 MJ' 1 J 2J+1(-i)(J'-1+K) ?? ??K' 0 KJ' 1 J = εμ8pi2(-i)(J'-1+M+J'-1+K)?? ??M' 0 MJ' 1 J ?? ??K' 0 KJ' 1 J = εμ8pi2(-i)(M+K)?? ??M' 0 MJ' 1 J ?? ??K' 0 KJ' 1 J The 3-J symbols vanish unless: K' + 0 = K and M' + 0 = M. So, I = εμ8pi2(-i)(M+K)?? ??M 0 MJ' 1 J ?? ??K 0 KJ' 1 J δM'MδK'K. b. ?? ??M 0 MJ' 1 J and ?? ??K 0 KJ' 1 J vanish unless J' = J + 1, J, J - 1 ∴ ?J = ±1, 0 The K quantum number can not change because the dipole moment lies along the molecule's C3 axis and the light's electric field thus can exert no torque that twists the molecule about this axis. As a result, the light can not induce transitions that excite the molecule's spinning motion about this axis. 36. 96 a. B atom: 1s22s22p1, 2P ground state L = 1, S = 12 , gives a degeneracy ((2L+1)(2S+1)) of 6. O atom: 1s22s22p4, 3P ground state L = 1, S = 1, gives a degeneracy ((2L+1)(2S+1)) of 9. The total number of states formed is then (6)(9) = 54. b. We need only consider the p orbitals to find the low lying molecular states: 2pi 1pi 6σ 5σ 2p2p Which, in reality look like this: 5σ 6σ 1pi 2pi This is the correct ordering to give a 2Σ+ ground state. The only low-lying electron configurations are 1pi35σ2 or 1pi45σ1. These lead to 2Π and 2Σ+ states, respectively. 97 c. The bond orders in both states are 2.5. d. The 2Σ is + but g/u symmetry cannot be specified since this is a heteronuclear molecule. e. Only one excited state, the 2Π, is spin-allowed to radiate to the 2Σ+. Consider symmetries of transition moment operators that arise in the electric dipole contributions to the transition rate z → Σ+, x,y → Π, ∴ the 2Π → 2Σ+ is electric dipole allowed via a perpendicular band. f. Since ionization will remove a bonding electron, the BO+ bond is weaker than the BO bond. g. The ground state BO+ is 1Σ+ corresponding to a 1pi4 electron configuration. An electron configuration of 1pi3 5σ1 leads to a 3Π and a 1Π state. The 3Π will be lower in energy. A 1pi2 5σ2 configuration will lead to higher lying states of 3Σ-, 1?, and 1Σ+. h. There should be 3 bands corresponding to formation of BO+ in the 1Σ+, 3Π, and 1Π states. Since each of these involves removing a bonding electron, the Franck- Conden integrals will be appreciable for several vibrational levels, and thus a vibrational progression should be observed. 37. a. The bending (pi) vibration is degenerate. b. H---C≡N 98 ? bending fundamental c. H---C≡N ? stretching fundamental d. CH stretch (ν3 in figure) is σ, CN stretch is σ, and HCN (ν2 in figure) bend is pi. e. Under z (σ) light the CN stretch and the CH stretch can be excited, since ψ0 = σ, ψ1 = σ and z = σ provides coupling. f. Under x,y (pi) light the HCN bend can be excited, since ψ0 = σ, ψ1 = pi and x,y = pi provides coupling. g. The bending vibration is active under (x,y) perpendicular polarized light. ?J = 0, ±1 are the selection rules for ⊥ transitions. The CH stretching vibration is active under (z) || polarized light. ?J = ±1 are the selection rules for || transitions. 99 38. F φi = εi φj = h φi + ∑ j [ ]Jj - Kj φi Let the closed shell Fock potential be written as: Vij = ∑ k ?? ??2< >ik|jk - < >ik|kj , and the 1e- component as: hij = < φi| - 12 ?2 - ∑ A ZA|r - RA| |φj > , and the delta as: δij = < >i|j , so that: hij + Vij = δijεi. using: φi = ∑ μ Cμiχμ , φj = ∑ ν Cνjχν , and φk = ∑ γ Cγkχγ , and transforming from the MO to AO basis we obtain: Vij = ∑ kμγνκ CμiCγkCνjCκk?? ??2< >μγ|νκ - < >μγ|κν = ∑ kμγνκ (CγkCκk)(CμiCνj)?? ??2< >μγ|νκ - < >μγ|κν = ∑ μν (CμiCνj) Vμν where, Vμν = ∑ γκ Pγκ?? ??2< >μγ|νκ - < >μγ|κν , and Pγκ = ∑ k (CγkCκk) , hij = ∑ μν (CμiCνj) hμν , where 100 hμν = < χμ| - 12 ?2 - ∑ A ZA|r - RA| |χν > , and δij = < >i|j = ∑ μν (CμiSμνCνj) . So, hij + Vij = δijεj becomes: ∑ μν (CμiCνj) hμν + ∑ μν (CμiCνj) Vμν = ∑ μν (CμiSμνCνj) εj , ∑ μν (CμiSμνCνj) εj - ∑ μν (CμiCνj) hμν - ∑ μν (CμiCνj) Vμν = 0 for all i,j ∑ μν Cμi[ ]εjSμν - hμν - Vμν Cνj = 0 for all i,j Therefore, ∑ ν [ ]hμν + Vμν - εjSμν - Cνj = 0 This is FC = SCE in the AO basis. 39. The Slater Condon rule for zero (spin orbital) difference with N electrons in N spin orbitals is: 101 E = < >|H + G| = ∑ i N < >φi|h|φi + ∑ i>j N ?? ??< >φiφj|g|φiφj - < >φiφj|g|φjφi = ∑ i hii + ∑ i>j ( )gijij - gijji = ∑ i hii + 12∑ ij ( )gijij - gijji If all orbitals are doubly occupied and we carry out the spin integration we obtain: E = 2 ∑ i occ hii + ∑ ij occ ( )2gijij - gijji , where i and j now refer to orbitals (not spin-orbitals). 40. If the occupied orbitals obey Fφk = εkφk , then the expression for E in problem 39 can be rewritten as. E = ∑ i occ ?? ?? ?? ??hii + ∑ j occ ( )2gijij - gijji + ∑ i occ hii We recognize the closed shell Fock operator expression and rewrite this as: E = ∑ i occ Fii + ∑ i occ hii = ∑ i occ ( )εi + hii 102 41. I will use the QMIC software to do this problem. Lets just start from the beginning. Get the starting "guess" MO coefficients on disk. Using the program MOCOEFS it asks us for the first and second MO vectors. We input 1, 0 for the first mo (this means that the first MO is 1.0 times the He 1s orbital plus 0.0 times the H 1s orbital; this bonding MO is more likely to be heavily weighted on the atom having the higher nuclear charge) and 0, 1 for the second. Our beginning LCAO-MO array looks like: ??? ? ?? ??1.0 0.0 0.0 1.0 and is placed on disk in a file we choose to call "mocoefs.dat". We also put the AO integrals on disk using the program RW_INTS. It asks for the unique one- and two- electron integrals and places a canonical list of these on disk in a file we choose to call "ao_integrals.dat". At this point it is useful for us to step back and look at the set of equations which we wish to solve: FC = SCE. The QMIC software does not provide us with a so-called generalized eigenvalue solver (one that contains an overlap matrix; or metric), so in order to use the diagonalization program that is provided we must transform this equation (FC = SCE) to one that looks like (F'C' = C'E). We do that in the following manner: Since S is symmetric and positive definite we can find an S- 1 2 such that S- 1 2 S+ 1 2 = 1, S- 1 2 S = S+ 1 2 , etc. 103 rewrite FC = SCE by inserting unity between FC and multiplying the whole equation on the left by S- 1 2 . This gives: S- 1 2 FS- 1 2 S+ 1 2 C = S- 1 2 SCE = S+ 1 2 CE. Letting: F' = S- 1 2 FS- 1 2 C' = S+ 1 2 C, and inserting these expressions above give: F'C' = C'E Note, that to get the next iteration’s MO coefficients we must calculate C from C': C' = S+ 1 2 C, so, multiplying through on the left by S- 1 2 gives: S- 1 2 C' = S- 1 2 S+ 1 2 C = C This will be the method we will use to solve our fock equations. Find S- 1 2 by using the program FUNCT_MAT (this program generates a function of a matrix). This program will ask for the elements of the S array and write to disk a file (name of your choice ... a good name might be "shalf") containing the S- 1 2 array. Now we are ready to begin the iterative Fock procedure. a. Calculate the Fock matrix, F, using program FOCK which reads in the MO coefficients from "mocoefs.dat" and the integrals from "ao_integrals.dat" and writes the resulting Fock matrix to a user specified file (a good filename to use might be something like "fock1"). 104 b. Calculate F' = S- 1 2 FS- 1 2 using the program UTMATU which reads in F and S- 1 2 from files on the disk and writes F' to a user specified file (a good filename to use might be something like "fock1p"). Diagonalize F' using the program DIAG. This program reads in the matrix to be diagonalized from a user specified filename and writes the resulting eigenvectors to disk using a user specified filename (a good filename to use might be something like "coef1p"). You may wish to choose the option to write the eigenvalues (Fock orbital energies) to disk in order to use them at a later time in program FENERGY. Calculate C by using. C = S- 1 2 C'. This is accomplished by using the program MATXMAT which reads in two matrices to be multiplied from user specified files and writes the product to disk using a user specified filename (a good filename to use might be something like "mocoefs.dat"). c. The QMIC program FENERGY calculates the total energy: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν , and ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν . This is the conclusion of one iteration of the Fock procedure ... you may continue by going back to part a. and proceeding onward. d. and e. Results for the successful convergence of this system using the supplied QMIC software are as follows (this data is provided to give the student assurance that 105 they are on the right track; alternatively one could switch to the QMIC program SCF and allow that program to iteratively converge the Fock equations): The one-electron AO integrals: ??? ? ?? ??-2.644200 -1.511300 -1.511300 -1.720100 The two-electron AO integrals: 1 1 1 1 1.054700 2 1 1 1 0.4744000 2 1 2 1 0.5664000 2 2 1 1 0.2469000 2 2 2 1 0.3504000 2 2 2 2 0.6250000 The "initial" MO-AO coefficients: ??? ? ?? ??1.000000 0.000000 0.000000 1.000000 AO overlap matrix (S): ??? ? ?? ??1.000000 0.578400 0.578400 1.000000 S - 1 2 ?? ?? ?? ??1.168032 -0.3720709 -0.3720709 1.168031 106 ************** ITERATION 1 ************** The charge bond order matrix: ??? ? ?? ??1.000000 0.0000000 0.0000000 0.0000000 The Fock matrix (F): ??? ? ?? ??-1.589500 -1.036900 -1.036900 -0.8342001 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.382781 -0.5048679 -0.5048678 -0.4568883 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.604825 -0.2348450 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9153809 -0.4025888 -0.4025888 0.9153810 107 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.9194022 -0.8108231 -0.1296498 1.218985 The one-electron MO integrals: ?? ?? ?? ??-2.624352 -0.1644336 -0.1644336 -1.306845 The two-electron MO integrals: 1 1 1 1 0.9779331 2 1 1 1 0.1924623 2 1 2 1 0.5972075 2 2 1 1 0.1170838 2 2 2 1 -0.0007945194 2 2 2 2 0.6157323 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84219933 108 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.80060530 the difference is: -0.04159403 ************** ITERATION 2 ************** The charge bond order matrix: ??? ? ?? ??0.8453005 0.1192003 0.1192003 0.01680906 The Fock matrix: ??? ? ?? ??-1.624673 -1.083623 -1.083623 -0.8772071 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.396111 -0.5411037 -0.5411037 -0.4798213 The eigenvalues of this matrix (Fock orbital energies) are: 109 [ ]-1.646972 -0.2289599 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9072427 -0.4206074 -0.4206074 0.9072427 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.9031923 -0.8288413 -0.1537240 1.216184 The one-electron MO integrals: ?? ?? ?? ??-2.617336 -0.1903475 -0.1903475 -1.313861 The two-electron MO integrals: 1 1 1 1 0.9626070 2 1 1 1 0.1949828 110 2 1 2 1 0.6048143 2 2 1 1 0.1246907 2 2 2 1 0.003694540 2 2 2 2 0.6158437 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84349298 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.83573675 the difference is: -0.00775623 ************** ITERATION 3 ************** The charge bond order matrix: ??? ? ?? ??0.8157563 0.1388423 0.1388423 0.02363107 111 The Fock matrix: ??? ? ?? ??-1.631153 -1.091825 -1.091825 -0.8853514 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.398951 -0.5470731 -0.5470730 -0.4847007 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.654745 -0.2289078 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9058709 -0.4235546 -0.4235545 0.9058706 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.9004935 -0.8317733 -0.1576767 1.215678 The one-electron MO integrals: 112 ?? ?? ?? ??-2.616086 -0.1945811 -0.1945811 -1.315112 The two-electron MO integrals: 1 1 1 1 0.9600707 2 1 1 1 0.1953255 2 1 2 1 0.6060572 2 2 1 1 0.1259332 2 2 2 1 0.004475587 2 2 2 2 0.6158972 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84353018 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84225941 113 the difference is: -0.00127077 ************** ITERATION 4 ************** The charge bond order matrix: ??? ? ?? ??0.8108885 0.1419869 0.1419869 0.02486194 The Fock matrix: ??? ? ?? ??-1.632213 -1.093155 -1.093155 -0.8866909 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.399426 -0.5480287 -0.5480287 -0.4855191 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.656015 -0.2289308 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9056494 -0.4240271 -0.4240271 0.9056495 114 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.9000589 -0.8322428 -0.1583111 1.215595 The one-electron MO integrals: ?? ?? ?? ??-2.615881 -0.1952594 -0.1952594 -1.315315 The two-electron MO integrals: 1 1 1 1 0.9596615 2 1 1 1 0.1953781 2 1 2 1 0.6062557 2 2 1 1 0.1261321 2 2 2 1 0.004601604 2 2 2 2 0.6159065 The closed shell Fock energy from formula: 115 ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84352922 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84332418 the difference is: -0.00020504 ************** ITERATION 5 ************** The charge bond order matrix: ??? ? ?? ??0.8101060 0.1424893 0.1424893 0.02506241 The Fock matrix: ??? ? ?? ??-1.632385 -1.093368 -1.093368 -0.8869066 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.399504 -0.5481812 -0.5481813 -0.4856516 116 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.656219 -0.2289360 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9056138 -0.4241026 -0.4241028 0.9056141 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.8999892 -0.8323179 -0.1584127 1.215582 The one-electron MO integrals: ?? ?? ?? ??-2.615847 -0.1953674 -0.1953674 -1.315348 The two-electron MO integrals: 1 1 1 1 0.9595956 117 2 1 1 1 0.1953862 2 1 2 1 0.6062872 2 2 1 1 0.1261639 2 2 2 1 0.004621811 2 2 2 2 0.6159078 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84352779 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84349489 the difference is: -0.00003290 ************** ITERATION 6 ************** 118 The charge bond order matrix: ??? ? ?? ??0.8099805 0.1425698 0.1425698 0.02509460 The Fock matrix: ??? ? ?? ??-1.632412 -1.093402 -1.093402 -0.8869413 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.399517 -0.5482056 -0.5482056 -0.4856730 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.656253 -0.2289375 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9056085 -0.4241144 -0.4241144 0.9056086 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.8999786 -0.8323296 -0.1584283 1.215580 119 The one-electron MO integrals: ?? ?? ?? ??-2.615843 -0.1953846 -0.1953846 -1.315353 The two-electron MO integrals: 1 1 1 1 0.9595859 2 1 1 1 0.1953878 2 1 2 1 0.6062925 2 2 1 1 0.1261690 2 2 2 1 0.004625196 2 2 2 2 0.6159083 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84352827 from formula: 120 ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84352398 the difference is: -0.00000429 ************** ITERATION 7 ************** The charge bond order matrix: ??? ? ?? ??0.8099616 0.1425821 0.1425821 0.02509952 The Fock matrix: ??? ? ?? ??-1.632416 -1.093407 -1.093407 -0.8869464 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.399519 -0.5482093 -0.5482092 -0.4856761 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.656257 -0.2289374 Their corresponding eigenvectors (C' = S + 1 2 * C) are: 121 ?? ?? ?? ??-0.9056076 -0.4241164 -0.4241164 0.9056077 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.8999770 -0.8323317 -0.1584310 1.215580 The one-electron MO integrals: ?? ?? ?? ??-2.615843 -0.1953876 -0.1953876 -1.315354 The two-electron MO integrals: 1 1 1 1 0.9595849 2 1 1 1 0.1953881 2 1 2 1 0.6062936 2 2 1 1 0.1261697 2 2 2 1 0.004625696 2 2 2 2 0.6159083 122 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84352922 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84352827 the difference is: -0.00000095 ************** ITERATION 8 ************** The charge bond order matrix: ??? ? ?? ??0.8099585 0.1425842 0.1425842 0.02510037 The Fock matrix: ??? ? ?? ??-1.632416 -1.093408 -1.093408 -0.8869470 123 S - 1 2 F S - 1 2 ?? ?? ?? ??-1.399518 -0.5482103 -0.5482102 -0.4856761 The eigenvalues of this matrix (Fock orbital energies) are: [ ]-1.656258 -0.2289368 Their corresponding eigenvectors (C' = S + 1 2 * C) are: ?? ?? ?? ??-0.9056074 -0.4241168 -0.4241168 0.9056075 The "new" MO-AO coefficients (C = S - 1 2 * C'): ?? ?? ?? ??-0.8999765 -0.8323320 -0.1584315 1.215579 The one-electron MO integrals: ?? ?? ?? ??-2.615842 -0.1953882 -0.1953882 -1.315354 124 The two-electron MO integrals: 1 1 1 1 0.9595841 2 1 1 1 0.1953881 2 1 2 1 0.6062934 2 2 1 1 0.1261700 2 2 2 1 0.004625901 2 2 2 2 0.6159081 The closed shell Fock energy from formula: ∑ kl 2<k|h|k> + 2<kl|kl> - <kl|lk> + ∑ μ>ν ZμZνRμν = -2.84352827 from formula: ∑ k εk + <k|h|k> + ∑ μ>ν ZμZνRμν = -2.84352827 the difference is: 0.00000000 f. In looking at the energy convergence we see the following: 125 Iter Formula 1 Formula 2 1 -2.84219933 -2.80060530 2 -2.84349298 -2.83573675 3 -2.84353018 -2.84225941 4 -2.84352922 -2.84332418 5 -2.84352779 -2.84349489 6 -2.84352827 -2.84352398 7 -2.84352922 -2.84352827 8 -2.84352827 -2.84352827 If you look at the energy differences (SCF at iteration n - SCF converged) and plot this data versus iteration number, and do a 5th order polynomial fit, we see the following: 0 2 4 6 8 10 0.00 0.01 0.02 0.03 0.04 0.05 Iteration SCF(iter) - SCF(conv) y = 0.144 - 0.153x + 0.063x^2 - 0.013x^3 + 0.001x^4 R = 1.00 126 In looking at the polynomial fit we see that the convergence is primarily linear since the coefficient of the linear term is much larger than those of the cubic and higher terms. g. The converged SCF total energy calculated using the result of problem 40 is an upper bound to the ground state energy, but, during the iterative procedure it is not. Only at convergence does the expectation value of the Hamiltonian for the Hartree Fock determinant become equal to that given by the equation in problem 40. h. Yes, the 1σ2 configuration does dissociate properly because at at R→∞ the lowest energy state is He + H+, which also has a 1σ2 orbital occupancy (i.e., 1s2 on He and 1s0 on H+). 42. 2. At convergence the MO coefficients are: φ1 = ??? ? ?? ??-0.8999765 -0.1584315 φ2 = ?? ?? ?? ??-0.8323320 1.215579 and the integrals in this MO basis are: h11 = -2.615842 h21 = -0.1953882 h22 = -1.315354 g1111 = 0.9595841 g2111 = 0.1953881 g2121 = 0.6062934 g2211 = 0.1261700 g2221 = 004625901 g2222 = 0.6159081 a. H = ??? ? ?? ??<1σ2|H|1σ2> <1σ2|H|2σ2> <2σ2|H|1σ2> <2σ2|H|2σ2> = ?? ?? ?? ??2h11 + g1111 g1122 g1122 2h22 + g2222 127 = ??? ? ?? ??2*-2.615842 + 0.9595841 0.1261700 0.1261700 2*-1.315354 + 0.6159081 = ??? ? ?? ??-4.272100 0.126170 0.126170 -2.014800 b. The eigenvalues are E1 = -4.279131 and E2 = -2.007770. The corresponding eigenvectors are: C1 = ??? ? ?? ??-.99845123 0.05563439 , C2 = ?? ?? ?? ??0.05563438 0.99845140 c. 1 2 ? ?? ?????? ?????? ???a12φ 1 + b 1 2φ 2 α? ?? ???a12φ 1 - b 1 2φ 2 β + ? ?? ?????? ???a12φ 1 - b 1 2φ 2 α? ?? ???a12φ 1 + b 1 2φ 2 β = 12 2 ?? ? ?? ? ?? ? ?? ? a 1 2φ 1 + b 1 2φ 2 ? ?? ???a12φ 1 - b 1 2φ 2 + ? ?? ???a12φ 1 - b 1 2φ 2 ? ?? ???a12φ 1 + b 1 2φ 2 (αβ - βα) = 12( )aφ1φ1 - bφ2φ2 (αβ - βα) = a| |φ1αφ1β - b| |φ2αφ2β . d. The third configuration |1σ2σ| = 12[ ]|1α2β| - |1β2α| , Adding this configuration to the previous 2x2 CI results in the following 3x3 'full' CI: H = ?? ?? ?? ??<1σ 2|H|1σ2> <1σ2|H|2σ2> <1σ2|H|1σ2σ> <2σ2|H|1σ2> <2σ2|H|2σ2> <2σ2|H|1σ2σ> <1σ2σ|H|1σ2> <2σ2|H|1σ2σ> <1σ2σ|H|1σ2σ> 128 = ? ? ? ? ? ? ? ?2h11 + g1111 g1122 1 2[ ]2h12 + 2g2111 g1122 2h22 + g2222 12[ ]2h12 + 2g2221 1 2[ ]2h12 + 2g2111 1 2[ ]2h12 + 2g2221 h11 + h22 + g2121 + g2211 Evaluating the new matrix elements: H13 = H31 = 2 *(-0.1953882 + 0.1953881) = 0.0 H23 = H32 = 2 *(-0.1953882 + 0.004626) = -0.269778 H33 = -2.615842 - 1.315354 + 0.606293 + 0.126170 = -3.198733 = ?? ?? ?? ??-4.272100 0.126170 0.00.126170 -2.014800 -0.269778 0.0 -0.269778 -3.198733 e. The eigenvalues are E1 = -4.279345, E2 = -3.256612 and E3 = -1.949678. The corresponding eigenvectors are: C1 = ?? ?? ?? ??-0.998252800.05732290 0.01431085 , C2 = ?? ?? ?? ??-0.02605343-0.20969283 -0.97742000 , C3 = ?? ?? ?? ??-0.05302767-0.97608540 0.21082004 f. We need the non-vanishing matrix elements of the dipole operator in the MO basis. These can be obtained by calculating them by hand. They are more easily obtained by using the TRANS program. Put the 1e- AO integrals on disk by running the 129 program RW_INTS. In this case you are inserting z11 = 0.0, z21 = 0.2854, and z22 = 1.4 (insert 0.0 for all the 2e- integrals) ... call the output file "ao_dipole.ints" for example. The converged MO-AO coefficients should be in a file ("mocoefs.dat" is fine). The transformed integrals can be written to a file (name of your choice) for example "mo_dipole.ints". These matrix elements are: z11 = 0.11652690, z21 = -0.54420990, z22 = 1.49117320 The excitation energies are E2 - E1 = -3.256612 - -4.279345 = 1.022733, and E3 - E1 = -1.949678.- -4.279345 = 2.329667. Using the Slater-Conden rules to obtain the matrix elements between configurations we obtain: Hz = ?? ?? ?? ??<1σ 2|z|1σ2> <1σ2|z|2σ2> <1σ2|z|1σ2σ> <2σ2|z|1σ2> <2σ2|z|2σ2> <2σ2|z|1σ2σ> <1σ2σ|z|1σ2> <2σ2|z|1σ2σ> <1σ2σ|z|1σ2σ> = ? ? ? ? ? ? ? ?2z11 0 1 2[ ]2z12 0 2z22 12[ ]2z12 1 2[ ]2z12 1 2[ ]2z12 z11 + z22 = ?? ?? ?? ??0.233054 0 -0.7696290 2.982346 -0.769629 -0.769629 -0.769629 1.607700 Now, <Ψ1|z|Ψ2> = C1THzC2, (this can be accomplished with the program UTMATU) 130 = ?? ? ?? ?-0.99825280 0.05732290 0.01431085 T ?? ? ?? ?0.233054 0 -0.769629 0 2.982346 -0.769629 -0.769629 -0.769629 1.607700 ?? ? ?? ?-0.02605343 -0.20969283 -0.97742000 = -.757494 and, <Ψ1|z|Ψ3> = C1THzC3 = ?? ? ?? ?-0.99825280 0.05732290 0.01431085 T ?? ? ?? ?0.233054 0 -0.769629 0 2.982346 -0.769629 -0.769629 -0.769629 1.607700 ?? ? ?? ?-0.05302767 -0.97608540 0.21082004 = 0.014322 g. Using the converged coefficients the orbital energies obtained from solving the Fock equations are ε1 = -1.656258 and ε2 = -0.228938. The resulting expression for the PT first-order wavefunction becomes: |1σ2>(1) = - g22112(ε2 - ε1) |2σ2> |1σ2>(1) = - 0.1261702(-0.228938 + 1.656258) |2σ2> |1σ2>(1) = -0.0441982|2σ2> h. As you can see from part c., the matrix element <1σ2|H|1σ2σ> = 0 (this is also a result of the Brillouin theorem) and hence this configuration does not enter into the first-order wavefunction. i. |0> = |1σ2> - 0.0441982|2σ2>. To normalize we divide by: [ ]1 + (0.0441982)2 = 1.0009762 131 |0> = 0.999025|1σ2> - 0.044155|2σ2> In the 2x2 CI we obtained: |0> = 0.99845123|1σ2> - 0.05563439|2σ2> j. The expression for the 2nd order RSPT is: E(2) = - |g2211| 2 2(ε2 - ε1) = - 0.1261702 2(-0.228938 + 1.656258) = -0.005576 au Comparing the 2x2 CI energy obtained to the SCF result we have: -4.279131 - (-4.272102) = -0.007029 au 43. STO total energy: -2.8435283 STO3G total energy -2.8340561 3-21G total energy -2.8864405 The STO3G orbitals were generated as a best fit of 3 primitive Gaussians (giving 1 CGTO) to the STO. So, STO3G can at best reproduce the STO result. The 3-21G orbitals are more flexible since there are 2 CGTOs per atom. This gives 4 orbitals (more parameters to optimize) and a lower total energy. 44. R HeH+ Energy H2 Energy 1.0 -2.812787056 -1.071953297 1.2 -2.870357513 -1.113775015 132 1.4 -2.886440516 -1.122933507 1.6 -2.886063576 -1.115567684 1.8 -2.880080938 -1.099872589 2.0 -2.872805595 -1.080269098 2.5 -2.856760263 -1.026927710 10.0 -2.835679293 -0.7361705303 Plotting total energy vs. geometry for HeH+: 0 2 4 6 8 10 12 -2.90 -2.88 -2.86 -2.84 -2.82 -2.80 R (au) Total Energy (au) Plotting total energy vs. geometry for H2: 133 0 2 4 6 8 10 12 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 Internuclear Distance (au) Total Energy (au) For HeH+ at R = 10.0 au, the eigenvalues of the converged Fock matrix and the corresponding converged MO-AO coefficients are: -.1003571E+01 -.4961988E+00 .5864846E+00 .1981702E+01 .4579189E+00 -.8245406E-05 .1532163E-04 .1157140E+01 .6572777E+00 -.4580946E-05 -.6822942E-05 -.1056716E+01 -.1415438E-05 .3734069E+00 .1255539E+01 -.1669342E-04 .1112778E-04 .7173244E+00 -.1096019E+01 .2031348E-04 Notice that this indicates that orbital 1 is a combination of the s functions on He only (dissociating properly to He + H+). 134 For H2 at R = 10.0 au, the eigenvalues of the converged Fock matrix and the corresponding converged MO-AO coefficients are: -.2458041E+00 -.1456223E+00 .1137235E+01 .1137825E+01 .1977649E+00 -.1978204E+00 .1006458E+01 -.7903225E+00 .5632566E+00 -.5628273E+00 -.8179120E+00 .6424941E+00 .1976312E+00 .1979216E+00 .7902887E+00 .1006491E+01 .5629326E+00 .5631776E+00 -.6421731E+00 -.8181460E+00 Notice that this indicates that orbital 1 is a combination of the s functions on both H atoms (dissociating improperly; equal probabilities of H2 dissociating to two neutral atoms or to a proton plus hydride ion). 45. The H2 CI result: R 1Σg+ 3Σu+ 1Σu+ 1Σg+ 1.0 -1.074970 -0.5323429 -0.3997412 0.3841676 1.2 -1.118442 -0.6450778 -0.4898805 0.1763018 1.4 -1.129904 -0.7221781 -0.5440346 0.0151913 1.6 -1.125582 -0.7787328 -0.5784428 -0.1140074 1.8 -1.113702 -0.8221166 -0.6013855 -0.2190144 2.0 -1.098676 -0.8562555 -0.6172761 -0.3044956 135 2.5 -1.060052 -0.9141968 -0.6384557 -0.4530645 5.0 -0.9835886 -0.9790545 -0.5879662 -0.5802447 7.5 -0.9806238 -0.9805795 -0.5247415 -0.5246646 10.0 -0.980598 -0.9805982 -0.4914058 -0.4913532 0 2 4 6 8 10 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 State 1 State 2 State 3 State 4 Internuclear Distance (au) Total Energy (au) For H2 at R = 1.4 au, the eigenvalues of the Hamiltonian matrix and the corresponding determinant amplitudes are: determinant -1.129904 -0.722178 -0.544035 0.015191 136 |1σgα1σgβ| 0.99695 0.00000 0.00000 0.07802 |1σgβ1σuα| 0.00000 0.70711 0.70711 0.00000 |1σgα1σuβ| 0.00000 0.70711 -0.70711 0.00000 |1σuα1σuβ| -0.07802 0.00000 0.00000 0.99695 This shows, as expected, the mixing of the first 1Σg+ (1σg2) and the 2nd 1Σg+ (1σu2) determinants in the first and fourth states, and the 3Σu+ = ( 1 2( )|1σgβ1σuα| + |1σgα1σuβ| ), and 1Σu+= ( 12( )|1σgβ1σuα| - |1σgα1σuβ| ) states as the second and third states. Also notice that the first 1Σg+ state has coefficients (0.99695 - 0.07802) (note specifically the + - combination) and the second 1Σg+ state has the opposite coefficients with the same signs (note specifically the + + combination). The + + combination always gives a higher energy than the + - combination. 46. F atoms have 1s22s22p5 2P ground electronic states that are split by spin-orbit coupling into 2P3/2 and 2P1/2 states that differ by only 0.05 eV in energy. 137 a. The degeneracy of a state having a given J is 2J+1, and the J=3/2 state is lower in energy because the 2p orbital shell is more than half filled (I learned this in inorganic chemistry class), so qel = 4 exp(-0/kT) + 2 exp(-0.05 eV/kT). 0.05 eV is equivalent to k(500 K), so 0.05/kT = 500/T, hence qel = 4 exp(-0/kT) + 2 exp(-500/T). b. Q = qN/N! so, ln Q = N lnq – lnN! E =kT2 ?lnQ/?T = NkT2 ?lnq/?T = Nk{1000 exp(-500/T)/[4 + 2 exp(-500/T)]} c. Using the fact that kT=0.03eV at T=300°K, make a (qualitative) graph of E /N vs T for T ranging from 100°K to 3000°K. 138 T E/N T = 100 K T = 3000 K 1000k/6 At T = 100 K, E/N is small and equal to 1000k exp(-5)/(4 + 2 exp(-5)). At T = 3000 K, E/N has grown to 1000k exp(-1/6)/(4 + 2 exp(-1/6)) which is approximately 1000k/6. 47. a. The difference between a linear and bent transition state would arise in the vibrational and rotational partition functions. For the linear TS, one has 3N-6 vibrations (recall that 139 one loses one vibration as a reaction coordinate), but for the bent TS, one has 3N-7 vibrations. For the linear TS, one has 2 rotational axes, and for the bent TS, one has 3. So the ratio of rate constants will reduce to ratios of vibration and rotation partition functions. In particular, one will have klinear/kbent = (qvib3N-6 qrot2/qvib3N-7qrot3) = (qvib/qrot). b. Using qt ~ 108, qr ~ 102, qv ~ 1, I would expect klinear/kbent to be of the order of 1/102 = 10-2. 140 48. Constructing the Slater determinant corresponding to the "state" 1s(α)1s(α) with the rows labeling the orbitals and the columns labeling the electron gives: |1sα1sα| = 12!??? ? ?? ??1sα(1) 1sα(2) 1sα(1) 1sα(2) = 12 ( )1sα(1)1sα(2) - 1sα(1)1sα(2) = 0 49. Starting with the MS=1 3S state (which in a "box" for this ML=0, MS=1 case would contain only one product function; |1sα2sα|) and applying S- gives: S- 3S(S=1,MS=1) = 1(1 + 1) - 1(1 - 1) h～ 3S(S=1,MS=0) = h～ 2 3S(S=1,MS=0) = ( )S-(1) + S-(2) |1sα2sα| = S-(1)|1sα2sα| + S-(2)|1sα2sα| = h～ 12?? ? ?? ?1 2 + 1 - 1 2?? ? ?? ?1 2 - 1 |1sβ2sα| + h～ 12?? ? ?? ?1 2 + 1 - 1 2?? ? ?? ?1 2 - 1 |1sα2sβ| 141 = h～ ( )|1sβ2sα| + |1sα2sβ| So, h～ 2 3S(S=1,MS=0) = h～ ( )|1sβ2sα| + |1sα2sβ| 3S(S=1,MS=0) = 12 ( )|1sβ2sα| + |1sα2sβ| The three triplet states are then: 3S(S=1,MS=1)= |1sα2sα|, 3S(S=1,MS=0) = 1 2 ( )|1sβ2sα| + |1sα2sβ| , and 3S(S=1,MS=-1) = |1sβ2sβ|. The singlet state which must be constructed orthogonal to the three singlet states (and in particular to the 3S(S=1,MS=0) state) can be seen to be: 1S(S=0,MS=0) = 1 2 ( )|1sβ2sα| - |1sα2sβ| . Applying S2 and Sz to each of these states gives: Sz |1sα2sα| = ( )Sz(1) + Sz(2) |1sα2sα| = Sz(1)|1sα2sα| + Sz(2))|1sα2sα| = h～ ?? ? ?? ?1 2 |1sα2sα| + h～ ?? ? ?? ?1 2 |1sα2sα| = h～ |1sα2sα| S2 |1sα2sα| = (S-S+ + Sz2 + h～ Sz) |1sα2sα| = S-S+|1sα2sα| + Sz2|1sα2sα| + h～ Sz|1sα2sα| = 0 + h～ 2 |1sα2sα| + h～ 2|1sα2sα| = 2h～ 2 |1sα2sα| 142 Sz 12 ( )|1sβ2sα| + |1sα2sβ| = ( )Sz(1) + Sz(2) 12 ( )|1sβ2sα| + |1sα2sβ| = 12 ( )Sz(1) + Sz(2) |1sβ2sα| + 12 ( )Sz(1) + Sz(2) |1sα2sβ| = 12 ?? ? ?? ? h～ ?? ? ?? ? -12 + h～ ?? ? ?? ?1 2 |1sβ2sα| + 12 ?? ? ?? ? h～ ?? ? ?? ?1 2 + h～ ?? ? ?? ? -12 |1sα2sβ| = 0h～ 12 ( )|1sβ2sα| + |1sα2sβ| S2 12 ( )|1sβ2sα| + |1sα2sβ| = (S-S+ + Sz2 + h～ Sz) 12 ( )|1sβ2sα| + |1sα2sβ| = S-S+ 12 ( )|1sβ2sα| + |1sα2sβ| = 12( )S-(S+(1) + S+(2))|1sβ2sα| + S-(S+(1) + S+(2))|1sα2sβ| = 12( )S- h～ |1sα2sα| + S- h～ |1sα2sα| = 2 h～ 12( )(S-(1) + S-(2))|1sα2sα| = 2 h～ 12( )h～|1sβ2sα| + h～|1sα2sβ| = 2 h～ 2 12( )|1sβ2sα| + |1sα2sβ| Sz |1sβ2sβ| = ( )Sz(1) + Sz(2) |1sβ2sβ| = Sz(1)|1sβ2sβ| + Sz(2))|1sβ2sβ| 143 = h～ ?? ? ?? ? -12 |1sβ2sβ| + h～ ?? ? ?? ? -12 |1sβ2sβ| = -h～ |1sβ2sβ| S2 |1sβ2sβ| = (S+S- + Sz2 - h～ Sz) |1sβ2sβ| = S+S-|1sβ2sβ| + Sz2|1sβ2sβ| - h～ Sz|1sβ2sβ| = 0 + h～ 2 |1sβ2sβ| + h～ 2|1sβ2sβ| = 2h～ 2 |1sβ2sβ| Sz 12 ( )|1sβ2sα| - |1sα2sβ| = ( )Sz(1) + Sz(2) 12 ( )|1sβ2sα| - |1sα2sβ| = 12 ( )Sz(1) + Sz(2) |1sβ2sα| - 12 ( )Sz(1) + Sz(2) |1sα2sβ| = 12 ?? ? ?? ? h～ ?? ? ?? ? -12 + h～ ?? ? ?? ?1 2 |1sβ2sα| - 12 ?? ? ?? ? h～ ?? ? ?? ?1 2 + h～ ?? ? ?? ? -12 |1sα2sβ| = 0h～ 12 ( )|1sβ2sα| - |1sα2sβ| S2 12 ( )|1sβ2sα| - |1sα2sβ| = (S-S+ + Sz2 + h～ Sz) 12 ( )|1sβ2sα| - |1sα2sβ| = S-S+ 12 ( )|1sβ2sα| - |1sα2sβ| = 12( )S-(S+(1) + S+(2))|1sβ2sα| - S-(S+(1) + S+(2))|1sα2sβ| = 12( )S- h～ |1sα2sα| - S- h～ |1sα2sα| 144 = 0 h～ 12( )(S-(1) + S-(2))|1sα2sα| = 0 h～ 12( )h～|1sβ2sα| - h～|1sα2sβ| = 0 h～ 2 12( )|1sβ2sα| - |1sα2sβ| 50. As shown in problem 22c, for two equivalent pi electrons one obtains six states: 1? (ML=2); one state (MS=0), 1? (ML=-2); one state (MS=0), 1Σ (ML=0); one state (MS=0), and 3Σ (ML=0); three states (MS=1,0, and -1). By inspecting the "box" in problem 22c, it should be fairly straightforward to write down the wavefunctions for each of these: 1? (ML=2); |pi1αpi1β| 1? (ML=-2); |pi-1αpi-1β| 1Σ (ML=0); 1 2( )|pi1βpi-1α| - |pi1αpi-1β| 3Σ (ML=0, MS=1); |pi1αpi-1α| 3Σ (ML=0, MS=0); 1 2( )|pi1βpi-1α| + |pi1αpi-1β| 145 3Σ (ML=0, MS=-1); |pi1βpi-1β| 51. We can conveniently couple another s electron to the states generated from the 1s12s1 configuration: 3S(L=0, S=1) with 3s1(L=0, S=12 ) giving: L=0, S=32 , 12 ; 4S (4 states) and 2S (2 states). 1S(L=0, S=0) with 3s1(L=0, S=12 ) giving: L=0, S=12 ; 2S (2 states). Constructing a "box" for this case would yield: 146 ML MS 0 3 2 |1sα2sα3sα| 1 2 |1sα2sα3sβ|, |1sα2sβ3sα|, |1sβ2sα3sα| One can immediately identify the wavefunctions for two of the quartets (they are single entries): 4S(S=32 ,MS=32 ): |1sα2sα3sα| 4S(S=32 ,MS=-32 ): |1sβ2sβ3sβ| Applying S- to 4S(S=32 ,MS=32 ) yields: S-4S(S=32 ,MS=32 ) = h～ 32(32 + 1) - 32(32 - 1) 4S(S=32 ,MS=12 ) = h～ 3 4S(S=32 ,MS=12 ) S-|1sα2sα3sα| = h～ ( )|1sβ2sα3sα| + |1sα2sβ3sα| + |1sα2sα3sβ| So, 4S(S=32 ,MS=12 ) = 13 ( )|1sβ2sα3sα| + |1sα2sβ3sα| + |1sα2sα3sβ| Applying S+ to 4S(S=32 ,MS=-32 ) yields: S+4S(S=32 ,MS=-32 ) = h～ 32(32 + 1) - -32(-32 + 1) 4S(S=32 ,MS=-12 ) = h～ 3 4S(S=32 ,MS=-12 ) 147 S+|1sβ2sβ3sβ| = h～ ( )|1sα2sβ3sβ| + |1sβ2sα3sβ| + |1sβ2sβ3sα| So, 4S(S=32 ,MS=-12 ) = 13 ( )|1sα2sβ3sβ| + |1sβ2sα3sβ| + |1sβ2sβ3sα| It only remains to construct the doublet states which are orthogonal to these quartet states. Recall that the orthogonal combinations for systems having three equal components (for example when symmetry adapting the 3 sp2 hybrids in C2v or D3h symmetry) give results of + + +, +2 - -, and 0 + -. Notice that the quartets are the + + + combinations and therefore the doublets can be recognized as: 2S(S=12 ,MS=12 ) = 1 6 ( )|1sβ2sα3sα| + |1sα2sβ3sα| - 2|1sα2sα3sβ| 2S(S=12 ,MS=12 ) = 1 2 ( )|1sβ2sα3sα| - |1sα2sβ3sα| + 0|1sα2sα3sβ| 2S(S=12 ,MS=-12 ) = 1 6 ( )|1sα2sβ3sβ| + |1sβ2sα3sβ| - 2|1sβ2sβ3sα| 2S(S=12 ,MS=-12 ) = 1 3 ( )|1sα2sβ3sβ| - |1sβ2sα3sβ| + 0|1sβ2sβ3sα| 52. As illustrated in problem 24, a p2 configuration (two equivalent p electrons) gives rise to the term symbols: 3P, 1D, and 1S. Coupling an additional electron (3d1) to this p2 configuration will give the desired 1s22s22p23d1 term symbols: 3P(L=1,S=1) with 2D(L=2,S=12 ) generates; L=3,2,1, and S=32 , 12 with term symbols 4F, 2F,4D, 2D,4P, and 2P, 148 1D(L=2,S=0) with 2D(L=2,S=12 ) generates; L=4,3,2,1,0, and S=12 with term symbols 2G, 2F, 2D, 2P, and 2S, 1S(L=0,S=0) with 2D(L=2,S=12 ) generates; L=2 and S=12 with term symbol 2D. 53. The notation used for the Slater Condon rules will be as follows: (a.) zero (spin orbital) difference; < >|F + G| = ∑ i < >φi|f|φi + ∑ i>j ?? ??< >φiφj|g|φiφj - < >φiφj|g|φjφi = ∑ i fii + ∑ i>j ( )gijij - gijji (b.) one (spin orbital) difference (φp ≠ φp'); < >|F + G| = < >φp|f|φp' + ∑ j≠p;p' ?? ??< >φpφj|g|φp'φj - < >φpφj|g|φjφp' = fpp' + ∑ j≠p;p' ( )gpjp'j - gpjjp' (c.) two (spin orbital) differences (φp ≠ φp' and φq ≠ φq'); < >|F + G| = < >φpφq|g|φp'φq' - < >φpφq|g|φq'φp' = gpqp'q' - gpqq'p' (d.) three or more (spin orbital) differences; < >|F + G| = 0 149 i. 3P(ML=1,MS=1) = |p1αp0α| < >|p1αp0α|H|p1αp0α| = <| 10| H | 10|> Using the Slater Condon rule (a.) above (I will denote these SCa-SCd): < >|10|H|10| = f11 + f00 + g1010 - g1001 ii. 3P(ML=0,MS=0) = 12( )|p1αp-1β| + |p1βp-1α| < >3P(ML=0,MS=0)|H|3P(ML=0,MS=0) = 12(< >|p1αp-1β|H|p1αp-1β| + < >|p1αp-1β|H|p1βp-1α| + < >|p1βp-1α|H|p1αp-1β| + < >|p1βp-1α|H|p1βp-1α| ) Evaluating each matrix element gives: < >|p1αp-1β|H|p1αp-1β| = f1α1α + f-1β-1β + g1α-1β1α-1β - g1α-1β-1β1α (SCa) = f11 + f-1-1 + g1-11-1 - 0 < >|p1αp-1β|H|p1βp-1α| = g1α-1β1β-1α - g1α-1β-1α1β (SCc) = 0 - g1-1-11 < >|p1βp-1α|H|p1αp-1β| = g1β-1α1α-1β - g1β-1α-1β1α (SCc) = 0 - g1-1-11 < >|p1βp-1α|H|p1βp-1α| = f1β1β + f-1α-1α + g1β-1α1β-1α - g1β-1α-1α1β (SCa) 150 = f11 + f-1-1 + g1-11-1 - 0 Substitution of these expressions give: < >3P(ML=0,MS=0)|H|3P(ML=0,MS=0) = 12 (f11 + f-1-1 + g1-11-1 - g1-1-11 - g1-1-11 + f11 + f-1-1 + g1-11-1) = f11 + f-1-1 + g1-11-1 - g1-1-11 iii. 1S(ML=0,MS=0); 13(|p0αp0β| - |p1αp-1β| - |p-1αp1β|) < >1S(ML=0,MS=0)|H|1S(ML=0,MS=0) = 13(< >|p0αp0β|H|p0αp0β| - < >|p0αp0β|H|p1αp-1β| - < >|p0αp0β|H|p-1αp1β| - < >|p1αp-1β|H|p0αp0β| + < >|p1αp-1β|H|p1αp-1β| + < >|p1αp-1β|H|p-1αp1β| - < >|p-1αp1β|H|p0αp0β| + < >|p-1αp1β|H|p1αp-1β| + < >|p-1αp1β|H|p-1αp1β| ) Evaluating each matrix element gives: < >|p0αp0β|H|p0αp0β| = f0α0α + f0β0β + g0α0β0α0β - g0α0β0β0α (SCa) = f00 + f00 + g0000 - 0 < >|p0αp0β|H|p1αp-1β| = < >|p1αp-1β|H|p0αp0β| = g0α0β1α-1β - g0α0β-1β1α (SCc) 151 = g001-1 - 0 < >|p0αp0β|H|p-1αp1β| = < >|p-1αp1β|H|p0αp0β| = g0α0β?1α1β - g0α0β1β?1α (SCc) = g00-11 - 0 < >|p1αp-1β|H|p1αp-1β| = f1α1α + f-1β-1β + g1α-1β1α-1β - g1α-1β-1β1α (SCa) = f11 + f-1-1 + g1-11-1 - 0 < >|p1αp-1β|H|p-1αp1β| = < >|p-1αp1β|H|p1αp-1β| = g1α-1β-1α1β - g1α-1β1β-1α (SCc) = g1-1-11 - 0 < >|p-1αp1β|H|p-1αp1β| = f-1α?1α + f1β1β + g-1α1β?1α1β - g-1α1β1β?1α (SCa) = f-1-1 + f11 + g-11-11 - 0 Substitution of these expressions give: < >1S(ML=0,MS=0)|H|1S(ML=0,MS=0) = 13(f00 + f00 + g0000 - g001-1 - g00-11 - g001-1 + f11 + f-1-1 + g1-11-1 + g1-1-11 - g00-11 + g1-1-11 + f-1-1 + f11 + g-11-11) = 13(2f00 + 2f11 + 2f-1-1 + g0000 - 4g001-1 + 2g1-11-1 + 2g1-1-11) iv. 1D(ML=0,MS=0) = 16( )2|p0αp0β| + |p1αp-1β| + |p-1αp1β| 152 Evaluating < >1D(ML=0,MS=0)|H|1D(ML=0,MS=0) we note that all the Slater Condon matrix elements generated are the same as those evaluated in part iii. (the signs for the wavefunction components and the multiplicative factor of two for one of the components, however, are different). < >1D(ML=0,MS=0)|H|1D(ML=0,MS=0) = 16(4f00 + 4f00 + 4g0000 + 2g001-1 + 2g00-11 + 2g001-1 + f11 + f-1-1 + g1-11-1 + g1-1-11 + 2g00-11 + g1-1-11 + f-1-1 + f11 + g-11-11) = 16(8f00 + 2f11 + 2f-1-1 + 4g0000 + 8g001-1 + 2g1-11-1 + 2g1-1-11) 54. i. 1?(ML=2,MS=0) = |pi1αpi1β| < >1?(ML=2,MS=0)|H|1?(ML=2,MS=0) = < >|pi1αpi1β|H|pi1αpi1β| = f1α1α + f1β1β + g1α1β1α1β - g1α1β1β1α (SCa) = f11 + f11 + g1111 - 0 = 2f11 + g1111 153 ii. 1Σ(ML=0,MS=0) = 12( )|pi1αpi-1β| - |pi1βpi-1α| < >3Σ(ML=0,MS=0)|H|3Σ(ML=0,MS=0) = 12(< >|pi1αpi-1β|H|pi1αpi-1β| - < >|pi1αpi-1β|H|pi1βpi-1α| - < >|pi1βpi-1α|H|pi1αpi-1β| + < >|pi1βpi-1α|H|pi1βpi-1α| ) Evaluating each matrix element gives: < >|pi1αpi-1β|H|pi1αpi-1β| = f1α1α + f-1β-1β + g1α-1β1α-1β - g1α-1β-1β1α (SCa) = f11 + f-1-1 + g1-11-1 - 0 < >|pi1αpi-1β|H|pi1βpi-1α| = g1α-1β1β-1α - g1α-1β-1α1β (SCc) = 0 - g1-1-11 < >|pi1βpi-1α|H|pi1αpi-1β| = g1β-1α1α-1β - g1β-1α-1β1α (SCc) = 0 - g1-1-11 < >|pi1βpi-1α|H|pi1βpi-1α| = f1β1β + f-1α-1α + g1β-1α1β-1α - g1β-1α-1α1β (SCa) = f11 + f-1-1 + g1-11-1 - 0 Substitution of these expressions give: < >3Σ(ML=0,MS=0)|H|3Σ(ML=0,MS=0) = 12 (f11 + f-1-1 + g1-11-1+ g1-1-11+ g1-1-11 + f11 + f-1-1 + g1-11-1) = f11 + f-1-1 + g1-11-1+ g1-1-11 iii. 3Σ(ML=0,MS=0) = 12( )|pi1αpi-1β| + |pi1βpi-1α| 154 < >3Σ(ML=0,MS=0)|H|3Σ(ML=0,MS=0) = f11 + f-1-1 + g1-11-1 - 0 < >|pi1αpi-1β|H|pi1βpi-1α| = g1α-1β1β-1α - g1α-1β-1α1β (SCc) = 0 - g1-1-11 < >|pi1βpi-1α|H|pi1αpi-1β| = g1β-1α1α-1β - g1β-1α-1β1α (SCc) = 0 - g1-1-11 < >|pi1βpi-1α|H|pi1βpi-1α| = f1β1β + f-1α-1α + g1β-1α1β-1α - g1β-1α-1α1β (SCa) = f11 + f-1-1 + g1-11-1 - 0 Substitution of these expressions give: < >3Σ(ML=0,MS=0)|H|3Σ(ML=0,MS=0) = 12 (f11 + f-1-1 + g1-11-1- g1-1-11- g1-1-11 + f11 + f-1-1 + g1-11-1) = f11 + f-1-1 + g1-11-1- g1-1-11 55. The order of the answers is J, I, G. K, B, D, E, A, C, H, F 56. 155 p = N/(V-Nb) – N2 a/(kTV2) but p/kT = (?lnQ/?V)T,N so we can integrate to obtain ln Q lnQ = ∫ (p/kT) dV = ∫ [N/(V-Nb) – N2 a/(kTV2)] dV = N ln(V-Nb) + N2a/kT (1/V) So, Q = {(V-Nb)exp[(a/kT) (N/V)]}N 57. a. MD because you need to keep track of how far the molecule moves as a function of time and MC does not deal with time. b. MC is capable of doing this although MD is also. However, MC requires fewer computational steps, so I would prefer to use it. 156 c. MC can do this, as could MD. Again, because MC needs fewer computational steps, I’d use it. Suppose you are carrying out a Monte-Carlo simulation involving 1000 Ar atoms. Further suppose that the potentials are pairwise additive and that your computer requires approximately 50 floating point operations (FPO's) (e.g. multiply, add, divide, etc.) to compute the interaction potential between any pair of atoms d. For each MC move, we must compute only the change in potential energy. To do this, we need to compute only the change in the pair energies that involve the atom that was moved. This will require 999x50 FPOs (the 99 being the number of atoms other than the one that moved). So, for a million MC steps, I would need 106 x 999 x 50 FPOs. At 100 x106 FPOs per second, this will require 495 seconds, or a little over eight minutes. e. 157 Because the statistical fluctuations in MC calculations are proportional to (1/N)1/2, where N is the number of steps taken, I will have to take 4 times as many steps to cut the statistical errors in half. So, this will require 4 x 495 seconds or 1980 seconds. f. If we have one million rather than one thousand atoms, the 495 second calculation of part d would require 999,999/999 times as much time. This ratio arises because the time to compute the change in potential energy accompanying a MC move is proportional to the number of other atoms. So, the calculation would take 495 x (999,999/999) seconds or about 500,000 seconds or about 140 hours. g. We would be taking 10-9s/(10-15 s per step) = 106 MD steps. Each step requires that we compute all forces(-?V?RI,J) between all pairs of atoms. There are 1000x999/2 such pairs. So, to compute all the forces would require (1000x999/2)x 50 FPOs = 2.5 x107 FPOs. So, we will need 158 2.5 x107 FPOs/step x 106 steps/(100 FPOs per second) = 2.5 x105 seconds or about 70 hours. h. The graduate student is 108 times slower than the 100 Mflop computer, so it will take her/him 108 times as long, so 495 x108 seconds or about 1570 years. 58. First, Na has a 2S ground state term symbol whose degeneracy is 2S + 1 = 2. Na2 has a 1Σ ground state whose degeneracy is 1. The symmetry number for Na2 is σ = 2. The D0 value given is 17.3 kcal mol-1. The Kp equilibrium constant would be given in terms of partial pressures as (and then using pV=NkT) Kp = pNa2/pNa2 = (kT)-1 (qNa/V)2/(qNa2/V) in terms of the partition functions. 159 a. qNa = (2pimkT/h2)3/2 V qel qNA2 = (2pim’kT/h2)3/2 V (8pi2IkT/h2) 1/2 [ exp-hν/2kT) (1- exp-hν/kT))-1 exp(De/kT) We can combine the De and the –hν/2kT to obtain the D0 which is what we were given. b. For Na (I will use cgs units in all cases): q/V = (2pi 23 1.66x10-24 1.38 x10-16 1000)3/2 2 = (6.54 x1026) x 2 = 1.31 x1027 For Na2: q/N = 23/2 x (6.54 x1026) (1000/0.221) (1/2) (1-exp(-229/1000))-1 exp(D0/kT) = 1.85 x1027 (2.26 x103) (4.88) (5.96 x103) = 1.22 x1035 So, Kp = [1.22 x1035]/[(1.38 x10-16)(1000) (1.72 x1054) = 0.50 x10-6 dynes cm-2 = 0.50 atm-1. 160 59. The differences in krate will arise from differences in the number of translational, rotational, and vibrational partition functions arising in the adsorbed and gas-phase species. Recall that krate = (kT/h) exp(-E*/kT) [qTS/V]/[(qNO/V) (qCl2/V)] In the gas phase, NO has 3 translations, two rotations, and one vibration Cl2 has 3 translations, two rotations, and one vibration the NOCl2 TS, which is bent, has 3 translations, three rotations, and five vibrations (recall that one vibration is missing and is the reaction coordinate) In the adsorbed state, NO has 2 translations, one rotation, and three vibrations Cl2 has 2 translations, one rotation, and three vibrations the NOCl2 TS, which is bent, has 2 translations, one rotation, and eight vibrations (again, one vibration is missing and is the reaction coordinate). 161 So, in computing the partition function ratio: [qTS/V]/[(qNO/V) (qCl2/V)] for the adsorbed and gas-phase cases, one does not obtain the same number of translational, rotational, and vibrational factors. In particular, the ratio of these factors for the adsorbed and gas-phase cases gives the ratio of rate constants as follows: kad/kgas = (qtrans/V)/qvib which should be of the order of 108 (using the ratio of partition functions as given). Notice that this result suggests that reaction rates can be altered by constraining the reacting species to move freely in lower dimensions even if one does not alter the energetics (e.g., activation energy or thermochemistry).

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