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William A. Goddard, III, wag@kaist.ac.kr

Lecture 7, September 24, 2009. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday.

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William A. Goddard, III, wag@kaist.ac.kr

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  1. Lecture 7, September 24, 2009 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday William A. Goddard, III, wag@kaist.ac.kr WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Senior Assistant: Dr. Hyungjun Kim: linus16@kaist.ac.kr Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist.ac.kr Special assistant: Tod Pascal:tpascal@wag.caltech.edu EEWS-90.502-Goddard-L04

  2. Schedule changes There was no lecture on Sept. 22 because of the EEWS conference Goddard will be traveling Oct 2-11 and will not give the lectures scheduled for Oct. 6 and 8 Consequently an extra lecture will be added at 2pm on Wednesday Sept. 30 and another at 2pm Wednesday Oct 14 L8: Sept. 29, as scheduled L9: Sept. 30, new replaces Oct 6 L10: Oct. 1, as scheduled L11: Oct. 13, as scheduled L12: Oct. 14, new replaces Oct 8 L13: Oct. 15, as scheduled EEWS-90.502-Goddard-L04

  3. Last time EEWS-90.502-Goddard-L04

  4. Contour plots of 3s, 3p, 3d hydrogenic orbitals EEWS-90.502-Goddard-L04

  5. The ground state of He atom Put both electrons in 1s orbitals ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2)(ab-ba) Φ1s = exp(-z r) EHe = 2(½ z2) – 2Zz + (5/8)z, whereJ1s,1s = (5/8) z Applying the variational principle, dE/dz = 0 get z = (Z – 5/16) = 1.6875 E= 2(-½z2) = - z2 = -2.8477 h0 Interpretation: two electrons move independently in the orbital Φ1s= exp(-zr) which has been adjusted to account for the average shielding due to the other electron in this orbital. On the average this other electron is closer to the nucleus about 31% of the time so that the effective charge seen by each electron is 2-0.31=1.69 The total energy is just the sum of the individual energies. EEWS-90.502-Goddard-L04

  6. Now 3rd electron to form Li ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)] = 0 by Pauli principle Thus the 3rd electron must go into 2s or 2p atomic orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2sa)] ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py) Li+: Φ1s= exp(-zr) with z = Z-0.3125 = 2.69  R1s = 1/z = 0.372 a0 = 0.2A EEWS-90.502-Goddard-L04

  7. Add 3rd electron to the 2p orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py) The 2p orbital sees effective charge of Zeff = 3 – 2 = 1, since it goes to zero at z=0, so that there is no shielding of 1s Get size: R2p = n2/Zeff = 4 a0 = 2.12A Energy: e2p = -(Zeff)2/2n2 = -1/8 h0 = -3.40 eV 1s 0.2A 2p 2.12A EEWS-90.502-Goddard-L04

  8. Add the 3rd electron to the 2s orbital 1s 2s 2.12A 0.2A R~0.2A ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py) The 2s orbital must be orthogonal to the 1s, which means that it must have a spherical nodal surface at ~ 0.2A, the size of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0 so that it is not completely shielded by the 1s orbitals. The result is Zeff2s = 3 – 1.72 = 1.28 Size: R2s = n2/Zeff = 3.1 a0 = 1.65A Energy: e2s = -(Zeff)2/2n2 = -0.205 h0 = 5.57 eV EEWS-90.502-Goddard-L04

  9. Li atom excited states Energy MO picture State picture zero DE = 2.2 eV 17700 cm-1 564 nm 1st excited state -0.125 h0 = -3.4 eV (1s)2(2p) 2p -0.205 h0 = -5.6 eV (1s)2(2s) 2s Ground state Exper 671 nm DE = 1.9 eV -2.723 h0 = 74.1 eV 1s EEWS-90.502-Goddard-L04

  10. Aufbau principle for atoms Uuo, 118 Rn, 86 Particularly stable atoms, closed shells Xe, 54 Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, 2 EEWS-90.502-Goddard-L04

  11. Many-electron configurations General aufbau ordering Particularly stable EEWS-90.502-Goddard-L04

  12. General trends along a row of the periodic table As we fill a shell, thus B(2s)2(2p)1 to Ne (2s)2(2p)6 Zeff increases leading to a decrease in the radius ~ n2/Zeff And an increase in the IP ~ (Zeff)2/2n2 Example Zeff2s= 1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne Thus R(2s Li)/R(2s Ne) ~ 4.64/1.28 = 3.6 EEWS-90.502-Goddard-L04

  13. General trends along a column of the periodic table As we go down a colum Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s Things get more complicated The radius ~ n2/Zeff And the IP ~ (Zeff)2/2n2 But the Zeff tends to increase, partially compensating for the change in n so that the atomic sizes increase only slowly as we go down the periodic table and The IP decrease only slowly (in eV): 5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs (13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At 24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn EEWS-90.502-Goddard-L04

  14. EEWS-90.502-Goddard-L04

  15. Transition metals; consider [Ar] plus one electron [IP4s = (Zeff4s )2/2n2 = 4.34 eV  Zeff4s = 2.26; 4s<4p<3d IP4p = (Zeff4p )2/2n2 = 2.73 eV  Zeff4p = 1.79; IP3d = (Zeff3d )2/2n2 = 1.67 eV  Zeff3d = 1.05; IP4s = (Zeff4s )2/2n2 = 11.87 eV  Zeff4s = 3.74; 4s<3d<4p IP3d = (Zeff3d )2/2n2 = 10.17 eV  Zeff3d = 2.59; IP4p = (Zeff4p )2/2n2 = 8.73 eV  Zeff4p = 3.20; IP3d = (Zeff3d )2/2n2 = 24.75 eV  Zeff3d = 4.05; 3d<4s<4p IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04; IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 4.47; K Ca+ Sc++ As the net charge increases the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4 Thus charged system prefers 3d vs 4s EEWS-90.502-Goddard-L04

  16. Transition metals; consider Sc0, Sc+, Sc2+ 3d: IP3d = (Zeff3d )2/2n2 = 24.75 eV  Zeff3d = 4.05; 4s: IP4s = (Zeff4s )2/2n2 = 21.58 eV  Zeff4s = 5.04; 4p: IP4p = (Zeff4p )2/2n2 = 17.01 eV  Zeff4p = 4.47; Sc++ (3d)(4s): IP4s = (Zeff4s )2/2n2 = 12.89 eV  Zeff4s = 3.89; (3d)2: IP3d = (Zeff3d )2/2n2 = 12.28 eV  Zeff3d = 2.85; (3d)(4p): IP4p = (Zeff4p )2/2n2 = 9.66 eV  Zeff4p = 3.37; Sc+ (3d)(4s)2: IP4s = (Zeff4s )2/2n2 = 6.56 eV  Zeff4s = 2.78; (4s)(3d)2: IP3d = (Zeff3d )2/2n2 = 5.12 eV  Zeff3d = 1.84; (3d)(4s)(4p): IP4p = (Zeff4p )2/2n2 = 4.59 eV  Zeff4p = 2.32; Sc For high charge (Sc++) 3d more stable than 4s  (3d)1 For neutral system (Sc) fill 4s:  (3d)(4s)2 This is because increased charge makes the differential shielding for 4s vs 3d less important than the in n quantum number 3 vs 4. EEWS-90.502-Goddard-L04

  17. Implications on transition metals The simple Aufbau principle puts 4s below 3d But increasing the charge tends to prefers 3d vs 4s. Thus Ground state of Sc 2+ , Ti 2+ …..Zn 2+ are all (3d)n For all neutral elements K through Zn the 4s orbital is easiest to ionize. This is because of increase in relative stability of 3d for higher ions The character of transition metals Sc-Cu columns depends on circumstances EEWS-90.502-Goddard-L04

  18. Transtion metal orbitals EEWS-90.502-Goddard-L04

  19. Next More detailed description of first row atoms Li: (2s) Be: (2s)2 B: [Be](2p)1 C: [Be](2p)2 N: [Be](2p)3 O: [Be](2p)4 F: [Be](2p)5 Ne: [Be](2p)6 EEWS-90.502-Goddard-L04

  20. Consider the ground state of B: [Be](2p)1 x Ψ(1,2,3,4,5) = A[Φ Ignore the [Be] core then Can put 1 electron in 2px, 2py, or 2pz each with either up or down spin. Thus get 6 states. We will depict these states by simplified contour diagrams in the xz plane, as at the right. Of course 2py is zero on this plane. Instead we show it as a circle as if you can see just the front part of the lobe sticking out of the paper. 2px z 2pz 2py Because there are 3 degenerate states we denote this as a P state. Because the spin can be +½ or –½, we call it a spin doublet and we denote the overall state as 2P EEWS-90.502-Goddard-L04

  21. New material EEWS-90.502-Goddard-L04

  22. Consider the ground state of C: [Be](2p)2 x x x z z z Ignore the [Be] core then Can put 2 electrons in 2px, 2py, or 2pz each with both up and down spin. Or can put one electron in each of two orbitals: (2px)(2py), (2px)(2px), (2py)(2pz), We will depict these states by simplified contour diagrams in the xz plane, as at the right. (2px)2 (2pz)2 (2px)(2pz) Which state is better? The difference is in the electron-electron repulsion: 1/r12 Clearly two electrons in the same orbital have a much smaller average r12 and hence a much higher e-e repusion. Thus the ground state has each electron in a different 2p orbital EEWS-90.502-Goddard-L04

  23. Consider the states of C: formed from (x)(y), (x)(z), (y)(z) x y Consider first (x)(y): can form two spatial products: Φx(1)Φy(2) and Φy(1)Φx(2) These are not symmetric, thus must combine Φ(1,2)s=φx(1) φy(2) + φy(1) φx(2) Φ(1,2)a= φx(1) φy(2) - φy(1) φx(2) (2px)(2py) Which state is better? The difference is in the electron-electron repulsion: 1/r12 To analyze this, expand the orbitals in terms of the angular coordinates, r,θ,φ Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)+(sinφ1) (cosφ2)] =f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)] Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[(cosφ1)(sinφ2)-(sinφ1) (cosφ2)] =f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)] EEWS-90.502-Goddard-L04

  24. Consider the symmetric and antisymmetric combinations of (x)(y) x (2px)(2py) y Φ(1,2)s= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ1+φ2)] Φ(1,2)a= f(r1)f(r2)(sinθ1)(sinθ2)[sin(φ2 -φ1)] The big difference is that Φ(1,2)a = 0 when φ2 = φ1 and is a maximum for φ2 and φ1 out of phase by p/2. But for Φ(1,2)s the probability of φ2 = φ1 is comparable to that of being out of phase by p/2. Thus the best combination is Φ(1,2)a Combining with the spin parts we get [φx(1) φy(2) + φy(1) φx(2)](ab-ba) or spin = 0 [φx(1) φy(2) - φy(1) φx(2)](ab+ba), also aa and bb or spin = 1 Thus for 2 electrons in orthogonal orbitals, high spin is best because the electrons can never be at same spot at the same time EEWS-90.502-Goddard-L04

  25. Summarizing the states for C atom x (2px)(2py) y Ground state: three triplet states=2L=1. Thus L=1, denote as 3P (xy-yx) ≡ [x(1)y(2)-y(1)x(2)] (xz-zx) (yz-zy) Next state: five singlet states=2L+1. Thus L=2, denote as 1D (xy+yx) (xz+zx) (yz+zy) (xx-yy) (2zz-xx-yy) Highest state: one singlet=2L+1. thus L=0. Denote as 1S (zz+xx+yy) Hund’s rule. Given n electrons distributed among m equivalent orgthogonal orbitals, the ground state is the one with the highest possible spin. Given more than one state with the highest spin, the highest orbital angular momentum is the GS EEWS-90.502-Goddard-L04

  26. Calculating energies for C atom x (2px)(2py) y The energy of xy is Exy = hxx + hyy + Jxy = 2hpp +Jxy Thus the energy of the 3P state is E(3P) = Exy – Kxy = 2hpp +Jxy - Kxy For the (xy+yx) component of the 1D state, we get E(1D) = Exy + Kxy = 2hpp +Jxy + Kxy Whereas for the (xx-yy) component of the 1D state, we get E(1D) = Exx - Kxy = 2hpp +Jxx - Kxy This means that Jxx - Kxy = Jxy + Kxy so that Jxx = Jxy + 2Kxy Also for (2zz-xx-yy) we obtain E = 2hpp +Jxx - Kxy For (zz+xx+yy) we obtain E(1S) = 2hpp + Jxx + 2 Kxy EEWS-90.502-Goddard-L04

  27. Summarizing the energies for C atom E(1S) = 2hpp + Jxx + 2Kxy 3Kxy E(1D) = 2hpp +Jxx - Kxy = 2hpp +Jxy + Kxy 2Kxy E(3P) = 2hpp + Exy – Kxy = 2hpp +Jxy - Kxy EEWS-90.502-Goddard-L04

  28. Comparison with experiment E(1S) 3Kxy E(1D) 2Kxy E(3P) C Si Ge Sn Pb TA’s look up data and list excitation energies in eV and Kxy in eV. Get data from Moore’s tables or paper by Harding and Goddard Ann Rev Phys Chem ~1985) EEWS-90.502-Goddard-L04

  29. Summary ground state for C atom x z (2px)(2pz) x z x z Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pxa)(2pza)] = = A[(1s)2(2s)2(2pxa)(2pza)] =A[(Be)(2pxa)(2pza)] = = A[(xa)(za)] = which we visualize as Ψ(1,2,3,4,5,6)xy = A[(xa)(ya)] which we visualize as Ψ(1,2,3,4,5,6)yz = A[(ya)(za)] which we visualize as Note that we choose to use the xz plane for all 3 wavefunctions, so that the py orbitals look like circles (seeing only the + lobe out of the plane EEWS-90.502-Goddard-L04

  30. Consider the ground state of N: [Be](2p)3 Ignore the [Be] core then Can put one electron in each of three orbitals: (2px)(2py)(2pz) Or can put 2 in 1 and 1 in another: (x)2(y), (x)2(z), (y)2(x), (y)2(z), (z)2(x), (z)2(y) As we saw for C, the best state is (x)(y)(z) because of the lowest ee repulsion. xyz can be combined with various spin functions, but from Hund’s rule we expect A[(xa)(ya)(za)] = [Axyz]aaa to be the ground state. (here Axyz is the antisymmetric combination of x(1)y(2)x(3)] The four symmetric spin functions are aaa, aab + aba + baa, abb + bba + bab, bbb With Ms = 3/2, 1/2, -1/2 -3/2, which refer to as S=3/2 or quartet Since there is only one xyz state = 2L+1 with L=0, we denote it as L=0, leading to the 4S state. EEWS-90.502-Goddard-L04

  31. Energy of the ground state of N: A[(xa)(ya)(za)] = [Axyz]aaa E(2P) 2Kxy E(2D) 4Kxy E(4S) x z Simple product xyz leads to Exyz = 3hpp + Jxy + Jxz + Jyz E(4S) = <xyz|H|A[xyz]>/<xyz|A[xyz]> Denominator = <xyz|A[xyz]> = 1 Numerator = Exyz - Kxy - Kxz – Kyz = E(4S) = 3hpp + (Jxy - Kxy) + (Jxz - Kxz) + (Jyz – Kyz) Pictorial representation of the N ground state =3hpp + 2Jxy + Jxx + 1Kxy TA’s check this =3hpp + 3Jxy + 1Kxy = 3hpp + 2Jxy + Jxx - 1Kxy Since Jxy =Jxz =Jyz and Kxy= Kxz = Kyz =3hpp + 3Jxy - 3Kxy EEWS-90.502-Goddard-L04

  32. Comparison with experiment E(1S) 2Kxy E(1D) 4Kxy E(3P) N P As Sb Bi TA’s look up data and list excitation energies in eV and Kxy in eV. Get data from Moore’s tables or paper by Harding and Goddard Ann Rev Phys Chem ~1985) EEWS-90.502-Goddard-L04

  33. Consider the ground state of O: [Be](2p)4 x x z z Only have 3 orbitals, x, y, and z. thus must have a least one doubly occupied Choices: (z)2(x)(y), (y)2(x)(z), (x)2(z)(y) and (z)2(x)2, (y)2(x)2, (y)2(z)2 Clearly it is better to have two singly occupied orbitals. Just as for C atom, two singly occupied orbitals lead to both a triplet state and a singlet state, but the high spin triplet with the same spin for the two singly occupied orbitals is best (2px)2(2py)(2pz) (2pz)2(2px)2 EEWS-90.502-Goddard-L04

  34. Summary ground state for O atom x z x x z z Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)] = A[(1s)2(2s)2(2pya)(2pyb)(2pxa)(2pza)] = = A[(ya)(yb)(xa)(za)] = which we visualize as (2py)2(2px)(2pz) Ψ(1,2,3,4,5,6)yz = A[(xa)(xb)((ya)(za)] which we visualize as (2px)2(2py)(2pz) Ψ(1,2,3,4,5,6)xy = A[(za)(zb)(((xa)(ya)] which we visualize as (2pz)2(2px)(2py) We have 3 = 2L+1 equivalent spin triplet (S=1) states that we denote as L=1 orbital angular momentum, leading to the 3P state EEWS-90.502-Goddard-L04

  35. Calculating energies for O atom (2py)2(2px)(2pz) The energy of Ψ(1,2,3,4,5,6)xz= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)] = A[[Be](ya)(yb)(xa)(za)] is Exz = E(Be) + 2hyy + hxx + hzz + Jyy+ 2Jxy+ 2Jyx+ Jxz – Kxy – Kyz – Kxz Check: 4 electrons, therefore 4x3/2 = 6 coulomb interactions 3 up-spin electrons, therefor 3x2/2 = 3 exchange interactions Other ways to group energy terms Exz = 4hpp + Jyy + (2Jxy – Kxy) + (2Jyx – Kyz) + (Jxz– Kxz) Same energy for other two components of 3P state EEWS-90.502-Goddard-L04

  36. Comparison of O states with C states x x z z Ne (2p)61S C (2p)23P O (2p)43P Compared to Ne, we have Hole in x and z Hole in y and z Hole in x and y Compared to Be, we have Electron in x and z Electron in y and z Electron in x and y Thus holes in O map to electrons in C EEWS-90.502-Goddard-L04

  37. Summarizing the energies for O atom E(1S) O S Se Te Po 3Kxy TA’s look up data and list excitation energies in eV and Kxy in eV. Get data from Moore’s tables or paper by Harding and Goddard Ann Rev Phys Chem ~1985) E(1D) 2Kxy E(3P) EEWS-90.502-Goddard-L04

  38. Consider the ground state of F: [Be](2p)5 x z Only have 3 orbitals, x, y, and z. thus must have two doubly occupied Choices: (x)2(y)2 (z), (x)2(y)(z)2, (x)(y)2(z)2 Clearly all three equivalent give rise to spin doublet. Since 3 = 2L+1 denote as L=1 or 2P Ψ(1,2,3,4,5,6,7)z= A[(1sa)(1sb)(2sa)(2sb)(2pya)(2pyb)(2pxa)(2pza)] = A[(1s)2(2s)2(2pya)(2pyb)(2pxa)(2pza)] = = A[[Be](xa)(xb)(ya)(yb)(za)] which we visualize as EEWS-90.502-Goddard-L04

  39. Comparison of F states with B states x x z z Ne (2p)61S B (2p)12P F (2p)52P Compared to Ne, we have Hole in z Hole in x Hole in y Compared to Be, we have Electron in z Electron in x Electron in y Thus holes in F map to electrons in B EEWS-90.502-Goddard-L04

  40. Calculating energies for F atom (2px)2(2py)2(2pz) The energy of Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)] is Exz = 5hpp + Jxx+ Jyy+ (4Jxy – 2Kxy) + (2Jxz – Kxz) + (2Jyx– Kyz) Check: 5 electrons, therefore 5x4/2 = 10 coulomb interactions 3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions 2 down-spin electrons, therefore 2x1/2 = 1 exchange interaction Other ways to group energy terms Same energy for other two components of 2P state EEWS-90.502-Goddard-L04

  41. Comparison of F states with B states x x z z Ne (2p)61S B (2p)12P F (2p)52P Compared to Ne, we have Hole in z Hole in x Hole in y Compared to Be, we have Electron in z Electron in x Electron in y Thus holes in F map to electrons in B EEWS-90.502-Goddard-L04

  42. Consider the ground state of Ne: [Be](2p)6 x z Only have 3 orbitals, x, y, and z. thus must have all three doubly occupied Choices: (x)2(y)2(z)2 Thus get spin singlet, S=0 Since just one spatial state, 1=2L+1  L=0. denote as 1S Ψ(1-10)z = A[[Be](xa)(xb)(ya)(yb)(za)(zb)] which we visualize as Ne (2p)61S EEWS-90.502-Goddard-L04

  43. Calculating energy for Ne atom (2px)2(2py)2(2pz)2 The energy of Ψ(1-9)z= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} is Exz = 6hpp +Jxx+Jyy+ Jzz + (4Jxy – 2Kxy) + (4Jxz – 2Kxz) + (4Jyx– 2Kyz) Check: 6 p electrons, therefore 6x5/2 = 15 coulomb interactions 3 up-spin electrons, therefore 3x2/2 = 3 exchange interactions 3 down-spin electrons, therefore 3x2/2 = 3 exchange interaction Since Jxx = Kxx we can rewrite this as Exz = 6hpp +(2Jxx-Kxx ) +(2Jyy-Kyy ) +(2Jzz-Kzz ) + 2(2Jxy – Kxy) + 2(2Jxz – Kxz) + 2(2Jyx– Kyz) Which we will find later to be more convenient for calculating the wavefunctions using the variational principle EEWS-90.502-Goddard-L04

  44. Summary of ground states of Li-Ne Li 2S (2s)1 N 4S (2p)3 Be 1S (2s)2 O 3P (2p)4 Ignore (2s)2 B 2P (2p)1 F 2P (2p)5 C 3P (2p)2 Ne 1S (2p)6 EEWS-90.502-Goddard-L04

  45. Bonding H atom to He Starting with the ground state of He, (1s)2 = A(He1sa)(He1sb) and bringing up an H atom (H1sa), leads to HeH: A[(He1sa)(He1sb)(H1sa)] But properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the He 1s since both have an a spin. Consequently H1s must get a nodal plane, increasing its KE. The smaller R the larger increase in KE. Get a repulsive interaction, no bond R EEWS-90.502-Goddard-L04

  46. Bonding H atom to Ne Starting with the ground state of Ne, (1s)2(2s)2(2p)6Ψ(Ne)= A{[Be](2pxa)(2pxb)(2pya)(2pyb)(2pza)(2pzb)} and bringing up an H atom (H1sa) along the z axis, leads to A{[Be](2px)2(2py)2(Ne2pza)(Ne2pzb)(H1sa)} Where we focus on the Ne2pz orbital that overlaps the H atom The properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the Ne 2pz since both have an a spin. R Consequently H1s must get a nodal plane, increasing its KE. The smaller R the larger increase in KE. Get a repulsive interaction, no bond EEWS-90.502-Goddard-L04

  47. x z Now consider Bonding H atom to all 3 states of F Bring H1s along z axis to F and consider all 3 spatial states. R EEWS-90.502-Goddard-L04

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