© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Course number: KAIST EEWS Room E Hours: Tuesday and Thursday Senior Assistant: Dr. Hyungjun Kim: Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: Special assistant: Tod Lecture 7, September 24, 2009

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 3 Last time

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 4 Contour plots of 3s, 3p, 3d hydrogenic orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 5 The ground state of He atom Put both electrons in 1s orbitals Ψ He (1,2) = A [(Φ 1s  )(Φ 1s  Φ 1s (1)Φ 1s (2) (  Φ 1s = exp(-  r) E He = 2(½  2 ) – 2Z  (5/8)  where  J 1s,1s = (5/8)  Applying the variational principle, dE/d  = 0 get  = (Z – 5/16) = E= 2(-½  2 ) = -  2 = h 0 Interpretation: two electrons move independently in the orbital Φ 1s   exp(-  r) 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 =1.69 The total energy is just the sum of the individual energies.

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 6 Now 3 rd electron to form Li Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  (Φ 1s  by Pauli principle Thus the 3 rd electron must go into 2s or 2p atomic orbital Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2s  Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2pz  (or 2px or 2py) Li + : Φ 1s   exp(-  r) with  = Z = 2.69  R 1s = 1/  = a 0 = 0.2A

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 7 Add 3 rd electron to the 2p orbital Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2pz  (or 2px or 2py) The 2p orbital sees effective charge of Z eff = 3 – 2 = 1, since it goes to zero at z=0, so that there is no shielding of 1s Get size: R 2p = n 2 /Z eff = 4 a 0 = 2.12A Energy: e 2p = -(Z eff ) 2 /2n 2 = -1/8 h 0 = eV 0.2A 1s 2.12A 2p

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 8 Add the 3 rd electron to the 2s orbital Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2pz  (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 Z eff 2s = 3 – 1.72 = 1.28 Size: R 2s = n 2 /Z eff = 3.1 a 0 = 1.65A Energy: e 2s = -(Z eff ) 2 /2n 2 = h0 = 5.57 eV 0.2A 1s 2.12A 2s R~0.2A

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 9 Li atom excited states 1s 2s h 0 = -3.4 eV 2p Energy zero h 0 = -5.6 eV h0 = 74.1 eV MO picture State picture (1s) 2 (2s) (1s) 2 (2p)  E = 2.2 eV cm nm Ground state 1 st excited state Exper 671 nm  E = 1.9 eV

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 10 He, 2 Ne, 10 Ar, 18 Zn, 30 Kr, 36 Aufbau principle for atoms Particularly stable atoms, closed shells Xe, 54 Rn, 86 Uuo, 118

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 11 Many-electron configurations General aufbau ordering Particularly stable

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 ~ n 2 /Zeff And an increase in the IP ~ (Zeff) 2 /2n 2 Example Z eff 2s = 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 ~ n 2 /Zeff And the IP ~ (Zeff) 2 /2n 2 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 14

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 15 Transition metals; consider [Ar] plus one electron [IP 4s = (Z eff 4s ) 2 /2n 2 = 4.34 eV  Z eff 4s = 2.26; 4s<4p<3d IP 4p = (Z eff 4p ) 2 /2n 2 = 2.73 eV  Z eff 4p = 1.79; IP 3d = (Z eff 3d ) 2 /2n 2 = 1.67 eV  Z eff 3d = 1.05; IP 4s = (Z eff 4s ) 2 /2n 2 = eV  Z eff 4s = 3.74; 4s<3d<4p IP 3d = (Z eff 3d ) 2 /2n 2 = eV  Z eff 3d = 2.59; IP 4p = (Z eff 4p ) 2 /2n 2 = 8.73 eV  Z eff 4p = 3.20; IP 3d = (Z eff 3d ) 2 /2n 2 = eV  Z eff 3d = 4.05; 3d<4s<4p IP 4s = (Z eff 4s ) 2 /2n 2 = eV  Z eff 4s = 5.04; IP 4p = (Z eff 4p ) 2 /2n 2 = eV  Z eff 4p = 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 16 Transition metals; consider Sc 0, Sc +, Sc 2+ 3d: IP 3d = (Z eff 3d ) 2 /2n 2 = eV  Z eff 3d = 4.05; 4s: IP 4s = (Z eff 4s ) 2 /2n 2 = eV  Z eff 4s = 5.04; 4p: IP 4p = (Z eff 4p ) 2 /2n 2 = eV  Z eff 4p = 4.47; 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. (3d)(4s): IP 4s = (Z eff 4s ) 2 /2n 2 = eV  Z eff 4s = 3.89; (3d) 2 : IP 3d = (Z eff 3d ) 2 /2n 2 = eV  Z eff 3d = 2.85; (3d)(4p): IP 4p = (Z eff 4p ) 2 /2n 2 = 9.66 eV  Z eff 4p = 3.37; Sc + (3d)(4s) 2 : IP 4s = (Z eff 4s ) 2 /2n 2 = 6.56 eV  Z eff 4s = 2.78; (4s)(3d) 2 : IP 3d = (Z eff 3d ) 2 /2n 2 = 5.12 eV  Z eff 3d = 1.84; (3d)(4s)(4p): IP 4p = (Z eff 4p ) 2 /2n 2 = 4.59 eV  Z eff 4p = 2.32; Sc

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 18 Transtion metal orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 20 Consider the ground state of B: [Be](2p) 1 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. 2p x 2p z 2p y z x 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 2 P

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 21 New material

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 22 Consider the ground state of C: [Be](2p) 2 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. Which state is better? The difference is in the electron-electron repulsion: 1/r 12 z x z x z x (2p x ) 2 (2p z ) 2 (2p x )(2p z ) Clearly two electrons in the same orbital have a much smaller average r 12 and hence a much higher e-e repusion. Thus the ground state has each electron in a different 2p orbital

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 23 Consider the states of C: formed from (x)(y), (x)(z), (y)(z) Consider first (x)(y): can form two spatial products: Φ x (1)Φ y (2) and Φ y (1)Φ x (2) These are not symmetric, thus must combine Φ(1,2) s =φ x (1) φ y (2) + φ y (1) φ x (2) Φ(1,2) a = φ x (1) φ y (2) - φ y (1) φ x (2) y x (2p x )(2p y ) Which state is better? The difference is in the electron-electron repulsion: 1/r 12 To analyze this, expand the orbitals in terms of the angular coordinates, r,θ,φ Φ(1,2) s = f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[(cosφ 1 )(sinφ 2 )+(sinφ 1 ) (cosφ 2 )] =f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[sin(φ 1 +φ 2 )] Φ(1,2) a = f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[(cosφ 1 )(sinφ 2 )-(sinφ 1 ) (cosφ 2 )] =f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[sin(φ 2 -φ 1 )]

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 24 Consider the symmetric and antisymmetric combinations of (x)(y) y x (2p x )(2p y ) Φ(1,2) s = f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[sin(φ 1 +φ 2 )] Φ(1,2) a = f(r 1 )f(r 2 )(sinθ 1 )(sinθ 2 )[sin(φ 2 -φ 1 )] The big difference is that Φ(1,2) a = 0 when φ 2 = φ 1 and is a maximum for φ 2 and φ 1 out of phase by  /2. But for Φ(1,2) s the probability of φ 2 = φ 1 is comparable to that of being out of phase by  /2. Thus the best combination is Φ(1,2) a Thus for 2 electrons in orthogonal orbitals, high spin is best because the electrons can never be at same spot at the same time Combining with the spin parts we get [φ x (1) φ y (2) + φ y (1) φ x (2  or spin = 0 [φ x (1) φ y (2) - φ y (1) φ x (2  also  and  or spin = 1

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 25 Summarizing the states for C atom Ground state: three triplet states=2L=1. Thus L=1, denote as 3 P (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 1 D (xy+yx) (xz+zx) (yz+zy) (xx-yy) (2zz-xx-yy) Highest state: one singlet=2L+1. thus L=0. Denote as 1 S (zz+xx+yy) y x (2p x )(2p y ) 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 26 Calculating energies for C atom The energy of xy is E xy = h xx + h yy + J xy = 2h pp +J xy Thus the energy of the 3 P state is E( 3 P) = E xy – K xy = 2h pp +J xy - K xy For the (xy+yx) component of the 1 D state, we get E( 1 D) = E xy + K xy = 2h pp +J xy + K xy Whereas for the (xx-yy) component of the 1 D state, we get E( 1 D) = E xx - K xy = 2h pp +J xx - K xy This means that J xx - K xy = J xy + K xy so that J xx = J xy + 2K xy Also for (2zz-xx-yy) we obtain E = 2h pp +J xx - K xy For (zz+xx+yy) we obtain E( 1 S) = 2h pp + Jxx + 2 Kxy y x (2p x )(2p y )

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 27 Summarizing the energies for C atom E( 1 S) = 2h pp + J xx + 2K xy E( 1 D) = 2h pp +J xx - K xy = 2h pp +J xy + K xy E( 3 P) = 2h pp + E xy – K xy = 2h pp +J xy - K xy 2K xy 3K xy

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 28 Comparison with experiment E( 1 S) E( 1 D) E( 3 P) 2K xy 3K xy 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)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 29 Summary ground state for C atom z x (2p x )(2p z ) Ψ(1,2,3,4,5,6) xz = A [(1s  )(1s  )(2s  )(2s  )(2p x  )(2p z  )] = = A [(1s   (2s) 2 (2p x  )(2p z  )] = A [(Be)(2p x  )(2p z  )] = = A [(x  )(z  )] = which we visualize as Ψ(1,2,3,4,5,6) xy = A [(x  )(y  )] which we visualize as Ψ(1,2,3,4,5,6) yz = A [(y  )(z  )] which we visualize as z x 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 z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 [(x  )(y  )(z  )] = [Axyz]  to be the ground state. (here Axyz is the antisymmetric combination of x(1)y(2)x(3)] The four symmetric spin functions are  With Ms  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 4 S state.

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 31 Energy of the ground state of N: A [(x  )(y  )(z  )] = [Axyz]  Simple product xyz leads to E xyz = 3h pp + J xy + J xz + J yz E( 4 S) = <xyz|H| A[xyz]  <xyz| A[xyz]  Denominator = <xyz| A[xyz]  = 1 Numerator = E xyz - K xy - K xz – K yz = E( 4 S) = 3h pp + (J xy - K xy ) + (J xz - K xz ) + (J yz – K yz ) E( 2 P) E( 2 D) E( 4 S) 4K xy 2K xy =3h pp + 3J xy - 3K xy =3h pp + 3J xy + 1K xy = 3h pp + 2J xy + J xx - 1K xy =3h pp + 2J xy + J xx + 1K xy TA’s check this z x Pictorial representation of the N ground state Since J xy = J xz = J yz and K xy = K xz = K yz

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 32 Comparison with experiment E( 1 S) E( 1 D) E( 3 P) 4K xy 2K xy 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)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 33 Consider the ground state of O: [Be](2p) 4 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 z x z x (2p x ) 2 (2p y )(2p z ) (2p z ) 2 (2p x ) 2

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 34 Ψ(1,2,3,4,5,6) xz = A [(1s  )(1s  )(2s  )(2s  )(2p y  )(2p y  )(2p x  )(2p z  )] = A [(1s   (2s) 2 (2p y  )(2p y  )(2p x  )(2p z  )] = = A [(y  )(y  )(x  )(z  )] = which we visualize as Summary ground state for O atom z x Ψ(1,2,3,4,5,6) xy = A [(z  )(z  )(((x  )(y  )] which we visualize as Ψ(1,2,3,4,5,6) yz = A [(x  )(x  )((y  )(z  )] which we visualize as (2p x ) 2 (2p y )(2p z ) z x z x We have 3 = 2L+1 equivalent spin triplet (S=1) states that we denote as L=1 orbital angular momentum, leading to the 3 P state (2p y ) 2 (2p x )(2p z ) (2p z ) 2 (2p x )(2p y )

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 35 Calculating energies for O atom is E xz = E(Be) + 2h yy + h xx + h zz + J yy + 2J xy + 2J yx + J xz – K xy – K yz – K xz 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 E xz = 4h pp + J yy + (2J xy – K xy ) + (2J yx – K yz ) + (J xz – K xz ) Same energy for other two components of 3 P state (2p y ) 2 (2p x )(2p z ) The energy of Ψ(1,2,3,4,5,6) xz = A [(1s  )(1s  )(2s  )(2s  )(2p y  )(2p y  )(2p x  )(2p z  )] = A [[Be](y  )(y  )(x  )(z  )]

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 36 Comparison of O states with C states O (2p) 4 3 P C (2p) 2 3 PNe (2p) 6 1 S 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 z x Thus holes in O map to electrons in C z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 37 Summarizing the energies for O atom E( 1 S) E( 1 D) E( 3 P) 2K xy 3K xy O S Se Te Po 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)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 38 Consider the ground state of F: [Be](2p) 5 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 2 P Ψ(1,2,3,4,5,6,7) z = A [(1s  )(1s  )(2s  )(2s  )(2p y  )(2p y  )(2p x  )(2p z  )] = A [(1s   (2s) 2 (2p y  )(2p y  )(2p x  )(2p z  )] = = A [[Be](x  )(x  )(y  )(y  )(z  )] which we visualize as z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 39 Comparison of F states with B states F (2p) 5 2 P B (2p) 1 2 PNe (2p) 6 1 S 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 z x z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 40 Calculating energies for F atom is E xz = 5h pp + J xx + J yy + (4J xy – 2K xy ) + (2J xz – K xz ) + (2J yx – K yz ) 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 2 P state (2p x ) 2 (2p y ) 2 (2p z ) The energy of Ψ(1-9) z = A {[Be](2p x  )(2p x  )(2p y  )(2p y  )(2p z  )]

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 41 Comparison of F states with B states F (2p) 5 2 P B (2p) 1 2 PNe (2p) 6 1 S 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 z x z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 42 Consider the ground state of Ne: [Be](2p) 6 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 1 S Ψ(1-10) z = A [[Be](x  )(x  )(y  )(y  )(z  )(z  )] which we visualize as Ne (2p) 6 1 S z x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 43 Calculating energy for Ne atom is E xz = 6h pp +J xx +J yy + J zz + (4J xy – 2K xy ) + (4J xz – 2K xz ) + (4J yx – 2K yz ) 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 J xx = K xx we can rewrite this as E xz = 6h pp +(2J xx -K xx ) +(2J yy -K yy ) +(2J zz -K zz ) + 2(2J xy – K xy ) + 2(2J xz – K xz ) + 2(2J yx – K yz ) Which we will find later to be more convenient for calculating the wavefunctions using the variational principle (2p x ) 2 (2p y ) 2 (2p z ) 2 The energy of Ψ(1-9) z = A {[Be](2p x  )(2p x  )(2p y  )(2p y  )(2p z  )(2p z  )}

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 44 Summary of ground states of Li-Ne N 4 S (2p) 3 O 3 P (2p) 4 F 2 P (2p) 5 Ne 1 S (2p) 6 Li 2 S (2s) 1 Be 1 S (2s) 2 B 2 P (2p) 1 C 3 P (2p) 2 Ignore (2s) 2

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 45 Bonding H atom to He Starting with the ground state of He, (1s) 2 = A (He1s  )(He1s  ) and bringing up an H atom (H1s  ), leads to R HeH: A [(He1s  )(He1s  )(H1s  )] 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  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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 46 Bonding H atom to Ne Starting with the ground state of Ne, (1s) 2 (2s) 2 (2p) 6 Ψ(Ne)= A {[Be](2p x  )(2p x  )(2p y  )(2p y  )(2p z  )(2p z  )} and bringing up an H atom (H1s  ) along the z axis, leads to A {[Be](2p x ) 2 (2p y ) 2 (Ne2p z  )(Ne2p z  )(H1s  } Where we focus on the Ne2pz orbital that overlaps the H atom 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 The properties of A or alternatively the Pauli Principle tells us that the H1s must get orthogonal to the Ne 2p z since both have an  spin.

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 47 Now consider Bonding H atom to all 3 states of F R Bring H1s along z axis to F and consider all 3 spatial states. z x