Lecture 3, October 3, 2016 aufbau principle atoms-bonding H to to NeFON

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1 Lecture 3, October 3, 2016 aufbau principle atoms-bonding H to to NeFON Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistant: Shane Flynn sflynn@caltech.edu Recitation: Mon 3pm

2 The excited states of H atom z hφ k = e k φ k for all excited states k We will use spherical polar coordinates, r, θ, φ where z=r cosθ, x=r sinθ cosφ, y=r sinθ sinφ h = - ½ 2 Z/r is independent of θ and φ which can be seen by noting that 2 = d 2 /dx 2 + d 2 /dy 2 + d 2 /dy 2 is independent of θ and φ (it transforms like r 2 = x 2 + y 2 + z 2 ). Consequently the eigenfunctions of h can be written as a factor depending only on r and a factor depending only on θ and φ φ nlm = R nl (r) Z lm (θ,φ) where the reason for the numbers nlm will become apparent later x θ φ y 2

3 The ground state of H atom z h = - ½ 2 Z/r φ nlm = R nl (r) Z lm (θ,φ) θ For the ground state, R 1s = exp(-zr) and x φ y Z 1s = Z 00 = 1 (a constant), where l=0 and m=0 again ignoring normalization. The radial wavefunction is nodeless, as expected The angular function is a constant, so that it leads to no new nodes, which is clearly the best for KE. 3

4 The p excited angular states of H atom φ nlm = R nl (r) Z lm (θ,φ) Now lets consider excited angular functions, Z lm. They must have nodal planes to be orthogonal to Z 00 The simplest would be Z 10 =z = r cosθ, which is zero in the xy plane. Exactly equivalent are Z 11 =x = rsinθcosφ which is zero in the yz plane, and Z 1-1 =y = rsinθsinφ, which is zero in the xz plane Also it is clear that these 3 angular functions with one angular nodal plane are orthogonal to each other. Thus the integrand of xz has nodes in both the xy and zy planes, leading to a zero integral ight 2016 William A. Goddard III, a p z copyr ll rights reserved z z z p z p x + x x p x p z x 4

5 More p functions? So far we have the s angular function Z 00 = 1 with no angular nodal planes And three p angular functions: Z 10 =p z, Z 11 =p x, Z 1-1 =p y, each with one angular nodal plane Can we form any more angular functions with one nodal plane orthogonal to the above 4 functions? For example we might rotate p x by an angle α z p x about the y axis to form p x. However multiplying, α say by p z, leads to the integrand pzp x which + clearly does not integrate to zero - z p z p x. x Thus there are exactly three pi functions, Z 1m, α - + with m=0,+1,-1, all of which have the same KE. + - x copyright 2016 WilliamA. Goddard III, all rights reserved 5 Since the p functions have nodes, they lead to a higher KE than the s function

6 More angular functions? So far we have the s angular function Z 00 = 1 with no angular nodal planes And three p angular functions: Z 10 =p z, Z 11 =p x, Z 1-1 =p y, each with one angular nodal plane Next in energy will be two angular nodal planes, which we call d functions. We can easily construct three d functions z d xy = xy =r 2 (sinθ) 2 cosφ sinφ d yz = yz =r 2 (sinθ)(cosθ) sinφ - + d zx = zx =r 2 (sinθ)(cosθ) cosφ where d xz is shown here + - d xz x Each of these is orthogonal to each other (and to the s and the three p functions) 6

7 More d angular functions? In addition we can construct three other functions with two nodal planes d x2-y2 = x 2 y 2 = r 2 (sinθ) 2 [(cosφ) 2 (sinφ) 2 ] z d y2-z2 = y 2 z 2 = r 2 [(sinθ) 2 (sinφ) 2 (cosθ) 2 ] + d z2-x2 = z 2 x 2 = r 2 [(cosθ) 2 -(sinθ) 2 (cosφ) 2 ] - where d z2-x2 is shown here, + Each of these three is orthogonal to the previous three d functions (and to the s and the three p functions) This leads to six functions with 2 nodal planes - d z2-x2 x 7

8 More d angular functions? In addition we can construct three other functions with two nodal planes d x2-y2 = x 2 y 2 = r 2 (sinθ) 2 [(cosφ) 2 (sinφ) 2 ] z d y2-z2 = y 2 z 2 = r 2 [(sinθ) 2 (sinφ) 2 (cosθ) 2 ] + d z2-x2 = z 2 x 2 = r 2 [(cosθ) 2 -(sinθ) 2 (cosφ) 2 ] - where d z2-x2 is shown here, + Each of these three is orthogonal to the previous three d functions (and to the s and the three p functions) This leads to six functions with 2 nodal planes However (x 2 y 2 ) + (y 2 z 2 ) + (z 2 x 2 ) = 0 Which indicates that there are only two independent such functions. We combine the 2 nd two as (z 2 x 2 ) - (y 2 z 2 ) = [2 z 2 x 2 - y 2 ] = [3 z 2 x 2 - y 2 z 2 ] = = [3 z 2 r 2 ] which we denote as d z2 - d z2-x2 x 8

9 Summarizing we get 5 d angular functions Z 20 = d z2 = [3 z 2 r 2 ] m=0, dσ Z 21 = d zx = zx =r 2 (sinθ)(cosθ) cosφ 2 Z 2-1 = d yz = yz =r (sinθ)(cosθ) sinφ Z 22 = d x2-y2 = x 2 y 2 = r 2 (sinθ) 2 [(cosφ) 2 (sinφ) 2 ] Z 22 = d xy = xy =r 2 (sinθ) 2 cosφ sinφ We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2π = 360º When this number, m=0 we call it a σ function When m=1 we call it a π function m = 1, dπ z - 57º m = 2, dδ d z2 x When m=2 we call it a δ function When m=3 we call it a φ function 9

10 So far we have Summarizing the angular functions one s angular function (no angular nodes) called l=0 three p angular functions (one angular node) called l=1 five d angular functions (two angular nodes) called l=2 Continuing we can form seven f angular functions (three angular nodes) called l=3 nine g angular functions (four angular nodes) called l=4 where l is referred to as the angular momentum quantum number And there are (2l+1) m values for each l 10

11 The real (Z lm ) and complex (Y lm ) momentum functions copyright 2016 William A. Goddard III, all rights res Here the bar over m è negative erved 11

12 Excited radial functions Clearly the KE increases with the number of angular nodes so that s < p < d < f < g Now we must consider radial functions, R nl (r) The lowest is R 10 = 1s = exp(-zr) All other radial functions must be orthogonal and hence must have one or more radial nodes, as shown here Zr = 7.1 Zr = 2 Zr = 1.9 Note that we are plotting the cross section along the z axis, but it would look exactly the same along any other axis. Here copyright 2016 William A. Goddard III, all rights reserve R 20 = 2s = [Zr/2 1]exp(-Zr/2) and R 30 = 3s = [2(Zr) 2/27 2(Zr)/3 + 1]exp(-Zr/3) 12 d

13 name total nodal planes radial nodal planes angular nodal planes Size (a0) 1s s p s p d s p Combination of radial and angular nodal planes Combining radial and angular functions gives the various excited states of the H atom. They are named as shown where the n quantum number is the total number of nodal planes plus 1 The nodal theorem does not specify how 2s and 2p are related, but it turns out that the total energy depends only on n. E nlm = - Z 2 /2n 2 The potential energy is given by PE = - Z 2 /2n 2 = -Z/ Rˉ, where Rˉ =n 2 /Z Thus Enlm = - Z/2Rˉ 4d f

14 Sizes hydrogen orbitals Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48 Rˉ =(n 2 /Z) a 0 = 0.53 (n 2 /Z) A H--H H C H H H H H 4.8 H H H H 14

15 Energy zero Hydrogen atom excited states E nlm = - Z 2 /2n 2 = -(1/2n 2 )h 0 = 13.6 ev/n h 0 = -0.9 ev h 0 = -1.5 ev 4s 4p 4d 4f 3s 3p 3d h 0 = -3.4 ev 2s 2p -0.5 h 0 = ev 1s 15

16 Plotting of orbitals: line cross-section vs. contour A. Goddard III, all rights reserved 16 copyright 2016 William

17 Contour plots of 1s, 2s, 2p hydrogenic orbitals -Goddard-L03 1s, no nodes 2s, one radial node 2p, one angular node Ch120a 17

18 Contour plots of 3s, 3p, 3d hydrogenic orbitals 3s, two radial nodes 3p, one angular node one radial node copyright 2016 WilliamA. Goddard III, all rights reserved 18 3d, two angular nodes

19 Contour plots of 4s, 4p, 4d hydrogenic orbtitals 4s, 3 radial nodes 4p, one angular node two radial nodes copyright 2016 WilliamA. Goddard III, all rights reserved 19 4d, two angular nodes and one radial node

20 20a-Goddard-L03 Contour plots of hydrogenic 4f orbitals All seven 4f have three angular nodes and no radial nodes Ch1 20

21 He atom describe using He+ orbitals With 2 electrons, we can form the ground state of He by putting both electrons in the He + 1s orbital, just like the MO state of H 2 Ψ He (1,2) = A[(Φ 1s α)(φ 1s β)]= Φ 1s (1)Φ 1s (2) (αβ βα) Ε He = 2<1s h 1s> + J 1s,1s J 1s,1s =< Φ 1s(1)Φ 1s(2) 1/r 12 Φ 1s(1)Φ 1s(2)> two one-electron energies one Coulomb repulsion 21

22 He atom describe using He+ orbitals With 2 electrons, we can form the ground state of He by putting both electrons in the He + 1s orbital, just like the MO state of H 2 Ψ He (1,2) = A[(Φ 1s α)(φ 1s β)]= Φ 1s (1)Φ 1s (2) (αβ βα) Ε He = 2<1s h 1s> + J 1s,1s First lets review the energy for He +. Writing Φ 1s = exp(-ζr) we determine the optimum ζ for He + as follows <1s KE 1s> = + ½ ζ 2 (goes as the square of 1/size) <1s PE 1s> = - Zζ (linear in 1/size) 22

23 He atom describe using He+ orbitals With 2 electrons, we can form the ground state of He by putting both electrons in the He + 1s orbital, just like the MO state of H 2 Ψ He (1,2) = A[(Φ 1s α)(φ 1s β)]= Φ 1s (1)Φ 1s (2) (αβ βα) Ε He = 2<1s h 1s> + J 1s,1s First lets review the energy for He +. Writing Φ 1s = exp(-ζr) we determine the optimum ζ for He + as follows <1s KE 1s> = + ½ ζ 2 (goes as the square of 1/size) <1s PE 1s> = - Zζ (linear in 1/size) Applying the variational principle, the optimum ζ must satisfy de/dζ = ζ - Z = 0 leading to ζ = Z, This value of ζ leads to KE = ½ Z 2, PE = -Z 2, E=-Z 2 /2 = -2 h 0. write PE=-Z/R 0, so that the average radius is R 0 =1/ζ 23

24 J 1s,1s = e-e energy of He atom He + orbitals e 1 Now consider He atom: E He = 2(½ ζ 2 ) 2Zζ + J 1s,1s How can we estimate J 1s,1s Assume that each electron moves on a sphere With the average radius R 0 = 1/ζ And assume that e 1 at along the z axis (θ=0) Neglecting correlation in the electron motions, e 2 will on the average have θ=90º so that the average e1-e2 distance is ~sqrt(2)*r 0 Thus J 1s,1s ~ 1/[sqrt(2)*R0] = 0.71 ζ A rigorous calculation (notes chapter 3, appendix 3-C page 6) Gives J 1s,1s = (5/8) ζ R 0 e 2 24

25 The optimum atomicorbital for He atom He atom: E He = 2(½ ζ 2 ) 2Zζ + (5/8)ζ Applying the variational principle, the optimum ζ must satisfy de/dζ = 0 leading to 2ζ - 2Z + (5/8) = 0 Thus ζ = (Z 5/16) = KE = 2(½ ζ 2 ) = ζ 2 PE = - 2Zζ + (5/8)ζ = -2 ζ 2 E= - ζ 2 = h 0 Thus in He the 2 nd electron in the 1s orbital partially shields the nuclear charge Z by the equivalent of charges at the nucleu Ignoring e-e interactions the energy would have been E = -4 h 0 The exact energy is E = h 0 (from memory, TA please check). Thus this simple independent electron approximation accounts for 98.1% of the exact result. 25

26 Interpretation: The optimum atomic orbital for He atom Ψ He (1,2) = Φ 1s (1)Φ 1s (2) with Φ 1s = exp(-ζr) We find that the optimum ζ = (Z 5/16) = With this value of ζ, the total energy is E= - ζ 2 = h 0 This wavefunction can be interpreted as two electrons moving independently in the orbital Φ 1s = exp(-ζr) that 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. Ionizing the 2 nd electron, the 1 st electron readjusts to ζ = Z = 2 With E(He + ) = -Z 2 /2 = - 2 h 0. thus the ionization potential (IP) is h 0 = 23.1 ev (exact value = 24.6 ev) 26

27 Now lets add a 3 rd electron to form Li Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β)(φ 1s γ)] Problem: with either γ = α or γ = β, we get Ψ Li (1,2,3) = 0 This is an essential result of the Pauli principle. Only 2 electrons in same orbital, one of each spin Thus the 3 rd electron must go into an excited orbital Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β) )(Φ 2s α)] or Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β) )(Φ 2pz α)] (or 2px or 2py) First consider Li + with Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β)] Here Φ 1s = exp(-ζr) with ζ = Z = 2.69 and E = -ζ 2 = h 0. Since the E (Li 2+ ) = -9/2 = -4.5 h 0 the IP = h 0 = 74.1 ev The size of the 1s orbital is R 0 = 1/ζ = a 0 = 0.2A 27

28 Consider adding the 3 rd electron to the 2p orbital Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β) )(Φ 2pz α)] (or 2px or 2py) Since the 2p orbital goes to zero at z=0, there is very little shielding so that it sees an effective charge of Z eff = 3 2 = 1, leading to a size of R 2p = n 2 /Z eff = 4 a 0 = 2.12A And an energy of e = -(Z eff ) 2 /2n 2 = -1/8 h0 = ev 1s 2p 0.2A 2.12A 28

29 Consider adding the 3 rd electron to the 2s orbital Ψ Li (1,2,3) = A[(Φ 1s α)(φ 1s β) )(Φ 2pz α)] (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 = = 1.28 This leads to a size of R 2s = n 2 /Z eff = 3.1 a 0 = 1.65A And an energy of e = -(Z eff ) 2 /2n 2 = h0 = ev 1s 0.2A copyright 2016 WilliamA. Goddard III, all rights reserved 29 2s 2.12A R~0.2A

30 Energy MO picture Li atom excited states State picture zero h 0 = -3.4 ev h 0 = -5.6 ev h0 = 74.1 ev 2s 1s 2p 1 st excited state 2 (1s) (2p) (1s) 2 (2s) Ground state copyright 2016 William A. Goddard III, all rights rese ΔE = 2.2 ev cm nm rved 30 Exper 671 nm E = 1.9 ev

31 Energy Aufbau principlefor atoms Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, s 3s 2s 1s p 3p 2p copyright 2016 Willi d 3d 14 4f ama. Goddard III, all rights reserved 31 Including shielding, 2s<2p 3s<3p<3d, 4s<4p<4d<4f Get generalized energy spectrum for filling in the electrons to explain the periodic table. Full shells at 2, 10, 18, 30, 36 electrons

32 copyright 2016 WilliamA. Goddard III, all rights reserved 32 Ar, 18 Kr, 36 Zn, 30 He, 2 Ne, 10

33 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 For each atom we add one more proton to the nucleus and one more electron to the valence shell But the valence electrons only partially shield each other. Thus 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 (2s Li)/(2s Ne) ~ 4.64/1.28 =

34 Many-electron configurations General aufbau ordering Particularly stable 34

35 General trends along a column of the periodic table As we go down a column 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 35

36 Ch120a-Goddard -L03 copyright 2016 WilliamA. Goddard III, all rights reserved 36

37 Ch120a-Goddar d-l03 copyright 2016 WilliamA. Goddard III, all rights reserved 37

38 opyright 2016 WilliamA. Goddard III, all rights reserved 38 c

39 K Ca + Transition metals; consider [Ar] plus one electron [IP 4s = (Z eff 4s ) 2 /2n 2 = 4.34 ev è Z eff 4s = 2.26; IP 4p = (Z eff 4p ) 2 /2n 2 = 2.73 ev è Z eff 4p = 1.79; 4s<4p<3d IP 3d = (Z eff 3d ) 2 /2n 2 = 1.67 ev è Z eff 3d = 1.05; IP 4s = (Z eff 4s ) 2 /2n 2 = ev è Z eff 4s = 3.74; 4s<3d<4p IP 3d = (Z eff 3d ) 2 /2n 2 = ev è Z eff 3d = 2.59; IP 4p = (Z eff 4p ) 2 /2n 2 = 8.73 ev è Z eff 4p = 3.20; IP 3d = (Z eff 3d ) 2 /2n 2 = ev è Z eff 3d = 4.05; 3d<4s<4p Sc ++ IP 4s = (Z eff 4s ) 2 /2n 2 = ev è Z eff 4s = 5.04; IP 4p = (Z eff 4p ) 2 /2n 2 = ev è Z eff 4p = 4.47; copyright 2016 WilliamA. Goddard III, all rights reserved 39 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

40 Sc ++ Sc + Transition metals; consider Sc 0, Sc +, Sc 2+ 3d: IP 3d = (Z eff 3d ) 2 /2n 2 = ev è Z eff 3d = 4.05; 4s: IP 4s = (Z eff 4s ) 2 /2n 2 = ev è Z eff 4s = 5.04; 4p: IP 4p = (Z eff 4p ) 2 /2n 2 = ev è Z eff 4p = 4.47; (3d)(4s): IP 4s = (Z eff 4s ) 2 /2n 2 = ev è Z eff 4s = 3.89; (3d) 2 : IP 3d = (Z eff 3d ) 2 /2n 2 = ev è Z eff 3d = 2.85; (3d)(4p): IP 4p = (Z eff 4p ) 2 /2n 2 = 9.66 ev è Z eff 4p = 3.37; Sc (3d)(4s) 2 : IP 4s = (Z eff 4s ) 2 /2n 2 = 6.56 ev è Z eff 4s = 2.78; (4s)(3d) 2 : IP 3d = (Z eff 3d ) 2 /2n 2 = 5.12 ev è Z eff 3d = 1.84; (3d)(4s)(4p): IP 4p = (Z eff 4p ) 2 /2n 2 = 4.59 ev è Z eff 4p = 2.32; As increase net charge the increases in the differential shielding for 4s vs 3d is less important than the difference in n quantum number 3 vs 4. 40

41 Implications for 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 41

42 Transtion metal orbitals 42

43 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 43

44 Consider the ground state of B: [Be](2p) 1 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 x z copyright 2016 WilliamA. Goddard III, all rights reserved 44 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

45 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)(2pz), (2py)(2pz), We will depict these states by simplified contour diagrams in the xz plane, as at the right. Which state is better? (2p x ) 2 (2p z ) 2 (2p x )(2p z ) x x x z z z 45

46 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)(2pz), (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 (2p x ) 2 (2p z ) 2 (2p x )(2p z ) x x x z z z 46

47 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)(2pz), (2py)(2pz), We will depict these states by simplified contour diagrams in the xz plane, as at the right. (2p x ) 2 (2p z ) 2 (2p x )(2p z ) x x x z z Which state is better? The difference is in the electron-electron repulsion: 1/r 12 Clearly 2 electrons in the same orbital has higher e-e repulsion than 1 in px and 1 in py z 47

48 Consider the states of C formed from (x)(y), (x)(z), (y)(z) x l products: Consider first (x)(y): can form two spatia Φ x (1)Φ y (2) and Φ y (1)Φ x (2) (2p x )(2p y ) These are not spatially symmetric, th us 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) 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 x (1) = f(r 1 )(sinθ 1 )(cosφ 1 ) φ y (2) = f(r 2 )(sinθ 2 )(sinφ 2 ) In 2-electron phi space get node when φ 1 = π/2 (and 3π/2) and when φ 2 = 0 and π π φ x (1) φ y (2) φ π φ 2 y 48

49 Consider the states of C formed from (2p x )(2p y ) π φ x (1) φ y (2) φ π φ 1 φ y (1) φ x (2) π φ φ 2 π Get max when φ 1 = φ 2 Hence bad e-e- energy Get 0 when φ 1 = φ 2 Hence good e-e- energy Φ(1,2) s =φ x (1) φ y (2) + φ y (1) φ x (2) π φ φ 2 Φ(1,2) a = φ x (1) φ y (2) - φ y (1) φ x (2) π φ φ 2 + π π For H2 found gu-ug good but gu+ug bad, same thing 49

50 Consider the symmetric and antisymmetric combinations of (x)(y) 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) Combining with the spin parts we get bad good [φ x (1) φ y (2) + φ y (1) φ x (2)](αβ βα) è spin = S = 0 [φ x (1) φ y (2) - φ y (1) φ x (2)](αβ+βα), also αα and ββ è spin = S = 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. This is part of Hund s rule 50

51 Exchange energy How much better is triplet [φ x (1) φ y (2) - φ y (1) φ x (2)] compared to open shell singlet φ x (1) φ y (2) + φ y (1) φ x (2)] 3E = <xy-yx h(1)+h(2)+1/r 12 xy-yx>/<xy-yx xy-yx> Normalization <xy-yx xy-yx>=2<xy xy-yx>=2[<xy xy> - <xy yx>] Denominator [<x x><y y>] [<x y><y x>] Numerator 1 0 < xy-yx h(1)+h(2)+1/r 12 xy-yx >=2<xy h(1)+h(2)+1/r 12 xy-yx> <xy h(1)+h(2)+1/r 12 xy>=<xy h(1) xy>+ <xy h(2) xy>+ <xy 1/r 12 xy> No exchange = [<x h x><y y>]+[<x x><y h y>] + J xy = [<x h x> + <y h y>] + J xy <xy h(1)+h(2)+1/r 12 yx>=<xy h(1) yx>+ <xy h(2) yx>+ <xy 1/r 12 yx> [<x h y><y x>]+[<x y><y h x>] + K xy exchange = K xy E = 2<p h p>+ J xy - K xy Thus triplet and singlet differ by 2K xy 1E = 2<p h p>+ J xy + K xy just twice the exchange energy, Kxy>0 1

52 Summarizing the states for C atom Ground state: three triplet states=2l=1. Thus L=1, denoxte as 3 P (xy-yx) [x(1)y(2)-y(1)x(2)] (2p x )(2p y ) (xz-zx) (yz-zy) y Next state: five singlet states=2l+1. (xy+yx) Thus L=2, denote as 1 D (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) copyright 2016 WilliamA. Goddard III, all rights reserved 5 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 2

53 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 (2p x )(2p y ) 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 x y 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 CS h ) 12 = 0a- 2 G h od pdpard+ -L0 J 3 xx + 2 K co x py y right 2016 William A. Goddard III, all rights reserved 53

54 Summarizing the energies for C atom E( 1 S) = 2h pp + J xx + 2K xy 3K xy E( 1 D) = 2h pp +J xx - K xy = 2h pp +J xy + K xy 2K xy E( 3 P) = 2h pp + E xy K xy = 2h pp +J xy - K xy 54

55 Comparison with experiment E( 1 S) 3K xy E( 1 D) 2K xy E( 3 P) 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) 55

56 Summary ground state for C atom Ψ(1,2,3,4,5,6) xz = A[(1sα)(1sβ)(2sα)(2sβ)(2p x α)(2p z α)] = = A[(1s) 2 (2s) 2 (2p x α)(2p z α)] =A[(Be)(2p x α)(2p z α)] = = A[(xα)(zα)] = which we visualize as (2p x )(2p z ) Ψ(1,2,3,4,5,6) xy = A[(xα)(yα)] x x z which we visualize as Ψ(1,2,3,4,5,6) yz = A[(yα)(zα)] which we visualize as x z z 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 56

57 Consider the ground state of N: [Be](2p) 3 copyright 2016 WilliamA. Goddard III, all rights reserved 57 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[(xα)(yα)(zα)] = [Axyz]ααα to be the ground state. (hereaxyz is the antisymmetric combination of x(1)y(2)x(3)] The four symmetric spin functions are ααα, (ααβ + αβα + βαα), (αββ + ββα + βαβ), βββ 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 4 S state.

58 Energy of the ground state of N: A[(xα)(yα)(zα)] = [Axyz]ααα 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 ) 2K xy E( 2 P) =3h pp + 2J xy + J xx + 1K xy TA s check this x z Pictorial representation of the N ground state E( 2 D) =3h pp + 3J xy + 1K xy = 3h pp + 2J xy + J xx - 1K xy 4K xy E( 4 S) =3h pp + 3J xy - 3K xy copyright 2016 William A. Go ddard III, all rights reserved Since J xy = J xz = J yz and K xy = K xz = K yz 58

59 E( 1 S) Comparison with experiment 2K xy E( 1 D) 4K xy E( 3 P) 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) 59

60 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 (2p x ) 2 (2p y )(2p z ) x z Clearly it is better to have two singly occupied orbitals. (2p z ) 2 (2p x ) 2 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 x z 60

61 Summary ground state for O atom Ψ(1,2,3,4,5,6) xz = A[(1sα)(1sβ)(2sα)(2sβ)(2p y α)(2p y β)(2p x α)(2p z α)] = A[(1s) 2 (2s) 2 (2p y α)(2p y β)(2p x α)(2p z α)] = = A[(yα)(yβ)(xα)(zα)] = which we visualize as (2p y ) 2 (2p x )(2p z ) Ψ(1,2,3,4,5,6) yz = A[(xα)(xβ)((yα)(zα)] which we visualize as (2p x ) 2 (2p y )(2p z ) Ψ(1,2,3,4,5,6) xy = A[(zα)(zβ)(((xα)(yα)] x which we visualize as (2p ) 2 (2p )(2p ) z x y z x x z z copyright 2016 WilliamA. Goddard III, all rights reserved 61 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

62 Calculating energies for O atom The energy of Ψ(1,2,3,4,5,6) xz = A[(1sα)(1sβ)(2sα)(2sβ)(2p y α)(2p y β)(2p x α)(2p z α)] = A[[Be](yα)(yβ)(xα)(zα)] 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 ) 62

63 O (2p) 4 3 P Comparison of O states with C states Ne (2p) 6 1 S C (2p) 2 3 P Compared to Ne, we have x z 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 x 63 z

64 Summarizing the energies for O atom E( 1 S) 3K xy O S Se Te Po E( 1 D) 2K xy 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( 3 P) 64

65 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(1-9) z = A[(1sα)(1sβ)(2sα)(2sβ)(2p y α)(2p y β)(2p x α)(2p z α)] = A[(1s) 2 (2s) 2 (2p y α)(2p y β)(2p x α)(2p z α)] = = A{[Be](xα)(xβ)(yα)(yβ)(zα)} which we visualize as reserved x z copyright 2016 William A. Goddard III, all rights 65

66 The energy of Calculating energies for F atom Ψ(1-9) z = A{[Be](2p x α)(2p x β)(2p y α)(2p y β)(2p z α)] 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 ) 66

67 F (2p) 5 2 P Comparison of F states with B states Ne (2p) 6 1 S B (2p) 1 2 P x z 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 x 67 z

68 The energy of Calculating energies for F atom Ψ(1-9) z = A{[Be](2p x α)(2p x β)(2p y α)(2p y β)(2p z α)] 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 ) 68

69 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 è L=0. denote as 1 S Ψ(1-10) z = A[[Be](xα)(xβ)(yα)(yβ)(zα)(zβ)] x which we visualize as Ne (2p) 6 1 S z 69

70 Calculating energy for Ne atom (2p x ) 2 (2p y ) 2 (2p z ) The energy of 2 )} Ψ(1-9) z = A{[Be](2p x α)(2p x β)(2p y α)(2p y β)(2p z α)(2p z β copyright 2016 WilliamA. Goddard III, all rights reserved 70 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

71 Summary of ground states of Li-Ne erved 7 Li 2 S (2s) 1 N 4 S (2p) 3 Be 1 S (2s) 2 Ignore (2s) 2 B 2 P (2p) 1 O 3 P (2p) 4 F 2 P (2p) 5 C 3 P (2p) 2 Ne 1 S (2p) 6 copyright 2016 William A. Goddard III, all rights res 1

72 Bonding H atom to He Starting with the ground state of He, (1s) 2 = A(He1sα)(He1sβ) and bringing up an H atom (H1sα), leads to HeH: A[(He1sα)(He1sβ)(H1sα)] But properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the He 1s since both have an α spin. A[(He1sα)(He1sβ)(θα)] Where θ = H 1s S He 1s Consequently θ has a nodal plane, increasing its KE. Smaller R è larger S è larger increase in KE. Get a repulsive interaction, no bond 72 R

73 Bonding H atom to Ne Starting with the ground state of Ne, (1s) 2 (2s) 2 (2p) 6 Ψ(Ne)= A{(2p x α)(2p x β)(2p y α)(2p y β)(2p z α)(2p z β)} (omitting the Be) and bringing up an H atom (H1sα) along the z axis, leads to A{(2p x ) 2 (2p y ) 2 (Ne2p z α)(ne2p z β)(h1sα)} Where we focus on the Ne2pz orbital that overlaps the H atom The properties of A (Pauli Principle) tell us that the H1s must get orthogonal to the Ne R 2p z since both have an α spin. copyr ight 2016 W θ = H 1s S Ne 2pz θ has a nodal plane, increasing its KE. Smaller R è larger S è larger increase in KE. Get a repulsive interaction, no bond illiama. Goddard III, all rights reserved 73

74 Now consider Bonding H atom to all 3 states of F x z Bring H1s along z axis to F and consider all 3 spatial states. F 2p z doubly occupied, thus H1s must get orthogonal è repulsive A{(2p x α) 1 (2p y ) 2 (F2p z α)(f2p z β)(h1sα)} F 2p z doubly occupied, thus H1s must get orthogonal è repulsive A{(2p x ) 2 (2p y α) 1 (F2p z α)(f2p z β)(h1sα)} F 2p z singly occupied, Now H1s need not get orthogonal if it has opposite spin, can get bonding R 74

75 x Now consider Bonding H atom to x 2 y 2 z 1 state of F Focus on 2p z and H1s singly occupied energy orbitals 3Σ + z (P z H - H P z ) Antibonding state (S=1, triplet) [φ pz (1) φ H (2) - φ H (1) φ pz (2)](αα) Just like H 2. Bonding state (S=0, singlet) Full [φ pz wavefunction (1) φ H (2) + φ H (1) for bond φ pz (2)](αβ βα) becomes A{(F2p x ) 2 (F2p y ) 2 [(Fp z )(H)+(H)(Fp z )](αβ βα)} 1Σ + (Pz H + H P z) R copyright 2016 William A. Goddard III, all rights r Full wavefunction for antibond becomes A{(F2p x ) 2 (F2p y ) 2 [(Fp z )(H)-(H)(Fp z )](αα)} eserved 75

76 Schematic depiction of HF Denote the ground state of HF as Where the line connecting the two singly occupied orbitals è covalent bonding We will not generally be interested in the antibonding state, but if we were it would be denoted as 76

77 Bond a 2 nd Starting with the ground state of HF as H atom to the ground state of HF? can a 2nd H can be bonded covalently, say along the x axis? antibond bond A{(F2p x α)(f2p x β)(h x α)(f2p y ) 2 [(Fp z )(H z )+(H z )(Fp z )](αβ βα)} This leads to repulsive interactions just as for NeH. Since all valence orbitals are paired, there are no other possible covalent bonds and H 2 F is not stable. 77

78 Now consider Bonding H atom to all 3 states of O x Bring H1s along z axis to O and consider all 3 spatial states. O 2p z doubly occupied, thus H1s must get orthogonal è repulsive z O 2p z singly occupied. Now H1s need not get orthogonal if it has opposite spin, can get bonding Get S= ½ state, Two degenerate states, denote as 2 Π R 78

79 Bonding H atom to x 1 y 2 z 1 and x 2 y 1 z 1 states of O The full wavefunction for the bonding state 2Π y bond A{(O2p x ) 2 (O2p y α) 1 [(Op z )(H)+(H)(Op z )](αβ βα)} 2Π x bond x z A{(O2p x α) 1 (O2p y ) 2 [(Op z )(H)+(H)(Op z )](αβ βα)} 2Π R copyright 2016 William A. Goddard II (P z H + H P z ) I, all rights reserved 79

80 Bond a 2 nd H atom to the ground state of OH Starting with the ground state of OH, we can ask whether a 2 nd H can be bonded covalently, say along the x axis. x z 2Π y 2Π x Bonding a 2nd H along the x axis to the 2 Π y state leads to repulsive interactions just as for NeH. No bond. bond antibond A {(O2p x α)(o2p x β)(h x α) (O2p y α) 1 [(Op z )(H)+(H)(Op z )](αβ βα)} Bonding a 2nd H along the x axis to the 2 Π state x leads to a covalent bond bond bond copyright 2016 WilliamA. Goddard III, all rights reserved 80 A{(O2p y ) 2 [(Op x )(H x )+(H x )(Op x )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)}

81 Analize Bond in the ground state of H 2 O bond bond A{(O2p y ) 2 [(Op x )(H x )+(H x )(Op x )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)} This state of H 2 O is a spin singlet state, which we denote as 1 A 1. x z For optimum bonding, the pz orbital should point at the H z while the px orbital should point at the H x Thus the bond angle should be 90º. θ e R e In fact the bond angle is far from 90º for H 2 O, but it does approach 90º for S à Se à Te 81

82 What is origin of large distorsion in bond angle of H 2 O A{(O2p y ) 2 [(Op x )(H x )+(H x )(Op x )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)} x OHx bond OHz bond z Bonding Hz to pz leaves the Hz orbital orthogonal to px and py while bonding Hx to px leaves the Hx orbital orthogonal to py and pz, so that there should be little interference in the bonds, except that the Hz orbital can overlap the Hx orbital. copyright 2016 WilliamA. Goddard III, all rights reserved 82 Since the spin on Hx is α half the time and β the other half while the same is true for Hz, then ¼ the time both are α and ¼ the time both are β. Thus the Pauli Principle (the antisymmetrizer) forces these orbitals to become orthogonal. This increases the energy as the overlap of the 1s orbitals increases. Increasing the bond angle reduces this repulsive interaction

83 Testing the origin of bond angle distorsion in H 2 O x z If the increase in bond angle is a reponse to the overlap of the Hz orbital with the Hx orbital, then it should increase as the H---H distance decreases. In fact: θ e R e Thus the distortion increases as Re decreases, becoming very large for R=1A (H H of 1.4A, which leads to large overlap ~0.5). To test this interpretation, Emily Carter and wag (1983) carried out calculations of the optimum θ e as a function of R for H 2 O and found that increasing R from 0.96A to 1.34A, decreases in θ e by 11.5º (from º to 95º). 83

84 Validation of concept that the bond angle increase is due to H---H overlap Although the driving force for distorting the bond angle from 90º to 104.5º is H H overlap, the increase in the bond angle causes many changes in the wavefunction that can obscure the origin. Thus for the H s to overlap the O orbitals best, the pz and px orbtials mix in some 2s character, that opens up the angle between them. This causes the O2s orbital to build in p character to remain orthonal to the bonding orbitals. copyright WilliamA. Goddard III, all rights reserved 84

85 Bond a 3 rd H atom to the ground state of H 2 O? Starting with the ground state of H 2 O We can a 3rd H along the y axis. This leads to A{(O2p y α)(o2p y β)(h y α) [(Op x )(H x )+(H x )(Op x )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)} OHy antibond OHx bond OHz bond This leads to repulsive interactions just as for NeH. Since all valence orbitals are paired, there are no other possible covalent bonds and H 3 O is not stable. 85

86 Stopped Monday October 3, 2016

87 Now consider Bonding H atom to the ground state of N Bring H1s along z axis to O and N 2p z singly occupied, forms bond to Hz x R z A{(N2p x α)(n2p y α)[(np z )(H z )+(H z )(Np z )](αβ βα)} NHz bond Two unpaired spins, thus get S=1, triplet state Denote at 3 Σ - 87

88 Bond a 2 nd H atom to the ground state of NH Starting with the ground state of NH, bring a 2 nd H along the x axis. This leads to a 2 nd covalent bond. A{(N2p y α)[(np x )(H x )+(H x )(Np x )](αβ βα)[(np z )(H z )+(H z )(Np z )](αβ βα)} bond Denote this as 2 B 1 state. bond Again expect 90º bond angle. θ e R e x z Indeed as for H 2 O we find big deviations for the 1st row, but because N is bigger than O, the deviations are smaller. s reserved copyright 2016 William A. Goddard III, all right 87

89 Bond a 3 rd Starting with the ground state of H 2 N Ch120a-Goddard-L H atom to the ground state of H 2 N We can a 3rd H along the y axis. This leads to A{[(Np y )(H y )+(H y )(Np y )](αβ βα)[(np x )(H x )+(H x )(Np x )](αβ βα)[(np z )(H z )+(H z )(Np z )] (αβ βα)} NHy bond Denote this as 1 A 1 state. Again expect 90º bond angle. NHx bond Indeed as for H 2 N we find big deviations for the 1st row, but because there are not 3 bad H---H interactions the deviations are larger. θ e R e illiam A. Goddard III, all rights res NHz bond 03 copyright 2016 W erved 88

90 Bond a 4 th H atom to the ground state of H 3 N? The ground state of H 3 N has all valence orbitals are paired, there are no other possible covalent bonds and H 4 N is not stable. 89