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Lecture 4, January 14, 2011 aufbau principle atoms

Lecture 4, January 14, 2011 aufbau principle atoms. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093

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Lecture 4, January 14, 2011 aufbau principle atoms

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  1. Lecture 4, January 14, 2011 aufbau principle atoms Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy 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 Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>

  2. The excited states of H atom z θ y φ x hφk = ekφk 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 = d2/dx2 + d2/dy2 + d2/dy2 is independent of θ and φ (it transforms like r2 = x2 + y2 + z2). Consequently the eigenfunctions of h can be written as a factor depending only on r and a factor depending only on θ and φ φnlm = Rnl(r) Zlm(θ,φ) where the reason for the numbers nlm will become apparent later

  3. The ground state of H atom z θ y φ x h = - ½ 2– Z/r φnlm = Rnl(r) Zlm(θ,φ) For the ground state, R1s = exp(-Zr) and Z1s = Z00 = 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, which is clearly the best for KE.

  4. The p excited angular states of H atom z z pz + px - + x - x z pxpz - + x + - φnlm = Rnl(r) Zlm(θ,φ) Now lets consider excited angular functions, Zlm. They must have nodal planes to be orthogonal to Z00 The simplest would be Z10=z= r cosθ, which is zero in the xy plane. Exactly equivalent are Z11=x= rsinθcosφ which is zero in the yz plane, and Z1-1=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 ∫yz has nodes in both the xy and xz planes, leading to a zero integral pz

  5. More p functions? px’ z a + - x So far we have the s angular function Z00 = 1 with no angular nodal planes And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal plane Can we form any more with one nodal plane orthogonal to the above 4 functions? For example we might rotate px by an angle a about the y axis to form px’. However multiplying, say by pz, leads to the integrand pzpx’ which clearly does not integrate to zero pzpx’ . z Thus there are exactly three pi functions, Z1m, with m=0,+1,-1, all of which have the same KE. Since the p functions have nodes, they lead to a higher KE than the s function a + - - x +

  6. More angular functions? z dxz - + x + - So far we have the s angular function Z00 = 1 with no angular nodal planes And three p angular functions: Z10 =pz, Z11 =px, Z1-1 =py, each with one angular nodal plane Next in energy will be the d functions with two angular nodal planes. We can easily construct three functions dxy = xy =r2 (sinθ)2 cosφ sinφ dyz = yz =r2 (sinθ)(cosθ)sinφ dzx = zx =r2 (sinθ)(cosθ)cosφ where dxz is shown here Each of these is orthogonal to each other (and to the s and the three p functions)

  7. More d angular functions? z dz2-x2 + - - x + In addition we can construct three other functions with two nodal planes dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2] dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2] dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2] where dz2-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

  8. More d angular functions? z dz2-x2 + - - x + In addition we can construct three other functions with two nodal planes dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2] dy2-z2 = y2 – z2 = r2 [(sinθ)2(sinφ)2 – (cosθ)2] dz2-x2 = z2 – x2 = r2 [(cosθ)2 -(sinθ)2(cosφ)2] where dz2-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 (x2 – y2) + (y2 – z2) + (z2 – x2) = 0 Which indicates that there are only two independent such functions. We combine the 2nd two as (z2 – x2) - (y2 – z2) = [2 z2 – x2 - y2 ] = [3 z2 – x2 - y2 –z2] = = [3 z2 – r2 ] which we denote as dz2

  9. Summarizing we get 5 d angular functions Z20 = dz2 = [3 z2 – r2 ] m=0, ds Z21 = dzx = zx =r2 (sinθ)(cosθ)cosφ Z2-1 = dyz = yz =r2 (sinθ)(cosθ)sinφ Z22 = dx2-y2 = x2 – y2 = r2 (sinθ)2 [(cosφ)2 – (sinφ)2] Z22 = dxy = xy =r2 (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 2p = 360º When this number, m=0 we call it a s function When m=1 we call it a p function When m=2 we call it a d function When m=3 we call it a f function z dz2 + m = 1, dp - - x + m = 2, dd 57º

  10. Summarizing the angular functions • So far we have • 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

  11. The real (Zlm) and complex (Ylm) momentum functions Here the bar over m  negative

  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, Rnl(r) The lowest is R10 = 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 R20 = 2s = [Zr/2 – 1]exp(-Zr/2) and R30 = 3s = [2(Zr)2/27 – 2(Zr)/3 + 1]exp(-Zr/3)

  13. Combination of radial and angular nodal planes R ˉ ˉ ˉ R R 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. Enlm = - Z2/2n2 The potential energy is given by PE = - Z2/2n2 = -Z/ , where =n2/Z Thus Enlm = - Z/2 angular nodal planes Size (a0) radial nodal planes total nodal planes name 1s 0 0 0 1.0 2s 1 1 0 4.0 2p 1 0 1 4.0 3s 2 2 0 9.0 3p 2 1 1 9.0 3d 2 0 2 9.0 4s 3 3 0 16.0 4p 3 2 1 16.0 4d 3 1 2 16.0 4f 3 0 3 16.0

  14. Sizes hydrogen orbitals ˉ =(n2/Z) a0 = 0.53 n2 A R Hydrogen orbitals 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f Angstrom (0.1nm) 0.53, 2.12, 4.77, 8.48 1.7 H 4.8 H C H--H H H H H 0.74 H H H H

  15. Hydrogen atom excited states -0.033 h0 = -0.9 eV 4s 4p 4d 4f -0.056 h0 = -1.5 eV 3s 3p 3d -0.125 h0 = -3.4 eV 2s 2p -0.5 h0 = -13.6 eV 1s Energy Enlm = - Z2/2n2 = -(1/2n2)h0 = 13.6 eV/n2 zero

  16. Plotting of orbitals: line cross-section vs. contour

  17. Contour plots of 1s, 2s, 2p hydrogenic orbitals 1s, no nodes 2s, one radial node 2p, one angular node

  18. Contour plots of 3s, 3p, 3d hydrogenic orbitals 3p, one angular node one radial node 3s, two radial nodes 3d, two angular nodes

  19. Contour plots of 4s, 4p, 4d hydrogenic orbtitals 4p, one angular node two radial nodes 4s, 3 radial nodes 4d, two angular nodes and one radial node

  20. Contour plots of hydrogenic 4f orbitals All seven 4f have three angular nodes and no radial nodes

  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 H2 ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2)(ab-ba) EHe= 2<1s|h|1s> + J1s,1s two one-electron energies one Coulomb repulsion

  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 H2 ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2)(ab-ba) EHe= 2<1s|h|1s> + J1s,1s First lets review the energy for He+. Writing Φ1s= exp(-zr) we determine the optimum z for He+ as follows <1s|KE|1s> = + ½ z2 (goes as the square of 1/size) <1s|PE|1s> = - Zz (linear in 1/size)

  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 H2 ΨHe(1,2) = A[(Φ1sa)(Φ1sb)]= Φ1s(1)Φ1s(2)(ab-ba) EHe= 2<1s|h|1s> + J1s,1s First lets review the energy for He+. Writing Φ1s= exp(-zr) we determine the optimum z for He+ as follows <1s|KE|1s> = + ½ z2 (goes as the square of 1/size) <1s|PE|1s> = - Zz (linear in 1/size) Applying the variational principle, the optimum z must satisfy dE/dz = z -Z = 0 leading to z =Z, This value of z leads to KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0. write PE=-Z/R0, so that the average radius is R0=1/z

  24. J1s,1s = e-e energy of He atom – He+ orbitals e1 R0 e2 Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s How can we estimate J1s,1s Assume that each electron moves on a sphere With the average radius R0 = 1/z And assume that e1 at along the z axis (θ=0) Neglecting correlation in the electron motions, e2 will on the average have θ=90º so that the average e1-e2 distance is ~sqrt(2)*R0 Thus J1s,1s ~ 1/[sqrt(2)*R0] = 0.71 z A rigorous calculation (notes chapter 3, appendix 3-C page 6) Gives J1s,1s = (5/8) z

  25. The optimum atomic orbital for He atom He atom: EHe = 2(½ z2) – 2Zz + (5/8)z Applying the variational principle, the optimum z must satisfy dE/dz = 0 leading to 2z - 2Z + (5/8) = 0 Thus z = (Z – 5/16) = 1.6875 KE = 2(½ z2) = z2 PE = - 2Zz + (5/8)z = -2 z2 E= - z2 = -2.8477 h0 Ignoring e-e interactions the energy would have been E = -4 h0 The exact energy is E = -2.9037 h0 (from memory, TA please check). Thus this simple approximation accounts for 98.1% of the exact result.

  26. Interpretation: The optimum atomic orbital for He atom ΨHe(1,2) = Φ1s(1)Φ1s(2) with Φ1s= exp(-zr) We find that the optimum z = (Z – 5/16) = 1.6875 With this value of z, the total energy is E= - z2 = -2.8477 h0 This wavefunction can be interpreted as two electrons moving 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 2.00-0.31=1.69 The total energy is just the sum of the individual energies. Ionizing the 2nd electron, the 1st electron readjusts to z = Z = 2 With E(He+) = -Z2/2 = - 2 h0. thus the ionization potential (IP) is 0.8477 h0 = 23.1 eV (exact value = 24.6 eV)

  27. Now lets add a 3rd electron to form Li ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)(Φ1sg)] Problem with either g = a or g = b, we get ΨLi(1,2,3) = 0 This is an essential result of the Pauli principle Thus the 3rd electron must go into an excited orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2sa)] or ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (or 2px or 2py) First consider Li+ with ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb)] Here Φ1s= exp(-zr) with z = Z-0.3125 = 2.69 and E = -z2 = -7.2226 h0. Since the E (Li2+) = -9/2 = -4.5 h0 the IP = 2.7226 h0 = 74.1 eV The size of the 1s orbital is R0 = 1/z = 0.372 a0 = 0.2A

  28. Consider adding the 3rd electron to the 2p orbital ΨLi(1,2,3) = A[(Φ1sa)(Φ1sb) )(Φ2pza)] (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 Zeff = 3 – 2 = 1, leading to a size of R2p = n2/Zeff = 4 a0 = 2.12A And an energy of e = -(Zeff)2/2n2 = -1/8 h0 = -3.40 eV 1s 0.2A 2p 2.12A

  29. Consider adding 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 This leads to a size of R2s = n2/Zeff = 3.1 a0 = 1.65A And an energy of e = -(Zeff)2/2n2 = -0.205 h0 = -5.57 eV

  30. 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

  31. Aufbau principle for atoms Energy 14 10 4f Kr, 36 4d 6 4p Zn, 30 2 10 4s Ar, 18 3d 6 3p 2 3s Ne, 10 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 6 2p 2 2s He, 2 2 1s

  32. Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, 2

  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 ~ 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 (2s Li)/(2s Ne) ~ 4.64/1.28 = 3.6

  34. Many-electron configurations General aufbau ordering Particularly stable

  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 ~ 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

  36. 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

  37. 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 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.

  38. 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

  39. Transtion metal orbitals

  40. 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

  41. Consider the ground state of B: [Be](2p)1 x 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

  42. 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

  43. 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 spatially symmetric, thus 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)]

  44. 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 = S = 0 [φx(1) φy(2) - φy(1) φx(2)](ab+ba), also aa and bb or 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

  45. 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

  46. 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

  47. 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

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