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

Lecture 6, September 17, 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 6, September 17, 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. Last time EEWS-90.502-Goddard-L04

  3. Energy for 2 electron product of spinorbitals For H2, H(1,2) = h(1) + h(2) +1/r12 + 1/R Product wavefunction Ψ(1,2) = ψa(1)ψb(2) E = haa + hbb + Jab + 1/R haa ≡ <ψa(1)|h(1)|ψa(1)> ≡ <a|h|a> hbb ≡ <ψb(2)|h(2)|ψb(2)> ≡ <b|h|b> Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> = ∫| ψa(1)|2 |ψb(2)|2 /r12 is the total Coulomb interaction between the electron density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Thus Jab > 0 Jab 1/R if ra is centered on atom a while rb is centered on atom b a large distance R far from a. For shorter distances at which the densities overlap, the Jab decreases (shielding) EEWS-90.502-Goddard-L04

  4. Energy for antisymmetrized product of 2 spinorbitals A ψa(1)ψb(2) E = haa + hbb + (Jab –Kab) + 1/R Kab = < ψaψb|1/r12|ψb ψa > = ∫[ψa*(1)ψb(1)][ψb*(2)ψa(2)] /r12 is the exchange integral for ψa and ψb Jab > Kab > 0 EEWS-90.502-Goddard-L04

  5. Both electrons have the same spin ψa(1) = Φa(1)a(1) ψb(2) = Φb(2)a(2) <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb> the spatial orbitals for same spin must be orthogonal Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)] = [Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)] The total energy is Eaa = haa + hbb + (Jab –Kab) + 1/R haa≡ <Φa(1)|h(1)|Φa(1)> and similarly for hbb Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)> Kab≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)> Jab > Kab > 0 EEWS-90.502-Goddard-L04

  6. Assume electrons have the opposite spin ψa(1) = Φa(1)a(1) ψb(2) = Φb(2)b(2) Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]= We obtain <ψa|ψb>= 0 = < Φa| Φb><a|b> = 0 Independent of the overlap of the spatial orbitals. Thus spatial orbitals can overlap, <Φa|Φb> = S The exchange term for spin orbitals with opposite spin is zero; get exchange only between spinorbitals with the same spin Thus the total energy is Eab = haa + hbb + Jab + 1/R Just as for the simple product wavefunction, Φa(1)Φb(2) EEWS-90.502-Goddard-L04

  7. For spinorbitals with opposite spin, must combine Slater Determinants to obtain full permutational symmetry The antisymmetrized wavefunction leads to Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]= =[Φa(1) a(1)][Φb(2)b(2)] – [Φb(1) b(1)][Φa(2)a(2)] Interchanging the spins leads to [Φa(1) b(1)][Φb(2)a(2)] – [Φb(1) a(1)][Φa(2)b(2)] = = A[Φa(1)Φb(2)][b(1)a(2)] Which is neither + or – times the starting wavefunction. Thus must combine to obtain proper spatial and spin symmetry EEWS-90.502-Goddard-L04

  8. Correct space and spin symmetry for ab wavefunctions [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)] Is symmetric in spin coordinates (the MS = 0 component of the S=1 triplet) [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(2)][a(1)b(2)] Is antisymmetric in spin coordinates (the MS = 0 component of the S=0 triplet) Thus for the ab case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry EEWS-90.502-Goddard-L04

  9. The triplet state for 2 electrons The wavefunction [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)] Leads directly to 3Eab = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)] [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)] These three states are collectively referred to as the triplet state and denoted as having spin S=1. It has 3 components MS = +1: [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)] MS = 0 : [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)] MS = -1: [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)] where < Φa|Φb> = 0 EEWS-90.502-Goddard-L04

  10. The singlet state for two electrons The other combination of MS=0 determinants leads to the singlet state and is denoted as having spin S=0 [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] We will analyze the energy for this wavefunction next. It is more complicated since < Φa|Φb> ≠ 0 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)> 1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2) This is the general energy for the ab+ba singlet, but most relevant to us is for H2, where Φa=XL and Φb=XR In this case it is convenient to write the triplet state in terms of the same overlapping orbitals, even though they could be orthogonalized for the triplet state EEWS-90.502-Goddard-L04

  11. Analysis of singlet and triplet energies for H2 Taking Φa=XL and Φb=XRfor H2, the VB energy for the bonding state (g, singlet) is 1Eg = <ab|H|(ab+ba)>/<ab|(ab+ba)> 1Eg = {(haa + hbb + (hab + hba) S + Jab + Kab + (1+S2)/R}/(1 + S2) Similary for the VB triplet we obtain 3Eu = <ab|H|(ab-ba)>/<ab|(ab-ba)> 3Eu = {(haa + hbb - (hab + hba) S + Jab - Kab + (1-S2)/R}/(1 - S2) We find it useful to define a classical energy, with no exchange or interference or resonance Ecl = <ab|H|ab>/<ab|ab> = haa + hbb + Jab + 1/R Then we can define the energy as Egx = +Ex/(1 + S2) Eux = - Ex/(1 - S2) Ex = {(hab + hba) S + Kab –EclS2} 1Eg = Ecl + Egx 3Eu = Ecl + Eux where EEWS-90.502-Goddard-L04

  12. The VB exchange energies for H2 For H2, the classical energy is slightly attractive, but again the difference between bonding (g) and anti bonding (u) is essentially all due to the exchange term. 1Eg = Ecl + Egx 3Eu = Ecl + Eux -Ex/(1 - S2) Each energy is referenced to the value at R=∞, which is -1 for Ecl, Eu, Eg 0 for Exu and Exg +Ex/(1 + S2) EEWS-90.502-Goddard-L04

  13. Analysis of the VB exchange energy, Ex where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2 Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St Where t = (hab – Shaa) contains the 1e part T2 ={Kab –S2Jab} contains the 2e part Clearly the Ex is dominated by T1 and clearly T1 is dominated by the kinetic part, TKE. T2 T1 Ex Thus we can understand bonding by analyzing just the KE part if Ex TKE EEWS-90.502-Goddard-L04

  14. Re-examine the energy for H2+ The same kinetic term important for H2 is also the critical part of the energy for H2+ the VB wavefunctions are Φg= (хL + хR) and Φu= (хL - хR) (ignoring normalization) where H = h + 1/R. This leads to eg = <L+R|H|L+R>/ <L+R|L+R> = 2 <L|H|L+R>/ 2<L|L+R> = (hLL + hLR)/(1+S) + 1/R = (hLL+ShLL+hLR -ShLL)/(1+S) + 1/R = (hLL + 1/R) + (hLR-ShLL)/(1+S) eg = ecl + t/(1+S) eu = ecl - t/(1-S) ecl = (hLL + 1/R) is the energy for bringing a proton up to H atom t = (hLR - ShLL) contains the terms that dominate the bonding and antibonding character of these 2 states EEWS-90.502-Goddard-L04

  15. The VB interference or resonance energy for H2+ The VB wavefunctions for H2+Φg= (хL + хR) and Φu= (хL - хR) lead to eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux ecl = (hLL + 1/R) is the classical energy t = (hLR - ShLL) is the VB interference or resonance energy that dominates the bonding and antibonding of these states t/(1+S) t/(1+S) EEWS-90.502-Goddard-L04

  16. Contragradience хL хR The above discussions show that the interference or exchange part of KE dominates the bonding, tKE=KELR –S KELL This decrease in the KE due to overlapping orbitals is dominated by tx = ½ [< (хL). ((хR)> - S[< (хL)2> Dot product is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience) EEWS-90.502-Goddard-L04

  17. Comparison of exchange energies for 1e and 2e bonds E(hartree) Eu1x Eg1x R(bohr) H2 Eg1x ~ +2St /(1 + S2) Eu1x ~ -2St /(1 - S2) H2+ case egx ~ +t/(1 + S) eux ~ -t/(1 - S) For R=1.6bohr (near Re), S=0.7 Eg1x ~ 0.94t vs. egx ~ 0.67t Eu1x ~ -2.75t vs. eux ~ -3.33t For R=4 bohr, S=0.1 Eg1x ~ 0.20t vs. egx ~ 0.91t Eu1x ~ -0.20t vs. eux ~ -1.11t EEWS-90.502-Goddard-L04

  18. H atom, excited states r +Ze In atomic units, the Hamiltonian h = - (Ћ2/2me)2– Ze2/r Becomes h = - ½ 2– Z/r Thus we want to solve hφk = ekφk for all excited states k φnlm = Rnl(r) Zlm(θ,φ) product of radial and angular functions Ground state: R1s = exp(-Zr), Zs = 1 (constant) EEWS-90.502-Goddard-L04

  19. The excited angular states of H atom, 1 nodal plane z z pz + px - + x - x φnlm = Rnl(r) Zlm(θ,φ) the excited angular functions, Zlm must have nodal planes to be orthogonal to Zs 3 cases with one nodal plane Z10=pz= r cosθ (zero in the xy plane) Z11=px= rsinθcosφ (zero in the yz plane) Z1-1=py= rsinθsinφ (zero in the xz plane) We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2p = 360º If m=0, denote it as s: pz = ps If m=1, denote it as p: px, py = pp If m=2 we call it a d function If m=3 we call it a f function EEWS-90.502-Goddard-L04

  20. The excited angular states of H atom, 2 nodal planes z dz2 + - - x + 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 cos2φ Z22 = dxy = xy =r2 (sinθ)2 sin2φ m = 1, dp 57º m = 2, dd Where we used [(cosφ)2 – (sinφ)2]=cos2φ and 2cosφ sinφ=sin2φ Summarizing: 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 seven f angular functions (three angular nodes) called ℓ=3 nine g angular functions (four angular nodes) called ℓ=4 ℓ is referred to as the angular momentum quantum number There are (2ℓ+1) m values, Zℓm for each ℓ EEWS-90.502-Goddard-L04

  21. 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) EEWS-90.502-Goddard-L04

  22. Combination of radial and angular nodal planes ˉ ˉ ˉ R 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 EEWS-90.502-Goddard-L04

  23. New material EEWS-90.502-Goddard-L04

  24. 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 1.7A 0.74A H 4.8 H C H--H H H H H H H H H EEWS-90.502-Goddard-L04

  25. 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 zero EEWS-90.502-Goddard-L04

  26. Plotting of orbitals: line cross-section vs. contour EEWS-90.502-Goddard-L04

  27. Contour plots of 1s, 2s, 2p hydrogenic orbitals EEWS-90.502-Goddard-L04

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

  29. Contour plots of 4s, 4p, 4d hydrogenic orbtitals EEWS-90.502-Goddard-L04

  30. Contour plots of hydrogenic 4f orbitals EEWS-90.502-Goddard-L04

  31. He atom With 2 electrons, the ground state has both in the He+ 1s orbital 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, KE = ½ Z2, PE = -Z2, E=-Z2/2 = -2 h0. writing PE=-Z/R0, we see that the average radius is R0=1/z Now consider He atom: EHe = 2(½ z2) – 2Zz + J1s,1s EEWS-90.502-Goddard-L04

  32. e-e energy of He atom e1 R0 e2 How can we estimate J1s,1s Assume that each electron moves on a sphere withthe 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 EEWS-90.502-Goddard-L04

  33. 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. EEWS-90.502-Goddard-L04

  34. 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 2-0.31=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) EEWS-90.502-Goddard-L04

  35. 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 EEWS-90.502-Goddard-L04

  36. 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 EEWS-90.502-Goddard-L04

  37. 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 orbtials. 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 EEWS-90.502-Goddard-L04

  38. 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

  39. 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 6 2p Get generalized energy spectrum for filling in the electrons to explain the periodic table. Full shells at 2, 10, 18, 30, 36 electrons 2 2s He, 2 2 1s EEWS-90.502-Goddard-L04

  40. Kr, 36 Zn, 30 Ar, 18 Ne, 10 He, 2 EEWS-90.502-Goddard-L04

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

  42. 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 EEWS-90.502-Goddard-L04

  43. 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

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  46. EEWS-90.502-Goddard-L04

  47. 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

  48. 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. EEWS-90.502-Goddard-L04

  49. 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 EEWS-90.502-Goddard-L04

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

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