© 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 6, September 17, 2009

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 2 Last time

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 3 For H 2, H(1,2) = h(1) + h(2) +1/r /R Product wavefunction Ψ(1,2) = ψ a (1)ψ b (2) E = h aa + h bb + J ab + 1/R h aa ≡ ≡ h bb ≡ ≡ J ab ≡ = ∫| ψ a (1)| 2 |ψ b (2)| 2 /r 12 is the total Coulomb interaction between the electron density  a (1)=| ψ a (1)| 2 and  b (2)=| ψ b (2)| 2 Thus J ab > 0 J ab 1/R if  a is centered on atom a while  b is centered on atom b a large distance R far from a. For shorter distances at which the densities overlap, the J ab decreases (shielding) Energy for 2 electron product of spinorbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 4 Energy for antisymmetrized product of 2 spinorbitals A ψ a (1)ψ b (2) E = h aa + h bb + ( J ab – K ab ) + 1/R K ab = = ∫[ψ a * (1)ψ b (1)][ψ b * (2)ψ a (2)] /r 12 is the exchange integral for ψ a and ψ b J ab > K ab > 0

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 5 Both electrons have the same spin ψ a (1) = Φ a (1)  (1) ψ b (2) = Φ b (2)  (2) = 0 = = the spatial orbitals for same spin must be orthogonal A ψ a (1)ψ b (2)= A [Φ a (1)Φ b (2)][  (1)  (2)] = [Φ a (1)Φ b (2)- Φ b (1)Φ a (2)][  (1)  (2)] The total energy is E  = h aa + h bb + (J ab –K ab ) + 1/R h aa ≡ and similarly for h bb J ab = K ab ≡ J ab > K ab > 0

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 6 ψ a (1) = Φ a (1)  (1) ψ b (2) = Φ b (2)  (2) A ψ a (1)ψ b (2)= A [Φ a (1)Φ b (2)][  (1)  (2)]= We obtain = 0 = = 0 Independent of the overlap of the spatial orbitals. Thus spatial orbitals can overlap, = 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 E  = h aa + h bb + J ab + 1/R Just as for the simple product wavefunction, Φ a (1)Φ b (2) Assume electrons have the opposite spin

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 7 For spinorbitals with opposite spin, must combine Slater Determinants to obtain full permutational symmetry The antisymmetrized wavefunction leads to A ψ a (1)ψ b (2)= A [Φ a (1)Φ b (2)][  (1)  (2)]= =[Φ a (1)  (1)][Φ b (2)  (2)] – [Φ b (1)  (1)][Φ a (2)  (2)] Interchanging the spins leads to [Φ a (1)  (1)][Φ b (2)  (2)] – [Φ b (1)  (1)][Φ a (2)  (2)] = = A [Φ a (1)Φ b (2)][  (1)  (2)] Which is neither + or – times the starting wavefunction. Thus must combine to obtain proper spatial and spin symmetry

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 8 Correct space and spin symmetry for  wavefunctions [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)+  (1)  (2)]= A [Φ a (1)Φ b (2)][  (1)  (2)]- A [Φ b (1)Φ a (2)][  (1)  (2)] Is symmetric in spin coordinates (the M S = 0 component of the S=1 triplet) [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][  (1)  (2)-  (1)  (2)]= A [Φ a (1)Φ b (2)][  (1)  (2)]+ A [Φ b (1)Φ a (2)][  (1)  (2)] Is antisymmetric in spin coordinates (the M S = 0 component of the S=0 triplet) Thus for the  case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 9 The triplet state for 2 electrons The wavefunction [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)+  (1)  (2)] Leads directly to 3 E  = h aa + h bb + (J ab –K ab ) + 1/R Exactly the same as for [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)] [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)] These three states are collectively referred to as the triplet state and denoted as having spin S=1. It has 3 components M S = +1: [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)] M S = 0 : [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)+  (1)  (2)] M S = -1: [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][  (1)  (2)] where = 0

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 10 The singlet state for two electrons The other combination of M S =0 determinants leads to the singlet state and is denoted as having spin S=0 [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][  (1)  (2)-  (1)  (2)] We will analyze the energy for this wavefunction next. It is more complicated since ≠ 0 1 E = / 1 E = {(h aa + h bb + (h ab + h ba ) S 2 + J ab + K ab + (1+S 2 )/R}/(1 + S 2 ) This is the general energy for the ab+ba singlet, but most relevant to us is for H 2, where Φ a =X L and Φ b =X R 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 11 Analysis of singlet and triplet energies for H 2 Taking Φ a =X L and Φ b =X R for H 2, the VB energy for the bonding state (g, singlet) is 1 E g = / 1 E g = {(h aa + h bb + (h ab + h ba ) S + J ab + K ab + (1+S 2 )/R}/(1 + S 2 ) Similary for the VB triplet we obtain 3 E u = / 3 E u = {(h aa + h bb - (h ab + h ba ) S + J ab - K ab + (1-S 2 )/R}/(1 - S 2 ) We find it useful to define a classical energy, with no exchange or interference or resonance E cl = / = h aa + h bb + J ab + 1/R E g x = + E x /(1 + S 2 ) E u x = - E x /(1 - S 2 ) E x = {(h ab + h ba ) S + K ab –E cl S 2 } where Then we can define the energy as 1 E g = E cl + E g x 3 E u = E cl + E u x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 12 The VB exchange energies for H 2 1 E g = E cl + E g x 3 E u = E cl + E u x For H 2, 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. Each energy is referenced to the value at R=∞, which is -1 for E cl, E u, E g 0 for E x u and E x g + E x /(1 + S 2 ) - E x /(1 - S 2 )

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 13 Analysis of the VB exchange energy, E x where E x = {(h ab + h ba ) S + K ab –E cl S 2 } = T   T  Here T   {(h ab + h ba ) S –(h aa + h bb )S 2 } = 2S  Where  = (h ab – Sh aa ) contains the 1e part T    {K ab –S 2 J ab } contains the 2e part Clearly the E x is dominated by T  and clearly T  is dominated by the kinetic part, T . ExEx T2T2 T1T1 T  Thus we can understand bonding by analyzing just the KE part if E x

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 14 e cl = (h LL + 1/R) is the energy for bringing a proton up to H atom  = (h LR - Sh LL ) contains the terms that dominate the bonding and antibonding character of these 2 states Re-examine the energy for H 2 + The same kinetic term important for H2 is also the critical part of the energy for H 2 + the VB wavefunctions are Φ g = (х L + х R ) and Φ u = (х L - х R ) (ignoring normalization) where H = h + 1/R. This leads to e g = / = 2 / 2 = (h LL + h LR )/(1+S) + 1/R = (h LL +Sh LL +h LR - Sh LL )/(1+S) + 1/R = (h LL + 1/R) + (h LR -Sh LL )/(1+S) e g = e cl +  /(1+S) e u = e cl -  /(1-S)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 15 The VB interference or resonance energy for H 2 + The VB wavefunctions for H 2 + Φ g = (х L + х R ) and Φ u = (х L - х R ) lead to e g = (h LL + 1/R) +  /(1+S) ≡ e cl + E g x e u = (h LL + 1/R) -  /(1-S) ≡ e cl + E u x e cl = (h LL + 1/R) is the classical energy  = (h LR - Sh LL ) is the VB interference or resonance energy that dominates the bonding and antibonding of these states  /(1+S)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 16 Contragradience The above discussions show that the interference or exchange part of KE dominates the bonding,  KE =KE LR –S KE LL This decrease in the KE due to overlapping orbitals is dominated by Dot product is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience)  x = ½ [ - S [ хLхL хRхR

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 17 Comparison of exchange energies for 1e and 2e bonds H 2 E g 1x ~ +2S  /(1 + S 2 ) E u 1x ~ -2S  /(1 - S 2 ) H 2 + case e g x ~ +  /(1 + S) e u x ~ -  /(1 - S) E u 1x E g 1x R(bohr) E(hartree) For R=1.6bohr (near R e ), S=0.7 E g 1x ~ 0.94  vs. e g x ~  E u 1x ~  vs. e u x ~  For R=4 bohr, S=0.1 E g 1x ~ 0.20  vs. e g x ~  E u 1x ~  vs. e u x ~ 

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 18 H atom, excited states In atomic units, the Hamiltonian h = - (Ћ 2 /2m e )    – Ze 2 /r Becomes h = - ½    – Z/r Thus we want to solve hφ k = e k φ k for all excited states k r +Ze φ nlm = R nl (r) Z lm (θ,φ) product of radial and angular functions Ground state: R 1s = exp(-Zr), Z s = 1 (constant)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 19 The excited angular states of H atom, 1 nodal plane φ nlm = R nl (r) Z lm (θ,φ) the excited angular functions, Z lm must have nodal planes to be orthogonal to Z s 3 cases with one nodal plane x z + - pzpz Z 10 =p z = r cosθ (zero in the xy plane) Z 11 =p x = rsinθcosφ (zero in the yz plane) Z 1-1 =p y = rsinθsinφ (zero in the xz plane) x z +- pxpx We find it useful to keep track of how often the wavefunction changes sign as the φ coordinate is increased by 2  = 360º If m=0, denote it as  : p z = p  If m=1, denote it as  : p x, p y = p  If m=2 we call it a  function If m=3 we call it a  function

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 20 The excited angular states of H atom, 2 nodal planes x z + - d z º Z 20 = d z2 = [3 z 2 – r 2 ] m=0, d  Z 21 = d zx = zx =r 2 (sinθ)(cosθ) cosφ Z 2-1 = d yz = yz =r 2 (sinθ)(cosθ) sinφ Z 22 = d x2-y2 = x 2 – y 2 = r 2 (sinθ) 2 cos2φ Z 22 = d xy = xy =r 2 (sinθ) 2 sin2φ m = 1, d  m = 2, d  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, Z ℓm for each ℓ Where we used [(cosφ)2 – (sinφ)2]=cos2φ and 2cosφ sinφ=sin2 φ

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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, 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 Note that we are plotting the cross section along the z axis, but it would look exactly the same along any other axis. Here R 20 = 2s = [Zr/2 – 1]exp(-Zr/2) and R 30 = 3s = [2(Zr) 2 /27 – 2(Zr)/3 + 1]exp(-Zr/3) Zr = 2 Zr = 1.9 Zr = 7.1

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 22 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/, where =n 2 /Z Thus E nlm = - Z/(2 ) 1s s p s p d s p d f name total nodal planesradial nodal planesangular nodal planes ˉ R ˉ R ˉ R Size (a 0 )

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 23 New material

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 H--H C 0.74A H H H H 1.7A HH HH HH 4.8

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 25 Hydrogen atom excited states 1s -0.5 h 0 = eV 2s h 0 = -3.4 eV 2p3s h 0 = -1.5 eV 3p3d4s h 0 = -0.9 eV 4p4d4f Energy zero

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 26 Plotting of orbitals: line cross-section vs. contour

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 27 Contour plots of 1s, 2s, 2p hydrogenic orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 28 Contour plots of 3s, 3p, 3d hydrogenic orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 29 Contour plots of 4s, 4p, 4d hydrogenic orbtitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 30 Contour plots of hydrogenic 4f orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 31 He atom With 2 electrons, the ground state has both in the He + 1s orbital Ψ He (1,2) = A [(Φ 1s  )(Φ 1s  Φ 1s (1)Φ 1s (2) (   He  + J 1s,1s Now consider He atom: E He = 2(½  2 ) – 2Z  J 1s,1s First lets review the energy for He +. Writing Φ 1s   exp(-  r) we determine the optimum  for He + as follows = + ½  2 (goes as the square of 1/size) = - Z  (linear in 1/size) Applying the variational principle, the optimum  must satisfy dE/d  =  -  Z = 0 leading to  =  Z, KE = ½ Z 2, PE = -Z 2, E=-Z 2 /2 = -2 h 0. writing PE=-Z/R 0, we see that the average radius is R 0 =1/ 

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 32 e-e energy of He atom How can we estimate J 1s,1s Assume that each electron moves on a sphere withthe average radius R 0 = 1/  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  A rigorous calculation (notes chapter 3, appendix 3-C page 6) Gives J 1s,1s = (5/8)  R0R0 e1e1 e2e2

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 33 The optimum atomic orbital for He atom He atom: E He = 2(½  2 ) – 2Z  (5/8)  Applying the variational principle, the optimum  must satisfy dE/d  = 0 leading to 2  - 2Z + (5/8) = 0 Thus  = (Z – 5/16) = KE = 2(½  2 ) =  2 PE = - 2Z  (5/8)  = -2  2 E= -  2 = h 0 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 approximation accounts for 98.1% of the exact result.

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 34 Interpretation: The optimum atomic orbital for He atom Ψ He (1,2) = Φ 1s (1)Φ 1s (2) with Φ 1s   exp(-  r) We find that the optimum  = (Z – 5/16) = With this value of , the total energy is E= -  2 = h 0 This wavefunction can be interpreted as two electrons moving 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. Ionizing the 2 nd electron, the 1 st electron readjusts to  = Z = 2 With E(He + ) = -Z 2 /2 = - 2 h 0. thus the ionization potential (IP) is h0 = 23.1 eV (exact value = 24.6 eV)

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 35 Now lets add a 3 rd electron to form Li Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  (Φ 1s  Problem with either  or , we get Ψ Li (1,2,3) = 0 This is an essential result of the Pauli principle Thus the 3 rd electron must go into an excited orbital Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2s  or Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2pz  (or 2px or 2py) First consider Li + with Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  Here Φ 1s   exp(-  r) with  = Z = 2.69 and E = -  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/  = a 0 = 0.2A

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 36 Consider adding the 3 rd electron to the 2p orbital Ψ Li (1,2,3) = A [(Φ 1s  )(Φ 1s  )(Φ 2pz  (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 = 3.40 eV 0.2A 1s 2.12A 2p

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 37 Consider adding 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 orbtials. The result is Z eff 2s = 3 – 1.72 = 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 = 5.57 eV 0.2A 1s 2.12A 2s R~0.2A

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 38 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 39 Aufbau principle for atoms 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f Energy He, 2 Ne, 10 Ar, 18 Zn, 30 Kr, 36 Get generalized energy spectrum for filling in the electrons to explain the periodic table. Full shells at 2, 10, 18, 30, 36 electrons

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 40 He, 2 Ne, 10 Ar, 18 Zn, 30 Kr, 36

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 41 Many-electron configurations General aufbau ordering Particularly stable

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 ~ 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 = 3.6

© copyright 2009 William A. Goddard III, all rights reservedEEWS 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 ~ 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 44

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 45

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 46

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 47 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 48 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 ++ 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. (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 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

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 50 Transtion metal orbitals

© copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 51 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 52 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 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 53