Lecture 3, January 9, 2015 Bonding in H2+

Transcription

1 Lecture 3, January 9, 2015 Bonding in H2+ Elements of Quantum Chemistry with Applications to Chemical Bonding and Properties of Molecules and Solids Course number: Ch125a; Room 147 Noyes Hours: 11-11:50am Monday, Wednesday, Friday William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Special Instructor: Julius Su Teaching Assistants: Hai Xiao Mark Fornace 1

2 Now consider H 2 + molecule Bring a proton up to an H atom to form H 2 + Is the molecule bound? That is does it have a lower energy at finite R than at R = Several possibilities Electron is on the left proton, L L r L e r R R Electron is on the right proton, R Or we could combine them At R = these are have the same energy, but not for finite R In QM we always want the wavefunction with the lowest energy. Question: which combination is lowest? Ch125a-Goddard-L01 2

3 Combine Atomic Orbitals for H 2 + molecule Symmetric Two extreme combination possibilities Antisymmetric combination Which is best (lowest energy)? the D g = Sqrt[2(1+S)] and D u = Sqrt[2(1-S)] factors above are the constants needed to ensure that <Φ g Φ g > = Φ g Φ g dxdydz = 1 (normalized) <Φ u Φ u > = Φ u Φ u dxdydz = 1 (normalized) I will usually eschew writing such factors, leaving them to be understood Ch125a-Goddard-L01 3

4 Energies of of H 2 + Molecule g state is bound since starting the atoms at any distance between arrows, the molecule will stay bonded, with atoms vibrating forth and back Ungood state: u Good state: g LCAO = Linear Combination of Atomic Orbitals Ch125a-Goddard-L01 4

5 But WHY is the g state bound? Ch125a-Goddard-L01 5

6 Common rational : But WHY is the g state bound? Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region. This negative charge in the bond region attracts the two positive nuclei leading to the bond Sounds reasonable Ch125a-Goddard-L01 6

7 Common rational : But WHY is the g state bound? Superimposing two orbitals and squaring to get the probability leads to moving charge into the bond region. This negative charge in the bond region attracts the two positive nuclei leading to the bond Sounds reasonable, but increasing the density in bond region è decrease density near atoms, thus moves electrons from very attractive region near nuclei to less attractive region near bond midpoint, this INCREASES the PE Ch125a-Goddard-L01 7

8 Compare change in density with local PE function The local PE for the electron is PE(r) = -1/r a 1/r b lowered at the bond midpoint from the value of a single atom Since the best local PE is still near the nucleus The Φ g = χ L + χ R wavefunction moves charge to the bond region AT THE EXPENSE of the charge near the nuclei, causing an increase in the PE, and opposing bonding 8 Ch125a-Goddard-L01

9 The PE of H 2 + for g and u states The total PE of H 2 + for the Φ g = χ L + χ R and Φ u = χ L - χ R wavefunctions (relative to the values of V g = V u = -1 h 0 at R = ) Ch125a-Goddard-L01 9

10 If the bonding is not due to the PE, then it must be KE Ch125a-Goddard-L01 10

11 If the bonding is not due to the PE, then it must be KE The shape of the Φ g = χ L + χ R and Φ u = χ L - χ R wavefunctions compared to the pure atomic orbital (all normalized to a total probability of one). Ch125a-Goddard-L01 We see a dramatic decrease in the slope of the g orbital along the bond axis compared to the atomic orbital. This leads to a dramatic decrease in KE compared to the atomic orbital This decrease arises only in the bond region. It is this decrease in KE that is responsible for the bonding in H

12 The KE of g and u wavefunctions for H 2 + Use top part of 2-7 The change in the KE as a function of distance for the g and u wavefunctions of H 2 + (relative to the value at R= of KE g =KE u =+0.5 h 0 ) Ch125a-Goddard-L01 Comparison of the g and u wavefunctions of H + 2 (near the optimum bond distance for the g state), showing why g is so bonding and u is so antibonding 12

13 Why does KEg has an optimum? Ch125a-Goddard-L01 R too short leads to a big decrease in slope but over a very short region, è little bonding R is too large leads to a decrease in slope over a long region, but the change in slope is very small è little bonding Optimum bonding occurs when there is a large region where both atomic orbitals have large slopes in the opposite directions (contragradient). This leads to optimum bonding 13

14 KE dominates PE Ch125a-Goddard-L01 Ungood state: u Good state: g Changes in the total KE and PE for the g and u wavefunctions of H 2 + (relative to values at R= of KE :+0.5 h 0 PE: -1.0 h 0 E: -0.5 h 0 The g state is bound between R~1.5 a 0 and (starting the atoms at any distance in this range leads to atoms vibrating forth and back. Exciting to the u state leads to dissociation 14

15 KE dominates PE, leading to g as ground state Calculations show this, but how could we have predicted that g is better than u without calculations? Answer: the nodal theorem: The ground state of a QM systems has no nodes. Thus g state lower E than u state Ch125a-Goddard-L01 15

16 The nodal Theorem The ground state of a system has no nodes (more properly, the ground state never changes sign). This is often quite useful in reasoning about wavefunctions. For example the nodal theorem immediately implies that the g wavefunction for H 2 + is the ground state (not the u state) Ch125a-Goddard-L01 16

17 The nodal Theorem 1D Schrodinger equation, H Ψ k = E k Ψ k One dimensional: H =- ½ d 2 /dx 2 + V(x) Consider the best possible eigenstate of H with a node, Ψ 1 and construct a nonnegative function Ө 0 = Ψ 1 as in b For every value of x, V(x)[Ψ 1 ] 2 = V(x)[Ө 0 ] 2 so that a b Φ 1 Ө 0 V 0 = [Ө 0 ] * V(x)[Ө 0 ] = [Ψ 1 ] * V(x)[Ψ 1 ] 2 = V 1 c Φ 0 Also dө 0 /dx 2 = d Ψ 1 /dx 2 for every value of x except the single point at which the node occurs. Thus T 0 = ½ dө 0 /dx 2 = ½ dψ 1 /dx 2 = T 1. Hence E 0 = T 0 + V 0 = T 1 + V 1 = E 1. Ch125a-Goddard-L01 17

18 The nodal Theorem 1D We just showed that for the best possible eigenfunction of H with a node, H Ψ 1 = E 1 Ψ 1 a Φ 1 Ө 0 = Φ 1 has the same energy as Ψ 1 E 0 = T 0 + V 0 = T 1 + V 1 = E 1. However Ө 0 is just a special case of a nodeless wavefunction that happens to go to 0 at one point. In general not requiring it so go to zero will lower the energy. Thus we could smooth out Ө 0 in the region of the node as in c, decreasing the KE and lowering the energy. Thus the optimum nodeless wavefunction Ψ 0 leads to E 0 < E 1. Only for a potential so repulsive at some point, that all wavefunctions are 0, do we get E 0 = E 1 b Ө 0 c Φ 0 Ch125a-Goddard-L01 18

19 The nodal Theorem for excited states in 1D For one-dimensional finite systems, we can order all eigenstates by the number of nodes E 0 < E 1 < E 2... E n < E n+1 (where a sufficiently singular potential can lead to an = sign ) The argument is the same as for the ground state. Consider best wavefunction Ψ n with n nodes and flip the sign at one node to get a wavefunction Ө n-1 that changes sign only n-1 times. Show that E n-1 = E n But Ө n-1 is not the best with n-1 sign changes. Thus we can smooth out Ө n-1 in the region of the extra node to decrease the KE and lower the energy for the Ψ n-1,. Thus the optimum n-1 node wavefunction leads to E n-1 < E n. Ch125a-Goddard-L01 19

20 The nodal Theorem 3D In 2D a wavefunction that changes size once will have a line of points with Ψ 1 =0 (a nodal line) For 3D there will be a 2D nodal surface with Ψ 1 =0. In 3D the same argument as for 1D shows that the ground state is nodeless. We start with Ψ 1 the best possible eigenstate with a nodal surface and construct a nonnegative function Ө 0 = Ψ 1 For every value of x,y,z, V(x,y,z)[Ψ 1 ] 2 = V(x,y.z)[Ө 0 ] 2 so that V 0 = [Ө 0 ] * V(xyz)[Ө 0 ] = [Ψ 1 ] * V(xyz)[Ψ 1 ] 2 = V 1 Also Ө 0 2 = Ψ 1 2 everywhere except along a 2D plane (0 vol.) Thus T 0 = ½ Ө 0 2 = ½ Ψ 1 2 = T 1. Hence E 0 = T 0 + V 0 = T 1 + V 1 = E 1. As before E 1 is the best possible energy for an eigenstate with a nodal plane. However Ө 0 can be improved by smoothing Thus the optimum nodeless wavefunction Φ 0 leads to E 0 < E 1. Ch125a-Goddard-L01 20

21 The nodal Theorem for excited states in 3D For 2D and 3D, one cannot order all eigenstates by the number of nodes. Thus consider the 2D wavefunctions Φ 00 Φ Φ 01 Φ 20 Φ 11 Φ It is easy to show as in the earlier analysis that E 00 < E 10 < E 20 < E 21 E 00 < E 01 < E 11 < E 21 But the nodal argument does not indicate the relative energies of E 10 and E 20 versus E 01 Ch125a-Goddard-L01 21

22 Back to H 2 + g state u state Nodal theorem è The ground state must be the g wavefunction Ch125a-Goddard-L01 22

23 Ch120a-Goddard-L02 Symmetry Considerations The operation of inversion (denoted as ^ I ) through the origin of a coordinate system changes the coordinates as x è -x y è -y z è -z Taking the origin of the coordinate system as the bond midpoint, inversion changes the electronic coordinates as illustrated. After inversion the electron is remains a distance of r a from one of the nuclei and r b from the other, but their identities are transposed. Thus the potential energy, v(r) is unchanged by inversion, v(-x,-y,-z) v( ^ I r) = v(r) where r is considered as the 3D vector with components x,y,z Thus we say that v(r) is invariant under inversion 23

24 Considering symmetry Under inversion the kinetic energy terms in the Hamiltonian are also unchanged Hence the full Hamiltonian is invariant under inversion h(-x,-y,-z) h( ^ I r) = h(x,y,z) = h(r) Now consider that we had solved hφ=εφ for the exact wavefunction φ and apply the inversion to both sides ^ I hφ= ε ^ I φ Which we can rewrite as h(-r)φ(-r)=εφ(-r) But h(-r) = h(r) because of inversion symmetry Thus h(r)φ(-r)=εφ(-r) Ch120a-Goddard-L02 24

25 4 th postulate of QM Consider the exact eigenstate of a system HΦ = EΦ and multiply the Schrödinger equation by some CONSTANT phase factor (independent of position and time) exp(iα) = e iα e iα HΦ = H (e iα Φ) = E (e iα Φ) Thus Φ and (e iα Φ) lead to identical properties and we consider them to describe exactly the same state. 4th Postulate: wavefunctions differing only by a constant phase factor describe the same state Ch120a-Goddard-L02 25

26 Continue with symmetry discussion We just derived that because h(-r) = h(r) Then for any eigenfunction φ(r) of h h(r)φ(r) = εφ(r) It must be that φ(-r) also is an eigenfunction of the same h with the same energy, ε. h(r)φ(-r) = εφ(-r) There are two distinct possibilities here. (1) There is only ONE state with energy ε or (2) There is more than ONE state with energy ε. Ch120a-Goddard-L02 26

27 Case 1: Nondegenerate states Consider case 1 first in which only one state has energy ε. By the 4th postulate of QM, then at most, φ(-r) and φ(r) can differ by a phase factor, φ(-r) = e iα φ(r) Now consider what happens if the inversion is applied twice x è -x è x; y è -y è y z è -z è z Nothing changes! Thus applying inversion twice is equivalent to doing nothing. This do-nothing operator is called einheit (German for identity or unit) and denoted as e ^. We write ^ I ^ I = ( ^ I ) 2 = e ^ and say that the inversion operator is of order two Since ^ I φ(r) = φ(-r) = e iα φ(r) then applying a second inversion leads to ( ^ I ) 2 φ(r) = ^ I φ(-r) = ^ I e iα φ(r) = e i2α φ(r) But ( ^ I ) 2 φ(r) = e ^ φ(r) = φ(r); Thus e i2α φ(r) =φ(r) Hence e i2α = 1 or e iα = ±1 Ch120a-Goddard-L02 27

28 Case 1: Nondegenerate states: conclusion Since ^ I φ(r) = φ(-r) = e iα φ(r) and e iα = ±1 We conclude that either ^ I φ g (r) = + φ g (r) g for gerade or even or ^ I φ u (r) = - φ u (r) u for ungerade or odd Thus for a system with inversion, each nondegenerate eigenstate is of either g or u inversion symmetry. Indeed for H + φ 2 we found g (r) φ u (r) Ch120a-Goddard-L02 28

29 Case 2: Now consider the case where the state is degenerate In this case φ and ^ I φ can be independent states (not proportional to each other). Thus we can combine them to form two new wavefunctions and Since ^ I ( ) = ^ I φ + φ ^ and I ^ ( ) = I φ - φ Both are eigenfunctions of h with the same energy) we see that Since h(r)φ(r) = εφ(r) and h(r)φ(-r) = εφ(-r) We see that h(r)φ g (r) = εφ g (r) and h(r)φ u (r) = εφ u (r) Thus every degenerate eigenfunction of h can be rewritten so that each state is either g or u We say that all eigenstates of systems with inversion symmetry are either g or u Ch120a-Goddard-L02 29

30 Apply symmetry principle to H 2 + Ch120a-Goddard-L02 φ g (r) = + φ g (r) g for gerade or even φ u (r) = - φ u (r) u for ungerade or odd 30

31 More quantitative description of H2+ (from Ch chapter 5 (chapter 7) The Hamiltonian for H2+ is (1) (2) 31

32 Cylindrical Coordinates for H2+ where 32

33 Elliptic coordinates The solutions are where 33

34 Bates numerical solutions 34

35 Angular eigenfunctions for diatomic molecules 35

36 Inversion symmetry 36

37 The eigenstates of H2+ 37

38 Figure The total energy for various states of H2+ [DR Bates, K Ledsham, and AL Stewart; Phil Trans. Roy. Soc. (London), 246, 215 (1953)] The 2pπ u and 3dσ g states have very small minima (~0.01 ev at large R (~7a0 and 9 a0)) H2s, 2p D e =0.6 h 0 H1s R e =2.0 a 0 38

39 More quantitative description of the bonding in H2+ See (really chapter 5) Ch The Linear Combination of Atomic Orbitals (LCAO) description of bonding is where χ a and χ b are atomic orbitals on the left and right and S is their overlap The charge densities can be partitioned into a classical part (superposition of densities) and the exchange part (4) 39

40 Classical and exchange density 40

41 Example numbers With normalization we get 41

42 Relation of bonding to classical and exchange density Table 1 shows that (12) which causes 42

43 Note ΔT is negative Note ΔV is positive 43

44 The scale optimized LCAO wavefunction The Virial Theorem paradox with ζ = 1 as for the H atom. Now we will optimize ζ For H2+ the potential energy (PE) operator is So that the scaled operator is if we also scale R. and hence the new PE is V ζ = ζ V 1 (note V is negative) and the new KE is T ζ = ζ 2 T 1 Thus the total scaled E = ζ 2 T 1 + ζ V 1 leading to an optimum value of 2ζ opt T 1 + V 1 = 0 or ζ opt= - V 1 / 2T 1 = Thus the optimum T opt = V 12 /4T 1 and Vopt = - V 12 /2T 1 Satisfying the Virial theorem, Eopt = -T opt = V opt /2, as expected. Starting with R = 2.5 a 0 (the Re for LCAO) this leads to a new R e = 2.02 a 0 which is nearly the exact value 44

45 Table Energies for LCAO description of H2+ (atomic units) Re-LCAO Exact R e R exact = 2.0 D=Bond E D 1 = D opt = D exact =

46 Discussion of bonding of orbitals Optimizing the scale of the LCAO wavefunction leads to an excellent Re=2.02 a0 compared to the exact value of 2.0a0 The Bond energy, D, increases by 32% to D=0.086 compared to Dexact = 0.10 h0 Adding a 2p z basis function on each nucleus would lead to a further substantial improvement The fact that the Virial theorem must be satisfied at Re and at R= shows that the total bond energy D = -ΔT = ½ ΔV Confuses some to believe that the bonding is due to the PE In fact as shown clearly in Fig. 4, it is the large decrease in KE due to superposition of orbitals on two centers that is responsible for the bonding This big decrease in KE allows the contraction at shorter R of orbitals about each center until the Virial Theorem is satisfied at Re. This is illustrated in Fig

47 Energies for H2+ (Fig of Ch ) T exact V LCAO E exact T LCAO E LCAO V exact For R>5a0, the LCAO and exact values of T, V, E are similar with the big decrease in T LCAO dominating the bond For short R<4a0, the wavefunction contracts about each nucleus (ζ>1) Leading finally to the Virial Theorem values at R e 47

48 LCAO orbitals Figure Contributions (h0) to bonding in H2+ Evaluated at R=2a0 The decrease of 0.117h0 in T for the LCAO wavefunction at Re=2a0 dominates the final bond of 0.1 ho

49 Must have two hands to clap In trying to understand the origin of binding it is important to discuss simultaneously both the states that bind and the states that antibind (It takes 2 hands to clap) Our analysis using LCAO wavefunction does this since we examine both the g and the u states as in Fig. 7 Here no matter which atomic-like orbitals that we use, the g state is below the u state and the energy change from R= is dominated by the decrease in KE, showing that it is this decrease in KE that is responsible for the bond Fig. 7 (Fig from Ch =-73) 49

50 Classical and exchange contributions (14) where (15) For LCAO, T cl = 0.5, thus it is T x which gets more negative as R decreases that is responsible for bonding in H2+ 50

51 51

52 T x so that However the dot product quantity (19) 52

53 T x which Region over which negative, enhancing bonding is 53

54 contragradience (20) for all R. where The remainder of Tx is (21) (28) (29) LCAO (30) 54

55 55

56 Fig. 7,2-10 Fig. 7,2-9 Plots along the bond axis 56

57 Analytic analysis of bonding and antibonding in H2+ (5) 57

58 Analysis of Eg and Eu of H2+ where 58

59 We saw above Thus So that τ dominates the bonding and antibonding Put in analytic form for S and τ here Both S and τ go to zero exponentially with R so we can consider the bonding to be proportional to S 59

60 Energies (hartree) for the LCAO wavefunctions of H2+ 60

61 Energies (hartree) for the LCAO wavefunctions of H2+ 61

62 Higher excited states of H2+ 62

63 Splitting of degeneracies (7.3.2a Ch ) where (13b) (14) Hence 63

64 Since Breaking the 2s-2pz degeneracy and We have However (19) with energies and (20) 64

65 we obtain we obtain 65

66 2s,2p States of H perturbed by a proton at distance R. 66

67 where E a = E cl τ/(1-s) 67

68 68

69 Figure 3. Molecular orbitals from 2s,2p states 69

70 Total Energies of 2s,2p excited states of H2+ Fig from Ch

71 Electronic Energies of 2s,2p excited states of H2+ Fig a from Ch , same as Fig except omit 1/R nuclear term This shows the correlation from R~ to R=0 71

72 Enlargement of Fig. 5a 72

73 Energies for H2+ (Hartree) Table

74 The Molecular orbitals of 2s,2p states H c Fig. 6a Fig. 6a 74

75 Fig. 7 nodal patterns and names of 2s,2p derived states of H2+ United atom name at the right, separated atom name at the left and standard MO name 75

76 Energies near the united atom limit 7.3.2d 76

77 Correlation Diagrams

78 Correlation of MOs with Separated atom states 78

79 Fig. 9 Correlation diagram for H2+ 79

80 Heteronuclear Molecules 80

81 81

82 82

83 (6) 83

84 which leads to (7) 84

85 (11) (12) 85

86 Summary

87 The Role of Kinetic Energy in Chemical Binding: II. Contragradience W. A. Goddard III and C. W. Wilson, Jr. Theor. Chim. Acta. 26, 211 (1972) wag46 The Role of Kinetic Energy in Chemical Binding: I. The Nonclassical or Exchange Kinetic Energy C. W. Wilson, Jr., and W. A. Goddard III Theor. Chim. Acta. 26, 195 (1972) wag45 has not been widely adopted 87