Lecture 12 February 8, 2013 Symmetry, Homonuclear diatomics

Transcription

1 Lecture 12 February 8, 2013 Symmetry, Homonuclear diatomics 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, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Ross Fu Fan Liu Ch120a-Goddard-L11 Ch120a-1

2 Inversion Symmetry for H 2 For multielectron systems, the inversion symmetry inverts all electron coordinates simultaneously (x k,y k,z k ) (-x k,-y k,-z k ) for k=1,..n But I*I takes (x k,y k,z k ) (+x k,+y k,+z k ) for k=1,..n That is I*I=e the identity operator Because I*I=e, we showed in L2 that the inversion symmetry of H 2 leads to the result that every eigenstate of H 2 is either g or u Ch120a-Goddard-L11

3 Ch120a-Goddard-L11 Permutational symmetry from L2 For a two-electron system the Hamiltonian is invariant (unchanged) upon transposition of the electrons (changing both spatial and spin coordinates simultaneously) Tspace-spin H(1,2) = H(2,1) = H(1,2) But the Tspace-spin * Tspace-spin = e (identity) Thus for every eigenstate of the Hamiltonian we obtain either Ψs(1,2) = +1 Ψs(1,2) Ψa(1,2) = -1 Ψa(1,2) But our Hamiltonian does not depend on spin. Hence Ψ(1,2) = Φ(1,2)χ(1,2) Also H(1,2) is separately unchanged by transposing either just the spatial coordinates or the spin coordinates Thus either Φ(2,1) = +Φ(1,2) or Φ(2,1) = -Φ(1,2) and either χ(2,1) = +χ(1,2) or χ(2,1) = -χ(1,2) 3

4 Our Hamiltonian for H 2, Permutational symmetry, summary H(1,2) =h(1) + h(2) + 1/r12 + 1/R Does not involve spin This it is invariant under 3 kinds of permutations Space only: ρ 1 ρ 2 Spin only: σ 1 σ 2 Space and spin simultaneously: (ρ 1,σ 1 ) (ρ 2,σ 2 ) Since doing any of these interchanges twice leads to the identity, we know from previous arguments that Ψ(2,1) = ± Ψ(1,2) symmetry for transposing spin and space coord Φ(2,1) = ± Φ(1,2) symmetry for transposing space coord Χ(2,1) = ± Χ(1,2) symmetry for transposing spin coord Ch120a-Goddard-L11 Ch120a-4

5 The 6 th postulate of QM: the Pauli Principle For every eigenstate of an electronic system H(1,2, i j N)Ψ(1,2, i j N) = EΨ(1,2, i j N) The electronic wavefunction Ψ(1,2, i j N) changes sign upon transposing the total (space and spin) coordinates of any two electrons Ψ(1,2, j i N) = - Ψ(1,2, i j N) We can write this as τ ij Ψ = - Ψ for all i and j Ch120a-Goddard-L11 Ch120a-5

6 The role of symmetry in QM In this course we are concerned with the solutions of the Schrodinger equation, HΨ=EΨ, but we do not actually want to solve this equation. Instead we want to extract the maximum information about the solutions without solving it. Symmetry provides a powerful tool for doing this. Some transformation R 1 is called a symmetry transformation if it has the property that R 1 (HΨ)=H(R 1 Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group. Ch120a-Goddard-L11 6

7 The definition of a Group 1). Closure: If R 1,R 2 ε G (both are symmetry transformations) then R 2 R 1 is also a symmetry transformation, R 2 R 1 ε G 2. Identity. The do-nothing operator or identity, R 1 = e ε G is clearly is a symmetry transformation 3. Associativity. If (R 1 R 2 )R 3 =R 1 (R 2 R 3 ). 4. Inverse. If R 1 ε G then the inverse, (R 1 ) -1 ε G,where the inverse is defined as (R 1 ) -1 R 1 = e. For the case of the inversion symmetry, the group is {e, I} Since I*I = e we see that (I) -1 = I For the case of the transposition symmetry, the group is {e, τ} Since τ * τ = e we see that (τ) -1 = I Ch120a-Goddard-L11 7

8 The degenerate eigenfunctions of H form a representation If HΨ=EΨ then H(R 1 Ψ)= E(R 1 Ψ) for all symmetry transformations of H. Thus the transformations amount the n denegerate functions, {S=(R i Ψ), where R i Ψ i ε G} lead to a set of matrices that multiply in the same way at the group operators. The Mathematicians say that these functions form a basis for a representation of G. Of course the functions in S may not all be different, so that this representation can be reduced. The mathematicians went on to show that one could derive a set of irreducible representations that give all possible symmetries for the H. reorientations from which one can construct any possible. For the inversion and transposition groups, all representations are nondegenate, just symmetric and antisymmetric Ch120a-Goddard-L11 8

9 Example, an atom. For an atom any rotations about any axis passing through the nucleus is a symmetry transformation. This leads to the group denoted as SO(3) by the mathematicians [O(3) indicates 3 three-dimensional real space, S because the inversion is not included). The irreducible representations of O(3) are labeled as S (non degenerate) and referred to as L=0 P (3 fold degenerate) and referred to as L=1 D (5 fold degenerate) and referred to as L=2 F (7 fold degenerate) and referred to as L=3 G (9 fold degenerate) and referred to as L=4 Ch120a-Goddard-L11 9

10 H 2 O, an example of C 2v consider the nonlinear H 2 A molecule, with equal bond lengths, e.g. H 2 O, CH 2, NH 2 The symmetry transformations are 1. e for einheit (unity) x x, y y, z z 2. C 2z, rotation about the z axis by 2π/2=180º, x -x, y -y, z z 3. σ xz, reflection in the xz plane, x x, y -y, z z 4. σ yz, reflection in the yz plane, x -x, y y, z z Which is denoted as the C 2v group. But C 2z * C 2z = e and σ xz * σ xz = e and σ yz * σ yz = e Hence the eigenfunctions of H are each sym or antisym with respect to each symmetry Ch120a-Goddard-L11 10

11 The character table for C 2v Since (C 2z ) 2 = e, (σ xz ) 2 = e, (σ yz ) 2 = e We expect wavefunctions to be ±1 under each operation name 1-e N-e a 1 a 2 b 1 b 2 Since C 2z σ xz = σ yz the symmetries for σ yz are already implied by C 2z σ xz. Thus there are only 4 possible symmetries. Ch120a-Goddard-L11 11

12 Stereographic projections Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive x are circles and negative x are squares. y x e σ yz C 2z C 2z Start with a general point, denoted as e and follow where it goes on various symmetry operations. This make relations between the symmetry elements transparent. e.g. C 2z σ xz = σ yz Combine these as below to show the relationships σ xz σ yz C 2z σ yz σ xz C 2z σ yz y x e σ xz σ xz Ch120a-Goddard-L11 C 2v 12

13 Symmetries for NH 2 Ψ 1 = A{(N2p y α)[(np L )(H L )+(H L )(Np L )](αβ βα)[(np R )(H R )+(H R )(Np R )](αβ βα)} NH L bond Applying σ xz to the wavefunction leads to NH R bond Ψ 2 = A{(N2p y ) 0 [(Np R )(H R )+(H R )(Np R )](αβ βα)[(np L )(H L )+(H L )(Np L )](αβ βα)} Each term involves transposing two pairs of electrons, e.g., 1 3 and 2 4 as interchanging electrons. Since each interchange leads to a sign change we find that σ xz Ψ 1 = Ψ 2 = Ψ 1 L R Thus interchanging a bond pair leaves Ψ invariant Also the N(1s) 2 and (2s) 2 pairs are invariant under all operations Ch120a-Goddard-L11 13

14 NH 2 symmetry continued Ψ 1 = A{(N2p y α)[(np L )(H L )+(H L )(Np L )](αβ βα)[(np R )(H R )+(H R )(Np R )](αβ βα)} NH L bond NH R bond Consider now the singly occupied Np y orbital C 2z changes the sign but, σ xz does not. Thus Np x transforms as b 1 and The total wavefunction transforms as B 1. Include the S= 1/2, leads to The 2 B 1 state 1-e N-e a 1 a 2 b 1 b 2 Ch120a-Goddard-L11 14

15 H 2 O Symmetries for H 2 O and CH 2 NH 2 CH 2 H 2 O CH 2 A{(O2p y ) 2 [(Op x )(H x )+(H x )(Op x )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)} OHx bond OHz bond A{(C2p y ) 0 [(Cp x )(H x )+(H x )(Cp x )](αβ βα)[(cp z )(H z )+(H z )(Cp z )](αβ βα)} Since have enen number electrons in 2py, wavefunction is invarient under all symmetry transformations, thus must be 1 A 1. Ch120a-Goddard-L11 15

16 Now do triplet state of CH 2 Soon we will consider the triplet state of CH 2 in which one of the 2s nonbonding electrons (denoted as σ to indicate symmetric with respect to the plane of the molecule) is excited to the 2p x orbital (denoted as π to indicate antisymmetric with respect to the plane) A{(C2sα) 1 (2p x α) 1 [(Cp L )(H L )+(H L )(Cp L )](αβ βα)[(cp R )(H R )+(H R )(Cp R )](αβ βα)} CH L bond Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as A{[(CH L ) 2 (CH R ) 2 ](Cσα) 1 (Cπα) 1 } Here σ is invariant (a 1 ) while π transforms as b 1. Since both s and p are unpaired the ground state is triplet or S=1 CH R bond Thus Ch120a-Goddard-L11 the symmetry of copyright triplet 2011 CHWilliam 2 is 3 A. BGoddard 1 III, all rights reserved y σ=2s π=2p x z 16

17 Second example, C 3v, with NH 3 as the prototype 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)} NH b bond NH x bond We will consider a system such as NH 3, with three equal bond lengths. Here we will take the z axis as the symmetry axis and will have one H in the xz plane. The other two NH bonds will be denoted as b and c. NH c bond z x Ch120a-Goddard-L11 17

18 Symmetry elements for C 3v Take the z axis out of the plane. The six symmetry operations are: b b b z C 3 x e x x x c c c C 3 2 = C 3-1 b b b σ xz C 3 2 c x x x σ xz σ xz C 3 c c Ch120a-Goddard-L11 18

19 b C 3 σ xz C 3 c σ xz C 3 2 C 32 = C 3-1 The C 3v symmetry group e σ xz x The σ xz C 3 transformation corresponds to a reflection in the bz plane (which is rotated by C 3 from the xz plane) and the σ xz C 3 2 transformation corresponds to a reflection in the cz plane (which is rotated by C 3 2 from the xz plane). Thus these 3 reflections are said to belong to the same class. Since {C 3 and C 3-1 do similar things and are converted into each other by σ xz we say that they are in the same class. Ch120a-Goddard-L11 19

20 The character table for C 3v The E symmetry (irreducible representation) is of degree 2, which means that if φ px is an eigenfunction of the Hamiltonian, the so is φ py and they are degenerate. This set of degenerate functions would be denoted as {e x,e y } and said to belong to the E irreducible representation. The characters in this table are used to analyze the symmetries, but we will not make use of this until much later in the course. Thus an atom in a P state, say C( 3 P) at a site with C 3v symmetry, would generally split into 2 levels, { 3 P x and 3 P y } of 3 E symmetry and 3 P z of 3 A 1 symmetry. Ch120a-Goddard-L11 20

21 Application for C 3v, NH 3 z σ LP x We will write the wavefunction for NH 3 as x Ψ=A{(σ LP ) 2 [(NH b bond) 2 (NH c bond) 2 (NH x bond) 2 ]} b c where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as σ LP Consider first the effect of σ xz. This leaves the NH x bond pair invariant but it interchanges the NH b and NH c bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the σ LP orbital is invariant. Ch120a-Goddard-L11 21

22 Consider the effect of C 3 Next consider the C 3 symmetry operator. It does not change σ LP. It moves the NH x bond pair into the NH b pair, moves the NH b pair into the NH c pair and moves the NH c pair into the NH x pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example. (135) φ(1)φ(3)φ(5) φ(3)φ(5)φ(1) (say this as e1 is replaced by e3 is replaced by e5 is replace by 1) This is the same as (35) (13) φ(1)φ(3)φ(5) φ(1)φ(5)φ(3) φ(3)φ(5)φ(1) The point is that this is equivalent to two transpostions. Hence by the PP, the wavefunction will not change sign. Since C 3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change. Thus the wavefunction for NH 3 has 1 A 1 symmetry Ch120a-Goddard-L11 22

23 Linear molecules, C v symmetry O=C=O z Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x 2 +y 2 ), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series, φ(ρ,α,z)=f(ρ,z){a 0 + Σ m=1 m= [a m cos mα + b m sin mα] x Ch120a-Goddard-L11 23

24 Linear molecules, C v symmetry O=C=O z Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x 2 +y 2 ), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series, φ(ρ,α,z)=f(ρ,z){a 0 + Σ m=1 m= [a m cos mα + b m sin mα] Clearly the kinetic energy will increase with m, so that for the same f(ρ,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc x Ch120a-Goddard-L11 24

25 Linear molecules, C v symmetry O=C=O z Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x 2 +y 2 ), α, z (axis along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be expanded as a Fourier series, φ(ρ,α,z)=f(ρ,z){a 0 + Σ m=1 m= [a m cos mα + b m sin mα] Clearly the kinetic energy will increase with m, so that for the same f(ρ,z), we expect m=0 lowest, then m=±1, then m=±2, m=±3, etc Also if we rotate the molecule about the z axis by some angle β, the states with the same m get recombined [cos m(α+β)] = (cos mα)(cosmβ) (sin mα)(sin mβ) [sin m(α+β)] = (sin mα)(cosmβ) + (cos mα)(sin mβ) Which means that the wavefunctions with the same m are degenerate Ch120a-Goddard-L11 x 25

26 C v symmetry group The symmetry operators are: x O=C=O z R z (α): counterclockwise rotation by an angle α about the z axis σ xz : reflection in the xz plane (this takes +α into α) Ch120a-Goddard-L11 26

27 C v symmetry group The symmetry operators are: R z (α): counterclockwise rotation by an angle α about the z axis σ xz : reflection in the xz plane (this takes +α into α) σ = R z (α) σ xz R z (-α); reflection in a plane rotated by an angle α from the xz plane (there are an infinite number of these) x O=C=O z Ch120a-Goddard-L11 27

28 σ = R z (α) σ xz R z (-α) is a reflection in the plane rotated by an angle α from the xz plane Take the z axis out of the plane. The six symmetry operations are: y y y y e x α e x α e x α e x C -α σ xz C α y y e x σ α e x σ Ch120a-Goddard-L11 28

29 The symmetry operators are: R z (α): counterclockwise rotation by an angle α about the z axis σ xz : reflection in the xz plane (this takes +α into α) σ = R z (α) σ xz R z (-α); reflection in a plane rotated by an angle α from the xz plane (there are an infinite number of these) e: einheit (unity) This group is denoted as C v,the character table (symmetries) are name 1-e N-e σ C v symmetry group x O=C=O z π δ φ γ Ch120a-Goddard-L11 29

30 The symmetry functions for C v Lower case letters are used to denote one-electron orbitals Ch120a-Goddard-L11 30

31 Application to FH The ground state wavefunction of HF is A{(F2p x ) 2 (F2p y ) 2 [(Fp z )(H)+(H)(Fp z )](αβ βα)} In C v symmetry, the bond pair is σ (m=0),while the p x and p y form a set of π orbitals (m=+1 and m=-1). Consider the case of up spin for both π x and π y Ψ(1,2) = A{φ x αφ y α}=(φ x φ y φ y φ x ) αα Rotating by an angle γ about the z axis leads to φ a = cosγ φ x + sinγ φ y and φ b = cosγ φ y - sinγ φ x This leads to (φ a φ b φ b φ a ) = [(cosγ) 2 +(sinγ) 2 ] }=(φ x φ y φ y φ x ) Thus (φ x φ y φ y φ x ) transforms as Σ. Ch120a-Goddard-L11 31

32 Continuing with FH Thus the (p x ) 2 (p y ) 2 part of the HF wavefunction A{(F2p x ) 2 (F2p y ) 2 [(Fp z )(H)+(H)(Fp z )](αβ βα)} Since both αα and ββ transform like Σ; the total wavefunction transforms as Σ The symmetry table, demands that we also consider the symmetry with respect to reflection in the xz plane. Here px is unchanged while py changes sign. Since there are two electrons in py the wavefunction is invariant. Thus the ground state of FH has 1 Σ + symmetry Ch120a-Goddard-L11 32

33 Next consider the ground state of OH We write the two wavefunctions for OH as Ψ x =A{(σ OH bond) 2 [(π x α)(π y α)(π x β)]} 2 Π x Ψ y =A{(σ OH bond) 2 [(π x α)(π y α)(π y β)]} We saw above that A{(π x α)(π y α)} transforms like Σ. thus we need examine only the transformations of the downspin orbital. But this transforms like π. Thus the total wavefunction is 2 Π. 2 Π y x z Another way of describing this is to note that A{(π x ) 2 (π y ) 2 } transforms like Σ and hence one hole in a (π) 4 shell, (π) 3 transforms the same way as a single electron, (π) 1 Ch120a-Goddard-L11 33

34 Now consider the ground state of NH A{(NH bond) 2 (N2p x α)(n2p y α)} We saw earlier that up-spin in both x and y leads to Σ symmetry. Considering now the reflection, σ xz, we see that with just one electron in π y, we now get Σ -. Thus the ground state of NH is 3 Σ -. x z Ch120a-Goddard-L11 34

35 Now consider Bonding H atom to all 3 states of C (2p x )(2p z ) (2p y )(2p z ) (2p x )(2p y ) Bring H1s along z axis to C and consider all 3 spatial states. O 2p z singly occupied. H1s can get bonding Get S= ½ state, Two degenerate states, denote as 2 Π No singly occupied orbital for H to bond with x z Ch120a-Goddard-L11 35

36 Ground state of CH ( 2 Π) The full wavefunction for the bonding state 2 Π x A{(2s) 2 (OHσ bond) 2 (O2p x α) 1 } 2 Π y x z A{(2s) 2 (OHσ bond) 2 (O2p y α) 1 } Ch120a-Goddard-L11 36

37 Bond a 2 nd H atom to the ground state of CH Starting with the ground state of CH, we bring a 2 nd H along the x axis. Get a second covalent bond This leads to a 1 A 1 state. No unpaired orbtial for a second covalent bond. x z Ch120a-Goddard-L11 37

38 x z Analyze Bond in the ground state of CH 2 Ground state has 1 A 1 symmetry. For optimum bonding, the p z orbital should point at the H z while the p x orbital should point at the H x. Thus the bond angle should be 90º. As NH 2 (103.2º) and OH 2 (104.5º), we expect CH 2 to have bond angle of ~ 102º θ e R e Ch120a-Goddard-L11 38

39 But, the Bending potential surface for CH 2 1 B 1 1 g 1 A kcal/mol 3 B 1 3 Σ g - The ground state of CH 2 is the 3 B 1 state not 1 A 1. Thus something is terribly wrong in our analysis of CH 2 Ch120a-Goddard-L11 39

40 Short range Attractive interaction sz with H 2 ev 1 ev BeH Compare bonding in BeH+ and BeH Long range Repulsive interaction with H TA s check numbers, all from memory BeH + has long range attraction no short range repulsion 3 ev Ch120a-Goddard-L11 BeH + 1eV Repulsive orthogonalization of zs with sz H 40

41 Compare bonding in BeH and BeH 2 BeH ev R=1.31A 2.03 ev R=1.34A MgH ev R=1.65 A 1.34 ev R=1.73A 1 Σ + 2 Σ + ~3.1 ev ~2.1 e 1 Σ + linear Expect linear bond in H-Be-H and much stronger than the 1 st bond Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz. TA s check numbers, all from memory Ch120a-Goddard-L11 Cannot bind 3 rd H because no singly occupied orbitals left. 41

42 Homonuclear Diatomics 42

43 Homonuclear Diatomics Molecules the valence bond view Consider bonding two Ne atoms together Clearly there will be repulsive interactions as the doubly occupied orbitals on the left and right overlap, leading to repulsive interactions and no bonding. In fact as we will consider later, there is a weak attractive interaction scaling as -C/R 6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here The symmetry of this state is 1 Σ g + 43

44 Halogen dimers Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2 P state) leading to 9 possible configurations for F 2. Of these only one leads to strong chemical binding This also leads to a 1 Σ g + state. Spectroscopic properties are listed below. Note that the bond energy decreases for Cl 2 to Br 2 to I 2, but increases from F 2 to Cl 2. we will get back to this later. 44

45 Di-oxygen or O 2 molecule Next consider bonding of two O atoms. Each O has 3 possible configurations (It is a 3 P state) leading to 9 possible configurations for O 2. Of these one leads to directly to a double bond This suggests that the ground state of O 2 is a singlet state. At first this seemed plausible, but by the late 1920 s Mulliken established experimentally that the ground state of O 2 is actually a triplet state, which he had predicted on the basis of molecular orbital (MO) theory. This was a fatal blow to VB theory, bringing MO theory to the fore, so we will consider next how Mulliken was able to figure this out in the 1920 s without the aid of computers. 45

46 The homonuclear diatomic correlation diagram Mulliken knew the ordering of the atomic orbitals and considered how combinations of the atomic orbitals would change as the nuclei were pushed together to eventually form a united atom. First consider the separate atoms limit where there is a large but finite distance R separating the atoms. The next slide shows the combinations formed from 1s, 2s, and 2p orbitals. 46

47 Separated atoms limit Note that in each case we get one bonding combination (no new nodal plane) and one antibonding combination (new nodal plane, red lines) 47

48 Splitting of levels General nodal arguments allow us to predict that But which is lower of and which is lower of Here the nodal plane arguments do not help 48

49 At large R 2pσ better bonding than 2pπ In earlier lectures we considered the strength of one-electron bonds where we found that Since the overlap of pσ orbitals is obviously higher than pπ We expect that bonding antibonding 49

50 MO notation Separated atom limit Separated atoms notation 50

51 United atom limit Next consider the limit in which the two nuclei are fused together to form a united atom For N2 this would lead to a Si atom. Here we get just the normal atomic aufbau states 1s < 2s < 2p < 3s < 3p < 4s,3d < 4p etc But now we consider an itty bity elongation of the Si nucleus toward two N nuclei and how the atomic states get perturbed For the 1s orbital all that happens is that the energy goes up (less electron density on the nuclei) and the symmetry becomes σ g 51

52 2s and 2p united atom orbitals Similarly 2s just goes to 2σ g (and a lower binding) But the 2p case is more interesting For the 2pσ state the splitting of the nuclei lead to increased density on the nuclei and hence increased binding while for 2pπ there is no change in density Thus 2pσ u < 2pπ u 52

53 Summarizing united atom limit Note for 3d, the splitting is 3dσ < 3dπ < 3dδ Same argument as for 2p 53

54 Summary more united atom levels 54

55 Correlation diagram for Carbon row homonuclear diatomics C 2 N 2 O 2 United atom limit O + 2 N + copyright 2011 William A. Goddard III, all rights 2 reserved F 2 separated atom limit 55

56 Homonuclear Diatomics Molecules the valence bond view Consider bonding two Ne atoms together Clearly there will be repulsive interactions as the doubly occupied orbitals on the left and right overlap, leading to repulsive interactions and no bonding. In fact as we will consider later, there is a weak attractive interaction scaling as -C/R 6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here The symmetry of this state is 1 Σ g + 56

57 Halogen dimers Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2 P state) leading to 9 possible configurations for F 2. Of these only one leads to strong chemical binding This also leads to a 1 Σ g + state. Spectroscopic properties are listed below. Note that the bond energy decreases for Cl 2 to Br 2 to I 2, but increases from F 2 to Cl 2. we will get back to this later. 57

58 Using the correleation diagram In order to use the correlation diagram to predict the states of diatomic molecules, we need to have some idea of what effective R to use (actually it is the effective overlap with large R small S and small R large S). Mulliken s original analysis [Rev. Mod. Phys. 4, 48 (1932)] was roughly as follows. 1. N2 was known to be nondegenerate and very strongly 2 bound with no low-lying excited states Choices for N

59 N2 MO configurations This is compatible with several orderings of the MOs Largest R Smallest R

60 N 2 + But the 13 electron molecules BeF, BO, CO+, CN, N2+ Have a ground state with 2S symmetry and a low lying 2S sate. In between these two 2 Σ states is a 2 Π state with spin orbital splitting that implies a π 3 configuration This implies that Is the ground configuration for N2 and that the low lying states of N2+ are This agrees with the observed spectra 60

61 Correlation diagram for Carbon row homonuclear diatomics C 2 N 2 O 2 United atom limit O + 2 N + copyright 2011 William A. Goddard III, all rights 2 reserved F 2 separated atom limit 61

62 1s and 2s cases B A B A 62

63 Bond Anti BO

64 More about O2 64

65 O 2 MO configuration For O2 the ordering of the MOs Is unambiguous (1π g ) Next consider states of (1π g )

66 Have 4 spatial combinations Which we combine as where x and y denote π x and π y φ 1, φ 2 denote the angle about the axis States based on (π) 2 and F is independent of φ 1, φ 2 Rotating about the axis by an angle γ, these states transform as Σ Σ

67 States arising from (π) 2 Adding spin we get MO theory explains the triplet ground state and low lying singlets O 2 Energy (ev) (π) Ground state

68 First excited configuration skip (1π g ) 2 Ground configuration (1π (1π g ) 3 u ) 3 excited configuration 1 Σ u + 1 u (1π u ) 3 (1π g ) 3 3 Σ u - 3 Σ u + Only dipole allowed transition from 3 Σ g - 1 Σ u - 3 u Strong transitions (dipole allowed) S=0 (spin) Σ g Σ u or Π u but Σ - Σ - 68

69 The states of O 2 molecule Moss and Goddard JCP 63, 3623 (1975) (π u ) 3 (π g ) 3 (π u ) 4 (π g ) 2 69

70 Exitation energies (ev) to O 2 excited states vertical 70

71 Role of O 2 in atmosphere Moss and Goddard JCP 63, 3623 (1975) Strong Get 3 P + 1 D O atom Weak Get 3 P + 3 P O atom 71

72 Implications UV light > 6 ev (λ < 1240/6 = 207 nm) can dissociate O 2 by excitation of 3 Σ u + which dissociates to two O atom in 3 P state UV light > ~7.2 ev can dissociate O 2 by excitation of 3 Σ u - which dissociates to one O atom in 3 P state and one in 1 D (maximum is at ~8.6 ev, Schumann-Runge bands) Net result is dissociation of O 2 into O atoms 72

73 Regions of the atmosphere O + hν O + + e - Heats from light O 2 + hν O + O stratopause O + O 2 O 3 tropopause altitude (km) O 3 + hν O + O 2 Heats from light Heated from earth mesosphere 300 stratosphere troposphere 73

74 ionosphere night D layer day Heaviside-Kennelly layer Reflects radio waves to allow long distance communications 74

75 nightglow At night the O atoms created during the day can recombine to form O 2 The fastest rates are into the Herzberg states, 3 Σ + 1 Σ - u u 3 u Get emission at ~2.4 ev, 500 nm Called the nightglow (~ 90 km) 75

76 Problem with MO description: dissociation 3 Σ g - state: [(π gx )(π gy )+ (π gy ) (π gx )] As R (π gx ) (x L x R ) and (π gy ) (y L y R ) Get equal amounts of {x L y L and x R y R } and {x L y R and x R y L } Ionic: [(O - )(O + )+ (O + )(O - )] covalent: (O)(O) But actually it should dissociate to neutral atoms 76

77 Back to valence bond (and GVB) Four ways to combine two 3P states of O to form a σ bond bad Open shell Each doubly occupied orbital overlaps a singly occupied orbital, not so repulsive Closed shell Looks good because make p bond as in ethene, BUT have overlapping doubly occupied orbitals antibonding 77

78 Analysis of open shell configurations Each can be used to form a singlet state or a triplet state, e.g. Singlet: A{(x L ) 2 (y R ) 2 [(y L )(x R ) + (x R )(y L )](αβ βα)} Triplet: A{(x L ) 2 (y R ) 2 [(y L )(x R ) - (x R )(y L )](αβ+βα)} and αα, ββ Since (y L ) and (x R ) are orthogonal, high spin is best (no chance of two electrons at same point) as usual 78

79 GVB wavefunction of triplet O 2 : sigma orbitals (O2s L ) 2 O2pz L bond O2pz R R=4 bohr (O2s R ) 2 R=3 bohr R e =2.28 bohr Get orthogonal to O2s on other center Causes some (2s-lpz) to stay orthogonal to bond pair Moss, Bobrowicz, Goddard JCP 63, 4632 (1975) 79

80 GVB wavefunction of triplet O 2 : pi orbitals (Opx L ) 2 O2px R Spin paired O2py L R=4 bohr (Opy R ) 2 R=3 bohr R e =2.28 bohr Get orthogonal to O2pπ on other center Doubly occupied orbtial delocalizes (bonding) Moss, Bobrowicz, Goddard JCP 63, 4632 (1975) 80

81 GVB orbitals at R e Problem: one VB configuration not enough + 81

82 VB description of O Must have resonance of two VB configurations 82

83 Back to valence bond (and GVB) Four ways to combine two 3P states of O to form a σ bond bad Open shell Each doubly occupied orbital overlaps a singly occupied orbital, not so repulsive Closed shell Looks good because make p bond as in ethene, BUT have overlapping doubly occupied orbitals antibonding 83

84 Bond energies 5.2 ev 84

85 Bring H toward px on Left O Bond H to O 2 2 A state Overlap doubly occupied (π xl ) 2 thus repulsive Get HOO bond angle ~ 90º S=1/2 (doublet) Overlap singly occupied (π xl ) 2 thus bonding Antisymmetric with respect to plane: A irreducible representation (Cs group) Bond weakened by ~ 51 kcal/mol due to loss in O 2 resonance 85

86 Bond 2 nd H to HO 2 to form hydrogen peroxide Bring H toward py on right O Expect new HOO bond angle ~ 90º Expect HOOH dihedral ~90º Indeed H-S-S-H: HSS = 91.3º and HSSH= 90.6º But H-H overlap leads to steric effects for HOOH, net result: HOO opens up to ~94.8º HOOH angle 111.5º trans structure, 180º only 1.2 kcal/mol higher 86

87 Rotational barriers HOOH 7.6 kcal/mol Cis barrier 1.19 kcal/mol Trans barrier HSSH: 5.02 kcal/mol trans barrier 7.54 kcal/mol cis barrier 87

88 Compare bond energies (kcal/mol) O 3 2 Σ - g HO-O H-O HO-OH 51.1 HOO-H 85.2 Interpretation: OO σ bond = 51.1 kcal/mol OO π bond = =67.9 kcal/mol (resonance) Bonding H to O 2 loses 50.8 kcal/mol of resonance Bonding H to HO 2 loses the other 17.1 kcal/mol of resonance Intrinsic H-O bond is =102.3 compare CH 3 O-H: HO bond is

89 Add material for O2 + C2H4 (sing and trip) 89

90 Bond O 2 to O to form ozone Require two OO σ bonds get States with 4, 5, and 6 pπ electrons Ground state is 4π case Get S=0,1 but 0 better Goddard et al Acc. Chem. Res. 6, 368 (1973) 90

91 sigma GVB orbitals ozone 91

92 Pi GVB orbitals ozone Some delocalization of central Opπ pair Increased overlap between L and R Opπ due to central pair 92

93 Bond O 2 to O to form ozone lose O-O π resonance, 51 kcal/mol New O-O σ bond, 51 kcal/mol Gain O-Oπ resonance,<17 kcal/mol,assume 2/3 New singlet coupling of π L and π R orbitals Total splitting ~ 1 ev = 23 kcal/mol, assume ½ stabilizes singlet and ½ destabilizes triplet Expect bond for singlet of = 23 kcal/mol, exper = 25 Expect triplet state to be bound by = -1 kcal/mol, probably between +2 and -2 93

94 Alternative view of bonding in ozone Start here with 1-3 diradical Transfer electron from central doubly occupied pπ pair to the R singly occupied pπ. Now can form a π bond the L singly occupied pπ. Hard to estimate strength of bond 94

95 Ring ozone Form 3 OO sigma bonds, but pπ pairs overlap Analog: cis HOOH bond is =43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O 2. Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane. Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol, But if formed it might be rather stable with respect various chemical reactions. Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN

96 Photochemical smog High temperature combustion: N 2 + O 2 2NO Thus Auto exhaust NO 2 NO + O 2 2 NO 2 NO 2 + hν NO + O O + O 2 + M O 3 + M O 3 + NO NO 2 + O 2 Get equilibrium Add in hydrocarbons NO 2 + O 2 + HC + hν Me(C=O)-OO-NO 2 peroxyacetylnitrate 96

97 More on N 2 The elements N, P, As, Sb, and Bi all have an (ns)2(np)3 configuration, leading to a triple bond Adding in the (ns) pairs, we show the wavefunction as This is the VB description of N2, P2, etc. The optimum orbitals of N2 are shown on the next slide. The MO description of N2 is Which we can draw as 97

98 GVB orbitals of N2 R e =1.10A R=1.50A R=2.10A 98

99 Hartree Fock Orbitals N 2 99

100 100

101 The configuration for C

102 The configuration for C2 Si 2 has this configuration From the 3 Π u was thought to be the ground state Now 1 Σ + is ground state

103 Stopped L8, Jan

104 Ground state of C 2 MO configuration Have two strong π bonds, but sigma system looks just like Be 2 which leads to a bond of ~ 1 kcal/mol The lobe pair on each Be is activated to form the sigma bond. The net result is no net contribution to bond from sigma electrons. It is as if we started with HCCH and cut off the Hs 104

105 C 2, Si 2, 105

106 106

107 Low-lying states of C2 107

108 108

109 109

110 Include B2, Be2, Li2, Li2+ 110

111 Van der Waals interactions For an ideal gas the equation of state is given by pv =nrt where p = pressure; V = volume of the container n = number of moles; R = gas constant = N A k B N A = Avogadro constant; k B = Boltzmann constant Van der Waals equation of state (1873) [p + n 2 a/v 2 )[V - nb] = nrt Where a is related to attractions between the particles, (reducing the pressure) And b is related to a reduced available volume (due to finite size of particles) 111

112 Noble gas dimers No bonding at the VB or MO level Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R 6 Ar 2 σ R e D e LJ 12-6 Force Field E=A/R 12 B/R 6 = D e [ρ 12 2ρ 6 ] = 4 D e [τ 12 τ 6 ] ρ= R/Re τ= R/σ where σ = R e (1/2) 1/6 =0.89 R e 112

113 London Dispersion The weak binding in He2 and other noble gas dimers was explained in terms of QM by Fritz London in 1930 The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R 3, but since the average dipole is zero the first nonzero contribution is from 2 nd order perturbation theory, which scales like -C/R 6 (with higher order terms like 1/R 8 and 1/R 10 ) 113

114 London Dispersion The weak binding in He2 and other nobel gas dimers was explained in terms of QM by Fritz London in 1930 The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R 3, but since the average dipole is zero the first nonzero contribution is from 2 nd order perturbation theory, which scales like -C/R 6 (with higher order terms like 1/R 8 and 1/R 10 ) Consequently it is common to fit the interaction potentials to functional forms with a long range 1/R 6 attraction to account for London dispersion (usually referred to as van der Waals attraction) plus a short range repulsive term to account for short Range Pauli Repulsion) 114

115 MO and VB view of He dimer, He 2 MO view Ψ MO (He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 VB view Net BO=0 Ψ VB (He 2 ) = A[(χ L α)(χ L β)(χ R α)(χ R β)]= (χ L ) 2 (χ R ) 2 Substitute σ g = χ R + χ L and σ g = χ R - χ L Get Ψ MO (He 2 ) Ψ MO (He 2 ) Pauli orthog of χ R to χ L repulsive 115

116 Remove an electron from He 2 to get He 2 + MO view Ψ(He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 Two bonding and two antibonding BO= 0 Ψ(He 2+ ) = A[(σ g α)(σ g β)(σ u α)]= (σ g ) 2 (σ u ) BO = ½ Get 2 Σ + u symmetry. Bond energy and bond distance similar to H 2+, also BO = ½ 116

117 MO view Remove an electron from He 2 to get He 2 + Ψ(He 2 ) = A[(σ g α)(σ g β)(σ u α)(σ u β)]= (σ g ) 2 (σ u ) 2 Two bonding and two antibonding BO= 0 Ψ(He 2+ ) = A[(σ g α)(σ g β)(σ u α)]= (σ g ) 2 (σ u ) BO = ½ Get 2 Σ u + symmetry. Bond energy and bond distance similar to H 2+, also BO = ½ VB view Substitute σ g = χ R + χ L and σ g = χ L - χ R Get Ψ VB (He 2 ) A[(χ L α)(χ L β)(χ R α)] - A[(χ L α)(χ R β)(χ R α)] = (χ L ) 2 (χ R ) - (χ R ) 2 (χ L ) - 117

118 He Σ g + (σ g ) 1 (σ u ) 2 2 Σ u + (σ g ) 2 (σ u ) - BO=0.5 MO good for discuss spectroscopy, VB good for discuss chemistry Check H2 and H2+ numbers He 2 R e =3.03A D e =0.02 kcal/mol No bond H 2 R e =0.74xA D e =110.x kcal/mol BO = 1.0 H 2 + R e =1.06x A D e =60.x kcal/mol BO =

119 Re-examine the energy for H 2 + For H 2 + the VB wavefunctions were Φ g = (х L + х R ) and Φ u = (х L - х R ) (ignoring normalization) where H = h + 1/R. This leads to the energy for the bonding state e g = <L+R H L+R>/ <L+R L+R> = 2 <L H L+R>/ 2<L L+R> = (h LL + h LR )/(1+S) + 1/R And for the antibonding state e u = (h LL - h LR )/(1-S) + 1/R We find it convenient to rewrite as e g = (h LL + 1/R) + τ/(1+s) e u = (h LL + 1/R) - τ/(1-s) where τ = (h LR - Sh LL ) includes the terms that dominate the bonding and antibonding character of these 2 states 119

120 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 ε g = (h LL + 1/R) + τ/(1+s) e cl + E g x ε u = (h LL + 1/R) - τ/(1-s) e cl + E u x where τ = (h LR - Sh LL ) is the VB interference or resonance energy and ε cl = (h LL + 1/R) is the classical energy As shown here the τ dominates the bonding and antibonding of these states 120

121 stop 121