Lecture 14 February 7, 2014 Symmetry

Transcription

1 Lecture 14 February 7, 2014 Symmetry 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, 3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants:Sijia Dong Samantha Johnson Ch120a-Goddard-L11 Ch120a- 1 Goddard-

2 symmetry material 2

3 Inversion Symmetry for H 2 For multielectron systems, the inversion symmetry inverts all electron coordinates simultaneously ( k,y k,z k )è (- k,-y k,-z k ) for k=1,..n But I*I takes ( k,y k,z k )è (+ 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

4 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) 4

5 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- 5 Goddard-

6 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- 6 Goddard-

7 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 etract the maimum 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. 7

8 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 8

9 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 9

10 Eample, an atom. For an atom any rotations about any ais 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 10

11 H 2 O, an eample 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) è, yè y, zè z 2. C 2z, rotation about the z ais by 2π/2=180º, è -, yè -y, zè z 3. σ z, reflection in the z plane, è, yè -y, zè z 4. σ yz, reflection in the yz plane, è -, yè y, zè z Which is denoted as the C 2v group. But C 2z * C 2z = e and σ z * σ z = e and σ yz * σ yz = e Hence the eigenfunctions of H are each sym or antisym with respect to each symmetry 11

12 The character table for C 2v Since (C 2z ) 2 = e, (σ z ) 2 = e, (σ yz ) 2 = e We epect wavefunctions to be ±1 under each operation name 1-e N-e a 1 a 2 b 1 b 2 Since C 2z σ z = σ yz the symmetries for σ yz are already implied by C 2z σ z. Thus there are only 4 possible symmetries. 12

13 Consider the stereographic projection of the points on the surface of a sphere onto a plane, where positive are circles and negative are squares. y e σ yz C 2z Stereographic projections 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 z = σ yz Combine these as below to show the relationships σ z σ yz C 2z σ yz σ z C 2z σ yz y e σ z σ z C 2v 13

14 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 z 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., 1ç è 3 and 2ç è 4 as interchanging electrons. Since each interchange leads to a sign change we find that σ z 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 14

15 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 z does not. Thus Np 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 15

16 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 )(H )+(H )(Op )](αβ βα)[(op z )(H z )+(H z )(Op z )](αβ βα)} OH bond OHz bond A{(C2p y ) 0 [(Cp )(H )+(H )(Cp )](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. 16

17 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 ecited to the 2p orbital (denoted as π to indicate antisymmetric with respect to the plane) A{(C2sα) 1 (2p 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 the symmetry of copyright triplet 2011 CHWilliam 2 is 3 A. BGoddard 1 III, all rights reserved y σ=2s π=2p z 17

18 Second eample, C 3v, with NH 3 as the prototype A{[(Np y )(H y )+(H y )(Np y )](αβ βα)[(np )(H )+(H )(Np )](αβ βα)[(np z )(H z )+(H z )(Np z )] (αβ βα)} NH b bond NH bond We will consider a system such as NH 3, with three equal bond lengths. Here we will take the z ais as the symmetry ais and will have one H in the z plane. The other two NH bonds will be denoted as b and c. NH c bond z 18

19 Symmetry elements for C 3v Take the z ais out of the plane. The si symmetry operations are: b b b z C 3 e c c c C 3 2 = C 3-1 b b b σ z C 3 2 c σ z σ z C 3 c c 19

20 b C 3 σ z C 3 c σ z C 3 2 C 32 = C 3-1 The C 3v symmetry group e σ z The σ z C 3 transformation corresponds to a reflection in the bz plane (which is rotated by C 3 from the z plane) and the σ z C 3 2 transformation corresponds to a reflection in the cz plane (which is rotated by C 3 2 from the z 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 z we say that they are in the same class. 20

21 The character table for C 3v The E symmetry (irreducible representation) is of degree 2, which means that if φ p 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,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 and 3 P y } of 3 E symmetry and 3 P z of 3 A 1 symmetry. 21

22 Application for C 3v, NH 3 We will write the wavefunction for NH 3 as Ψ=A{(σ LP ) 2 [(NH b bond) 2 (NH c bond) 2 (NH 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 z. This leaves the NH 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. z σ LP 22

23 Consider the effect of C 3 Net consider the C 3 symmetry operator. It does not change σ LP. It moves the NH 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 pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For eample. (135) φ(1)φ(3)φ(5) è 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) è φ(1)φ(5)φ(3) è 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 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( 2 +y 2 ), α, z (ais along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be epanded 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] 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( 2 +y 2 ), α, z (ais along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be epanded 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 epect m=0 lowest, then m=±1, then m=±2, m=±3, etc 25

26 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( 2 +y 2 ), α, z (ais along z) Since the wavefunction has period of 2π in α, the α dependence of any wavefunction can be epanded 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 epect m=0 lowest, then m=±1, then m=±2, m=±3, etc Also if we rotate the molecule about the z ais 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 26

27 C v symmetry group The symmetry operators are: O=C=O z R z (α): counterclockwise rotation by an angle α about the z ais σ z : reflection in the z plane (this takes +α into α) 27

28 C v symmetry group The symmetry operators are: O=C=O z R z (α): counterclockwise rotation by an angle α about the z ais σ z : reflection in the z plane (this takes +α into α) σ = R z (α) σ z R z (-α); reflection in a plane rotated by an angle α from the z plane (there are an infinite number of these) 28

29 σ = R z (α) σ z R z (-α) is a reflection in the plane rotated by an angle α from the z plane Take the z ais out of the plane. The si symmetry operations are: y y y y e α e α e α e C -α σ z C α y y e σ α e σ 29

30 C v symmetry group The symmetry operators are: O=C=O z R z (α): counterclockwise rotation by an angle α about the z ais σ z : reflection in the z plane (this takes +α into α) σ = R z (α) σ z R z (-α); reflection in a plane rotated by an angle α from the z 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 π δ φ γ 30

31 The symmetry functions for C v Lower case letters are used to denote one-electron orbitals 31

32 Application to FH The ground state wavefunction of HF is A{(F2p ) 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 and p y form a set of π orbitals (m=+1 and m=-1). Consider the case of up spin for both π and π y Ψ(1,2) = A{φ αφ y α}=(φ φ y φ y φ ) αα Rotating by an angle γ about the z ais leads to φ a = cosγ φ + sinγ φ y and φ b = cosγ φ y - sinγ φ This leads to (φ a φ b φ b φ a ) = [(cosγ) 2 +(sinγ) 2 ] }=(φ φ y φ y φ ) Thus (φ φ y φ y φ ) transforms as Σ. 32

33 Continuing with FH Thus the (p ) 2 (p y ) 2 part of the HF wavefunction A{(F2p ) 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 z plane. Here p 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 33

34 Net consider the ground state of OH We write the two wavefunctions for OH as Ψ =A{(σ OH bond) 2 [(π α)(π y α)(π β)]} 2 Π Ψ y =A{(σ OH bond) 2 [(π α)(π y α)(π y β)]} We saw above that A{(π α)(π y α)} transforms like Σ. thus we need eamine only the transformations of the downspin orbital. But this transforms like π. Thus the total wavefunction is 2 Π. 2 Π y z Another way of describing this is to note that A{(π ) 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 34

35 Now consider the ground state of NH A{(NH bond) 2 (N2p α)(n2p y α)} We saw earlier that up-spin in both and y leads to Σ symmetry. Considering now the reflection, σ z, we see that with just one electron in π y, we now get Σ -. Thus the ground state of NH is 3 Σ -. z 35

36 Now consider Bonding H atom to all 3 states of C (2p )(2p z ) Bring H1s along z ais to C and consider all 3 spatial states. O 2p z singly occupied. H1s can get bonding (2p y )(2p z ) Get S= ½ state, Two degenerate states, denote as 2 Π (2p )(2p y ) No singly occupied orbital for H to bond with z 36

37 Ground state of CH ( 2 Π) The full wavefunction for the bonding state 2 Π A{(2s) 2 (OHσ bond) 2 (O2p α) 1 } 2 Π y z A{(2s) 2 (OHσ bond) 2 (O2p y α) 1 } 37

38 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 ais. Get a second covalent bond This leads to a 1 A 1 state. No unpaired orbtial for a second covalent bond. z 38

39 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 orbital should point at the H. Thus the bond angle should be 90º. As NH 2 (103.2º) and OH 2 (104.5º), we epect CH 2 to have bond angle of ~ 102º θ e R e 39

40 But, the Bending potential surface for CH 2 1 B 1 1 Δ 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 40

41 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 BeH + 1eV Repulsive orthogonalization of zs with sz H 41

42 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 + linear Epect linear bond in H-Be-H and much stronger than the 1 st bond Epect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz. TA s check numbers, all from memory ~3.1 ev ~2.1 e 1 Σ + Cannot bind 3 rd H because no singly occupied orbitals left. 42