Lecture 14 February 7, 2014...

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L11 Ch120a-Goddard-

L01

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Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

William A. Goddard, III, [email protected] 316 Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Lecture 14 February 7, 2014

Symmetry

Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday

Teaching Assistants:Sijia Dong <[email protected]> Samantha Johnson <[email protected]>

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L12

symmetry material

2

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L12

Inversion Symmetry for H2

Because I*I=e, we showed in L2 that the inversion symmetry of H2 leads to the result that every eigenstate of H2 is either g or u

For multielectron systems, the inversion symmetry inverts all electron coordinates simultaneously (xk,yk,zk)è (-xk,-yk,-zk) for k=1,..N But I*I takes (xk,yk,zk)è (+xk,+yk,+zk) for k=1,..N That is I*I=e the identity operator

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L12 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)


L02

5

Permutational symmetry, summary Our Hamiltonian for H2, 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: ρ1 ⎜⎝ ρ2

Spin only: σ1 ⎜⎝ σ2 Space and spin simultaneously: (ρ1,σ1) (ρ2,σ2) 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


L02

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The 6th 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

τij Ψ = - Ψ for all i and j


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 R1 is called a symmetry transformation if it has the property that R1 (HΨ)=H(R1Ψ) The set of all possible symmetries transformations of H are collected into what is called a Group.


The definition of a Group

1). Closure: If R1,R2 ε G (both are symmetry transformations) then R2 R1 is also a symmetry transformation, R2 R1 ε G 2. Identity. The do-nothing operator or identity, R1 = e ε G is clearly is a symmetry transformation

3. Associativity. If (R1R2)R3 =R1(R2R3).

4. Inverse. If R1 ε G then the inverse, (R1)-1 ε G ,where the inverse is defined as (R1)-1R1

= 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


The degenerate eigenfunctions of H form a representation

If HΨ=EΨ then H(R1Ψ)= E(R1Ψ) for all symmetry transformations of H.

Thus the transformations amount the n denegerate functions, {S=(RiΨ), where Ri Ψi 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


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


H2O, an example of C2v

consider the nonlinear H2A molecule, with equal bond lengths, e.g. H2O, CH2, NH2

The symmetry transformations are

1. e for einheit (unity) xèx, yèy, zèz

2. C2z, rotation about the z axis by 2π/2=180º, xè-x, yè-y, zèz

3.  σxz, reflection in the xz plane, xèx, yè-y, zèz

4.  σyz, reflection in the yz plane, xè-x, yèy, zèz

Which is denoted as the C2v group.

But C2z* C2z = e and σxz* σxz = e and σyz* σyz = e

Hence the eigenfunctions of H are each sym or antisym with respect to each symmetry


The character table for C2v

Since C2zσxz= σyz the symmetries for σyz are already implied by C2zσxz. Thus there are only 4 possible symmetries.

Since (C2z)2 = e, (σxz)2 = e, (σyz)2 = e We expect wavefunctions to be ±1 under each operation name

1-e N-e a1

a2

b1

b2


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.

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. C2zσxz= σyz

Combine these as below to show the relationships

xy e

C2z

σxz

σyz

C2z

σyz

σxz

σxz

C2z

σyz

xy e

σxz

C2z σyz

C2v


Symmetries for NH2

Ψ1 = A{(N2pyα)[(NpL )(HL)+(HL)(NpL)](αβ-βα)[(NpR)(HR)+(HR)(NpR)](αβ-βα)}

NHR bond NHL bond

Applying σxz to the wavefunction leads to Ψ2 = A{(N2py)0[(NpR)(HR)+(HR)(NpR)](αβ-βα)[(NpL)(HL)+(HL)(NpL)](αβ-βα)}

Each term involves transposing two pairs of electrons, e.g., 1çè3 and 2çè4 as interchanging electrons.

Since each interchange leads to a sign change we find that σxzΨ1 = Ψ2 = Ψ1 Thus interchanging a bond pair leaves Ψ invariant

Also the N(1s)2 and (2s)2 pairs are invariant under all operations

L

R


NH2 symmetry continued

Consider now the singly occupied Npy orbital C2z changes the sign but, σxz does not. Thus Npx transforms as b1 and The total wavefunction transforms as B1. Include the S= 1/2 , leads to The 2B1 state

1-e N-e a1

a2

b1

b2

Ψ1 = A{(N2pyα)[(NpL )(HL)+(HL)(NpL)](αβ-βα)[(NpR)(HR)+(HR)(NpR)](αβ-βα)}

NHR bond NHL bond


Symmetries for H2O and CH2 H2O

NH2

A{(O2py)2[(Opx )(Hx)+(Hx)(Opx)](αβ-βα)[(Opz)(Hz)+(Hz)(Opz)](αβ-βα)}

OHx bond OHz bond

H2O

A{(C2py)0[(Cpx )(Hx)+(Hx)(Cpx)](αβ-βα)[(Cpz)(Hz)+(Hz)(Cpz)](αβ-βα)} CH2

CH2

Since have enen number electrons in 2py, wavefunction is invarient under all symmetry transformations, thus must be 1A1.


Now do triplet state of CH2

Soon we will consider the triplet state of CH2 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 2px orbital (denoted as π to indicate antisymmetric with respect to the plane)

A{(C2sα)1(2pxα)1[(CpL )(HL)+(HL)(CpL)](αβ-βα)[(CpR)(HR)+(HR)(CpR)](αβ-βα)}

CHR bond CHL bond

Thus the symmetry of triplet CH2 is 3B1

Since we know that the two CH bonds are invariant under all symmetry operations, from now on we will write the wavefunction as A{[(CHL)2(CHR)2](Cσα)1(Cπα)1}

Here σ is invariant (a1) while π transforms as b1. Since both s and p are unpaired the ground state is triplet or S=1

σ=2s

z y

π=2px


Second example, C3v, with NH3 as the prototype

We will consider a system such as NH3, 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.

NHb bond

A{[(Npy )(Hy)+(Hy)(Npy)](αβ-βα)[(Npx )(Hx)+(Hx)(Npx)](αβ-βα)[(Npz)(Hz)+(Hz)(Npz)] (αβ-βα)}

NHx bond NHc bond

z x


x

b

c

x

b

c

Symmetry elements for C3v

Take the z axis out of the plane. The six symmetry operations are:

e C3

x

b

c C32= C3

-1

x

b

c

x

b

c σxz σxzC3

x

b

c

σxzC32

z x


The C3v symmetry group

x

b

c

e

C3

C32= C3

-1

σxz

σxzC3

σxzC32

The σxzC3 transformation corresponds to a reflection in the bz plane (which is rotated by C3 from the xz plane) and the σxzC3

2

transformation corresponds to a reflection in the cz plane (which is rotated by C3

2 from the xz plane). Thus these 3 reflections are said to belong to the same class.

Since {C3 and C3-1 do similar

things and are converted into each other by σxz we say that they are in the same class.


The E symmetry (irreducible representation) is of degree 2, which means that if φpx is an eigenfunction of the Hamiltonian, the so is φpy and they are degenerate.

This set of degenerate functions would be denoted as {ex,ey} 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.

The character table for C3v

Thus an atom in a P state, say C(3P) at a site with C3v symmetry, would generally split into 2 levels, {3Px and 3Py} of 3E symmetry and 3Pz of 3A1 symmetry.


Application for C3v, NH3

Ψ=A{(σLP)2[(NHb bond)2(NHc bond)2(NHx bond)2]}

z x

Consider first the effect of σxz. This leaves the NHx bond pair invariant but it interchanges the NHb and NHc bond pairs. Since the interchange two pairs of electrons the wavefunction does not change sign. Also the σLP orbital is invariant.

We will write the wavefunction for NH3 as

where we combined the 3 Valence bond wavefunctions in 3 pair functions and we denote what started as the 2s pair as σLP

x

c b

σLP


Next consider the C3 symmetry operator. It does not change σLP. It moves the NHx bond pair into the NHb pair, moves the NHb pair into the NHc pair and moves the NHc pair into the NHx pair. A cyclic permutation on three electrons can be written as (135) = (13)(35). For example. 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 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 C3 does this cyclic permutation on 6 electrons,eg (135)(246)= (13)(35)(24)(46), we still get no sign change.

Consider the effect of C3

(135)

(35) (13)

Thus the wavefunction for NH3 has 1A1 symmetry


Linear molecules, C∞v symmetry

Consider the wavefunction for one electron in a linear molecule. Here we use polar coordinates, ρ = sqrt(x2+y2), 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){a0 + Σ m=1m=∞ [am cos mα + bm sin mα]

O=C=O z x





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

O=C=O z x





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)

O=C=O z x

Which means that the wavefunctions with the same m are degenerate


The symmetry operators are: Rz(α): counterclockwise rotation by an angle α about the z axis σxz: reflection in the xz plane (this takes +α into –α)

C∞v symmetry group O=C=O z

x


The symmetry operators are: Rz(α): counterclockwise rotation by an angle α about the z axis σxz: reflection in the xz plane (this takes +α into –α) σ’ = Rz(α) σxz Rz(-α); reflection in a plane rotated by an angle α from the xz plane (there are an infinite number of these)


x


σ’ = Rz(α) σxz Rz(-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:

x

y

e

σxz C-α

x

y

e -α x

y

e α

Cα

x

y

e α

x

y

e x

y

e α

σ’ σ’


The symmetry operators are: Rz(α): counterclockwise rotation by an angle α about the z axis σxz: reflection in the xz plane (this takes +α into –α) σ’ = Rz(α) σxz Rz(-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


x

name

σ

π

δ

φ

γ

1-e N-e


The symmetry functions for C∞v

Lower case letters are used to denote one-electron orbitals


Application to FH The ground state wavefunction of HF is

A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](αβ-βα)}

In C∞v symmetry, the bond pair is σ (m=0),while the px and py form a set of π orbitals (m=+1 and m=-1). Consider the case of up spin for both πx and πy Ψ(1,2) = A{φxαφyα}=(φxφy- φyφx) αα

Rotating by an angle γ about the z axis leads to φa = cosγ φx + sinγ φy and φb = cosγ φy - sinγ φx This leads to (φaφb- φbφa) = [(cosγ)2 +(sinγ)2] }=(φxφy- φyφx)

Thus (φxφy- φyφx) transforms as Σ.


Continuing with FH Thus the (px)2(py)2 part of the HF wavefunction

A{(F2px)2(F2py)2[(Fpz)(H)+(H)(Fpz)](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 1Σ+ symmetry


Next consider the ground state of OH

Ψx=A{(σOH bond)2[(πxα)(πyα)(πxβ)]}

Ψy=A{(σOH bond)2[(πxα)(πyα)(πyβ)]}

We write the two wavefunctions for OH as

2Πy

2Πx

z

x

We saw above that

A{(πxα)(πyα)} transforms like Σ. thus we need examine only the transformations of the downspin orbital. But this transforms like π.

Another way of describing this is to note that A{(πx)2(πy)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

Thus the total wavefunction is 2Π.


Now consider the ground state of NH

z

x

A{(NH bond)2(N2pxα)(N2pyα)}

We saw earlier that up-spin in both x and y leads to Σ symmetry.

Considering now the reflection, σxz, we see that with just one electron in πy, we now get Σ-. Thus the ground state of NH is 3Σ-.


Now consider Bonding H atom to all 3 states of C

No singly occupied orbital for H to bond with

(2px)(2py)

(2py)(2pz)

Bring H1s along z axis to C and consider all 3 spatial states.

(2px)(2pz)

O 2pz singly occupied.

H1s can get bonding

Get S= ½ state,

Two degenerate states, denote as 2Π

z x


Ground state of CH (2Π)

z

x

A{(2s)2(OHσ bond)2(O2pxα)1}

The full wavefunction for the bonding state 2Πx

2Πy

A{(2s)2(OHσ bond)2(O2pyα)1}


Bond a 2nd H atom to the ground state of CH

z

x

Starting with the ground state of CH, we bring a 2nd H along the x axis.

Get a second covalent bond

This leads to a 1A1 state.

No unpaired orbtial for a second covalent bond.


Analyze Bond in the ground state of CH2

z

x

Ground state has 1A1 symmetry. For optimum bonding, the pz orbital should point at the Hz while the px orbital should point at the Hx. Thus the bond angle should be 90º.

As NH2 (103.2º) and OH2 (104.5º), we expect CH2 to have bond angle of ~ 102º

θe Re


But, the Bending potential surface for CH2

3B1

1A1

1B1

3Σg-

1Δg

9.3 kcal/mol

The ground state of CH2 is the 3B1 state not 1A1.

Thus something is terribly wrong in our analysis of CH2


Compare bonding in BeH+ and BeH

BeH+

BeH

3 eV

2 eV

1 eV

Long range Repulsive interaction

with H

Short range Attractive interaction sz with H

1eV Repulsive orthogonalization of

zs with sz H

BeH+ has long range attraction no short range repulsion

TA’s check numbers, all from memory


Compare bonding in BeH and BeH2

BeH+

TA’s check numbers, all from memory

MgH+

3.1 eV R=1.31A

1.34 eV R=1.73A

Expect linear bond in H-Be-H and much stronger than the 1st bond

Expect bond energy similar to BeH+, maybe stronger, because the zs orbital is already decoupled from the sz.

1Σ+

2Σ+

1Σ+

linear

Cannot bind 3rd H because no singly occupied orbitals left.

2.1 eV R=1.65 A

2.03 eV R=1.34A

~3.1 eV ~2.1 e

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