Chemistry 431Lecture 14

Wave functions as a basisDiatomic molecules

Polyatomic moleculesHuckel theory

NC State University

Wave functions as the basis forirreducible representations

The energy of the system will not change when symmetry Operations are applied. If we consider atomic orbitals onEach atom as the basis vectors we state generally that

RΨ = +Ψ or –ΨS-orbitals are always + since they are sphericalP-orbitals can be either + or -.

RΨ = +Ψ RΨ = –Ψ

Using the concepts developed in the hydrogen moleculeion we can consider combinations of all of the atomic orbitals that are important for diatomic molecules.

Only orbitals that have the same irreducible representation can mix. Mathematically this statement is the same assaying the the overlap integrals will be non-zero only ifthey contain the totally symmetric representation.

Mixing of s- and p-orbitals in diatomic molecules

Homonuclear diatomic molecules belong to the D∞h.

We will determine the reducible representations of eachof the combinations shown below.

Homonuclear diatomic molecules

+ +

+

The D∞h symmetry operations are E, ∞ σv, C∞, i, 2S∞, ∞ C2.

The reducible representations are:

E` ∞ σv C∞ i 2S∞ ∞ C2.s 2 2 2 0 0 0pz 2 2 2 0 0 0px, py 4 4cosφ 0 0 0 0

The s irrep reduces to Σg+ to Σu

+ .The pz irrep reduces to Σg

+ to Σu+ .

The px, py irreps reduce to Πg to Πu.

Homonuclear diatomic molecules

The D∞h symmetry operations are E, ∞ σv, C∞, i, 2S∞, ∞ C2.

Homonuclear diatomic molecules

For polyatomic molecules appropriate atomic orbitals mustbe combined into symmetry adapted linear combinations(SALCs).

H2O belongs to the C2v point group.Valence orbitals

2 H(1s) + O(2s) + O(2px) + O(2py) + O(2pz) Since the oxygen is unmoved by every symmetry operationthe character can be determined directly from the charactertable. However, the hydrogens need to be formed as SALCsfirst.

Polyatomic molecules

The reducible representations for hydrogen SALCs area1 and b1. The oxygen orbitals (read from the character Table are:2s = a12pz = a12px = b12py = b2

Polyatomic molecules

x

z

y

Note that the energies are obtainedfrom a calculation. However, the forms of the orbitals can be predicted based on symmetry considerations.

Molecules with π-systems are particularly important in Organic chemistry. The simplest of these is ethylene.

Ethylene

Ethylene belongs to the D2h point group. The carbons are symmetrically equivalent. The orbital set on each consists of

C(2s) + C(2px) + C(2py) + C(2pz)

H

CH

C

H

H

Character Table for the D2h Point Group

E C2 (z) C2 (y) C2 (x) i σ (xy) σ (xz) σ (yz) linear,rotations quadratic

Ag 1 1 1 1 1 1 1 1 x2, y2, z2

B1g 1 1 -1 -1 1 1 -1 -1 Rz xyB2g 1 -1 1 -1 1 -1 1 -1 Ry xzB3g 1 -1 -1 1 1 -1 -1 1 Rx yzAu 1 1 1 1 -1 -1 -1 -1B1u 1 1 -1 -1 -1 -1 1 1 zB2u 1 -1 1 -1 -1 1 -1 1 yB3u 1 -1 -1 1 -1 1 1 -1 x

The atomic orbitals are shown below:

The reducible representations of the symmetrically equivalentsets formed for the valence ao’s of ethylene can be determined and decomposed in the following groups.

Ethylene

Since the members of all of the sets can be exchanged completely with only the rotational operations, only the C2characters of the above character table need be considered.

Ethylene

The normalized SALCs can now be constructed:

Ethylene

The normalized SALCs can now be constructed:

Ethylene

1. Determine the molecular point group.2. Determine the normalized symmetry adapted linear

combinations (SALCs) using the projection operator method.If there is only one set of SALCs then these are also the MOs.

3. Use the SALCs to determine the elements of the Huckelsecular determinant

4. Simplify the mathematical formalism by letting Hii = α andHij = β. Divide each element by β and let x = (α - E)/β

5. For each eigenvalue solve for the eigenvectors (coefficients).These are the contributions of each SALC to the MO.

Huckel MO Theory

det |Hij – SijEo| = 0

|Hij – SijEo|CijΣj = 1

N= 0

Cij2Σ

j = 1

N= 1

Butadiene1

2 3

4 2 and 3 are symmetry related1 and 4 are symmetry relatedBoth transform as

E C2v σ(xz) σ(yz)Γ 2 0 2 0

ButadieneE C2v σ(xz) σ(yz)

Γ 2 0 2 0A1 1 1 1 1Total 2 + 0 + 2 + 0 = 4

E C2v σ(xz) σ(yz)Γ 2 0 2 0A2 1 1 -1 -1Total 2 + 0 - 2 + 0 = 0

E C2v σ(xz) σ(yz)Γ 2 0 2 0B1 1 -1 1 -1Total 2 + 0 + 2 + 0 = 4

E C2v σ(xz) σ(yz)Γ 2 0 2 0B1 1 -1 -1 1Total 2 + 0 - 2 + 0 = 0

The order is 4 so the character sums are divided by 4. The irreducible representations are:

Γ = a1 + b1

Butadiene

SALC14(b1) = p1 – p4 – p4 + p1

SALC14(a1) = p1 + p4 + p4 + p1

SALC14(b 2) = 12

p1 + p4

SALC14(a2) = 12

p1 – p4

SALC14(a1)

SALC14(b1)

Butadiene

SALC23(b1) = p2 – p3 – p3 + p2

SALC23(a1) = p2 + p3 + p3 + p2

SALC23(a2) = 12

p2 – p3

SALC23(b 2) = 12

p2 + p3SALC23(a1)

SALC23(b1)

Construct symmetrized Huckel DeterminantThe unsymmetrized Huckel determinant is a 4x4 matrix.The solution is a fourth order polynomial. By factoring according to symmetry the determinant can be simplified into2 2x2 matrices.

H22 = 12 p2 + p3 H p2 + p3 dτ

= 12 p2Hp2dτ + 1

2 p3Hp3dτ + 12 p2Hp3dτ + 1

2 p3Hp2dτ

= α + β

H11 = 12 p1 + p4 H p1 + p4 dτ

= 12 p1Hp1dτ + 1

2 p4Hp4dτ + 12 p1Hp4dτ + 1

2 p4Hp1dτ

= α

Construct symmetrized Huckel Determinant

H12 = 12 p1 + p4 H p2 + p3 dτ

= 12 p1Hp2dτ + 1

2 p4Hp2dτ + 12 p1Hp3dτ + 1

2 p4Hp3dτ

= β

H21 = H12

H33 = 12 p1 – p4 H p1 – p4 dτ

= 12 p1Hp1dτ – 1

2 p4Hp1dτ – 12 p4Hp1dτ + 1

2 p4Hp4dτ

= α

Construct symmetrized Huckel Determinant

H34 = 12 p1 – p4 H p2 – p3 dτ

= 12 p1Hp2dτ – 1

2 p4Hp2dτ – 12 p1Hp3dτ + 1

2 p4Hp3dτ

= β

H43 = H34

H44 = 12 p2 – p3 H p2 – p3 dτ

= 12 p2Hp2dτ + 1

2 p3Hp3dτ – 12 p2Hp3dτ – 1

2 p3Hp2dτ

= α – β

H13 = H31 = H24 = H42 = H14 = H41 = H23 = H32 = 0

Construct symmetrized Huckel Determinant

H11 – E H12 H13 H14

H21 H22 – E H23 H24

H31 H32 H33 – E H34

H41 H42 H43 H44 – E

Divide by β and let x =α – E

β

α – E β 0 0β α+β – E 0 00 0 α – E β0 0 β α–β – E

Solve each symmetry block separately

x 1 0 01 x + 1 0 00 0 x 10 0 1 x – 1

= 0det

x 11 x – 1

= x x – 1 – 1 = x2 – x – 1 , x = 1 ± 1 + 42 = 5 ± 1

2

x 11 x + 1

= x x + 1 – 1 = x2 + x – 1 , x = – 1 ± 1 + 42 = – 5 ± 1

2

a1 block

b1 block

Determine the orbital energies

5 – 12 β = 0.62β

– 5 + 12 β = – 0.62β

– 5 – 12 β = – 1.62β

5 + 12 β = 1.62β

Determine the eigenvectorsOnly the SALCs from the same irreducible representationCan contribute to the same MO.For the b2 block and x = -1.62 we have:

– 1.62 11 – 1.62 + 1

c11

c21,

– 1.62 11 – 0.62

c11

c21= 0

– 1.62c11 +c21 = 0 c11 – 0.62c21 = 0c21 = 1.62c11 c11 = 0.62c21

and c112 + c21

2 = 1 so c112 + 1.622c11

2 = 3.62c112 = 1

c112 = 1/3.62 = 0.276 and finally c11 = 0.525

this implies that c21 = 1.62(0.525) = 0.85

The MO is 0.53 0.707 p1 + p4 +0.85 0.707 p2 + p3

= 0.37 p1 + p4 + 0.60 p2 + p3

= 0.37p1 +0.60p2 + 0.60p3 + 0.37p4

Determine the eigenvectorsOnly the SALCs from the same irreducible representationCan contribute to the same MO.For the b2 block and x = 0.62 we have:

0.62 11 0.62 + 1

c11

c21,

0.62 11 1.62

c11

c21= 0

0.62c11 +c21 = 0 c11 + 1.62c21 = 0c21 = –0.62c11 c11 = –1.62c21

and c112 + c21

2 = 1 so c112 + 0.622c11

2 = 2.38c112 = 1

c112 = 1/2.38 = 0.722 and finally c11 = 0.85

this implies that c21 = –0.62(0.85) = – 0.53

The MO is 0.85 0.707 p1 + p4 – 0.53 0.707 p2 + p3

= 0.60 p1 + p4 – 0.37 p2 + p3

= 0.60p1 –0.37p2 – 0.37p3 + 0.60p4

SUMMARY

E = α + 1.62β

E = α + 0.62β

E = α − 0.62β

E = α − 1.62β

1b2

1a2

2b2

2a2

ψ1

ψ2

ψ3

ψ4

0.37 0.37

0.37 0.37

0.60 0.60

-0.60 -0.60

0.37 -0.37

0.60 -0.60

0.37-0.37

0.60 -0.60

1a1

1b1

2a1

2b1E = α + 1.62β

E = α + 0.62β

E = α - 0.62β

E = α - 1.62β

Bond orderThe bond order can be calculated from the MO coefficients.The p-bond order is the product of the coefficients of two adjacent atoms times the number of electrons in the respectiveorbitals.

Where nk is the number of electrons in the MO and cik and cjkAre the coefficients.

For butadiene the bond order for p1-p2 and p2-p3 is:

The bond order for p2-p3 is:

Bond order = nkcikc jkΣk = 1

occup

Bond order = 2(0.37)(0.60) + 2(0.60)(0.37) = 0.888

Bond order = 2(0.60)(0.60) + 2(0.37)( –0.37) = 0.446

Bond orderThe bond order of the butadiene double bond is 88.8% aslarge as that of ethylene.

However, the total π-bonding energy is 2(0.888)+0.446 = 2.22,which is greater than two ethylene molecules. This is an example of stabilization by delocalization.

The total energy is the energy of each MO times its occupancy.

For butadiene this is:(units of β)

Total energy = nkEkΣk = 1

occup

E = 2( – 1.62) + 2( – 0.62) = – 4.48

Most of the spectroscopy and reactivity of benzene is attributable to its π system. To a good approximation theσ and π orbitals can be separated.

We consider how the 6 p-orbitalstransform under D6h symmetry.

In this case we do not need toconstruct SALCs since there isonly one type of orbital.

Benzene

Γ

The operation required to carry the reference p1 orbital intoany of the others.

Projection operator approach

C6b2g

The operation required to carry the reference p1 orbital intoany of the others.

Projection operator approach

C3b2g

The operation required to carry the reference p1 orbital intoany of the others.

Projection operator approach

C2b2g

D6h character table

Molecular orbitals constructed using the projection operator approach

Mn

OH2

OH2H2O

H2O

OH2

H2O

In the crystal field picture the repulsionof the oxygen lone pairs causes thed orbitals to split. We consider eachwater as a point in the Oh point group.

E 6C3 0C2 0C4 2C2d 2i 0S4 0S6 0σh 4σd 2

The reducible representation can be decomposed as follows:

a1g = (1)(1)(6) + (6)(1)(2) + (3)(1)(2) + (3)(1)(4) + (6)(1)(2) = 48a2g = (1)(1)(6) + (6)(-1)(2) + (3)(1)(2) + (3)(1)(4) + (6)(-1)(2) = 0eg = (1)(2)(6) + (6)(0)(2) + (3)(2)(2) + (3)(2)(4) + (6)(0)(2) = 48t1g = (1)(3)(6) + (6)(1)(2) + (3)(-1)(2) + (3)(-1)(4) + (6)(-1)(2) = 0t2g = (1)(3)(6) + (6)(-1)(2) + (3)(-1)(2) + (3)(-1)(4) + (6)(1)(2) = 0a1u = (1)(1)(6) + (6)(1)(2) + (3)(1)(2) + (3)(-1)(4) + (6)(-1)(2) = 0a2u = (1)(1)(6) + (6)(-1)(2) + (3)(1)(2) + (3)(-1)(4) + (6)(1)(2) = 0eu = (1)(2)(6) + (6)(0)(2) + (3)(2)(2) + (3)(-2)(4) + (6)(0)(2) = 0t1u = (1)(3)(6) + (6)(1)(2) + (3)(-1)(2) + (3)(1)(4) + (6)(1)(2) = 48t2u = (1)(3)(6) + (6)(-1)(2) + (3)(-1)(2) + (3)(1)(4) + (6)(-1)(2) = 0

The irreps are a1g + eg + t1u.

The symmetries of the metal orbitals are obtained from the character table to be: a1g (s) + t1u (px, py, pz) + eg (z2, x2-y2) + t2g (xz,yz,xy)

The a1g ligand SALC is:

a1g =16

σ1 +σ 2 +σ 3 +σ 4 +σ 5 +σ 6( )

which has the correct symmetry to interact with the metal s-orbital

Metal-based σ* orbital

Ligand-based σ orbital

To construct the eg and t1u combinations for a group of order 48 would be quite tedious! Consider the nature of the z2 and x2-y2 orbitals to construct the two eg SALCs. Remember that the amplitude of the z2 torus in the xy plane is one-half that along the z-axis. Then the following two eg bonding combinations can be constructed.

eg (a) =112

−σ1 − σ2 − σ3 − σ4 + 2σ 5 + 2σ 6( ) & eg b( ) =12

−σ 2 −σ 4 + σ5 + σ6( )

The eg metal orbitals have the correct symmetry to interact with ligand σ SALCs, they are referred to as the dσ orbitals, but since they are destabilized in the process, the eg orbitals are also referred to as the dσ* orbitals

The three t1u SALCs must combine with the metal p-orbitals. Since there are only two lobes on each of the p's, then only two ligand AOs will be involved in each

t1u (x) =12

σ2 −σ 4( ); t1u (y) =12

σ5 − σ6( ); t1u (z) =12

σ1 −σ 3( )

The t2g metal orbitals are non-bonding with respect to σ-interactions.

The energy level diagram for σ-bonding in Oh complexes is shown below. MO theory predicts the same picture as CFT. In CFT, the metal eg orbitals are forced to higher energy by Coulombicrepulsions with ligand electrons while MO theory indicates that the higher energy eg orbitals mix with ligand σ-SALCs and are destabilized.

Approximate MO diagram for σ mixing in octahedral field

egLigand σ-SALCs

t2g

a1g

t1u

t1uega1g

Metal-ligand σ-

MOsMetal

orbitalsps

d{