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Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 1 -

5. Organic Molecules: Electronic Structure and Chemical Reactions

5.1 Brief reminder: LCAO approximation

We describe the molecular wave functions as a Linear Combination of Atomic Orbitals

(LCAO)

∑ ϕ= i

iikk cψ

and use this ansatz in the Schrödinger equation:

0=− EψψĤ

and obtain

( ) 0=∑ ϕ− i

ikik EHc

Multiplication by jϕ from left and integration:

( ) ( ) 0=∑ −=∑ −=∑ − i

jikjiik i

kik i

kik SEHcijEiHjciEHjc ˆˆ

with the Hamilton matrix iHjH ji ˆ=

and the overlap matrix ijS ji = .

Secular equations:

( ) 0=∑ − i

jikjiik SEHc or ( ) 0=− kk cSEH

Solution: Secular determinant: 0=− SEH ⇒ eigenvectors kc , eigenvalues kE .

(5.1 Example: secular equation for butadien, C-2p orbitals only).

5.2 Brief: reminder: Hückel molecular orbital model (HMO)

We consider only the π-system of a conjugated hydrocarbon using the approximations:

= =

else ; 0 neighborsnext ji,for ;

jifor ; β α

H ji

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 2 -

=

= else ; 0

ji ; 1 jiS

(5.2 Comments to HMO theory, example: butadiene).

5.3 The connection to symmetry

The LCAO ansatz in a basis set of n orbitals leads in general to a couples set of n secular

equations. With increasing number of basis function the numerical effort to solve the system

becomes becomes large quite rapidly. Even if some of the matrix elements of the H or S

matrix are zero, this does not necessarily simplify the problem.

The only way to drastically reduce the (numerical) effort for solving the system is to find a

basis transformation which blockdiagonalizes the H, S matrizes. A basis of symmetry adapted

fulfils this requirement:

We assume a set of ∑ i

il basis functions jϕ which can be transformed to a i sets of il functions,

i.e. ij lk iχ .. ..

1 1

= = is the k-th function belonging to irrep j. As '' jjS jj ≠= for 0 and '' jjH jj ≠= for 0 ,

the '' jj

kkH and ' '

jj kkS matrices are blockdiagonal:

0

00

00

00

00

00

00

2222 1

22 1

22 11

1111 1

11 1

11 11

2222 1

22 1

22 11

1111 1

11 1

11 11

111

1

111

1

111

1

111

1

=

−

c SS

SS SS

SS

E HH

HH HH

HH

lll

l

lll

l

lll

l

lll

l

O

L

MOM

K

L

MOM

K

O

L

MOM

K

L

MOM

K

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

, ,

Recipe:

1. The AO basis is a n dimensional reducible representation of the symmetry group.

Determine the number and symmetry type of contained irreps (section 3.13).

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 3 -

2. Construction of a set of orthonormal SALC using projection operators (section 4.6).

3. Reformulate secular equation in SALC basis.

(5.3 Example: butadiene).

5.4 Cyclic electron systems and cyclic groups

As an example we investigate the electronic structure of the π-system of benzene:

The symmetry group of the molecules is

D6h: E 2C6 2C3 C2 3C2’ 3C2’’ i 2S3 2S6 σh 3σd 3σvT

The characters can be determined by applying the operations and counting the number of

unaffected (+1) and inverted (-1) AOs (only those AOs which are moved in space contribute

to the character):

Γπ : 6 0 0 0 -2 0 0 0 0 -6 2 0

The representation can be analysed according to section 3.13, yielding:

Γπ = A2u + B2g + E1g + E2u

Now the SALCs can be constructed applying the projection operators. However, there is a

simpler procedure using the properties of cyclic groups.

We do not necessarily have to take into account all symmetry elements of the group, but can

choose any subgroup as well (accepting only partial symmetry adaptation if the choice of

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 4 -

subgroup is bad). A good choice for cyclic electron systems is the corresponding pure

rotational group (C6 in this case).

The uniaxial rotational groups are cyclic groups (section 3.4). All cyclic groups are Abelian

(section 3.4). In an Abelian group each elements belongs to its own class (AB=BA,

A=B-1AB, i.e. all elements are only self-conjugated). Therefore, the order of the group is

equal to the number of classes and all representations must be one-dimensional (section 3.12).

There is a general formulation for the j-th irrep of a cyclic group:

Cyclic group Cn: Cn1, Cn2, ..., Cnn = E

j-th irreducible representation: ( ) jmmnj εC =Γ with n πi

eε 2

=

For the benzene case we obtain the character table:

C6 C6 C62 C63 C64 C65 E=C66

Γ1 ε1 ε2 ε3 ε4 ε5 ε6

Γ2 ε2 ε4 ε6 ε8 ε10 ε12

Γ3 ε3 ε6 ε9 ε12 ε15 ε18

Γ4 ε4 ε8 ε12 ε16 ε20 ε24

Γ5 ε5 ε10 ε15 ε20 ε25 ε30

Γ6 ε6 ε12 ε18 ε24 ε30 ε36

It can be shown that the representations fulfil the orthogonality requirements in section 3.12.

The character table can be rearranged as follows:

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 5 -

C6 E C6 C3 C2 C32 C65

Α Γ6 1 1 1 1 1 1

Β Γ3 1 −1 1 −1 1 −1

Γ1 1 ε −ε∗ −1 −ε ε∗

Ε1 Γ5 1 ε∗ −ε −1 −ε∗ ε

Γ2 1 −ε∗ −ε −1 −ε∗ −ε

Ε2 Γ4 1 −ε −ε∗ −1 −ε −ε∗

There are 6 1-dimensional irreps. Γ1 and Γ5 (Γ2 and Γ4) can be combined to form two

dimensional reducible representations with non-complex characters. If some additional

symmetry is introduced such as e.g. mirror planes, which makes C6 / C65 and C3 / C32

conjugate, Γ1 / Γ5 and Γ2 / Γ4 merge to two 2-dimensional irreps E1 and E2.

An analysis of the benzene AO’s in terms of the group C6 yields:

Γπ = Γ6 + Γ3 + Γ1 + Γ5 + Γ2 + Γ4

Application of the projection operators of C6 to 1ϕ yields automatically the orthogonal set of

SALCs. After normalization:

6Γχ = 1/√6( +1 1ϕ +1 2ϕ +1 3ϕ +1 4ϕ +1 5ϕ +1 6ϕ )

3Γχ =1/√6( +1 1ϕ -1 2ϕ +1 3ϕ -1 4ϕ +1 5ϕ -1 6ϕ )

1Γχ =1/√6( +1 1ϕ +ε 2ϕ −ε ∗

3ϕ -1 4ϕ −ε 5ϕ +ε ∗

6ϕ )

5Γχ =1/√6( +1 1ϕ +ε ∗

2ϕ −ε 3ϕ -1 4ϕ −ε ∗

5ϕ +ε 6ϕ )

2Γχ =1/√6( +1 1ϕ −ε ∗

2ϕ −ε 3ϕ -1 4ϕ −ε ∗

5ϕ −ε 6ϕ )

4Γχ =1/√6( +1 1ϕ −ε 2ϕ −ε ∗

3ϕ -1 4ϕ −ε 5ϕ −ε ∗

6ϕ )

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 6 -

If real coefficients are preferred, one may choose suitable linear combinations (although this

is not necessary in principal):

6ΓΑ = χχ ; 3ΓΒ = χχ ; 51 ΓΓΕ1,1 += χχχ ; ( )5Γ1ΓΕ1,2 −= χχiχ ; 4Γ2ΓΕ2,1 += χχχ ; ( )42 ΓΓΕ2,2 −= χχiχ

Αχ = 1/√6( +1 1ϕ +1 2ϕ +1 3ϕ +1 4ϕ +1 5ϕ +1 6ϕ )

Βχ = 1/√6( +1 1ϕ -1 2ϕ +1 3ϕ -1 4ϕ +1 5ϕ -1 6ϕ )

Ε1,1χ = 1/√12( +2 1ϕ +1 2ϕ −1 3ϕ -2 4ϕ −1 5ϕ +1 6ϕ )

Ε1,2χ = 1/2( +1 2ϕ +1 3ϕ -1 4ϕ -1 5ϕ )

Ε2,1χ = 1/√12( +2 1ϕ −1 2ϕ -1 3ϕ +2 4ϕ -1 5ϕ -1 6ϕ )

Ε2,2χ = 1/2( +1 2ϕ −1 3ϕ +1 4ϕ −1 5ϕ )

(5.4 Energy levels of benzene in HMO model)

(5.5 Energy levels of cyclic unsaturated hydrocarbons in HMO model, geometric construction)

(5.6 Example: naphthalene, C60)

5.5 Symmetry of many electron states

How can we derive the symmetry of many electron states from the symmetry of the single

electron wave functions discussed so far? In the simplest approximation we assume that the

Hamiltonian is the sum of single electron Hamiltonians:

∑=Η i

iĥˆ

yielding wave functions which are the product of the single electron wave functions