Nankai University Hückel http://struchem.nankai.edu.cn

12

34

5

6

6

54

3

211

2

34

5

6

6

54

3

211

2

34

5

6

6

54

3

21

+

++

++

+

+

++ +

II

Hückel Molecular Orbital Theory

Nankai University Hückel http://struchem.nankai.edu.cn

Hückel Molecular Orbital Method—HMO1931 Hückel

• HMOHMO

• HMO ()

• HMO

Erich Hückel(1896-1980)

Nankai University Hückel http://struchem.nankai.edu.cn

4.1 4.1.1

pp-p

nm (m n )

— m<2n

4.1.2 m=nm > n

• (N, O, S, Cl 2 p•

m < n

4.1.3

1010

66

44

C CClH

H H34

COO

O 2-

CO3=

46

CC

C

H

H

HH

H+

32

Nankai University Hückel http://struchem.nankai.edu.cn

4.2

4.2.1 HMO1.

(MO)

LCAO-MO

Pauli Hund

Nankai University Hückel http://struchem.nankai.edu.cn

01iji jS i j

Hii=

0 ,,ij

i jH

i j

2. Hückel

Nankai University Hückel http://struchem.nankai.edu.cn

4.2.2 HMO1.

4 2pz 1 2 3 4

1 1 2 2 3 3 4 4c c c c

1 11 11 2 12 12 3 13 13 4 14 14

1 21 21 2 22 22 3 23 23 4 24 24

1 31 31 2 32 32 3 33 33 4 34 34

1 41 41 2 42 42 3 43 43 4 44 44

0000

c H ES c H ES c H ES c H ESc H ES c H ES c H ES c H ESc H ES c H ES c H ES c H ESc H ES c H ES c H ES c H ES

1 2

1 2 3

2 3 4

3 4

0000

c E cc c E c

c c E cc c E

Nankai University Hückel http://struchem.nankai.edu.cn

Ex

1 0 01 1 0 00 1 10 0 1

xxxx

4 2 2 2

1 0 0 1 0 1 1 01 1 0 1 1 0 1 3 1 ( 1)( 1) 00 1 1 0 1 0 10 0 1

x xx x x x x x x x x xx x xx

x4 3x2 + 1 = (x2 + x 1)(x2 x 1) = 0x = 1.618, 0.618

E1= + 1.618 (x1 = 1.618)E2= + 0.618 (x2 = 0.618)E3= 0.618 (x3 = 0.618)E4= 1.618 (x4 = 1.618)

1 2

1 2 3

2 3 4

3 4

0000

c x cc c x c

c c x cc c x

Nankai University Hückel http://struchem.nankai.edu.cn

E1= + 1.618 (x1 = 1.618)E2= + 0.618 (x2 = 0.618)E3= 0.618 (x3 = 0.618)E4= 1.618 (x4 = 1.618)

Ex E= x

2pz

E1 =

E2 =

E3 =

E4 =

Nankai University Hückel http://struchem.nankai.edu.cn

1 = 0.372 1 + 0.602 2 + 0.602 3 + 0.372 4

2= 0.602 1 + 0.372 2 0.372 3 0.602 4

3= 0.602 1 0.372 2 0.372 3 + 0.602 4

4 = 0.372 1 0.602 2 + 0.602 3 0.372 4

E1= + 1.618E2= + 0.618E3= 0.618E4= 1.618

x1= 1.618c1=c4; c2=c3=1.618c1

1=c1 1+1.618c1 2+1.618c1 3+c1 4

1

c1=0.372c2 c3 c4 c1

c2=0.602 c3=0.602 c4=0.372

1 = 0.372 1 + 0.602 2 + 0.602 3 + 0.372 4

161816181 24321

21 d..c

1 2

1 2 3

2 3 4

3 4

000

0

c x cc c x cc c x cc c x

Nankai University Hückel http://struchem.nankai.edu.cn

2pz

E1 =

E2 =

E3 =

E4 =

12

22

EL = 4( + )

ED = 2E1 + 2E2 = 4 + 4.472

E = 4 (4 + 4.472 ) = 4.472

2pz

E1 =

E2 =

(Delocalization Energy)

ED = EL ED = 0.472

Nankai University Hückel http://struchem.nankai.edu.cn

-

+ + + +- - - -

+ ++ +

----

+ +- -+ + --

++

++-

-

-

— —E4= 1.618

E3= 0.618

E2= + 0.618

E1= + 1.618

1 = +0.372 1 + 0.602 2 + 0.602 3 + 0.372 4

2= + 0.602 1 + 0.372 2 0.372 3 0.602 4

3= + 0.602 1 0.372 2 0.372 3 + 0.602 4

4 = + 0.372 1 0.602 2 + 0.602 3 0.372 4

Nankai University Hückel http://struchem.nankai.edu.cn

c1

c2c3

c4

c4

c3c2

c1

C2

A A

A A

S

SS

Sc1 c2 c3 c4

Nankai University Hückel http://struchem.nankai.edu.cn

1 2

1 2

0( 1) 0

c x cc c x

S: c1 = c4; c2 = c3

x = 1.618x = 0.618

A: c1 = c4; c2 = c3

x = 0.618

x = 1.618

1 01 1xx

1 01 1xx

c1 c2 c3 c4

1 2

1 2 3

2 3 4

3 4

000

0

c x cc c x cc c x cc c x

1 2

1 2

0( 1) 0

c x cc c x

Nankai University Hückel http://struchem.nankai.edu.cn

4.2.3 C

n nn Aii x, x=( E)/

i, j Aij Aji 1 0

12

34

5

60

100011100001100001100001110001

xx

xx

xx

43

2

1 0

001001001111

xxxx

Nankai University Hückel http://struchem.nankai.edu.cn

4.2.4 HMOn C n

n

1 2

1 0 0 0 1 0 0 1 1 0 01 1 0 0 1 1 0 0 1 00 1 0 0 0 1 0 0 1 0

0 0 0 1 0 0 0 0 0 00 0 0 1

n n n

x xx x xxD x xD Dx x

x x xx

D1= xD2= x2 1D3=xD2 D1=x3 2xD4=xD3 D2=x4 3x2 1……

2cos1k

kxn

k=1, 2, …n

2 sin1 1ki

ikcn n

2 cos1kkEn1.

k = 1, 2, ...n n

k=1, 2, …n

Nankai University Hückel http://struchem.nankai.edu.cn

2.

C E=C

+2 ~ 2

-2

-1

0

1

2

n=2 nn n=6 n=4 n=7n=5n=3

(

E) /

Nankai University Hückel http://struchem.nankai.edu.cn

3. (FEMO)

l = (2k+1)d d 1 C C CC

d d

5 d

2( ) sin5 5n

n xx

n =1x(C1) = 1d

x(C2) = 2d

x(C3) = 3d

x(C4) = 4d

1 1(C ) 2 5 sin 5 0.372

1 3(C ) 2 5 sin 3 5 0.602

1 4(C ) 2 5 sin 4 5 0.372

1 2(C ) 2 5 sin 2 5 0.602

C HMO

Nankai University Hückel http://struchem.nankai.edu.cn

4.2.5 HMO

c1 = c4c2 = c3 = c5 = c6

6

54

3

21

SxSy

x = 2 x = 1

1 2 6

1 2 3

2 3 4

3 4 5

4 5 6

1 5 6

000000

c x c cc c x c

c c x cc c x c

c c x cc c c x

1 2

1 2

2 0( 1) 0

c x cc c x

22 2 01 1x x xx

SxAy c1= c4=0c2= c3= c5= c6

x + 1 = 0 x = 1

x

y Sx c1 = c4c2 = c3c5 = c6

c1 = c4c2 = c3c5 = c6

Ax

Sy c2 = c6c3 = c5

c2 = c6c3 = c5c1 = c4 =0

Ay

Nankai University Hückel http://struchem.nankai.edu.cn

AxSy c1= c4c2= c3= c5= c6

0112xx x = 1

x = 2

AxAy c1= c4=0c2= c3= c5= c6 x 1 =0 x = 1

E1 = + 2

E2 = +

E3 = +

E4 =

E5 =

E6= 2

SxSy

AxSy

SxAy

AxAy

SxSy

AxSy

1 1 2 3 4 5 61 6

2 1 2 3 4 5 61 12 2 2

3 2 3 5 612

4 2 3 5 612

5 1 2 3 4 5 61 12 2 2

1 1 2 3 4 5 61 6

Nankai University Hückel http://struchem.nankai.edu.cn

12

34

5

6+

6

54

3

21

+

12

34

5

6+

6

54

3

21

+

+

12

34

5

6++

6

54

3

21

++ +

Nankai University Hückel http://struchem.nankai.edu.cn

4.2.6 HMO

k = 0, 1, 2, ..., n 1

1 0 0 11 1 0 00 1 0 0 0

0 0 0 11 0 0 1

xxx

xx

22 coskkEn

• |2 |E=

•• (

)•

+ 2

+

2

Nankai University Hückel http://struchem.nankai.edu.cn

Nankai University Hückel http://struchem.nankai.edu.cn

2

2 2

2

22

0

2

2

2

1

1

0

0.618

1.618

n , , n=4N+2, ,n=4N, ;n , , n=4N+1, ,

; n = 4N+3, , ;, ,

, 4N+2, Hückel 4n+2

32

66

56

76

+

Nankai University Hückel http://struchem.nankai.edu.cn

4.3

4.3.1

1=0.372 1+0.602 2+0.602 3+0.372 4

1 10.3722 1 0.3722

HMOi = c i 1 1 + c i 2 2+ +c i n n

cik2 k

(Population Analysis) HMO

Nankai University Hückel http://struchem.nankai.edu.cn

• k

mi i

12

22

1=0.372 1+0.602 2+0.602 3+0.372 4

2=0.602 1+0.372 2-0.372 3-0.602 4

1 ( )q1= 2 0.3722+2 0.6022 =1.00

q2= q3 = q4 =1.00=

2

1

occ

k i iki

q m c

Nankai University Hückel http://struchem.nankai.edu.cn

4.3.2

12

22

1=0.372 1+0.602 2+0.602 3+0.372 4

2=0.602 1+0.372 2-0.372 3-0.602 4

1 1 20.372*0.602 = 0.224

12

22

P12 =2(c11c12+c21c22) =2(0.372 0.602+0.602 0.372)=0.896

P23 =2(c12c13+c22c23)=0.448P34 =2(c13c14+c23c24)=0.896

Nankai University Hückel http://struchem.nankai.edu.cn

occ

kkjkikij ccnP

nk k

cki k i

1

0

Nankai University Hückel http://struchem.nankai.edu.cn

12

21

31

12

22

-

+ + + +- - - -

+ ++ +

----

+ +- -+ + --

++

++-

-

-

Nankai University Hückel http://struchem.nankai.edu.cn

4.3.3 1. i

( , )

jiji PN )( j i

1N1=3+0.896=3.896

C C

C C

H

H

H

H H

H0.896

0.896

0.448

CH2

C

CH2H2C

0.577

0.577 0.577

C (Nmax) CNmax = 0.577 3 +3 = 4.732

2. Fi=Nmax-Ni

F1=4.732-3.896=0.836F2=4.732-(3+0.896+0.448)=0.388F3=F2=0.388 F4=F1=0.836

Nankai University Hückel http://struchem.nankai.edu.cn

4.3.4

( )

1.000

0.667

0.399

HMO

C C C C1.000 1.000

0.896 0.448

0.388 0.836

Nankai University Hückel http://struchem.nankai.edu.cn

2.0.622

1.047

1.093

1.073

0.758

0.449

0.778

0.520

0.974

0.076

0.505

0.434

()

12

21

31 C C C C

0.7240.447 0.447

C C C C0.4470.894 0.8941-2 3-4 2-3

Nankai University Hückel http://struchem.nankai.edu.cn

87

6

54

3

2

1

1.027

0.8550.986

0.870

1.173

1.047

0.150

0.4820.4290.480

0.420 0.454

( NO2+)

1,3 ;

4,8 ;

9,10 ,

( CN-)

C C C C1.00 1.00

0.894 0.447

0.838 0.391

1,4

Nankai University Hückel http://struchem.nankai.edu.cn

4.4 *

4.4.1 1951

(Frontier Orbital)

HOMO LUMOHOMO LUMO

SMO HOMO LUMOHOMO LUMO

HOMO LUMO6 eV

Nankai University Hückel http://struchem.nankai.edu.cn

HOMO LUMO HOMOLUMO HOMO

LUMO HOMO LUMO

()

1.

BA

DC

C

D

B

A

D

C

B

A

h

CDA

B

A

B

D

C

A

B

C

D

h

Nankai University Hückel http://struchem.nankai.edu.cn Nankai University Hückel http://struchem.nankai.edu.cn

43

HOMO

LUMO

SOMO

SOMO ( )( )

HOMO

LUMO

LUMO

HOMO

hv

LUMO

HOMO HOMO

LUMO

Nankai University Hückel http://struchem.nankai.edu.cn

H2+I2 2HI

HOMOLUMOa. ( 6eV )b.c.

d.

2. H2 + I2

H

I I

H

HOMO

LUMO

ue

g H2

LUMO

HOMO

I2

I2 g

HI

g

e

u H2

I2

HOMOLUMO

, , ,H2

eH2

I

I2 2I

I + H2 IH2

I + IH2 2HI

Nankai University Hückel http://struchem.nankai.edu.cn

4.4.2 (Theory of Conservation of Orbital Symmetry)

Nankai University Hückel http://struchem.nankai.edu.cn Nankai University Hückel http://struchem.nankai.edu.cn

1

2

3

4

*

*A

S

A

S

A

A

S

S1

2

3

4

*

*S

A

S

A

A

S

A

S

C2