(2) +

cyclotron.tamu.edu

(2) +

Electrostatic forces in the H 2+

molecule

e − A + B + e −

A + B +

antibinding

region

binding

region


Potential energy diagrams for H and H 2

+

10

0

(a)

V = -e 2 /r

10

0

(b) V = -e 2 /r A

- e 2 /r B

+ e 2 /R eq

Energy (eV)

-10

-20

-30

-40

Energy (eV)

-10

-20

-30

-40

2.77 eV

-50

-50

-60

-4 -3 -2 -1 0 1 2 3 4

X Coordinate (Angstroms)

H atom

-60

-4 -3 -2 -1 0 1 2 3 4

X coordinate (Angstroms)

H

+

2

molecule


Quantum description of H 2

+


The Born-Oppenheimer Approximation

The electrons move so much faster than the nuclei that the

electronic energy depends only on the relative separation

of the nuclei and is unaffected by their motion. Therefore,

the electronic energy at a particular internuclear distance

will be the same as that for stationary nuclei.

e −

r Ae rBe


V

A + R B +

= −e

2




1

r

Ae

+

1

r

Be


1

R



The Schrödinger equation for H 2

+




T

nuc

+

H

el

+

2

e

R

⎞ r r

⎟Ψ(R,

r)


=

r r

EΨ(R,

r)

When the Born-Oppenheimer approximation is applied, H el

becomes

H

and the electronic motion can be separated from the nuclear motion;



H


el

Once ψ el

and E(R) have been determined as a function of R, the

Nuclear motion may be treated;

[ T + E(

R)

] ψ (R) = Eψ

(R)

nuc

el

2

h

= − ∇

2m

+

2

e ⎤ r

ψ (R, r) =

el

R ⎥


2

e

nuc

r

(

)

r


e

2




1

r

Ae

r

E(R) ψ (R, r)

nuc

+

el

r

1

r

Be




E(R) as a function of R for a stable bond

E(R)

0

R eq

D e

R


Molecular orbital theory

Instead of solving the Schrödinger equation to obtain the wavefunctions,

one “constructs” wavefunctions using linear combinations of orbitals from

each atom of the molecule. This is called the LCAO method.

In the case of H 2+

, the molecular orbital wavefunctions are

σ g

= N g

[ ϕ 1s

(A) + ϕ 1s

(B) ]

σ u

= N u

[ ϕ 1s

(A) − ϕ 1s

(B) ]


Formation of a sigma bonding molecular orbital

ϕ 1s

A

B

ψ A = φ 1s (A)

ψ B = φ 1s (B)

-8 -6 -4 -2 0 2 4 6 8

z/a 0

Ψ g

A

ψ g (AB) = N g [φ 1s (A) + φ 1s (B)]

B

-8 -6 -4 -2 0 2 4 6 8

z/a 0


Formation of an antibonding sigma molecular orbital

ψ B = - φ 1s (B)

ϕ 1s

A

ψ A = φ 1s (A)

B

-8 -6 -4 -2 0 2 4 6 8

z/a 0

ψ u (AB) = N u [φ 1s (A)−φ 1s (B)]

Ψ u

A

B

-8 -6 -4 -2 0 2 4 6 8

z/a 0


Polar plot representation of the H 2+

molecular orbitals

• • +


+ •

1s A

1s B

1s σ g

• − •

+ −

• •

1s A

1s B

1s σ u

*


The variation principle

When an approximate wavefunction is used to calculate the energy E a

,

E

a

=

*

∫ ψ Hψ dτ

a a

*

∫ ψ ψ dτ

a a

,

E a

is always greater than E exact

.


Exercise: Determine the ground state energy of a particle in

a 1-D box using the approximate wavefunction

Ψ a = N(xL − x 2 ).

0.7

0.6

ψ a

0.5

ψ exact

0.4

ψ

0.3

0.2

0.1

0.0

0 1 2 3 4 5

x


Application of the variation method to the H 2+

molecule

(α A − E)c 1 + (β − ES)c 2 = 0

(β − ES)c 1 + (α B − E)c 2 = 0

secular equations

where

α

A

*

= ∫ϕ


dτ,

A A

β

*

= ∫ϕ



A B

and

S

=


*

ϕ

A

ϕ dτ

B

α − E β − ES

β − ES α − E

= 0 secular determinant

E

u

=

α − β

1 − S

E

g

=

α

1

+

+

β

S


Integrals for calculating the energies of the H 2+

σ g

and σ u*

orbitals

*

S = ∫ϕ

ϕ dτ

A B

2

α = E − J + e /R,

1s

where J

β = S(E

2

+ e /R) − K where

1s

* 2

= ∫ϕ

(e /r ) ϕ dτ

A B A

* 2

K = ∫ϕ

(e /r ) ϕ dτ

A A B

Evaluate these integrals using spheroidal coordinates:

2

-R⎛

R ⎞

S = e ⎜1

+ R + ⎟

⎝ 2 ⎠

1 1 27.21

α 27.21

⎛ ⎞

−2R

= ⎜-

+ ⎟ + 1 − e (1 + R)

⎝ 2 R ⎠ R

⎛ 1 1 ⎞ R

β 27.21S - ⎟ −


= ⎜ + 27.21e (1 + R)

⎝ 2 R ⎠

[ ]


E(R) curves for the H 2+

σ g

and σ u*

orbitals

-6

E(R) (eV)

-9

-12

σ u

g

1 S

σ g

E H

E

E

u

α − β

=

1 − S

=

α

+

+

β

-15

-18

1 2 3 4 5 6

R/a 0


Exercise: Set up the secular determinant for the wavefunction

ψ = c 1 φ A + c 2 φ B + c 3 φ C

A

B

C

(α A

− E) (β(

AB − ES AB ) (β(

AC − ES

(β BA

ES AC )

BA

− ES BA ) (α(

B

− E) (β(

BC − ES BC ) = 0

(β CA

A B C

CA

− ES CA ) (β(

CB − ES CB ) (α(

C

− E)


Pictorial illustration showing the formation of H 2+

molecular orbitals

+ −

σ u

= N u

(φ 1sA

−φ 1sB

)

+

+

φ 1sA

φ 1sB

+

σ g

= N g

(φ 1sA

+ φ 1sB

)


Molecular orbitals formed from 2s and 2p atomic orbitals

+ +

+

+ +


Relative energies of molecular orbitals for

2 nd row homonuclear diatomic molecules


Molecular orbital energy level diagrams for second row

homonuclear diatomic molecules

Li 2

through N 2

O 2

through Ne 2


Exercise: Draw a molecular orbital energy level diagram for

the O 2− molecule, write its molecular orbital configuration,

and determine whether this molecule is paramagnetic. Which

molecule would you expect to have a larger bond energy, O 2

or O 2− ?

O O 2


O −

O 2−

: 2sσ g2

2sσ u2

2pσ g2

2pπ u

4

2pπ g

3


Paramagnetism

Liquid O 2 Liquid N 2


N

N 2

N

N 2

: 2sσ g2

2sσ u2

2pπ u4

2pσ g

2


Molecular orbital configurations and properties of the second row

homonuclear diatomic molecules


Metal-Metal Bonding

x

y

z

σ ∗

π ∗ M M

δ ∗

d xy

d xy

d z 2

d z 2

d xz,yz

δ

d xz,yz

π

σ


Term symbols for diatomic molecules

2S+

1

Λ

g /u

Λ =

M

M

L

S

M

= ∑ m

= ∑m

i

i

L

li

si

orbital m l

value

σ 0

π ±1

δ ±2

φ ±3

value of Λ symbol

0 Σ

1 Π

2 Δ

3 Φ


Exercises: Determine the term symbols for the following

configurations of H 2 :

a). 1sσ g2 (gs)

b). 1sσ g 1sσ u

c). 1sσ g 2sσ g


Potential energy curves for H 2

1sσ g 2sσ g

1sσ g 1sσ u

1sσ 2

2

g


Exercises: Determine the term symbols for the ground state

electron configuration of O 2


Potential energy curves for O 2

π g

π u

(2pπ g

1

2pπ g1

)


Molecular orbital energy level diagram for HF

HF

hydrogen

σ *

fluorine

nb

2p z

2p y

2p x

σ = c 1

ϕ 1s

(H) + c 2

ϕ 2pz

(F)

σ * = c 3

ϕ 1s

(H) − c 4

ϕ 2pz

(F)

1s

σ

2s


Molecular orbital energy level diagram for CO


The valence bond method

VB method – construct wavefunctions for electron pair bonds

using atomic orbitals of the two bonded atoms.

The resulting molecular orbitals are localized.

MO method – construct molecular wavefunctions using atomic

orbitals of all atoms of the molecule. The resulting

molecular orbitals are delocalized.

Example: Wavefunctions for the ground state of H 2

.

ψ VB

= N VB

[ φ A

(1) φ B

(2) + φ A

(2) φ B

(1)] 2 −½ [ α(1)β(2) − α(2)β(1)]

ψ MO

= σ g

(1) σ g

(2) 2 −½ [ α(1)β(2) − α(2)β(1)]

where

σ g

(1) σ g

(2) = N MO

[φ A

(1) φ A

(2) + φ B

(1) φ B

(2) + φ A

(1) φ B

(2) + φ A

(2) φ B

(1)]

ionic ionic covalent covalent


Molecular orbitals for linear methylene

H C H CH 2


Molecular orbital energy level diagrams for methylene

Walsh diagram for methylene

C CH 2

H H

C CH 2

H H


Formation of hybridized atomic orbitals in carbon

s + p z

p x

p y

sp hybridization

sp

sp

s + p x

+ p y

p z

p x p y

p z

s

sp 2 sp 2 sp 2

sp 2 hybridization

s + p x

+ p y

+ p y

sp 3 sp 3 sp 3 sp 3 sp 3 hybridization


sp hybrid orbitals

sp 2 hybrid orbitals


sp 3 hybrid orbitals

dsp 3 hybrid orbitals

d 2 sp 3 hybrid orbitals


Homework problem 5-7. Make a polar plot in the xy plane of the

wavefunction

Ψ sp

2 = −(1/3) 1/2 ϕ 2s

− (1/6) 1/2 ϕ 2px

− (1/2) 1/2 ϕ 2py

where the ϕ’s are H-atom wavefunctions. (Note: you may set r/a = 1

since only the angular dependence is of interest in this problem).

ϕ 2s

= N(2 − r/a)e −2r/a = 2Ne −2 − Ne −2 = Ne −2 = C

ϕ 2px

= N(r/a)e −2r/a sin θ cos ϕ = Ne −2 sin(π/2) cos ϕ = C cos ϕ

ϕ 2py

= N(r/a)e −2r/a sin θ sin ϕ = Ne −2 sin(π/2) sin ϕ = C sin ϕ

y

y = |ψ|sinφ

φ

x

x = |ψ|cosφ


Valence-bond structure of H 2 O

H

sp 3

H

O


Directional characteristics of hybridized orbitals

Hybrid type Coord. No. Geometry Bond angle(s)

sp 2 linear 180 o

sp 2 3 trigonal planar 120 o

sp 3 4 tetrahedral 109.5 o

dsp 2 4 square planar 90 o

dsp 3 5 trigonal bipyr. 120 o & 90 o

d 2 sp 3 6 octahedral 90 o


Molecular orbitals for linear CO 2

O

C O

CO 2


Molecular orbital energy level diagram for CO 2

σ p

*

2p

σ s

*

π x

*

π y

*

π x

(nb)

π y

(nb)

2p

2p

2s

π x

π y

σ p

σ s

Carbon atom

2s(nb)

2s(nb)

CO 2

molecule

2s 2s

Oxygen atoms


Walsh diagram for CO 2

and similar triatomic molecules.

σ p

*

σ s

*

π y

*

p x

, p x

π y

(nb)

π y

σ

*

p

σ

*

s

π x*

, π y

*

16 electrons

π x

(nb), π y

(nb)

π x

, π y

σ p

σ p

σ s

σ p

σ s

2s, 2s 2s, 2s

90 o 180 o

Exercise: Determine the geometry (linear or bent) and MO configuration

of each of the following molecules – NO 2

, O 3

, CS 2

, OF 2

, C 3

,


Valence-bond structure of CO 2

Lewis structure:

O C O

C has two σ bonds and two π bonds – use sp orbitals

each O has one π bond – use sp 2 orbitals

σ

σ

sp 2

sp 2

O

C

O

π

Exercise: Draw pictures representing the valence-bond structure of

NH 3

, CH 4

, C 2

H 4

, C 2

H 2

, SF 6

, IF 5

, ClF 3

.

sp

π


Bonding in the ethene molecule, C 2

H 4


Bonding in the ethyne molecule, C 2

H 2

sp

sp


Exercise: Draw pictures representing the valence-bond structure of

NH 3

, CH 4

, C 2

H 4

, C 2

H 2

, SF 6

, IF 5

, ClF 3

.


Delocalized π bonding – conjugated hydrocarbons

(− CH CH − CH CH −) n

pi- electron MO wavefunction: ψ = ∑

π

i


(i)

i

2p

secular determinant for π orbitals:

α 11

−E β 12

−S 12

E …….. β 1n

−S 1n

E

β 21

−S 21

E α 22

−E …….. β 2n

−S 2n

E

. .

. . = 0

. .

β n1

−S n1

E β n1

−S n1

E …….. α nn

−E


The Hückel approximation

a). Overlap integrals S ij

are set to zero unless i = j, in which case S ij

= 1.

b). All Coulomb integrals α ii

are assumed to be equal.

c). All exchange integrals except those on neighboring atoms are

set to zero.

α 11

−E β 12

−S 12

E …….. β 1n

−S 1n

E

β 21

−S 21

E α 22

−E …….. β 2n

−S 2n

E

. .

. .

. .

β n1

−S n1

E β n1

−S n1

E …….. α nn

−E

α−E β 0 …… 0 0

β α−E β …… 0 0

. .

. . = 0

. .

0 0 0 …… β α−E


Schematic representation of π molecular orbitals for butadiene.

π 4

π 3

energy

π 2

+

+ + +


− − −

π 1


Secular determinant for benzene


Energy level diagram for the π molecular orbitals of benzene.


Buckminsterfullerine C 60

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