Lecture 25: Introduction to Molecular Orbital Theory The material in this lecture covers the...

Lecture 25: Introduction to Molecular Orbital Theory The material in this lecture covers the following in Atkins.

14 Molecular structure Molecular Orbital Theory (a) Linear combinations of atomic orbitals (b) bonding orbitals (c) anti-bonding orbitals

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Molecular Orbital Theory H2+

We shall now discuss ways toapproximately solve :He (re,RN)ψ(re,RN) =Ee(RN )ψ(re,RN)

and represent the many - electronwave - function ψ(re,RN)

Hereˆ H e = ˆ T e + ˆ VNe + ˆ Vee + ˜ VNN

We shall this time use :


ξ1 =[A(1)B(2)+A(2)B(1)]×[α(1)β(2)−β(1)α(2)]BA

In valence bond theory we localized one electron to each atom

el 1. el 2.

or BA

In molecular orbital theory each electron moves over the whole molecule

BA

el 1.el 2.

BAand

el 2.el 1.

The two theories are limiting cases (see later) of a more exact theory

Both electrons can be on the same nuclei


We used the orbitals of theone - electron hydrogen to build upwavefunctions for many - electronatoms

We shall use the orbitals of

the one-electron H2+ molecule

to describe many electrondiatomic molecules


The hamiltonian for the H2+

molecule:is given by

H = −

h2

2me∇e

2

Kinetic energy of electron

−e2

4πεo

1

rA1

Attraction of el. 1 by A

−e2

4πεo

1

rB1

Attraction of el. 1 by B

+e2

4πεo

1

R

rA1

A B

1

rB1

R

Repulsion between A and B


We could now solve : ˆ H (r1,R)ϕ(r1,R) =E(R)ϕ(r1,R)With

V =−e2

4πεo[1

rA1+

1rB1

−1R

Where

H =−h2

2me∇e2 +V(r1,R)

To this end we write ϕ(r1,R) as a linear combination ( )of atomic orbitals LCAO

This is possible but tedious

We shall instead find approximate soluions

The atomic orbitals are in general those centered onthe atoms of our molecule

Was first done by a Danish astronomer in 1926

1sA (r1A ) =A(r1A ) =e−r1A /ao

(πao3 )3 / 2


For H2+ that is :

1sA (r1B ) =A(r1B ) =e−r1B /ao

(πao3 )3 / 2

Where r1A and r1B are related by:

r1B = r1A2+R2−2r1ARcosθ

rA1

A B

1

rB1

R

θ

We now assume the electron to be equallylikely to be on each nuclei and write the molecular orbitals as :

ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)

Note : we have as many linear combinationsas we have atomic orbitals


ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)

We can normalize our orbitals for easy energycalculation and use in probability density

ϕ+(1)ϕ+(1)∫ dv=N2 (A(1)+B(1))(A(1)+B(1))dv=1∫

ϕ+(1)ϕ+(1)∫ dv=N2 (A(1)A(1)dv∫ +N2 (B(1)B(1)dv∫+2N2 (A(1)B(1)dv∫ =1

A normalized B normalized overlap S > 1

ϕ+(1)ϕ+(1)∫ dv = +1×N2 +1×N2

+2S×N2 =1

N =1

2(1+S)


ϕ +(1) = A(1) +B(1) ϕ−(1) = A(1) −B(1)

After normalization

ϕ +(1) =1

2(1+ S)[A(1) +B(1)] ϕ−(1) =

1

2(1− S)[A(1) −B(1)]

The electron probability density:

ϕ +(1)ϕ+ (1) =1

2(1+ S)[A(1) +B(1)]2

=1

2(1+ S)A(1)2 +

1

2(1+ S)B(1)2 + 2

1

2(1+ S)AB(1)

Density at A(Reduced)

Density at B(Reduced)

Density between A and B(Increased)

(a)The amplitude of the bonding molecular orbital in a hydrogen molecule-ion in a plane containing the two nuclei and

(b) a contour representation of the amplitude.


ϕ +(1) =1

2(1+ S)[A(1) +B(1)]


A representation of the constructive interference that occurs when two H 1s orbitals overlap and form a bonding orbital. Compare this illustration with Fig.14.14.

ϕ +(1) =1

2(1+ S)[A(1) +B(1)]


The boundary surface of a orbital encloses the region where the electrons that occupy the orbital are most likely to be found. Note that the orbital has cylindrical symmetry.

ρ(1) =1

2(1+ S)A(1)2 +

1

2(1+ S)B(1)2 + 2

1

2(1+ S)AB(1)


The electron density calculated by forming the square of the wavefunction used to construct Fig.14.14. Note the accumulation of electron density in the internuclear region.

ρ(1) =1

2(1+ S)A(1)2 +

1

2(1+ S)B(1)2 + 2

1

2(1+ S)AB(1)


H =−

h2

2me∇e

2−e2

4πεo[

1rA1

+1

rB1−

1R

] ϕ+(1)=1

2(1+S)[A(1)+B(1)]

We can now evaluate the energy : E+ = ϕ+(1) ˆ H(1)∫ ϕ+ (1)dv

E+=1

2(1+S)[A(1)+B(1)]ˆ H (1)∫ [A(1)+B(1)]dv

=

12(1+S)

] A(1)∫ [−h2

2me∇e

2−e2

4πεo

1rA1

]A(1)dv

+

12(1+S)

] B(1)∫ [−h2

2me∇e

2−e2

4πεo

1rB1

]B(1)dv

+1

2(1+S)] A(1)∫ [

e2

4πεo

1rB1

]A(1)dv

+1

2(1+S)] B(1)∫ [

e2

4πεo

1rA1

]B(1)dv

+

22(1+S)

] A(1)∫ [−h2

2me∇e

2−e2

4πεo[

1rA1

+1

rB1]]B(1)dv+

e2

4πεo

1R

rA1

A B

1

rB1

R

θ


E+ =1

2(1+ S)[A(1) +B(1)] ˆ H(1)∫ [A(1) +B(1)]dv

=

1

2(1+S)] A(1)∫ [−

h2

2me∇e

2 −e2

4πεo

1rA1

]A(1)dv

+

1

2(1+S)] B(1)∫ [−

h2

2me∇e

2 −e2

4πεo

1rB1

]B(1)dv

+1

2(1+S)] A(1)∫ [−

e2

4πεo

1rB1

]A(1)dv

+1

2(1+S)] B(1)∫ [−

e2

4πεo

1rA1

]B(1)dv

+

2

2(1+S)] A(1)∫ [−

h2

2me∇e

2 −e2

4πεo[1

rA1+

1rB1

]]B(1)dv

rA1

A B

1

rB1

R

θ+e2

4πεo

1

R

E1sH

2(1+ S)

E1sH

2(1+ S)

J2(1+S)

J2(1+S)

K'(1+S)


- J

2(1+S)=

12(1+S)

] A(1)∫ A(1)[−e2

4πεo

1rB1

]dv

-J

2(1+S)=

12(1+S)

] B(1)B(1∫ [−e2

4πεo

1rA1

])dv

A B

rB1

R

1

2(1+ S)] A (1)A(1)dv

Interaction (attraction) between nucleus B and the charge cloud A(1)A(1)/2(1+S)

Interaction (attraction) between nucleus A and the charge cloud B(1)B(1)/2(1+S)


K'=

22(1+S)

] A(1)∫ [−h2

2me∇e

2−e2

4πεo[

1rA1

+1

rB1]]B(1)dv

rA1

A B

1

rB1

R

θ

=2

2(1+S)] A(1)∫ [−

h2

2me∇e

2−e2

4πεo

1rB1

]]B(1)dv

−2

2(1+S)] A(1)B(1)∫ [

e2

4πεo

1rA1

]dv

SE1sH/(1+S)

−K

Interaction (attraction) between nucleus A and the charge cloud A(1)B(1)/(1+S)


+e2

4πεo

1

RE+ =

E1sH

2(1+ )S+

E1sH

2(1+ S)−

J2(1+S)

−J

2(1+S)−

K -SE1sH(1+S)

E+=E1sH+SE1sH

(1+S)−

J +K(1+S)

+e2

4πεo

1R

We get in a similar way that

E−=E1sH−J -K(1-S)

+e2

4πεo

1R

E+=E1sH−J +K(1+S)

+e2

4πεo

1R


The calculated and experimental molecular potential energy curves for a hydrogen molecule-ion.

E+=E1sH+J +K(1+S)

+e2

4πεo

1R

E−=E1sH−J -K(1-S)

+e2

4πεo

1R


A representation of the destructive interference that occurs when two H1s orbitals overlap and form an antibonding * orbital. Compare this illustration with Fig.14.20.

ϕ−(1) =1

2(1−S)[A(1)−B(1)]


(a) The amplitude of the antibonding molecular orbital in a hydrogen molecule-ion in a plane containing the two nuclei and

(b) a contour representation of the amplitude. Note the internuclear node.

ϕ−(1) =1

2(1−S)[A(1)−B(1)]


The electron density calculated by forming the square of the wavefunction used to construct Fig.14.20. Note the elimination of electron density from the internuclear region.

ρ−(1) =1

2(1− S)A(1)2 +

12(1− S)

B(1)2 − 21

2(1− S)AB(1)


(b) In an antibonding orbital, the nuclei are attracted to an accumulation of electron density outside the internuclear region.

A partial explanation of the origin of bonding and antibonding effects.

(a) In a bonding orbital, the nuclei are attracted to the accumulation of electron density in the internuclear region.

ϕ +(1) =1

2(1+ S)[A(1) +B(1)] ϕ−(1) =

12(1−S)

[A(1)−B(1)]

A molecular orbital energy level diagram for orbitals constructed from the overlap of H1s orbitals; the separation of the levels corresponds to that found at the equilibrium bond length. The ground electronic configuration of H2 is obtained by accommodating the two electrons in the lowest available orbital (the bonding orbital).


rA1

A B

1

rB1

R

What you should learn from this lecture

Be able to construct the Hamiltonian

H =−h2

2me∇e

2−e2

4πεo

1rA1

−e2

4πεo

1rB1

+e2

4πεo

1R

Understand the justification for constructingthe two molecular orbitals as linear combinationsof atomic orbitals ϕ+(1)=N+(A(1)+B(1)) and ϕ−(1)=N−(A(1)−B(1))and be able to evaluate the normalizationconstants from the overlap integral S= A(1)B(1)dv∫

You are not asked to derive the energy expressions

for ϕ+: E+=E1sH−J +K(1+S)

+e2

4πεo

1R

or ϕ- :E−=E1sH−J -K(1-S)

+e2

4πεo

1R

What you should learn from this lecture

However you should understand the meening of

J = A(1)∫ A(1)[−e2

4πεo

1rB1

]dv and

K= A(1)B(1)∫ [e2

4πεo

1rA1

]dv and undertand how K

stabilize ϕ+ and destabalizes ϕ− .

rA1

A B

1

rB1

R

It is expected that you can derive the expressionfor the density due to ϕ+ as:

ρ(1)=1

2(1+S)A(1)2+

12(1+S)

B(1)2+21

2(1+S)A(1)B(1)

You shoule further understand the meening of

A(1)2, B(1)2, and A(1)B(1). Also you should understand

how +21

2(1+S)A(1)B(1) influence the energy E+ as well

as the features of Fig 14.16

What you should learn from this lecture It is expected that you can derive the expressionfor the density due to ϕ- as:

ρ(1)=1

2(1−S)A(1)2+

12(1−S)

B(1)2−21

2(1−S)A(1)B(1)

You shoule further understand the meening of

A(1)2, B(1)2, and A(1)B(1). Also you should understand

how −21

2(1−S)A(1)B(1) influences the energy E - as well

as the features of Fig 14.21

Understand the features of the plots in Fig. 14.17 and 14.19for ϕ+(1)=N+(A(1)+B(1)) and ϕ−(1)=N−(A(1)−B(1)),respectively.

Make note of the potential energy curve in Fig 14.18 and observe that E+ is lowered less in energy (compared to E1s than E- is increased (see also Fig 14.24

Lecture 25: Introduction to Molecular Orbital Theory The material in this lecture covers the...

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