• The chemical bond is at the heart of chemistry.
• A qualitative molecular orbital (MO) model is suggested.
• The MO model helps to get a good understanding on ; electronic structure, bond order, bond energy, bond length of diatomic molecule.
MS310 Quantum Physical Chemistry

2
+
Discussion of the bond : starting with the simplest molecule, H
2
+
Hamiltonian of H
2
+ molecule is given by
 2
 
2 m p
 2
(
 a
2   2 b
)

2 m e
 e
2  e 2
4
 
0
1
( r a

1
)
 r b e 2
4
 
0
1
R
MS310 Quantum Physical Chemistry

How can solve it? ‘ Born-Oppenheimer approximation ’
: Nucleus and electron motion separated.
Why this approximation is physically true?
→ proton (2000 times) heavier than electron, motion of proton : slower than electron.
(detail in ch 14, and the two motions can be decoupled. We can solve S.E. for a fixed nuclear separation)
Experimentally, H
2
+ ion is stable
→ solve the Schrödinger equation, exist at least 1 bound state
Zero of energy : distance of H atom and H + ion becomes infinity
→ negative energy for H
2
+ molecule
→ minimum energy at a distance R e
(eq. bond length)
MS310 Quantum Physical Chemistry

Interaction between 2 H atoms
MS310 Quantum Physical Chemistry

2
+
The relative energies of 2 H atoms
2 H atoms : more stable 2624 kJ/mol than 4 separated charges
H
2 molecule : more stable 436 kJ/mol than infinitely separated 2 H
→ bond energy is small part of total energy charge distribution of molecule : not so much different from a superposition of charge distribution of atom
Approximate molecular wave function
  c a

H 1 s a
 c b

H 1 s b
φ
H1s
: atomic orbital(AO) using the variational parameter ζ

H 1 s

1


( a
0
) 2
3 e
  r / a
0
MS310 Quantum Physical Chemistry

Probability : Not change with interchange of nuclei a and b
→ |c a
| = |c b
| or c a
=
± c b
Therefore, 2 molecular orbitals

 g u

 c g
(

H 1 s a c u
(

H 1 s a
 
H 1 s b
 
H 1 s b
)
)
ψ g
: symmetric ψ u
: antisymmetric
Overlap of 2 atomic orbital
MS310 Quantum Physical Chemistry

Value of c g and c u
 c
* g
(

*
H 1 s a
: normalization condition
 
*
H 1 s b
) c g
(

H 1 s a
 
H 1 s b
) d

 c g
2
(
 
*
H 1 s a

H 1 s a d
   
*
H 1 s b

H 1 s b d
 
2
 
*
H 1 s b

H 1 s a d

)
 c g
2
( 1

1

2
 
*
H 1 s b

H 1 s a d

)
Overlap integral S ab
S ab
  
*
H 1 s b

H 1 s a d

Calculated value of c g by the overlap integral
Similarly, c u
1
 c g
2
( 2

2 S ab
), c g
 given by c u

1
2

2 S ab
1
2

2 S ab
MS310 Quantum Physical Chemistry

g
u
Energy corresponding to ψ g is
E g





* g
 g d

* g
 g d



1
2 ( 1

S ab
)
H
1 aa


H
S ab ab
(
 
*
H 1 s a
H ij
   i
*
H
 j d


H 1 s a d
   
*
H 1 s b

H 1 s a d
   
*
H 1 s a
Similarly, energy corresponding to ψ u is
E u

 
 
* u
 u d

* u
 u d


H
1 aa


H
S ab ab
MS310 Quantum Physical Chemistry

H 1 s b d
   
*
H 1 s b

H 1 s b d

)

Why E g is lower than E u
?
Using the Born-Oppenheimer approximation
H aa
  
*
H 1 s a
(

 2
2 m
 2  e
2
4

0 r a
)

H 1 s a d
  e
2
4

0
R
 
*
H 1 s a

H 1 s a d
   
H
*
1 s a e
2
4

0 r b

H 1 s a d

φ
H1sa
: eigenfunction of

 2
2 m
 2  e
2
4
 
0 r a aa
= E
1 s
+ e
2
4

0
R
J , where J = *
H 1 s a e
2
4

0 r b

H 1 s a d

J : coulomb integral
H aa
: total energy of undisturbed H atom separated from a bare proton by the distance R (non bonded energy)
MS310 Quantum Physical Chemistry

Also, H ba
Similarly,
= H ab
.
H ba
  
*
H 1 s b
(

 2
2 m
 2  e 2
4

0 r a
)

H 1 s a d
  e 2
4

0
R
 
*
H 1 s b

H 1 s a d
   
*
H 1 s b e 2
4

0 r b

H 1 s a d

Evaluate it,
 
*
H 1 s b
(

 2
2 m
 2  e
2
4
 
0 r a
)

H 1 s a d
 
E
1 s
 
*
H 1 s b

H 1 s a d
 
E
1 s
S ab e 2
4
 
0
R
 
*
H 1 s b

H 1 s a d
  e 2
4
 
0
R
S ab
Therefore, H ba
= S ab
( E
1 s
+ e
2
4

0
R
) K , where K =
*
H 1 s b e
2
4

0 r b

H 1 s a d

K : exchange integral(resonance integral)
No simple physical interaction available but consequence of interference of 2 atomic orbitals
J, K > 0 → H aa
, H ab
< 0 at R=R e
MS310 Quantum Physical Chemistry

 E g
Difference ∆E g
 E g
H aa
 and ∆E u is given by
K  S ab
J
1  S ab
,  E u
Go to page 13 and comeback!
 E u
H aa

∆E g
ψ g
<0 and ∆E u
>0 by a quantitative calculation
: stable state and ψ u
: unstable state, |∆E u
| > |∆E g
|
K -
1 -
S ab
S ab
J
Schrödinger equation solution by effective nuclear charge ζ
(it means E=E(R, ζ))
→ ζ=1.24 for ψ g
, ζ=0.90 for ψ u
Minimum energy of ψ g
: at R e
= 2.00 a
0
, S ab
Also, E u
(R) > 0 for all R : ψ u
= 0.46
is not bound state
Bonding energy D e simplest model : 2.36 eV exact value : 2.70 eV
Finally, ψ g
: bonding orbital and ψ u
: antibonding orbital
MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

g
u
Shape of ψ g and ψ u
, and difference
Dashed line : unbound state( ζ=1, same as the H1s AOs)
MS310 Quantum Physical Chemistry

Contour map of bonding and antibonding orbital
MS310 Quantum Physical Chemistry

Probability of bonding and antibonding orbitals
Dashed line : unbound state( ζ=1, same as the H1s AOs)
We can see the increase of probability along the internuclear axis in case of bonding orbital, and decrease of probability along the same in case of antibonding orbital.
MS310 Quantum Physical Chemistry

Probability of bonding and antibonding
Light blue line : ∆ψ g
2 , ∆ψ u
2
 
2 g
 
2 g

1
2
(

2
H 1 s a
)

1
2
(

2
H 1 s b
)
  u
2   u
2 
1
2
(

2
H 1 s a
)

1
2
(

2
H 1 s b
)
MS310 Quantum Physical Chemistry

For both ψ g and ψ u
, the electronic change is delocalized over the whole molecule. However, the change is also localized between the nuclei ( for ψ g
), behind the nuclei ( for ψ u
).
→ Charge build up between the nuclei is the key of a chemical bonding.
Effect of charge redistribution to KE and PE
→ Virial theorem : <E potential
> = -2 <E kinetic
> for coulomb potential
(In fact, quantum mechanical virial theorem is given by

2 p m
2
 
1
2
 r   V ( r ) 
In coulomb potential, V(r) = 1/r and ∇ V(r) = - 1/r 2 .
Therefore, <E potential
> = -2 <E kinetic
> is obtained.)
Use E total
= E kinetic
+E potential
, It follows that
<E total
> = - <E kinetic
> = <E potential
>/2
MS310 Quantum Physical Chemistry

<∆E total
> = <∆E kinetic
> = <∆E potential
>/2
For stable molecule, <∆E total
> < 0
→ <∆E kinetic
> > 0 and <∆E potential
> < 0
→ How this effect affects to ψ g and ψ u as far as bond formation is concerned?
When we bring the proton and H atom to a distance R e and let them interact, in case of ζ=1(Atomic orbital of H), e delocalization occurs.
Then, kinetic energy ↓? Consider particle in a box
If box length ↑, kinetic energy ↓.
Therefore, if the electron is delocalized over the whole molecule, kinetic energy decrease. → bond formation?
MS310 Quantum Physical Chemistry

However, Optimal value of ζ=1.24
In this situation, some of charge redistribution around 2 nuclei
→ decrease of size of box
→ kinetic energy ↑
However, potential energy ↓
( ζ : 1 →1.24 because of the increase of coulomb interaction)
Result : <∆E potential
> lowered more than <∆E kinetic
> raised
→ <∆E total
> decrease further in second step.
Although <∆E kinetic
> and <∆E potential
> large, <∆E total
> is small for
ζ increase 1 to 1.24.
MS310 Quantum Physical Chemistry

For H
2
+, we will obtained two MOs, with different energies.
MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

Orbital energy diagram of H
2 and HF
MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

All MOs for homonuclear diatomics can be divided into two groups with each of two ‘symmetry operations’
1) rotation about the molecular axis after this operation, MO unchanged : σ symmetry
1 nodal plane containing the molecular axis : π symmetry
2) inversion through center of molecule : σ(x,y,z)→σ(-x,-y,-z)
If σ(x,y,z)=σ(-x,-y,-z), MO unchanged : g symmetry
If σ(x,y,z)=-σ(-x,-y,-z) : u symmetry
Example of this symmetric operations in H
2
1 σ g
,1 π u
: bonding orbital, 1 σ u
*,1 π g
+ molecule.
*: antibonding orbital
MS310 Quantum Physical Chemistry

Horizontal axis : molecular axis
Arrow : inversion operation
MS310 Quantum Physical Chemistry

 2 different notations.
- MOs are classified according to symmetry and increasing energy.
ex) 2 σ g orbital has same symmetry but higher energy than 1 σ g
Integer indicating the relative energy is omitted and the AOs from which the MOs are generated are listed instead.
ex) σ g
(2s) MO has higher energy than σ g
(1s) MO
* : denote antibonding
• With s orbitals, only σ MO exists.
• With 2p orbitals, 2 MOs exist.
1) axis of the 2p orbital lies on the intermolecular axis(by convention, z axis) : σ orbital generated.
It called as 3 σ g or σ g
(2p z
) orbital.
2) combining 2p x or 2p y orbitals, π orbital generated because of nodal plane containing the molecular axis. These 2 MOs are degenerated and called 1 π u or π u
(2p x
) and π u
(2p y
)
MS310 Quantum Physical Chemistry

 In principle, we should take linear combination of all the basis functions(basis set).
However, we can reduce the number of AOs for which c ij nonzero by the energy of AO.
is
 Mixing between 1s and 2s : neglect in this level.
 Mixing between 2s and 2p z
: both have σ symmetry
→ ‘ s-p mixing ’, but it decreases for increase of atomic number
(it means, Li
2
→F
2
) because energy difference between 2s and
2p z increases.
→ Separated MOs from the 2s and 2p orbitals, combine the MOs of s-p mixing
MS310 Quantum Physical Chemistry

Use HF calculation, for H
2 by 1 σ g
For O
2
<1 σ u
*<2 σ g and F
2
, order of 1 π u to N
<2 σ u
*<1 π u
<3 σ
2
, order of energy level is given g
<1 and 3 σ g
π g
*<3 σ u
* is changed.
MS310 Quantum Physical Chemistry

MOs of H
2
+ molecule
MS310 Quantum Physical Chemistry

In above slide, including only major AO in each case.
(no s-p mixing) : no optimization of orbital exponent( ζ=1)
See the shape of MOs in H
2
+
* u
1 σ g
: no nodal plane 2
All σ u
σ g
: 1 nodal plane 3 σ g
: 2 nodal planes
* orbitals have a nodal plane perpendicular to the internuclear axis
Amplitude for all the antibonding σ MOs : zero at middle
Case of F
2
, ζ values of F
2 are greater than H
2
+ (8.65 for 1s, 5.1 for 2p) and it makes rapidly decrease of probability
See the shape of MOs in F
2
.
→ difference between H
2
+ and F
2
1 σ g
: to small to overlap(so localized) → very small contribution to bonding
3 σ u
1 π u
*: more nodal plane than case of H
2
+

MOs of F
2 molecule(1 σ g
,3 σ u
*, 1 π u
)
MS310 Quantum Physical Chemistry

Many-electron molecules : configuration is useful
First, see H
2 and He
2
.
In this case, only 1s orbitals used for making MOs.
We must consider
1. Energy of molecule is not a sum of energy of MOs.
2. Bonding and antibonding information is given by relative sign of AO coeffiencits, but it does not convey whether electron is
‘bound’ to the molecule.
ex) case of O
3
is stable compared to separated O
2 and electron
Even though the additional electron is placed in an antibonding
MO
MS310 Quantum Physical Chemistry

H
2
: both electrons in 1 σ g
, lower than 1s
AO
→ Total energy is lowered by putting electrons in the 1 σ g
MO
He
2
: 2 electrons in 1 and 2 electrons in 1 σ
σ u g
, lower than 1s AO
*, higher than 1s AO
→ Total energy is increased by puttomg electrons in the MOs and He
2 is not stable.
(In fact, He
2 is stable ~5K, by VDW interaction-not chemical bond)
MS310 Quantum Physical Chemistry

After, F
2
F
2 and N
2
.
: neglect s-p mixing(2s AO below 21.6eV to 2p AO)
Configuration is given by
(1 σ g
) 2 (1 σ u
*) 2 (2 σ g
) 2 (2 σ u
*) 2 (3 σ g
) 2 (1 π u
) 2 (1 π u
) 2 (1 π g
*) 2 (1 π g
*) 2
2 σ : well described by 2s AO, 3σ : well described by 2p z
See 1 π u and 1 π g orbital is doubly degenerated.
AO,
N
2
: cannot neglect s-p mixing(2s AO below 12.4eV to 2p AO)
Configuration is given by
(1 σ g
) 2 (1 σ u
*) 2 (2 σ g
) 2 (2 σ u
*) 2 (1 π u
) 2 (1 π u
) 2 (3 σ g
) 2
Mixing changes shape of 2 σ and 3σ MO
2 σ g
2 σ u
: bonding character
: less antibonding character
3 σ g
: less bonding character
→ making triplet bond with the pair of 1π u
MOs
MS310 Quantum Physical Chemistry

Molecular orbital diagram of F
2
MS310 Quantum Physical Chemistry

Molecular orbital diagram of N
2
MS310 Quantum Physical Chemistry

MO formalism can extended to all first and second period.
Relative MO energy is given by
MS310 Quantum Physical Chemistry

This figure shows
1) Energy of MO decrease when atomic number increases
: by ζ increase when across the periodic table.(affects of large effective nuclear charge and smaller atomic size)
2) energy of 3 σ g decreases more rapidly than 1 π u
: by decreasing of s-p mixing.(2p x and 2p y
AO don’t mix with
2s AO, 1 π u orbital energy remains constant.)
→ inversion of order of MO energy occurs between N
2 and O
2
.
MS310 Quantum Physical Chemistry

By MO theory, we can predict magnetic moment of second period diatomic molecules.
bond order, bond energy, bond length, and vibrational force constant for series of H
2
→ Ne
2
 Bond energy : peak at N
2 and smaller peak at H
2
 Force constant : similar than bond energy trend(but it is more complicated trend for lighter molecules)
 Bond length : increased as bond energy and force constant decreases
These data can be qualitatively understood using MO theory
MS310 Quantum Physical Chemistry

MS310 Quantum Physical Chemistry

Approximately, energy of molecule is given by sum of each orbital energies.
→ put electron into bonding orbitals, molecule becomes stable and put electron into antibonding orbitals, molecule becomes unstable(easy to dissociation)
Define ‘ bond order ’
 Bond order = ½[total bonding electrons – total antibonding electrons]
Bond energy ↑ when bond order ↑.
Using bond order, easily understand electron configuration, why
He
2,
Be
2,
Ne
2 are unstable and bond of N
2 is so strong.
MS310 Quantum Physical Chemistry

Example Problem 12.4
Arrange the following in terms of increasing bond energy and bond length on the basis of their bond order:
N
2
 N
2
N
2
 N
2
2 
Bond order
N
2
  ( 1 σ g
) 2 ( 1 σ * u
) 2 ( 2 σ g
) 2 ( 2 σ * u
) 2 ( 1 π u
) 2 ( 1 π u
) 2 ( 3 σ g
) 1 0.5*(9-4)=2.5
N
2
 ( 1 σ g
) 2 ( 1 * u
) 2 ( 2 σ g
) 2 ( 2 σ * u
) 2 ( 1 π u
) 2 ( 1 π u
) 2 ( 3 σ g
) 2
N
2
  ( 1 g
) 2 ( 1 * u
) 2 ( 2 σ g
) 2 ( 2 σ * u
) 2 ( 1 u
) 2 ( 1 u
) 2 ( 3 σ g
) 2 ( 1 * g
) 1
0.5*(10-4)=3
0.5*(10-5)=2.5
N
2
2   ( 1 g
) 2 ( 1 * u
) 2 ( 2 σ g
) 2 ( 2 σ * u
) 2 ( 1 u
) 2 ( 1 π u
) 2 ( 3 σ g
) 2 ( 1 * g
) 1 ( π * g
) 1
0.5*(10-6)=2
Therfore, the bond energy is predicted to follow the order
N
2
> N
2

, N
2

> N
2
2 
Bond length decreases as the bond strength increases (opposite order)
N
2
< N
2

, N
2

< N
2
2 
MS310 Quantum Physical Chemistry

Now, we see heteronuclear diatomic molecule.
In this case, concept of bonding MO and antibonding MO preserves, but g and u symmetry breaks.( inversion operation )
Although breaks down of g and u symmetry, σ and π symmetry preserves and * uses for antibonding orbital.
MO notation changes by
Homonuclear 1 σ g
1 σ u
* 2 σ g
2 σ u
* 1 π u
3 σ g
1 π g
* 3 σ
Heteronuclear 1 σ 2σ 3 σ 4σ 1π 5σ 2π 6 σ u
*
MS310 Quantum Physical Chemistry

MO diagram of HF
MS310 Quantum Physical Chemistry

 2s electron of F : almost completely localized on F atom
 1 π electrons : completely localized on F atom
→ no overlap between 2p x
, 2p y
AO of F and 1s AO of H
 s-p mixing : 4 σ and 5σ* MO changes electron distribution in HF
→ 4σ MO has more antibonding character and 5σ* MO has more bonding character : bond order is 1(3 σ : largely localized on F,
4 σ : not totally bonding, 1π : completely localized on F)
 Charge on H : +0.51, F : -0.51
 Calculated dipole moment : 2.24 Debye is reasonable to experimental data, 1.91 Debye
However, in the antibonding 3 σ* orbital, this polarity is reversed by
‘bonding’ character of 3σ* orbital.(It makes delocalization of electrons) → dipole moment decrease when excited state of HF
MS310 Quantum Physical Chemistry

MO of 3 σ, 4σ, 1π of HF
MS310 Quantum Physical Chemistry

 Charge on atom in molecule : not observable
→ It means atomic charge cannot be assigned uniquely.
However, we know charge is not uniformly distributed.
 How can know this distribution?
→ introduce ‘molecular electrostatic potential’
 Molecular electrostatic potential : consider the contribution of the valence electrons and the atomic nuclei separately.
 Consider the nuclei first. For point charge q, the electrostatic potential is given by

( r )
 q
4

0 r
Therefore, contribution to the molecular electrostatic from the atomic nuclei is given by
 nuclei
( x
1
, y
1
, z
1
)
  i q
4
 i
0 r i
MS310 Quantum Physical Chemistry

Electron in the molecule : continuous charge distribution with a density at a point (x,y,z), related to n-electron wave function

( x , y , z )
  e

...

(

( x , y , z ; x
1
, y
1
, z
1
;...; x n
, y n
, z n
))
2 dx
1 dy
1 dz
1
...
dx n dy n dz n
Combining the contribution of nuclei and electrons, molecular electrostatic potential is given by

( x
1
, y
1
, z
1
)
  i q i
4

0 r i
 e


( x , y ,
4

0 r e z ) dxdydz
It can be calculated by HF and other methods and discussed in ch
16(computational chemistry), and we can see region of electron rich and poor.
MS310 Quantum Physical Chemistry

Electrostatic potential of HF molecule
MS310 Quantum Physical Chemistry

Solving the Schrödinger equation for the diatomic molecule : LCAO-MO model
How to solve it?
‘Secular determinant’
Study the molecular orbital diagram, electronic structure, and bond order, energy, and length
MS310 Quantum Physical Chemistry