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Ch 12. Chemical Bond in
Diatomic Molecules
• 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
12.1 The simplest 1-electron molecule : H2+
Discussion of the bond : starting with the simplest molecule, H2+
Hamiltonian of H2+ molecule is given by
2
2
2
2


e
1
1
e
1
2
2
2
ˆ
H 
( a   b ) 
e 
(  )
2m p
2m e
4 0 ra rb
4 0 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, H2+ 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 H2+ molecule
→ minimum energy at a distance Re(eq. bond length)
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Interaction between 2 H atoms
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12.2 The molecular wave function for
ground-state H2+
The relative energies of 2 H atoms
2 H atoms : more stable 2624 kJ/mol than 4 separated charges
H2 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
  ca H 1 s  cb H 1 s
a
b
φH1s : atomic orbital(AO)
using the variational parameter ζ
 H 1s 
1
(

 a0
3
2
) e  r / a 0
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Probability : Not change with interchange of nuclei a and b
→ |ca| = |cb| or ca = ± cb
Therefore, 2 molecular orbitals
 g  c g ( H 1 s   H 1 s )
a
b
 u  cu ( H 1 s   H 1 s )
a
b
ψg : symmetric ψu : antisymmetric
Overlap of 2 atomic orbital
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Value of cg and cu : normalization condition
*
*
*
c
(



 g H 1 sa H 1 sb )c g (  H 1 sa   H 1 sb )d
 c g2 (   *H 1 s  H 1 s d    *H 1 s  H 1 s d  2   *H 1 s  H 1 s d )
a
a
b
b
 c g2 ( 1  1  2   *H 1 sb  H 1 sa d )
Overlap integral Sab
Sab    *H 1 sb  H 1 sa d
Calculated value of cg by the overlap integral
1
1  c g2 ( 2  2 S ab ), c g 
2  2 S ab
1
c

Similarly, cu given by
u
2  2 S ab
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b
a
12.3 The energy corresponding to the molecular
wave functions ψg and ψu
Energy corresponding to ψg is
Eg
 Hˆ  d


   d
*
g
g
*
g
g

1
(   H* 1 sa Hˆ  H 1 sa d    H* 1 sb Hˆ  H 1 sa d    H* 1 sa Hˆ  H 1 sb d    H* 1 sb Hˆ  H 1 sb d )
2(1  S ab )

H aa  H ab
1  S ab
H ij   i* Hˆ  j d
Similarly, energy corresponding to ψu is
Eu 
* ˆ

u
 H ud
 
*
u
u
d

H aa  H ab
1  Sab
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Why Eg is lower than Eu?
Using the Born-Oppenheimer approximation
H aa   
*
H 1sa
2 2
e2
e2
e2
*
*
(  
)H 1sa d 
H 1sa H 1sa d   H 1sa
H 1sa d

2m
40 ra
40 R
40 rb
2 2
e2
 
φH1sa : eigenfunction of 
2m
4 0 ra
Therefore,
H aa = E1s +
e2
40 R
- J , where J = ∫
 H 1sa
*
e2
40 rb

H 1sa
d
J : coulomb integral
Haa : total energy of undisturbed H atom separated from a bare
proton by the distance R (non bonded energy)
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Also, Hba = Hab.
Similarly,
H ba   
*
H 1 sb
2 2
e2
e2
e2
*
*
(
 
) H 1 sa d 
 H 1 sb  H 1 sa d    H 1 sb
 H 1 sa d

2m
4 0 ra
4 0 R
4 0 rb
Evaluate it,

*
H 1 sb
e2
2 2
e2
(
 
) H 1 sa d  E 1 s   H* 1 sb  H 1 sa d  E 1 s S ab
2m
4 0 ra


4  R
0
*
H 1 sb
 H 1 s d 
a
e2
4 0 R
Therefore, H ba = S ab ( E1 s +
S ab
e2
40 R
) - K , where K = ∫
H* 1sb
e2
40 rb
H 1s d
K : exchange integral(resonance integral)
No simple physical interaction available but consequence
of interference of 2 atomic orbitals
J, K > 0 → Haa, Hab < 0 at R=Re
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a
Difference ∆Eg and ∆Eu is given by
E g  E g - H aa 
-K  S ab J
K - S ab J
, E u  E u - H aa 
1  S ab
1 - S ab
Go to page 13 and comeback!
∆Eg <0 and ∆Eu >0 by a quantitative calculation
ψg : stable state and ψu : unstable state, |∆Eu| > |∆Eg|
Schrödinger equation solution by effective nuclear charge ζ
(it means E=E(R,ζ))
→ ζ=1.24 for ψg , ζ=0.90 for ψu
Minimum energy of ψg : at Re = 2.00 a0, Sab = 0.46
Also, Eu(R) > 0 for all R : ψu is not bound state
Bonding energy De
simplest model : 2.36 eV
exact value : 2.70 eV
Finally, ψg : bonding orbital and ψu : antibonding orbital
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12.4 A closer look at the molecular wave functions
ψg and ψu
Shape of ψg and ψu, and difference
Dashed line : unbound state(ζ=1, same as the H1s AOs)
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Contour map of bonding and antibonding orbital
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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 : ∆ψg2, ∆ψu2
1
1
 g2   g2  ( H2 1 sa )  ( H2 1 sb )
2
2
1 2
1 2
2
2
 u   u  ( H 1 sa )  ( H 1 sb )
2
2
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For both ψg and ψu, the electronic change is delocalized over the
whole molecule. However, the change is also localized between
the nuclei ( for ψg ), behind the nuclei ( for ψu ).
→ Charge build up between the nuclei is the key of a chemical
bonding.
Effect of charge redistribution to KE and PE
→ Virial theorem : <Epotential> = -2 <Ekinetic> for coulomb potential
(In fact, quantum mechanical virial theorem is given by
p2
1

   r  V (r) 
2m
2
In coulomb potential, V(r) = 1/r and ∇V(r) = - 1/r2.
Therefore, <Epotential> = -2 <Ekinetic> is obtained.)
Use Etotal = Ekinetic +Epotential, It follows that
<Etotal> = - <Ekinetic> = <Epotential>/2
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<∆Etotal> = - <∆Ekinetic> = <∆Epotential>/2
For stable molecule, <∆Etotal> < 0
→ <∆Ekinetic> > 0 and <∆Epotential> < 0
→ How this effect affects to ψg and ψu as far as bond formation
is concerned?
When we bring the proton and H atom to a distance Re 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?
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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 : <∆Epotential> lowered more than <∆Ekinetic> raised
→ <∆Etotal> decrease further in second step.
Although <∆Ekinetic> and <∆Epotential> large, <∆Etotal> is small for
ζ increase 1 to 1.24.
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12.5 Combining atomic orbitals to form
molecular orbitals
For H2+, we will obtained two MOs,
with different energies.
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Orbital energy diagram of H2 and HF
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12.6 Molecular orbitals for homonuclear diatomic
molecules
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 H2+ molecule.
1σg ,1πu : bonding orbital, 1σu*,1πg*: antibonding orbital
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Horizontal axis : molecular axis
Arrow : inversion operation
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 2 different notations.
- MOs are classified according to symmetry and increasing energy.
ex) 2σg orbital has same symmetry but higher energy than 1σg
- Integer indicating the relative energy is omitted and the AOs from
which the MOs are generated are listed instead.
ex) σg(2s) MO has higher energy than σg(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 3σg or σg(2pz) orbital.
2) combining 2px or 2py orbitals, π orbital generated because of
nodal plane containing the molecular axis. These 2 MOs are
degenerated and called 1πu or πu(2px) and πu(2py)
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 In principle, we should take linear combination of all the
basis functions(basis set).
However, we can reduce the number of AOs for which cij is
nonzero by the energy of AO.
 Mixing between 1s and 2s : neglect in this level.
 Mixing between 2s and 2pz : both have σ symmetry
→ ‘s-p mixing’, but it decreases for increase of atomic number
(it means, Li2→F2) because energy difference between 2s and
2pz increases.
→ Separated MOs from the 2s and 2p orbitals, combine
the MOs of s-p mixing
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Use HF calculation, for H2 to N2, order of energy level is given
by 1σg<1σu*<2σg<2σu*<1πu<3σg<1πg*<3σu*
For O2 and F2, order of 1πuand 3σg is changed.
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MOs of H2+ molecule
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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 H2
+
σ u*
1σg: no nodal plane 2σg: 1 nodal plane 3σg: 2 nodal planes
All σu* orbitals have a nodal plane perpendicular to the
internuclear axis
Amplitude for all the antibonding σ MOs : zero at middle
Case of F2, ζ values of F2 are greater than H2+(8.65 for 1s, 5.1
for 2p) and it makes rapidly decrease of probability
See the shape of MOs in F2.
→ difference between H2+ and F2
1σg: to small to overlap(so localized) → very small contribution
to bonding
3σu*: more nodal plane than case of H2+
1πu: significant delocalization
→ large contribution
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MOs of F2 molecule(1σg,3σu*, 1πu)
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12.7 The electronic structure of many-electron
molecules
Many-electron molecules : configuration is useful
First, see H2 and He2.
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 O3- is stable compared to separated O2 and electron
Even though the additional electron is placed in an antibonding
MO
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H2 : both electrons in 1σg, lower than 1s
AO
→ Total energy is lowered by putting
electrons in the 1σg MO
He2 : 2 electrons in 1σg, lower than 1s AO
and 2 electrons in 1σu*, higher than 1s AO
→ Total energy is increased by puttomg
electrons in the MOs and He2 is not stable.
(In fact, He2 is stable ~5K, by VDW
interaction-not chemical bond)
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After, F2 and N2.
F2 : neglect s-p mixing(2s AO below 21.6eV to 2p AO)
Configuration is given by
(1σg)2(1σu*)2(2σg)2(2σu*)2(3σg)2(1πu)2(1πu)2(1πg*)2(1πg*)2
2σ : well described by 2s AO, 3σ : well described by 2pz AO,
See 1πu and 1πg orbital is doubly degenerated.
N2 : cannot neglect s-p mixing(2s AO below 12.4eV to 2p AO)
Configuration is given by
(1σg)2(1σu*)2(2σg)2(2σu*)2(1πu)2(1πu)2(3σg)2
Mixing changes shape of 2σ and 3σ MO
2σg : bonding character
2σu : less antibonding character
3σg : less bonding character
→ making triplet bond with the pair of 1πu MOs
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Molecular orbital diagram of F2
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Molecular orbital diagram of N2
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MO formalism can extended to all first and second period.
Relative MO energy is given by
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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 3σg decreases more rapidly than 1πu
: by decreasing of s-p mixing.(2px and 2py AO don’t mix with
2s AO, 1πu orbital energy remains constant.)
→ inversion of order of MO energy occurs between N2 and O2.
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12.8 Bond order, bond energy, and bond length
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 H2 → Ne2
 Bond energy : peak at N2 and smaller peak at H2
 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
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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
He2, Be2, Ne2 are unstable and bond of N2 is so strong.
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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 22 
Bond order
N 2  (1σ g )2 (1σ u* )2 (2σ g )2 (2σ u* )2 (1π u )2 (1π u )2 (3σ g )1
0.5*(9-4)=2.5
N 2  (1σ g )2 (1σ u* )2 (2σ g )2 (2σ u* )2 (1π u )2 (1π u )2 (3σ g )2
0.5*(10-4)=3
N 2  (1σ g )2 (1σ u* )2 (2σ g )2 (2σ u* )2 (1π u )2 (1π u )2 (3σ g )2 (1π *g )1
0.5*(10-5)=2.5
N 22   (1σ g )2 (1σ u* )2 (2σ g )2 (2σ u* )2 (1π u )2 (1π u )2 (3σ g )2 (1π *g )1(1π *g )1 0.5*(10-6)=2
Therfore, the bond energy is predicted to follow the order
N 2 > N 2 , N 2 > N 22 
Bond length decreases as the bond strength increases (opposite order)
N 2 <N 2 , N 2 < N 22 
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12.9 Heteronuclear diatomic molecules
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 1σg 1σu* 2σg 2σu* 1πu 3σg 1πg* 3σu*
Heteronuclear 1σ 2σ 3σ 4σ 1π 5σ 2π 6σ
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MO diagram of HF
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 2s electron of F : almost completely localized on F atom
 1π electrons : completely localized on F atom
→ no overlap between 2px, 2py AO of F and 1s AO of H
 s-p mixing : 4σ 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(3σ : largely localized on F,
4σ : 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 3σ* orbital, this polarity is reversed by
‘bonding’ character of 3σ* orbital.(It makes delocalization of
electrons) → dipole moment decrease when excited state of HF
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MO of 3σ, 4σ, 1π of HF
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12.10 The molecular electrostatic potential
 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
q
 (r ) 
4 0 r
Therefore, contribution to the molecular electrostatic from the
atomic nuclei is given by
nuclei ( x1 , y1 , z1 )  
i
qi
4 0 ri
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; x1 , y1 , z1;...; xn , yn , zn ))2 dx1dy1dz1 ...dxndyndzn
Combining the contribution of nuclei and electrons, molecular
electrostatic potential is given by
 ( x1 , y1 , z1 )  
i
qi
4 0 ri
 e 
 ( x, y, z )
dxdydz
4 0 re
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.
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Electrostatic potential of HF molecule
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Summary
- 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
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