Ch 12. Chemical Bond in Diatomic Molecules MS310 Quantum Physical Chemistry The chemical bond is at...

Ch 12. Chemical Bond in Ch 12. Chemical Bond in Diatomic MoleculesDiatomic Molecules

MS310 Quantum Physical Chemistry

• 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.

12.1 The simplest 1-electron molecule : H12.1 The simplest 1-electron molecule : H22++

Discussion of the bond : starting with the simplest molecule, H2+

Hamiltonian of H2+ molecule is given by

R

e

rr

e

mmH

bae

eba

p

1

4)11

(42

)(2

ˆ0

2

0

22

222

2



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)


Interaction between 2 H atoms

12.2 The molecular wave function for 12.2 The molecular wave function for

ground-state Hground-state H22++


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

φH1s : atomic orbital(AO)using the variational parameter ζ

ba sHbsHa cc 11

0/2

3

01 )(

1 arsH e

a


Probability : Not change with interchange of nuclei a and b

→ |ca| = |cb| or ca = ± cb

Therefore, 2 molecular orbitals

ψg : symmetric ψu : antisymmetric

)(

)(

11

11

ba

ba

sHsHuu

sHsHgg

c

c

Overlap of 2 atomic orbital


Value of cg and cu : normalization condition

)d211(c

)d2dd(c

d)(c)(c

as1Hbs1H

as1Hbs1Hbs1Hbs1Has1Has1H

bs1Has1Hbs1Has1H

*2g

***2g

g***

g

dSasHbsHab 11

*

Overlap integral Sab

Calculated value of cg by the overlap integral

ab

gabgS

cSc22

1),22(1 2

Similarly, cu given byab

uS

c22

1

12.3 The energy corresponding to the molecular 12.3 The energy corresponding to the molecular

wave functions wave functions ψψgg and and ψψuu


Energy corresponding to ψg is

ab

abaa

sHsHsHsHsHsHsHsHab

gg

gg

g

S

HH

dHdHdHdHS

d

dHE

bbbaabaa

1

)ˆˆˆˆ()1(2

1

ˆ

1*11

*11

*11

*1

*

*

dHH jiij ˆ*

Similarly, energy corresponding to ψu is

ab

abaa

uu

uu

u S

HH

d

dHE

1

ˆ

*

*


Why Eg is lower than Eu?

Using the Born-Oppenheimer approximation

dr

ed

R

ed

r

e

mH

aaaaaa sHb

sHsHsHsHa

sHaa 10

2*11

*1

0

2

10

22

2*1 44

)42

(

φH1sa : eigenfunction of ar

e

m 0

22

2

42

Therefore,

dr

eJwhere,J

R

eEH

asHasHb

*saa 11

0

2

0

2

1 4=

4+= ∫-

J : coulomb integralHaa : total energy of undisturbed H atom separated from a bare proton by the distance R (non bonded energy)


Also, Hba = Hab.Similarly,

dr

ed

R

ed

r

e

mH

ababab sHb

sHsHsHsHa

sHba 10

2*11

*1

0

2

10

22

2*1 44

)42

(

Evaluate it,

ab0

2

s1H*s1H

0

2

abs1s1H*s1Hs1s1H

a0

22

2*s1H

SR4

ed

R4

e

SEdEd)r4

e

m2(

ab

abab

Therefore,

dr

eKwhere,K)

R

eE(SH

ab sHb

*sHsabba 1

0

2

10

2

1 44 ∫- =+=

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


Difference ∆Eg and ∆Eu is given by

ab

abaauu

ab

abaagg S

JSKHEE

SJSK

HEE-1-

-,1-

-

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


12.4 A closer look at the molecular wave functions 12.4 A closer look at the molecular wave functions

ψψgg and and ψψuu

Shape of ψg and ψu, and differenceDashed line : unbound state(ζ=1, same as the H1s AOs)


Contour map of bonding and antibonding orbital


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.


Probability of bonding and antibondingLight blue line : ∆ψg

2, ∆ψu2

)(2

1)(

2

1

)(2

1)(

2

1

21

21

22

21

21

22

ba

ba

sHsHuu

sHsHgg


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

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

)(21

2

2

rr Vmp


<∆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?


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.

12.5 Combining atomic orbitals to form 12.5 Combining atomic orbitals to form molecular orbitalsmolecular orbitals


For H2+, we will obtained two MOs, with different energies.


Orbital energy diagram of H2 and HF

12.6 Molecular orbitals for homonuclear diatomic 12.6 Molecular orbitals for homonuclear diatomic

moleculesmolecules


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


Horizontal axis : molecular axisArrow : inversion operation


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)


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


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.


MOs of H2+ molecule


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+

1σg: no nodal plane 2σg: 1 nodal plane 3σg: 2 nodal planesAll σu* orbitals have a nodal plane perpendicular to the internuclear axisAmplitude 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 bonding3σu*: more nodal plane than case of H2

+

1πu: significant delocalization → large contribution

*

uσ


MOs of F2 molecule(1σg,3σu*, 1πu)


12.7 The electronic structure of many-electron 12.7 The electronic structure of many-electron

moleculesmoleculesMany-electron molecules : configuration is useful

First, see H2 and He2.In this case, only 1s orbitals used for making MOs.

We must consider1. 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


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)


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 MO2σg : bonding character2σu : less antibonding character3σg : less bonding character → making triplet bond with the pair of 1πu MOs


Molecular orbital diagram of F2


Molecular orbital diagram of N2


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


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.


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

12.8 Bond order, bond energy, and bond length12.8 Bond order, bond energy, and bond length


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.


1222*22*22 )3()1()1()2()2()1()1( guuugugN σππσσσσ

2222*22*22 )3()1()1()2()2()1()1( guuugugN σππσσσσ

1*2222*22*22 )1()3()1()1()2()2()1()1( gguuugugN πσππσσσσ

1*1*2222*22*222 )1()1()3()1()1()2()2()1()1( ggguuugugN ππσππσσσσ

Example Problem 12.4

Arrange the following in terms of increasing bond energy and bond

length on the basis of their bond order:2N 2N

2N

22N

Bond order

0.5*(9-4)=2.5

0.5*(10-4)=3

0.5*(10-5)=2.5

0.5*(10-6)=2

Therfore, the bond energy is predicted to follow the order

2N2N

2N

22N> >,

Bond length decreases as the bond strength increases (opposite order)

2N2N

2N

22N< <,


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

12.9 Heteronuclear diatomic molecules12.9 Heteronuclear diatomic molecules


MO diagram of HF


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


MO of 3σ, 4σ, 1π of HF


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

12.10 The molecular electrostatic potential12.10 The molecular electrostatic potential

r

qr

04)(

Therefore, contribution to the molecular electrostatic from the atomic nuclei is given by

i i

inuclei r

qzyx

0111 4),,(


Electron in the molecule : continuous charge distribution with a density at a point (x,y,z), related to n-electron wave function

nnnnnn dzdydxdzdydxzyxzyxzyxezyx ...)),,;...;,,;,,((...),,( 1112

111

Combining the contribution of nuclei and electrons, molecular electrostatic potential is given by

dxdydzr

zyxe

r

qzyx

ei i

i

00111 4

),,(

4),,(

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.


Electrostatic potential of HF molecule


- 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

Summary Summary

Ch 12. Chemical Bond in Diatomic Molecules MS310 Quantum Physical Chemistry The chemical bond is at...

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