1

This molecular ion exits and has been experimentally measured ; its dissociation equals 2.791 eV and its H-H

distance is 2.0 a0 (1.06Å).

There is no other 1 e - 2 nuclei stable system than H2+

Hydrogenoids exist even if they might be exoticHeH2+ is unstable relative to dissociation into He+ + H+.

H2+, the one electron system

2

Write the Schrodinger Equation for H2+

Tell whether terms are simple or difficult

3Simple, 1/R does not depend on the electron position !

4

There are exact solutions of the equation.We will consider an approximate one, open to generalization

is a one-electron wave function: a molecular orbital

We will consider that is Linear Combination of Atomic Orbitals.

For H2+, it is possible to find them using symmetry.

Mirror or Inversion center: A single atomic function is not a solution

I a = b and I b = a The Molecular orbitals must have the molecular symmetry.

g = a + b and u = a - b are solutions:

I g = a + b = g

I u = a - b = u

g = a + b = g

u = a - b = u

Molecular orbitals -LCAO

gerade

ungerade

5

Normalization

gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S

uIu 1sa-1sbI1sa-1sb1saI1sa- 1saI1sb- 1sbI1sa1sbI1sb2 - 2S

HBHASAB

6

Normalization

gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S

uIu 1sa-1sbI1sa-1sb 1saI1sa- 1saI1sb- 1sbI1sa1sbI1sb2 - 2S

Neglecting S: g√1sa+1sb) and u√1sa-1sb)

With S: g√S1sa+1sb) and u√ S1sa-1sb)

7

Density partition

gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S

¼ On atom A ¼ On atom B½ On the AB bond

½ On atoms

8

Neglecting S:1g√1sa+1sb) With S: 1g√S1sa+1sb)

No node, the whole space is in-phaseSymmetric with respect to h v C∞ C2 and I

9

Neglecting S:1g√1sa+1sb) With S: 1g√S1sa+1sb)

10

Neglecting S:1u√1sa-1sb) With S: 1u√S1sa-1sb)

Nodal plane

11

Charge, Bond index: Without S

gIg 1sa+1sbI1sa+1sb1saI1sa1sbI1sb1saI1sb1sbI1sa√

1/2 On atom A

D = C2 = 1/2

Q = 1 – D = +1/2

1/2 On atom B

By symmetry Half on each atoms

= 0 = 0

Square of the coefficient, square of amplitude

L = 1/√2 1/√2 = 1/2

LAB = CACB

12

Charge, Bond index: With S

gIg 1sa+1sbI1sa+1sb1saI1sa1sbI1sb1saI1sb1sbI1sa√S

1/2 On atom A

DA = CA 2 + CACB SAB = 1/2

Q = 1 – D = +1/2

1/2 On atom B

1/(2+2S) 1/(2+2S) S/(2+2S) S/(2+2S)

Half of the contributionFor bonds

L = 1/√2 1/√2 S = S/2

LAB = CACB SAB

13

Energies Eg and Eu

From the number of nodal planes, it follows that g is below u

Eg=()/(1+S) Eu=()/(1-S)

14

Eg and Eu, bonding and antibonding states

15

Eg and Eu, bonding and antibonding states

2s

2s 2s

1s1s

Rydberg States

Valence states

The bonding and antibonding levels are referred to “dissociation”

Not to the “free electron” ; an antibonding level could be higher than a bonding one if referred to a higher reference level.

’

16

From the number of nodal planes, it follows that g is below u

The atomic energy level

Remember !

17

This term represents the difference between Hmol and Hat.• Either the electron is close to A:

R and rb are nearly the same and [1/R - 1/rb] is small• Or the electron is far from A and 1sa

2 is small

<1saIHmolecularI1sa> ~ <1saIHatomicI1sa>

or Haa, the atomic level

-13.6 eV for H

18

This is a natural reference for a bond formation.For a system involving similar AOs, = 0

This is not the usual reference (free electron)For conjugated systems of unsaturated hydrocarbon It is the Atomic energy of a 2p orbital

or Haa, the atomic level

-11.4 eV for C (2p level)

19

or Hab,the bond interaction From the number of nodal planes, it follows that g is below u

represents the interaction energy between A and B

2 represents half of the energy gap (Eg - Eu )

is the resonance integral (~ -3 eV) negative

It should be roughly proportional to the overlap

Eg = ()/(1+S) Eu = ()/(1-S) Eg - Eu = (1-S)/(1-S2) - ()(1+S)/(1-S2) Eg - Eu = (- 2S) /(1-S2) ~

20

This is a natural unit for a bond formation.For a system involving similar bonds, is the unit

We define the unit including the negative sign.For conjugated systems of unsaturated hydrocarbon It represents half of a C=C bond (2 electrons gain the energy of the splitting)

or Hab, the value of the splitting

A C=C bond is 2

21

Eg and Eu, with S

S

S

The gap is ~2; the average EM value is close to above it.

22

The average EM value is above

Emean = (Eg+Eu)/2= [()/(1+S) +()/(1-S)]/2Emean = [ ()(1-S)/(1-S2) +()(1+S)/(1-S2)]/2

Emean =(S)/(1-S2)

S > 0 The mean value corresponds to <0 >0 a destabilization (energy loss)

The antibonding level is more antibondingthan the bonding level is bonding!

Small

Emoyen

1+S

1-S

Emean

23

g

udistanceinternucléaire

EnergieEnergy

A-B distance

The bonding level is stable for the equilibrium distance

Electron in the antibonding level should lead to dissociation

At small d, e2/R dominates

24

The molecule with several electrons

The orbitalar approximation: Molecular configurations.

H2

1g2Ground state

1u2

1g1u

2g2

Excited states

Rydberg states

Three rules: Pauli, Stability and Hund diagram of states

25

The molecule with several electrons

The orbitalar approximation: Molecular configurations.

H2

1g2Ground state

1u2

1g1u

2g2

Excited states

Rydberg states

Three rules: Pauli, Stability and Hund diagram of states

2s1s 2s2s

26

Diagram of orbitals

S

S

AO

right

AO

left

MO center

27

Diagram of orbitals

2e : best situation

1e

2e

3e

4e 0 (-4S)

# e energy gain

Positive (4e - repulsion)

28

Etatdiexcité

Etat excitésinguletou triplet

Etat Fondamental

*

*

*

udiexcited state S=0 E=

First excited states: gu ↓ ± ↓ ; and ↓↓ E=

One is alone =singlet state S=0 E=S

3 are degenerate = triplet spin S=1

gGround state S=0 E=

orbitals:

Diagram of States:

29

Mulliken charge, Bond index: ground state

gIg 1sa+1sbI1sa+1sb21saI1sa21sbI1sb21saI1sb21sb

I1sa

1 On atom A

DA = ii CA 2 +i i CACB SAB =

1

Q = 1 – D = 0

1 On atom B

Half of the contributionFor bonds

L = 2 1/√2 1/√2 S = 1 S

L = 2 1/√2 1/√2 = 1

LAB = i i CACB SAB

i : occupancy of orbital i

30

Mulliken charge, Overlap population: excited states

DA = ii CA 2 +i i CACB SAB =

1

Q = 1 – D = 0

First Excited States

OP = 2 1/√2 (-1)/√2 S = -1 S

OPAB = i i CACB SAB

i : orbital i occupancy

diexcited State

DA = ii CA 2 +i i CACB SAB =

1

Q = 1 – D = 0

OP = 1 1/√2 1/√2 S + 1 1/√2 (-1)/√2 S = 0

OPAB = i i CACB SAB

i : orbital i occupancy

31

Rydberg states, from 2s and 2p

To find M.O.s First construct Symmetry orbitals

Each atom A or B does not have the molecular symmetry,

It is necessary to pair atomic orbitals between symmetry related atoms.

32

Rydberg states

33

The drawings or the symmetry labels are unambiguous

Mathematic expression is Ambiguous; it requires defining S

+ for positive S, good for pedagogy- for similar direction on the z axis, better for generalizationbetter for computerization.

34

Sigma overlap

S-S S-d

p-ds-p

35

u g

36

u g

overlap is lateral; it concerns p or d orbitals that have a nodal plane

p-d in-phased-d in-phase

p-d out-of-phased-d out-of-phase

37

Bonding and antibonding d-d orbitals

38

overlap for d orbitals

39

Rydberg states

40

Rydberg states

41

There is no interaction no overlap no mixing between orbital of different symmetry

S = 0Les recouvrementsse compensent deux à deux

S > 0

z

2px

2pz

s is symmetric relative to z

p is antisymmetric relative to z

42

and separation

Linear molecule: symmetry relative to C∞ : SYM and

Planar molecule: symmetry relative to : SYM and

orbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.

Degenerate for H, not for C∞ not for V ; appropriate combination

shows symmetry.

WARNING! Do not confuse and orbitals and and overlaps.

43

orbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.

*

Bonding Antibonding

Rea

l

C

ompl

ex

44

Euler transformationComplex real

45

Orbitale antiliante

g

Orbitale liante

u

Bonding Antibonding

orbitals.

The overlap (the resonance integral) is weaker than the one.

46

Bonding Antibonding

orbitals: Lateral overlap.

47

orbitals.

2S and 2PZ mix

3u

3g

2g

antiliant

niveaux non liants

liant

Symétrie u

Symétrie g

Orbitales Moléculaires Orbitales de Symétrie NIVEAUX 2S-2P

2u

non bonding

Bonding

Antibonding

M. O. Symmetry Orbitals g are bondings-p hybrization u are antibonding

Symmetry g

Symmetry u

48

hybridization

Mixing 2s and 2p: requires degeneracy to maintain eigenfunctions of AOs.

Otherwise, the hybrid orbital is an average value for the atom, not an exact solution.

This makes sense when ligands impose directionality: guess of the mixing occurring in OMs.

49

Hybrid orbitals on A

non bonding

Bonding

Antibonding

M. O.

Hybrid orbitals on B

The non bonding hybrids Can be symmetryzed

50

Method to build M.O.s

• Determine the symmetry elements of the molecule

• Make the list of the functions involved (valence atomic orbitals)

• Classify them according to symmetry (build symmetry orbitals if necessary by mixing in a combination the set of orbitals related by symmetry)

• Combine orbitals of the same symmetry (whose overlap is significant and whose energy levels differ by less than 10 eV).

51

LCAO

a

This is a unitary transformation;

n AO → n MO

MO AO

52

Combination of 2 AOs of same symmetry

They mix to generate a bonding combination and an antibonding one.

The bonding orbital

is the in-phase combination

looks more like the orbital of lowest energy

(larger coefficient of mixing)

has an energy lower than this orbital

The antibonding orbital is the out-of-phase combination looks more to the orbital of highest energy (larger coefficient of mixing) has an energy higher than this orbital

A B

A B

B

A

Niveau liant

Niveau AntiliantAntibonding

Bonding

53

Combination of 3 AOs of same symmetry

One bonding combination, one non-bonding and one antibonding.

Either 2 bonding and 2 antibonding, or 1 bonding, 2 non-bonding and an antibonding

In general,

Combination of 4 AOs of same symmetry

54

Populating MOs

1. Fill the in increasing order, respecting the Pauli principle.

2. Do not consider where the electron originate ! This is a different problem « correlating » the « initial distribution » to the final one. To determine the ground state just respect rule 1!

2 CH2 → H2C=CH2 may be a fragment analysis to build ethene in the ground state, not an easy reaction leading directly to the ground state! C-C

p

C-C

p

CH2 H2C=CH2 CH 2

55

A-A Homonuclear diatomic molecules

Generalization of the LCAO approach:Build the symmetry orbitals and classify them by symmetryIf E2s-E2p < 10 eV combine orbitals of same symmetry

If E2s-E2p > 10 eV do not

0

Energy

Z

NLi-C O-F

2p

2s

10 eV

56

For Homonuclear diatomics

E2s(A) = E2s(B) E2p(A) = E2p(B)

Making symmetry orbitals, we combine symmetry related orbital first!

57Orbitales : les niveaux 2 u et 3u se combinent

si les niveaux 2s et 2p sont proches en énergie

3u

3g

2u

2g

Symmetry orbitals type

Hybridization: 2sg and 3sg may mix; 2su and 3su may mix

58

Symmetry orbitals 2

Place relative des Orbitales 3 g et des Orbitales u:

L'importance du relèvement de 3 g décroît

avec l'électronégativité de l'atome

3u

3g

1u

1g

Due to hybridization, 3g goes upThe relative order of E3g and E1u may change

59

Cas du lithium à L'azote. 3g en dessous de 1u.

Li-N: 3g above 1u

.

60

Lithium: Li2 2 valence electrons : one occupied MOConfiguration: (core)2g

2. Li-Li single bond . Beryllium: Be2 4 valence electrons configuration : Configuration: ((core)2g

22u2. 2 occupied MOs

no bond (excepting weak polarization). Boron: B2 6 valence electrons configuration: (core)2g

22u22u2’u.

2 occupied MOs + 2 unpaired electrons (Hund’s rule). B2 is paramagnetic. Bonding equivalent to a single bondCarbon: C2 8 valence electrons configuration: (core)2g

22u22u

22’u2. 4 occupied MOs 3 bonding, one antibondingStrong bonding :C=C: 2 bonds.

61

Nitrogen N2 10 valence electrons : 5 occupied MOConfiguration: (core)2g

22u22u

22’u23g2.

1 bond and 2 bonds: This is the most stable case with the maximum of bonding electrons.It corresponds to the shortest distance and to the largest dissociation energy. It is very poorly reactive, inert most of the time (representing 80% of atmosphere). 3g close but above 1u . A N-N elongation weakens the bonding and the hybridization; 3 3g passes below 1u

.

N N

62

Cas de l'oxygène et du fluor. 3g en dessous de 1u.

O-F: 3g below 1u

63

Oxygen O2 10 valence electrons : 5 occupied MO plus 2 unpaired electrons Configuration:1 s bond and 1 p bond (2 halves). paramagnetic.

Fluor F2 12 valence electrons : 7 occupied MO (core)2g

22u22u

22’u23g23g

23’g2 A single bond Neon Ne2. No bond

.

O O° °

64

Li-H

Li H

Li H

1sH

2sLi

H

Li

Niveau liant

Niveau Antiliant

-5.4 eV

-13.6 eV

Li Li-H H

Antibonding

Bonding

 =  0.9506 (2sLi) - 0.3105(1sH). = 0.3105 (2sLi) +0.9506 (1sH). dH= 1.807 QH= -.807 dLi= 0.193 QLi= +.807 Li-H is 80.7% ionic, 19.7% covalent. There is a dipole momentLi+-H-

.

Large coefficient on Li in antibonding

Large coefficient on H in bonding

65

HF

FH-FH

2s

2p

1s

*

Only one s bond

Large coefficient on 2pZ(F)

DipoleH+–F-

66

CO

-11.4 eV

-21.4 eV

-14.8 eV

C CO O

2p

2s

2p

E(eV)

C OCO

N2CO

67

orbitals of CO

• Antibonding: 2/3 2pZ(C) -1/3 2pZ(O)

• Non bonding: 1/3 2pZ(C)-1/3 2pZ(C)+1/3 2pZ(O)

It accounts for the electron pair on C

• Bonding: 2/3 2s(C) +1/3 2pZ(O)

+ =C O C O

+ =C O C O

+ =C O C O