H2+, the one electron system
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 exotic
HeH2+ is unstable relative to dissociation into He+ + H+.
1
e
ra
Ha
rb
R
Hb
Write the Schrodinger Equation for H2+
Tell whether terms are simple or difficult
2
e
ra
Ha
Simple, 1/R does not depend on
the electron position !
rb
R
Hb
3
Molecular orbitals -LCAO
There are exact solutions of the equation.
We will consider an approximate one, open to generalization
Y is a one-electron wave function: a molecular orbital
We will consider that Y 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 fa = fb and I fb = fa The Molecular orbitals must have the molecular
symmetry.
e
Yg = fa + fb and Yu = fa - fb are solutions:
I Yg = fa + fb = Yg
I Yu = fa - fb = Yu
s Yg = fa + fb = Yg
s Yu = fa - fb = Yu
ra
ungerade
gerade
Ha
rb
R
Hb
4
Y Normalization
<YgIYg > = <1sa+1sbI1sa+1sb>
= <1saI1sa> + <1saI1sb> + <1sbI1sa> + <1sbI1sb>
= 2 + 2S
<YuIYu > = <1sa-1sbI1sa-1sb>
= <1saI1sa> - <1saI1sb> - <1sbI1sa> + <1sbI1sb>
= 2 - 2S
SAB
HA
HB
5
Y Normalization
<YgIYg > = <1sa+1sbI1sa+1sb>
= <1saI1sa> + <1saI1sb> + <1sbI1sa> + <1sbI1sb>
= 2 + 2S
<YuIYu > = < 1sa - 1sbI1sa-1sb >
= <1saI1sa> - <1saI1sb> - <1sbI1sa> + <1sbI1sb>
= 2 - 2S
Neglecting S: Yg = 1/√2(1sa+1sb) and Yu = 1/√2(1sa-1sb)
With S: Yg = 1/√(2+2S) (1sa+1sb) and Yu = 1/√ (2-2S) 2(1sa-1sb)
6
Density partition
<YgIYg > = <1sa+1sbI1sa+1sb>
=<1saI1sa> + <1saI1sb> + <1sbI1sa> + <1sbI1sb>
= 2 + 2S
¼ On atom A
½ On the AB bond
¼ On atom B
½ On atoms
7
Neglecting S:1sg = 1/√2(1sa+1sb) With S: 1sg = 1/√(2+2S) (1sa+1sb)
No node, the whole space is in-phase
Symmetric with respect to sh sv C∞ C2 and I
8
Neglecting S:1sg = 1/√2(1sa+1sb) With S: 1sg = 1/√(2+2S) (1sa+1sb)
9
Neglecting S:1su = 1/√2(1sa-1sb) With S: 1su = 1/√(2-2S) (1sa-1sb)
Nodal
plane
10
Charge, Bond index: Without S
<YgIYg > = N2 <1sa+1sbI1sa+1sb> = 1
=N2 <1saI1sa> + N2 <1sbI1sb> + N2 <1saI1sb> + N2 <1sbI1sa>
N= 1/√2
=0
1/2 On atom A
By symmetry
1/2 On atom B Half on each
atoms
=0
L = 1/√2 1/√2 = 1/2
D = C2 = 1/2
Q = 1 – D = +1/2
Square of the coefficient,
square of amplitude
LAB = CACB
11
Charge, Bond index: With S
<YgIYg > = N2 <1sa+1sbI1sa+1sb> = 1
=N2 <1saI1sa> + N2 <1sbI1sb> + N2 <1saI1sb> + N2 <1sbI1sa>
N= 1/√2(1+S)
1/(2+2S)
1/2 On atom A
1/(2+2S)
S/(2+2S)
1/2 On atom B
S/(2+2S)
L = 1/√2 1/√2 S = S/2
DA = CA 2 + CACB SAB = 1/2
Q = 1 – D = +1/2
LAB = CACB SAB
Half of the contribution
For bonds
12
Energies Eg and Eu
From the number of nodal planes, it follows that Eg is below Eu
a
Eg=(a+b)/(1+S)
b
Eu=(a-b)/(1-S)
13
Eg and Eu, bonding and antibonding states
a-b
-b
a
+b
a
a+b
14
Eg and Eu, bonding and
antibonding states
ba’
Rydberg States
2s
2s
2s
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.
a
1s
1s
15
a
From the number of nodal planes, it follows that sg is below su
The atomic energy level
Remember !
16
a or H , the atomic level
aa
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 1sa2 is small
a = -13.6 eV for H
<1saIHmolecularI1sa> ~ <1saIHatomicI1sa>
17
a or H , the atomic level
aa
This is a natural reference for a bond formation.
For a system involving similar AOs, a = 0
This is not the usual reference (free electron)
For conjugated systems of unsaturated hydrocarbon
It is the Atomic energy of a 2p orbital
a = -11.4 eV for C (2p level)
18
b or H
ab,the
bond interaction
From the number of nodal planes, it follows that sg is below su
Eg = (a+b)/(1+S)
Eu = (a-b)/(1-S)
Eg - Eu = (a+b)(1-S)/(1-S2) - (a-b )(1+S)/(1-S2)
Eg - Eu = (+2b - 2Sa) /(1-S2) ~ 2b
b represents the interaction energy between A and B
2b represents half of the energy gap (Eg - Eu )
b is the
resonance integral (~ -3 eV) negative
It should be roughly proportional to the overlap
19
b or H
ab,
the value of the splitting
This is a natural unit for a bond formation.
For a system involving similar bonds, b 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)
b
A C=C bond is 2
20
Eg and Eu, with S
(a-b)/1-S
a
a
(a+b)/1+S
The gap is ~2b; the average EM value is close to a above it.
21
(a -b)/( 1-S )
a
The average EM value is above a
Emean
E
moyen
a
(a +b)/( 1+S )
Emean = (Eg+Eu)/2= [(a+b)/(1+S) +(a-b)/(1-S)]/2
Emean = [ (a+b)(1-S)/(1-S2) +(a-b)(1+S)/(1-S2)]/2
Emean =(a-bS)/(1-S2)
Small
- b S > 0 The mean value corresponds to
<0 >0
a destabilization (energy loss)
The antibonding level is more antibonding
than the bonding level is bonding!
22
The bonding level is stable for the equilibrium distance
Electron in the antibonding level should lead to dissociation
Energy
Energi e
Yu
di st ance
A-B
distance
i nternu cléai
re
Yg
At small d, e2/R dominates
23
The molecule with several electrons
The orbitalar approximation: Molecular configurations.
H2
Rydberg states
2sg2
1su2
Excited states
Ground state
Three rules: Pauli, Stability and Hund
1sg1su
1sg2
diagram of states
24
The molecule with several electrons
The orbitalar approximation: Molecular configurations.
H2
Rydberg states
2sg2
1su2
Excited states
Ground state
Three rules: Pauli, Stability and Hund
E = 2a2s - 2a1s + 2b2s2s
E=-2b
1sg1su
E=0
1sg2
E=2b
diagram of states
25
Diagram of orbitals
(a-b)/1-S
AO
left
AO
a
a
(a+b)/1+S
MO center
right
26
Diagram of orbitals
2e : best situation
#e
energy gain
1e
b
2e
2b
3e
b
4e
0 (-4bS)
a
Positive (4e - repulsion)
27
orbitals:
s*
s*
s*
s
s
s
Diagram of States:
Etat
Fondamental
Etat excit é
su
2 diexcited
singulet
Etat
diexcité
state S=0 E=2a-2b
ou trip let
First excited states: sgsu  ↓ ± ↓  ;   and ↓↓ E=2a
One is alone =singlet state S=0
E=2a-2aS
3 are degenerate = triplet spin S=1
sg2 Ground state S=0
E=2a+2b
28
Mulliken charge, Bond index: ground state
<YgIYg > = N2 <1sa+1sbI1sa+1sb> = 1
=1/2 <1saI1sa> + 1/2 <1sbI1sb> + 1/2 <1saI1sb> + 1/2 <1sbI1sa>
L = 2 1/√2 1/√2 S = 1 S
L = 2 1/√2 1/√2 = 1
1 On atom A
1 On atom B
DA = Sini CA 2 +Si ni CACB SAB
=1
ni : occupancy of orbital i
LAB = Si ni CACB SAB
Q=1–D=0
Half of the contribution
For bonds
29
Mulliken charge, Overlap population: excited states
First Excited States
OP = 1 1/√2 1/√2 S + 1 1/√2 (-1)/√2 S = 0
ni : orbital i occupancy
DA = Sini CA 2 +Si ni CACB SAB
=1
Q=1–D=0
OPAB = Si ni CACB SAB
diexcited State
OP = 2 1/√2 (-1)/√2 S = -1 S
ni : orbital i occupancy
DA = Sini CA 2 +Si ni CACB SAB
=1
Q=1–D=0
OPAB = Si ni CACB SAB
30
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.
31
Rydberg states
32
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 generalization
33
better for computerization.
Sigma overlap
S-S
s-p
S-d
p-d
34
u
g
35
P overlap is lateral; it concerns p or d orbitals that have a nodal plane
p-d in-phase
d-d in-phase
u
p-d out-of-phase
d-d out-of-phase
g
36
Bonding and antibonding d-d orbitals
37
d overlap for d orbitals
38
Rydberg states
39
Rydberg states
40
There is no interaction no overlap no mixing
between orbital of different symmetry
2p x
2p z
z
S=0
S>0
Les recouvrements
se comp ensent
s is symmetric
deux à deux relative to z
p is antisymmetric relative to z
41
s and p separation
Linear molecule: symmetry relative to C∞ : s SYM and p ANTI
Planar molecule: symmetry relative to s : s SYM and p ANTI
porbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.
Degenerate for H, not for C∞ not for
shows symmetry.
WARNING! Do not confuse s
p overlaps.
sV ; appropriate combination
and p orbitals and s and
42
p orbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.
Antibonding
*
Real
Complex
Bonding
43
Euler transformation
Complex
real
44
p orbitals.
Bonding
Orbit
ale liante
pu
Antibonding
Orbit
ale antiliante
pg
The p overlap (the b resonance integral) is weaker than the s
one.
45
p orbitals:
Lateral overlap.
Bonding
Antibonding
46
s orbitals.
Antibonding
ant iliant
3 su
2S and 2PZ mix
Symmetry u
Sy métrie u
2 su
nonniveaux
bonding
non liants
3 sg
Symmetry
g
Sy métrie g
Bonding
liant
2 sg
Orbit ales M oléculaires
M. O.
NIVEAUX 2S-2P
s-p hybrization
Orbitales de Sy mét rie
Symmetry Orbitals
g are bonding
u are antibonding
47
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
48
OMs.
Antibonding
The non bonding hybrids
Can be symmetryzed
non bonding
Hybrid orbitals on A
Bonding
Hybrid orbitals on B
M. O.
49
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).
50
LCAO
Y=Saf
MO
AO
This is a unitary transformation;
n AO
→ n MO
51
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
Antibonding
Niveau Antiliant
looks more like the orbital of lowest energy
A
(larger coefficient of mixing)
B
A
has an energy lower than this orbital
B
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
Niveau liant
Bonding
52
Combination of 3 AOs of same symmetry
In general,
One bonding combination, one non-bonding and one antibonding.
Combination of 4 AOs of same symmetry
Either 2 bonding and 2 antibonding, or 1 bonding, 2 non-bonding and an
antibonding
53
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!
CH 2
H
2 C=CH 2
p
CH
2
p
p C-C
s
s
p C-C
54
A-A Homonuclear diatomic molecules
Generalization of the LCAO approach:
Build the symmetry orbitals and classify them by symmetry
If E2s-E2p < 10 eV combine orbitals of same symmetry
If E2s-E2p > 10 eV do not
0
2p
E n erg y
1 0 eV
2s
Z
L i-C
N
O -F
55
For Homonuclear diatomics
E2s(A) = E2s(B)  E2p(A) = E2p(B)
Making symmetry orbitals,
we combine symmetry related orbital first!
56
Symmetry orbitals
s -type
3 su
3 sg
2 su
2 sg
Orbit ales
niveaux 2
s et2s
3 s se combinent
Hybridization: 2sg and
3ss :glesmay
mix;
u and 3su may mix
u
si les niveaux 2s et 2p sont proches en énergie
u
57
Symmetry orbitals
2p
3 su
1 pg
1 pu
3 sg
Place relative des Orbitales 3
sg
et des Orbitales
1p u :
Due L'importance
to hybridization,
3sg goes
du relèvement de 3
s décroît up
l'élect ronégativit é de l'atome
The relativeavecorder
of E3sg and E1pu may change
g
58
.
Li-N: 3s above 1p
Cas du lithium à L'azote.g3sg en dessous de u
1pu.
3 su
1 pg
3 sg
1 pu
8
6,7
5
3,4
2 su
2
2 sg
1
59
Lithium: Li2 2 valence electrons : one occupied MO
Configuration: (core)2sg2. Li-Li single bond s.
Beryllium: Be2 4 valence electrons configuration :
Configuration: ((core)2sg22su2. 2 occupied MOs
no bond (excepting weak polarization).
Boron: B2 6 valence electrons configuration:
(core)2sg22su22pu2p’u.
2 occupied MOs + 2 unpaired electrons (Hund’s rule). B2 is
paramagnetic. Bonding equivalent to a single p bond
Carbon: C2 8 valence electrons configuration:
(core)2sg22su22pu22p’u2.
4 occupied MOs 3 bonding, one antibonding
Strong bonding :C=C: 2 p bonds.
60
Nitrogen N2 10 valence electrons :
5 occupied MO
Configuration:
(core)2sg22su22pu22p’u23sg2.
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).
3sg close but above 1pu .
A N-N elongation weakens the bonding
and
. the hybridization; 3 3sg passes
below 1pu
N
N
61
O-F: 3sg below 1pu
Cas de l'oxygène et du fluor. 3sg en dessous de 1pu.
3 su
1 pg
1 pu
3 sg
8
6,7
4,5
3
2 su
2
2 sg
1
62
Oxygen O2 10 valence electrons : 5 occupied MO plus 2
unpaired electrons
Configuration:
1 s bond and 1 p bond (2 halves). paramagnetic.
O
°O
°
.
Fluor F2 12 valence electrons : 7 occupied MO
(core)2sg22su22pu22p’u23sg23pg23p’g2 A single bond
Neon Ne2. No bond
63
Li-H
Large coefficient on Li in s* antibonding
Niveau Antiliant
s = 0.9506 (2sLi) - 0.3105(1sH).
Antibonding
Li
H
s = 0.3105 (2sLi) +0.9506 (1sH).
dH= 1.807 QH= -.807
Li
2s
dLi= 0.193 QLi= +.807
-5.4 eV
1s
Li-H is 80.7% ionic, 19.7%
-13.6 eV
H
covalent.
Li
H
There is a dipole moment
Niveau liant
Lid+-Hd-.
Bonding
Li
H
Li
Li-H
H
Large coefficient on H in s bonding
64
HF
s*
Only one s bond
Large coefficient on 2pZ(F)
1s
Dipole
Hd+–Fd-
2p
s
2s
H
H-F
F
65
s
CO
CO
E(eV)
N2
1p -  1p -
2p
-11.4 e V
s
2p
-14.8 e V
1p +  1p +
2s
-21.4 e V
2s
C
C
CO
CO
O
O
66
s orbitals of CO
• Antibonding: 2/3 2pZ(C) -1/3 2pZ(O)
O
C
O
C
+
=
• Non bonding: 1/3 2pZ(C)-1/3 2pZ(C)+1/3 2pZ(O)
C
C
O
+
O
=
It accounts for the electron pair on C
• Bonding: 2/3 2s(C) +1/3 2pZ(O)
C
O
C
+
=
O
67