Transcript Molecular Orbitals several atoms

Molecular Orbitals
several atoms
Any group of atoms:
•Molecules,
•Molecular ions,
•Fragments,
•Supermolecules
1
Describing molecular properties as
a whole
Molecular orbital theory is a method for
determining molecular structure in which
electrons are not assigned to individual
bonds between atoms, but are treated as
moving under the influence of the nuclei in
the whole molecule.
2
The orbitalar approximation
Molecular orbital is a function that
describes the wave-like behavior of a
single electron in a molecule. A
polyelectronic
wave-function
is
expressed in terms (as a product or a
determinant) of MOs. The use of the
term "orbital" was first used in English
by Robert S. Mulliken in 1925 as the
English translation of Schrödinger's
use
of
the
German
word,
'Eigenfunktion'.
Robert Sanderson
Mulliken
1996-1986
Nobel 1966
3
Generalizing the
LCAO approach:
A linear combination of
atomic orbitals or LCAO
Sir John Lennard-Jones
1894-1954
Linus Carl Pauling
1901-1994 Nobel 1962
It was introduced in 1929 by Lennard-Jones
with the description of bonding in the
diatomic molecules of the first main row of
the periodic table, but had been used earlier
by Pauling for H2+.
4
Any calculation starts by inputs and provides outputs
Input
Definition of the system
Total number of electrons
Choice of state (ground state or
excited state)
Nature of atoms involved (Z)
Atomic Orbitals
Geometry to define Potential
Output
Energy,
Molecular Orbitals
Electronic density
Spin, Magnetism
Search for minimization of energy
The calculated geometry is the minimum found by optimization
5
Structure and Reactivity
How the energy of a system respond to a structural variation?
Changing distance or angle.
6
Symmetry
To describe a molecule is describing
•Its geometry and its symmetry
•Its energy
MOs are eigenfunctions of the symmetry operator and of H
For H2, we have use symmetries without solving the Schrodinger
equation,
For a larger system, symmetry helps simplifying or analyzing.
One common symmetry is the plane for planar molecules; this
causes s and p separation.
To be considered when some parameter varies (structure
modification, structure optimization, reactivity) symmetry has to
be preserved during the variation
7
Valence orbitals
size of the orbitals
Usually, the atomic orbitals that participate to the MOs
are valence orbitals.
These are the valence orbitals of the neutral atom. For a
cation in its highest oxidation state, this might be the
unoccupied shell whereas the size of the cation is given
by the outermost occupied shell
See the case of Li+
8
Valence orbitals
size of the orbitals, the Li+ example
Size of the cation Li+ according to Slater?
The last occupied shell is the 1s
Z=(3-0.31) <r> = 1.5/2.69 a0 = 0.558 a0 = 0.295 Ả
This is small (The Pauling ionic radius is also small, 0.60 Ả)
The valence orbital 2s is the one involved in forming bonds
Z=(3-2*0.85) <r> = 5/1.3 a0 = 3.846 a0 = 2.034
Ả
Atomic radius (empirical): 1.45 Ả
Atomic radius (calculated): 1.67 Ả
Covalent radius (2008 values): 1.28 Ả
Covalent radius (empirical): 1.34 Ả
van der Waals radius: 1.82 Ả
The covalent radius for the 2s orbital is such that the orbitals overlap;
it is less than the <r> value
9
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).
10
Valence orbitals of H2O
yOz
Molecular
plane
xOz
symmetry
plane
C2
z axis
2sO
S
S
S
sigma
2pxO
A
S
A
pi
2pyO
S
A
A
sigma
2pzO
S
S
S
sigma
1s1
S
-
-
sigma
1s2
S
-
-
sigma
Symmetry relative to the molecular plane, sigma-pi separation
11
Valence orbitals of H2O
yOz
Molecular
plane
xOz
symmetry
plane
C2
z axis
2sO
S
S
S
2pxO
A
S
A
2pyO
S
A
A
2pzO
S
S
S
1s1
S
-
-
1s2
S
-
-
Symmetry relative to xOz, necessitates building symmetry orbitals
Build them ! What are their energy levels?
12
Valence orbitals of H2O
yOz
Molecular
plane
xOz
symmetry
plane
C2
z axis
2sO
S
S
S
A1
2pxO
A
S
A
B1
2pyO
S
A
A
B2
2pzO
S
S
S
A1
1 (1s1+1s2)
S
S
S
A1
S
A
A
B2
2
1 (1s1-1s2)
2
Redundancy: Three groups; A2 is not present
13
C2v
E
yOz
molecular
plane
xOz
symmetry
plane
C2
axis
A1
1
1
1
1
A2
1
-1
-1
1
B1
1
1
-1
-1
B2
1
-1
1
-1
14
Symmetry orbitals
1
2
1
2
(1s1+1s2)
(1s1-1s2)
The H-H distance is
2*sin(105/2)*1.09 Ả = 1.73 Ả
This is long relative to 0.74 Ả !
The energy level of the symmetry orbital
is the atomic level for H, -13.6 eV
15
Values from Extended Hückel 1962
(-eV)
2s
H
13.6
Li
5.4
He
24.25
Be
10
B
15.2
C
21.4
N
26
O
32.3
F
40
2p
3.5
6
8.5
11.4
13.4
14.8
18.1
Na
Mg
Al
Si
P
S
Cl
3s
5.1
9
12.3
17.3
18.6
20
30
3p
3.
4.5
6.5
9.2
14
13.3
15
1s
Roald Hoffmann
Nobel 1981
William Nunn Lipscomb, Jr.
American 1919Nobel 1976
16
1 Valence orbital of H2O
Symmetry B1
1
2
2sO
A1
2pxO
B1
2pyO
B2
2pzO
A1
1 (1s1+1s2)
A1
It is a molecular orbital
It has p symmetry.
2
(1s1-1s2)
The atomic orbital
2pxO is a lone in its
group,
B2
E2p(O)= -14.8 eV
17
2 Valence orbitals of H2O
Symmetry B2
1
2sO
A1
2pxO
B1
2pyO
B2
2pzO
A1
(1s1+1s2)
A1
E2p(O)= -14.8 eV
and
E1S(H) = -13.6 eV
s* A
2
1 (1s -1s )
1
2
2
2py
B2
-13.6eV
s A-s
0-14.8 eV
B
sA
18
2sO
A1
2pxO
B1
2pyO
B2
2pzO
A1
3 Valence orbitals of H2O
Symmetry A1
E2s(O)= -32.3 eV
E2p(O)= -14.8 eV
And E1S(H) = -13.6 eV
E2s is low in energy an d does
not mix
s* S
1
(1s1+1s2)
A1
2pz
2
1 (1s -1s )
1
2
2
0
B2
-13.6eV
-14.8eV
s A+s B
sS
19
6 AO
MOs
→ 6
3 A1
→ 3 A1
2B2
→ 2B2
1 B1
→ 1 B1
What is the ordering of the levels?
20
y
y
°
z
z
Overlap involved
in B2
Overlap involved
in A1
At 45° same overlap
angle HOH/2 = 105/2 = 52°5 > 45°
Splitting in B2 > Splitting in A1
21
2B2
6 AO
→ 6 MOs
s* A
O
3 A1
H
4A1
→ 3 A1
2B2
→ 2B2
1 B1
→ 1 B1
s* S
H
2p
x
sS
In the Lewis formula, two electron pairs, two bonds
3A1
The electron pairs are 2s and 2p
Spectroscopy shows different electron levels
sA
1B2
The bonds are the result of two contributing MO
A1 and B2
Orbital numbering: 1A1 for the core?
2 s
0
2A1
22
Hybridization
MOs
sp3 agreement with VSEPR
and Total density
sp2 closer to MOs and spectroscopy:
two different pairs
23
The Walsh diagram
E(eV)
2px
s A
s S
-13.6
O
-14.8
sS
sA
2s
O
-33.3
 90°
  °
24
Reduction of symmetry
25
Typical Bond Angles in AH2 Molecules
molecule electronic configuration H-A-H bond angle
BeH2
(1a1)2(1b2)2
180°
BH2
(1a1)2(1b2)2(2a1)1
127°
CH2
(1a1)2(1b2)2(2a1)1(b1)1
134°
CH2
(1a1)2(1b2)2(2a1)2
102°
NH2
(1a1)2(1b2)2(2a1)2(b1)1
103°
OH2
(1a1)2(1b2)2(2a1)2(b1)2
104°
MgH2
(1a1)2(1b2)2
180°
AlH2
(1a1)2(1b2)2(2a1)1
119°
SiH2
(1a1)2(1b2)2(2a1) 1(b1)1
118°
SiH2
(1a1)2(1b2)2(2a1)2
93°
PH2
(1a1)2(1b2)2(2a1)2(b1)1
92°
SH2
(1a1)2(1b2)2(2a1)2(b1)2
92°
1a1= 1sg ; 1b2= 1su
26
F. Walsh diagram for H2S
27
This is called an orbital correlation diagram:
28
C
2sC
A1
2pxC
B1
2pyC
B2
2pzC
A1
1 (1s1+1s2)
A1
3 Valence orbitals of CH2
Symmetry A1
E2s(C)= -21.4 eV
E2p(C)= -11.4 eV
And E1S(H) = -13.6 eV
E2sC – E1sH < 10 eV
Hybrid orbitals
A
Not atomic eigenfunctions
But part of the MOs eigenfunctions for
the molecule
2
1 (1s -1s )
1
2
2
B2
2s-2pZ
2A1 is bonding 3A1 is non-bonding 4A1 is antibonding
2s+2pZ
29
C
2sC
A1
2pxC
B1
2pyC
B2
2pzC
A1
1 (1s1+1s2)
3 Valence orbitals of CH2
Symmetry A1
E2s(C)= -21.4 eV
E2p(C)= -11.4 eV
And E1S(H) = -13.6 eV
E2sC – E1sH < 10 eV
A
s*
A1
2 pz
2
-1 3 .6 eV
1 (1s -1s )
1
2
2
S
s A +s
-1 1.4 eV
C
B
B2
Among the 3 A1 valence MO, the lowest one is bonding
and the middle one non-bonding; this differs from H2O!
sS
30
hydridization
Justification:
• s and pZ both are of same symmetry and appear in the same linear
combinations (MOs); hybridization allows anticipating.
• hybridization aims combining AOs, each hybrid maximizing the
interaction with a partner (here the symmetry orbital on Hs) and
minimizing those with others. The idea is that we can thus separate
the interactions with different partners. If correct, this is useful and
allows transferability (replacing a substituent by another one
involving the same hybrid).
• Mathematically a set of hybrids is equivalent to a set of canonic
orbitals
Failure:
• It is not possible to rigorously separate the interactions with different
partners. Hybrids interact with each other.
• Existing symmetries reappear if interactions between hybrids is
included.
31
Methane
The MO orbital description conflicts with sp3 hybridization
32
33
sp3 hybridization
2s
2px
2py
2pz
t1
1/2
1/2
1/2
1/2
t2
1/2
1/2
- 1/2
-1/2
t3
1/2
-1/2
1/2
-1/2
t4
1/2
-1/2
-1/2
1/2
<tiItj>= δij
<tiIHItj>= (E2s-E2p)/4
Eti = <tiIHIti> = (E2s+3E2p)/4
34
sp3 hybridization
x
b
b
b
b
x
b
b
b
b
x
b
(E2s+3E2p)/4
b
b
b
x
=0
With b =(E2s-E2p)/4
E2p
1(E2s-E2p)/4
3 (E2s-E2p)/4
E2s
35
sp3 hybrids along C3
¼ of 2s
3/4 of 2p
2s
2px
2py
2pz
t1
1/2
0
0
√/2
t2
1/2
√(2/3)
0
-1/√12
t3
1/2
-1/√6
1/√2
-1/√12
t4
1/2
-1/√6 -1/√2 -1/√12
36
•
•
•
•
•
Interest of hybridization
Pictorial
Anticipating AO combinations within MOs
Global density
Transferable: Analysis through substituents
Writing VB structures
What is bad with hybridization?
• Not describing symmetry
• Not good for spectroscopy
• Lack of orthogonality between hybrid orbitals
belonging to the same center (tails of localized
orbitals)
37
sp3 hybrids along C2
2s
2px
2py
2pz
t1
1/2
1/√2
0
1/2
t2
1/2
-1/√2
0
1/2
t3
1/2
0
1/√2
-1/2
t4
1/2
0
-1/√2
-1/2
38
39
40
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
41
OMs.
Angular dependence
Expression along C2 (z is the main axis)
t = a s + √(1-a2) [pzcosq ± px sinq]
t = a’s + √(1-a’2) [pzcosq’ ± py sinq’]
2a2+2a’2=1
√(1-a2) sinq  /√2
√(1-a’2) sinq’  /√2
q
q’
:
:
q 5.5/ (<9°)  5.75° → a = 0.459
a’ = √(1/2-a2) = 0.548
sinq’  (/√2)/(1/ √(1-a’2)=0.8453
q’  57°  4.7/ > 9°
42
Angular dependence
q
q’
:
:
a2+a’2= ½
√(1-a2) sinq  /√2
√(1-a’2) sinq’  /√2
(1-a2) sinq2  /2 (1-a’2) sinq’ 2
1-/2sinq2  a2
1-/2sinq’2  a’2
1-/2sinq2 +1-/2sinq’2  a2 + a’2
2 = /2sinq2 +/2sinq’2 + ½
3 = /sinq2 +/sinq’2
When q decreases q’ increases!
43
Weight of s orbital
q = 105°5 water
a is smaller than ½
a’ is larger than ½
The weight of s is larger in the lone pairs.
The s level is lower in energy than the p level. It has to participate to the
stabilization of lone pairs.
L-, CH3+
90°
pure p
sp3 ligand
109°28’
¼s¾p
CH3-, L+
>109° 28’
>1/4 s
44
For the reconstructed Silicon(100)
surface, the s character is higher
for the dangling bonds of the
outmost atoms.
Si
Si
Si
-
Si
Si
Si
Si
+
Si
Si
Si
S
i
Si
45
H3
The orbitals of H3 (equilateral
triangle in the xy plane)
Are symmetry orbitals of NH3
and CH3 matching p orbitals
1/√2 -1/√2
-b
2px
2py
b
2pz
1/√3
46
H3
1/√2 -1/√2
The orbitals of H3 (equilateral
triangle in the xy plane) are
symmetry orbitals of NH3 and
CH3 matching p orbitals
-b
2px
2py
b
2pz
1/√3
47
Pseudo symmetry: for H3 the
H3
1/√2 -1/√2
OM are degenerate.
They remain degenerate interacting
2/3 = 1/√2 + c2 with the 2p orbitals (E)
-b
2px
√(2/3)
2py
Normalization
1 = (2/√3)2 +2 c2
48
H3
1/√2 -1/√2 -1/√6 -1/√6
The orbitals of H3 (equilateral
triangle in the xy plane) are
symmetry orbitals of NH3 and
CH3 matching p orbitals
-b
√(2/3)
2px
2py
b
2pz
1/√3
49
NH3
s system
Lone pair
Bonding
50
NH3
1/√2
p system
-1/√2
-1/√6 -1/√6
2px
2py
√(2/3)
51
Ammonia
52
Free rotation of a methyl group
adjacent to a p system
Hyperconjugation
CH3 always has one orbital conjugating with
the p system
53
Mirror symmetry
pseudosymmetry
Free rotation of a methyl group
X
Z
X
x
x
p
y
s
s
y
Z
p
54
Orientation of CH2D conjugated
with a CH2+
D
+
or
+
D
55
D
+
+
A
D
x
CH2
With a substituent, the 2
orbitals are not equivalent.
That of highest energy level
interacts more strongly with
the CH2 group
56
Ethane
57
Ethene (ethylene)
58
Ethyne (acetylene)
59
BF3
60
Octet rule, eighteen electron rule
N in NH3
4 atomic orbitals
4 bonding orbitals or non bonding
accommodating 8 electrons
3 antibonding orbitals
C in CH4
4 atomic orbitals
4 bonding orbitals accommodating 8
electrons
4 antibonding orbitals
W in W(CO)6
9 atomic orbitals
9 bonding orbitals or non bonding
accommodating 8 electrons
6 antibonding orbitals
61
Ligand field for octahedral environment
W in W(CO)6
9 atomic orbitals
9 bonding orbitals or non bonding
accommodating 8 electrons
6 antibonding orbitals
The s interaction raises the dX2 and dX2-dY2
levels
The p with p *CO stabilizes the “non bonding
levels”
62
Ligand field for octahedral environment
W in W(CO)6
9 atomic orbitals
9 bonding orbitals or non
bonding accommodating 8
electrons
6 antibonding orbitals
The s interaction raises the
dX2 and dX2-dY2 levels
The p with p *CO stabilizes
the “non bonding levels”
63
Ligand field for octahedral environment
Ti in Ti(H2O)2(HO)4
Or Ti in TiO2
Ti in 9 atomic orbitals
9 bonding orbitals or non
bonding accommodating 8
electrons
6 antibonding orbitals
The s interaction raises
the dX2 and dX2-dY2 levels
The p interaction with pO
(oxygen pairs) is a
stabilizing interaction.
The same occurs with
PH3 ligands
Rutile TiO2
These 6
electrons
count
for the
eighteen
electron64
rule
Ligand field for various environments
65
nickel (Z= 28)
Ni(PH3)2...CH2=CH2
Z
Z
D
D
Ni
D
D
Ni
NiD2 where D is a donor substituent (group PH3 ). Ni will be only
represented by the set of d (x2-y2, z2, xy, xz et yz) orbitals taking E(3d) = 12 eV. D is modeled by an s orbital with 2 electrons: E(D)= - 15 eV.
What is the best orientation for the C=C bond?
1) Draw an energy diagram for NiD2
2) Why the ethylene molecule is represented above by its p* orbital?
3) Draw an energy diagram for the interaction and tell which orientation is
the best.
66
Ni(PH3)2...CH2=CH2
2
x -y
z
2
Z
Z
2
D
D
Ni
D
D
Ni
xy, xz, yz
Interaction with dx2-y2
Largest interaction
Interaction
with d(x+y),z
weaker
67
s CH , A
 sA
z
 sS
s CH , A
y
2p
1
1
s CO , A
1
p CO , B
2
x
2p
y
,B
p CO , B
*
 sS
1
2p
2
s CO , A
1
 sA
s CH , B
1
sS
s CH , A
Formaldehyde
2s
CH2
CH
O,
A
1
1
CH2O
2
H
2
2s
O
C=O
O
O
68
B
s orbital
With p overlap
69