perturbation theory h 0 is the hamiltonian of for a known system for which we have the solutions:...
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Perturbation Theory
H0 is the Hamiltonian of for a known system for which we have the solutions: the energies, e0, and the wavefunctions, f0.
H0f0 = e0f0
We now change the system slightly (change a C into a N, create a bond between two atoms). The Hamiltonian will be changed slightly.
For the changed system H = H0 + H1
H1 is the change to the Hamiltonian.
We want to find out what happens to the molecular orbital energies and to the MOs.
How are energies and wave functions affected by a change?
Energy to ei0:
Zero order (no change, no correction): ei0
First Order correction: 1,
010iiii HdvfHf
Wave functions corrections to f0i
Zero order (no correction): f0i
First order correction: 000
1,
jij ji
ij fee
H
Example: Creating allyl system out of ethylene plus methyl radical
Pi system only:
“Unperturbed” system: ethylene + methyl radical1
2 31
2 3
2/)( 210
3 uuf 03e
30
2 uf 02e
2/)( 210
1 uuf 01e
00
0
00H
0
0
H
00
00
00001
HHH
“Perturbed” system: allyl system
3
000
3
13,0
33j
jj
j fee
Hff
2
000
2
12,0
22j
jj
j fee
Hff
1
000
1
11,0
11j
jj
j fee
Hff
13,3
033 Hee
12,2
022 Hee
11,1
011 Hee
03
02
03
02
01
03
03
01
11,3
02
02
01
11,2
011
02
1
))()/((
0
21
21
00
00
000
02
12
1
))/((
0
21
21
00
00
000
100
)/()/(
ff
f
ff
feeHfeeHff
+
0
0
21
21
00
00
000
02
1
2
1 01
011
eee
Working out of the predicted values for the “perturbed” system. MO 1 only.
Energy e1
Orbital 1
Note the stabilizing interaction. Bonding!
Mixes in bonding
Mixes in bonding
Mixes in anti-bonding
Mixes in anti-bonding
Projection Operator
Algorithm to create an object forming a basis of an irreducible rep from an arbitrary function.
^^
RRh
lP
jj
jj
Where the projection operator results from using the symmetry operators, R, multiplied by characters of the irreducible reps. j indicates the desired symmetry.
lj is the dimension of the irreducible rep (number in the E column).
1sA 1sB
z
y
Starting with the 1sA create a function of A1 sym
¼(E1sA + C21sA + v1sA + v’1sA) = ¼ (1sA + 1sB + 1sA + 1sB) = ( ½(1sA + 1sB)
Consider the bonding in NF3, C3v. We can see what Irreducible Reps can be obtained from the atomic orbitals. Then get the Symm Adapted Linear Combinations by using Projection Op.
A 3 0 -1
A B C D
B 3 0 1
C 3 0 1
D 3 0 1
A = A2 + E
B = C = D = A1 + E
1
23
}
N
F
F F
N
F
F F
N
F
F F
N
F
F FThese p orbitals are set-up as sigma and pi with respect to the internuclear axis.
Now we know which irreducible reps we can get out of each kind of atomic orbital. Have to get the correct linear combination of the AOs. The E irred. rep is the bad one.
Now construct SALCs first A2 then E.
^^
RRh
lP
jj
jj
A = A2 + E
PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2)
N
F
FF C3N
F
FF
N
F
FF C3'
N
F
FF
p1
p1
N
F
FF vN
F
FF
p1-p1
N
F
FF
N
F
FFv'
p1
-p3
N
F
FF
N
F
FFv''p1 -p2
No AO on N is A2
N
F
FF
Now construct an SALC of A2 sym
See what the different symmetry operations do to the one of the types of p orbitals.
Now get the SALC of E symmetry. Note that E is doubly degenerate.
PE(p1) = (2p1 - p2 - p3) = E1
Apply projection operator to p1
But since it is two dimensional, E, there should be another SALC. Apply PE to p2.
N
F
FF
PE(p2) = (2p2 - p3 - p1) = E’
But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero.
Create a linear combination of
E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3
Have to choose k such that they are orthogonal.
0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1))
k = ½
E2 = (3/2 p2 - 3/2 p3) = p2 – p3
N
F
FF
N
F
FF
N
F
FF
N
F
FF
N
F
FF
For now we show the interaction of the N p orbitals (px and py) of E symmetry with the SALC of the F p orbitals of E symmetry.
We will return to C3v molecules later on….
Molecular orbitals of
heteronuclear diatomic molecules
The general principle of molecular orbital theoryInteractions of orbitals (or groups of orbitals) occur when
the interacting orbitals overlap. the energy of the orbitals must be similarthe interatomic distance must be short enough but not too short
A bonding interaction takes place when:
regions of the same sign overlapAn antibonding interaction takes place when:
regions of opposite sign overlap
Combinations of two s orbitals in a homonuclear molecule (e.g. H2)
Antibonding
Bonding
In this case, the energies of the A.O.’s are identical
More generally:ca(1sa)cb(1sb)]
n A.O.’s n M.O.’s
The same principle is applied to heteronuclear diatomic molecules
But the atomic energy levels are lower for the heavier atom
Orbital potential energies (see also Table 5-1 in p. 134 of textbook)
Average energies for all electrons in the same level, e.g., 3p(use to estimate which orbitals may interact)
The molecular orbitals of carbon monoxide
CO
cc(C)co(O)]
2s 2p
C -19.43 -10.66
O -32.38 -15.85
E(eV)
Each MO receives unequal contributions from C and O (cc ≠ co)
Group theory is used in building molecular orbitals
CO C°v (use C2v)
z
s pz
px py
A1
A1
A1
A1
B1 B2
B1 B2
“C-like MO”
“O-like MO”
Frontier orbitals
“C-like MO’s”
“O-like MO’s”
mixing
Larger homo lobe on C
Bond order 3
A related example: HF
H F C°v (use C2v)
s (A1) 2p(A1, B1, B2)
H -13.61 (1s)
F -40.17 (2s) -18.65
No s-s int.(E > 13 eV)
Non-bonding(no E match)
Non-bonding(no symmetry match)
Extreme cases: ionic compounds (LiF)
Li transfers e- to F
Forming Li+ and F-
A1
A1
Molecular orbitals for larger molecules
1. Determine point group of molecule (if linear, use D2h and C2v instead of D∞h or C∞v)
2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom)
3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1)
4. Find the reducible representations (they correspond to the symmetry of group orbitals, also called Symmetry Adapted Linear Combinations SALC’s of the orbitals).
5. Find AO’s in central atom with the same symmetry
6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E
F-H-F-
D∞h, use D2h
1st consider combinations of2s and 2p orbitals from F atoms
8 GROUP ORBITALSDEFINED
Group orbitals can now be treated as atomic orbitalsand combined with appropriate AO’s from H
1s(H) is Ag so it matches two group orbitals 1 and 3
Both interactions are symmetry allowed, how about energies?
-13.6 eV
-40.2 eV
-13.6 eV
Good E matchStrong interaction
Poor E matchweak interaction
Bonding e
Non-bonding e
Lewis structureF-H-F-
implies 4 e around H !
MO analysisdefines 3c-2e bond
(2e delocalized over 3 atoms)
CO2
D∞h, use D2h
(O O) group orbitals the same as for F F
But C has more AO’s to be considered than H !
CO2
D∞h, use D2h
No match
Carbon orbitals
Ag-Ag interactions B1u-B1u interactions
All four are symmetry allowed
Primary Ag interaction
Primary B1u interaction
Bonding
Bonding
Non-bonding
Non-bonding
4 bondsAll occupied MO’s are 3c-2e
LUMO
HOMO
The frontier orbitals of CO2
Molecular orbitals for larger molecules: H2O
1. Determine point group of molecule: C2v
2. Assign x, y, z coordinates (z axis is higher rotation axis; if non-linear y axis in outer atoms point to central atom - not necessary for H since s orbitals are non-directional)
3. Find the characters of the representation for the combination of 2s orbitals on the outer atoms, then for px, py, pz. (as for vibrations, orbitals that change position = 0, orbitals that do not change =1; and orbitals that remain in the same position but change sign = -1)
4. Find the irreducible representations (they correspond to the symmetry of group orbitals,also called Symmetry Adapted Linear Combinations SALC’s of the orbitals).
5. Find AO’s in central atom with the same symmetry
6. Combine AO’s from central atom with those group orbitals of same symmetry and similar E
For H H group orbitals
v’ two orbitals interchanged
E two orbitals unchanged
C2 two orbitals interchanged
2 20 0
v two orbitals unchanged
No match
pz
bonding
slightlybonding
antibonding
px
bonding
antibonding
py
non-bonding
3 10
Find reducible representation for 3H’s
Irreducible representations:
Molecular orbitals for NH3
pz
bonding
Slightlybonding
anti-bonding
bonding
anti-bonding
LUMO
HOMO
Projection OperatorAlgorithm of creating an object forming a basis for an irreducible rep from an arbitrary function.
^^
RRh
lP
jj
jj
Where the projection operator sums the results of using the symmetry operations multiplied by characters of the irreducible rep. j indicates the desired symmetry.
lj is the dimension of the irreducible rep. h the order order of the group.
1sA 1sB
z
y
Starting with the 1sA create a function of A1 sym
¼(E1sA + C21sA + v1sA + v’1sA) = ¼ (1sA + 1sB+ 1sB + 1sA)
Consider the bonding in NF3
N
F
F F
N
F
F F
N
F
F F
N
F
F F
A 3 0 -1
A B C D
B 3 0 1
C 3 0 1
D 3 0 1
A = A2 + E
B = C = D = A1 + E
1
23
}
Now construct SALC
^^
RRh
lP
jj
jj
A = A2 + E
PA2(p1) = 1/6 (p1 + p2 + p3 + (-1)(-p)1 + (-1)(-1p3) + (-1)(-p2)
N
F
FF C3N
F
FF
N
F
FF C3'
N
F
FF
N
F
FF vN
F
FF
p1-p1
N
F
FF
N
F
FFv'
p1
-p3
N
F
FF
N
F
FFv''p1 -p2
No AO on N is A2
N
F
FF
Now construct an SALC of A2 sym
See what the different symmetry operations do to the one of the types of p orbitals.
E:
PA2(p1) = (2p1 - p2 - p3) = E1
Apply projection operator to p1
But since it is two dimensional, E, there should be another SALC
N
F
FF
PA2(p2) = (2p2 - p3 - p1) = E’
But E1 and E’ should be orthogonal. We want sum of products of coefficients to be zero.
Create a linear combination of
E2 = E’ + k E1.= (-1 +k*2) p1 + (2 + k(-1)) p2 + (-1 + k(-1))p3
Have to choose k such that they are orthogonal.
0 = (2(-1 + k*2) -1 (2 + k(-1)) -1 (-1 + k(-1))
k = ½
E2 = (3/2 p2 - 3/2 p3) = p2 – p3
N
F
FF
N
F
FF
N
F
FF
N
F
FF
N
F
FF
The geometriesof electron domains
Molecular shapes:When we discussed
VSEPR theory
Can this be describedin terms of MO’s?
Hybrid orbitals
s + p = 2 sp hybrids (linear)
s + 2p = 3 sp2 hybridstrigonal planar
s + 3p = 4 sp3 hybridstetrahedral
s + 3p + d = 5 dsp3 hybridstrigonal bipyramidal
s + 3p + 2d = 6 d2sp3 hybridsoctahedral
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