remember what these matrices mean! transformed d orbitals = matrix * original d orbitals
DESCRIPTION
We know that each irreducible rep is 1 dimensional but let’s assume that we do not know the dimensions of the irreducible reps. - PowerPoint PPT PresentationTRANSCRIPT
We know that each irreducible rep is 1 dimensional but let’s assume that we do not know the dimensions of the irreducible reps.
We do know that the d orbitals on an isolated atom are equivalent. Let’s guess that we need all five for a representation. It will turn out to be reducible but again assume that we do not know that.
We have to find out how each d orbital transforms under the C2v operations. Formulate the matrices of the representation based on the d orbitals in this order dz2, dxz, dyz, dxy, dx2-y2.
Operation E C2 v(xz) v(yz)
Representation,
Trace 5 1 1 1
F
SF x
z
10000
01000
00100
00010
00001
Remember what these matrices mean!
Transformed d orbitals = matrix * original d orbitals
As the distance between atoms decreases
Atomic orbitals overlap
Bonding takes place if:
the orbital symmetry must be such that regions of the same sign overlapthe energy of the orbitals must be similarthe interatomic distance must be short enough but not too short
If the total energy of the electrons in the molecular orbitalsis less than in the atomic orbitals, the molecule is stable compared with the atoms
Combinations of two s orbitals (e.g. H2)
Antibonding
Bonding
More generally:ca(1sa)cb(1sb)]
n A.O.’s n M.O.’s
Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together(total energy is lowered)
Electrons in antibonding orbitals increase the energy of the system.
Bothand notation indicates symmetric with respect to rotation about internuclear axis
zC2
zC2 zC2
zC2Not
p orbitals can combine in two ways, and .
and notation meanschange of sign upon 180 rotation
and notation means nochange of sign upon rotation
Combinations of two p orbitals
zC2
zC2
Both sigma and pi interactions of p orbitals
Combination of s and p orbitals
Combination of two misaligned p orbitals in mode.
No overlap, no bonding
d orbitals overlappingNo interaction – different symmetry type overlap: symmetric for any rotation
type overlap: antisymmetric for 180 rotation,
type overlap. Changes sign for rotation by 90
NO NOYES
Is there a net interaction?
Relative energies of interacting orbitals must be similar
Strong interaction Weak interaction
Molecular orbitalsfor diatomic molecules
From H2 to Ne2
Electrons are placedin molecular orbitals
following the same rulesas for atomic orbitals:
Fill from lowest to highestMaximum spin multiplicity
Electrons have different quantum numbers including spin (+ ½, - ½)
Bond order = # of electrons
in bonding MO's# of electrons in antibonding MO's
12
-
O2 (2 x 8e)
1/2 (10 - 6) = 2A double bond
Or counting onlyvalence electrons:
1/2 (8 - 4) = 2
Note subscriptsg and u
symmetric/antisymmetricupon i
Place labels g or u in this diagram
g
g
u
u
g
u
g
u
u
g
g or u?
Orbital mixingIn this diagram some of the MOs have the same symmetry.
This allows for interaction to take place. Mixing can occur.
Orbitals of g symmetry can mix together to yield a more stable and a less stable combination.
Likewise for u
sum
dif ference
For the orbitals of g symmetry.
Overall result: orbital mixing
When two MO’s of the same symmetry mixthe one with higher energy moves higher and the one with lower energy moves lower.
This revised ordering occurs Li N, but not O Ne
Notice the change in ordering here caused by the mixing.
O Ne Li N
The electronic configuratations
Li2 bond order 1
Be2 bond order ?
B2 bond order 1C2 bond order 2
N2 bond order 3
O2 bond order 2 F2 bond order 1
Expect the shortest bond length for N2
(But remember the atomic radii decrease with F smallest.)
Bond lengths in diatomic molecules
Filling bonding orbitals
Filling antibonding orbitals
In Li2 through N2 g above u. O2 through Ne2 u above g.-The orbitals are more affected by effective nuclear charge than are the -Less mixing for O, F and Ne due to smaller orbitals and
greater bond distance. Paramagneticdue to mixing
Neutral: C2 u2 u
2 (double bond)
Dianion: C22- u
2 u2 g
2(triple bond)
Neutral: O2 u2 u
2 g1 g
1 (double bond, paramagnetic)Dianion: O2
2- u2 u
2 g2 g
2 (single bond, diamagnetic)
Photoelectron Spectroscopy
h(UV o X rays) e-
Ionization energy
hphotons
kinetic energy of expelled electron= -
Ionization is a transition between states
• Initial State: Neutral (or anion)
• Final State: Atom/Molecule/Anion after an electron is removed, plus the ejected electron
• M → M+ + e-
init = M ; final = M+ + e-
• For direct photoionization, transition probability is always > 0
h + Molecule
= Ionization Energy
e- + Molecule+
Eh - Ee- = EM+ - EM
IE = Difference in energy between states of M, M+
Ionization Energy (eV)
161718 15
↿⇂↿ ⇂H 1sH 1s
σ
σ*
What about molecules?
molecular rotationslower energy microwave radiation
electron transitionshigher energy visible and UV radiation
molecular vibrationsmedium energy IR radiation
Ground State
Excited StateDuring an electronic transition
the complex absorbs energy
electrons change orbital
the complex changes energy state
Timescale : ≈10-15 sec
Timescale of geometry changes (vibrations): ≈10-12 sec
As a result, observe vertical (Franck-Condon) transitions
In other words, we assume that we only have to consider the electronic portion of the ground- and excited-state wavefunctions to understand these transitions: Born-Oppenheimer approximation
Transitions between molecular potential energy surfaces
18
17
16
15
0
Ioni
zatio
n E
nerg
y (e
V)
H2+
r (Å)0 1 2
H2
Potential Energy Surface Description of the Ionization of Dihydrogen
Much more on this next time!!
Ground state (X) = 1g+
Ionization Energy (eV)151617181920
:N≡N:
2p
2s
2p
2s
1g+
2g+
1u+
2u
1u
1g
First ion state (X) = 2g+
Second ion state (A) = 2u
Third ion state (B) = 2u+
Consider Dinitrogen
N21g
2g+
2u+
2u+
Ground state (X) = 1g+
:N≡N:
2p
2s
2p
2s
1g+
2g+
1u+
2u
1u
1g
Potential Well Description
Models to describe molecular electronic structure
MO Theory compared to
Valence Bond Theory
Consider methane.
VSEPR gives 4 sp3 hybrid orbitals.
24 22 20 18 16 14 12
Ionization Energy (eV)
CH 4
24 22 20 18 16 14 12
Ionization Energy (eV)
CH 4
Photoelectron Spectroscopy
2s
2p
sp3
So why are there two valence ionizations separated by almost 10 eV?
Use of reducible representations in M.O. theory
Consider transformation properties ofvectors aligned with the 4 C-H bonds.
Td E 8C3 3C2 6S4 6σd
σ 4 1 0 0 2
1]120084[24
11
Aa
0]120084[24
12
Aa
0]00088[24
1Ea
0]1200012[24
11
Ta
1]1200012[24
12
Ta
C-H = A1 + T2
http://www.mpip-mainz.mpg.de/~gelessus/group.html
Apply Reduction Formula:
This is what the bonding MOs orbitals should be.
24 22 20 18 16 14 12
Ionization Energy (eV)
CH4
24 22 20 18 16 14 12
Ionization Energy (eV)
CH4
CH4
t2 (1, 2, 3)
a1 (1)
2p (t2)
2s (a1)
C H4
LCAO Description of Methane, Td
Obtained using the carbon AOs as basis objects. Obtained using the hydrogen
AOs as basis objects.Orbitals of same symmetry combine and bond.
N2 O2
*u (2s)
u (2p)
g (2p)
*u (2s)
g (2p)u (2p)
u (2p)
(Energy required to remove electron, lower energy for higher orbitals)
Very involved in bonding(vibrational fine structure)
Now back to our ordering of the MOs of diatomics.
Orbital EnergyOrbital Energy
Very Simple Molecular Orbital Theory
A molecular orbital, f, is expressed as a linear combination of atomic orbitals, holding two electrons.
The multi-electron wavefunction and the multi-electron Hamiltonian are
)...3()2(1...)3,2,1( 211 fffF
electrons
iihH ...)3,2,1(
Where hi is the energy operator for electron i and involves only electron i
AO
llluaf
Seek F such that...)3,2,1(...)3,2,1(...)3,2,1( EFFH
Divide by F(1,2,3…) recognizing that hi works only on electron i.
electrons
i j
ji Eif
ifh
)(
)(
jjj fefh Since each term in the summation depends on the coordinates of a different electron then each term must equal a constant.
)...3()2()1()...3()2()1()...3()2()1( 332132213211 fhffffhffffh
)...3()2()1( 321 ffEf
)...3()2()1( 321 fffhi
jjj fefh
Multiply equation at top by one of the atomic orbitals, uk k = 1,2,…#AOs, and integrate.
dvhuuh jkjk ,
dvfuedvhfu jkjjk
AO
lllj uaf
Recall the expansion of a molecular orbital in terms of the atomic orbitals.
Now use the expansion of the MO in the AOs.
dvuuS jkjk ,
0)( ,, AO
llklkl eSha
These integrals are fixed numerical values.
Now our attention is on the equation that only involves one electron.
AO
lllj uaf
Define
The al, l = 1, 2, 3…#AOs are unknowns.
These values are known numerical quantities.
AO
l
AO
lllkjllk dvuauedvuahu
AO
llkljl
AO
lkl dvuuaedvuhua
There are #AO such equations. Secular equations.
0)( ,, AO
llklkl eSha For k = 1 to AO
These are the secular equations. The number of such equations is equal to the number of atomic orbitals, AO.
For there to be a nontrivial solution (all al not equal to zero) to the set of secular equations then the determinant below must equal zero
The number of equations is equal to the number of AOs equations. The number of unknowns, the al, is also equal to the number of AOs.
0
)()(
)()(
,,1,1,
,1,11,11,1
AOAOAOAOAOAO
AOAO
eSheSh
eSheSh
0
)()(
)()(
,,1,1,
,1,11,11,1
AOAOAOAOAOAO
AOAO
eSheSh
eSheSh
Drastic assumptions can now be made. We will use the simple Huckel approximations.
hi,i = if orbital i is on a carbon atom.
Si,i = 1, normalized atomic orbitals,
hi,j = , if atom i bonded to atom j, zero otherwise
Sij = 0 if I is not equal to j
•Expand the secular determinant into a polynomial of degree AO in e, the energy.
•Obtain the allowed values of e by finding the roots of the polynomial.
•Choose one particular value of e, substitute into the secular equations.
•Obtain the coefficients of the atomic orbitals within the molecular orbital.
ExampleThe allyl pi system. 1
2 3
The secular equations:
(-e)a1 + a2 + 0 a3 = 0
a1 + (-e) a2 + a3 = 0
0 a1 + a2 + (-e) a3 = 0
Simplify by dividing every element by and setting (-e)/ = x
0)0(1)1(
10
11
012
xxx
x
x
x
2,0 x
For x = -sqrt(2)
e = sqrt(2)
0210
0121
0012
321
321
321
aaa
aaa
aaa
321 2 uuuf normalized
)2()121(
1321
5.0222uuuf
12 3
For x = 0; e =
Substitute into the secular equations
0010
0101
0010
321
321
321
aaa
aaa
aaa
31
2 0
aa
a
2/1
2/1
2/1
)2(
2/1
2/1
2/1
0
0
Verify that
h f = e f
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.
Changes (approximate)
Energy
Zero order (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
ExamplePi system only:
Perturbed system: allyl system
Unperturbed system: ethylene + methyl radical
12 3
12 3
2/)( 210
3 uuf 03e
30
2 uf 02e
2/)( 210
1 uuf 01e
00
0
00H
H
0
0
H
00
00
00001
HHH
0
0
21
21
00
00
000
02
1
2
111
e
03
02
03
2
03
03
01
11,3
02
02
01
11,2
11
02
1
))()/((
0
21
21
00
00
000
02
12
1
))/((
0
21
21
00
00
000
100
)/()/(
ff
f
f
feehfeehf
+
0
0
21
21
00
00
000
02
1
2
111
e
03
02
03
2
03
03
01
11,3
02
02
01
11,2
11
02
1
))()/((
0
21
21
00
00
000
02
12
1
))/((
0
21
21
00
00
000
100
)/()/(
ff
f
f
feehfeehf
+
Mixes in bonding
Mixes in bonding
Mixes in anti-bonding
Mixes in anti-bonding
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 results from using the symmetry operations multiplied by characters of the irreducible reps. j indicates the desired symmetry.
lj is the dimension of the irreducible rep.
1sA 1sB
z
y
Starting with the 1sA create a function of A1 sym
¼(E1sA + C21sA + v1sA + v’1sA) = ½(1sA + 1sB)