1
Molecular Orbital Theory (MO)
Bonding MO in an H2 molecule
Combination of atomic orbitals on
all atoms in a molecule
• Suitable symmetry
• Similar energy
n AOs forms n MOs
Diatomic molecules: H2, F2, CO,....
Polyatomic molecules: BF3, CH4,....
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Wave Interference
+
– –
–
–
–
++
+Constructive
Destructive
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LCAO = Linear Combination of Atomic
Orbitals
Combination of 2 wave
functions (orbitals) with
the SAME sign
Combination of 2 wave
functions (orbitals) with
OPPOSITE sign
Ψ = c1 ΨA + c2 ΨB
Ψ* = c3 ΨA − c4 ΨB
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LCAO = Linear Combination of Atomic
Orbitals
Ψ Bonding MO
Ψ* Antibonding MO
Atomic orbitals
1 s1 s
Number of MO = Number of AO
ADD
SUBSTRACT
Ψ = c1 ΨA + c2 ΨB
Ψ* = c3 ΨA − c4 ΨB
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LCAO = Linear Combination of Atomic
Orbitals
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Ψ Bonding MO
Ψ* Antibonding MO
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Differences between VB and MO
H H
VB
MO
H H
Localized bonds Delocalized bonds
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stabilization
destabilization
Energetic
In comparison
with free atoms
Y Bonding MO
Y* Antibonding MO
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π MO by Combination of p AO
Antibonding π MO
Bonding π MO
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H3
+
Number of nodal planes incr.
Energy incr. stability decr.
Bonding MO
Antibonding MO
Nonbonding MO
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H3
+
Bonding MO
Antibonding MO
Energy incr.
stability decr.
Number of nodal planes incr.
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LCAO = Linear Combination of AO
LCAO
n atoms with m orbitals
Ψi = c1 Ψ1 + c2 Ψ2 + c3 Ψ3 +...+ cn Ψn
6 AOs (s+2p+3s) will form 6 MOs
•MO with the lowest energy
•no nodal plane
•the most bonding
•combination of one AO on each atom
•all AOs with the same sign
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Antibonding MO
Bonding MO
Filling Electrons to MOs
Aufbau
Hund
Pauli
Rules for filling electrons to MOs
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Bond Order
Bond Order = ½ (number of e in bonding MOs
− number of e in antibonding MOs)
2
*
eMOeMO
BO
−
=
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Bond Order
1s 1s
σ1s
σ1s∗
H2
+ cation
1s 1s
σ1s
σ1s∗
1s 1s
σ1s
σ1s∗
1s 1s
σ1s
σ1s∗
He2
+ cation
H2 molecule
He2 molecule
0.5
0.5 0.0
1.0
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Bond Order
---
1.08
0.74
1.06
Bond length,
Å
2300.512He2
+
0022He2
432102H2
2550.501H2
+
Bond energy,
kJ mol−1
Bond
Order
Antibond.
electrons
Bond.
electrons
Molecule
One-electron bond: 1 bonding electron forms a stronger
bond than 2 bonding and 1 antibonding electrons
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z
z
x
x
y
y
z
z
x
x
y
y
+
+
+
−
−
−
MOs by Combination of p AOs
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MOs by Combination of p AOs
Antibonding MO σ2pz*
Bonding MO σ2pz
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MOs by Combination of p AOs
Antibonding MO π2px*
Bonding MO π2px
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Pi Bond in Ethene
HOMO = highest occupied MO
LUMO = lowest unoccupied MO
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Types of MO
Better overlap decreases energy of bonding MO
and increses energy of antibonding MO:
σ > π > δ
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d(z2) d(z2) d(x2−y2) d(x2−y2)
d(xy) d(xy)
d(xz) d(xz)
d(yz) d(yz)
σ π δ
σ∗ π∗ δ∗
MO from d orbitals
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Mixing of s-p orbitals
Mixing s-p
No mixing s-p
Energetically similar orbitals on the same atom can mix
Small s-p gap
Large s-p gap
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Energy
from H2 to N2from O2 to Ne2
Mixing s-pNo mixing s-p
s2s
s*2s
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Diatomic Molecules
π2
p
Increasing s-p mixing
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2s
2px 2py 2pz 2px 2py 2pz
2s
σ2s
σ*2s
σ2p
σ*2p
π2p π2p
π*2p π*2p
E
Interaction Diagram for Li2 to N2
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Diatomic Molecules
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Diatomic Molecules in the Gas Phase
Bond length (pm) Ebond(kJ mol-1)
Li-Li σ2s
2 267 110
Be...Be σ2s
2 σ*2s
2 ? ?
B-B σ2s
2 σ*2s
2 π2p
2 159 290
C=C σ2s
2 σ*2s
2 π2p
4 124 602
N≡N σ2s
2 σ*2s
2 π2p
4 σ2p
2 110 942
O=O σ2s
2 σ*2s
2 σ2p
2 π2p
4 π*2p
2 121 494
F-F σ2s
2 σ*2s
2 σ2p
2 π2p
4 π*2p
4 142 155
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N2 Triple bond O2 paramagnetic molecule
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Oxygen and its Molecular Ions
1.01.52.02.5Bond order
↑↓↑↑
Bond length, pm
Occupation of
HOMO
πx* a πy*
Number of valence
electrons
149126121112
↑↓↑↓↑↑
14131211
O2
2−O2
−O2O2
+
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Multiplicity
M = 2 S + 1
S = sum of unpaired spins (½) in an atom or a molecule
2½
2
1½
1
½
0
S
↑↑↑quartet4
↑↑↑↑quintet5
↑↑triplet3
↑↑↑↑↑sextet6
↑dublet2
↑↓singlet1
M
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Triplet Oxygen 3Σ
Singlet Oxygen 1Δ
95 kJ mol-1
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Isoelectronic Molecules
O2
−, Cl2
+, ⋅ClO13
F2, O2
2−, ClO−14
O2
+, ⋅NO, SO+11
O2, SO12
N2, CO, CN−, BF, NO+, TiO, SiO10
BO, CN, CP, CO+9
Diatomic SpeciesNumber of
valence electrs
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MO in Polar Molecules
Ψ∗ = c3 ΨA − c4 ΨΒ
Ψ = c1 ΨA + c2 ΨΒ
χ(A) = χ(B) a nonpolar bond
c1 = c2 c3 = c4
Same contribution from A and B
χ(A) < χ(B) a polar bond
c1 < c2 bonding MO has higher
contribution from B
c3 > c4 antibonding MO has higher
contribution from A
χ(A) << χ(B) ionic bond
c1 → 0 bonding MO = ΨΒ
c4 → 0 antibonding MO = ΨA
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MO in Polar Molecules - HF
antibonding MO
nonbonding MO
bonding MO
weakly bonding MO
Bonding MO concentrated on an atom with high electronegativity - F
Antibonding MO concentrated on an atom with low electronegativity - H
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Molecular Orbitals in CH4
C + 4H
8 electrons
Atomic orbitals
used
C s + 3×p
4H 4×s
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Molecular Orbitals in CH4
AO carbon
MO of methane
AO hydrogen
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PES in Agreement with the MO Model
Area = 3 Area = 1
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Benzene
H
H
H
H
H
H
Separate sigma and pi system
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Bonding MOs in Benzene
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C3H3
+
C4H4
2+
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C5H5¯
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C6H6
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H
H
H
H
H
H
1,3-butadiene
HOMO
LUMO
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Molecular Ions
O2 → O2
+ + e−
Splitting off the weakest
bound e in HOMO
IE
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Molecular Ions
CN + e− → CN−
Adding e to HOMO
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Excitation of Molecules
Etot = E(electronic) + E(vibrat) + E(rotat) + Eother
Individual parts of Etot are independent – very different
magnitudes (Bornova-Oppenheimmer approximation)
E(electron) 100 kJ mol−1 UV and visible
E(vibrat) 1.5 – 50 kJ mol−1 Infrared (IR)
E(rotat) 0.1 – 1.5 kJ mol−1 Microwave and far IR
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Rotational Energy
Quantization of rotational energy
E(rotat) = (ħ2/2I) J(J +1)
J = rotat quantum number
I = moment of inertia (m r2)
m = m1m2/(m1 + m2)
Selection Rule ΔJ = ±1
At normal temperature, molecules are in many
excited rotational states, rotational energy is
comparable to thermal energy of molecular motion
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Vibrational Energy
Quantization of vibrational energy
E(vibrat) = k ħ2 (v + ½)
v = vibrat quantum number
Selection Rule Δv = ±1
Zero Point Energy:
for v = 0 E(vibrat) = ½ k ħ2
H2 E(disoc) = 432 kJ mol−1
E(v = 0) = 25 kJ mol−1
At normal temperature, molecules are in ground
vibrational state v = 0
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Vibrational Energy
2173H2
+
2990.3D2
4159.2H2
Vibrational energy,
cm−1
Molecule
ΰ = 1/2π (k /m)½
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Rotational – Vibrational Spectrum of HCl(g)
IR region
ν0 = 2886 cm−1
ν0
Δv = ±1
ΔJ = ±1
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Types of Vibrations
Valence
Deformation
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Infrared and Raman Spectroscopies
Infrared spectroscopy
Vibration must change dipole moment of a
molecule (HCl, H2O)
Raman spectroscopy
Vibration must change polarization of a molecule (H2)
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ΰ = 1/2π (k /m)½