1. Molecular Orbital Theory
Tutorial 8
Molecular Orbital Theory

2. Aim of Molecular Orbital Theory
Valence bond theory cannot fully explain observed properties of molecules  e.g it shows that O2 is “Diamagnetic”, however O2 shows to be “Paramagnetic” in fact !
MO theory completes the picture of covalent bond formation

3. Molecular Orbital Theory
Molecular orbital theory describes covalent bonds in terms of molecular orbitals, which result from interaction of the atomic orbitals of the bonding atoms and are associated with the entire molecule.
Atomic Orbital = belongs to only one atom
Molecular Orbital = belongs to the entire molecule

4. Molecular Orbital Theory
MO theory considers the wave nature of electrons (electrons have a dual nature “particles and waves”)
Waves interact in 2 ways
Constructive interaction
Destructive interaction

5. Molecular Orbital Theory
Therefore, Electrons interact in 2 ways
Bonding Molecular orbitals
Lower energy
(More stable)
A constructive way
&
Anti-bonding Molecular orbitals
Higher energy
(Less stable)
A destructive way
N.B Bonding MO is of lower energy than individual atomic orbitals, that’s why bond formation is exothermic

6. H2 H σ 1s2 σ* 1s + σ 1s The molecular orbital diagram of H2 will be:
destructive
constructive H
σ* 1s
Higher energy
(Less stable)
σ 1s
Lower energy
(More stable) +
The molecular orbital diagram of H2 will be:
σ* 1s
σ 1s
The electronic configuration of H2 molecule:
σ 1s2

7. MO diagram : P orbitals e.g N2
1s2 2s2 2p3
N N
2px + 2px
σ* 2px
σ 2px
1s + 1s
σ* 1s
σ 1s
2py + 2py
п 2py
п* 2py
2s + 2s
σ* 2s
σ 2s
2pz + 2pz
п 2pz
п * 2pz

8. P orbitals

9. P orbitals overlap

10. How are molecular orbitals filled
They are filled in order of increasing energy
σ 1s
σ* 1s
σ 2s
σ* 2s
σ 2px
п 2py
п* 2py
п 2pz
п* 2pz
σ* 2px
Orbitals with equal energy are half-filled first
“Hund’s Rule”

11. How are molecular orbitals filled
Notes:
The number of molecular orbitals formed is always equal to the number of atomic orbitals combined.
Each molecular orbital can accommodate up to two electrons with opposite spins
When there are several orbitals of equal energy, electrons enter singly with parallel spins, where no pairing occurs until the orbitals are half-filled. “Hund’s Rule”
The number of electrons in the molecular orbitals is equal to the sum of all the electrons on the bonding atoms.

12. Molecular orbital model can predict the magnetic properties of the molecule
All electrons paired
Unpaired electrons
Diamagnetic
Paramagnetic

13. Molecular Orbital Energy Level diagram of O2
8O : 1s2 2s2 2p4
σ 1s
σ* 1s
σ 2s
σ* 2s
σ 2px
п 2py
п* 2py
п 2pz
п* 2pz
σ* 2px
O + O
8 + 8 = 16 electrons
Paramagnetic
HOMO = п* 2py & п* 2pz
LUMO = σ* 2px
N.B
The valence MO diagram of O2 starts from σ 2s

14. HOMO : Highest Occupied Molecular Orbital
LUMO : Lowest Unoccupied Molecular Orbital

15. Bond Order
A calculation that gives an indication about the stability of the molecule
3 : triple bond
2 : double bond
1 : single bond
0 : molecule doesn’t exist

16. Bond Order Li2 Diamagnetic HOMO : σ 2s LUMO : σ* 2s
3 + 3 = 6
σ 1s
σ* 1s
σ 2s
σ* 2s
Diamagnetic
HOMO : σ 2s
LUMO : σ* 2s
Bond order = ½ (4 – 2) = 1
Electronic configuration of Li2 :
σ 1s2 σ* 1s2 σ 2s2

17. Exceptions B2, C2, N2 Light molecules in period 2
Have flipped σ 2p & п 2p molecular orbitals
П 2py
σ* 2s
σ 2s
σ* 1s
σ 1s
П* 2pz
σ 2px
П 2pz
σ* 2px
П * 2py

18. Question 1: B2 5 + 5 = 10 Bond order = ½ (6 – 4) = 1 П 2py σ* 2s σ 2s
П* 2pz
σ 2px
П 2pz
σ* 2px
П* 2py
Bond order = ½ (6 – 4) = 1

19. Question 2: NO+ 7 + 8 – 1 = 14 П 2py σ* 2s σ 2s σ* 1s σ 1s П* 2pz
σ 2px
П 2pz
σ* 2px
П* 2py
Bond order = ½ (10 – 4) = 3
Diamagnetic
HOMO: σ 2px
LUMO: П* 2py & П* 2pz

20. Question 2: CN- 6 + 7 + 1 = 14 П 2py σ* 2s σ 2s σ* 1s σ 1s П* 2pz
σ 2px
П 2pz
σ* 2px
П* 2py
Bond order = ½ (10 – 4) = 3
Diamagnetic
HOMO: σ 2px
LUMO: П* 2py & П* 2pz

21. Question 2: BN 5 + 7 = 12 П 2py σ* 2s σ 2s σ* 1s σ 1s П* 2pz σ 2px
Bond order = ½ (8 – 4) = 2
Diamagnetic
HOMO: П 2py & П 2pz
LUMO: σ 2px

22. Question 2: F2 9 + 9 = 18 П 2py σ* 2s σ 2s σ* 1s σ 1s П* 2pz σ 2px
Bond order = ½ (10 – 8) = 1
Diamagnetic
HOMO: П* 2py & П* 2pz
LUMO: σ* 2px