CHAPTER 4: MOLECULAR ORBITAL THEORY
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CHAPTER 4: MOLECULAR ORBITAL THEORY CHEM210/Chapter 4/2014/01 It is clear that hybridized valence bond theory does not explain much of the observed phenomena for small molecules such as paramagnetism or UV absorption and a better theory needs to be introduced to rationalize these experimental observations. In molecular orbital (MO) theory, electrons occupy orbitals each of which spans the entire molecule. Molecular orbitals each hold up to two electrons and obey Hund’s rule, just like atomic orbitals. Ground-rule of MO theory: number of MOs that can be formed must equal the number of atomic orbitals of the constituent atoms. Each MO has an associated energy and to derive the electronic ground state of a molecule, the available electrons are placed in MOs according to the aufbau principle, beginning with the lowest energy.
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CHAPTER 4: MOLECULAR ORBITAL THEORY. It is clear that hybridized valence bond theory does not explain much of the observed phenomena for small molecules such as paramagnetism or UV absorption and a better theory needs to be introduced to rationalize these experimental observations. - PowerPoint PPT Presentation
Transcript of CHAPTER 4: MOLECULAR ORBITAL THEORY
CHAPTER 4: MOLECULAR ORBITAL THEORY
CHEM210/Chapter 4/2014/01
It is clear that hybridized valence bond theory does not explain much of the observed phenomena for small molecules such as paramagnetism or UV absorption and a better theory needs to be introduced to rationalize these experimental observations.
In molecular orbital (MO) theory, electrons occupy orbitals each of which spans the entire molecule.
Molecular orbitals each hold up to two electrons and obey Hund’s rule, just like atomic orbitals.
Ground-rule of MO theory: number of MOs that can be formed must equal the number of atomic orbitals of the constituent atoms.
Each MO has an associated energy and to derive the electronic ground state of a molecule, the available electrons are placed in MOs according to the aufbau principle, beginning with the lowest energy.
CHEM210/Chapter 4/2014/02
BONDING IN HYDROGEN
An approximate description of the MOs in hydrogen can be obtained by considering them a linear combinations of atomic orbitals (LCAOs).
Each of the H atoms has 1s atomic orbital with associated wave functions, Ψ1 and Ψ2 and the signs of the wavefunction associated with the 1s orbital may be either + or -.The possible combinations of the two 1s orbitals are given by equations:
ΨMO (in-phase) = ΨMO = N[Ψ1 + Ψ2] ΨMO (out-of-phase) = Ψ*MO = N*[Ψ1 - Ψ2]
where N and N*are the normalization factors, ΨMO is an in-phase (bonding) interaction and Ψ*
MO is an out-of-phase (antibonding) interaction
1sA + 1sB = MO1 1sA – 1sB = MO2
constructive interference destructive interference
CHEM210/Chapter 4/2014/03
The interaction between the H 1s AOs on forming H2 may be represented by the MO diagram shown below.
The label for a molecular orbital tells us three things:
• its shape.• the parent atomic orbitals from which it was formed.• its stability (bonding or anti-bonding): antibonding character is designated
with an asterisk (*)
CHEM210/Chapter 4/2014/04
CHEM210/Chapter 4/2014/05
The ground state electronic configuration of H2 may be written as using the notation; σg (1s2).
The orbital interaction diagram can be used to predict several properties of the H2 molecule:
• H2 molecule is diamagnetic.• Bond order = 1.
Bond order = ½[(number of bonding electrons) – ( number of antibonding electrons)]
CHEM210/Chapter 4/2014/06
HELIUM (He2)
σg (1s2) σu* (1s2)
The bonding effect of the σg (1s2) is cancelled by the antibonding effect of σu
* (1s2)
Electrons in the anti-bonding MO (1sA-1sB) offset the energy gained by
placing electrons in the bonding orbitals.The He2 molecule is not a stable species
CHEM210/Chapter 4/2014/07
Bond order = 0
A high bond order indicates high bond energy and short bond length.
Consider H2+, H2, He2
+, He2: first row diatomic molecules and ions
H2 H2+ He2
+ He2
Magnetism Dia- Para- Para- -
Bond order 1 ½ ½ 0
Bond energy (kJ/mol)
436 225 251 -
Bond length (pm)
74 106 108 -
Eσg (1s2)
σu* (1s2) ↑↓ ↑ ↑↓
↑↑↓↑↓
CHEM210/Chapter 4/2014/08
LITHIUM (Li2)
Remember Li: 1s22s1
Both the 1s and 2s overlap to produce s bonding and anti-bonding orbitals
BO = ½(nb - na) = ½(4 - 2) = 1
= a single bond
σg (1s2) σu* (1s2) σg (2s2)
CHEM210/Chapter 4/2014/09
BERYLLIUM (Be2)Remember Be: 1s22s2
σg (2s2) σu* (2s2)
BO = ½(nb - na) = ½(4 - 4) = 0 No bond! The molecule is not stable!
What about p orbitals?If the overlap lies along the major bond axis then must constitute a σ bond.If they overlap perpendicular to the axis they will form π bonds.FLUORINE (F2) AND OXYGEN (O2)
The valence shell of a fluorine atom contains 2s and 2p AOs and the formation of the F2 molecule involves 2s – 2s and 2p – 2p interactions.
CHEM210/Chapter 4/2014/10
CHEM210/Chapter 4/2014/01
For F: 1s22s22p5
We expect F to use 2p orbitals this time (valence electrons).
subtraction
subtraction
addition
addition
CHEM210/Chapter 4/2014/12
F2 - It has 14 valence
electrons
O2 – has the same MO
configuration but has two electrons less.
This makes O2
paramagnetic as you will have unpaired electrons
Determine the bond orders
CHEM210/Chapter 4/2014/13
Fluorine and oxygen follow the simple logic of LCAO as they are both very electronegative and the energy difference between the 2s and 2p is large.
With boron, carbon and nitrogen a slightly different story emerges: for these elements, a degree of “s and p orbital interaction” occurs, whereby a change in the energy of the relative molecular orbitals occurs
This leads to another arrangement of the molecular orbitals.
π2p π2p
σ2p
π2p* π2p*
σ2p*
2p2p
E
CHEM210/Chapter 4/2014/14
HETERONUCLEAR DIATOMICS
Even when the atoms in a diatomic molecule are different, we use the homonuclear diatomic diagram with the s-p interaction as an approximation.
NITRIC OXIDE (NO)
Valence electrons = 5 + 6 = 11
σ2s
σ2s*
π2p
σ2p
π2p*
σ2p*
↑↓
↑↓
↑↓ ↑↓
↑↓
↑Molecule is stable and paramagnetic – agrees with experimental data.
Bond order = 2.5
CHEM210/Chapter 4/2014/15
NO+: Number of valence electrons: 5 + 6 - 1 = 10CN–: Number of valence electrons: 4 + 5 + 1 = 10These structures are isoelectronic
σ2s
σ2s*
π2p
σ2p
π2p*
σ2p*
↑↓
↑↓
↑↓ ↑↓
↑↓ Bond order = 3.0
CHEM210/Chapter 4/2014/16
Question: Can NeO exist?Number of valence electrons: 8 + 6 = 14
σ2s
σ2s*
π2p
σ2p
π2p*
σ2p*
↑↓
↑↓
↑↓ ↑↓
↑↓
↑↓ ↑↓
Bond order = 1.0
What about NeO2-?
