adv chem chapt 9
TRANSCRIPT
Covalent Bonding Orbitals
Hybridization and the Localized Electron Model
Localized Electron Model
The arrangement of valence electrons is represented by the Lewis structure or structures, and the molecular geometry can be predicted from the VSEPR model.
Atomic orbitals are used to share electrons and form bonds
Hybridization
In general we assume that bonding involves only the valence orbitals.
The mixing of atomic orbitals to form special bonding orbitals is called hybridization. Carbon is said to undergo sp3 hybridization or is sp3
hybridized because is uses one s orbital and three p orbitals to form four identical bonding orbitals.
The four sp3 orbitals are identical in shape each one having a large lobe and a small lobe. The four orbitals are oriented in space so that the large lobes form a tetrahedral arrangement.
Hybridization
Hybridization
sp3 example: Methane
sp3 example: Ammonia
sp2 Hybridization
sp2 Hybridization
sp2 Hybridization
Ethylene (C2H4) is commonly used in plastics and has a C=C double bond. Each carbon uses sp2 hybridization in this molecule because a double bond acts as one effective pair.
In forming the sp2 orbitals, one 2p orbital on carbon has not been used. This remaining p orbital is oriented perpendicular to the plane of the sp2 orbitals.
The double bond utilizes one sigma bond that is hybridized and one pi bond with the unhybridized p orbital.
sp2 Hybridization
sp2 Hybridization: Ethylene
Multiple Bonds
Single bonds are sigma bonds (σ) and the electron pair is shared in an area centered on a line running between the atoms. These are hybridized bonding orbitals.
With multiple bonds, a sigma bond is formed and then one or two pi bond (π) form. These electrons occupy the space above and below the sigma bond and use unhybridized orbitals.
sp2 Hybridization: Ethylene
sp Hybridization
sp hybridization involves one s orbital and one p orbital. Two effective pairs will always require sp hybridization.
CO2 is sp hybridized
sp Hybridization
sp Hybridization: CO2
sp Hybridization: CO2
sp Hybridization: CO2
sp Hybridization: N2
sp3d Hybridization
When a molecule exceeds the octet rule, hybridization occurs using d orbitals. Also called dsp3.
PCl5 has sp3d hybridization and is trigonal bipyramidal.
sp3d Hybridization: PCl5
Each chlorine atom displays a tetrahedral arrangement around the atom.
sp3d2 Hybridization
An octahedral arrangement requires six effective pairs around the central atom.
SF6 has sp3d2 hybridization.
Example Problem
How is the xenon atom in XeF4 hybridized?
sp3d2
Localized Electron Summary
Draw the Lewis Structure
Determine the arrangement of electron pairs using the VSEPR model.
Specify the hybrid orbitals needed to accommodate the electron pairs. Do not overemphasize the characteristics of
the separate atoms. It is not where the valence electrons originate that is important; it is where they are needed in the molecule to achieve stability.
Effective Pairs and Their Spatial Arrangement
The Molecular Orbital Model
Molecular Orbital Model
The localized electron model works very well with the prediction of structure and bonding of molecules, but the electron correlation problem still exists.
Since we do not know the details of the electron movements, we cannot deal with the electron-electron interactions in a specific way
The Molecular Orbital model helps us to deal with the molecular problem.
Molecular Orbitals
Molecular orbitals (MOs) have many of the same characteristics as atomic orbitals. Two of the most important are:
MOs can hold two electrons with opposite spins.
The square of the MO’s wave function indicates electron probability.
MOs
For simplicity we will first look at the H2 molecule.
The combination of hydrogen 1s atomic orbitals results in 2 molecular orbitals.
The wave phases of the atomic orbitals combine/overlap. Since electrons move in wave functions, this causes constructive and destructive interference in the wave pattern.
When the orbitals are added, the matching phases produce constructive interference and the opposite phases produce destructive interference.
MOs
A constructive combination gives a bonding MO. This gives an enhanced electron probability between the nuclei.
The destructive combination gives an antibonding MO. This interference produces a node between the nuclei.
MOs
Two MOs exist for H2:
MO1= 1sH1 + 1sH2
MO1 is constructive and therefore a bonding MO
MO1 is lower energy
MO2 = 1sH1 – 1sH2
MO2 is destructive and therefore an antibonding MO
MO2 is higher energy
MOs
The type of electron distribution described in these MOs is called sigma as in the localized electron model. MO1 and MO2 are sigma (σ) molecular orbitals.
In this molecule only the molecular orbitals are available for occupation by electrons. The 1s atomic orbitals of the hydrogen atoms no longer exist, because the H2 molecule – a new entity – has its own set of new orbitals.
MOsThe energy level of the bonding MO is lower and more stable than that of the antibonding MO. Since molecule formation favors the lowest energy state, this provides the driving force for molecule formation of H2.
This is called probonding.
If two electrons were forced to occupy the higher-energy MO2
this would be anti-bonding and the lower energy of the separated atoms would be favored.
Bonding and Antibonding
MOs
Labels are given to MOs indicate their symmetry (shape), the parent atomic orbitals, and whether they are bonding or antibonding.
Antibonding character is indicated by an asterisk.
Subscripts indicate parent orbitals
σ and π indicate shape.
H2 has the following MOs: MO1 = σ1s
MO2 = σ1s*
MOs
Molecular electron configurations can be written in much the same way as atomic (electron) configurations. Since the H2 molecule has two electrons in the σ1s molecular orbital, the electron configuration is: σ1s
2
Each molecular orbital can hold two electrons, but the spins must be opposite.
Orbitals are conserved. The number of molecular orbitals will always be the same as the number of atomic orbitals used to construct them.
MOs
From this molecular electron configuration, we can determine a molecules stability.
Would H2- be stable?
(σ1s )2 ( σ1s*) 1
The key idea is that H2- would exist if it were a lower
energy than its separated parts. Two electrons are in bonding and one is in antibonding. Since more electrons favor bonding H2
- is formed.
This also is a good indicator of bond strength. H2 has a stronger bond than H2
-. The net lowering of the bonding electrons by one is a direct relationship to bond strength. H2 is twice as strong.
Bond Order
To indicate bond strength, we use the concept of bond order.
Example: H2 has a bond order of 1
H2- has a bond order of ½
Bond Order =#of bondinge− −#of antibondinge−
2
Bond Order
Bond order is an indication of bond strength because it reflects the difference between the number of bonding electrons and the number of antibonding electrons.
Larger bond order means greater bond strength.
Bond order of 0 gives us a molecule that doesn’t exist.
Bonding in Homonuclear Diatomic Molecules
Homonuclear Bonding
When looking at bonding beyond energy level 1, we need to consider what orbitals are overlapping and therefore bonding.
Li2 has electrons in the 1s and 2s orbitals; the 2s orbitals are much larger and overlap, but the 1s orbitals are smaller and do not overlap.
To participate in molecular orbitals, atomic orbitals must overlap in space. This means that only the valence orbitals of the atoms contribute significantly to the molecular orbitals of a particular molecule.
Li2 MO
What is the molecular electron configuration and bond order of Li2?
σ2s2 with a bond order of 1
Li2 is a stable molecule because the overall energy of the molecule is lower than the separate atoms.
Be2 MO
What is the molecular electron configuration and bond order of Be2?
(σ2s)2 (σ2s*)2 with a bond order of 0
Be2 has 2 bonding and 2 antibonding electrons and is not more stable than the individual atoms. Be2 does not form.
MOs from p orbitals
p orbitals must overlap in such a way that the wave patterns produce constructive interference. As with the s orbitals, the destructive interference produces a node in the wave pattern and decreases the probability of bonding.
MOs from p orbitals
When the parallel p orbitals are combined with the positive and negative phases matched, constructive interference occurs, giving a bonding π orbital. When orbitals have opposite phases, destructive interference results in an antibonding π orbital.
MOs from p orbitals
Since the electron probability lies above and below the line between the nuclei (with parallel p orbitals), the stability of a π molecular bonding orbital is less than that of a σ bonding orbital. Also, the antibonding π MO is not as unstable as the antibonding σ MO. The energies associated with the orbitals reflect this stability.
B2 Example
1s2 2s2 2p1
1s2 does not bond
2s2 and 2p1 bond
(σ2s)2 (σ2s*)2 (σ2p)2
Bond order: 1
Exceptions
B2, C2, and N2 molecules use the same set of molecular orbitals that we expect but some mixing of orbital energies occurs. The s and p atomic orbitals mix or hybridize in a way that changes some MO energy states. This affects filling order and pairing of electrons.
Paramagnetism
Most materials have no magnetism until they are placed in a magnetic field. However, in the presence of such a field, magnetism of two types can be induced:
Paramagnetism – causes the substance to be attracted into the magnetic field.
Diamagnetism – causes the substance to be repelled from the magnetic field.
Paramagnetism
Paramagnetism
Paramagnetism is associated with unpaired electrons and diamagnetism is associated with paired electrons.
Any substance that has both paired and unpaired electrons will exhibit a net paramagnetism since the effect of paramagnetism is much stronger than that of diamagnetism.
Paramagnetism Video
Summary
There are definite correlations between bond order, bond energy, and bond length. As bond order increases so does bond energy and bond length decreases.
Comparison of bond orders between different molecules cannot predict bond energies of different molecules. B2 and F2 both have bond order of 1 but bond energies
are very different. B-B is a much stronger bond.
N2 has a bond order of 3 and has a very large bond energy. N2 is a very stable molecule and is used to drive powerful reactions.
Example Problem
For O2, O2+, and O2
-, give the MO electron configuration and the bond order for each. Which has the strongest bond?
O2 : (σ2s)2 (σ2s*)2 (σ2p)2 (π2p)4 (π2p*)2 BO=2
O2+: (σ2s)2 (σ2s*)2 (σ2p)2 (π2p)4 (π2p*)1 BO=2.5
O2- : (σ2s)2 (σ2s*)2 (σ2p)2 (π2p)4 (π2p*)3 BO=1.5
B2 C2 N2 O2 F2
σ2p* σ2p* π2p* π2p* σ2p π2p π2p σ2p σ2s* σ2s* σ2s σ2s σ1s* σ1s* σ1s σ1s
Example Problem
Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules: Ne2 and P2
Ne2 bond order is 0: does not exist
P2 bond order is 3 and diamagnetic
B2 C2 N2 O2 F2
σ2p* σ2p*
π2p* π2p* σ2p π2p π2p σ2p σ2s* σ2s* σ2s σ2s σ1s* σ1s* σ1s σ1s
Bonding in Heteronuclear Diatomic Molecules
Heteronuclear Molecules
When dealing with different atoms within diatomic molecules we can still use the MO model to determine bond order and magnetism
NO example
The valence electrons from both atoms fill in the order expected by the model.
The bond order is 2.5 and is paramagnetic.
Example Problem
Use the MO model to predict the magnetism and bond order of the NO+ and CN- ions.
Both ions are diamagnetic and have the same configuration. Their bond order is 3
Heteronuclear Diatomics
What happens with the diatomic molecules are very different?
A molecular orbital forms between two different atomic orbitals. HF example Note: energy level
difference vs. electronegativity
Combining the Localized Electron and MO models
Resonance
When a molecule has resonance. It is usually a double bond that can have different positions around the molecule.
The single σ bonds remain localized and the π bonds are said to be delocalized.
Resonance
Benzene: All C-C bonds are known to be equivalent and the molecule has resonance
Resonance
Benzene: The σ bonds remain centered (on the plane) between C atoms.
Resonance
Benzene: The p orbitals are perpendicular to the plane and form π bonds above and below the plane. The electrons in the π bonds delocalize and give six equivalent C-C bonds that give the structure true resonance.
This is called delocalized π bonding.
NO3-
NO3- ion also displays
delocalized π bonding.
The End
MO Worksheet
B2 C2 N2 O2 F2
σ2p* σ2p*
π2p* π2p*
σ2p π2p
π2p σ2p
σ2s* σ2s*
σ2s σ2s
σ1s* σ1s*
σ1s σ1s