Covalent Bonding Orbitals Adapted from bobcatchemistry

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 sp 3 hybridization or is sp 3 hybridized because is uses one s orbital and three p orbitals to form four identical bonding orbitals.  The four sp 3 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

sp 3 example: Methane

sp 3 example: Ammonia

sp 2 Hybridization

Ethylene (C2H4) is commonly used in plastics and has a C=C double bond. Each carbon uses sp 2 hybridization in this molecule because a double bond acts as one effective pair.  In forming the sp 2 orbitals, one 2p orbital on carbon has not been used. This remaining p orbital is oriented perpendicular to the plane of the sp 2 orbitals.  The double bond utilizes one sigma bond that is hybridized and one pi bond with the unhybridized p orbital.

sp 2 Hybridization

sp 2 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.

sp 2 Hybridization: Ethylene

sp Hybridization  sp hybridization involves one s orbital and one p orbital. Two effective pairs will always require sp hybridization.  CO 2 is sp hybridized

sp Hybridization

sp Hybridization: CO 2

sp Hybridization: N 2

sp 3 d Hybridization  When a molecule exceeds the octet rule, hybridization occurs using d orbitals. Also called dsp 3.  PCl 5 has sp 3 d hybridization and is trigonal bipyramidal.

sp 3 d Hybridization: PCl 5 Each chlorine atom displays a tetrahedral arrangement around the atom.

sp 3 d 2 Hybridization  An octahedral arrangement requires six effective pairs around the central atom.  SF 6 has sp 3 d 2 hybridization.

Example Problem How is the xenon atom in XeF 4 hybridized? sp 3 d 2

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 H 2 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 H 2 :  MO 1 = 1s H1 + 1s H2  MO 1 is constructive and therefore a bonding MO  MO 1 is lower energy  MO 2 = 1s H1 – 1s H2  MO 2 is destructive and therefore an antibonding MO  MO 2 is higher energy

MOs The type of electron distribution described in these MOs is called sigma as in the localized electron model. MO 1 and MO 2 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 H 2 molecule – a new entity – has its own set of new orbitals.

MOs The 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 H 2. This is called probonding. If two electrons were forced to occupy the higher-energy MO 2 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.  H 2 has the following MOs:  MO 1 = σ 1s  MO 2 = σ 1s *

MOs Molecular electron configurations can be written in much the same way as atomic (electron) configurations. Since the H 2 molecule has two electrons in the σ 1s molecular orbital, the electron configuration is: σ 1s 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 H 2 - be stable?  (σ 1s ) 2 ( σ 1s *) 1  The key idea is that H 2 - 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 H 2 - is formed.  This also is a good indicator of bond strength. H 2 has a stronger bond than H 2 -. The net lowering of the bonding electrons by one is a direct relationship to bond strength. H 2 is twice as strong.

Bond Order To indicate bond strength, we use the concept of bond order. Example: H 2 has a bond order of 1 H 2 - has a bond order of ½

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.  Li 2 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.

Li 2 MO What is the molecular electron configuration and bond order of Li 2 ?  σ 2s 2 with a bond order of 1  Li 2 is a stable molecule because the overall energy of the molecule is lower than the separate atoms.

Be 2 MO What is the molecular electron configuration and bond order of Be 2 ?  (σ 2s ) 2 (σ 2s *) 2 with a bond order of 0  Be 2 has 2 bonding and 2 antibonding electrons and is not more stable than the individual atoms. Be 2 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.

B 2 Example  1s 2 2s 2 2p 1  1s 2 does not bond  2s 2 and 2p 1 bond  (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2  Bond order: 1

Exceptions  B 2, C 2, and N 2 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 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 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.  B 2 and F 2 both have bond order of 1 but bond energies are very different. B-B is a much stronger bond.  N 2 has a bond order of 3 and has a very large bond energy. N 2 is a very stable molecule and is used to drive powerful reactions.

Example Problem For O 2, O 2 +, and O 2 -, give the MO electron configuration and the bond order for each. Which has the strongest bond? O 2 : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 2 BO=2 O 2 + : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 1 BO=2.5 O 2 - : (σ 2s ) 2 (σ 2s *) 2 (σ 2p ) 2 (π 2p ) 4 (π 2p *) 3 BO=1.5 B 2 C 2 N 2 O 2 F 2 σ 2p * π 2p * σ 2p π 2p σ 2p σ 2s * σ 2s σ 1s * σ 1s

Example Problem Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules: Ne 2 and P 2 Ne 2 bond order is 0: does not exist P 2 bond order is 3 and diamagnetic B 2 C 2 N 2 O 2 F 2 σ 2p * π 2p * σ 2p π 2p σ 2p σ 2s * σ 2s σ 1s * σ 1s

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.

NO 3 -  NO 3 - ion also displays delocalized π bonding.

The End

MO Worksheet B 2 C 2 N 2 O 2 F 2 σ 2p * π 2p * σ 2p π 2p σ 2p σ 2s * σ 2s σ 1s * σ 1s