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# Chapter 9 Covalent Bonding: Orbitals

Shaun Williams, PhD

## Hybridization and the Localized Electron Model

### Exercise 1

Draw the Lewis structure for methane, $$\chem{CH_4}$$.

• What is the shape of a methane molecule? $$\phantom{\text{tetrahedral}}$$
• What are the bond angles? $$\phantom{109.5^\circ}$$

## Hybridization and the Localized Electron Model

Draw the Lewis structure for methane, $$\chem{CH_4}$$.

• What is the shape of a methane molecule? $$\color{green} \text{tetrahedral}$$
• What are the bond angles? $$\color{green} 109.5^\circ$$

### Concept Check 1

What is the valence electron configuration of a carbon atom?

$$\phantom{s^2p^2}$$

Why can't the bonding orbitals for methane be formed by an overlap of atomic orbitals?

### Concept Check 1 - Answer

What is the valence electron configuration of a carbon atom?

$$s^2p^2$$

Why can't the bonding orbitals for methane be formed by an overlap of atomic orbitals?

Because this would lead to two different types of C-H bonds and we know that methane has four identical C-H bonds that are 109.5° apart from each other (not 90° from each other).

### Bonding in Methane

• Assume that the carbon atom has four equivalent atomic orbitals, arranged tetrahedrally.

### Hybridization

• Mixing of the native atomic orbitals to form special orbitals for bonding.

### $$sp^3$$ Hybridization

• Combination of one s and three p orbitals.
• Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, the localized electron model assumes that the atom adopts a set of $$sp^3$$ orbitals; the atom becomes $$sp^3$$ hybridized.
• The four orbitals are identical in shape.

### Exercise 2

Draw the Lewis structure for $$\chem{C_2H_4}$$ (ethylene)?

• What is the shape of an ethylene molecule? $$\phantom{\text{trigonal planar around each carbon atom}}$$
• What are the approximate bond angles around the carbon atoms? $$\phantom{120^\circ}$$

Draw the Lewis structure for $$\chem{C_2H_4}$$ (ethylene)?

• What is the shape of an ethylene molecule? $$\color{green} \text{trigonal planar around each carbon atom}$$
• What are the approximate bond angles around the carbon atoms? $$\color{green} 120^\circ$$

### $$sp^2$$ Hybridization

• Combination of one $$s$$ and two $$p$$ orbitals.
• Gives a trigonal planar arrangement of atomic orbitals.
• One $$p$$ orbital is not used.
• Oriented perpendicular to the plane of the $$sp^2$$ orbitals.

### Sigma ($$\Sigma$$) Bond

• Electron pair is shared in an area centered on a line running between the atoms.

### Pi ($$\Pi$$) Bond

• Forms double and triple bonds by sharing electron pair(s) in the space above and below the $$\sigma$$ bond.
• Uses the unhybridized p orbitals.

### $$sp$$ Hybridization

• Combination of one $$s$$ and one $$p$$ orbital.
• Gives a linear arrangement of atomic orbitals.
• Two $$p$$ orbitals are not used.
• Needed to form the $$\pi$$ bonds.

### Exercise 3

Draw the Lewis structure for $$\chem{PCl_5}$$?

• What is the shape of phosphorus pentachloride? $$\phantom{\text{trigonal bipyramidal}}$$
• What are the bond angles? $$\phantom{90^\circ \text{ and } 120^\circ}$$

Draw the Lewis structure for $$\chem{PCl_5}$$?

• What is the shape of phosphorus pentachloride? $$\color{green} \text{trigonal bipyramidal}$$
• What are the bond angles? $$\color{green} 90^\circ \text{ and } 120^\circ$$

### $$dsp^3$$ Hybridization

• Combination of one $$d$$, one $$s$$, and three $$p$$ orbitals.
• Gives a trigonal bipyramidal arrangement of five equivalent hybrid orbitals.

### Exercise 3

Draw the Lewis structure for $$\chem{XeF_4}$$?

• What is the shape of an xenon tetrafluoride? $$\phantom{\text{octahedral}}$$
• What are the bond angles? $$\phantom{90^\circ \text{ and } 180^\circ}$$

Draw the Lewis structure for $$\chem{XeF_4}$$?

• What is the shape of an xenon tetrafluoride? $$\color{green} \text{octahedral}$$
• What are the bond angles? $$\color{green} 90^\circ \text{ and } 180^\circ$$

### $$d^2sp^3$$ Hybridization

• Combination of two $$d$$, one $$s$$, and three $$p$$ orbitals.
• Gives an octahedral arrangement of six equivalent hybrid orbitals.

### Using the Localized Electron Model

• Draw the Lewis structure(s).
• Determine the arrangement of electron pairs using the VSEPR model.
• Specify the hybrid orbitals needed to accommodate the electron pairs.

## The Molecular Orbital Model

• Regards a molecule as a collection of nuclei and electrons, where the electrons are assumed to occupy orbitals much as they do in atoms, but having the orbitals extend over the entire molecule.
• The electrons are assumed to be delocalized rather than always located between a given pair of atoms.
• The electron probability of both molecular orbitals is centered along the line passing through the two nuclei.
• Sigma ($$\sigma$$) molecular orbitals (MOs)
• In the molecule only the molecular orbitals are available for occupation by electrons.

### Relative Stabilities

• $$\chem{MO_1}$$ is lower in energy than the $$s$$ orbitals of free atoms, while $$\chem{MO_2}$$ is higher in energy than the $$s$$ orbitals.
• Bonding molecular orbital – lower in energy
• Antibonding molecular orbital – higher in energy

### Description of Bonding

• The molecular orbital model produces electron distributions and energies that agree with our basic ideas of bonding.
• The labels on molecular orbitals indicate their symmetry (shape), the parent atomic orbitals, and whether they are bonding or antibonding.
• Molecular electron configurations can be written in much the same way as atomic electron configurations.
• Each molecular orbital can hold 2 electrons with opposite spins.
• The number of orbitals are conserved.

### Bond Order

• Larger bond order means greater bond strength. $$\text{Bond order} = \frac{\left( \text{# of bonding }e^- \right) - \left( \text{# of antibonding }e^- \right)}{2}$$

### Example: $$\chem{H_2}$$

$$\text{Bond order}=\frac{2-0}{2} = 1$$

### Example: $$\chem{H_2^-}$$

$$\text{Bond order}=\frac{2-1}{2} = \frac{1}{2}$$

## Bonding on Homonuclear Diatomic Molecules

### Homonuclear Diatomic Molecules

• Composed of 2 identical atoms.
• Only the valence orbitals of the atoms contribute significantly to the molecular orbitals of a particular molecule.

### Paramagnetism

• Paramagnetism – substance is attracted into the inducing magnetic field.
• Unpaired electrons ($$\chem{O_2}$$)
• Diamagnetism – substance is repelled from the inducing magnetic field.
• Paired electrons ($$\chem{N_2}$$)

## Bonding in Heteronuclear Diatomic Molecules

### Heteronuclear Diatomic Molecules

• Composed of 2 different atoms.

### Heteronuclear Diatomic Molecule: $$\chem{HF}$$

• The $$2p$$ orbital of fluorine is at a lower energy than the $$1s$$ orbital of hydrogen because fluorine binds its valence electrons more tightly.
• Electrons prefer to be closer to the fluorine atom.
• Thus the $$2p$$ electron on a free fluorine atom is at a lower energy than the $$1s$$ electron on a free hydrogen atom.

### Orbital Energy-Level Diagram for the $$\chem{HF}$$ Molecule

• The diagram predicts that the $$\chem{HF}$$ molecule should be stable because both electrons are lowered in energy relative to their energy in the free hydrogen and fluorine atoms, which is the driving force for bond formation.

### Heteronuclear Diatomic Molecule: HF (cont.)

• The $$\sigma$$ molecular orbital containing the bonding electron pair shows greater electron probability close to the fluorine.
• The electron pair is not shared equally.
• This causes the fluorine atom to have a slight excess of negative charge and leaves the hydrogen atom partially positive.
• This is exactly the bond polarity observed for $$\chem{HF}$$.

## Combining the Localized Electron and Molecular Orbital Models

### Delocalization

• Describes molecules that require resonance.
• In molecules that require resonance, it is the $$\pi$$ bonding that is most clearly delocalized, the $$\sigma$$ bonds are localized.
• $$p$$ orbitals perpendicular to the plane of the molecule are used to form $$\pi$$ molecular orbitals.
• The electrons in the $$\pi$$ molecular orbitals are delocalized above and below the plane of the molecule.

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