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Chapter 4
Molecular Structure and Orbitals
Chapter 4Table of Contents
Copyright ©2016 Cengage Learning. All Rights Reserved.
(4.1) Molecular structure: The VSEPR model
(4.2) Bond polarity and dipole moments
(4.3) Hybridization and the localized electron model
(4.4) The molecular orbital model
(4.5) Bonding in homonuclear diatomic molecules
(4.6) Bonding in heteronuclear diatomic molecules
(4.7) Combining the localized electron and molecular orbital models
Chapter 4
Copyright ©2016 Cengage Learning. All Rights Reserved.
Question to Consider
The spicy flavor of chili peppers is attributed to a complex molecule with multiple hybridizations
What is the name of this molecule?
Section 4.1Molecular Structure: The VSEPR Model
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The VSEPR Model
Molecular structure is the three-dimensional arrangement of atoms in a molecule
Valence shell electron-pair repulsion (VSEPR) model
The structure around a given atom is determined principally by minimizing electron pair repulsions
Section 4.1Molecular Structure: The VSEPR Model
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Electron Structures
Linear structure can be observed in BeCl2 Each electron pair on Be is shared with a Cl atom
BF3 shows a trigonal planar structure
Each electron pair is shared with a fluorine atom
Section 4.1Molecular Structure: The VSEPR Model
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Electron Structures
When there are four pairs of electrons around an atom, they take up a tetrahedral structure
The bond angle for such a structure is 109.5 degrees
In the presence of a lone pair, the molecular structure is a trigonal pyramid
Section 4.1Molecular Structure: The VSEPR Model
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Electron Structures
Consider the structure of NH3, which has one lone pair
The arrangement of electron pairs is tetrahedral, but the arrangement of atoms is not
Section 4.1Molecular Structure: The VSEPR Model
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Lone Pair Trends
Bonding pairs are shared between two nuclei
Electrons can be close to either nucleus
Lone pairs center around just one nucleus, and both electrons choose that nucleus
Lone pairs need more space than bonding pairs
They compress the angles between bonding pairs
Section 4.1Molecular Structure: The VSEPR Model
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Electron Structures
When there are five electron pairs, the structure that produces minimal repulsion is a trigonal bipyramid
It consists of two trigonal-based pyramids that share a common base
The best arrangement for six pairs of electrons around a given atom is the octahedral structure
This structure has 90-degree bond angles
Section 4.1Molecular Structure: The VSEPR Model
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Table 4.2 - Structures of Molecules that have Four Electron Pairs Around the Central Atom
Section 4.1Molecular Structure: The VSEPR Model
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Table 4.3 - Structures of Molecules with Five Electron Pairs Around the Central Atom
Section 4.1Molecular Structure: The VSEPR Model
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Table 4.3 - Structures of Molecules with Five Electron Pairs Around the Central Atom (Contd)
Section 4.1Molecular Structure: The VSEPR Model
Copyright ©2016 Cengage Learning. All Rights Reserved.
Concept Check
Determine the shape and bond angles for each of the following molecules:
HCN
PH3
SF4
O3
KrF4
Section 4.1Molecular Structure: The VSEPR Model
Copyright ©2016 Cengage Learning. All Rights Reserved.
Problem Solving Strategy - Steps to Apply the VSEPR Model
Draw the Lewis structure for the molecule
Count the electron pairs and arrange them in the way that minimizes repulsion
Put the pairs as far apart as possible
Determine the positions of the atoms from the way the electron pairs are shared
Determine the name of the molecular structure from the positions of the atoms
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.2 - Prediction of Molecular Structure II
When phosphorus reacts with excess chlorine gas, the compound phosphorus pentachloride (PCl5) is formed. In the gaseous and liquid states, this substance consists of PCl5molecules, but in the solid state it consists of a 1:1 mixture of PCl4
+ and PCl6– ions. Predict the geometric structures of PCl5,
PCl4+ , and PCl6
– .
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.2 - Prediction of Molecular Structure II
Solution
The Lewis structure for PCl5 is shown
Five pairs of electrons around the phosphorus atom require a trigonal bipyramidal arrangement
When the chlorine atoms are included, a trigonal bipyramidal molecule results:
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.2 - Prediction of Molecular Structure II
The Lewis structure for the PCl4+ ion [5 + 4(7) – 1 = 32 valence
electrons] is shown below
There are four pairs of electrons surrounding the phosphorus atom in the PCl4
+ ion, which requires a tetrahedral arrangement of the pairs
Since each pair is shared with a chlorine atom, a tetrahedral PCl4
+ cation results
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.2 - Prediction of Molecular Structure II
The Lewis structure for PCl6– [5 + 6(7) + 1 = 48 valence
electrons] is shown below
Since phosphorus is surrounded by six pairs of electrons, an octahedral arrangement is required to minimize repulsions, as shown below in the center
Since each electron pair is shared with a chlorine atom, an octahedral PCl6
– anion is predicted
Section 4.1Molecular Structure: The VSEPR Model
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The VSEPR Model and Multiple Bonds
While using the VSEPR model, a double bond must be considered as one effective pair
The two pairs involved in the double bond are not independent pairs
The double bond acts as one center of electron density that repels other electron pairs
With molecules that exhibit resonance, any one of the resonance structures can be used to predict its molecular structure using the VSEPER model
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.4 - Structures of Molecules with Multiple Bonds
Predict the molecular structure of the sulfur dioxide molecule. Is this molecule expected to have a dipole moment?
Solution
First, determine the Lewis structure for the SO2 molecule, which has 18 valence electrons
The expected resonance structures are:
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.4 - Structures of Molecules with Multiple Bonds
To determine the molecular structure, count the electron pairs around the sulfur atom
In each resonance structure the sulfur has one lone pair, one pair in a single bond, and one double bond
Counting the double bond as one pair yields three effective pairs around the sulfur
A trigonal planar arrangement is required, which yields a V-shaped molecule
Section 4.1Molecular Structure: The VSEPR Model
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Interactive Example 4.4 - Structures of Molecules with Multiple Bonds
Thus the structure of the SO2 molecule is expected to be V-shaped, with a 120-degree bond angle
The molecule has a dipole moment directed as shown:
Since the molecule is V-shaped, the polar bonds do not cancel
Section 4.1Molecular Structure: The VSEPR Model
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Molecules Containing No Single Central Atom
The VSEPR model can accurately determine the structure of complicated molecules such as methanol
Lewis structure:
There are four pairs of electrons around the C and O atoms, which give rise to a tetrahedral arrangement
Space requirements of the lone pairs distort the arrangement
Section 4.1Molecular Structure: The VSEPR Model
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Figure 4.8 - The Molecular Structure of Methanol
(a) The arrangement of electron pairs and atoms around the carbon
(b) The arrangement of bonding and lone pairs around oxygen (c) The molecular structure
Section 4.1Molecular Structure: The VSEPR Model
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Accuracy of the VSEPR Model
It aptly predicts the molecular structures of most molecules formed from non-metallic elements
It can be used to predict the structures of molecules with hundreds of atoms
It fails to determine the molecular structure in certain instances
Phosphine (PH3) and ammonia (NH3) have similar Lewis structures but different bond angles—94 degrees and 107 degrees, respectively
Section 4.2Bond Polarity and Dipole Moments
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Dipole Moment
A molecule that has a center of positive charge and a center of negative charge is said to be dipolar or to possess dipole moment
It is represented by an arrow pointing to the negative charge center
The tail indicates the positive charge center
Section 4.2Bond Polarity and Dipole Moments
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Dipole Moment
Electrostatic potential diagrams can also be used to represent dipole moment
The colors of visible light are used to show variation in distribution of charge
Red - Most electron-rich region
Blue - Most electron-poor region
Section 4.2Bond Polarity and Dipole Moments
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Bond Polarity Trends
Any diatomic molecule with polar bonds will exhibit dipole moments
This behavior can also be exhibited by polyatomic molecules
Few molecules possess polar bonds but lack dipole moment
Occurs when the individual bond polarities are arranged in a manner that they cancel each other out
Section 4.2Bond Polarity and Dipole Moments
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Table 4.4 - Types of Molecules with Polar Bonds but No Resulting Dipole Moment
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
For each of the following molecules, show the direction of the bond polarities and indicate which ones have a dipole moment:
HCl
Cl2 SO3 (planar molecule with the oxygen atoms spaced evenly
around the central sulfur atom)
CH4 (tetrahedral with the carbon atom at the center)
H2S (V-shaped with the sulfur atom at the point)
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
Solution
The HCl molecule:
The electronegativity of chlorine is greater than that of hydrogen
Thus the chlorine will be partially negative, and the hydrogen will be partially positive
The HCl molecule has a dipole moment:
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
The Cl2 molecule:
The two chlorine atoms share the electrons equally
No bond polarity occurs and the Cl2 molecule has no dipole moment
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
The SO3 molecule:
The electronegativity of oxygen is greater than that of sulfur
This means that each oxygen will have a partial negative charge, and the sulfur will have a partial positive charge
The bond polarities arranged symmetrically as shown cancel, and the molecule has no dipole moment
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
The CH4 molecule:
Carbon has a slightly higher electronegativity than does hydrogen
This leads to small partial positive charges on the hydrogen atoms and a small partial negative charge on the carbon:
The bond polarities cancel, and the molecule has no dipole moment
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
The H2S molecule:
Since the electronegativity of sulfur is slightly greater than that of hydrogen, the sulfur will have a partial negative charge, and the hydrogen atoms will have a partial positive charge, which can be represented as:
Section 4.2Bond Polarity and Dipole Moments
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Example 4.5 - Bond Polarity and Dipole Moment
This case is analogous to the water molecule, and the polar bonds result in a dipole moment oriented as shown:
Section 4.3Hybridization and the Localized Electron Model
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Hybridization
It refers to the mixing of the native atomic orbitals to form special orbitals for bonding
Atoms may adopt a different set of atomic orbitals or hybrid orbitals from those in the free state
Section 4.3Hybridization and the Localized Electron Model
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sp3 Hybridization
It can be observed upon combination of one 2s and three 2porbitals
Whenever an atom requires a set of equivalent tetrahedral atomic orbitals, this model assumes that the atom adopts a set of sp3 orbitals
The atom becomes sp3 hybridized
Section 4.3Hybridization and the Localized Electron Model
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Figure 4.15 - The Formation of sp3 Hybrid Orbitals
Section 4.3Hybridization and the Localized Electron Model
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Concept Check
What is the valence electron configuration of a carbon atom?
Why can’t the bonding orbitals for methane be formed by an overlap of atomic orbitals?
Section 4.3Hybridization and the Localized Electron Model
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Concept Check
Why can’t sp3 hybridization account for the ethylene molecule?
Section 4.3Hybridization and the Localized Electron Model
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sp2 Hybridization
Gives a trigonal planar arrangement of atomic orbitals with bond angles of 120 degrees
It occurs on the combination of one 2s and two 2p orbitals
One p orbital is not used
It is oriented perpendicular to the plane of the sp2 orbitals
Section 4.3Hybridization and the Localized Electron Model
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Types of sp2 Hybridized Bonds
Sigma () bond
Electron pair is shared in an area centered on a line running between the atoms
Pi () bond
Forms double and triple bonds by sharing electron pair(s) in the space above and below the σ bond using the unhybridized porbitals
A double bond always consists of one bond and one bond
If an atom is surrounded by three effective pairs, a set of sp2
hybrid orbitals is required
Section 4.3Hybridization and the Localized Electron Model
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Figure 4.24 - A Carbon–Carbon Double Bond
Section 4.3Hybridization and the Localized Electron Model
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sp Hybridization
It occurs upon combination of one s and one p orbital
Two effective pairs around an atom always require sphybridization of that atom
It follows a linear arrangement of atomic orbitals
p orbitals that remain unchanged upon hybridization are used
in the formation of bonds
Section 4.3Hybridization and the Localized Electron Model
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Figure 4.31 - Bonding in CO2 Part (a)
Section 4.3Hybridization and the Localized Electron Model
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Concept Check
Draw the Lewis structure for HCN
Which of the hybrid orbitals are used?
Draw HCN and:
Show all the bonds between the atoms
Label each or bond
Section 4.3Hybridization and the Localized Electron Model
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dsp3 Hybridization
It is a combination of one d, one s, and three p orbitals
It results in a trigonal bipyramidal arrangement of five equivalent hybrid orbitals
The image illustrates hybrid
orbitals in a phosphorus atom
Section 4.3Hybridization and the Localized Electron Model
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d2sp3 Hybridization
An atom is d2sp3 hybridized when there is a combination of two d, one s, and three p orbitals
It results in an octahedral arrangement of six equivalent hybrid orbitals
The image illustrates the
orbitals in a sulfur atom
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.9 - The Localized Electron Model IV
How is the xenon atom in XeFe4 hybridized?
Solution
XeFe4 has six pairs of electrons around xenon that are arranged octahedrally to minimize repulsions
An octahedral set of six atomic orbitals is required to hold these electrons, and the xenon atom is d2sp3 hybridized
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.9 - The Localized Electron Model IV
Section 4.3Hybridization and the Localized Electron Model
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Concept Check
For each of the following molecules, determine:
(a) Bond angle
(b) Expected hybridization of the central atom
NH3 SO2 KrF2 CO2 ICl5
Section 4.3Hybridization and the Localized Electron Model
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Figure 4.36 - The Relationship of the Number of Effective Pairs, their Spatial Arrangement, and the Hybrid Orbital set Required
Section 4.3Hybridization and the Localized Electron Model
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Figure 4.36 - The Relationship of the Number of Effective Pairs, their Spatial Arrangement, and the Hybrid Orbital set Required (Contd)
Section 4.3Hybridization and the Localized Electron Model
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Problem Solving Strategy: 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
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
For each of the following molecules or ions, predict the hybridization of each atom, and describe the molecular structure
a. CO b. BF4– c. XeF2
Solution
a. CO
The CO molecule has 10 valence electrons, and its Lewis structure is
Each atom has two effective pairs, which means that both are sp hybridized
: C O :
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
The triple bond consists of a σ bond produced by the overlap of
an sp orbital from each atom and two bonds produced by the overlap of 2p orbitals from each atom
The lone pairs are in the sp orbitals
Since the CO molecule has only two atoms, it must be linear
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
b. BF4–
The BF4– ion has 32 valence electrons
The Lewis structure shows four pairs of electrons around the boron atom, which means a tetrahedral arrangement:
This requires sp3 hybridization of the boron atom
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
Each fluorine atom also has four electron pairs and can be assumed to be sp3 hybridized
The BF4– ion’s molecular structure is tetrahedral
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
c. XeF2
The XeF2 molecule has 22 valence electrons
The Lewis structure shows five electron pairs on the xenon atom, which requires trigonal bipyramidal arrangement:
Note that the lone pairs are placed in the plane where they are 120 degrees apart
Section 4.3Hybridization and the Localized Electron Model
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Interactive Example 4.10 - The Localized Electron Model V
To accommodate five pairs at the vertices of a trigonal bipyramid requires that the xenon atom adopt a set of five dsp3 orbitals
Each fluorine atom has four electron pairs and can be assumed to be sp3 hybridized
The XeF2 molecule has a linear arrangement of atoms
Section 4.4The Molecular Orbital Model
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Properties of Molecular Orbitals
The electron probability of both MOs is centered along the line passing through the two nuclei
MO1 - Greatest electron probability is between the nuclei
MO2 - Greatest electron probability is on either sides of the nuclei
MO1 and MO2 are referred to as sigma (σ) molecular orbitals
In the molecule, only the molecular orbitals are available for occupation by electrons
Section 4.4The Molecular Orbital Model
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Properties of Molecular Orbitals
MO1 is lower in energy than the 1s orbitals of free atoms, while MO2 is higher in energy than the 1s orbitals
A bonding molecular orbital is lower in energy than the atomic orbitals from which it is composed
An antibonding molecular orbital is higher in energy than the atomic orbitals from which it is composed
Each molecular orbital can hold 2 electrons with opposite spins
Section 4.4The Molecular Orbital Model
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Figure 4.39 - Bonding and Antibonding Molecular Orbitals (MOs)
Section 4.4The Molecular Orbital Model
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Properties of Molecular Orbitals
The MO model is physically reasonable
In bonding MOs, electrons have a higher probability of being between the nuclei
In antibonding MOs, electrons are mainly outside the space between the nuclei
The labels on MOs indicate their symmetry, the parent atomic orbitals, and whether they are bonding or antibonding
Molecular electron configurations can be written in a manner similar to atomic electron configurations
Section 4.4The Molecular Orbital Model
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Properties of Molecular Orbitals
The number of orbitals are conserved
The number of MOs will always be equal to the number of atomic orbitals used to construct them
Section 4.4The Molecular Orbital Model
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Fig 4.40 - Molecular Orbital Energy-Level Diagram of the H2 Molecule
Section 4.4The Molecular Orbital Model
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Bond Order
It refers to the difference between the number of bonding electrons and the number of antibonding electrons divided by 2
2 is considered given the fact that electrons are taken in the form of pairs
Larger bond order is generally related to greater bond strength
number of bonding electrons number of antibonding electronsBond order =
2
Section 4.4The Molecular Orbital Model
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Calculation of Bond Order for a H2– ion
2 1 1Bond order = =
1 2
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Homonuclear Diatomic Molecules
These molecules are composed of two identical atoms
The valence orbitals significantly contribute to the MO of a particular molecule
Electron probability in such molecules is high above and below the line between the nuclei
Both orbitals are pi () molecular orbitals
Bonding MO - 2p
Antibonding MO - 2p*
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Bonding and Antibonding in Homonuclear Diatomic Molecules
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Paramagnetism
Paramagnetism causes a substance to be attracted into the inducing magnetic field
Associated with unpaired electrons
Diamagnetism causes a substance to be repelled from the inducing magnetic field
Associated with paired electrons
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Interactive Example 4.12 - The Molecular Orbital Model II
Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules
(a) Ne2 (b) P2
Solution
(a) Ne2
The valence orbitals for Ne are 2s and 2p
Construct the MO structure
The Ne2 molecule has 16 valence electrons (8 from each atom)
Placing these electrons in the appropriate molecular orbitals produces the following diagram:
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Interactive Example 4.12 - The Molecular Orbital Model II
σ2p*
2p*
2p
E σ2p
σ2s*
σ2s
The bond order is (8 – 8)/2 = 0, and Ne2 does not exist
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Interactive Example 4.12 - The Molecular Orbital Model II
(b) P2
The P2 molecule contains phosphorus atoms from the third row of the periodic table
Assume that the diatomic molecules of the Period 3 elements can be treated in a way very similar which has been seen so far
Draw the MO diagram for P2 analogous to that for N2
The only change will be that the molecular orbitals will be formed from 3s and 3p atomic orbitals
The P2 molecule has 10 valence electrons (5 from each phosphorus atom)
Section 4.5Bonding in Homonuclear Diatomic Molecules
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Interactive Example 4.12 - The Molecular Orbital Model II
The resulting molecular orbital diagram is:
σ3p*
3p*
σ3p
E 3p
σ3s*
σ3s
The molecule has a bond order of 3 and is expected to be diamagnetic
Section 4.6Bonding in Heteronuclear Diatomic Molecules
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Heteronuclear Diatomic Molecules
These molecules are composed of two different atoms
They may also involve, in special cases, molecules that contain atoms adjacent to each other in the periodic table
They are best described by the MO model
Section 4.6Bonding in Heteronuclear Diatomic Molecules
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Energy-Level Diagram of Heteronuclear Diatomic Molecules - An Example (HF)
Consider the HF molecule and assume that fluorine uses only its 2p orbitals to bond to hydrogen
The MOs for HF are composed of fluorine 2p and hydrogen 1sorbitals
The fluorine 2p orbital is lower in energy than the hydrogen 1s orbital
The σ MO holding the binding electron pairs will show higher electron probability closer to fluorine
Electron pairs are not equally shared
Section 4.6Bonding in Heteronuclear Diatomic Molecules
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Energy-Level Diagram of Heteronuclear Diatomic Molecules - An Example (HF)
Partial MO energy-level diagram Electron probability distribution
Section 4.6Bonding in Heteronuclear Diatomic Molecules
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Interactive Example 4.13 - The Molecular Orbital Model III
Use the molecular orbital model to predict the magnetism and bond order of the NO+ and CN– ions
Solution
The NO+ ion has 10 valence electrons (5 + 6 – 1)
The CN– ion also has 10 valence electrons (4 + 5 + 1)
Both ions are therefore diamagnetic and have a bond order derived from the equation
8 2 = 3
2
Section 4.6Bonding in Heteronuclear Diatomic Molecules
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Interactive Example 4.13 - The Molecular Orbital Model III
The molecular orbital diagram for these two ions is the same
σ2p*
2p*
σ2p
E 2p
σ2s*
σ2s
Section 4.7Combining the Localized Electron and Molecular Orbital Models
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An Introduction
The ideal bonding model must contain the following:
Simplicity of the localized electron model
Delocalization feature of the molecular orbital model
Combining both models will help describe molecules that require resonance
σ bond can be described as being localized
bonding can be treated as being delocalized
Section 4.7Combining the Localized Electron and Molecular Orbital Models
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Figure 4.59 - (a) The Molecular Orbital System in Benzene