chapter 4 molecular structure and orbitals - hsbr1.com · example 4.5 - bond polarity and dipole...

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


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


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



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



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



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



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



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



Table 4.2 - Structures of Molecules that have Four Electron Pairs Around the Central Atom



Table 4.3 - Structures of Molecules with Five Electron Pairs Around the Central Atom



Table 4.3 - Structures of Molecules with Five Electron Pairs Around the Central Atom (Contd)



Concept Check

Determine the shape and bond angles for each of the following molecules:

HCN

PH3

SF4

O3

KrF4



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



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

– .




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:




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




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



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



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:




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




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



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



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



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


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



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



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



Table 4.4 - Types of Molecules with Polar Bonds but No Resulting Dipole Moment



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)




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:




The Cl2 molecule:

The two chlorine atoms share the electrons equally

No bond polarity occurs and the Cl2 molecule has no 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




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




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:




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


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



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



Figure 4.15 - The Formation of sp3 Hybrid Orbitals



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?



Concept Check

Why can’t sp3 hybridization account for the ethylene molecule?



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



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



Figure 4.24 - A Carbon–Carbon Double Bond



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



Figure 4.31 - Bonding in CO2 Part (a)



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



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



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



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



Interactive Example 4.9 - The Localized Electron Model IV



Concept Check

For each of the following molecules, determine:

(a) Bond angle

(b) Expected hybridization of the central atom

NH3 SO2 KrF2 CO2 ICl5



Figure 4.36 - The Relationship of the Number of Effective Pairs, their Spatial Arrangement, and the Hybrid Orbital set Required



Figure 4.36 - The Relationship of the Number of Effective Pairs, their Spatial Arrangement, and the Hybrid Orbital set Required (Contd)



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



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 :




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




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




Each fluorine atom also has four electron pairs and can be assumed to be sp3 hybridized

The BF4– ion’s molecular structure is tetrahedral




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




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


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




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



Figure 4.39 - Bonding and Antibonding Molecular Orbitals (MOs)




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




The number of orbitals are conserved

The number of MOs will always be equal to the number of atomic orbitals used to construct them



Fig 4.40 - Molecular Orbital Energy-Level Diagram of the H2 Molecule



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



Calculation of Bond Order for a H2– ion

2 1 1Bond order = =

1 2

Section 4.5Bonding in Homonuclear Diatomic Molecules


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*



Bonding and Antibonding in Homonuclear Diatomic Molecules



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



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:




σ2p*

2p*

2p

E σ2p

σ2s*

σ2s

The bond order is (8 – 8)/2 = 0, and Ne2 does not exist




(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)




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


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



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



Energy-Level Diagram of Heteronuclear Diatomic Molecules - An Example (HF)

Partial MO energy-level diagram Electron probability distribution



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



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


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


Figure 4.59 - (a) The Molecular Orbital System in Benzene

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