Chapter 8 Chemical Bonding. Bonds Forces that hold groups of atoms together and make them function...

Chapter 8Chemical Bonding

Bonds• Forces that hold groups of atoms together and

make them function as a unit.• We will consider three major categories of

chemical bonds: Ionic Bonds, Covalent Bonds, and Metallic Bonds.

Examples of Bonding Types

Lewis Structures

• Since bonds hold atoms close together, then the valence electrons are responsible for bonding since they are on the outside of an atom.

• It has been recognized for a long time that the noble gases have great chemical stability. With few exceptions they are unreactive or inert.

• The noble gases have 8 valence electrons with the exception of He which has 2.

He 1s2

Ne 1s22s22p6

Ar 1s22s22p63s23p6

Kr 1s22s22p63s23p64s23d104p6

Xe 1s22s22p63s23p64s23d104p65s24d105p6

Lewis Structures

The electronic configuration of the noble gases is described as being energetically stable.

We can draw a Lewis diagram to illustrate the number of valence electrons an atom has.

In a Lewis diagram valence electrons are represented by dots placed above, below and to the left and right of the atoms symbol.

Ee.g. element with 4 valence electrons

Lewis Structures

There are two simple rules to keep in mind when drawing Lewis diagrams:

• Place one dot in each of the four locations before doubling up.• There can be only a maximum of 2 dots in any one location.

E E E E

Lewis Structures

What is the Lewis diagram for H?

H

1. First write the electron configuration:

1s1

2. Identify the number of valence electrons. 1 valence electron.

For a representative element it is easy to identify the number of valence electrons as this is equal to the group number.

Lewis Structures

What is the Lewis diagram for S?

1. First write the electron configuration: [Ne]3s23p4

2. Identify the number of valence electrons. 6 valence electrons

SAlternatively you can recognize that S is in group VIA so has six valence electrons

Lewis Structures

What is the Lewis diagram for S?

1. First write the electron configuration: [Ne]3s23p4

2. Identify the number of valence electrons. 6 valence electrons

S Alternatively you can recognize that S is in group VIA so has six valence electrons

Lewis Structures

LEWIS STRUCTURES OF THE ELEMENTS

IA IIA

SKIP

B’S IIIA IVA VA VIA VIIA VIIA

H

Li

Na

Be

Mg

B

Al

C

Si P

N O F

ClS

He

Ne

Ar

LEWIS STRUCTURES OF IONS

(AFTER REMOVAL OR ADDITION OF ELECTRONS)

IA IIA

SKIP

B’S IIIA IVA VA VIA VIIA VIIA

2+ 4- 3- 2- 1-

H

Li

Na

Be

Mg

B

Al

C

Si P

N O F

ClS

He

Ne

Ar

1+

3+

The octet rule states that:

“Atoms interact in order to obtain a stable octet of eight valence electrons”

The octet rule works extremely well at describing the interactions of the representative elements.

Lewis Structures

One way in which atoms can interact to satisfy the octet rule is by transferring electrons between each other.

Transferring of electrons results in the atoms acquiring net positive and negative charges.

When an atom loses or gains electrons a simple ion is formed.Cations have more protons than electrons and are positive.Anions have more electrons than protons and are negative.

Lewis Structures

Ionic Bonds• Formed from electrostatic attractions of

closely packed, oppositely charged ions.

• Formed when an atom that easily loses electrons reacts with one that has a high electron affinity.

Na[Ne]3s1

Na+ + 1e-

Cl-+ 1e-

[Ne]3s23p5

Cl

Consider a Na atom what happens if it loses one electron?

Consider a Cl atom would you expect it to lose or gain electrons?

[Ne]

[Ne]3s23p6

11 P and 11 e- 11 P and 10 e-

I.E.

E.A.

17 P and 17 e- 17 P and 18 e-

Ionization A Review

Metals tend to lose electrons forming positively charged ions

called cations.

• A representative metal will lose its group number of

electrons to obtain a stable octet.

Na → Na+ + 1e- ( Isoelectronic with Ne)

Mg → Mg2+ + 2e- (isoelectronic with Ne)

What would the charge be of the ion formed by a Li atom?

And which Noble gas is it isoelectronic with?

+1 The ion formed would be Li+

Isoelectronic with He

Non-metals tend to gain electrons forming negatively charged

ions called anions.

• A representative non-metal will gain (8 - group number)

electrons to obtain a stable octet.

O + 2e- → O2- (isoelectronic with Ne)

S + 2e- → S2- (isoelectronic with Ar)

What would the charge be of the ion formed by a I atom?

Which Noble gas is it isoelectronic with?

-1 The ion formed would be I-

Isoelectronic with Xe

Noble Stability

Lewis Structure of NaCl

Na+Cl-Na+Cl-Na+Cl-

Cl-Na+Cl-Na+Cl-Na+

Forces between oppositely charged ions are calledIonic bonds. Each ion is surrounded by an octet of Electrons, thus making the ions stable.

Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.

Ions combine in the ratio

that results in zero

charge to form ionic

compounds.

Which ions are the

smaller ones?

Crystal Lattice of NaCl


Ionic compounds do not exist as discrete molecules. Instead they exist as crystals where ions of opposite charges occupy positions known as lattice sites.

Ions combine in the ratio

that results in zero

charge to form ionic

compounds.

Which ions are the

smaller ones? Sodium



Sodium Chloride Lattice

In our early lectures we defined a molecule as “as a compoundMade of nonmetals.”

Molecules exist as particles containing the number of atoms specified by their formula.

e.g. a water molecule is a particle containing 2 hydrogen atoms and one oxygen atom and has the formula H2O.

Molecular Compounds

Non-metals may also complete their octets by sharing electrons.

This may occur between non-metal atoms of the same type:

e.g. H2, O2, N2, Cl2, F2, I2, etc

Or between different types of non-metal atoms:

e.g. CO2, H2O, CH4, etc

Molecular Compounds

Consider two hydrogen atoms separated by a large distance. Each has 1 electron in a 1s atomic orbital.

+ -

+-

Now lets bring the two atoms together so there orbitals overlap.

Why does the electron stay around the nucleus?

Covalent Bond Formation

+

-

+-

The atomic orbitals overlap to form a new molecular orbital.

This is a stable configuration as each H atom can have a full 1s

susbshell (like He) where the electrons spend most of their time

shared between the atoms. In this arrangement each nucleus

feels an inwards attraction to the two electrons. This is called

covalent bonding.

+

-

+-

This new arrangement of protons and electrons is more stable than separate hydrogen atoms since the attraction of a proton to two electrons is a stronger attraction compared to one proton to one electron of a hydrogen atom.

Lewis Structures• A single bond results when two atoms share

one pair of electrons.

• A lone pair, or unshared pair, of electrons is a pair of electrons that is not shared.

• A bonding pair of electrons is a pair of electrons shared between two atoms.

Multiple Bonds• A double bond results when two atoms

share two pairs of electrons.

• A triple bond results when two atoms share three pairs of electrons

• Bond length is the distance between the nuclear centers of the two atoms jointed together in a bond.

We can draw Lewis diagrams showing the arrangement of valence electrons in covalent compounds. In these diagrams we represent each pair of electrons between atoms as a line.

So for the H2 molecule discussed previously the Lewis diagram would be:

H – HAll other electrons are represented by dots as described previously.

Covalent Bond Formation

Molecular Compounds

Draw Lewis Structures of the following molecular compounds

Ha. H2O O

H

HHO

Note each element has a Noble gas structure by electron sharing

b. NH3 NH H

H

NH

HH

Covalent bonding e’s

Nonbondingelectons

Simplified Lewis Structures

OH H

Straight lines are used to indicate a shared pair, or a covalent bond.

Nonbonding electrons

Lewis Structure ConstructionStep 1 Connect each element with a single line

Step 2 Use the “P” formula to determine the number of extra bonds

Step 3 Insert the extra bonds, to make double or triple bonds.

Step 4 Give each atom an octet of electrons, except hydrogen

Step 5 Determine the formal charge of each element

Examples: Give Lewis Structures for the following

P = 8(n-q) +2q - 2(n-1) - vN = number of atoms in moleculeQ = number of hydrogen atomsV = total number of valence electrons

CO2

H2CO3 SO3 NO2+

Lewis Structure of Carbon Dioxide

CO O

First, connect atoms with lines

Second, use “p” formula to determine the number of extra bonds.

P = 8(n-q) + 2q – 2(n-1) - v

P = 8(3-0) + 2(0) – 2(3-1) - 16P = 24 + 0 – 4 - 16

P = 44 extra bonding electrons2 extra bonds2 extra lines


CO O

Third, add extra lines

Fourth, give each atom an octet of electrons

O C OO C O


CO O



O C OO C O

O C O OOO OC C


CO O



O C OO C O

O C O OOO OC C

Fifth, give each an atom a formal chargeIf the element owns less than its valence, then it is positiveIf the element has more than its valence, then it is negative

OOOOOO C C C


CO O



O C OO C O

O C O OOO OC C


OOOOOO C C C-


CO O



O C OO C O

O C O OOO OC C


OOOOOO C C C+-


CO O



O C OO C O

O C O OOO OC C


OOOOOO C C C+- -


CO O



O C OO C O

O C O OOO OC C


OOOOOO C C C+ +- -

Practice Draw the most stable Lewis structure for CO2.

Practice Draw the most stable Lewis structure for CO2.

OO C

There are actually three possible Lewis structures for SO3.

SO O

O

S OO

O

SO O

O

Each of these three structures is equivalent. We say they are in “resonance” or that they are “resonance structures”.

-

-

-

-

- -

Sulfur Trioxide Lewis Structure

2+ 2+ 2+

Resonance Form Rules

Resonance Structures

Practice Determine the most stable structure for theDetermine the most stable structure for the phosphite ion by calculating formal charge for each phosphite ion by calculating formal charge for each

ion.ion.

OO

O P OO P O

3-3- OO

O P OO P O

3-3-

- -

-

- -

-

Practice ProblemGive the most stable resonance form of H2SO4


S

Step #1 Connect atoms with single bonds in the most symmetrical arrangement possible

O

O

O

OH H


S

Step #2 Find the number of extra bonding electrons with the “P” formula

O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines


S

Step #3 Give each atom an octet of electrons

O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines


S

Step #4 Give each atom a formal charge

O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines

+

-

-

+


S


O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines

+

-

-

+

According to resonance rules this should not be very stable


S


O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines

+

-

-

+

Moving electrons around might improve stability


S


O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines-

+

This arrangement is better, but still not the best.


S


O

O

O

OH H

P=8(7-2)+2(2)-2(7-1)-32=0 NO extra lines

Since sulfur is a period three element, then it can accommodate more than eight electrons. This is the most stable arrangement.

Exceptions to the Octet Rule

Some molecules have less than eight electrons in a Lewis structure.

BF3 is an example

These are known as electron-deficient compoundsOther molecules have more than an octet.

SF6 is an example

Some molecules have an odd number of electrons when summing up the total number of electrons. These don’t obey the octet rule.

NO is an example

Free radicals are molecules having an odd number of valence electrons.

Example

What is the Lewis dot structure of SO32- ion? Does

this ion have resonance structures?

Metallic Bonds

A metallic bond consists of the nuclei of metal atoms surrounded by a “sea” of evenly spaced shared electrons. The nuclei is then attracted to each electron by the same amount, but in different directions, thus making the nuclei stay in a fixed position.

Electron Sea Model

Electronegativity• Electronegativity is a measure of an element’s ability to

attract bonding electrons. It is used for predicting the degree to which bonding pairs of electrons are shared unequally.

Electronegativity Trends

Atomic Size Relationship• Changes in electronegativity are related to

increasing atomic size.• The size of the valence orbitals increases as

the value of the principle quantum number (n) increases.

• Therefore, the atomic size increases and electronegativity decreases as you go down a group.

Bond Types

Bond PolarityBond polarity is a measure of the extent to which bonding electrons are shared between two atoms in a covalent bond.

Polar Covalent Bonds• Compounds that contain two or more different

elements may contain polar covalent bonds.

:C O:• A polar molecule contains bonds that have an uneven

distribution of charge because electrons in the bonds are not shared equally by the two atoms.

Ionization vs. Electronegativities

Practice

Which of the following bonds in each pair are more polar?

C-S or C-O

Cl-Cl or O=O

N-H or C-H

Different Elemental Forms

• Oxygen is found in two forms: oxygen, O2, and ozone, O3.

• Carbon is found in many different forms: soot, diamond, graphite, nanotubes, etc.

• These different forms of an element are known as allotropes.

Evidence for Resonance Structures• Ozone bond lengths were both found to be

128pm• A single bond would have a longer bond

length than a double bond, but in ozone we find the bond lengths are equal meaning they are a mixture of single and double bonds.

Bond LengthsBond length depends on the identity of the atoms as well as on the number of bonds between them.

The bond order is the number of bonds between two atoms.

1 for a single bond

2 for a double bond

3 for a triple bond

Covalent Bond Lengths

Examples of Bond Length

Bond Strengths

Bond Energies• The energy needed to break 1 mole of

covalent bonds in the gas phase is the bond energy of that bond.

• Breaking bonds consumes energy whereas forming bonds releases energy.

• Bond energies can be used to estimate Hrxn.

Hrxn = Hbond breaking + Hbond forming

Energy of Reaction CalculationCalculation ΔH for the following reaction.

CH4 + O2 CO2 + HOH


CH4 + 2 O2 CO2 + 2 HOH

First energy must be added to break the reactant bonds. Bonds to break are 4 C-H bonds and 2 O=O double bond. Next energy is released when 2 C=O bonds are formed and 4 H-O bonds are formed.


CH4 + 2 O2 CO2 + 2 HOH

First energy must be added to break the reactant bonds. Bonds to break are 4 C-H bonds and 2 O=O double bond. Next energy is released when 2 C=O bonds are formed and 4 H-O bonds are formed.

Bonds broken (endothermic) Bonds formed (exothermic)

4 C-H bonds 4(414) = 1656 kj

2 O=O bonds 2(498) = 996 kj

2 C=O 2(799) = 1598 kj

4 H-O 4(464) = 1856 kj

2652 kj 3452 kj

ΔH = 2652 -3452 = 800 kj

Example

The End

Review Questions

ChemTour: Bonding

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This ChemTour shows how ionic and covalent bonds form.

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ChemTour: Lewis Dot Structures


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Students learn to draw and use Lewis dot structures to visualize and represent molecular structures and the locations of valence electrons. Includes Practice Exercises.

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ChemTour: The Periodic Table


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This ChemTour offers a guided tour of the trends summarized by the periodic table (metallic properties, subshells, electronegativity, and atomic radius), and explains how to use this tool to predict an element’s characteristics, including its bonding capacity.

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ChemTour: Resonance


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Molecules that cannot be represented by a single Lewis dot structure are said to exhibit resonance. This ChemTour describes resonance and explains how to represent it visually.

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ChemTour: Expanded Valence Shells


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In this ChemTour, students explore the exceptions to the octet rule, and learn to identify the conditions under which an element will expand its outer electron shell to hold more than 8 electrons. Includes Practice Exercises.

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ChemTour: Estimating Enthalpy Changes


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In this ChemTour, students learn how to use average bond energies to estimate the energy released during a combustion reaction. Includes Practice Exercises.

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Molecular Potential Energy of N2 and O2

The plot to the left shows the molecular potential energy curve for N2. Which of the plots below shows the correct molecular potential energy curve for O2 (blue solid line) compared to N2 (red dashed line)?

A) B) C)

Molecular Potential Energy of N2 and O2

Consider the following arguments for each answer and vote again:

A. The O2 double bond is less stable than the N2 triple bond, so the dissociation energy of O2 should be less than that of N2.

B. An oxygen atom is smaller than a nitrogen atom, so O2 should have a smaller bond length than N2.

C. The fact that O2 has two electrons in antibonding Π* orbitals is reflected in the potential energy curve always being above zero.

Bond Order of Oxygen

The oxygen molecule, O2, has a bond order of 2 in its ground state.

How many electrons can be removed from O2 without altering the bond order? Assume the electrons are always removed from the highest-occupied molecular orbital.

A) 2 B) 4 C) 6

Bond Order of Oxygen

Consider the following arguments for each answer

and vote again:

A. Removing 2 electrons from antibonding orbitals will not affect the bond order.

B. To retain a bond order of 2, a total of 2 electrons must be removed from a bonding orbital and 2 must be removed from an antibonding orbital.

C. Removing 6 electrons will leave 4 electrons in 2 bonding orbitals.

Which of the following three molecules would be made less stable by removing a single electron from the highest-occupied molecular orbital?

A) He2 B) NO C) CN

Stability of He2, CN, and NO

Stability of He2, CN, and NO

Consider the following arguments for each answer and vote again:

A. The 4 electrons in He2 are incapable of holding 2 helium atoms together, so removing 1 electron will only make matters worse.

B. Removing an electron from NO would decrease its bond order from 2.5 to 2, thus decreasing its stability.

C. The highest-occupied molecular orbital for CN is a bonding orbital, so removing an electron will decrease CN's stability.

Lewis Electron Dot Structure of Formaldehyde

Lewis electron dot structures depict the arrangement of valence electrons around the atoms of a molecule, as shown below for water, H2O.

Which of the following is the correct Lewis electron dot structure for formaldehyde, CH2O?

A) B) C)

Lewis Electron Dot Structure of Formaldehyde

Please consider the following arguments for each

answer and vote again:

A. The octet rule is satisfied, and the formal charges of all of the atoms are zero.

B. The carbon-oxygen triple bond lends extra stability to this structure.

C. The octet rule is satisfied without the need to form double bonds.

Bonding Structure of NO2-

How many bonds would it take to complete the structure of the nitrite ion, pictured to the left?

A) 0 B) 1 C) 2

Bonding Structure of NO2-



A. All of the atoms are attached, so more bonds are not needed.

B. The correct overall charge and the octet rule are satisfied by 1 single bond and 1 double bond.

C. Both oxygens are double bonded to the nitrogen in order for their formal charges to be zero.

Molecular Structures of Pentane

Which of the following depicts a molecule that is different from the one shown to the left?

A) B) C)

Molecular Structures of Pentane



A. This molecule, known as isopentane, has a unique T-shaped arrangement.

B. In this molecule, only two carbon atoms are bonded to the second carbon.

C. This molecule is unbranched, whereas the one pictured in the question is not.

Chapter 8 Chemical Bonding. Bonds Forces that hold groups of atoms together and make them function...

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