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© 2014 Pearson Education, Inc.

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 9

Chemical

Bonding I: The

Lewis Model

(2)

© 2014 Pearson Education, Inc.

Pg. 471

(3)

Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine

Two possible skeletal structures of formaldehyde (CH2O) 3

H C O H H

C O H

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis

structure.

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

(4)

Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine

4

H C O H

C – 4 e- O – 6 e- 2H – 2x1 e-

12 e-

2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12

formal charge

on C = 4 - 2 - ½ x 6 = -1

formal charge

on O = 6 - 2 - ½ x 6 = +1

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

-1 +1

(5)

Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine

5 C – 4 e-

O – 6 e- 2H – 2x1 e-

12 e-

2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12

H

C O H

formal charge

on C = 4 - 0 - ½ x 8 = 0

formal charge

on O = 6 - 4 - ½ x 4 = 0

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

0 0

(6)

Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine

Formal Charge and Lewis Structures

6

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which

formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1 H

C O H

0 0

(7)

© 2014 Pearson Education, Inc.

9.1 Bonding Models and AIDS Drugs

• HIV-protease is a protein synthesized by the human immunodeficiency virus (HIV).

• This particular protein is crucial to the virus’s ability to multiply and cause AIDS.

• Pharmaceutical companies designed molecules that would disable HIV- protease by sticking to the molecule’s active site—protease inhibitors.

• To design such a molecule, researchers used bonding theories to simulate the shape of potential drug molecules and how they would interact with the protease molecule.

(8)

© 2014 Pearson Education, Inc.

Lewis Bonding Theory

• One of the simplest bonding theories is called Lewis theory.

• Lewis theory emphasizes valence electrons to explain bonding.

• Using Lewis theory, we can draw models, called Lewis structures.

– Also known as electron dot structures

• Lewis structures allow us to predict many properties of molecules.

– Such as molecular stability, shape, size, and polarity

(9)

© 2014 Pearson Education, Inc.

Bonding Theories

• Explain how and why atoms attach together to form molecules

• Explain why some combinations of atoms are stable and others are not

– Why is water H2O, not HO or H3O?

• Can be used to predict the shapes of molecules

• Can be used to predict the chemical and physical properties of compounds

(10)

© 2014 Pearson Education, Inc.

Why Do Atoms Bond?

• Chemical bonds form because they lower the potential

energy between the charged particles that compose atoms.

• A chemical bond forms when the potential energy of the bonded atoms is less than the potential energy of the separate atoms.

• To calculate this potential energy, you need to consider the following interactions:

– Nucleus-to-nucleus repulsions – Electron-to-electron repulsions – Nucleus-to-electron attractions

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© 2014 Pearson Education, Inc.

9.2 Types of Bonds

• We can classify bonds based on the kinds of

atoms that are bonded together.

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© 2014 Pearson Education, Inc.

Types of Bonding

INSERT FIGURE 9.1

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© 2014 Pearson Education, Inc.

Ionic Bonds

• When a metal atom loses electrons it becomes a cation.

– Metals have low ionization energy, making it relatively easy to remove electrons from them.

• When a nonmetal atom gains electrons it becomes an anion.

– Nonmetals have high electron affinities, making it advantageous to add electrons to these atoms.

• The oppositely charged ions are then attracted to each other, resulting in an ionic bond.

(14)

© 2014 Pearson Education, Inc.

Covalent Bonds

• Nonmetal atoms have relatively high ionization energies, so it is difficult to remove electrons from them.

• When nonmetals bond together, it is better in terms of

potential energy for the atoms to share valence electrons.

– Potential energy is lowest when the electrons are between the nuclei.

• Shared electrons hold the atoms together by attracting nuclei of both atoms.

(15)

© 2014 Pearson Education, Inc.

Metallic Bonds

• The relatively low ionization energy of metals allows them to lose electrons easily.

• The simplest theory of metallic bonding involves the metal atoms releasing their valence electrons to be shared as a pool by all the atoms/ions in the metal.

– An organization of metal cation islands in a sea of electrons

– Electrons delocalized throughout the metal structure

• Bonding results from attraction of cation for the delocalized electrons.

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© 2014 Pearson Education, Inc.

Metallic Bonding

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© 2014 Pearson Education, Inc.

9.3 Representing Valence Electrons with Dots Valence Electrons and Bonding

9.1

Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that

particpate in chemical bonding.

價電子是原子的外殼層電子。

價電子是參與化學鍵結的電子。

1A ns1 1

2A ns2 2

3A ns2np1 3

4A ns2np2 4

5A ns2np3 5

6A ns2np4 6

7A ns2np5 7

Group e- configuration # of valence e-

(18)

9.1

Octet rule

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© 2014 Pearson Education, Inc.

Octet Rule

• When atoms bond, they tend to gain, lose, or share electrons to result in eight valence electrons.

• ns2np6

– Noble gas configuration

• Many exceptions

– H, Li, Be, B attain an electron configuration like He.

• He = two valence electrons, a duet.

• Li loses its one valence electron.

• H shares or gains one electron.

o Though it commonly loses its one electron to become H+

• Be loses two electrons to become Be2+.

o Though it commonly shares its two electrons in covalent bonds, resulting in four valence electrons

• B loses three electrons to become B3+.

o Though it commonly shares its three electrons in covalent bonds, resulting in six valence electrons

– Expanded octets for elements in period 3 or below

• Using empty valence d orbitals

(20)

Q:The general electron configuration for noble gas atoms is:

A. ns

2

np

6

B. ns

2

np

5

C. ns

2

np

4

D. ns

2

np

3

E. ns

2

Answer: A

(21)

© 2014 Pearson Education, Inc.

9.4 Ionic Bonding: Lewis Symbols and Lattice Energies

9.2

Li + F Li+ F

- The Ionic Bond

離子鍵

1s22s1 1s22s22p5 1s2 1s22s22p6 [He] [Ne]

Li Li+ + e- e- + F F

-

F

-

Li+ + Li+ F

-

(22)

Q:Which one of the following

compounds is most likely to be an ionic ompound?

A. KF B. CCl

4

C. CS

2

D. CO

2

E. ICl

Answer: A

(23)

9.3

Lattice energy (E) increases as Q increases and/or as r decreases.當 Q 增加 和 / 或 r 減少可以增加靜電能

cmpd lattice energy MgF2

MgO LiF LiCl

2957 3938 1036

853

Q= +2,-1 Q= +2,-2

r F < r Cl

Ion bond and Electrostatic (Lattice) Energy 靜電(晶格)能

E = k Q+Q- r

Q+ is the charge on the cation陽離子電荷 Q- is the charge on the anion陰離子電荷

r is the distance between the ions 離子間的距離

Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.晶格(

靜電

)

(E)是完全分離一莫耳的固態離子化合物轉為氣態離子所需的能量

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© 2014 Pearson Education, Inc.

Determining Lattice Energy: The Born–Haber Cycle

• The Born–Haber cycle is a hypothetical series of reactions that represents the formation of an ionic compound from its constituent elements.

• The reactions are chosen so that the change in enthalpy of each reaction is known except for the last one, which is the lattice energy.

9.3

決定靜電能

DHoverallo = DHo 1 + DHo 2 + DHo 3 + DHo 4 + DHo 5

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© 2014 Pearson Education, Inc.

Na(s) + ½ Cl2(g) → NaCl(s) Enthalpy of standard formation:

ΔHfo = -411 kJ

(26)

Prentice Hall © 2005

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine

26

Born-Haber Cycle Explanation

1. Enthalpy of sublimation:

Na(s) → Na(g) ΔH1 = +107 kJ 2. Bond dissociation energy:

½ Cl2 (g) → Cl (g) ΔH2 = +122 kJ 3. First ionization energy:

Na(g) → Na+(g) + e- ΔH3 = +496 kJ 4. Electron affinity:

Cl(g) + e- → Cl-(g) ΔH4 = -349 kJ 5. Lattice energy:

Na+(g) + Cl-(g) → NaCl(s)(計算才知道)ΔH5 = -787 kJ 6. Enthalpy of standard formation:

Na(s) + ½ Cl2(g) → NaCl(s) ΔHfo = -411 kJ ΔHfo = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 = -411 kJ

(27)

© 2014 Pearson Education, Inc.

Born–Haber Cycle for NaCl

DH°f(NaCl, s) = DH°f(Na atoms,g) + DH°f(Cl atoms, g) + DH°f(Na+,g) + DH°f(Cl,g) + DH°(NaCl lattice)

DH°f(NaCl, s) = DH°f(Na atoms,g) + DH°f(Cl–Cl bond energy) + Na 1st ionization energy + Cl electron affinity + NaCl lattice energy

Na(s) → Na(g) DHf(Na,g)

½ Cl2(g) → Cl(g) DHf(Cl,g) Na(g) → Na+(g) DHf(Na+,g) Cl (g) → Cl(g) DHf(Cl,g)

Na+ (g) + Cl(g) → NaCl(s) DH (NaCl lattice) Na(s) + ½ Cl2(g) → NaCl(s) DHf (NaCl, s)

Na(s) → Na(g) DHf(Na,g)

½ Cl2(g) → Cl(g) DHf(Cl,g) Na(g) → Na+(g) DHf(Na+,g) Cl (g) → Cl(g) DHf(Cl,g)

Na+ (g) + Cl(g) → NaCl(s) DH (NaCl lattice)

Na(s) → Na(g) +108 kJ

½ Cl2(g) → Cl(g) +½ (244 kJ) Na(g) → Na+(g) +496 kJ Cl (g) → Cl(g) −349 kJ

Na+ (g) + Cl(g) → NaCl(s) DH (NaCl lattice) Na(s) + ½ Cl2(g) → NaCl(s) −411 kJ

NaCl lattice energy = DH°f(NaCl, s) − [DH°f(Na atoms, g) + DH°f(Cl–

Cl bond energy) + Na 1st ionization energy + Cl electron affinity ]

NaCl lattice energy = (−411 kJ)

− [(+108 kJ) + (+122 kJ) + (+496 kJ) + (−349 kJ) ]

= −788 kJ

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© 2014 Pearson Education, Inc.

Trends in Lattice Energy: Charge and Ion Size

E = k Q+Q- r

9.3

Lattice energy (E) increases as Q increases and/or as r decreases.當 Q 增加 或 r 減少可以增加靜電能

cmpd lattice energy NaF

CaO LiCl NaCl

910 3414

834 788

Q= +1,-1 Q= +2,-2

r Li < r Na Q+ is the charge on the cation陽離子電荷 Q- is the charge on the anion陰離子電荷

r is the distance between the ions 離子間的距離

(29)

© 2014 Pearson Education, Inc.

• Hard and brittle crystalline solids

– All are solids at room temperature

• Melting points generally > 300 C

• The liquid state conducts electricity

– The solid state does not conduct electricity

• Many are soluble in water

– The solution conducts electricity well

Properties of Ionic Compounds

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© 2014 Pearson Education, Inc.

Conductivity of NaCl

In NaCl(aq), the ions are separated and allowed to

move to the charged rods.

In NaCl(s), the ions are stuck in position and not allowed to move to the charged rods.

(31)

© 2014 Pearson Education, Inc.

9.5 Lewis Theory of Covalent Bonding

The covalent bonding is a chemical bond in which two or more electrons are shared by two atoms.共價鍵是兩個原子共 用兩個或更多電子的化學鍵

Why should two atoms share electrons?

為什麼兩個原子能夠分享電子?

F + F

7e- 7e-

F F

8e- 8e-

F F

F F

Lewis structure of F2

lone pairs lone pairs

lone pairs lone pairs

single covalent bond

single covalent bond

9.4

(32)

© 2014 Pearson Education, Inc.

Covalent Bonding: Bonding and Lone Pair Electrons

• Electrons that are shared by atoms are called bonding pairs.

• Electrons that are not shared by atoms but belong to a particular atom are called

lone pairs.

Also known as nonbonding pairs

(33)

8e-

H + O + H H O H or H O H 2e- 2e-

Lewis structure of water水的路易士結構

Double bond – two atoms share two pairs of electrons

雙鍵

– 兩個原子分享兩個電子對

single covalent bonds

O C O or O C O

8e-8e-8e-

double bonds double bonds

Triple bond – two atoms share three pairs of electrons

三鍵

– 兩個原子分享三個電子對

N N 8e- 8e-

N N

triple bond triple bond

or

Single bond – two atoms share one pairs of electrons

單鍵

– 兩個原子分享一個電子對

(34)

© 2014 Pearson Education, Inc.

Predictions of Molecular Formulas by Lewis Theory

Oxygen is more stable when it is singly bonded to two other atoms.

• Molecular compounds have low melting points and boiling points.

– MP generally < 300 °C

– Molecular compounds are found in all three states at room temperature.

(35)

© 2014 Pearson Education, Inc.

Intermolecular Attractions versus Bonding

(36)

© 2014 Pearson Education, Inc.

9.6 Electronegativity and Bond Polarity Polar Covalent Bonding

Polar covalent bond is a covalent bond with greater electron density around one of the two atoms

極性共價鍵

極性鍵結

一個共價鍵中兩個原子的一方原子周圍有較高電子密度。

H F

electron rich region

電子富有區域

electron poor region電子缺乏區域

e- rich e- poor

d+ d-

EN 2.1 EN 4.0

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© 2014 Pearson Education, Inc.

Electronegativity

L

L

Large: atomic radius(原子半徑)

Metallic (金屬)

Large: Ionization energy游離能

Electron affinity電子親和力 Nonmetallic 非金屬

Electronegativity 電負度

The ability of an atom to attract bonding electrons to itself is called electronegativity.電負度是在一個化學鍵中一個 原子吸引接近它自己的電子的能力

(38)

© 2014 Pearson Education, Inc.

Bond Polarity

Classification of bonds by difference in electronegativity 鍵結的分類是依照電負度的差異

離子

Covalent share e- 共享電子

Polar Covalent

partial transfer of e- 電子的局部轉移

Ionic transfer e- 電子轉移 Increasing difference in electronegativity增加電負度的差異

共價 極性共價

(39)

© 2014 Pearson Education, Inc.

(40)

© 2014 Pearson Education, Inc.

Bond Dipole Moments

• Dipole moment, m, is a measure of bond polarity.

m = (q)(r) Measured in Debyes, D

(41)

© 2014 Pearson Education, Inc.

9.7 Lewis Structures of Molecular Compounds and Polyatomic Ions

Writing Lewis Structures for Molecular Compounds 1. Write the correct skeletal structure for the molecule.

Put least electronegative element in the center.

2. Calculate total number of valence e

-

. Add 1 for each negative charge. Subtract 1 for each positive charge.

3. Complete an octet for all atoms except hydrogen 4. If any atoms lack an octet, form double or triple

bonds as necessary to give them octets.

(42)

Write the Lewis structure of nitrogen trifluoride (NF3).

Step 1 – N is less electronegative than F, put N in center

F N F

F

Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5) 5 + (3 x 7) = 26 valence electrons

Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. 26 - (3 x 2) = 20 valence electrons Step 4 - Check, are # of e- in structure equal to number of valence e- ?

3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons

9.6

(43)

Write the Lewis structure of the carbonate ion (CO32-).

Step 1 – C is less electronegative than O, put C in center

O C O

O

Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4) -2 charge – 2e-

4 + (3 x 6) + 2 = 24 valence electrons

Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. 24 - (3 x 2) = 18 valence electrons Step 4 - Check every atom satisfied octet rule,

9.6 Step 5 - central atom C is not for octet rule, we move a lone

pair from one of the O atoms to form another bond with C. Now the octet rule also satisfied for te C atom.

2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24

(44)

O C O O

- -

O C O

O

-

-

O C

O

O -

- 9.8 What are the resonance structures of the

carbonate (CO32-) ion?

9.8 Resonance and Formal Charge

Resonance

A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.

O O + O -

O O

O + -

(45)

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Resonance

. . . .

O S

. O

. . . . . . . . . . .

. . . . . .

O S

. O

. . . . . . . . . . . . .

• The actual molecule is a combination of the resonance forms—a resonance hybrid.

The molecule does not resonate between the two

forms, though we often draw it that way.

(46)

Q:Within each of the following species all bonds are equivalent. To show this, for which one of these compounds must we draw two resonance structures?

A. CO

2

B. O

3

C. CO D. NO E. H

2

S

Answer: B

O O O O O O

+1

-1 -1

+1

(47)

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Ozone Layer

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© 2014 Pearson Education, Inc.

Formal Charge

• During bonding, atoms may end with more or fewer electrons than the valence electrons they brought in order to fulfill

octets.

• His results in atoms having a formal charge.

FC = valence e − nonbonding e − ½ bonding e

(49)

© 2014 Pearson Education, Inc.

(50)

9.7

Two possible skeletal structures of formaldehyde (CH2O)

H C O H H

C O H

An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis

structure.

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

(51)

H C O H

C – 4 e- O – 6 e- 2H – 2x1 e-

12 e-

2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12

formal charge

on C = 4 - 2 - ½ x 6 = -1

formal charge

on O = 6 - 2 - ½ x 6 = +1

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

-1 +1

9.7

(52)

C – 4 e- O – 6 e- 2H – 2x1 e-

12 e-

2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12

H

C O H

formal charge

on C = 4 - 0 - ½ x 8 = 0

formal charge

on O = 6 - 4 - ½ x 4 = 0

formal charge on an atom in a Lewis

structure

= 1

2

total number of bonding electrons

( )

total number of valence electrons in the free atom

-

total number of nonbonding electrons

-

0 0

9.7

(53)

Formal Charge and Lewis Structures

9.7

1. For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which

formal charges are present.

2. Lewis structures with large formal charges are less plausible than those with small formal charges.

3. Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.

Which is the most likely Lewis structure for CH2O?

H C O H

-1 +1 H

C O H

0 0

(54)

Q:What is the formal charge on the bromine atom in BrO

3-

, drawn with three single bonds?

A. -2 B. -1 C. 0 D. +1 E. +2

Answer: E

Br O

O

-1

+2 -1

O

-1

-1

(55)

© 2014 Pearson Education, Inc.

9.9

Exceptions to the Octet Rule: Odd-Electron Species, Incomplete Octets, and Expanded Octets

• Odd number electron species (e.g., NO)

– Will have one unpaired electron – Free-radical

– Very reactive

• Incomplete octets

– B, Al

• Expanded octets

– Elements with empty d orbitals can have

more than eight electrons.

(56)

Exceptions to the Octet Rule

The Incomplete Octet (2A,3A)

H Be H Be – 2e-

2H – 2x1e- 4e- BeH2

BF3

B – 3e- 3F – 3x7e- 24e-

F B F

F

3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24

9.9

(57)

Exceptions to the Octet Rule

Odd-Electron Molecules N – 5e-

O – 6e- 11e-

NO N O

The Expanded Octet (central atom with principal quantum number n > 2)

SF6

S – 6e- 6F – 42e- 48e-

S F

F

F F F

F

6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48

9.9

(58)

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Drawing Resonance Structures

1. Draw the first Lewis structure that maximizes octets.

2. Assign formal charges.

3. Move electron pairs from atoms with (−) formal charge toward atoms with (+) formal charge.

4. If (+) formal charge atom second row, only move in electrons if

you can move out electron pairs from multiple bond.

5. If (+) formal charge atom third row or below, keep bringing in electron pairs to reduce the formal charge, even if get expanded octet.

−1

− 1

+2

(59)

Q:

Which response includes all the molecules below that do not

follow the octet rule?

(1) H

2

S (2) BCl

3

(3) PH

3

(4) SF

4

A. 2,4 B. 2,3 C. 1,2 D. 3,4 E. 1,4

Answer: A

(60)

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9.10 Bond Energies and Bond Lengths

Bond Energies

The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.

H2 (g) H (g) + H (g) DH0 = 436.4 kJ Cl2 (g) Cl (g) + Cl (g) DH0 = 242.7 kJ HCl (g) H (g) + Cl (g) DH0 = 431.9 kJ

O2 (g) O (g) + O (g) DH0 = 498.7 kJ O O N2 (g) N (g) + N (g) DH0 = 941.4 kJ N N

Bond Energy

Bond Energies

Single bond < Double bond < Triple bond

9.10

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Trends in Bond Energies

• In general, the more electrons two atoms share, the stronger the covalent bond.

– Must be comparing bonds between like atoms – C≡C (837 kJ) > C═C (611 kJ) > C—C (347 kJ) – C≡N (891 kJ) > C ═ N (615 kJ) > C—N (305 kJ)

• In general, the shorter the covalent bond, the stronger the bond.

– Must be comparing similar types of bonds

– Br—F (237 kJ) > Br—Cl (218 kJ) > Br—Br (193 kJ) – Bonds get weaker down the column.

– Bonds get stronger across the period.

(62)

Average bond energy in polyatomic molecules H2O (g) H (g) + OH (g) DH0 = 502 kJ

OH (g) H (g) + O (g) DH0 = 427 kJ Average OH bond energy = 502 + 427

2 = 464 kJ

9.10

(63)

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Using Bond Energies to Estimate D

rxn

• The actual bond energy depends on the surrounding atoms and other factors.

• We often use average bond energies to estimate the DH rxn .

– Works best when all reactants and products in gas state

• Bond breaking is endothermic, DH(breaking) = +.

• Bond making is exothermic, DH(making) = −.

DH

rxn

= ∑ (DH(bonds broken)) + ∑ (DH(bonds formed))

(64)

Bond Energies (BE) and Enthalpy changes in reactions

DH0 = total energy input – total energy released

= SBE(reactants) – SBE(products)

Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.

9.10

Using Bond Energies to Estimate D

rxn

DH

rxn

= ∑ (DH(bonds broken)) + ∑ (DH(bonds formed))

• endothermic,

-DH(breaking) = +.

• exothermic,

-DH(making) = −.

(65)
(66)

9.10

H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g)

(67)

Use bond energies to calculate the enthalpy change for:

H2 (g) + F2 (g) 2HF (g) DH0 = SBE(reactants) – SBE(products)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

DH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

9.10

(68)

Q

:Estimate the enthalpy change for the reaction 2SO

2

+ O

2

→2SO

3

given the following bond energies

BE(S - O) = 347 kJ BE(O=O) = 499 kJ A. -152 kJ

B. +152 kJ C. -195 kJ D. +195 kJ E. +846 kJ

Answer: C

DH0 = SBE(reactants) – SBE(products)

DH0 = 2 x 2 x 347 + 499 – 2 x 3 x 347 = -195 kJ

(69)

© 2014 Pearson Education, Inc.

Bond Lengths

• The distance between the nuclei of

bonded atoms is called the bond length.

• Because the actual bond length depends on the other atoms around the bond we often use the average bond length.

– Averaged for similar bonds from many compounds

(70)

© 2014 Pearson Education, Inc.

Trends in Bond Lengths

• In general, the more electrons two atoms share, the shorter the covalent bond.

– Must be comparing bonds between like atoms – C≡C (120 pm) < C═C (134 pm) < C—C (154 pm) – C≡N (116 pm) < C═N (128 pm) < C—N (147 pm)

• Generally, bond length decreases from left to right across period.

– C—C (154 pm) > C—N (147 pm) > C—O (143 pm)

• Generally, bond length increases down the column.

– F—F (144 pm) > Cl—Cl (198 pm) > Br—Br (228 pm)

• In general, as bonds get longer, they also get

weaker.

(71)

© 2014 Pearson Education, Inc.

Bond Lengths

(72)

© 2014 Pearson Education, Inc.

9.11 Bonding in Metals: The Electron Sea Model

• Metallic solids do conduct heat well.

• As temperature increases, the electrical conductivity of metals decreases.

• Metallic solids do conduct electricity well.

• Metallic solids are malleable and ductile.

• Metallic solids do reflect light.

• Metals generally have high melting points and boiling points.

– All but Hg are solids at room temperature.

• Melting points of metals generally decrease down column.

-Li (180.54 ºC) > Na (97.72 ºC) > K (63.38 ºC)

• Melting points of metal generally increase left to right across period. (溶點右上角最大)

-Na (97.72 ºC) < Mg (650 ºC) < Al (660.32 ºC)

https://www.youtube.com/watch?v=Bjf9gMDP47s

(73)

© 2014 Pearson Education, Inc.

Drawing Resonance Structures

1. Draw the first Lewis structure that maximizes octets.

2. Assign formal charges.

3. Move electron pairs from atoms with (−) formal charge toward atoms with (+) formal charge.

4. If (+) formal charge atom second row, only move in

electrons if you can move out electron pairs from multiple bond.

5. If (+) formal charge atom third row or below, keep bringing in electron pairs to reduce the formal charge, even if get expanded octet.

+1

−1

−1

+1

−1

−1

(74)

© 2014 Pearson Education, Inc.

Evaluating Resonance Structures

• Better structures have fewer formal charges.

• Better structures have smaller formal charges.

• Better structures have the negative formal charge on the more

electronegative atom.

參考文獻

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