© 2014 Pearson Education, Inc.
Sherril Soman
Grand Valley State University
Lecture Presentation
Chapter 9
Chemical
Bonding I: The
Lewis Model
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Pg. 471
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.
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
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
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Nine
Formal Charge and Lewis Structures
61. 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
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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.
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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
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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
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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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9.2 Types of Bonds
• We can classify bonds based on the kinds of
atoms that are bonded together.
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Types of Bonding
INSERT FIGURE 9.1
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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.
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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.
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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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Metallic Bonding
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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-
9.1
Octet rule
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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
Q:The general electron configuration for noble gas atoms is:
A. ns
2np
6B. ns
2np
5C. ns
2np
4D. ns
2np
3E. ns
2Answer: A
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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
-
Q:Which one of the following
compounds is most likely to be an ionic ompound?
A. KF B. CCl
4C. CS
2D. CO
2E. ICl
Answer: A
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
)能
(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© 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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Na(s) + ½ Cl2(g) → NaCl(s) Enthalpy of standard formation:
ΔHfo = -411 kJ
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
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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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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
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• 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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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.
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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
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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
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© 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.
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Intermolecular Attractions versus Bonding
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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
或極性鍵結
是 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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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
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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
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Bond Dipole Moments
• Dipole moment, m, is a measure of bond polarity.
m = (q)(r) Measured in Debyes, D
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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.
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
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
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 + -
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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.
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
2B. O
3C. CO D. NO E. H
2S
Answer: B
O O O O O O
+1
-1 -1
+1
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Ozone Layer
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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−
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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.
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
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
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
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
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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.
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
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
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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
Q:
Which response includes all the molecules below that do not
follow the octet rule?
(1) H
2S (2) BCl
3(3) PH
3(4) SF
4A. 2,4 B. 2,3 C. 1,2 D. 3,4 E. 1,4
Answer: A
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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.
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
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Using Bond Energies to Estimate D H°
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))
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 H°
rxnDH
rxn= ∑ (DH(bonds broken)) + ∑ (DH(bonds formed))
• endothermic,
-DH(breaking) = +.
• exothermic,
-DH(making) = −.
9.10
H2 (g) + Cl2 (g) 2HCl (g) 2H2 (g) + O2 (g) 2H2O (g)
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
Q
:Estimate the enthalpy change for the reaction 2SO
2+ O
2→2SO
3given 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
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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
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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.
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Bond Lengths
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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
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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
−1
+1
−1
−1
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