Transcript Chapter 9 Chemical Bonding I lewis Theory

Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 9
Chemical
Bonding I:
Lewis Theory
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Bonding Theories
• explain how and why atoms attach together
• explain why some combinations of atoms are stable
and others are not
 why is water H2O, not HO or H3O
• one of the simplest bonding theories was developed by
•
•
G.N. Lewis and is called Lewis Theory
Lewis Theory emphasizes valence electrons to explain
bonding
using Lewis Theory, we can draw models – called
Lewis structures – that allow us to predict many
properties of molecules
 aka Electron Dot Structures
 such as molecular shape, size, polarity
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Why Do Atoms Bond?
• processes are spontaneous if they result in a system
•
•
•
with lower potential energy
chemical bonds form because they lower the potential
energy between the charged particles that compose
atoms
the potential energy between charged particles is
directly proportional to the product of the charges
the potential energy between charged particles is
inversely proportional to the distance between the
charges
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Potential Energy Between
Charged Particles
1  q1  q2 
E potential 


4 0  r 
• 0 is a constant
 = 8.85 x 10-12 C2/J∙m
• for charges with the same sign, Epotential is + and the
•
•
magnitude gets less positive as the particles get farther
apart
for charges with the opposite signs, Epotential is  and
the magnitude gets more negative as the particles get
closer together
remember: the more negative the potential energy, the
more stable the system becomes
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Potential Energy Between
Charged Particles
The attraction
repulsion between
like-charged particles
opposite-charged
increasesincreases
particles
as the as
particles
the
particles
get closer
get closer
together. To
Bringing
bring
them closer lowers
requiresthe
the addition
potential
energy
of more
of the
energy.
system.
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Bonding
• a chemical bond forms when the potential
energy of the bonded atoms is less than the
potential energy of the separate atoms
• have to consider following interactions:
nucleus-to-nucleus repulsion
electron-to-electron repulsion
nucleus-to-electron attraction
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Types of Bonds
Types of Atoms
metals to
nonmetals
nonmetals to
nonmetals
metal to
metal
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Type of Bond
Ionic
Covalent
Metallic
Bond
Characteristic
electrons
transferred
electrons
shared
electrons
pooled
7
Types of Bonding
8
Ionic Bonds
• when metals bond to nonmetals, some electrons
from the metal atoms are transferred to the
nonmetal atoms
metals have low ionization energy, relatively easy to
remove an electron from
nonmetals have high electron affinities, relatively
good to add electrons to
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Covalent Bonds
• nonmetals 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 lowest when the electrons are between the
nuclei
• shared electrons hold the atoms together by attracting
nuclei of both atoms
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Determining the Number of Valence
Electrons in an Atom
• the column number on the Periodic Table will tell you
how many valence electrons a main group atom has
 Transition Elements all have 2 valence electrons; Why?
1A
2A
3A
4A
5A
6A
7A
8A
Li
Be
B
C
N
O
F
Ne
1 e-1
2 e-1
3 e-1
4 e-1
5 e-1
6 e-1
7 e-1
8 e-1
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Lewis Symbols of Atoms
• aka electron dot symbols
• use symbol of element to represent nucleus and
•
inner electrons
use dots around the symbol to represent valence
electrons
 pair first two electrons for the s orbital
 put one electron on each open side for p electrons
 then pair rest of the p electrons

Li

Be


B

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


C




N





O






F






Ne



12
Lewis Symbols of Ions
• Cations have Lewis symbols without
valence electrons
Lost in the cation formation
• Anions have Lewis symbols with 8 valence
electrons
Electrons gained in the formation of the anion

Li•
Li+1


F

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

  
 F  




  
1
13
Stable Electron Arrangements
And Ion Charge
• Metals form cations by losing
enough electrons to get the
same electron configuration
as the previous noble gas
• Nonmetals form anions by
gaining enough electrons to
get the same electron
configuration as the next
noble gas
• The noble gas electron
configuration must be very
stable
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Na
Atom’s
Electron
Config
[Ne]3s1
Na+1
Ion’s
Electron
Config
[Ne]
Mg
[Ne]3s2
Mg+2
[Ne]
Al
[Ne]3s23p1
Al+3
[Ne]
O
[He]2s22p4
O-2
[Ne]
F
[He]2s22p5
F-1
[Ne]
Atom
Ion
15
Octet Rule
• when atoms bond, they tend to gain, lose, or share electrons to
•
result in 8 valence electrons
ns2np6
 noble gas configuration
• many exceptions
 H, Li, Be, B attain an electron configuration like He
 He = 2 valence electrons
 Li loses its one valence electron
 H shares or gains one electron
 though it commonly loses its one electron to become H+
 Be loses 2 electrons to become Be2+
 though it commonly shares its two electrons in covalent bonds, resulting in 4
valence electrons
 B loses 3 electrons to become B3+
 though it commonly shares its three electrons in covalent bonds, resulting in 6
valence electrons
 expanded octets for elements in Period 3 or below
 using empty valence d orbitals
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Lewis Theory
• the basis of Lewis Theory is that there are
certain electron arrangements in the atom that
are more stable
octet rule
• bonding occurs so atoms attain a more stable
electron configuration
more stable = lower potential energy
no attempt to quantify the energy as the calculation
is extremely complex
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Properties of Ionic Compounds
• hard and brittle crystalline solids
Melting an Ionic Solid
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
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Conductivity of NaCl
in NaCl(s), the
ions are stuck in
position and not
allowed to move
to the charged
rods
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in NaCl(aq), the
ions are
separated and
allowed to move
to the charged
rods
19
Lewis Theory and Ionic Bonding
• Lewis symbols can be used to represent the
transfer of electrons from metal atom to
nonmetal atom, resulting in ions that are
attracted to each other and therefore bond

Li

+


F

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

 F  



  



Li +
1
20
Predicting Ionic Formulas
Using Lewis Symbols
• electrons are transferred until the metal loses all its
•
valence electrons and the nonmetal has an octet
numbers of atoms are adjusted so the electron transfer
comes out even
Li



Li

O


 O 



  




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2 Li +
2
Li2O
21
Energetics of Ionic Bond Formation
• the ionization energy of the metal is endothermic
 Na(s) → Na+(g) + 1 e ─
DH° = +603 kJ/mol
• the electron affinity of the nonmetal is exothermic
 ½Cl2(g) + 1 e ─ → Cl─(g)
DH° = ─ 227 kJ/mol
• generally, the ionization energy of the metal is larger
•
than the electron affinity of the nonmetal, therefore the
formation of the ionic compound should be
endothermic
but the heat of formation of most ionic compounds is
exothermic and generally large; Why?
 Na(s) + ½Cl2(g) → NaCl(s)
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DH°f = -410 kJ/mol
22
Ionic Bonds
• electrostatic attraction is nondirectional!!
no direct anion-cation pair
• no ionic molecule
chemical formula is an empirical formula, simply
giving the ratio of ions based on charge balance
• ions arranged in a pattern called a crystal lattice
every cation surrounded by anions; and every anion
surrounded by cations
maximizes attractions between + and - ions
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Lattice Energy
• the lattice energy is the energy released when the
solid crystal forms from separate ions in the gas state
 always exothermic
 hard to measure directly, but can be calculated from
knowledge of other processes
• lattice energy depends directly on size of charges and
inversely on distance between ions
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Born-Haber Cycle
• method for determining the lattice energy of an
ionic substance by using other reactions
use Hess’s Law to add up heats of other processes
•
DH°f(salt) = DH°f(metal atoms, g) + DH°f(nonmetal atoms, g)
+ DH°f(cations, g) + DH°f(anions, g) + DH°f(crystal lattice)
 DH°f(crystal lattice) = Lattice Energy
 metal atoms (g)  cations (g), DH°f = ionization energy
don’t forget to add together all the ionization energies to get to the
desired cation
M2+ = 1st IE + 2nd IE
 nonmetal atoms (g)  anions (g), DH°f = electron affinity
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Born-Haber Cycle for NaCl
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Practice - Given the Information Below,
Determine the Lattice Energy of MgCl2
Mg(s)  Mg(g)
½ Cl2(g)  Cl(g)
Mg(g)  Mg+1(g)
Mg+1(g)  Mg+2(g)
Cl(g)  Cl-1(g)
Mg(s) + Cl2(g)  MgCl2(s)
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DH1°f = +147.1 kJ/mol
DH2°f = +121.3 kJ/mol
DH3°f = +738 kJ/mol
DH4°f = +1450 kJ/mol
DH5°f = -349 kJ/mol
DH6°f = -641.3 kJ/mol
27
Practice - Given the Information Below,
Determine the Lattice Energy of MgCl2
Mg(s)  Mg(g)
½ Cl2(g)  Cl(g)
Mg(g)  Mg+1(g)
Mg+1(g)  Mg+2(g)
Cl(g)  Cl-1(g)
Mg(s) + Cl2(g)  MgCl2(s)
DH1°f = +147.1 kJ/mol
DH2°f = +121.3 kJ/mol
DH3°f = +738 kJ/mol
DH4°f = +1450 kJ/mol
DH5°f = -349 kJ/mol
DH6°f = -641.3 kJ/mol
DH 6f  DH1f  2DH 2f  DH 3f  DH 4f  2DH 5f  DH f latticeenergy

DH f latticeenergy  DH 6 f  DH1f  2DH 2 f  DH 3f  DH 4 f  2DH 5f







DH f latticeenergy  (641.3 kJ) - (147.1 kJ)  2(121.3 kJ)  (738 kJ)  (1450 kJ)  2(-349 kJ) 
DH f latticeenergy  2521 kJ
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Trends in Lattice Energy
Ion Size
• the force of attraction between charged
particles is inversely proportional to the
distance between them
• larger ions mean the center of positive charge
(nucleus of the cation) is farther away from
negative charge (electrons of the anion)
larger ion = weaker attraction = smaller lattice
energy
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Lattice Energy vs.
Ion Size
Lattice Energy
Metal Chloride
(kJ/mol)
LiCl
-834
NaCl
-787
KCl
-701
CsCl
-657
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Trends in Lattice Energy
Ion Charge
• the force of attraction between
•
oppositely charged particles is
directly proportional to the product
of the charges
larger charge means the ions are
more strongly attracted
Lattice Energy =
-910 kJ/mol
 larger charge = stronger attraction =
larger lattice energy
• of the two factors, ion charge
generally more important
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Lattice Energy =
-3414 kJ/mol
31
Example 9.2 – Order the following ionic
compounds in order of increasing magnitude of
lattice energy.
CaO, KBr, KCl, SrO
First examine the ion charges and
order by product of the charges
Ca2+& O2-, K+ & Br─,
K+ & Cl─, Sr2+ & O2─
(KBr, KCl) < (CaO, SrO)
Then examine the ion sizes of
each group and order by radius;
larger < smaller
(KBr, KCl) same cation,
Br─ > Cl─ (same Group)
(CaO, SrO) same anion,
Sr2+ > Ca2+ (same Group)
KBr < KCl < SrO
(CaO,
< SrO)
CaO
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Ionic Bonding
Model vs. Reality
• ionic compounds have high melting points and boiling
points
 MP generally > 300°C
 all ionic compounds are solids at room temperature
• because the attractions between ions are strong,
breaking down the crystal requires a lot of energy
 the stronger the attraction (larger the lattice energy), the
higher the melting point
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Ionic Bonding
Model vs. Reality
• ionic solids are brittle and hard
• the position of the ion in the crystal is critical to
establishing maximum attractive forces –
displacing the ions from their positions results
in like charges close to each other and the
repulsive forces take over
+
-
-+
+
-
+
-+ +- -+ +- -+ +- -+ +
+ - + - + - + - + - + - + + + + +
- + - + - + -
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-
+
34
Ionic Bonding
Model vs. Reality
• ionic compounds conduct electricity in the liquid state
•
•
•
or when dissolved in water, but not in the solid state
to conduct electricity, a material must have charged
particles that are able to flow through the material
in the ionic solid, the charged particles are locked in
position and cannot move around to conduct
in the liquid state, or when dissolved in water, the ions
have the ability to move through the structure and
therefore conduct electricity
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Single Covalent Bonds
• two atoms share a pair of electrons
 2 electrons
• one atom may have more than one single bond
••
••
••
F
••
F
••
••
F
••
••
••
••
• F
••
••
••
H• • O
•H
••
••
H O H
••
•
F •
••
••
••
F
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Double Covalent Bond
• two atoms sharing two pairs of electrons
4 electrons
•
••
•O
••
•
••
•O
••
O •• O
••
···
·· O
··O
·
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Triple Covalent Bond
• two atoms sharing 3 pairs of electrons
6 electrons
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••
•N
•
•
•
••
•N
•
•
N• N
••
••
··N N ··
39
Covalent Bonding
Predictions from Lewis Theory
• Lewis theory allows us to predict the formulas of
•
molecules
Lewis theory predicts that some combinations should be
stable, while others should not
 because the stable combinations result in “octets”
• Lewis theory predicts in covalent bonding that the
attractions between atoms are directional
 the shared electrons are most stable between the bonding atoms
 resulting in molecules rather than an array
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Covalent Bonding
Model vs. Reality
• molecular compounds have low melting points and
boiling points
 MP generally < 300°C
 molecular compounds are found in all 3 states at room
temperature
• melting and boiling involve breaking the attractions
between the molecules, but not the bonds between
the atoms
 the covalent bonds are strong
 the attractions between the molecules are generally weak
 the polarity of the covalent bonds influences the strength of
the intermolecular attractions
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Intermolecular Attractions vs. Bonding
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Ionic Bonding
Model vs. Reality
• some molecular solids are brittle and hard, but
many are soft and waxy
• the kind and strength of the intermolecular
attractions varies based on many factors
• the covalent bonds are not broken, however, the
polarity of the bonds has influence on these
attractive forces
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Ionic Bonding
Model vs. Reality
• molecular compounds do not conduct electricity in the
•
•
•
liquid state
molecular acids conduct electricity when dissolved in
water, but not in the solid state
in molecular solids, there are no charged particles
around to allow the material to conduct
when dissolved in water, molecular acids are ionized,
and have the ability to move through the structure and
therefore conduct electricity
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Bond Polarity
• covalent bonding between unlike atoms results in
unequal sharing of the electrons
one atom pulls the electrons in the bond closer to its side
one end of the bond has larger electron density than the
other
• the result is a polar covalent bond
bond polarity
the end with the larger electron density gets a partial
negative charge
the end that is electron deficient gets a partial positive
charge
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HF

EN 2.1 H
F
 EN

4.0

d H •• F d
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Electronegativity
• measure of the pull an atom has on bonding
electrons
• increases across period (left to right) and
• decreases down group (top to bottom)
fluorine is the most electronegative element
francium is the least electronegative element
• the larger the difference in electronegativity,
the more polar the bond
negative end toward more electronegative atom
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Electronegativity Scale
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Electronegativity and Bond Polarity
• If difference in electronegativity between bonded atoms
is 0, the bond is pure covalent
 equal sharing
• If difference in electronegativity between bonded atoms
•
•
is 0.1 to 0.4, the bond is nonpolar covalent
If difference in electronegativity between bonded atoms
0.5 to 1.9, the bond is polar covalent
If difference in electronegativity between bonded atoms
larger than or equal to 2.0, the bond is ionic
4%
0
0.4
Percent Ionic Character
51%
2.0
Electronegativity Difference
“100%”
4.0
49
Bond Polarity
ENCl = 3.0
3.0 - 3.0 = 0
Pure Covalent
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ENCl = 3.0
ENH = 2.1
3.0 – 2.1 = 0.9
Polar Covalent
ENCl = 3.0
ENNa = 1.0
3.0 – 0.9 = 2.1
Ionic
50
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Bond Dipole Moments
• the dipole moment is a quantitative way of describing the
polarity of a bond
 a dipole is a material with positively and negatively charged ends
 measured
• dipole moment, m, is a measure of bond polarity
 it is directly proportional to the size of the partial charges and
directly proportional to the distance between them
 m = (q)(r)
 not Coulomb’s Law
 measured in Debyes, D
• the percent ionic character is the percentage of a bond’s
measured dipole moment to what it would be if full ions
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Dipole Moments
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Water – a Polar Molecule
stream of
water
attracted
to a
charged
glass rod
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stream of
hexane
not
attracted
to a
charged
glass rod
54
Example 9.3(c) - Determine whether an N-O
bond is ionic, covalent, or polar covalent.
• Determine the electronegativity of each element
•
•
N = 3.0; O = 3.5
Subtract the electronegativities, large minus small
(3.5) - (3.0) = 0.5
If the difference is 2.0 or larger, then the bond is
ionic; otherwise it’s covalent
difference (0.5) is less than 2.0, therefore covalent
• If the difference is 0.5 to 1.9, then the bond is
polar covalent; otherwise it’s covalent
difference (0.5) is 0.5 to 1.9, therefore polar covalent
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Lewis Structures
of Molecules
• shows pattern of valence electron distribution in
the molecule
• useful for understanding the bonding in many
compounds
• allows us to predict shapes of molecules
• allows us to predict properties of molecules and
how they will interact together
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Lewis Structures
• use common bonding patterns
 C = 4 bonds & 0 lone pairs, N = 3 bonds & 1 lone pair,
O= 2 bonds & 2 lone pairs, H and halogen = 1 bond, Be
= 2 bonds & 0 lone pairs, B = 3 bonds & 0 lone pairs
 often Lewis structures with line bonds have the lone
pairs left off
 their presence is assumed from common bonding patterns
• structures which result in bonding patterns
different from common have formal charges
B
C
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N
O
F
57
Writing Lewis Structures of Molecules
HNO3
O
1) Write skeletal structure
 H always terminal

in oxyacid, H outside attached to O’s
H
O
N
O
 make least electronegative atom central

N is central
2) Count valence electrons
 sum the valence electrons for each
atom
 add 1 electron for each − charge
 subtract 1 electron for each + charge
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N=5
H=1
O3 = 3∙6 = 18
Total = 24 e58
Writing Lewis Structures of Molecules
HNO3
3) Attach central atom to the surrounding atoms with
pairs of electrons and subtract from the total
O

H — O — N — O
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Electrons
Start 24
Used 8
Left 16
59
Writing Lewis Structures of Molecules
HNO3

4) Complete octets, outside-in
: O :
 H is already complete with 2


H — O — N — O
 1 bond



and re-count electrons
N=5
H=1
O3 = 3∙6 = 18
Total = 24 eTro, Chemistry: A Molecular Approach
Electrons
Electrons
Start 24
Start 16
Used 8
Used 16
Left 16
Left 0
60
:
Writing Lewis Structures of Molecules
HNO3
5) If all octets complete, give extra
electrons to central atom.
 elements with d orbitals can have
more than 8 electrons

Period 3 and below
:

O

|
octet, bring in electrons from
H — O — N
outside atoms to share
6) If central atom does not have
 follow common bonding patterns
if possible
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
:

O:
61
Practice - Lewis Structures
• CO2
• H3PO4
• SeOF2
• SO3-2
• NO2-1
• P2H4
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Practice - Lewis Structures
• CO2
• H3PO4
16 e-
32
:O::C::O:
• SeOF2
26 e- • ••
•F
• NO2-1
18 e-
••
••
•O
•
••
Se F ••
•• ••
Tro, Chemistry: A Molecular Approach
H
••
O
••
• SO3-2
••
•O•
• •
••
N
e-
••
O ••
••
26 e-
••
•O
•
••
• P2H4
14 e-
H
••
•O•
• •
P
•O
•
••
••
•O•
• •
S
••
H
H
P
••
P
••
••
O
••
H
H
••
O ••
••
H
63
Formal Charge
• during bonding, atoms may wind up with more
or less electrons in order to fulfill octets - this
results in atoms having a formal charge
FC = valence e- - nonbonding e- - ½ bonding eleft O
FC = 6 - 4 - ½ (4) = 0
•• •••• • •• •
O •• S • O •
S
FC = 6 - 2 - ½ (6) = +1 ••
••
right O
FC = 6 - 6 - ½ (2) = -1
• sum of all the formal charges in a molecule = 0
 in an ion, total equals the charge
Tro, Chemistry: A Molecular Approach
64
Writing Lewis Formulas of
Molecules (cont’d)
7) Assign formal charges to the atoms
a) formal charge = valence e- - lone pair e- - ½ bonding eb) follow the common bonding patterns




O

0


S

+1 

-1 
O


H
O 
|
||

H  C  C  O  H
|

H
Tro, Chemistry: A Molecular Approach
all 0
65
Common Bonding Patterns
B
B
-
C
N
O
+
C
+
N
+
O
-
-
-
C
Tro, Chemistry: A Molecular Approach
N
O
F
F
F
+
-
66
Practice - Assign Formal Charges
• CO2
• H3PO4
••
O
••
H
• SeOF2
••
•F
•
••
• NO2-1
••
•O
•
• SO3-2
••
•O•
• •
••
Se F ••
•• ••
••
N
Tro, Chemistry: A Molecular Approach
••
O ••
••
••
•O•
• •
••
•O
•
••
• P2H4
H
P
•O
•
••
••
•O•
• •
S
••
H
H
P
••
P
••
••
O
••
H
H
••
O ••
••
H
67
Practice - Assign Formal Charges
• CO2
• H3PO4
P = +1 H
rest 0
all 0
• SeOF2
Se = +1
•• -1
•O•
• •
••
••
O
P O
••
••
•O H
•
••
••
•F
•
••
• NO2-1
••
•O
•
• SO3-2
•• -1
•O•
• •
••
Se F ••
•• ••
••
N
Tro, Chemistry: A Molecular Approach
•• -1
O ••
••
S = +1
•• -1
•O•
• •
••-1
•O
S
•
••
••
• P2H4
H
H
P
••
P
••
H
•• -1
O ••
••
all 0
H
H
68
Resonance
• when there is more than one Lewis structure for a
molecule that differ only in the position of the
electrons, they are called resonance structures
• the actual molecule is a combination of the
resonance forms – a resonance hybrid
it does not resonate between the two forms,
though we often draw it that way
• look for multiple bonds or lone pairs
••
••
••
•• O ••
••
•• S •• O
••
Tro, Chemistry: A Molecular Approach
••
•• O
••
••
••
••S ••
•• O ••
69
Resonance
Tro, Chemistry: A Molecular Approach
70
Ozone Layer
Tro, Chemistry: A Molecular Approach
71
Rules of Resonance Structures
• Resonance structures must have the same connectivity
 only electron positions can change
• Resonance structures must have the same number of
•
•
•
•
•
electrons
Second row elements have a maximum of 8 electrons
 bonding and nonbonding
 third row can have expanded octet
Formal charges must total same
Better structures have fewer formal charges
Better structures have smaller formal charges
Better structures have − formal charge on more
electronegative atom
Tro, Chemistry: A Molecular Approach
72
Drawing Resonance Structures
1. draw 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 (+) fc atom 2nd row, only move
in electrons if you can move out
electron pairs from multiple
bond
5. if (+) fc atom 3rd row or below,
keep bringing in electron pairs to
reduce the formal charge, even if
get expanded octet.
Tro, Chemistry: A Molecular Approach
-1
··
··O ··
·· O
··
··
··O
··
-1
N
+1
-1
··
·· O ··
+1
N
··
O ··
··
-1
··
O
··
73
Exceptions to the Octet Rule
• expanded octets
elements with empty d orbitals can have more
than 8 electrons
• odd number electron species e.g., NO
will have 1 unpaired electron
free-radical
very reactive
• incomplete octets
B, Al
Tro, Chemistry: A Molecular Approach
74
Drawing Resonance Structures
1. draw 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 (+) fc atom 2nd row, only move
in electrons if you can move out
electron pairs from multiple bond
5. if (+) fc atom 3rd row or below,
keep bringing in electron pairs to
reduce the formal charge, even if
get expanded octet.
Tro, Chemistry: A Molecular Approach
H
H
··
O
··
··
O
··
-1
··
··O ··
+2
S
·
O ··
·
-1 ··
0
··
O ··
S
0
·· O 0
··
··
O
··
··
O
··
H
H
75
Practice - Identify Structures with Better or
Equal Resonance Forms and Draw Them
••
• CO2
• H3PO4
P = +1 H
all 0
• SeOF2
Se = +1
• O • -1
• •
••
••
O
P O
••
••
•O H
•
••
••
•F
•
••
• NO2-1
••
•O
•
• SO3-2
•• -1
•O•
• •
••
Se F ••
•• ••
••
N
Tro, Chemistry: A Molecular Approach
•• -1
O ••
••
S = +1
•• -1
•O•
• •
••-1
•O
S
•
••
••
• P2H4
H
H
P
••
P
••
H
•• -1
O ••
••
all 0
H
H
76
Practice - Identify Structures with Better or
Equal Resonance Forms and Draw Them
• CO2
•
H3PO4
-1 • ••
•
•O•
none
••
O
P
•• +1
•O
•
••
H
• SeOF2
••
• O • -1
• •
••
••
•F
Se F ••
•
••
•• ••
+1
••
N
•• -1
O ••
••
H
H
• SO3-2
all 0
••
•F
•
••
••
•O
•
••
•O
•
••
Se F ••
•• ••
• NO2-1
••
•O
•
••
O
••
H
••
•O•
• •
S
••
••
O•
•• •
••
•O
•
••
• P2H4
-1
••
•O
• ••
Tro, Chemistry: A Molecular Approach
••
N
••
O ••
H
••
•O
•
••
••
•O•
• •
S
••
•O•
• •
S
••
H
H
P
••
P
••
••
O•
•• •
••
O•
•• •
••
O
••
• ••
•O
P
•O
•
••
••
•O
•
••
all 0
••
O
••
H
••
•O•
• •
S
••
H
••
O ••
S=0
in all
res. forms
H none
77
Bond Energies
• chemical reactions involve breaking bonds in reactant
•
•
molecules and making new bond to create the products
the DH°reaction can be calculated by comparing the cost
of breaking old bonds to the profit from making new
bonds
the amount of energy it takes to break one mole of a
bond in a compound is called the bond energy
 in the gas state
 homolytically – each atom gets ½ bonding electrons
Tro, Chemistry: A Molecular Approach
78
Trends in Bond Energies
• the more electrons two atoms share, the stronger
the covalent bond
C≡C (837 kJ) > C=C (611 kJ) > C−C (347 kJ)
C≡N (891 kJ) > C=N (615 kJ) > C−N (305 kJ)
• the shorter the covalent bond, the stronger the
bond
Br−F (237 kJ) > Br−Cl (218 kJ) > Br−Br (193 kJ)
bonds get weaker down the column
Tro, Chemistry: A Molecular Approach
79
Using Bond Energies to Estimate DH°rxn
• the actual bond energy depends on the surrounding
•
atoms and other factors
we often use average bond energies to estimate the
DHrxn
 works best when all reactants and products in gas state
• bond breaking is endothermic, DH(breaking) = +
• bond making is exothermic, DH(making) = −
DHrxn = ∑ (DH(bonds broken)) + ∑ (DH(bonds formed))
Tro, Chemistry: A Molecular Approach
80
81
Estimate the Enthalpy of the Following Reaction
H
H
+
O
O
H
O
O
H
82
Estimate the Enthalpy of the Following Reaction
H2(g) + O2(g)  H2O2(g)
reaction involves breaking 1mol H-H and 1 mol O=O
and making 2 mol H-O and 1 mol O-O
bonds broken (energy cost)
(+436 kJ) + (+498 kJ) = +934 kJ
bonds made (energy release)
2(464 kJ) + (142 kJ) = -1070
DHrxn = (+934 kJ) + (-1070. kJ) = -136 kJ
(Appendix DH°f = -136.3 kJ/mol)
Tro, Chemistry: A Molecular Approach
83
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
Tro, Chemistry: A Molecular Approach
84
Trends in Bond Lengths
• the more electrons two atoms share, the shorter the
covalent bond
 C≡C (120 pm) < C=C (134 pm) < C−C (154 pm)
 C≡N (116 pm) < C=N (128 pm) < C−N (147 pm)
• decreases from left to right across period
 C−C (154 pm) > C−N (147 pm) > C−O (143 pm)
• 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
Tro, Chemistry: A Molecular Approach
85
Bond Lengths
Tro, Chemistry: A Molecular Approach
86
Metallic Bonds
• low ionization energy of metals allows them to
lose electrons easily
• the simplest theory of metallic bonding involves
the metals atoms releasing their valence electrons
to be shared by all to 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
Tro, Chemistry: A Molecular Approach
87
Metallic Bonding
Tro, Chemistry: A Molecular Approach
88
Metallic Bonding
Model vs. Reality
• metallic solids conduct electricity
• because the free electrons are mobile, it
allows the electrons to move through the
metallic crystal and conduct electricity
• as temperature increases, electrical
conductivity decreases
• heating causes the metal ions to vibrate
faster, making it harder for electrons to make
their way through the crystal
Tro, Chemistry: A Molecular Approach
89
Metallic Bonding
Model vs. Reality
• metallic solids conduct heat
• the movement of the small, light electrons
through the solid can transfer kinetic energy
quicker than larger particles
• metallic solids reflect light
• the mobile electrons on the surface absorb
the outside light and then emit it at the same
frequency
Tro, Chemistry: A Molecular Approach
90
Metallic Bonding
Model vs. Reality
• metallic solids are malleable and ductile
• because the free electrons are mobile, the
direction of the attractive force between the
metal cation and free electrons is adjustable
• this allows the position of the metal cation
islands to move around in the sea of
electrons without breaking the attractions
and the crystal structure
Tro, Chemistry: A Molecular Approach
91
Metallic Bonding
Model vs. Reality
• metals generally have high melting points and boiling
points
 all but Hg are solids at room temperature
• the attractions of the metal cations for the free electrons
•
•
•
•
is strong and hard to overcome
melting points generally increase to right across period
the charge on the metal cation increases across the
period, causing stronger attractions
melting points generally decrease down column
the cations get larger down the column, resulting in a
larger distance from the nucleus to the free electrons
Tro, Chemistry: A Molecular Approach
92