Chapter 9:
Models of Chemical Bonding
Chem 6A, Section D Oct 11, 2011
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Chem 6A Michael J. Sailor, UC San Diego
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Announcements:
• Practice Final is posted on the web:
http://sailorgroup.ucsd.edu/Chem6A_sailor/Final_Exam_MASTER.pdf
• Thurs Nov 10 quiz (#7) will be on Chapter 7
• Tues of thanksgiving week is review
• No office hours Weds Nov 23
Chem 6A Michael J. Sailor, UC San Diego
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3 Types of Chemical Bonds
Fig 9.2
This
Chapter
Chem 6A Michael J. Sailor, UC San Diego
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Bonding in Compounds
covalent
ionic
Na+
H
O
water molecule
H
Cl
Cl
Cl-
chlorine molecule
Covalent bond = neutral atoms
held together by sharing a
pair of electrons
Ionic bond = charged atoms
(ions) held together by
electrostatic forces
An assembly of atoms held together
by covalent bonds is a molecule
Coulomb’s law:
charge
z1z2q 2
E=
4 πεo r1−2
distance
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Chem 6A Michael J. Sailor, UC San Diego
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Ionic Bonding
Na+
Cl-
The rock salt lattice
Chem 6A Michael J. Sailor, UC San Diego
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Lattice enthalpies and ionic radius
Lattice Enthalpy vs 1/(Ionic radii)
Lattice enthalpy, kJ/mol
950
NaF
900
850
800
NaCl
750
NaBr
700
NaI
650
3
3.2
3.4
3.6
3.8
4
1/(Na-X) distance, Å
4.2
4.4
-1
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Lattice Enthalpies
(see problem 9.58)
The thermite reaction (shown below) is highly
exothermic, mainly due to the larger energy of
the Al2O3 crystal lattice relative to Fe2O3.
Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)
•Using the data in the table below, calculate the
lattice enthalpy of Al2O3 and Fe2O3.
•What is the main reason for the larger lattice
energy of Al2O3?
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Lattice Enthalpies
Calculate the lattice enthalpy of Al2O3(s) at 25°C from the
following data:
Process
Enthalpy (ΔH), kJ/mol
?
First ionization energy of Al(g)
+578
Second ionization energy of Al(g) +1820
Third ionization energy of Al(g) +2750
Enthalpy of formation of Al(g)
+294
Enthalpy of formation of O2(g)
0
Bond energy of O2(g)
+498
Electron affinity of O(g)
-141
Electron affinity of O-(g)
+844
-1676
Lattice enthalpy Al2O3(s)
Enthalpy of formation of Al2O3(s)
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Chem 6A Michael J. Sailor, UC San Diego
PROBLEM: Lattice Enthalpies
Calculate the lattice enthalpy of Fe2O3(s) at 25°C from the
following data:
Process
Enthalpy (ΔH), kJ/mol
?
First ionization energy of Fe(g)
+759
Second ionization energy of Fe(g) +1561
Third ionization energy of Fe(g) +2957
Enthalpy of formation of Fe(g)
+415
Enthalpy of formation of O2(g) 0
Bond energy of O2(g)
+498
Electron affinity of O(g)
-141
Electron affinity of O-(g)
+844
-826
Lattice enthalpy Fe2O3(s)
Enthalpy of formation of Fe2O3(s)
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Lattice Enthalpies
Rewrite process in terms of chemical equations:
Process
Enthalpy (ΔH), kJ/mol
Lattice
O3(s)
2Fe3+(g)enthalpy
+ 3O2-(g) Fe
→2Fe
2O3(s)
?
Fe(g)ionization
→ Fe+(g) +energy
eFirst
of Fe(g)
+759
Second
of Fe(g) Fe+(g) →ionization
Fe2+(g) + eenergy
+1561
Third
Fe2+(g)ionization
→ Fe3+(g) energy
+ e- of Fe(g) +2957
Enthalpy
of(g)formation of Fe(g)
Fe(s) → Fe
+415
Enthalpy of formation of O2(g) 0
Bond
energy
O2(g) →
2O(g)of O2(g)
+498
Electron
O(g) + e- affinity
→ O-(g) of O(g)
-141
O-(g) + e-affinity
→ O2-(g)of O-(g)
Electron
+844
-826
Enthalpy
of formation
of3(s)
Fe2O3(s)
2Fe(s)+ 3/2O
2(g) → Fe2O
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Chem 6A Michael J. Sailor, UC San Diego
SOLUTION: Lattice Enthalpies
Rearrange equations to add up to 2Fe3+(g)+ 3O2-(g) → Fe2O3(s):
Process
2Fe(s)+ 3/2O2(g) → Fe2O3(s) Enthalpy (ΔH), kJ/mol
-826
2Fe(g) → 2Fe(s) 2(-415)
2Fe+(g) + 2e- → 2Fe(g) 2(-759)
2Fe2+(g) + 2e- → 2Fe+(g) 2(-1561)
2Fe3+(g) + 2e- → 2Fe2+(g) 2(-2957)
3O(g) → 3/2 O2(g) 3/2(-498)
3O-(g) → 3O(g) + 3e- 3(+141)
3O2-(g) → 3O-(g) + 3e-
3(-844)
2Fe3+(g)+ 3O2-(g) → Fe2O3(s)
Chem 6A Michael J. Sailor, UC San Diego
-15,066
?
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SOLUTION: Lattice Enthalpies
Do the same for 2Al3+(g)+ 3O2-(g) → Al2O3(s):
Process
Enthalpy (ΔH), kJ/mol
2Al(s)+ 3/2O2(g) → Al2O3(s) -1676
2Al(g) → 2Al(s) 2(-294)
2Al+(g) + 2e- → 2Al(g) 2(-578)
2Al2+(g) + 2e- → 2Al+(g) 2(-1820)
2Al3+(g) + 2e- → 2Al2+(g) 2(-2750)
3O(g) → 3/2 O2(g) 3/2(-498)
3O-(g) → 3O(g) + 3e- 3(+141)
3O2-(g) → 3O-(g) + 3e-
3(-844)
-15,416
2Al3+(g)+ 3O2-(g) → Al2O3(s)
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Chem 6A Michael J. Sailor, UC San Diego
SOLUTION: Lattice Enthalpies
Summary:
Process
Enthalpy (ΔH), kJ/mol
2Al3+(g)+ 3O2-(g) → Al2O3(s)
-15,416
2Fe3+(g)+ 3O2-(g) → Fe2O3(s)
-15,066
Difference:
2Al3+(g)+ Fe2O3(s)→ Al2O3(s) + 2Fe3+(g)
-350
kJ/mol
Al2O3(s) lattice is more stable than Fe2O3(s) lattice by 350 kJ/mol
The enthalpy of the thermite reaction is:
2Al(s)+ Fe2O3(s)→ Al2O3(s) + 2Fe(s)
-850
kJ/mol
So 350/850, or 41% of the energy for this reaction comes from the
difference in lattice energies. Why is Al2O3(s) so much more stable?
Chem 6A Michael J. Sailor, UC San Diego
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SOLUTION: Lattice Enthalpies
2Al(s)+ Fe2O3(s)→ Al2O3(s) + 2Fe(s)
350 kJ/mol, or 41% of the energy for the
thermite reaction comes from the difference
in lattice energies. Why is Al2O3(s) so much
more stable than Fe2O3(s)?
Ion
ΔH = -850
kJ/mol
64 + 140 = 204
Fe3+
Ionic radius
(pm)
Fe3+ 64
Al3+ 54
O2- 140
O
54 + 140 = 194
Al3+
O
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Chem 6A Michael J. Sailor, UC San Diego
SOLUTION: Lattice Enthalpies
Calculate the electrostatic energy in one M-O bond:
Coulomb’s law:
charge
z1z2q 2
E=
4 πεo r1−2
64 + 140 = 204
distance
Fe-O:
)
3⋅ 2 ⋅ (1.602 ×10Al-O
Difference:
4 π (8.854 ×10 )(204 ×10 )
ionic bond is
= 6.78 x 10
stronger than Fe-O
Al-O:
)
3⋅ 2 ⋅ (1.602
×10 by
ionic
bond
E=
4 π (8.854 ×10 )(194 ×10 )
210 kJ/mol
= 7.13 x 10
E=
Fe3+
O
−19 2
−12
−12
-18
J
54 + 140 = 194
−19 2
−12
Al3+
−12
-18
O
J
Chem 6A Michael J. Sailor, UC San Diego
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Lattice enthalpies and hardness
7200
diamond
Hardness (Knoop scale)
Sapphire ring
Diamond solitaire
5400
3600
SiC
sapphire (Al O )
1800
2
3
hematite (Fe O )
2 3
halite (NaCl)
0
0
1000
2000
3000
Hematite stone bracelet
4000
5000
Lattice enthalpy density, kJ/mL
6000
Rock salt (halite)
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Lattice Enthalpies
A related (but simpler) problem:
Calculate the enthalpy of formation of AgF(s) from the following
data:
Process
Enthalpy (ΔH), kJ/mol
Lattice enthalpy of AgF(s)
-971
First ionization energy of Ag(g)
+731
+284
Enthalpy of formation of Ag(g)
Enthalpy of formation of F(g)
+79
+328
Electron affinity of F(g)
a) +451 kJ/mol
b) -284 kJ/mol
c) -205 kJ/mol
d) -246 kJ/mol
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e) none of the above
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Lattice Enthalpies
A related (but simpler) problem:
Calculate the enthalpy of formation of AgF(s) from the following
data:
Process
Enthalpy (ΔH), kJ/mol
Ag+(g) + F-(g) → AgF(s)
-971
Ag(g) → Ag+(g) + e-
+731
+284
Ag(s) → Ag(g)
½ F2(g) → F(g)
+79
-328
F(g) + e- → F-(g)
a) +451 kJ/mol
b) -284 kJ/mol
c) -205 kJ/mol
d) -246 kJ/mol
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e) none of the above
Chem 6A Michael J. Sailor, UC San Diego
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Lewis Dot Structures
Fig 9.4
Dots used to indicate covalent bonds
Chem 6A Michael J. Sailor, UC San Diego
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Drawing Lewis “Dot” Structures
• Count up all valence electrons
• Pair up electrons to form bonds or lone pairs
• Satisfy octet rule (every atom has 8
electrons, either as lone pairs or in shared
bonding pairs)
Examples: CH4, O3, NF3
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Chem 6A Michael J. Sailor, UC San Diego
PROBLEM: Drawing Lewis structures
Draw the Lewis structure for ozone, O3.
Valence electrons for O:
3x6
Total electrons:
18
Total pairs of electrons:
9
Satisfy octet rule with shared electron pairs
O
O
Chem 6A Michael J. Sailor, UC San Diego
O
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PROBLEM: Drawing Lewis structures
How many lone pairs of electrons are there in the Lewis
structure of NF3?
a. 1
b. 3
c. 6
d. 9
Lone pairs
e. 10
F
F
N
Total electrons:
26
Total pairs of electrons:
13
F
Total bonding pairs: 3
Total lone pairs: 10
ANSWER: e
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Chem 6A Michael J. Sailor, UC San Diego
Bonding in Compounds
covalent
ionic
Na+
H
O
water molecule
H
Cl
Cl
Cl-
chlorine molecule
Covalent bond = neutral atoms
held together by sharing a
pair of electrons
Ionic bond = charged atoms
(ions) held together by
electrostatic forces
An assembly of atoms held together
by covalent bonds is a molecule
Chem 6A Michael J. Sailor, UC San Diego
Coulomb’s law:
z1z2q 2
E=
4 πεo r1−2
charge
distance
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Covalent Bonds
The bond energy of F2 is 159 kJ/mol. The bond energy of H2 is
432 kJ/mol. What is the energy of the HF bond?
If it were a simple average, it would be 296 kJ/mol
Actual value: 565 kJ/mol
Why is the bond energy of HF so much larger (by 269 kJ/mol)?
Chem 6A Michael J. Sailor, UC San Diego
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Electronegativity
• Relative ability of an atom to attract shared
electrons in a bond
• Pauling scale is based on relative bond
energies (H-F bond compared to H-H and
F-F bonds)—used in Silberberg text
• Mulliken scale is based on difference in
electron affinity and ionization potential
Chem 6A Michael J. Sailor, UC San Diego
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Pauling Electronegativity Scale
Fig. 9.19
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Chem 6A Michael J. Sailor, UC San Diego
Using Electronegativity to
Classify Bonds
Fig. 9.21
Things that lead to increased covalent character:
For anions: Polarizable large, highly negative
For cations: Large polarizing power, small, highly positive
Examples:
Ba-Cl (Δ electronegativity) = 3.0 – 0.9 = 2.1 ionic
Bi-I (Δ electronegativity) = 2.5 – 1.9 = 0.6 polar covalent
Si-H (Δ electronegativity) = 2.1 – 1.8 = 0.3 covalent
O-H (Δ electronegativity) = 3.5 – 2.1 = 1.4 polar covalent
Chem 6A Michael J. Sailor, UC San Diego
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Water is a good solvent because
it has a dipole:
This is a POLAR molecule
Chem 6A Michael J. Sailor, UC San Diego
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PROBLEM: Using electronegativity
to predict ionic character
Which has more ionic character: NH3 or NO2?
N-H electronegativity difference = 3.0 – 2.1 = 0.9
(N carries partial negative charge)
N-O electronegativity difference = 3.0 – 3.5 = 0.5
(O carries partial negative charge)
ANSWER: NH3 is more ionic
Chem 6A Michael J. Sailor, UC San Diego
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Noble Gas Compounds
History: In 1962, Neil Bartlett noticed that platinum
hexafluoride ionized O2 to O2+:
O2(g) + PtF6(g) → O2PtF6(s)
(The salt, O2+PtF6-(s))
Ionization Energy:
O2 → O2+ + e-
Xe → Xe+ + e-
1165 kJ/mol
1170 kJ/mol
so he tried the reaction:
Xe(g) + PtF6(g) → XePtF6(s)
The first compound made from a noble gas
Chem 6A Michael J. Sailor, UC San Diego
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MEMS Display-Qualcomm’s Mirasol
Chem 6A Michael J. Sailor, UC San Diego
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MicroElectroMechanical Systems
(MEMS) Micro-engine and
transmission
Applications
• Digital projectors
• Medical Devices
• Lab-on-a-chip
Hinge
Mirrors for Digital
Light Projector
(DLP technology)
35
Chem 6A Michael J. Sailor, UC San Diego
XeF2 used to etch MEMS devices
Si(s)+ 2XeF2(g) → 2Xe(g) + SiF4(g)
XeF2 attacks Si very selectively. It doesn’t react with SiO2
-- “Controlled Pulse-Etching with Xenon Difluoride” Kristofer S. J. Pister,
Transducers '97, the Ninth Inter. Conf. Solid-State Sensors & Actuators,
Chicago, IL, June 1997.
MICROMECHANICAL
RESONANT MAGNETIC
SENSOR IN STANDARD
CMOS
Beverley Eyre and Kristofer S. J.
Pister TRANSDUCERS ’97
1997 lnternational Conference on
Solid-state Sensors and Actuators
Chicago, June 16-19, 1997
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Chem 6A Michael J. Sailor, UC San Diego
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XeF2 used to etch MEMS devices
Q: “How did you find XeF2 in the first place?”
PISTER: While discussing TMAH+silicic acid as a possible
CMOS post-process etchant, a colleague of mine, Eli
Yablonovitch, suggested that XeF2 might be just the thing we
were looking for. We spent several days calling chemical supply
and excimer laser companies, all of whom denied that XeF2
could exist, and we ultimately gave up. Several weeks later, in a
discussion on polysilicon stringer etching, Mike Hecht of JPL
mentioned that he had used XeF2 to etch 300 microns of silicon
and stop on 50 Angstroms of SiO2, and that XeF2 could be
purchased from PCR…Mike reassembled his old reactor at JPL,
and we etched the first CMOS chip together at UCLA.
http://www.memsnet.org/pipermail/mems-talk/1995-March/000170.html
Kris Pister, UCSD ‘86 (Warren)
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Chem 6A Michael J. Sailor, UC San Diego
XeF2
http://www.xactix.com
Physical Properties
Density
MW
MP
Vapor Pressure
temp.)
Chem 6A Michael J. Sailor, UC San Diego
4.32
169.290
130-135 °C
3.9 Torr (@ room
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