chapter 9: models of chemical bonding - sailor...

Chem 6A Michael J. Sailor, UC San Diego

Chapter 9:Models of Chemical Bonding

Chem 6A, Section D Oct 11, 2011

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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

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Quiz 6 score histogram


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F D C- C C+ B- B B+ A- A A+

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3 Types of Chemical BondsFig 9.2

ThisChapter

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Bonding in Compounds

covalent ionic

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

€

E =z1z2q

2

4πεor1−2

Coulomb’s law:charge

distance

Na+

Cl-H

H

O ClCl

water moleculechlorine molecule


Ionic Bonding

Na+

Cl-

The rock salt lattice

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Lattice enthalpies and ionic radius

650

700

750

800

850

900

950

3 3.2 3.4 3.6 3.8 4 4.2 4.4

Lattice Enthalpy vs 1/(Ionic radii)La

ttic

e en

thal

py, k

J/m

ol

1/(Na-X) distance, Å- 1

NaF

NaCl

NaBr

NaI


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?

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PROBLEM: Lattice EnthalpiesCalculate the lattice enthalpy of Al2O3(s) at 25°C from the following data: Process Enthalpy (ΔH), kJ/molLattice enthalpy Al2O3(s) ?First ionization energy of Al(g) +578Second ionization energy of Al(g) +1820

Third ionization energy of Al(g) +2750Enthalpy of formation of Al(g) +294Enthalpy of formation of O2(g) 0Bond energy of O2(g) +498

Electron affinity of O(g) -141Electron affinity of O-

(g) +844Enthalpy of formation of Al2O3(s) -1676


PROBLEM: Lattice EnthalpiesCalculate the lattice enthalpy of Fe2O3(s) at 25°C from the following data: Process Enthalpy (ΔH), kJ/molLattice enthalpy Fe2O3(s) ?First ionization energy of Fe(g) +759Second ionization energy of Fe(g) +1561

Third ionization energy of Fe(g) +2957Enthalpy of formation of Fe(g) +415Enthalpy of formation of O2(g) 0Bond energy of O2(g) +498

Electron affinity of O(g) -141Electron affinity of O-

(g) +844Enthalpy of formation of Fe2O3(s) -826

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PROBLEM: Lattice Enthalpies

Rewrite process in terms of chemical equations: Process Enthalpy (ΔH), kJ/molLattice enthalpy Fe2O3(s) ?

First ionization energy of Fe(g) +759Second ionization energy of Fe(g) +1561Third ionization energy of Fe(g) +2957Enthalpy of formation of Fe(g) +415

Enthalpy of formation of O2(g) 0Bond energy of O2(g) +498Electron affinity of O(g) -141

Electron affinity of O-(g) +844

Enthalpy of formation of Fe2O3(s) -8262Fe(s)+ 3/2O2(g) → Fe2O3(s)

O-(g) + e- → O2-

(g) O(g) + e- → O-

(g) O2(g) → 2O(g)

Fe(s) → Fe(g)

Fe2+(g) → Fe3+

(g) + e-

Fe+(g) → Fe2+

(g) + e-

Fe(g) → Fe+(g) + e-

2Fe3+(g)+ 3O2-

(g) → Fe2O3(s)


SOLUTION: Lattice EnthalpiesRearrange equations to add up to 2Fe3+

(g)+ 3O2-(g) → Fe2O3(s):

Process Enthalpy (ΔH), kJ/mol

2Fe(s)+ 3/2O2(g) → Fe2O3(s) -826

3O2-(g) → 3O-

(g) + 3e- 3(-844) 3O-

(g) → 3O(g) + 3e- 3(+141)

3O(g) → 3/2 O2(g) 3/2(-498)

2Fe(g) → 2Fe(s) 2(-415)

2Fe3+(g) + 2e- → 2Fe2+

(g) 2(-2957)

2Fe2+(g) + 2e- → 2Fe+

(g) 2(-1561)

2Fe+(g) + 2e- → 2Fe(g) 2(-759)

2Fe3+(g)+ 3O2-

(g) → Fe2O3(s) -15,066 ?

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SOLUTION: Lattice EnthalpiesDo the same for 2Al3+

(g)+ 3O2-(g) → Al2O3(s):

Process Enthalpy (ΔH), kJ/mol2Al(s)+ 3/2O2(g) → Al2O3(s) -1676

3O2-(g) → 3O-

(g) + 3e- 3(-844) 3O-

(g) → 3O(g) + 3e- 3(+141)

3O(g) → 3/2 O2(g) 3/2(-498)

2Al(g) → 2Al(s) 2(-294)

2Al3+(g) + 2e- → 2Al2+

(g) 2(-2750)

2Al2+(g) + 2e- → 2Al+

(g) 2(-1820)

2Al+(g) + 2e- → 2Al(g) 2(-578)

2Al3+(g)+ 3O2-

(g) → Al2O3(s) -15,416


SOLUTION: Lattice EnthalpiesSummary: 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

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?

The enthalpy of the thermite reaction is:

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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)?

SOLUTION: Lattice Enthalpies2Al(s)+ Fe2O3(s)→ Al2O3(s) + 2Fe(s) ΔH = -850 kJ/mol

Ion Ionic radius(pm)

Fe3+ 64Al3+ 54O2- 140

Fe3+ O

Al3+ O

64 + 140 = 204

54 + 140 = 194


SOLUTION: Lattice EnthalpiesCalculate the electrostatic energy in one M-O bond:

€

E =z1z2q

2

4πεor1−2


distance

Fe-O:

€

E =3 ⋅ 2 ⋅ (1.602 ×10−19)2

4π (8.854 ×10−12)(204 ×10−12)= 6.78 x 10-18 J

Al-O:

€

E =3 ⋅ 2 ⋅ (1.602 ×10−19)2

4π (8.854 ×10−12)(194 ×10−12)= 7.13 x 10-18 J

Difference: Al-O ionic bond is

stronger than Fe-O ionic bond by

210 kJ/mol

Fe3+ O

Al3+ O

64 + 140 = 204

54 + 140 = 194

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Lattice enthalpies and hardness

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3600

5400

7200

0 1000 2000 3000 4000 5000 6000

Har

dnes

s (K

noop

sca

le)

Lattice enthalpy density, kJ/mL

diamond

SiC

sapphire (Al2O

3)

hematite (Fe2O

3)

halite (NaCl)Hematite stone bracelet

Rock salt (halite)

Sapphire ringDiamond solitaire


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PROBLEM: Lattice EnthalpiesA 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 Enthalpy of formation of Ag(g) +284 Enthalpy of formation of F(g) +79 Electron affinity of F(g) +328 a) +451 kJ/mol b) -284 kJ/mol c) -205 kJ/mol d) -246 kJ/mol e) none of the above

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PROBLEM: Lattice EnthalpiesA 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

Ag(s) → Ag(g) +284 ½ F2(g) → F(g) +79 F(g) + e- → F-

(g) -328 a) +451 kJ/mol b) -284 kJ/mol c) -205 kJ/mol d) -246 kJ/mol e) none of the above


Lewis Dot StructuresFig 9.4

Dots used to indicate covalent bonds

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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


PROBLEM: Drawing Lewis structuresDraw the Lewis structure for ozone, O3.

Valence electrons for O: 3x6 Total electrons: 18Total pairs of electrons: 9

OOO

Satisfy octet rule with shared electron pairs

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PROBLEM: Drawing Lewis structuresHow many lone pairs of electrons are there in the Lewis structure of NF3? a. 1 b. 3 c. 6 d. 9 e. 10

ANSWER: e

N

F

F

FLone pairs

Total electrons: 26Total pairs of electrons: 13

Total bonding pairs: 3Total lone pairs: 10


Bonding in Compounds

covalent ionic

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

€

E =z1z2q

2

4πεor1−2


distance

Na+

Cl-H

H

O ClCl

water moleculechlorine molecule

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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)?


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

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Pauling Electronegativity ScaleFig. 9.19


Using Electronegativity to Classify Bonds Fig. 9.21

Things that lead to increased covalent character:For anions: Polarizable large, highly negativeFor cations: Large polarizing power, small, highly positiveExamples: Ba-Cl (Δ electronegativity) = 3.0 – 0.9 = 2.1 ionicBi-I (Δ electronegativity) = 2.5 – 1.9 = 0.6 polar covalentSi-H (Δ electronegativity) = 2.1 – 1.8 = 0.3 covalentO-H (Δ electronegativity) = 3.5 – 2.1 = 1.4 polar covalent

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Water is a good solvent because it has a dipole:

This is a POLAR molecule


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

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Noble Gas CompoundsHistory: In 1962, Neil Bartlett noticed that platinum hexafluoride ionized O2 to O2

+:

O2(g) + PtF6(g) → O2PtF6(s)

Ionization Energy:

O2 → O2+ + e- 1165 kJ/mol

Xe → Xe+ + e- 1170 kJ/molso he tried the reaction:Xe(g) + PtF6(g) → XePtF6(s)

(The salt, O2+PtF6

-(s))

The first compound made from a noble gas


MEMS Display-Qualcomm’s Mirasol

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MicroElectroMechanical Systems (MEMS)

Applications• Digital projectors• Medical Devices• Lab-on-a-chip

Hinge

Micro-engine and transmission

Mirrors for Digital Light Projector

(DLP technology)


XeF2 used to etch MEMS devices

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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 CMOSBeverley Eyre and Kristofer S. J. Pister TRANSDUCERS ’971997 lnternational Conference on Solid-state Sensors and Actuators Chicago, June 16-19, 1997

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XeF2 used to etch MEMS devicesQ: “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)


XeF2http://www.xactix.com

Physical PropertiesDensity 4.32 MW 169.290 MP 130-135 °C Vapor Pressure 3.9 Torr (@ room temp.)

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chapter 9: models of chemical bonding - sailor...

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