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Visible light is a form of electromagnetic radiation

Electromagnetic radiation is characterized by its wave nature

6.1 The wave nature of light

The electromagnetic

spectrum

or “radiant energy”

Chapter 6 Electronic structure of atoms

c = nl

6.2 Quantized Energy and Photons

1. Blackbody radiation

2. The photoelectric effect

3. Emission spectra

emission of light from hot objects

emission of electrons from metal surfaceson which light shines

emission of light from electronically excited gas atoms

Some phenomena cannot be explained using a wave model of light:

A quantum is the smallest amount of energy that can be emitted or absorbed as electromagnetic radiation

The relationship between energy, E, and frequency is:

E = hnwhere h is Planck’s constant = 6.626 × 10-34 joule-seconds (J.s)

Energy of one photon = E = hn = hc/l

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Photoelectric effect

Chem 101 3

Chem 101 4

Bohr’s model predicts:

= -2.18 x 10-18 J {(1/nf2)-(1/ni2)} = E photon = h

Allowed energy states

Ground state

Electron escapes the nucleus -ionization

E = -hcRH (1/n2)= -2.18 x 10-18 J (1/n2), where n is an integer between 1 and ∞

ΔE = Efinal – Einitial

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l = h / mv

6.4 The wave behavior of matter

λ = wavelength (m)h = Plank’s constant (s-1)m = mass (kg)v = velocity (m/s)

Electron diffraction

Electron microscope imagehttp://intranet.dalton.org/departments/science/Science5/microscopy.html

Chem 1016

www.grayfieldoptical.com/

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A. No matter waves are produced.B. No, because the mass of the baseball is too large.C. Yes; but too small to allow any way of observing them.D. Yes; and they can be observed.E. Let me ask YOU; what is the sound of one hand clapping?

A baseball pitcher throws a fastball at 150 km/h. Does that moving baseball generate matter waves? If so, can we observe them?

C. Yes; but too small to allow any way of observing them.

A baseball pitcher throws a fastball at 150 km/h. Does that moving baseball generate matter waves? If so, can we observe them?

λ = h/mv= (6.63 x 10-34 J s)/{(150 g)(150 km/hr)}

λ = {6.63 x 10-34 (kg m2 /s2) s}/{(0.150 kg)(150x103m/3600s)}

λ = 1.08x10-34m

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Chem 101 9

Δx • Δ(mv) ≥ h / 4π

Δx = uncertainty in position (m)Δ(mv) = uncertainty in momentum (kgms-1)h =Plank’s constant (s-1)

4π = 4π

Probability function (Ψ2)

analogy: compare probability of dart landing herevs. there

Chem 101 10

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11

For interest only: do not need to know

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Orbital shapes (www.quimica3d.com)

Chem 101 13

To memorise

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To be aware of (ie: draw a d orbital)

Orbital shapes

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Chem 101 17

s orbitals (ℓ = 0)

n = 1, ℓ = 0

1s orbital

n = 2, ℓ = 0

2s orbital

node

[4πr2Ψ(r)2]

pg 230-231 (a closer look)

Chem 101 18

describes the main energy level; specifies electron

shell

describes the shape; specifies subshell

designates specific orbital; specifies orientation

mℓ = (-ℓ),…,0,…,(+ ℓ)

must be a positive integer n = 1,2,3,4,…

maximum value is (n-1), i.e. ℓ = 0,1,2,3…(n-1)use letters for ℓ (s, p, d and f for ℓ = 0, 1, 2, and 3).

maximum value depends on ℓ, can take integral values from – ℓ to + ℓ

the principal quantum number, n

the angular momentum quantum number, ℓ

the magnetic quantum number, mℓ

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A. n=2, l=1,m=-1B. n=3, l=2,m=0C. n=3, l=1,m=2D. n=4, l=0,m=0E. n=1, l=0,m=0

What is not possible? (Once you’ve solved that pass the time by sketching and naming the others)

A. n=2, l=1,m=-1 2px(or y or z)

B. n=3, l=2,m=0 3dxy (or xz or yz or z2 or x2-y2)

C. n=3, l=1,m=2D. n=4, l=0,m=0 4sE. n=1, l=0,m=0 1s

What is not possible? (Once you’ve solved that pass the time by sketching and naming the others)

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Chem 101

6.7 Many electron atoms

1 electron system (H,or He+ etc..) Multi- electron system (all atoms but H)

6.9 Electron Configurations and the Periodic Table

The periodic table can be used as a guide for electron configurations.

s-blockalkali and

alkaline earth

metals

p-blockmain group elements

d-blocktransition metals

f-block lanthanides and actinides

the period number is the value of n

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6.9 Electron Configurations and the Periodic Table

Structure of the atom

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7.1 Development of the Periodic TableThe periodic table is the most significant tool that chemists use for organizing and recalling chemical facts.

arrangement reflects trends in chemical and physical properties

Reaction of the alkali metals with water

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7.2 Effective Nuclear ChargeLi Na

Zeff= Z-S

7.3 Sizes of atoms and ions

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++++

++++

+++ ++++

+++++

++++++

+++ ++++++++

+++++++++

+++++++++

+++++++ +++++++

+ ++

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31

radius

valence electrons have higher n values,higher shells

incr

ease

sincreases

higher Zeff pulls the valence electrons toward the nucleus

7.3 Sizes of atoms : trends

Ionic radii relative to metallic (or covalent) radii (in Å)

Brown, LeMay, Bursten & Murphy “Chemistry The Central Science” 11th Ed., Pearson 2009, Fig. 7.8, p. 263

7.3 Periodic trends in Ionic Radii

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isoelectronic species

ions with same charge:

7.3 Periodic trends in Ionic Radii

Note sharp increase in ionization energy when a core electron is removed.

7.4 Ionization Energy

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7.5 Electron Affinities

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Summary:

easy to form negative ions

easy to form positive ions

smallest atoms

largest atoms

Chapter 8 :Basic concepts of chemical bonding8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule

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39

K+: [Ar]

The Octet Rule

A. Radius of Mg2+ > radius of F-

B. Radius of Mg2+ < radius of F-

C. We can’t tell as they are not in the same row or column

Which has the larger radius: Mg2+ or F- ?

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B. Radius of Mg2+ (0.86 A) < radius of F- (1.19 A) as they both have the same number of electrons [Ne] but Mg2+ has the greater Z (12 vs 9).

Which has the larger radius: Mg2+ or F- ?

isoelectronic species

Grey are atoms, blue are anions, red are cations

495 kJ mol-1 to remove electron

from sodium

8.2 Ionic bonding

349 kJ mol-1 back by giving electron

to chlorine

yet reaction of sodium metal and chlorine gas to form sodium chloride is violently exothermic

Also need to account for: electrostatic attraction between the newly formed sodium cation and chloride anion

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Formation of NaCl from Na and Cl2

Chem 101 44

Lattice energy

dQQEel

21

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45

Energetics of Ionic bonds:Lattice energy

Electron pairs are usually unequally shared between different atoms:

8.4 Bond Polarity and Electronegativity

F2 HF

nonpolarcovalent

bond

polarcovalent

bondelectrons

shared equally

one atom attracts bonding electrons

more than the other

high electron density

low

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Chem 101 47

Ionic and covalent bonding

Chem 101 48

represent the two extremes of the continuum

ionic covalent

extended lattice structures

molecules

Properties:

o high melting solids, brittle

o soluble in water

o solutions and melts conductelectricity

o low melting and boiling points (often gases, liquids at room temp.)

o most are insoluble in water

o solutions and melts are non-conducting

Properties:

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8.5 Lewis Structures

Lewis structures

Chem 101 50

1. Find the total number of valence electrons (account for any charges) = TOTAL

2. decide connection between atoms, draw a line to represent 1 electron pair for each connection, count the electrons SHARED

3. calculate the remaining electrons = TOTAL – SHARED, assign these to the terminal atoms to make octet (or 2 for H atom)

4. any electrons left? – put them on the central atom

5. if central atom doesn’t have an octet, make multiple bonds from nonbondedelectron pairs on terminal atoms

Choose central atom correctly

least electronegative atom(not H)

oxygen rarely bonds to itself

PCl3

For PCl3: 5 (for P) + (3 × 7) (for Cl3) = 26

26 – 6 = 20

3 bonds= 6

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Multiple Bonds

Chem 101 51

Formal charges

Chem 101 52

often can make more than one Lewis structure

bookkeeping of electrons

which one is correct?

calculate the charge on atom IF all bonding electrons shared equally

assign to the atom all unshared (nonbonding) electrons

½ of all bonding (shared) electrons+

formal charge = number of valence electrons – total assigned electrons

Evaluate Lewis structures

more stable if there are small (or no) formal charges

the most electronegative atom has the most negative formal charge

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Example:

Chem 101 53

N HH

H

N: valence = 5, formal charge = 5 – [2 + ½{6}] = 0

H: valence = 1, formal charge = 1 – [0 + ½{2}] = 0

Resonance hybrids

Chem 101 54

hybrid is intermediate between the two “parent” structures

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Resonance has impact on bond lengths and strengths

55

compare NO+, NO2- and NO3

-

N O

N OO-

N OO-

N-O bond average = 1.5

N-O bond average = 1.3

N-O bond = 3

aromatic compounds show resonance

Chem 101 56

benzene

shorthand notation omits H atoms

C-C bond average = 1.5

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Exceptions to the octet rule

Chem 101 57

2. less than an octet

1. odd number of electrons

or?

3. more than an octet

Exceptions to the octet rule…

Chem 101 58

can expand valence shell to make a Lewis structure with lower formal charge

SO42-

experimental info:

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8.8 Strength of Covalent Bonds

Chem 101 59

CH4 + Cl2 CH3Cl + HCl

Chem 101 60

ΔHrxn = 413kJ + 242 kJ – {328 kJ + 431 kJ} = -104 kJ

ΔH is negative (rxn. is exothermic) whenweak bonds are broken and strong bonds are formed

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Bond lengths

Chem 101 61

also depend on nature of atom and type of bond

trends : shorter bonds are stronger

1.47 163 kJ/mol

1.24 418 kJ/mol

N N

N N

N N 1.10 941kJ/mol

also calculated as averages

9.1 Molecular Shapes

AB2AB3

linear bent trigonal planar

trigonal pyramidal

T-shaped

Chapter 9 Molecular Geometry and Bonding Theories

For molecules of the general form ABn there are 5 fundamental shapes:

180° 109.5°

linear trigonal planar

120°

tetrahedral

90°

120°

90°

90°

trigonal bipyramidal

octahedral

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1. Draw Lewis structure, count electron domains

We use the electron-domain geometry to help us predict the molecular geometry.

2. Arrange electron domains to minimize repulsion

3. Inspect arrangement of atoms to determine molecular geometry

3

tetrahedral

trigonal pyramidal

2

41

2 0

3 0

2 1

2

3

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4 0

3 1

4

5 05

4 1

PCl5

SF4lone pair always in least crowded position

Molecules with Expanded Valence Shells

Atoms that have expanded octets have five electron domains (trigonal bipyramidal) or six electron domains (octahedral) electron-domain geometries.

2 2

Effect of nonbonding electrons and multiple bonds on bond angles

bonding pair

non-bonding pair

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3 25

2 3

ClF3

XeF2

6 06

5 1

SF6

BrF5

4 2 XeF4

note position of lone pairs

lone pairs as farapart as possible

9.3 Molecular Shape and Molecular Polarity

Dipoles are a vector quantity

overall dipole moment

= 0

electron density models

(red = high, blue = low)

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9.4 Covalent Bonding and Orbital Overlap

HCl2

HHCl

overlap regions

The change in potential energy as two hydrogen atoms combine to form the H2 molecule:

H2

9.5 Hybrid Orbitals

s p2 × sp

large lobes of sp hybrid orbitals

overlap regions

F 2porbital

F 2porbital

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D. The unhybridized p-orbital is perpendicular to the plane of the sp2 orbitals.

In an sp2 hybridized atom, what is the orientation of the unhybridized p orbital relative to the three sp2 hybrid orbitals?

one sorbital

two porbitals

hybridize

three sp2

hybrid orbitals

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9.5 Hybrid Orbitals

s p2 × sp

large lobes of sp hybrid orbitals

overlap regions

F 2porbital

F 2porbital

one sorbital

two porbitals

hybridize

three sp2

hybrid orbitals

sp2 hybrid orbitals shown together

(large lobes only)

one sorbital three p orbitals

four sp3 hybrid orbitals

sp3 hybrid orbitals shown together

(large lobes only)

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For geometries involving expanded octets on the central atom, we use d orbitals in our hybrids:

octahedrals, p, p, p, d

five sp 3d

s, p, p, p, d, d

six sp 3d 2

trigonal bipyramidal

9.5 Hybrid Orbitals

one π bond

9.6 Multiple Bonds

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Delocalized π Bonding

When writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion

In reality, each of the four atoms in the nitrate ion has a p orbital

the π electrons are delocalized throughout the ion

The p orbitals on all three oxygens overlap with the p orbital on the central nitrogen

The organic molecule benzene has six σ bonds and a p orbital on each carbon atom

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Molecular orbital (MO) theory (only section 9.7)

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MO diagram (energy level diagram)

Chem 101 83

bonding MO lowers energy

antibonding MO raises energy

bonding electrons

Bond order

Bond order = ½ {no. bonding electrons – no. antibonding electrons}

antibonding electrons

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Metal bonding (Sections 23.5 and 12.2)

MO model

an infinite chain of atoms

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Metals, insulators and semiconductors

11.1 A molecular comparison of gases, liquids and solidsThe fundamental difference between states of matter is the distance between particles.

gasliquid

solid

incompressible

Chapter 11 Intermolecular forces, liquids, and solids

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Intermolecular forces: a summary

interacting molecules or ions

are ions involved?

are polar molecules and ions

both present?

are polar molecules involved?

are H atoms bonded to N, O or

F atoms?A) ionic bondinge.g. NH4NO3

YES

NO

B) ion-dipole forcese.g. NaCl in H2O

YES NO

E) dispersion forcesonly (induced dipoles)

e.g. Ar(l), I2(s)

D) dipole-dipole forces

e.g. H2S, CH2Cl2

C) hydrogen bondinge.g. H2O, NH3, HF

YES

NO

NO

YES

van der Waals forcesA > B > C > D > E

11.2 Intermolecular forces

covalent bond (strong)

intermolecular attraction (weak)

Dipole-Dipole Interactions

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Hydrogen bonding

H2O

CH4

H2SeH2Te

SiH4GeH4

SnH4H2S

London Dispersion Forces

e ―

e ―

He atom

momentarily polar

δ– δ+

e ―

e ―e ―

e ―

He atom 1 δ– δ+He atom 2 δ– δ+

n-pentanebp 309 K

neopentanebp 283 K

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Ion-dipole forces

Ion-dipole interactions, are an important force in solutions of ions.

cation-dipole

anion-dipole

11.8 Bonding in solidsMolecular Solids

Consist of atoms or molecules held together by intermolecular forces.

London dispersiondipole-dipole

hydrogen bonding

benzene toluene phenol 5 –95 43 80 111 182

high mp due to efficient packing

high bp due to larger intermolecular forces

high mp & bp due to hydrogen

bonding

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Covalent-Network Solids

Consist of atoms held together, in large networks or chains, with covalent bonds.

1.42 Å

3.41 Å

Ionic Solids

Consist of ions held together by ionic bonds (i.e. by electrostatic forces of attraction).

CsCl ZnS“zinc blende”

CaF2“fluorite”

Cs

Cl

Zn

SCa

F

Metallic Solids

Consist entirely of metal atoms;

not covalently bonded.

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Hydrocarbons

Chem 101 97

contain only C and H

four different types: defined by kind of carbon to carbon bonds

largest possible no. of H atoms saturated hydrocarbons

unsaturated hydrocarbons

Drawing hydrocarbon structures

CH3CH2CH2CH3

C4H10

C4H10

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example:

Chem 101 99

ethyl

methyl

1 2 3 4 5 6 7 88 7 6 5 4 3 2 1

4-ethyl-2,7-dimethyloctaneoctaneethyl- methyl

parent name

Chem 101 100

halogens

-F

-Cl

-Br

-I

fluoro

chloro

bromo

iodo

-OH

-NO2

hydroxy

nitro

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Cycloalkanes - general formula: CnH2n

CH2 CH2

CH2

Alkenes - general formula: CnH2n

Chem 101

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Chem 101

1

4

6

1

52

3

4

6

3

5

2

cis-3,4-dimethyl-3-hexenetrans-3,4-dimethyl-3-hexene

1

4

6

1

5

2

3

4

6

3

5

2

cis-3,4-dichloro-3-hexenetrans-3,4-dichloro-3-hexene

Reactions of alkenes

Chem 101 104

addition reactions add to the two atoms that form the double bond

sp2 hybridized C atoms become sp3 hybridized

Br2, Cl2

general reaction:

Cl2

halogenation

same rxn for Br2

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Chem 101 105

H2

H2O

addition rxns, cont’d

hydrogenation

Ni, 500 °CNi, 500 °C

hydration

H2OH2OH2SO4H2SO4

H2H2

Chem 101 106

addition rxns, cont’d

HBr

HBrHBr

addition reactions

2 H2

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Reactions of aromatics

Chem 101 107

substitution reactions

in general:

all 6 H atoms on the ring are equivalent,

HNO3

H2SO4

H2OH2O

Br2

FeBr3

HBrHBr

Chem 101 108

be able to identify ALL of these functional groups

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Carboxylic acids sp2 hybridized C atom

Chem 101 109

produced by oxidation of alcohols

CH3CH2CH2OH + [O] CH3CH2CH

O

+ H2O

CH3CH2C

O

OH

[O]alcohol +excess O2 carboxylic acid

Esters

Chem 101 110

sp2 hybridized C atom

formed by condensation reaction between alcohols and carboxylic acids

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ester reaction

Chem 101 111

saponification base catalyzed ester hydrolysis

add H2O

CH3CH2C

O

+ H2OO[OH-]

CH3

HO H

CH3CH2C

O

OH HO-CH3+

ester + water [OH-]

alcohol + acid

amides, cont’d

Chem 101 112

prepared by reaction of carboxylic acid with ammonia or amine

condensation rxn.

amides undergo hydrolysis

RC N

O

R1

R2

+ H2OH+

heatR

C HN

O

R1

R2

OH +

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Chapter 12 Modern Materialsfocus on polymers, semiconductors, liquid crystals, superconductors

For the readings focus on: 12.1 (all) 12.2 (all) 12.3 484-485,488, 489,492, 12.6:499-505 (all) 12.8(all)

10th edition sections 12.1-12.2 (up to p. 501), 12.5

As you read this material ask yourself the following questions:

How are the properties of these novel materials related to the bonding and the intermolecular forces in these compounds?

12.6 PolymersMaking polymers

Many synthetic polymers have a backbone of C–C bonds.

ethylene ethylene ethylene polyethylene

coupling of monomers through multiple bonds is addition polymerization

Chem 101 114

C CH

H

H

H

Initiator

C C

H

H

H

H

C CH

H

H

H+ C C C C

H

H

H

HH

H H

H

ethylene polyethylenemonomer

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12.6 PolymersMaking polymers

Many synthetic polymers have a backbone of C–C bonds.

ethylene ethylene ethylene polyethylene

coupling of monomers through multiple bonds is addition polymerization

Polyethylene (PE)

Polypropylene (PP)

Polystyrene (PS)

polyvinyl chloride (PVC)

Addition polymers: Monomer Polymer

+ H2Oamine carboxylic

acid

Condensation polymerization: two molecules are joined to form a larger molecule by the elimination of a small molecule. e.g. water

polymers formed from two different monomers are called copolymersamide

Ester linkage

OH

O

HO

O

+ HOOHn n

H2O

O

O

HO

O

OH + H2O

O

OO

O

n

n

+ n H2O

first step

terephthlaic acid and 1,2 ethylene glycol form Mylar

diacid dialcohol

Amide linkage

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Condensation polymerization: two molecules are joined to form a larger molecule by the elimination of a small molecule.

polyurethane

polyethylene terephthalate

nylon 6,6

polycarbonate

Condensation polymers:

… back to: Semiconductors (section 12.1)

n -typep -type

- dopant atom has more valence electrons than the host atom- adds electrons to the conduction band

- e.g. P into Si

- dopant atom has fewer valence electrons than the host atom

- leads to holes in the valence band- e.g. B into Si