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Chapter 6 Electronic structure of atoms
6.2 Quantized Energy and Photons
6.1 The wave nature of light
Visible light is a form of electromagnetic radiation
Some phenomena cannot be explained using a wave model of light:
or “radiant energy”
emission of light from hot objects
1. Blackbody radiation
2. The photoelectric effect
3. Emission spectra
The
electromagnetic
spectrum
emission of electrons from metal surfaces
on which light shines
emission of light from electronically
excited gas atoms
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:
Electromagnetic radiation is characterized by its wave nature
E = hn
c = nl
where h is Planck’s constant = 6.626 × 10-34 joule-seconds (J.s)
Energy of one photon = E = hn = hc/l
Bohr’s model predicts:
Photoelectric effect
Electron escapes the nucleus ionization
Allowed energy states
Ground state
E = -hcRH (1/n2)
= -2.18 x 10-18 J (1/n2), where n is an integer between 1 and ∞
ΔE = Efinal – Einitial
= -2.18 x 10-18 J {(1/nf2)-(1/ni 2)} = E photon = h
Chem 101
3
6.4 The wave behavior of matter
Chem 101
4
Electron diffraction
l = h / mv
λ = wavelength (m)
h = Plank’s constant (s-1)
m = mass (kg)
v = velocity (m/s)
www.grayfieldoptical.com/
Electron microscope image
http://intranet.dalton.org/departments/science/Science
5/microscopy.html
6
Chem 101
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07/12/2010
A baseball pitcher throws a fastball at 150 km/h. Does that
moving baseball generate matter waves? If so, can we
observe them?
A.
B.
C.
D.
E.
A baseball pitcher throws a fastball at 150 km/h. Does that
moving baseball generate matter waves? If so, can we
observe them?
No matter waves are produced.
No, because the mass of the baseball is too large.
Yes; but too small to allow any way of observing them.
Yes; and they can be observed.
Let me ask YOU; what is the sound of one hand clapping?
C.
Yes; but too small to allow any way of observing 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
Probability function (Ψ2)
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
analogy: compare probability of dart landing here
vs. there
Chem 101
9
Chem 101
10
For interest only: do not need to know
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07/12/2010
To memorise
Orbital shapes (www.quimica3d.com)
Chem 101
13
To be aware of (ie: draw a d orbital)
s orbitals
(ℓ = 0)
[4πr2Ψ(r)2]
Orbital shapes
the principal quantum number,
n
must be a positive integer n = 1,2,3,4,…
the angular momentum quantum number,
n = 1, ℓ = 0
1s orbital
ℓ
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).
the magnetic quantum number,
mℓ
maximum value depends on ℓ, can take integral
values from – ℓ to + ℓ
describes the main energy
level; specifies electron
shell
describes the shape;
specifies subshell
designates specific orbital;
specifies orientation
mℓ = (-ℓ),…,0,…,(+ ℓ)
node
n = 2, ℓ = 0
2s orbital
Chem 101
17
pg 230-231 (a closer look)
Chem 101
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07/12/2010
What is not possible? (Once you’ve solved that pass the
time by sketching and naming the others)
A.
B.
C.
D.
E.
n=2, l=1,m=-1
n=3, l=2,m=0
n=3, l=1,m=2
n=4, l=0,m=0
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.
B.
C.
D.
E.
6.7 Many electron atoms
n=2, l=1,m=-1 2px(or y or z)
n=3, l=2,m=0 3dxy (or xz or yz or z2 or x2-y2)
n=3, l=1,m=2
n=4, l=0,m=0 4s
n=1, l=0,m=0 1s
6.9 Electron Configurations and the Periodic Table
The periodic table can be used as a guide for electron configurations.
the period number is the value of n
d-block
s-block
transition metals
alkali and
alkaline
earth
metals
main
group
elements
f-block
1 electron system (H,or He+ etc..)
p-block
lanthanides and actinides
Multi- electron system (all atoms but H)
Chem 101
6.9 Electron Configurations and the Periodic Table
Structure of the atom
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7.1 Development of the Periodic Table
Reaction of the alkali metals with water
The periodic table is the most significant tool that chemists use for organizing
and recalling chemical facts.
arrangement reflects trends in chemical and physical properties
7.2 Effective Nuclear Charge
Li
Zeff= Z-S
7.3 Sizes of atoms and ions
Na
++
+
+
+++
++
++
++
++
++
+
++
++
++
+++
++++
+++
++++
++++
+++
+++
+++
+++
++++
+++
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7.3 Periodic trends in Ionic Radii
7.3 Sizes of atoms : trends
Ionic radii relative to metallic (or covalent) radii (in Å)
higher Zeff pulls the valence
electrons toward the
nucleus
increases
valence
electrons have
higher n values,
higher shells
increases
radius
31
Brown, LeMay, Bursten & Murphy “Chemistry The Central Science” 11th Ed., Pearson 2009, Fig. 7.8, p. 263
7.3 Periodic trends in Ionic Radii
7.4 Ionization Energy
isoelectronic species
ions with same charge:
Note sharp increase in ionization energy when a core electron is removed.
7.5 Electron Affinities
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Chapter 8 :Basic concepts of chemical bonding
smallest atoms
8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule
Summary:
easy to form
negative ions
easy to form
positive ions
largest atoms
Which has the larger radius: Mg2+ or F- ?
The Octet Rule
A. Radius of Mg2+ > radius of FB. Radius of Mg2+ < radius of FC. We can’t tell as they are not in the same row or column
39
K +:
[Ar]
Which has the larger radius: Mg2+ or F- ?
8.2 Ionic bonding
yet reaction of sodium metal and
chlorine gas to form sodium
chloride is violently exothermic
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).
isoelectronic species
495 kJ mol-1 to
remove electron
from sodium
349 kJ mol-1 back
by giving electron
to chlorine
Also need to account for:
electrostatic attraction between the newly
formed sodium cation and chloride anion
Grey are atoms, blue are anions, red are cations
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Lattice energy
Formation of NaCl from Na and Cl2
E el 
Q1Q 2
d
Chem 101
44
8.4 Bond Polarity and Electronegativity
Energetics of Ionic bonds:Lattice energy
Electron pairs are usually unequally shared between different atoms:
high electron density
nonpolar
covalent
bond
polar
covalent
bond
electrons
shared
equally
F2
HF
low
one atom attracts
bonding electrons
more than the other
45
Ionic and covalent
bonding
represent the two extremes of the continuum
ionic
extended lattice
structures
covalent
molecules
Properties:
Properties:
o high melting solids, brittle
o low melting and boiling points
(often gases, liquids at room temp.)
o soluble in water
o most are insoluble in water
o solutions and melts conduct
electricity
Chem 101
47
Chem 101
o solutions and melts are
non-conducting
48
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Lewis
structures
8.5 Lewis Structures
PCl3
1. Find the total number of valence
electrons (account for any charges) =
TOTAL
For PCl3: 5 (for P) + (3 × 7) (for Cl3) = 26
2. decide connection between atoms, draw
a line to represent 1 electron pair for
each connection, count the electrons
SHARED
3 bonds= 6
3. calculate the remaining electrons =
TOTAL – SHARED, assign these to the
terminal atoms to make octet (or 2 for H
atom)
26 – 6 = 20
Choose central atom
correctly
least electronegative atom
(not H)
oxygen rarely bonds to
itself
4. any electrons left? – put them on the
central atom
5. if central atom doesn’t have an octet,
make multiple bonds from nonbonded
electron pairs on terminal atoms
Chem 101
50
Multiple Bonds
Formal charges
often can make more than one Lewis structure
which one is
correct?
bookkeeping of electrons
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
Chem 101
51
Example:
H
N
Chem 101
52
Resonance hybrids
hybrid is
intermediate
between the two
“parent”
structures
H
H
N: valence = 5, formal charge = 5 – [2 + ½{6}] = 0
H: valence = 1, formal charge = 1 – [0 + ½{2}] = 0
Chem 101
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Chem 101
54
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aromatic compounds show resonance
Resonance has impact on bond lengths and strengths
compare
NO+,
NO2-
and
NO3-
benzene
N-O bond = 3
N
O
-
N-O bond
average = 1.5
O
N
O

O
N
O
shorthand notation omits H atoms
N-O bond
average = 1.3
C-C bond average = 1.5
55
Exceptions to the octet rule
Chem 101
56
Exceptions to the octet rule…
can expand valence shell to make a Lewis structure with
lower formal charge
1. odd number of electrons
SO422. less than an octet
experimental
info:
or?
3. more than an octet
Chem 101
57
Chem 101
58
8.8 Strength of Covalent Bonds
CH4 + Cl2
CH3Cl + HCl
ΔHrxn = 413kJ + 242 kJ – {328 kJ + 431 kJ} = -104 kJ
ΔH is negative (rxn. is
exothermic) when
weak bonds are
broken and strong
bonds are formed
Chem 101
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Chem 101
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Chapter 9 Molecular Geometry and Bonding Theories
Bond lengths
9.1 Molecular Shapes
also depend on nature of atom and type of bond
also calculated as averages
AB3
AB2
trends : shorter bonds are stronger
N
N
1.47
163 kJ/mol
N
N
1.24
418 kJ/mol
N
N
1.10
941kJ/mol
linear
bent
trigonal
planar
trigonal
pyramidal
T-shaped
For molecules of the general form ABn there are 5 fundamental shapes:
180°
linear
Chem 101
90°
90°
109.5°
120°
120°
trigonal
planar
tetrahedral
61
trigonal
bipyramidal
90°
octahedral
We use the electron-domain geometry to help us predict the molecular geometry.
4
1
1. Draw Lewis structure, count electron domains
3
2
2. Arrange electron domains to minimize
repulsion
tetrahedral
3. Inspect arrangement of atoms to determine
molecular geometry
2
2
0
3
3
0
2
1
trigonal
pyramidal
Molecules with Expanded Valence Shells
Atoms that have expanded octets have five electron domains (trigonal
bipyramidal) or six electron domains (octahedral) electron-domain geometries.
4
4
0
3
1
2
2
Effect of nonbonding electrons and multiple
bonds on bond angles
bonding
pair
5
lone pair always in least
crowded position
5
0
PCl5
4
1
SF4
non-bonding
pair
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9.3 Molecular Shape and Molecular Polarity
5
3
2
Dipoles are a
vector quantity
ClF3
=0
2
note position
of lone pairs
3
overall dipole
moment
XeF2
electron density models
6
6
0
SF6
5
1
BrF5
4
2
XeF4
(red = high, blue = low)
lone pairs as far
apart as possible
9.4 Covalent Bonding and Orbital Overlap
HCl
H
9.5 Hybrid Orbitals
Cl2
H
s
p
2 × sp
overlap regions
H2
large lobes of sp hybrid orbitals
The change in potential energy
as two hydrogen atoms combine
to form the H2 molecule:
F 2p
orbital
F 2p
orbital
overlap regions
In an sp2 hybridized atom, what is the orientation of the
unhybridized p orbital relative to the three sp2 hybrid
orbitals?
D.
The unhybridized p-orbital is perpendicular to the plane of
the sp2 orbitals.
one s
orbital
hybridize
two p
orbitals
three
sp2
hybrid
orbitals
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9.5 Hybrid Orbitals
one s
orbital
p
s
hybridize
two p
orbitals
2 × sp
three
sp2
hybrid
orbitals
sp2 hybrid orbitals
shown together
(large lobes only)
large lobes of sp hybrid orbitals
F 2p
orbital
F 2p
orbital
one s
orbital
three p orbitals
sp3 hybrid orbitals
shown together
(large lobes only)
overlap regions
four sp3 hybrid orbitals
9.6 Multiple Bonds
9.5 Hybrid Orbitals
For geometries involving expanded octets on the central atom, we use d orbitals in
our hybrids:
s, p, p, p, d
five
sp 3d
trigonal
bipyramidal
octahedral
s, p, p, p, d, d
one π bond
six sp 3d 2
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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 p orbitals on all three oxygens overlap with the
p orbital on the central nitrogen

the π electrons are delocalized throughout the ion
The organic molecule benzene has six σ bonds
and a p orbital on each carbon atom
Molecular orbital (MO) theory (only section 9.7)
MO diagram (energy level diagram)
Bond order
Bond order = ½ {no. bonding electrons – no. antibonding electrons}
antibonding MO
raises energy
antibonding electrons
bonding MO
lowers energy
bonding electrons
Chem 101
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Metal bonding (Sections 23.5 and 12.2)
MO model
an infinite chain of atoms
Chapter 11 Intermolecular forces, liquids, and solids
Metals, insulators and semiconductors
11.1 A molecular comparison of gases, liquids and solids
The fundamental difference between states of matter is the distance between
particles.
incompressible
gas
liquid
solid
Intermolecular forces: a summary
11.2 Intermolecular forces
interacting
molecules or ions
YES
are polar
molecules and ions
both present?
YES
are ions
involved?
NO
A) ionic bonding
e.g. NH4NO3
B) ion-dipole forces
e.g. NaCl in H2O
A>B>C>D>E
NO
are polar
molecules
involved?
NO
are H atoms
bonded to N, O or
F atoms?
D) dipoledipole forces
YES
e.g. H2S, CH2Cl2
YES
covalent bond
(strong)
intermolecular
attraction (weak)
NO
Dipole-Dipole Interactions
E) dispersion forces
only (induced dipoles)
e.g. Ar(l), I2(s)
C) hydrogen bonding
e.g. H2O, NH3, HF
van der Waals
forces
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Hydrogen bonding
London Dispersion Forces
momentarily polar
e
―
e
―
δ–
He atom
e
H2O
e
H2Te
H2Se
H2S
SiH4
CH4
GeH4
―
―
He atom 1
e
―
e
―
δ–
He atom 2
δ+
δ+
δ–
δ+
SnH4
n-pentane
bp 309 K
neopentane
bp 283 K
London dispersion
dipole-dipole
hydrogen bonding
11.8 Bonding in solids
Molecular Solids
Consist of atoms or molecules held together by intermolecular forces.
benzene
toluene
5
–95
43
80
111
182
Ion-dipole forces
Ion-dipole interactions, are an important
force in solutions of ions.
high mp due to
efficient packing
cationdipole
aniondipole
phenol
high mp & bp
due to hydrogen
bonding
high bp due to larger
intermolecular forces
Covalent-Network Solids
Ionic Solids
Consist of atoms held together, in large networks or chains, with covalent bonds.
Consist of ions held together by ionic bonds (i.e. by electrostatic forces of attraction).
Cl
Cs
CsCl
S
Zn
ZnS
“zinc blende”
Ca
F
CaF2
“fluorite”
Metallic Solids
Consist entirely of metal atoms;
1.42 Å
not covalently bonded.
3.41 Å
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Hydrocarbons
contain only C and H
four different types:
Drawing hydrocarbon structures
defined by kind of carbon to carbon bonds
C4H10
CH3CH2CH2CH3
C4H10
unsaturated hydrocarbons
largest possible no. of H atoms  saturated hydrocarbons
Chem 101
97
example:
8
1
7
6
5
4
3
2
1
2
3
4
5
6
7
8
halogens
ethyl
-F
fluoro
-Cl
chloro
-Br
bromo
-I
iodo
methyl
-OH
parent name
hydroxy
-NO2 nitro
ethyl- methyl octane
4-ethyl-2,7-dimethyloctane
Chem 101
Cycloalkanes - general
99
formula:
CnH2n
Chem 101
Alkenes - general
100
formula: CnH2n
CH2 CH2
CH2
Chem 101
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07/12/2010
1
6
Reactions of alkenes
6
2
5
4
addition reactions
4
5
3
add to the two atoms that form the double bond
general reaction:
3
2
1
cis-3,4-dimethyl-3-hexene
trans-3,4-dimethyl-3-hexene
sp2 hybridized C atoms become sp3 hybridized
 Br2, Cl2
1
6
6
5
halogenation
Chem 101
2
4
5
4
3
Cl2
3
2
same rxn for Br2
1
trans-3,4-dichloro-3-hexene
cis-3,4-dichloro-3-hexene
Chem 101
104
addition rxns, cont’d
addition rxns, cont’d
 H2
hydrogenation
 HBr
Ni, 500 °C
H2
 H2O
HBr
hydration
addition reactions
H2SO4
2 H2
H 2O
Chem 101
Reactions of aromatics
105
Chem 101
106
be able to identify
ALL of these
functional groups
substitution reactions
in general:
all 6 H atoms on the ring are equivalent,
H2SO4
HNO3
H 2O
FeBr 3
Br2
Chem 101
HBr
107
Chem 101
108
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Carboxylic acids
Esters
sp2 hybridized C atom
produced by oxidation of alcohols
CH3CH2CH2OH
+
sp2 hybridized C atom
formed by condensation reaction between alcohols and carboxylic acids
CH3 CH2CH
[O]
+ H2 O
O
[O]
alcohol +excess O2  carboxylic acid
CH3 CH2C
OH
O
Chem 101
Chem 101
110
109
amides, cont’d
ester reaction
add H2O
saponification
CH3CH2C
prepared by reaction of carboxylic acid with ammonia or amine
base catalyzed ester hydrolysis
O CH3
+ H2O
[OH- ]
CH3CH2C OH
+
HO-CH3
O
O
HO
ester + water
H
[OH-]
condensation rxn.
alcohol + acid
Chem 101
Chapter 12 Modern Materials
focus on polymers, semiconductors, liquid crystals,
superconductors
111
amides undergo hydrolysis
O
R2
C
+ H2O
N
R
R1
O
H+
C
heat
R2
+
OH
R
R1
Chem 101
112
12.6 Polymers
Making polymers
Many synthetic polymers have a backbone of C–C bonds.
coupling of
monomers
through multiple
bonds is addition
polymerization
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
HN
ethylene
ethylene
ethylene
polyethylene
Initiator
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?
H
H
C
H
C
H
ethylene
Chem 101
H H
C
C
H H
H
H
C
+
H
C
H
H
H
H
H
C
C
C
C
H
H
H
H
polyethylene
monomer
114
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07/12/2010
Condensation polymerization: two molecules are joined to form a larger
molecule by the elimination of a small molecule. e.g. water
Amide linkage
12.6 Polymers
Making polymers
Many synthetic polymers have a backbone of C–C bonds.
ethylene
ethylene
Addition
polymers:
ethylene
Monomer
polyethylene
coupling of
monomers
through multiple
bonds is addition
polymerization
+ H2O
amine
carboxylic
acid
terephthlaic acid and 1,2 ethylene glycol form Mylar
Ester linkage
Polymer
H2O
diacid
+ n
n
Polypropylene (PP)
dialcohol
OH
HO
Polyethylene (PE)
amide
polymers formed
from two different
monomers are called
copolymers
OH
HO
first step
O
O
Polystyrene (PS)
n
O
HO
O
O
O
OH
O
polyvinyl chloride
(PVC)
Condensation polymerization: two molecules are joined to form a larger
molecule by the elimination of a small molecule.
Condensation
polymers:
O
O
n
+
H2O
+ n H2O
… back to: Semiconductors (section 12.1)
- dopant atom has
fewer valence
electrons than the
host atom
polyurethane
- leads to holes in
the valence band
polyethylene
terephthalate
- e.g. B into Si
- dopant atom has
more valence
electrons than the
host atom
- adds electrons to
the conduction band
p -type
n -type
- e.g. P into Si
nylon 6,6
polycarbonate
20