AP CHEMISTRY EXAM
Fall review
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Energy and Chemistry
ENERGY is the capacity to do work or
transfer heat.
HEAT is the form of energy that flows
between 2 samples because of their
difference in temperature.
Other forms of energy —
• light
• electrical
• kinetic and potential
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Kinetic and Potential Energy
Potential
energy —
energy a body
has because
of its position.
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Kinetic and Potential Energy
Kinetic energy
— energy of
motion.
• Translation
• Rotation
• Vibration
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Thermodynamics
• Thermodynamics is the science
of heat (energy) transfer.
Heat energy is associated
with molecular motions.
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Energy and Chemistry
All of thermodynamics
depends on the law of
THE CONSERVATION
OF ENERGY.
• The total energy of a
system is constant.
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UNITS OF ENERGY
1 calorie = heat to raise
temp. of 1.00 g of H2O
by 1.0 oC.
1000 cal = 1 kilocalorie
= 1 kcal
But we use the unit
called the JOULE
1 cal = 4.184 joules
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James Joule
1818-1889
Specific Heat Capacity
The heat “lost” or “gained” is
related to
a) sample mass
b) change in T and
c) specific heat capacity by
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
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Specific Heat Capacity
Substance
H2O
Al
glass
Spec. Heat (J/g•K)
4.184
0.902
0.84
Aluminum
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Specific Heat Capacity
If 25.0 g of Al cool from 310
oC to 37 oC, how many
joules of heat energy are
lost by the Al?
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
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Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many
joules of heat energy are lost by the Al?
heat gain or loss = q = (SH)(mass)(DT)
q = (0.902 J/g•K)(25.0 g)(37 - 310)K
q = - 6160 J
The negative sign on q signals heat
“lost by” or transferred out of Al.
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Heat Transfer and
Changes of State
Changes of state involve energy
Ice -----> Water
333 J/g (fusion)
An ENDOTHERMIC process.
+ energy
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Heat and Changes of State
• 1 g of ethanol requires
850 J to evaporate at 25
oC.
• To drop air temp. from 55
oC to 25 oC in a car
requires air to give up 3.6
kJ.
• How much ethanol must
be evaporated to do this?
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Heat and Changes of State
3.6 x 103 J (1 g/850 J)
= 4.2 g of ethanol
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Heat and Changes of State
What quantity of heat is required
to melt 500. g of ice and heat
the water to steam at 100 oC?
Specific heat of water = 4.2 J/g•K
Heat of fusion of ice = 333 J/g
Heat of vaporization = 2260 J/g
+333 J/g
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+2260 J/g
Heating/Cooling Curve for Water
See Figure 6.9
3 Evaporate water
Heat water
1 2
Melt ice
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4
Heat and Changes of State
What quantity of heat is required to melt 500. g of ice and
heat the water to steam at 100 oC?
1.
To melt ice
q = (500. g)(333 J/g) = 1.67 x 105 J
2.
To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J
3.
To evaporate water at 100 oC
q = (500. g)(2260 J/g) = 1.13 x 106 J
4. Total heat energy = 1.51 x 106 J = 1510 kJ
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ENTHALPY
DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC
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Endo- and Exothermic
Surroundings
System
heat
Surroundings
System
heat
qsystem > 0
qsystem < 0
T(system) goes up
T(system) goes down
ENDOTHERMIC
EXOTHERMIC
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USING ENTHALPY
Calc. DH for S(s) + 3/2 O2(g) --> SO3(g)
knowing that
S(s) + O2(g) --> SO2(g) DH1 = -320.5 kJ
SO2(g) + 1/2 O2(g) --> SO3(g)
DH2 = -75.2 kJ
The two equations add to give the desired
equation (at top), so
DHnet = DH1 + DH2 = -395.7 kJ
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energy
S solid
direct path
+ 3/2 O2
DH =
-395.7 kJ
SO3 gas
+O2
DH 1 =
-320.5 kJ
SO2 gas
+ 1/2 O2
DH 2 = -75.2 kJ
 DH along one path =
 DH along another path
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DH along one path =
DH along another path
• This equation is valid
because DH is a STATE
FUNCTION
• These depend only on
the state of the system
and not how it got
there.
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Standard Enthalpy Values
Standard enthalpy: DHo
Measured under standard conditions
P = 1 bar (1 atmosphere)
Concentration = 1 mol/L
T = usually 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas
DHof = 0 for elements in their standard
states.
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Using Standard Enthalpy Values
Calculate ²H
of reaction?
In general, when ALL
enthalpies of formation
are known,
DHorxn =  DHof (products)
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-  DHof (reactants)
Electromagnetic Radiation
• Most subatomic particles behave as
PARTICLES and obey the physics of
waves.
• Wavelength, 
• Node
• Amplitude
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Electromagnetic Radiation
wavelength
Visible light
Amplitude
wavelength
Node
Ultraviolet radiation
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Electromagnetic Radiation
• Waves have a frequency
• Greek letter “nu”, , = frequency.
Units are “cycles per sec”
• All radiation:  •  = c
• c = velocity of light = 3.00 x 108
m/sec
• Note that long wavelength --> small
frequency
• Short wavelength --> high frequency
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Electromagnetic Radiation
Red light has  = 700 nm. Calculate
the frequency.
-9
1 x 10 m
-7
700 nm •
= 7.00 x 10 m
1 nm
8
Freq =
3.00 x 10 m/s
7.00 x 10
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-7
m
 4.29 x 1014 sec -1
Quantization of Energy
Max Planck (1858-1947)
Energy of radiation is proportional to
frequency.
E = h•
h = Planck’s constant
= 6.6262 x 10-34 J•s
Suggests light is a particle (PHOTON)
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Quantization of Energy
E = h•
Light with large  (small ) has a
small E.
Light with a short  (large ) has a
large E.
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Energy of Radiation
PROBLEM: Calculate the energy of
1.00 mol of photons of red light.
 = 700. nm
 = 4.29 x 1014 sec-1
E = h•
= (6.63 x 10-34 J•s)(4.29 x 1014
sec-1)
= 2.85 x 10-19 J per photon
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Atomic Line Spectra and
Niels Bohr
Niels Bohr
(1885-1962)
A great accomplishment.
Rec’d Nobel Prize, 1922
Problems with theory —
• only successful for H.
• introduced quantum idea
artificially.
• So, we go on to QUANTUM
or WAVE MECHANICS
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Quantum or Wave Mechanics
de Broglie (1924) proposed
that all moving objects
have wave properties.
For light: E = mc2
L. de Broglie
(1892-1987)
E = h = hc / 
Therefore, mc = h / 
and for particles
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(mass)(velocity) = h / 
Quantum or Wave Mechanics
E.
Schrodinger
1887-1961
Solution to wave equation
gives set of
mathematical
expressions called
WAVE FUNCTIONS, 
Each describes an
allowed energy state of
an eQuantization introduced
naturally.
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WAVE FUNCTIONS, 
• Each  corresponds to an
ORBITAL — the region of space
within which an electron is found.
•  does NOT describe the exact
location of the electron.
• 2 is proportional to the probability
of finding an e- at a given point.
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Uncertainty Principle
W. Heisenberg
1901-1976
Solved problem of defining
nature of electrons in
atoms.
Cannot define position and
momentum (= m•v) of an
electron simultaneously.
We define e- energy exactly
but accept limitation that
we do not know exact
position.
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QUANTUM NUMBERS
Electrons are arranged in shells
and subshells.
n --> shell (energy level)
l --> subshell (orbital shape)
ml --> designates an orbital within
a subshell (orintation on a
coordinate system)
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s Orbitals
n = 1, l = 0
In first shell, there is 1 orbital.
All s orbitals are spherical in shape.
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Typical p orbital
p Orbitals
When n = 2, then l = 0 and planar
1
node
in n = 2 shell there are 2 types of
orbitals — 2 subshells
For l = 0 ml = 0
this is a s subshell
For l = 1 ml = -1, 0, +1
this is a p subshell with 3 orbitals
each oriented on an axis.
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d Orbitals
When n = 3, l = 0, 1, 2
there are 3 subshells in the shell.
For l = 0, ml = 0
---> s subshell with single orbital
For l = 1, ml = -1, 0, +1
---> p subshell with 3 orbitals
For l = 2, ml = -2, -1, 0, +1, +2
---> d subshell with 5 orbitals
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d Orbitals
typical d orbital
planar node
planar node
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f Orbitals
When n = 4, l = 0, 1, 2, 3 so there are 4 subshells in
the shell.
For l = 0, ml = 0
---> s subshell with single orbital
For l = 1, ml = -1, 0, +1
---> p subshell with 3 orbitals
For l = 2, ml = -2, -1, 0, +1, +2
---> d subshell with 5 orbitals
For l = 3, ml = -3, -2, -1, 0, +1, +2, +3
---> f subshell with 7 orbitals
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Arrangement of Electrons
in Atoms
Each orbital can be assigned no
more than 2 electrons!
This is tied to the existence of a 4th
quantum number, the electron
spin quantum number, ms.
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Electron Spin Quantum Number
Diamagnetic: NOT attracted to a
magnetic field
Paramagnetic: substance is attracted
to a magnetic field. Substance has
unpaired electrons.
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Pauli Exclusion Principle
No two electrons in the
same atom can have
the same set of 4
quantum numbers.
That is, each electron has a
unique address.
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Electron
Filling
Order
Figure 8.7
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Writing Atomic Electron
Configurations
Two ways of writing econfigs. One is called
spectroscopic notation.
SPECTROSCOPIC NOTATION
for H, atomic number = 1
1
1s
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value of n
no. of
electrons
value of l
Writing Atomic Electron
Configurations
ORBITAL BOX NOTATION
for He, atomic number = 2
Arrows
2
depict
electron
spin
1s
1s
Two ways of writing
configs. Other is called
the orbital box notation.
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Lithium
Group 1A
Atomic number = 3
1s22s1 ---> 3 total electrons
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Boron
3p
3s
2p
2s
1s
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Group 3A
Atomic number = 5
1s2 2s2 2p1 --->
5 total electrons
Carbon
3p
3s
2p
2s
1s
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Group 4A
Atomic number = 6
1s2 2s2 2p2 --->
6 total electrons
HUND’S RULE. When
placing electrons in
orbitals of the same
energy, place them singly
as long as possible.
Oxygen
3p
3s
2p
2s
1s
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Group 6A
Atomic number = 8
1s2 2s2 2p4 --->
8 total electrons
Neon
3p
3s
2p
2s
1s
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Group 8A
Atomic number = 10
1s2 2s2 2p6 --->
10 total electrons
We have reached the
end of the 2nd
period, and the 2nd
shell is full!
Sodium
Group 1A
Atomic number = 11
1s2 2s2 2p6 3s1 or
“neon core” + 3s1
[Ne] 3s1 (uses rare gas notation)
All Group 1A elements have
[core]ns1 configurations.
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Relationship of Electron Configuration
and Region of the Periodic Table
• Gray = s block
• Orange = p
block
• Green = d
block
• Violet = f block
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Transition Element
Configurations
3d orbitals used for
Sc - Zn (Table 8.4)
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Lanthanide Element
Configurations
4f orbitals used for
Ce - Lu and 5f for
Th - Lr (Table 8.2)
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Ion Configurations
To form cations from elements
remove 1 or more e- from subshell
of highest n [or highest (n + l)].
P [Ne] 3s2 3p3 - 3e- ---> P3+ [Ne] 3s2 3p0
3p
3p
3s
3s
2p
2p
2s
2s
1s
1s
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General Periodic Trends
• Atomic and ionic
size
• Ionization energy
• Electron affinity
Higher Z*.
Electrons held
more tightly.
Larger orbitals.
Electrons held less
tightly.
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Atomic Size
SIZE
• Size goes UP going down a group.
• Electrons are added further from
the nucleus, there is less attraction.
• Size goes DOWN going across a
period.
• Due to the increase in Z*. Each
added electron feels a greater and
greater + charge.
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Ion Sizes
+
Li,152 pm
3e and 3p
Li +, 60 pm
2e and 3 p
Forming
a cation.
• CATIONS are SMALLER than the
atoms from which they come.
• The electron/proton attraction has
gone UP and so size DECREASES.
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Ion Sizes
F,64 pm
9e and 9p
F- , 136 pm
10 e and 9 p
Forming
an anion.
• ANIONS are LARGER than the
atoms from which they come.
• The electron-proton attraction has
gone DOWN so size INCREASES.
• Trends in ion sizes are the same as
atom sizes.
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Redox Reactions
Why do metals lose
electrons in their
reactions?
Why does Mg form Mg2+
ions and not Mg3+?
Why do nonmetals take
on electrons?
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Ionization Energy
See Screen 8.12
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + e-
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Ionization Energy
See Screen 8.12
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + eMg+ (g) + 1451 kJ ---> Mg2+ (g) + e-
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Ionization Energy
See Screen 8.12
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + eMg+ (g) + 1451 kJ ---> Mg2+ (g) + e-
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Ionization Energy
See Screen 8.12
IE = energy required to remove an electron from an
atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + eMg+ (g) + 1451 kJ ---> Mg2+ (g) + e-
Mg+ has 12 protons and only 11 electrons.
+ > Mg.
Therefore,
IE
for
Mg
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Ionization Energy
See Screen 8.12
Mg (g) + 735 kJ ---> Mg+ (g) + eMg+ (g) + 1451 kJ ---> Mg2+ (g) + eMg2+ (g) + 7733 kJ ---> Mg3+ (g) + e-
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Ionization Energy
IE = energy required to remove an
electron from an atom in the gas
phase.
Mg (g) + 735 kJ ---> Mg+ (g) + e-
Mg+ (g) + 1451 kJ ---> Mg2+ (g) + eMg2+ (g) + 7733 kJ ---> Mg3+ (g) + e-
Energy cost is very high to dip into a
shell of lower n. This is why ox. no. =
Group no.
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Trends in Ionization Energy
1st Ionization energy (kJ/mol)
2500
He
Ne
2000
Ar
1500
Kr
1000
500
0
1
H
3
Li
5
7
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9
11
Na
13
15
17
19
K
21
23
25
27
29
31
Atomic Number
33
35
Trends in Ionization Energy
• IE increases across a
period because Z*
increases.
• Metals lose electrons more
easily than nonmetals.
• Metals are good reducing
agents.
• Nonmetals do not lose
electrons.
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Trends in Ionization Energy
• IE decreases down a
group
• Because size
increases.
• Reducing ability
generally increases
down the periodic
table.
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Electron Affinity
A few elements GAIN
electrons to form anions.
Electron affinity is the energy
involved when an anion
loses an electron.
A-(g) ---> A(g) + eE.A. = DE
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Trends in Electron Affinity
• Affinity for electron
increases across a
period (EA becomes
more positive).
• Affinity decreases
down a group (EA
becomes less
positive).
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Atom EA
F
+328 kJ
Cl +349 kJ
Br +325 kJ
I
+295 kJ
Trends in Electron Affinity
F Cl
Br
35 0
30 0
S
Si
20 0
Se
15 0
S4
10 0
Ge P
S3
Period
50
S2
S1
0
K
1
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2
3
4
5
Group
6
7
Elec tron affinity (kJ/m ol)
C
H
25 0
O
Forms of Chemical Bonds
• There are 2 extreme forms of
connecting or bonding atoms:
• Ionic—complete transfer of
electrons from one atom to
another
• Covalent—electrons shared
between atoms
• Most bonds are somewhere in
between.
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Electron transfer from an
element of low IE (metal)
to an element of high EA
(nonmetal)
2 Na(s) + Cl2(g) --->
2 Na+ + 2 ClUsually between:
Groups 1A , 2A &
transition metals and
Nonmetals (O and
halogens).
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Ionic
Bonds
Covalent Bonding
Covalent bonds form by sharing
VALENCE ELECTRONS,
those at the outer edge of the atom.
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Valence Electrons
Electrons are divided between core
and valence electrons.
Na 1s2 2s2 2p6 3s1
Core = [Ne] and
valence = 3s1
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1A
Valence Electrons
2A
3A
4A
5A
6A
Number of valence electrons is
equal to Group number.
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8A
7A
Covalent Bonding
The bond arises from the mutual
attraction of 2 nuclei for the same
electrons.
Bond is a balance of attractive and
repulsive forces.
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Electron
Distribution in
Molecules
G. N. Lewis
1875 - 1946
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• Electron distribution is
shown by Lewis
electron dot structures
• Electrons are
distributed as shared
or BOND PAIRS and
unshared or LONE
PAIRS.
Bond and Lone Pairs
• Electrons are distributed as shared or BOND
PAIRS and unshared or LONE PAIRS.
••
H
Cl
•
•
••
shared or
bond pair
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lone pair (LP)
Bond Formation
A bond can result from a “head-tohead” overlap of atomic orbitals
on neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Overlap of H (1s) and Cl (2p)
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Cl
••
•
•
Bond Formation
A bond can result from a “head-to-head” overlap of
atomic orbitals on neighboring atoms.
••
H
+
Cl
••
••
•
•
H
Cl
•
•
••
Overlap of H (1s) and Cl (2p)
This overlap puts bonding electrons
in a MOLECULAR ORBITAL along
the line between the two atoms and
forms a SIGMA BOND ().
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Rules of the Game
•No.
of valence electrons of an atom
= Group number
•For Groups 1A-4A, no. of bond pairs
expected = group number
• For Groups 5A-7A, BP’s = 8 - Grp.
No.
•Except for H (and atoms of 3rd and
higher periods),
BP’s + LP’s = 4 (8 electrons)
This is the OCTET
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RULE
Building a Dot Structure
Ammonia, NH3
1. Decide on the central atom;
never H.
Central atom has lowest EA.
Therefore, N is central
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
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Building a Dot Structure
3. Form a sigma bond between the
central atom and surH
N
rounding atoms.
H
H
4. Remaining e- form LONE PAIRS to
complete octet as needed.
3 BOND PAIRS and 1 LONE PAIR.
N has a share in 4 pairs (8 e-),
while H shares 1 pair.
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Carbon Dioxide, CO2
1. Central atom = C
2. Valence electrons = 16 (8 pair)
3. Form sigma bonds, leaving 6
pair.
4. Place lone pairs on outer atoms.
••
••
•
•
O
••
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C
O
••
•
•
Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
5. So C has an octet, form DOUBLE
BONDS between C and O.
••
••
•
•
O
••
C
O
•
•
••
The second bonding pair forms a pi
(p) bond.
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•
•
O
••
C
O
••
•
•
Double and
even triple
bonds are
commonly
observed for C,
N, P, O, and S
H2CO
SO3
•
•
C2F4
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O
••
C
O
••
•
•
Sulfur Dioxide, SO2
OR bring in
right pair
bring in
left pair
••
••
•
•
O
••
S
••
O
•
•
••
This leads to the following structures.
••
•
•
O
••
S
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••
•
•
O
••
••
•
•
••
O
••
S
•
•
O
••
Sulfur Dioxide, SO2
18 electrons can be distributed in two
ways with the following structures.
••
•
•
••
•
•
O S O
••
••
••
••
•
•
••
•
•
O S O
••
These equivalent structures are
called
RESONANCE STRUCTURES.
The true electronic structure is a
HYBRID of the two.
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Violations of the Octet Rule
Usually occurs with B and elements
of higher periods.
BF3
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SF4
MOLECULAR GEOMETRY
VSEPR
Valence Shell Electron Pair
Repulsion theory.
Molecule
Most important factor
adopts the
in determining geometry shape that
minimizes the
is relative repulsion
electron pair
between electron pairs.
repulsions.
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No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180
F
3
F
planar
trigonal
B
F
120
H
4
C
H
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109
tetrahedral
H
H
No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180
F
3
F
planar
trigonal
B
F
120
H
4
C
H
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109
tetrahedral
H
H
No. of e- Pairs
Around Central
Atom
2
Example
F—Be—F
Geometry
linear
180
F
3
F
planar
trigonal
B
F
120
H
4
C
H
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109
tetrahedral
H
H
Structure Determination
by VSEPR
••
H N H
Ammonia, NH3
H
1. Draw electron dot structure
2. Count BP’s and LP’s = 4
3. The 4 electron pairs are at the
corners of a tetrahedron.
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Structure Determination by
VSEPR
Ammonia, NH3
••
H N H ELECTRON PAIR
GEOMETRY is tetrahedral.
H
MOLECULAR GEOMETRY
(position of the atoms) is
PYRAMIDAL.
lone pair of electrons
in tetrahedral position
N
H
H
H
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Structure Determination by
VSEPR
Formaldehyde, CH2O
•
•
O
•
•
The electron pair
geometry is
PLANAR TRIGONAL
C
H
H
The molecular
geometry is also
planar trigonal.
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Structure Determination
by VSEPR
Methanol, CH3OH
1. Draw electron dot structure
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Violations of the Octet Rule
Consider boron trifluoride, BF3
The B atom is surrounded by
only 3 electron pairs.
Bond angles are 120o
••
•
•
•
•
F
••
•
•
F
••
B
•
•
Geometry
described as
planar trigonal
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•
•
F
••
Compounds with 5 or More
Pairs Around the Central Atom
90
F
F
P
Trigonal bipyramid
F
120 5 electron pairs
F
F
90
6 electron pairs
F
F
S
F
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F
Octahedron
F
F
90
Bond Order
the number of bonds between a pair
triple, BO = 3
of atoms.
H
H
H
C
C
1  and 2 p
C
single
double, BO = 2
BO = 1
1  and 1 p
1 
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N
Bond Order
Bond order is proportional to two
important bond properties:
(a) bond strength
(b) bond length
110 pm
745 kJ
414 kJ
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123 pm
Bond Length
Bond length is the distance between
the nuclei of two bonded atoms.
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Bond Length
Bond length
depends on
size of
bonded
atoms and
bond order.
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Bond Strength
The energy needed to break a bond.
•
BOND
STRENGTH (kJ/mol)
H—H
436
C—C
346
C=C
602
CC
835
The GREATER the number of bonds
(bond order) the HIGHER the bond
strength, the SHORTER the bond.
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Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
Net energy = DHrxn =
= energy required to break bonds
- energy given off when
bonds are made
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Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
Net energy = DHrxn =
= energy required to break bonds
- energy evolved when bonds are made
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
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Using Bond Energies
Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—Cl
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Sum of H-H + Cl-Cl bond energies =
436 kJ + 242 kJ =
+678 kJ
2 mol H-Cl bond energies = +864 kJ
Net = DH = +678 kJ - 864 kJ = -186 kJ
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Bond Polarity
+
H
-
••
Cl ••
••
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HCl is POLAR because
it has a positive end
and a negative end.
Polarity arises because
Cl has a greater share
in bonding electrons
than does H.
+
H
-
••
Cl ••
Bond Polarity
••
Due to bond polarity, the H—Cl bond
energy is GREATER than expected for a
“pure” covalent bond.
BOND
ENERGY
“pure” bond 339 kJ/mol calculated
real bond
432 kJ/mol measured
Difference = 92 kJ, proportional to the
difference in ELECTRONEGATIVITY, .
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Electronegativity, 
is a measure of the ability of an
atom in a molecule to attract
electrons to itself.
• F has maximum .
• Atom with lowest  is the center
atom in most molecules.
• Values of  in bonded
atoms determine
Linus Pauling
BOND POLARITY.
1901-1994
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Bond Polarity
Which bond is more polar (or DIPOLAR)?
O—H
O—F

3.5 - 2.1
3.5 - 4.0
D
1.4
0.5
OH is more polar than OF
O—H
- + 
O—F
+  -
and polarity is “reversed.”
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Molecular Polarity
Molecules—such as HCl and H2O— can
be POLAR (or dipolar).
They have a DIPOLE MOMENT. The polar
HCl molecule will turn to align with an
electric field.
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Molecular Polarity
Molecules—such as HCl and H2O—
can be POLAR (or dipolar).
They have a DIPOLE MOMENT. The
polar HCl molecule will turn to align
with an electric field.
POSITIVE
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H—Cl


NEGATIVE
Molecular Polarity
Molecules will be polar if
a) bonds are polar
AND
b) the molecule is NOT “symmetric”
Symmetric molecules
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Molecular
Polarity,
H 2O
••
O
••
H
H
polar
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O
H
H
+
Carbon Dioxide
•
•
O
••
C
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O
••
•
•
• CO2 is NOT
polar even
though the CO
bonds are
polar.
Two Theories of Bonding
• MOLECULAR
ORBITAL THEORY
— Robert Mullikan
(1896-1986)
• valence electrons
are delocalized
• valence electrons
are in molecular
orbitals which
spread over entire
molecule.
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Two Theories of Bonding
• VALENCE BOND
THEORY — Linus
Pauling
• valence electrons are
localized between
atoms (or are lone
pairs).
• half-filled atomic
orbitals overlap to form
bonds.
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Sigma Bond Formation by
Orbital Overlap
H
H
• +
•
•

•
sigma bond ( )
Two p
orbitals
overlap
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


Two s
orbitals
overlap
Bonding in BF3
ORBITAL HYBRIDIZATION
mix available orbitals to form a
new set of orbitals — HYBRID
ORBITALS — that give the
maximum overlap in the
correct geometry.
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Bonding in BF3

2s
hydridize orbs.

2p
rearrange electrons
  
2
three sp
hybrid orbitals
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unused p
orbital
Bonding in a Tetrahedron —
Formation of Hybrid Atomic
Orbitals
4 C atom orbitals hybridize
to form four equivalent sp3
hybrid atomic orbitals.
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Orbital Hybridization
BONDS SHAPE HYBRID REMAIN
linear
3
trigonal sp2
planar
1p
4
tetrahedral sp3
none
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sp
2 p’s
2
Multiple Bonds
Consider ethylene, C2H4
H
H
120
o
C
H
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sp
C
H
2
p Bonding in C2H4
The unused p orbital on each C
contains an electron. The two p
orbitals overlap to form the p bond.

2s
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 
2p

 
3 sp 2
hybrid
orbitals

p
orb.
for š
bond
Multiple Bonding
in C2H4
Note that sp2 bonds give trigonal
planar shape at each C.
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Consequences of Multiple
Bonding
Restricted rotation around C=C bond.
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Importance of carbon
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The addition of various functional
groups to ethylene results in
ADDITION polymers.
H
H
X Y
\
/
| |
C==C + XY  H--C---C--H
/
\
| |
H
H
H H
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ADDITION polymers:
Polyvinyl chloride Vinyl acetate
H
Cl
Teflon
\
/
Polypropylene
C==C
Polymethyl
/
\
pentane
H
H
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HCl is removed
(in a CONDENSATION reaction)
Water is removed
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Amides
This is the carbon
portion of an
amine (R-NH3)
N
O
||
The C-N (amide group) is important in
natural polymers such as proteins, and
synthetic polymers like nylon.
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IDEAL GAS LAW
PV=nRT
Brings together gas
properties.
Can be derived from
experiment and theory.
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Boyle’s Law
If n and T are
constant, then
PV = (nRT) = k
This means, for
example, that P
goes up as V goes
down.
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Robert Boyle
(1627-1691).
Son of Early of
Cork, Ireland.
Charles’s
Law
If n and P are
constant, then
V = (nR/P)T = kT
V and T are
directly related.
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Jacques Charles (17461823). Isolated boron
and studied gases.
Balloonist.
Avogadro’s Hypothesis
Equal volumes of gases at the
same T and P have the same
number of molecules.
V = n (RT/P) = kn
V and n are directly related.
twice as many
molecules
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Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask
with a volume of 2.50 L. What is the
pressure of O2 at 25 oC? Of H2O?
Solution
Strategy:
• Calculate moles of H2O2 and then
moles of O2 and H2O.
• Finally, calc. P from n, R, T, and V.
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Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with
a volume of 2.50 L. What is the
pressure of O2 at 25 oC? Of H2O?
Solution
0.016 mol of O2
P of O2 = nRT/V
(0.016 mol)(0.0821 L• atm/K • mol)(298 K)
=
2.50 L
P of O2 = 0.16 atm
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Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Solution
What is P of H2O? Calculate as above,
or recall Avogadro’s hypothesis.
V  n at same T and P
P  n at same T and V
2 times as many moles of H2O as of
O2. P  n. P of H2O is 2X P of O2.
P of H2O = 0.32 atm
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Dalton’s Law of Partial Pressures
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
0.32 atm 0.16 atm
What is the total pressure in the
flask?
Ptotal in gas mixture = PA + PB + ...
Therefore,
Ptotal = P(H2O) + P(O2) = 0.48 atm
Dalton’s Law: total P is sum of
PARTIAL pressures.
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