All chem-notes

Chemistry:
The Study of Change

TEACHER: QBA MIGUEL ANGEL CASTRO RAMÍREZ

Introduction to
Chemistry
and the
Scientific Method

ask a
draw a question
conclusion do research

analyze
data design an
experiment
make observations
and collect data

Chemistry: A Science for the 21st Century

Health and Medicine
•Sanitation systems
• Surgery with anesthesia
• Vaccines and antibiotics

Energy and the environment
•Fossil fuels
• Solar energy
• Nuclear energy

1.1

Chemistry: A Science for the 21st Century
Materials Technology
•Polymers, ceramics, liquid crystals
• Room-temperature superconductors?
• Molecular computing?

Food Technology
•Genetically modified crops
• “Natural” pesticides
• Specialized fertilizers

1.1

The Study of Chemistry
Macroscopic Microscopic

Chemists study the microscopic properties of matter, which in turn
produce matter’s observable macroscopic properties – thus, we
often switch back and forth between microscopic and macroscopic
views of matter in this course.
1.2

The scientific method is a systematic approach
to research. Although it is systematic, it is not a
rigid series of steps that must be done in a particular
order.
ask a
question
draw a
do research
conclusion

researcher’s form a
hidden bias hypothesis

analyze design an
data experiment
make observations
and collect data

A hypothesis is a tentative tested modified
explanation for a set of
observations that can be tested.

A theory is a unifying
principle that explains a body
of facts and/or those laws that
Atomic Theory are based on them.

A law is a concise statement of a
relationship between phenomena Force = mass x
that is always the same under the acceleration
same conditions.
1.3

Substances
Matter is anything that has mass and occupies
space.
Matter that has a uniform and unchanging
composition is called a (pure) substance.
examples of pure substances include table salt, pure
water, oxygen, gold, etc.

States of Matter
Matter normally occupies one of three
phases, or states. These are:

P Solid
P Liquid
P Gas

* Plasma is a 4th state of matter in which the particles are
at extremely high temperatures (over 1,000,000 °C).

States of Matter

As we shall see in more detail later, the phase (or
state) of a substance is determined by the average
kinetic energy of the particles that make up the
substance, (i.e., temperature) and the strength of the
attractive forces holding the substance’s particles
together.
moderate
liquid

weak
gas strong
solid

States of Matter
Solids
 Solids have a definite shape and volume.

 The particles of a solid cannot exchange positions.
 Solids are incompressible.

States of Matter
Liquids
 Liquids have definite volumes
 Liquids do not have a fixed
shape
 Like solids, liquids are also
incompressible

States of Matter
Gases
 Gases take on the shape and
volume of their container
 Unlike solids and liquids,
gases are highly compressible

States of Matter

Technically, the word
“gas” refers to a
substance that is in the
gas phase at room
temperature.
The word “vapor” refers
to the gaseous state of a
substance that is normally
a solid or liquid at room
temperature.

Classification of Matter
Matter can be classified based on its characteristics
into the following categories and subcategories:

1. mixtures
• homogeneous (solution)
• heterogeneous

2. (pure) substances
• compounds
• elements


A pure substance is a form of matter that has a
definite composition and distinct properties.

examples: gold, salt, iron, pure water, sugar

A mixture is a combination of two or more
substances in which each substance retains its
own distinct identity.

examples: salt water, oil & vinegar dressing, granite, air

Classification Summary see page 13

Mixtures
Mixtures can be heterogeneous or homogeneous.
Heterogeneous mixture : the composition is not
uniform throughout. You can visibly see the
different components.
examples: cement, iron filings in
sand, granite, milk, oil and water,
etc.

Homogenous mixture (also called a solution): The
composition of the mixture is the same throughout.

Solutions are made up of two components:
(1) the solute which is dissolved in
(2) the solvent.

If the solvent is water, the solution
is called an aqueous solution
which is symbolized: (aq).

We often think of a solution as being a solid
dissolved in a liquid. However…

In a solution, both the solvent and solute can be in
any phase – solid, liquid or gas.

solvent solute example
liquid solid salt dissolved in water

liquid liquid gasoline (a mix of liquids)
gas gas air (O2 dissolved in nitrogen)
solid solid alloys (brass, bronze, etc.)

If a substance dissolves in another substance, we
say the first substance is soluble in the second. If
they do not dissolve, they are said to be insoluble.
example: carbon dioxide is soluble in air
gold is insoluble in water

In the case of liquids, we use a special term:
If two liquids completely dissolve in each other, they
are said to be miscible. If they do not, they are
immiscible.
example: alcohol and water are miscible
gasoline and water are immiscible

A mixture can be separated into its pure
components by simple physical methods.

Filtration is a means of separating a
solids from liquids. For example, we
can filter out the sand from a mix of
sand and water.

Magnetic substances can
be separated using a
magnet.

Separation of a Mixture
Fractional crystallization is a means of separating
two solids by adding a solvent that will dissolve one
of the solids but not the other; the mixture is then
filtered to separate out the insoluble solid. Finally,
the solvent is evaporated off to recover the
remaining solid.

For example, we can separate
salt from sand by adding hot
water to dissolve the salt, then
filter off the sand. The water is
then evaporated off, leaving the
salt behind.

Distillation is a means of separating two liquids
based on differences in their boiling points.
The substance with the lowest boiling point
“boils off” and is then cooled and condensed
back into a liquid. The liquid is collected in a
receiver flask.

This method is only effective for
substances that are liquids at
room temperature with
significant differences in their
boiling points.
distillation apparatus


Chromotography is the separation of a mixture
based on solubility in a “mobile” solvent coupled with
an adherence to a “stationary phase” medium, such
as paper or silica gel, etc.

column chromotography is a
common means of separating
components from a mixture

Thin Layer Chromotography
can be used to separate the
components of chlorophyll from
a crushed plant leaf.


Other means of separation:
Other techniques of separating a mixture include
sublimation, extraction, and leaching, etc.
If you had a jar containing both nails and marbles,
the only way to separate them would be by hand

speak to the
hand…

Example:

You are given a test tube which contains a mixture of water, methanol, aspirin,
acetanilide and aluminum oxide. (Acetanilide, aluminum oxide and aspirin are
all white, powdery solids at room temperature and are thus visibly
indistinguishable from each other. Water and methanol are both colorless
liquids at room temperature and are also visibly indistinguishable from each
other.) Assume your only source of heat is a Bunsen burner which can produce
a maximum temperature of 600C. Using the following information, devise a
method to separate this mixture. Be specific and complete in your answer.

substance melting point boiling point what it dissolves/does not dissolve in
water 0 C 100 C dissolves in cold or hot methanol
methanol – 97 C 65C dissolves in cold or hot water
aspirin 135 C decomposes at dissolves in methanol or water (if above
140C 10C)
aluminum 2072C 2980 C does not dissolve in either methanol or
oxide water at any temperature
acetanilide 114C 304C dissolves only in hot (50C) water or
warm (25°C) methanol

Pure Substances:
Elements and Compounds

 Mixtures are composed of two or more substances
physically combined.
 Recall that a (pure) substance is matter that has
a uniform, unchanging composition
 Pure substances may be elements or
compounds

Elements
An element is a substance that cannot be
separated into simpler substances by chemical
means.
carbon sulfur
• 114 elements have been identified

mercury • 82 elements occur naturally on Earth
examples include carbon, sulfur,
copper iron
copper, iron, and mercury

• 32 elements have been synthesized by scientists.
examples: technetium, americium, and seaborgium

Symbols for Elements
Elements are identified by a one or two-letter symbol.
The first letter, which is ALWAYS capitalized, is
typically the first letter in the name of the element.

eg, C = carbon, H = hydrogen

The second letter (which is only used if other elements
have the same first letter) is NEVER capitalized.

eg, Cl = chlorine, He = helium.

Some symbols are based on the Latin name

eg, iron is Fe (for ferrum) and sodium is Na (for natrium)

Symbols for Elements

http://www.privatehand.com/flash/elements.html

Compounds
A compound is a substance composed of atoms
of two or more different elements chemically
bonded in fixed proportions.

As such, they can be chemically decomposed
into their component elements.

table salt (NaCl) sugar

Water (H2O)

Sucrose (C12H22O11)


The properties of a compound are different from
the properties of its component elements

For example, table salt is composed of sodium and chlorine.
Sodium is a soft, silver colored metal that reacts violently
with water, and chlorine is a pale-green poisonous gas – yet
when chemically combined, they form table salt, a white
crystalline solid you put on your eggs in the morning!

+ =

Compounds

Compounds can only be separated into their pure
components (elements) by chemical means.

For example:
Iron is separated from iron ore (Fe2O3)
by heating the ore in a blast furnace and
reacting it with carbon monoxide and
elemental carbon (in the form of “coke”).
Water can be separated into its
elements, hydrogen and oxygen, by
passing an electric current through it,
a process called electrolysis.

Compounds
There are TWO kinds of compounds, depending on
the nature of the chemical bond holding the atoms
together.
Molecules form when two or more neutral atoms
form bonds between them by sharing electrons

Note that some elements exist as molecules.
For example,the following elements occur in
nature as molecular diatomic elements:
H2 O2
H2, N2, O2, F2, Cl2, Br2 and I2

They are molecules, but they are NOT N2
compounds, because they have only Cl2
one kind of element present.

Compounds

Ionic compounds are composed of ions, which are
atoms that have a (+) or (-) charge.

+ ions are called cations and form +
─
+ ─ +
─ + +
when an atom loses ─ + ─
─ ─
electrons + ─ +
+ ─
ions are called anions and form
when an atom gains electrons

Ionic compounds form when cations and anions
form electrostatic attractions between them
(opposite charges attract)

Classification
MATTER Summary

can it be separated
YES NO
by physical means?

MIXTURE PURE
SUBSTANCE

is the mixture uniform can the substance be
throughout? chemically decomposed into
simpler substances?

YES NO YES NO

heterogeneous
solution compound element
mixture

Physical & Chemical Properties

Physical Properties are measurable properties

• mass $ density
• boiling point $ solubility in water

Chemical Properties describe how a
substance reacts with other substances

• flammability $bonds with oxygen
• reacts with water $decomposes when
heated

Extensive and Intensive Properties
Physical properties can be classified as being either
extensive or intensive properties.

An extensive property of a material depends upon
how much matter is being considered. Extensive
properties are additive.
• mass
• length
• volume

Extensive and Intensive Properties

An intensive property of a material is independent
of the amount of matter is being considered, and is
not additive.
• density • melting point
• temperature •color

Note that ALL chemical properties
are intensive properties.

Physical & Chemical Changes

A physical change does not alter the composition
or identity of a substance.
sugar dissolving
ice melting
in water

A chemical change (reaction) alters the identity
or composition of the substance(s) involved.

hydrogen burns
in air to form
water

Physical & Chemical Changes
Evidence of a chemical reaction include:
1. Heat and light (both) produced
2. Gas produced (bubbles)
3. Solid precipitate forms
4. Color changes occur

Measurement
The SI System of Measurement
Scientists around the world use a unified system of
measurement (Le Systeme Internationale d’Unites,
or SI for short).

There are seven fundamental “quantities” that
can be measured:

Length Temperature Luminous intensity
Mass Electric Current

Time Chemical quantity

International System of Units (SI)

Each base quantity is given a unit with a
specific name and symbol

page 16

International System of Units (SI)
The SI units are based on metrics. Each power
of ten change is given a special prefix used with
the base unit.
You must know these prefixes

see page 17

Measurements with SI Units

Length (SI unit = meter) The meter is often
divided into cm and mm. (10 mm = 1 cm ).

Your little finger is about 1 cm in width.
A dime is about 1 mm thick.

English/Metric equivalencies

1 inch = 2.54 cm
1 meter = 39.37 inches


Volume (SI unit = m3) Volume is the amount of
space occupied by something.

A more common unit is the dm3 =1 liter.
A smaller unit that we will use frequently is the cm 3.

1 cm3 = 1 ml
1000 ml = 1 liter

1 liter = 1.057 quarts
1 ml ~ 15 drops

Measuring Volume
regular solids: volume = length x width x height
liquids—use a graduated cylinder. To read the
scale correctly, read the volume at the lowest part
of the meniscus - the curve of the liquid’s surface
in a container.
Your eye should be level with
the meniscus when reading the
volume

meniscus

Measuring Volume continued

irregular solids: volume is found by displacement.

Begin with a known volume of water. Add the solid.
The amount of water displaced is the volume of the
solid.
volume of solid =
6 6 volume displaced :
6.0 – 4.0 = 2.0 cm3
4 4

2 2


Mass (SI unit = kilogram): the amount of matter.
The mass of a given object is constant.

A kilogram is about 2.2 pounds -- this is too large a
unit for most chemistry labs, so we will use grams
instead.

Note that mass and weight are two different
things…


Weight is a measure of the force due to gravity
acting on a mass. The weight of an object
changes, depending on the gravitational force
acting on it.

For example, on the moon you would weigh only 1/6th what
you do on Earth, because the force of gravity on the moon
is only 1/6th that of Earth.

http://www.exploratorium.edu/ronh/weight/

The Importance of Units
On 9/23/99, the Mars Climate Orbiter entered Mar’s atmosphere 100 km (62
miles) lower than planned and was destroyed by heat because the
engineers that designed the rocket calculated the force provided by the
engines in pounds, but NASA engineers thought the force was given in the
units of Newtons (N) when they determined when to fire the rockets…

1 lb = 1 N
1 lb = 4.45 N

“This is going to be the
cautionary tale that will be
embedded into introduction
to the metric system in
elementary school, high
school, and college science
courses till the end of time.”

Measuring Mass
Triple beam balance

Electronic balance

We still use the term “weighing” even though we are finding
the mass of an object, not its weight…

1 kg = 2.203 lbs 1 paperclip  1 gram
1 lb = 453.6 grams

Temperature (SI unit = kelvin) is a measure of the
average kinetic energy (energy due to motion) of
the atoms and molecules that make up a substance.

There are three common temperature scales
Fahrenheit (oF) – English system, based on the freezing
point of salt water.
Centigrade (oC) – metric system, based on the freezing and
boiling points of pure water
Kelvin (K) – SI unit, also called the “Absolute” scale; 0 K
(Absolute Zero) is defined as the temperature at which all
motion stops (kinetic energy = 0).

Temperature

Conversions:
K = oC + 273.15
273 K = 0 oC
373 K = 100 oC

o
C = 5 (oF – 32)
9

9
o
F= (oC) + 32
5

32 oF = 0 oC
212 oF = 100 oC

Temperature
Examples
A thermometer reads 12o F. What would this be in oC ?
The conversion formula from oF to oC is: oC = 5/9(oF – 32)
Inserting the values gives: : oC = 5/9(12oF – 32)
o
C = 5/9(-20) = -11.1oC

A thermometer reads 315.3 K. What would this be in oF ?
First convert K to oC: 315.3 K – 273.15 = 42.15oC

The conversion formula from oC to oF is: oF = 9/5(oC) +32.
Inserting the values gives: : oC = 9/5(42.15 oC) + 32
o
C = (75.9) + 32 = 107.9 oF

Time (SI unit = second). This is the only non-
metric SI unit. We still use 1 day = 24 hours,
1 hour = 60 minutes, 1 minute = 60 seconds
We do use metric fractions of time, however, such
as milliseconds (1/1000th of a second), etc.

Chemical Quantity ( SI unit = mole). Since atoms
are so tiny, it takes a LOT of them to make even
one gram. In fact, you would have to put
602,200,000,000,000,000,000,000 atoms of carbon
(that’s 6.022 X 1023) on a balance to get just 12
grams of carbon!

The Mole continued

That huge number (6.022 X 1023) is given a special name; it is
called “Avogadro’s Number,” symbolized NA, after the Italian
physicist, Lorenzo Romano Amedeo Avogadro who lived
between 1776-1856.

Just like 1 dozen = 12 things, we define:

1 mole = 6.022 X 1023 things
Avogadro

JUST HOW BIG IS AVOGADRO’S NUMBER??

 1 mole of oranges would cover the surface of the earth to a
depth of 9 miles!
 If you stacked 1 mole of notebook paper, it would take you
5,800 years, traveling at the speed of light (186,000,000
miles per second) to reach the top of the stack!
 If you were given 1 mole of dollar bills when the universe
began 13 billion years ago, and you immediately began
spending money at the rate of one million dollars per second,
you would still have about 190 billion trillion dollars left !
 but 1 mole of Hydrogen atoms would only mass about 1
gram!

Working with Numbers: Scientific Notation

The number of atoms in 12 g of carbon:
602,200,000,000,000,000,000,000
6.022 x 1023
The mass of a single carbon atom in grams:
0.0000000000000000000000199
1.99 x 10-23
N is a number n is a positive or
between 1 and 10 N x 10 n
negative integer

Derived Units
Although we need only seven fundamental SI units,
we can combine different units to obtain new units,
called derived units.
For example, speed is distance per unit time, so we
must combine the unit for distance (m) and time
(sec) to get the SI unit for speed:
speed = meters per second (m/s)
We will be working with many different derived
units in this course. It is important to pay attention
to the individual units that make up derived units!!

Derived Units
Density is the mass per unit volume of a substance. It is
calculated using the equation:

mass m
density = volume d= V

SI derived unit for density is kg/m 3 . This is not a convenient
unit in chemistry, so we usually use the units g/cm3 or g/mL

Every substance has a unique density. For example:
substance density You need to know the density of water.
gasoline 0.70 g/cm3
Any object that is more dense than water
water 1.00 g/cm 3
will sink in water; if it is less dense, it will
aluminum 2.70 g/cm3 float in water
lead 11.35 g/cm3

Dimensional Analysis:
A problem solving
technique
desired unit
given unit x = desired unit
given unit

The Mathematics of Units
In algebra, we learn that:
u x u = u2 and… (2u)3 = 8 u3
u
u = 1 (the u’s cancel!) and u x a = a
u
If we let “u” = units, then every measured quantity is
a number x a unit. We can solve problems by
setting them up so that the unit we do NOT want
gets cancelled out by dividing u/u in the problem.
Thus, if a/u is a conversion (say 100 cm/1 m) then
we can convert cm to meters etc. using this
conversion factor so that the cm cancel…

Dimensional Analysis
Method of Solving Problems
1. Determine which unit conversion factor(s) are needed
2. Carry units through calculation
3. If all units cancel except for the desired unit(s), then the
problem was solved correctly.

given quantity x conversion factor = desired quantity

given unit x desired unit = desired unit
given unit

Dimensional Analysis
Method of Solving Problems

Example: How many μm are in 0.0063 inches?
Begin with what units you have “in hand,” then make a list of
all the conversions you will need.

conversion factors needed:
0.0063 in = ? 1 inch = 2.54 cm
106 μm = 1 m 1 m = 100 cm

2.54 cm 1m 106 μm
0.0063 inch x x x = 160 μm
1 inch 10 cm
2 1m

Dimensional Analysis Method
of Solving Problems
Example: The speed of sound in air is about 343 m/s. What
is this speed in miles per hour? (1 mile = 1609 meters)
conversion units

meters to miles 1 mi = 1609
m
seconds to hours 1 min = 60 s 1 hour = 60 min

m 1 mi 60 s 60 min mi
343 x x x = 767
s 1609 m 1 min 1 hour hour

Uncertainty,
Precision & Accuracy
in Measurements

Uncertainty, Precision and Accuracy in
Measurements

When you measure length using a meterstick, you
often have to estimate to the nearest fraction of a
line.
The uncertainty in a measured value is partly due
to how well you can estimate such fractional units.

The uncertainty also depends on how accurate the
measuring device, itself, is.

http://
www.mhhe.com/physsci/chemistry/chang7/esp/folder_structure/ch/

Precision and Accuracy

Accuracy – how close a measurement is to the true or
accepted value

To determine if a measured value is accurate, you would
have to know what the true or accepted value for that
measurement is – this is rarely known!

Precision – how close a set of measurements are to
each other; the scatter of repeated measurements
about an average.
We may not be able to say if a measured value is accurate,
but we can make careful measurements and use good
equipment to obtain good precision, or reproducibility.

A target analogy is often used to compare accuracy and
precision.

accurate precise not accurate
& but &
precise not accurate not precise


example: which is more accurate: 0.0002 g or 2.0 g?
answer: you cannot tell, since you don’t know what the
accepted value is for the mass of whatever object this is that
you are weighing!

example: which is more precise: 0.0002 g or 2.0 g?
answer: surprisingly, the most precise value is 2.0 g, not the
0.0002 g. The number of places behind the decimal is not
what determines precision! If that were so, I could increase
my precision by simply converting to a different metric prefix
for the same measurement:
Which is more precise: 2 cm or 0.00002 km? They are, in
fact, identical!


Precision is a measure of the uncertainty in a measured value.
Any measured value is composed of those digits of which you
are certain, plus the first estimated digit.

1 2 3 4
1

The length of the object is at least 1.7 cm, and we might
estimate the last digit to be half a unit, and say it is 1.75 cm
long. Others might say 1.74 or possibly 1.76 – the last digit
is an estimate, and so is uncertain.

We always assume an uncertainty of ±1 in the last digit.
The percent error in a measured value is defined as:

% error = ± uncertainty x 100
measured value

The smaller the percent error, the greater the precision – the
smaller the % error, the more likely two measurements will be
close together using that particular measuring instrument.

Thus: 2.0 ± 0.1 has a % error of (0.1/2.0) x 100 = ±5%
but 0.0002 has a % error of (0.0001/0.0002) x 100 = ± 50%

Percent Difference
When determining the accuracy of an experimentally
determined value, it must be compared with the “accepted
value.” One common method of reporting accuracy is called
the percent difference ( %) – this gives how far off your
value is, as a percent, from the accepted value:

Percent difference:

experimental value – accepted value
% = x 100
accepted value

example: In an experiment, a student determines the
density of copper to be 8.74 g/cm 3. If the accepted value is
8.96 g/cm3, determine the student’s error as a percent
difference.

% = experimental value – accepted value x 100
accepted value

8.74 – 8.96 x 100 = − 2.46 %
% =
8.96

The (-) sign indicates the experimental value is 2.46%
smaller than the accepted value; a (+) % means the
experimental value is larger than the accepted value.


We will be doing math operations involving measurements
with uncertainties, so we need a method of tracking how the
uncertainty will affect calculated values – in other words, how
many places behind the decimal do we really get to keep the
answer?

The method requires us to keep track of significant digits.

Significant digits (or significant figures) are all of
the known digits, plus the first estimated or
uncertain digit in a measured value.

Significant Figures: Rules
• Any digit that is not zero is significant
1.234 kg 4 significant figures
• Zeros between nonzero digits are significant
606 m 3 significant figures
• Zeros to the left of the first nonzero digit are not significant
0.08 L 1 significant figure
• If a number is greater than 1, then all zeros to the right of
the decimal point are significant
2.0 mg 2 significant figures
• If a number is less than 1, then only the zeros that are at
the end and in the middle of the number are significant
0.00420 g 3 significant figures


How many significant figures are in
each of the following measurements?
24 mL 2 significant figures

3001 g 4 significant figures

0.0320 m3 3 significant figures

6.4 x 104 molecules 2 significant figures

560 kg You cannot tell!!


Suppose you wanted to estimate the number of jellybeans in a
jar, and your best guess is around 400.

Now – is the uncertainty in your estimate ±1 jellybean, or is it
±10 jellybeans, or maybe even ±100 jellybeans (if you weren’t
very good at estimating jellybeans…)

We need a way to write 400 and indicate in some way
whether that was 400 ±1 vs 400±10 vs 400±100. The plain
number “400” is ambiguous as to where the uncertain digit is.

Use scientific notation to remove the ambiguity:

400 ± 1 = 4.00 x 102 = 3 sig figs 400 ± 100 = 4 x 102 = 1 sig fig
400 ± 10 = 4.0 x 102 = 2 sig figs

Rounding Numbers
Given the number 6.82 and asked to round to 2 sig digits we
would write 6.8. We write 6.8 because 6.82 is closer to 6.8
than it is to 6.9
Given the number 6.88 and asked to round to 2 sig digits, we
would write 6.9. We write 6.9 because 6.88 is closer to 6.9
than it is to 6.8
You were taught this long ago.
You were also probably taught that, given the number 6.85,
and asked to round this to 2 sig digits, you would write 6.9.

My question is, WHY did you round UP? 6.85 is JUST as
close to 6.8 as it is to 6.9! Since it is in the middle, it could be
rounded either way! And we should round it “either way.”

Rounding Numbers
Since the rounding is “arbitrarily” up, this can introduce
some round-off errors in chain calculations involving this
number – the final value will be too large if you always
round up when the next digit is exactly 5.
Because rounding is “arbitrary” when the next digit is
exactly 5, we introduce the following “odd-even rounding
rule:
When the next digit is exactly 5, round up or
down to make the number an even number.

e.g. : 4.65 ≈ 4.6 but 4.75 ≈ 4.8
Note however, that 4.651 is closer to 4.7 than 4.6, so we
round it to 4.7: only invoke the “odd-even rule” when the
next digit is exactly 5.

Math Operations with Significant Digits
We need a set of rules to determine how the
uncertainty or error will “propagate” or move through a
series of calculations and affect the precision of our
final answer.

There is one rule for addition
and subtraction, and one rule
for multiplication and division.
Do not mix them and match
them and confuse them!

Significant Figures
Addition or Subtraction
The answer cannot have more digits to the right of the
decimal point than any of the original numbers.

89.332
+1.1 one digit after decimal point
90.432 round off to 90.4

3.70 two digits after decimal point
-2.9133
0.7867 round off to 0.79

Significant Figures
Addition or Subtraction
We often encounter two numbers that must be added that
are in scientific notation. We cannot add them and
determine the number of places “behind the decimal” unless
they have the same power of 10 – we may have to convert!

Example: What is the sum of 2.4 x 102 + 3.77 x 103 ?

3.77 x 103
Always convert the smaller power of 10 to the
0.24 x 103 larger power of 10

4.01 x 103
The answer is good to 2 behind the decimal
when written as x103 -- that is, the uncertain
digit is in the “tens” place (± 10)

Significant Figures
To determine the power of 10, visualize a see-saw when you
move the decimal point:
10n Increasing the power of 10
+ n means you must move the
decimal to the LEFT one
n
place for each power of 10
increase

10n + n Moving the decimal determines
both the magnitude and the +/-
n value of 10n

Decreasing the power of 10
+n means you must move the
10 n
decimal to the RIGHT by one
n
place for each power of 10
decrease.

Significant Figures

Example: What is the answer to the following, to the correct
number of significant digits?

3.0268 x 10-2
- - 0.012 x 10-2
0 1.2 x 10-4

3.0148 x 10-2 = 3.015 x 10-2

Significant Figures

Multiplication or Division
The number of significant figures in the result is set by the
original number that has the smallest number of significant
figures
4.51 x 3.6666 = 16.536366 = 16.5

3 sig figs round to
3 sig figs

6.8 ÷ 112.04 = 0.0606926 = 0.061

round to
2 sig figs 2 sig figs

Significant Figures
Exact Numbers
Numbers from definitions or numbers of objects are
considered to have an infinite number of significant figures

Example: Find the average of three measured lengths:
6.64, 6.68 and 6.70 cm.
These values each have 3 significant figures

6.64 + 6.68 + 6.70
= 6.67333 = 6.67 =7
3

Because 3 is an exact number the answer is not rounded to 7, but
rather reported to be 6.67 cm (three sig figures).

1.8

Atoms, Molecules and Ions

Chapter 2

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Early Ideas

Our understanding of the structure of
matter has undergone profound changes
in the past century.

Nonetheless, what we know today did
not arrive on a sudden inspiration. We
can trace a fairly steady plodding
towards our current understanding,
starting as far back as 400 BCE…

Early Ideas
Democritus (c.a. 400 BCE)
All matter was composed of tiny, indivisible
particles called atoms (atomos = indivisible)
Each kind of matter had its own unique kind of
atom – ie., there were water atoms, air atoms, fire
atoms, bread atoms, etc.
The properties of matter could be explained by the
shape and size of its atoms.

Fire atoms water atoms
“ouch!” rolls & flows

Early Ideas
Most importantly, Democritus believed atoms
existed in a vacuum – that is, there was “nothing”
in the spaces between atoms…
Vacuum??

Aristotle, among others, refused
to believe in the existence of
“nothingness” that still occupied
space…

As a result, Democritus’ ideas were not very well
receieved. It would be some 1200 years before the
idea of atoms was revisited!

Early Ideas
Aristotle
Aristotle was the court philosopher to Alexander the
Great. Because of this, Aristotle’s ideas were given a
lot of weight .

Aristotle believed that all matter
was composed of four elements:

earth, air, fire and water.

Early Ideas
These elements could be “inter-converted” into each
other by exchanging the “properties” of hot, cold, dry
and wet. FIRE
hot dry

AIR EARTH

wet cold
WATER
example
Heating WATER exchanged “hot” for “cold” which created “AIR”
(which we see as steam…)
WATER (cold, wet) AIR (hot, wet)

Early Ideas
This idea that one kind of
element could be converted
into another eventually led
to the belief in Alchemy –
that one could turn lead into
gold by performing the right
chemical reaction!

Early Ideas
The “scientific method” of inquiry was developed
during the 17th and 18th centuries. The invention of
the balance and other instruments soon led to a new
understanding about the nature of matter.

The French chemist, Antoine-
Laurent Lavoisier (1743-1794),
presented two important ideas
which would later help lead to
a new, more developed atomic
theory of matter…

Lavoisier

1. The Law of Conservation of Matter: matter is not
created or destroyed in chemical reactions. Any
atomic theory would have to explain why matter is
not gained or lost in reactions.
2. Lavoisier defined element as any substance that
could not be chemically broken down into a simpler
substance.

Lavoisier was a meticulous experimenter. He also
helped develop the metric system of measurement.
He is often called the “Father of Modern Chemistry,”
in recognition of his pioneering works.

Lavoiser experimenting with respiration

Early Ideas

Joseph Proust, another 18th century French
scientist, proposed the Law of Definite
Proportion, which states that the mass ratios of
elements present in different samples of the same
compound do not vary.

For example, the percent by mass of the elements
present in sugar are always found to be:
53.3% oxygen, 40.0% carbon and 6.7% hydrogen.

John Dalton (1766-1824)
Dalton started out as an
apothecary's assistant (today,
we would call him a pharmacist).
He was also interested in both
meteorology and the study of
gases.

Dalton developed a new atomic theory of the nature
of matter based on several postulates. His theory
differed significantly from the early ideas of
Democritus, but they both agreed that the simplest
form of matter was the atom.

Dalton’s Atomic Theory (1808)
1. Elements are composed of extremely small particles
called atoms.
2. All atoms of a given element are identical, having the
same size, mass and chemical properties. The atoms of
a given element are different from the atoms of all other
elements.
3. Compounds are composed of atoms of more than one
element. In any compound, the ratio of the numbers of
atoms of any two of the elements present is either an
integer or a simple fraction.
4. A chemical reaction involves only the separation,
combination, or rearrangement of atoms; it does not
result in their creation or destruction.

Dalton’s Atomic Theory

Law of Conservation of Matter and
Definite Proportion Explained…

+ =

16 X + 8Y 8 X2Y

Law of Multiple Proportions
If Dalton’s ideas about atoms were correct, then he
proposed that the mass of a compound containing
different numbers of a given element (atom) would
vary by the mass of that one whole atom – that is:

If two elements can combine to form more than
one compound, then the masses of one element
that combine with a fixed mass of the other
element are in ratios of small, whole numbers.

Dalton’s Atomic Theory
Consider the mass ratio of oxygen to carbon in
the two compounds: CO and CO2

16
= 1.33
12
2.67 / 1.33 = 2

32 = 2.67
12

Note that the mass of oxygen that combines with 12 g of
carbon in carbon dioxide is 2 x greater than the mass of oxygen
that combines with 12 g of carbon in carbon monoxide.

Modern Ideas
In the late 19th and early 20th centuries, three
important experiments that shed light on the
nature of matter were conducted:

1. J.J. Thomson’s investigation of cathode rays that led
to the discovery of the electron.

2. Robert Millikan’s “Oil drop experiment” that
determined the charge and mass of the electron.

3. Ernest Rutherford’s “Gold foil experiment” that finally
gave us the current “nuclear” model of the atom.

Cathode rays, discovered by William Crookes,
are formed when a current is passed through an
evacuated glass tube. Cathode rays are
invisible, but a phosphor coating makes them
visible.

J.J. Thomson
The Electron is Discovered

J.J. Thomson helped show
that cathode rays are made up
of negatively charged particles
(based on their deflection by
magnetic and electric fields).
Sir Joseph John Thomson
1856-1940

N

S

J.J. Thomson

Thomson showed that all cathode rays are identical,
and are produced regardless of the type of metals
used for the cathode and anode in the cathode ray
tube.
Thomson was unable to determine either the actual
electric charge or the mass of these cathode ray
particles. He was, however, able to determine the
ratio of the electric charge to the mass of the
particles.

J.J. Thomson
To do this, he passed cathode rays simultaneously through
electric and magnetic fields in such a way that the forces
acting on the cathode ray particles (now called electrons)
due to the fields cancelled out. The ratio of the electric field
strength to the square of the magnetic field strength at this
point was proportional to the charge to mass ratio of the
electron.
Electric
field only +

Both

Magnetic
field only

_

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J.J. Thomson

The value he obtained, −1.76 x 108 C/g*, was always
the same, regardless of the source of the cathode
rays.
This value was nearly 2000 times larger than the
charge to mass ratio of a hydrogen ion (H+)!
This indicated that either the charge of the electron
was very large, or that the mass of the electron was
very, very small – much smaller than the mass of a
hydrogen atom, which was the lightest atom known.

*the SI unit of electric charge is the Coulomb (C)

J.J. Thomson

Thomson proposed that these electrons were
not just very small particles, but were actually a
sub-atomic particle present in all atoms.

We thus credit Thomson with the “discovery”
of the electron because of his work in determining
their physical characteristics, and his rather bold
hypothesis that they were present in all atoms
(which was later shown to be true).

The Plum Pudding Model

Since the atom is neutrally charged, if it has (-)
charged electrons, there must also be a (+) part to
the atom to cancel the negative electrons.

This showed that Dalton’s idea that atoms were
indivisible is NOT correct – instead, the atom is
composed of TWO oppositely charged parts.

Thomson thought the atom was a diffuse (+)
charged object, with electrons stuck in it, like
raisins in pudding (the plum pudding model).

Thomson’s Plum Pudding
Model of the Atom

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Millikan’s Oil Drop Experiment

Robert Millikan (1911) designed an experiment to
determine the actual charge of an electron.

He suspended charged oil
drops in an electric field.
The drops had become
charged by picking up free
electrons after passing
through ionized air.


FELEC = E x q when the downward force of
gravity on the drop was
balanced by the upward force
FGRAVITY = m x g of the electric field, then:
E x q = m x g or q = mg/E

Knowing the mass (m) of the oil drop, and the
strength of the electric field (E), he was able to find
the charge (q) on the oil drop.

To find the charge of the electron, he found the
smallest difference between the charges on any
two oil drops.

eg: Suppose you find three oil drops have the following
charges: 12.4, 7.6, 10.8. The differences between the
charges are:
12.4 – 10.8 = 1.6 10.8 – 7.6 = 3.2
12.4 – 7.6 = 4.8 4.8 – 3.2 = 1.6

You would conclude the charge of the electron was 1.6
charge units.


Using this technique, Millikan was able to determine
the charge of an electron to be:

e = C 1.602 x 10C 19 C

Using Thomson’s charge to mass ratio and the
charge for the electron, Millikan determined the
mass of the electron to be 9.11 x 10-31 kilogram.
For his work, Millikan received the 1923 Nobel Prize in
Physics.


Radioactivity was discovered in 1895
It was found that there are three distinct types of
radiation: (+) alpha particles, (-) beta particles, and
neutral gamma rays.

(Uranium compound)

Rutherford’s
Gold Foil
Experiment
(1908 Nobel Prize in Chemistry)

Rutherford designed an experiment using these
newly discovered alpha-particles to test if Thomson’s
plum pudding model was correct.
He fired (+) alpha particles at the gold foil. If the
Thomson model was correct, most of the alpha
particles would pass through the foil with little
deflection.

Rutherford’s Experiment
Expected Results of Rutherford’s Experiment
The force of repulsion is directly proportional to the product of
the charges of the alpha particle and nucleus and inversely
proportional to the square of the distance between the center
of the two charges. F = kQ1Q2/R2
A large, diffuse positive charge is not able to repel a (+) alpha
particle very strongly, because the alpha particle cannot make
a close approach, so the angle of deflection,θ, would be fairly
small.
θ
(+) -particle R

Rutherford’s
Gold Foil
Experiment

When Rutherford performed the experiment,
nearly all the alpha particles passed through the
foil without deflection, as expected…

However, some particles were deflected
significantly, and perhaps one in 2000 were
actually deflected nearly 180 degrees!


Rutherford was stunned. This would be like firing a
machine gun at an apple, and having most of the
bullets pass through -- but every once in a while one
of the bullets would bounce off the apple and come
back and hit you! Why would this happen???

DUCK,
ERNIE!

?!?

something small and massive must be in there that ?
deflects only those bullets that directly hit it…


Only a positive charge with a very, very small
radius would allow the alpha particle to approach
close enough to experience a significant repulsion.

Strong repulsion!
-particle
θ
nucleus
R

By carefully measuring the angles of deflection, θ,
Rutherford was able to determine the approximate
size of this positive core to the atom.


Next, by measuring the kinetic energy of the alpha
particle before and after the collision, Rutherford
was able to apply conservation of momentum and
determine the mass of the atom’s positive core.

Putting it all together, he was able to conclude that
all the positive charge -- and about 99.9% of the
mass -- of an atom was concentrated in a very tiny
area in the middle of the atom, which he called the
nucleus.


Only the very few (+) α-particles that passed very
near this incredibly tiny (+) nucleus were strongly
deflected; most α-particles never came near the
nucleus and so were not deflected significantly.

*note carefully that the
(+) α-particles never
actually collide with
the (+) nucleus – the
repulsive force between
the like charges is too
great for that to occur!

Rutherford’s Model
of the Atom
The estimated size of this nucleus was such a tiny fraction of
the total volume of the atom, that at first Rutherford doubted
his own conclusion.

atomic radius ~ 100 pm = 1 x 10-10 m
nuclear radius ~ 5 x 10-3 pm = 5 x 10-15 m

“If the atom is the Houston Astrodome, then the
nucleus is a marble on the 50-yard line.”

As another size comparison, if the nucleus were the size of a
basketball, placed at PHS, the atom would be over 20 km in
diameter, reaching Martin to the North, and just missing the
US 131 Business Loop exit to the South!

The basketball-
sized nucleus would
also mass about
70,000,000,000
tons! This is
equivalent to about
100,000 cruise ship
ocean liners!

• Rutherford fired (+) charged alpha particles at thin sheets of gold foil and measured the angles at
which the alpha particles were deflected.

• Rutherford was testing the validity of Thomson’s plum pudding model. If this model were correct,
the (+) alpha particles would not be deflected by the diffuse (+) charge of Thomson’s atom.

• When Rutherford performed the experiment, he found that the majority of alpha particles did, in
fact, pass without significant deflection. However, a small number were significantly deflected, and
a very few were strongly deflected nearly 180 degrees.

• By measuring the angles of deflection, Rutherford was able to calculate the size and mass of the (+)
center that could produce the observed deflections. He found that all the (+) charge and about
99.9% of the atom’s mass was concentrated in a tiny region (about 1/100,000 the volume of the
atom).

• Only those alpha particles that passed very close to the nucleus experienced a strong enough
repulsion to produced significant deflections – most particles never came near the nucleus, and so
were not deflected.

• AP Extras:
• The repulsive force depends on 1/R 2 between the (+) alpha particle and the (+) charge of the
nucleus.
• He also relied on conservation of momentum to help him determine the mass of the nucleus
which was repelling the alpha particles.

Chadwick’s Experiment (1932)
(1935 Noble Prize in Physics)
Discovery of the Neutron
•H atoms have 1 p; He atoms have 2 p

?? •ratio of mass He/mass H should be 2/1 = 2
•measured ratio of mass He/mass H = 4 ???
James Chadwick discovered that when 9Be was bombarded
with alpha particles, a neutral particle was emitted, which was
named the neutron.
α + 9Be 1
n + 12C + energy
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
Now the mass ratios can be explained if He has 2 neutrons
and 2 protons, and H has one proton with no neutrons

mass n  mass p  1,840 x mass e-

Atomic number, Mass number and Isotopes
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
also called the nucleon number = atomic number (Z) + number of neutrons

Isotopes are atoms of the same element (X) with different
numbers of neutrons in their nuclei
Mass Number A
ZX
Element Symbol nuclide
Atomic Number
1 2 3
examples
1H protium
1H (D) deuterium 1H (T) tritium
14 235
C Carbon-14 U Uranium-235
6 92

Atomic number, Mass number and Isotopes

Examples:
How many protons, neutrons, and electrons are in 14C ?
6

6 protons, 8 (14 - 6) neutrons, 6 electrons

How many protons, neutrons, and electrons are in 59 Fe ?
26

26 protons, 33 (59 - 26) neutrons, 26 electrons

The Periodic Table
of the Elements

We now understand that the number of protons in
the nucleus of the atom is what “defines” the
element and gives each element its unique
properties.

The Periodic Table of the Elements

Transition metals

group
period

Elements
Properties of Metals
• malleable and ductile
• lustrous
• good conductors
• lose e- to form cations

Properties of Non-metals
• brittle
• dull
• poor conductors
• gain e- to form anions

Elements
Properties of Metalloids
• properties are intermediate between those of metals
and nonmetals
• semi-conductors

Names of Families or Groups
1A = alkali metals 5A = pnictides
2A = alkaline earths 6A = chalcogens
3A = boron family 7A = halogens
4A = carbon family 8A = noble gases

The chemical properties of elements within a
Family or Group are similar

Elements

Natural abundance
of elements in the
Earth’s crust

Natural abundance
of elements in the
human body

Molecules & Ions

A molecule is an aggregate of two or more neutral
atoms in a definite arrangement held together by
chemical forces

Note that some elements exist as molecules. For
example,the following elements occur in nature as molecular
diatomic elements:
H2, N2, O2, F2, Cl2, Br2 and I2

H2 F2
They are molecules, but they are NOT
compounds, because they have only
one kind of element present. O2 N2

Molecules & Ions
A polyatomic molecule contains more than two atoms

O3, H2O, NH3, C3H6O

An allotrope is one of two or more distinct molecular forms of
an element, each having unique properties. For example, O 2
and O3 are allotropes of oxygen; diamond, graphite and
buckminster fullerene (C60) are all different allotropes of
carbon.


Ionic compounds are composed of ions, which are
atoms that have a (+) or (-) charge.

+ ions are called cations and form when
C +
C + +
C +C + C
an atom loses electrons
C C
+ C
-ions are called anions andC + when +
form +
an atom gains electrons C

Ionic compounds form when cations and anions
form electrostatic attractions between them
(opposite charges attract)

Molecules and Ions
A monatomic ion contains only one atom
Examples: Na+, Cl-, Ca2+, O2-, Al3+, N3-

note that the convention is to indicate the magnitude of the
charge first, and then the sign: e.g., Ca2+, not Ca+2

A polyatomic ion contains more than one atom
Examples: ClO3-, NO2- , CN- , SO42-

Molecules and Ions

Examples

27 3+
How many electrons are in 13 Al ?

13 protons, so there are 13 – 3 =10 electrons

78
How many electrons are in 34 Se2- ?

34 protons, so there are 34 + 2 = 36 electrons

Charges of common monatomic ions

see page 54

Note that some atoms, especially transition metals, have multiple charge states

Note also that metals typically form (+) charged ions, nonmetals form (-)
charged ions.

Also note the relation between the magnitude of the charge
and the group number (1A, 5A, etc) for most elements.

The charge of representative metals (group 1A, 2A and 3A)
is equal to the group number
The charge of representative nonmetals (group 4A-7A) is
equal to: (the group number – 8)

Chemical
Nomenclature
Determining the names and
formulas of chemical compounds
IUPAC = International Union of Pure and Applied Chemists.
This is the group that determines the official rules of
nomenclature for all chemical elements and compounds

Chemical Formulas
A chemical formula is a combination of element
symbols and numbers that represents the
composition of the compound.

Subscripts following an element’s symbol
indicate how many of that particular atom are
present. If no subscripts are given, it is
assumed that only one of that atom is present in
the compound.

NH3 C3H6S P4O10

1 N + 3 H atoms 3 C + 6 H + 1 S atoms 4 P + 10 O atoms

Chemical Formulas
A molecular formula shows the exact number of
atoms of each element in the smallest unit of a
substance

Chemical Formulas

An empirical formula shows the simplest whole-
number ratio of the atoms in a substance

molecular empirical
H2O H2O
N2H4 NH2
C2H8O2 CH2O

C6H12O6 CH2O

note that different molecular compounds
may have the same empirical formula

Ionic Formulas

For ionic compounds the formula is always the
same as the empirical formula.
The sum of the charges of the cation(s) and anion(s) in each
formula unit must equal zero. Thus, the ratio
of cations to anions can always be reduced to simple, whole
number ratios. The ionic compound NaCl

Na+500Cl-500 = NaCl

Naming Binary
Molecular Compounds

Naming Molecular Compounds

We will only consider naming binary molecules.

Binary molecular compounds typically form
between two non-metals, or a non-metal and a
metalloid.

Naming Molecules:

1st element + root of 2nd element + “-ide”

e.g. : HCl = hydrogen chloride

See page 62
If there is more than one of
a given element, we use
prefixes to indicate the
number of each kind of
atom present.

The prefix mono is only
used for atoms that can
form more than one
compound with the second
element. For this class,
oxygen is the main
element that does this.


Examples of naming molecules

HI hydrogen iodide

NF3 nitrogen trifluoride

SO2 sulfur dioxide

N2Cl4 dinitrogen tetrachloride

NO2 nitrogen dioxide

N2O dinitrogen monoxide (laughing gas)

If the second element begins with a vowel, the
terminal vowel of the prefix is allowed to be
dropped.
For example
N2O4 could be called dinitrogen tetroxide, rather
than dinitrogen tetraoxide.
CO would be called carbon monoxide, not carbon
monooxide
Note, however, that the official IUPAC rule states
that the vowel is only dropped for “compelling
linguistic reasons.”

Naming Compounds containing Hydrogen
Compounds containing hydrogen can be named using the
Greek prefixes, but most have common names that are
accepted by IUPAC. The most common examples are:
B2H6 diboron hexahydride diborane
CH4 carbon tetrahydride methane
SiH4 silicon tetrahydride silane

NH3 nitrogen trihydride ammonia
phosphorus trihydride phosphine
PH3
dihydrogen monoxide water
H2O
dihydrogen sulfide hydrogen sulfide
H2S

Determining the formula of molecules from the
name

The subscripts tell you the number of each type
of element present, so naming molecules from
the formula is straightforward.

e.g. sulfur hexafluoride = SF6
dichlorine heptoxide = Cl2O7

The order in which the atoms are listed in molecules is based
on something called electronegativity. For now, we can predict
the order using the chart on the next slide…

Chemical Formulas
Order of Elements in Writing Molecular Formulas
H

B C N O F

Si P S Cl

Ge As Se Br

Sb Te I

Organic chemistry is the branch of chemistry that
deals with carbon compounds

Carbon is unique among all the elements in its
ability to catenate, or form long or branching
chains of carbon atoms.
We usually write these chains as “condensed formulas” that
assumes carbons are bonded to each other as follows:

H H H
= CH3CH2CH3
H C C C H
note that we could also
H
write this as: C3H8
H H

Organic molecules that contain only carbon and
hydrogen are called hydrocarbons.

The first 10 simple hydrocarbons
Hydrocarbon
compounds are
named based on
the number of
carbon atoms in
the “backbone” or
chain of carbon
atoms.

Naming Ionic Compounds
Ionic Compounds
Ionic compounds are typically composed
of a metal cation and a non-metal anion

$ name of cation = simply the name of the element
$ name of anion = root of element’s name + - “ide”

Naming Ionic Compounds

Binary ionic compounds are named:
name of metal ion + root of non-metal + “-ide”

e.g. BaCl2 barium chloride
K2O potassium oxide
Na2S sodium sulfide
Mg3N2 magnesium nitride
Al2O3 aluminum oxide

Formula of Ionic Compounds

Determining the formula of ionic compounds from
the name is a little more involved – unlike
molecular compounds, the name does not give us
the subscripts. These must be determined based
on the charges of each ion.

Remember that the total number of (+)
and (-) charges in any ionic compound
must sum to zero.


2 x +3 = +6 3 x -2 = -6

aluminum oxide Al2O3
Al3+ O2-

1 x +2 = +2 2 x -1 = -2
calcium bromide CaBr2
Ca2+ Br-

1 x +2 = +2 1 x -2 = -2
magnesium sulfide MgS
Mg2+ S2-

Note that if you take the magnitude of the charge of
the cation, and make it the subscript on the anion,
and take the magnitude of the anion’s charge and
make it the subscript of the cation, the compound
will always end up with a net neutral charge. Now, if
possible, reduce the subscripts to a simpler ratio,
and you have the correct formula for the compound!

+3 -2
Al O Al2O3
2 3
Al3+ O2-

Pb
Cu
Multivalent ions:
W
The Non-Representative
Atoms
Fe
Mn
Co

Transition and other
multi-valent metal ions

Most elements form only ions with one charge.
However, most of the transition metals, as well as
Pb and Sn, have more than one possible charge
state. We say they are multi-valent.
e.g. : copper can exist in either a +1 or +2 charge
state: Cu+ or Cu2+
The formula or name of the compound must indicate
which charge state the metal cation is in.

Older method gives a common name for
each valence state
Cu+ cuprous Fe2+ ferrous
Cu2+ cupric Fe3+ ferric
Cr2+ chromous Hg22+ mercurous
Cr3+ chromic Hg2+ mercuric

e.g. CuCl = cuprous chloride Hg2I2 = mercurous iodide
Fe2O3 = ferric oxide PbO = plumbous oxide

To determine which charge state the cation is in, you
must look at the anion, and calculate the charge of
the cation…

CuS
S is always -2, and there is only one Cu to cancel this out,
so copper must be +2. Thus, this is cupric sulfide.

Fe2O3
Subscript on O is the charge of the iron! Thus, Fe is +3 and
this compound is ferric oxide.

Stock System:
We indicate charge on metal with Roman numerals

FeCl2 2 Cl- = -2 so Fe is 2+ iron(II) chloride
FeCl3 3 Cl- = -3 so Fe is 3+ iron(III) chloride
Cr2S3 3 S-2 = -6 so Cr is 3+ chromium(III) sulfide

2-
NH4 +
C2O 4

Polyatomic
Ions
SO4 2-

2-
C2H3O 2

Naming Polyatomic Ions
There are certain groups of neutral atoms that bond
together, and then gain or lose one or more
electrons from the group to form what is called a
polyatomic ion. Most polyatomic ions are
negatively charged anions.

Examples:
OH- = hydroxide ion CN- = cyanide ion
NO3- = nitrate ion NH4+ = ammonium ion
SO42- = sulfate ion SO32- = sulfite ion

Naming Polyatomic Ions

Naming ionic compounds containing polyatomic ions
is straightforward:

Name the cation + name the (polyatomic) anion

Examples:
NaOH = sodium hydroxide
K2SO4 = potassium sulfate
Fe(CN)2 = iron (II) cyanide
(NH4)2CO3 = ammonium carbonate

Compound
Summary

see page 64

NAMING ACIDS AND BASES

There are a different set of
rules for naming acids. Some
of the rules are based on a
much older system of
nomenclature, and so the
rules are not as simple as they
are for molecular and normal
ionic compounds.

Acids
An acid can be defined as a substance that yields
hydrogen ions (H+) when dissolved in water. These
H+ ions then bond to H2O molecules to form H3O+,
called the hydronium ion.

Many molecular gases, when
dissolved in water, become acids:

•HCl (g) = hydrogen chloride
•HCl (aq) = HCl dissolved in water
which forms (H3O+,Cl-) = hydrochloric acid

Acids

All acids have hydrogen as the first listed
element in the chemical formula.
For nomenclature purposes, there are two major
types of acids:

Oxoacids (also called oxyacids) = acids that
contain oxygen. eg: H2SO4, HC2H3O2

Non-oxo acids = acids that do not contain oxygen.
eg: HCl (aq), H2S (aq)

Acids
Rules for naming non-oxoacids
acid = “hydro-” + root of anion + “-ic acid”

see page 65

*

*note that we add an extra syllable for acids with sulfur and phosphorus:
it’s not hydrosulfic acid, but hydrosulfuric acid. Similarly, acids with
phosphorus will end in phosphoric, not phosphic acid.

Acids
An oxoacid is an acid that contains hydrogen,
oxygen, and another element –
That is, oxoacids are the protonated form of
those polyatomic ions that have oxygen in their
formulas.
examples:
HClO3 chloric acid

HNO2 nitrous acid

H2SO4 sulfuric acid

When naming oxoacids, NO “hydro” prefix is used.
Instead, the acid name is the root of the name of the
oxoanion + either “-ic” acid or “-ous” acid, as follows:

If the name of the polyatomic anion ends in
“ate,” drop the -ate and add “ic acid.”
eg: SO42- = sulfate anion H2SO4 = sulfuric acid
C2H3O2- = acetate anion HC2H3O2 = acetic acid

If the name of the polyatomic anion ends in
“ite,” drop the -ite and add “ous acid.”
eg: SO32- = sulfite anion H2SO3 = sulfurous acid
NO2- = nitrite anion HNO 2 = nitrous acid

Acids
Naming Oxoacids and Oxoanions

see page 66

Acids

As a mnemonic aid, I always use the following:
ic goes with ate because….”IC…I ATE it!
ite goes with ous like……tonsil-ITE-OUS, senior-ITE-OUS

Bases

A base can be defined as a substance
that yields hydroxide ions (OH-) when
dissolved in water.

NaOH sodium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide

Hydrates
Hydrates are compounds that have a specific
number of water molecules attached to them.

BaCl2•2H2O barium chloride dihydrate
LiCl•H2O lithium chloride monohydrate
MgSO4•7H2O magnesium sulfate heptahydrate
Sr(NO3)2 •4H2O strontium nitrate tetrahydrate

CuSO4•5H2O CuSO4
cupric sulfate anhydrous
pentahydrate cupric sulfate

Hydrates
Other terms associated with hydrates

Anhydrous: without water; this term describes
hydrated compounds after “drying.”
Hygroscopic: readily absorbs moisture directly
from the air.
Deliquescent: absorbs moisture from the air so
readily, that these compounds can take on enough
water to actually start to dissolve.
Water of hydration: the water absorbed and
incorporated into hygroscopic compounds

Mass Relationships in Chemical
Reactions

Chapter 3


Relative Masses of the Elements

Micro World Macro World
atoms & molecules grams

Atomic mass is the mass of an atom in atomic
mass units (amu). This is a relative scale
based on the mass of a 12C atom.
By definition: 1 atom 12C “weighs” 12 amu

On this scale 1H = 1.008 amu and 16O = 16.00 amu

Relative Masses of the Elements
How do we find the relative masses of the
other elements?
Imagine we have 66.00 grams of CO2. The compound
is decomposed and yields 18.00 grams of C and 48
grams of O. Since there are two oxygen atoms for
every 1 carbon atom, we can say that
48 g O xygen 2 × 24 gra m s O xyg en 24 g O
= so = 1 .3 3 3
18 g C arbon 1 8 g C a rbon 18 g C

This means that the relative mass of each oxygen atom is
1.333 x the mass of a carbon atom (12.00 amu) , or…
mass of oxygen = 1.333 x 12.00 amu = 16.00 amu

Average Atomic Mass

The average atomic mass of an element is the
weighted average mass of that element, reflecting
the relative abundances of its isotopes.

example: consider lithium (Li), which has two
isotopes with the following relative percent
abundances: 7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
The Average atomic mass of lithium would be:

 7.42 + 
92.58
  6.015 amu   7.016 amu = 6.941 amu
 100   100 

Average Atomic Mass

IA The masses reported at the
1 bottom of the “box” for each
H
1. 0079 IIA element in the Periodic Table
3
Li
4
Be is the average atomic mass
6.941 9. 012 for that element, (in amu).
11 12
Na Mg IIIB IVB
22.99 24.305
19 20 21 22
K Ca Sc Ti
39.098 40.08 44.956 47.90

Average Atomic Mass
see page 79

The Mole & Avogadro’s Number

The mole (mol) is the SI unit for the amount of a
substance that contains as many “things” as there
are atoms in exactly 12.00 grams of 12C.

This number, called Avogadro’s number (NA),
has been experimentally determined to be
approximately 6.0221367 X 1023 things.

1 mol = NA = 6.022 x 1023 “things”

We can have 1 mole of atoms, or molecules, or even dump
trucks. The mole refers only to a number, like the term
“dozen” means 12.

The Mole & Avogadro’s Number
JUST HOW BIG IS AVOGADRO’S NUMBER??
• If you stacked 1 mole of notebook paper, it would take you
5,800 years, traveling at the speed of light (186,000,000 miles
per second) to reach the top of the stack!
• If you were given 1 mole of dollar bills when the universe
began 13 billion years ago, and you immediately began
spending money at the rate of one million dollars per second,
you would still have about 190 billion trillion dollars left !
• 1 mole of oranges would cover the surface of the earth to a
depth of 9 miles!
• but 1 mole of Hydrogen atoms would only mass about 1
gram!

The Mole & Molar Mass
Molar mass is the mass, in grams, of exactly
1 mole of any object (atoms, molecules, etc.)
Note that because of the way we defined the mole :
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu

Thus, for any element

atomic mass (amu) = molar mass (grams)

For example: 1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li

One Mole of:
C = 12.01 g S = 32.06 g

Hg = 200.6 g

Cu = 63.55 g Fe = 55.85 g

Solving Mole Problems
We can now add the definitions of the mole, Avogadro’s
number, and molar mass to our repertoire of conversion
factors we can use in dimensional analysis problems.

Thus, given the mass, we can use the molar mass to
convert this to moles, and then use Avogadro’s number to
convert moles to particles, and vice versa…

M = molar mass in g/mol
NA = Avogadro’s number


How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K
conversion factors
1 mol K = 6.022 x 1023 atoms K

1 mol K 6.022 x 1023 atoms K
0.551 g K x x
39.10 g K 1 mol K

= 8.49 x 1021 atoms K

see page 81

see page 82

Molecular Mass

Molecular mass (or molecular weight) is the sum of
the atomic masses of the atoms in a molecule.
Example: consider SO2
1S 32.07 amu
2O + 2 x 16.00 amu
SO2 64.07 amu
SO2 1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
As was the case for atoms, for any molecule
molecular mass (amu) = molar mass (grams)

Formula Mass
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.

1Na 22.99 amu
NaCl
1Cl + 35.45 amu
NaCl 58.44 amu
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
For any ionic compound
formula mass (amu) = molar mass (grams)

Formula Mass

What is the formula mass of Ca3(PO4)2 ?

1 formula unit of Ca3(PO4)2

3 Ca 3 x 40.08
2P 2 x 30.97
8O + 8 x 16.00
310.18 amu
Since the formula mass, in grams (per mole), is
numerically equal to the molar mass, in amu, we find
that the formula mass of Ca3(PO4)2 = 310.18 grams
per mole of Ca3(PO4)2.

Molecular/Formula Masses
Using Molecular/Formula Masses in Dimensional
Analysis Problems
We can now add molecular & formula masses to our list of
conversion factors. They are used similarly to the way we
used the molar mass of the elements as conversion factors.
Example: How many H atoms are in 72.5 g of C3H8O ?

1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
conversion
factors

1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.022 x 1023 atoms H
1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms
72.5 g C3H8O x x x
60 g C3H8O 1 mol C3H8O 1 mol H atoms

= 5.82 x 1024 atoms H

see page 84

see page 85

The Mass Spectrometer

Atomic and molecular masses of unknown compounds are
determined using a mass spectrometer.
A gaseous sample of the unknown is bombarded with
electrons in an electron beam. This knocks electrons loose
from the unknown to produce cations. These cations are
then accelerated through perpendicular electric and
magnetic fields. The charge:mass ratio (e/m) of the
unknown ions determines the degree to which the particles
are deflected.

The greater the charge:mass ratio, the smaller the angle
through which the beam is deflected.

The Mass Spectrometer

We know the angle that a given e/m produces, so we can
identify the unknown ion when it registers on a special screen.

high e/m low e/m

Mass Spectrometer

Percent composition
Percent composition of an element in a compound is
the percent, by mass, of that element in the compound.
It can be calculated as follows:
n x molar mass of element
x 100%
molar mass of compound

where n is the number of moles of the
element in 1 mole of the compound
Knowing the percent composition, one can determine
the purity of a substance, (are there contaminants
present in the sample?) and you can even determine
the empirical formula of an unknown compound.

Percent composition
Example: What is the percent composition of
ethanol, which has the formula, C2H6O ?
First, we find the molecular mass of ethanol. This is found
to be: 2(12.01) + 6(1.008) + 1(16.00) = 46.07 grams/mole.

2 x (12.01 g)
% Composition: %C = x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H = x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O = x 100% = 34.73%
46.07 g

C2H6O
check: 52.14% + 13.13% + 34.73% = 100.0%

Percent composition
We can also determine the % by mass of groups
of atoms present in a compound in the same
manner.
Example: what is the percent water in epsom salts, which
has the formula: MgSO4 • 7 H2O ?
mass of water
% H2O = x 100
mass of compound
this is the molar
7(18.02) mass of water
= 24.31 + 32.07 + 4(16.00) + 7(18.02)

126.14 g H2O
= x 100 = 51.17% H2O
246.52 g cmpd

Percent composition
Example: How many grams of CaCl2 • 2 H2O must be weighed out
to obtain 12.20 grams of CaCl2?
There are two ways of solving this problem:
Method 1:
First determine the % CaCl2 in CaCl2 • 2 H2O:
110.98 g CaCl2
i. % CaCl2 = x 100 = 75.49%
147.02 g CaCl2 • 2 H2O

Then we note that the 12.20 g of CaCl2 desired must be
75.49% of the mass of the hydrate used:
ii. 75.49% of (X grams) of CaCl2•2 H2O = 12.20 g of CaCl2

 0.7549(X) = 12.20 or X = 12.20/0.7549 = 16.16 grams

Percent composition
Example: How many grams of CaCl2 • 2 H2O must be weighed out
to obtain 12.20 grams of CaCl2?
There are two ways of solving this problem:
Method 2:
Use dimensional analysis and molar masses:

1 mole CaCl2• 2 H2O
12.20 g CaCl2 x 1 mole CaCl2 x
110.98 g CaCl2 1 mole CaCl2

x 147.02 g CaCl2 • 2 H2O = 16.16 g
1 mole CaCl2• 2 H2O
note that, math-wise, both methods involve the exact same calculations
(i.e., the ratio of the molar mass of the hydrate to the molar mass of the
anhydrous form had to be determined). The only difference was the “logic”
you followed which led you to that calculation!

Percent Composition and
Empirical Formulas
Knowing the percent composition of a compound, one can
determine the empirical formula. It is essentially the same
process as finding the percent composition – only you work
backwards to find the molar mass of the compound…

1. First, you convert the % composition into grams. This is
easily done – suppose you had 100 grams of the
substance. Then, the mass, in grams, of each
component element is numerically the same as its
percent composition.
example: a sample of an iron ore is found to contain
69.94% Fe and 30.06% O. In 100 grams of the ore, there
would be 69.94 grams of Fe and 30.06 grams of oxygen.

Empirical Formulas
2. Next, knowing the mass of each element (in your 100
gram sample), determine the number of moles of that
element in your sample, by dividing the mass by the
molar mass of the element.

The number of moles of Fe and O in our sample of the iron
ore would be:
69.94 grams Fe x 1 mol Fe = 1.252 mol Fe
55.847 g

30.06 grams O x 1 mol O = 1.879 mol O
16.00 g

Empirical Formulas
3. To find the simplest mole ratio of the elements, divide
each by the smallest number:

in our iron ore sample, we would have:
1.879 mol O = 1.501 mol O per mole of Fe
1.252 mol Fe

4. If this ratio is a whole number, then you are done – if the
ratio is NOT a whole number, it must be converted to a
whole number ratio (we cannot have fractions of an
atom!)
Fe1.00O1.50 = Fe 2O 3 = Fe2O3
2 2

Empirical Formulas

The process is summarized in
Figure 3.5 on page 89 in your
textbook.

Empirical Formulas
Example: Determine the empirical formula of a compound
that has the following percent composition by mass:
K = 24.75%, Mn = 34.77%, and O = 40.51%

1 mol K
nK = 24.75 g K x = 0.6330 mol K
39.10 g K
1 mol Mn
nMn = 34.77 g Mn x = 0.6329 mol Mn
54.94 g Mn
1 mol O
nO = 40.51 g O x = 2.532 mol O
16.00 g O

Empirical Formulas

nK = 0.6330, nMn = 0.6329, nO = 2.532

0.6330 ~
K: ~ 1.0
0.6329
divide each element
0.6329
by the smallest mole Mn : = 1.0
value 0.6329
2.532 ~
O: ~ 4.0
0.6329

The empirical formula for the
compound is: KMnO4

Determination of Molecular Formulas

Experimental determination of a molecular formula

sample

Combust 10.0 g compound
Collect 24.078 g CO2 and 11.088 g H2O

1mole CO2 1mol C 12.011g C
24.078 g CO2 × × × = 6.571g C
44.01g 1mol CO 2 1mole C

1mole H2 O 2 mol H 1.008 g H
11.088 g H2O × × × = 1.241g H
18.016 g 1mol H2O 1m ole H

grams of O = 10.00 g - (6.571 + 1241) = 2.188 g O


mass of each moles of each mole ratios of empirical
element element the elements formula

1mole C 0.5471mole C
6.571g C × = 0.5471m ole C = 4.0 C
12.011g 0.1368 mol O

1mole H 1.231mole H
1.241g H × = 1.231m ole H = 9.0 H
1.008 g 0.1368 mol O

1mole O 0.1368 mol O
2.188 g O × = 0.1368 m ole O = 1.0 O
16.00 g 0.1368 mol O

empirical formula = C4H9O


The molecular weight of the compound was
determined experimentally to be 146.2 g/mol.
To determine the true molecular formula, divide the
molecular weight by the formula weight. This ratio
gives the number each subscript must be multiplied
by to give the molecular formula.
Formula weight of C4H9O = 73.1 g/mol
Molecular weight of compound = 146.2 g/mol
146.2 g / mol
= 2.0
73.1 g / mol

thus, the true molecular formula is (C4H9O)2 = C8H18O2

Working with
Chemical
Equations

Chemical Equations
A process in which one or more substances is changed into
one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction

For example, there are several ways of representing the
reaction of H2 with O2 to form H2O

How to “Read” Chemical Equations

2 Mg + O2 2 MgO
reactants form products

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
this is based on the molar masses of the
species and the coefficients in the reaction…

IT DOES NOT IMPLY THAT
2 grams Mg + 1 gram O2 makes 2 g MgO

Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left
side and the correct formula(s) for the product(s) on the
right side of the equation.
example: Ethane (C2H6) reacts with oxygen to form carbon
dioxide and water
 C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients)
to make the number of atoms of each element the same on
both sides of the equation. Do not change the subscripts.

2 C2H6 IS NOT = C4H12

Balancing Equations

3. Start by balancing those elements that appear in only one
reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbon 1 carbon  multiply CO2 by 2
on left on right

C2H6 + O2 2CO2 + H2O next balance H

6 hydrogen 2 hydrogen
on left on right  multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O next we will balance O

Balancing Equations

4. Balance those elements that appear in two or more
reactants or products.

C2H6 + O2 2CO2 + 3H2O multiply O2 by 7
2

2 oxygen 4 oxygen + 3 oxygen = 7 oxygen
on left (2x2) (3x1) on right

C2H6 + 7 O2 2CO2 + 3H2O to remove fraction
2 multiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

Balancing Equations

5. Check to make sure that you have the same number of
each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O 

4 C (2 x 2) 4C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)

Reactants Products
4C 4C
12 H 12 H
14 O 14 O

example: balance the following equations:

2 Fe + 3 S → Fe2S3

2 K + 2 H2O → 2 KOH + H2

16 Al + 3 S8 → 8 Al2S3

Pb(NO3)2 + 2KI → 2 KNO3 + PbI2

Stoichiometry
Stoichiometry is the quantitative study of reactants
and products in a chemical reaction.

One of the most important applications of balanced equations
is in determining the amount of one reactant required to react
completely with another, or in determining the theoretical
amount of product that should be formed in a given reaction.

These problems all follow the same set of “logic” steps –
indeed, almost any problem involving balanced equations
will always follow these same steps!

Stoichiometry

Stoichiometry
Flow Chart

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to determine the
mole:mole ratio between substances A and B, and from
this determine the number of moles of the sought quantity
4. Convert moles of sought quantity into desired units

Stoichiometry
Example: Methanol burns in air according to the equation:
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what
mass of water is produced?

The sequence of steps we follow in solving the problem are:

moles H2O
grams CH3OH moles CH3OH grams H2O
moles CH3OH

1 mol CH3OH 4 mol H2O 18.0 g H2O
209 g CH3OH x x x
32.0 g CH3OH 2 mol CH3OH 1 mol H2O

= 235 g H2O

Solution Stoichiometry see pages 142-150 in Chapter 4

It is often easier to work with solutions, rather than
solids. This means we also need a means of
quantitatively working with reactions in solution.

The molarity of a solution is the concentration of that
solution, expressed as the moles of solute present
in 1 liter of a solution

moles of solute
M = molarity =
liters of solution

read as, for example: 2M NaCl = 2 “molar” solution of NaCl

Solution Stoichiometry
example: what mass of KI is needed to prepare
500 mL of a 2.80 M solution of KI?

Solution plan: convert volume to moles using molarity, then
moles to mass using molar mass:

M KI MW KI
volume of KI solution moles KI grams KI

1L
500. mL x x 2.80 mol KI x 166 g KI
1000 mL 1 L soln 1 mol KI

= 232 g KI

known mass dissolve dilute to mark
of solute solute

Dilution is the procedure for preparing a less
concentrated solution from a more concentrated
solution.

Dilution
Add Solvent

Moles of solute Moles of solute
(concentrated) (c) = after dilution (d)

McVc = MdVd

example: How would you prepare 60.0 mL of 0.200 M
HNO3 from a stock solution of 4.00 M HNO3?

McVc = MdVd
Mc = 4.00 M Md = 0.200 M
Vc = ? L Vd = 0.0600 L

Md V d 0.200 M x 0.0600 L
Vc = = = 0.003 L = 3 mL
Mc 4.00 M

Thus, add 3 ml of acid to 57ml of water to form 60 ml of
solution (dilute the 3 ml of acid to a total volume of 60 ml)

Solving Solution Stoichiometry Problems
As with all stoichiometry problems, convert the starting unit to
moles. Note that we now have THREE methods of
converting to moles:

1. Use Avogadro’s number to convert particles to moles

# particles x NA = moles

2. Use the molar mass of the substance to convert grams
to moles

grams x mole = moles
molar mass


3. And now our third method is to use the molarity of
the solution to convert volume* to moles.

volume x moles = moles
L sol’n

Note also that we can convert moles to
volume by multiplying moles x 1/M

*the volumes must be in LITERS when converting
to moles using the molarity of the solution.

example: How many ml of 0.35 M Na3PO4 are needed to
react completely with 28.0 ml of a 0.42 M solution of
Ba(NO3)2, according to the balanced equation shown below:

3 Ba(NO3)2 + 2 Na3PO4 → Ba3(PO4)2 + 6 NaNO3

Solution plan: convert to moles, use mol : mol ratio from the
balanced equation, convert moles to liters using molarity, then
convert to mL.

0.42 mol Ba(NO3)2 2 mol Na3PO4
0.0280 L x x
Liter 3 mol Ba(NO3)2
1L 1000 mL
x x = 22.4 ml
0.35 mol Na3PO4 IL

Gravimetric Analysis
Gravimetric analysis is an analytical technique
based on the measurement of the mass (usually of
an ionic substance.)
The substance of interest is typically reacted in
solution and comes out as a precipitate. The
precipitate is then filtered off, dried and weighed.
Knowing the mass and chemical formula of the
precipitate that formed, we can calculate the mass
of a particular chemical component of the original
sample.

Gravimetric Analysis
1. Dissolve unknown substance in water
2. React unknown with known substance to form a precipitate
3. Filter and dry precipitate
4. Weigh precipitate
5. Use chemical formula and mass of precipitate to determine
amount of unknown ion

example: A 0.5662 gram sample of an unknown ionic
compound containing chloride ions is dissolved in water and
treated with an excess of AgNO3. If 1.0882 grams of AgCl
precipitated , what is the percent chlorine in the original
sample?

First determine the mass of Cl‾ ions in the AgCl ppt:

1.0882 g AgCl x 1 mol AgCl 1 mol Cl‾ ions x 35.45 g Cl
x
143.4 g AgCl 1 mol AgCl 1 mol Cl
= 0.2690 g
Now determine the % Cl in the original sample:

0.2690 g Cl
% Cl = x 100 = 47.51 %
0.5662 g unknown

Limiting Reagents
It is very rare that you mix reactants together in the
exact stoichiometric ratio needed for each to react
completely with the other.
Usually, you have a little “extra” of one of the
reactants compared to the other one, that is, one
reagent is present in excess.

The limiting reagent is the reactant that gets used
up completely -- that is, the one not present in
excess.

Limiting Reagents

The maximum amount of product that can be formed
is thus limited by the amount of the limiting reagent
present.
The reagents present in quantities greater than the
minimum amount necessary are called the reagents
in excess.

When the reaction is completed, there will be no
limiting reagent left over.
But there will be some of the reagents in excess left
over.

Limiting Reagents
Consider the reaction between NO and
O2 to form NO2. If we start with the mix
shown at the top right, and end with
the mix shown at the bottom right, we
see that oxygen was present in excess
(some is left over) which means that
NO was the limiting reagent.

2NO + 2O2 2NO2

NO is the limiting reagent
O2 is the excess reagent

Limiting Reagents
example: In one process, 124 g of Al are reacted with 601 g
of Fe2O3 according to the rxn: 2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed
OR
g Fe2O3 mol Fe2O3 mol Al needed g Al needed

1 mol Al 1 mol Fe2O3 160. g Fe2O3
124 g Al x x x = 367 g Fe2O3
27.0 g Al 2 mol Al 1 mol Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent

Limiting Reagents
Now…
Use limiting reagent (Al) to calculate amount of product that
can be formed.

g Al mol Al mol Al2O3 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

1 mol Al 1 mol Al2O3 102. g Al2O3
124 g Al x x x = 234 g Al2O3
27.0 g Al 2 mol Al 1 mol Al2O3

3.9

Reaction Yields
Theoretical Yield is the amount of product that
would result if all the limiting reagent reacted.

Actual Yield is the amount of product actually
obtained from a reaction.

Actual Yield
% Yield = x 100
Theoretical Yield

AP Inorganic Chemistry

Reactions in Aqueous Solution
Chapter 4


Symbols used in Equations

(s) solid yields

(l) liquid equilibrium

(g) gas  heated
20º C specified rxn
(aq) aqueous =
(dissolved in H2O) temperature

MnO2 MnO2 is a catalyst
solid precipitate
in the rxn
given off as a gas
arrows are only
used for products!

examples
NaCl (s) + AgNO3 (aq) AgCl ( ) + NaNO3 (aq)


Na2CO3 (s) Na2O (s) + CO2 (g)

MnO2
2 KClO3 (s) 2 KCl (s) + 3 O2 ( )


HCl (g) + H2O (l) H3O+ (aq) + Cl¯ (aq)

H2O or
HCl (g) H3O+ (aq) + Cl¯ (aq)

We often write H2O over the yields arrows when we
show that something is being dissolved in water.

General Properties of
Aqueous Solutions

Solutions

A solution is a homogenous mixture of 2 or more
substances

The solute is the substance(s) present in the
smaller amount.

The solvent is the substance present in the larger
amount

General Properties of Aqueous Solutions
During the solution process, the solute is first
surrounded by the solvent molecules, and the
attractions between the solvent and the solute help to
pull the solvent particles apart.

The isolated solute particles are in turn surrounded by
a “sphere” of solvent particles in a process called
solvation – in the case of aqueous solutions, the term
hydration is used.

Solutions
Hydration (or solvation) : the process in which a solute
particle, such as an ion or a neutral molecule, is surrounded
by water molecules arranged in a specific manner.

Hydration
“spheres”

Solutions

An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.

A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte

Solutions
To conduct electricity, a solution must contain
charged particles, that is, cations (+) and anions (-)

Strong Electrolyte – 100% dissociation
H2O
NaCl (s) Na+ (aq) + Cl- (aq)

Weak Electrolyte – not completely dissociated
H2O
CH3COOH CH3COO- (aq) + H3O+ (aq)

(all three species are present at equilibrium)

Solutions
Nonelectrolytes do not produce charged particles
in solution
H2O does not dissociate into
C6H12O6 (s) C6H12O6 (aq) ions when in solution

Classification of
Reactions

PbI2

Classes of Reactions
There are literally millions of known chemical
reactions. It would be impossible to learn or
memorize them all. Instead, we will learn five
fundamental “classes” of reactions.
1. Double displacement
2. Single displacement
3. Decomposition
4. Synthesis
5. Combustion

1. Double Displacement

Also called “metathesis” reaction. All double displacement
reactions follow a pattern in which two elements “trade
partners.”

AX + BY BX + AY

Double displacement reactions are typically one of two types:
1. precipitation reactions
2. neutralization reactions

Precipitation reactions involve the exchange of
cations between two ionic compounds that results
in the formation of an insoluble precipitate*:
eg: AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
*we will learn to determine which substances are
soluble and which are not later on in this chapter…

Neutralization reactions involve an acid and a
metal hydroxide; the acid’s H+ ion is exchanged
with the hydroxide’s metal cation to produce an
ionic “salt” and water.
eg: HCl + KOH KCl + HOH (which is just H2O!)


2. Single displacement:
Single displacement is a reaction in which one element
displaces another in a compound. The general pattern is:

A + BC B + AC

Single displacement reactions are typically one of three types:

Sr + 2H2O Sr(OH)2 + H2 Hydrogen Displacement

2 Mg + TiCl4 2MgCl2 + Ti Metal Displacement

Cl2 + 2KBr 2KCl + Br2 Halogen Displacement

3. Synthesis:
A synthesis reaction is one in which two substances react
and combine to form one substance. The general pattern of
the reaction is:

A + B AB

The reactants can be elements, compounds,
or one of each.
eg: 2 Al + 3 Br2 2 AlBr3
H2O + SO3 H2SO4

4. Decomposition:

Decomposition is the opposite of synthesis: one substance
decomposes (often by heating it) into two or more new
substances. The general pattern is:

AB A+ B

eg: 2 KClO3 Δ 2 KCl + 3 O2

2 NaHCO3 Δ Na2O + 2 CO2 + H2O

5. Combustion:

Technically, any reaction involving oxygen is a
combustion reaction.

A + O2 AOx

eg: 2 Mg + O2 2 MgO

Note that this reaction can also be
classified as a synthesis reaction.

Combustion reactions that involve hydrocarbon
compounds (which may or may not contain O or N)
reacting with oxygen gas will form CO2 and H2O.
Incomplete combustion forms CO (carbon monoxide).
The general reaction is:

CnHm + O2 CO2 + H2O

eg: CH4 + 2 O2 CO2 + 2 H2O

C2H6O + 3 O2 2 CO2 + 3 H2O

Classify the following reactions:
SD 1. 2 Na + 2 H2O 2 NaOH + H2

DD (P)
2. Ba(NO3)2 + Na2CrO4 BaCrO4 + 2NaNO3
C
3. C4H10 + 13 O2 8 CO2 + 10 H2O
DD (N)
4. Ca(OH)2 + H2S (aq) CaS + H2O
S
5. 4 Cu + O2 2 Cu2O
D
6. K2CO3 K2O + CO2
DD/D
7. NaHCO3 + HCl (aq) NaCl + H2O + CO2

We are now going to look at each class of
reaction in more detail.

We will begin with the sub-
categories of the double
displacement reaction

Precipitation Reactions
One of the most common types of double displacement
reaction is the precipitation reaction.
If a substance dissolves in a solvent, it is said to be soluble; if
it does not, it is insoluble. Substances that are insoluble
have a stronger attraction towards each other than they do
towards the solvent (or sometimes, the solvent molecules
have a stronger attraction towards themselves than towards
the solute particles).

Solubility is the maximum amount of solute that will dissolve
in a given quantity of solvent at a specific temperature.
Precipitate – insoluble solid that separates from solution


When aqueous solutions of Pb(NO3)2 and KI are mixed, a
bright yellow precipitate of PbI2 forms.

Pb(NO3)2 (aq) + 2KI (aq) PbI2 (s) + 2KNO3 (aq)

PbI2

Pb2+ and I- form strong attractions precipitate of PbI2

Solubility Rules
How do you know which substances are soluble
and which form precipitates in aqueous solutions?
There is a Table of general Solubility Rules on
page 123 in your textbook.

AP Chemistry students are required to memorize
this list. (You need to know it for the AP Test!)
Non-AP students are NOT required to memorize
this list; it will be provided to you on tests and
quizzes.


see page 123

and acetates (C2H3O2-)


While most double displacement reactions involve
the formation of a precipitate, some double
displacement reactions involve “dissolving” an
insoluble compound by forming a soluble salt.
These types of reactions most often involve the
reaction of an insoluble ionic compound with an
acid:
AgCl (s) + HC2H3O2 (aq) AgC2H3O2 (aq) + HCl (aq)

PbS (s) + 2 HNO3 (aq) Pb(NO3)2 (aq) + H2S (g)


Molecular and net ionic equations

The complete balanced equation, showing the formulas for
each reactant and species is called the “molecular equation.”
molecular equation
Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

We can also show how each species in the reaction will
dissociate or ionize when dissolved in water. Insoluble
compounds do not dissociate, but soluble compounds will. We
call this the “ionic equation.”
ionic equation
Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3-

Molecular and net ionic equations

Those species which appear unchanged on both sides of
the yields sign did not “participate” in the reaction. They
are said to be spectator ions.

Pb2+ + 2NO3− + 2Na+ + 2I− PbI2 (s) + 2Na+ + 2NO3−
In the above reaction, Na+ and NO3- are the spectator ions –
they do not participate in the net reaction.
The net ionic equation shows only those species which
actually participated in the reaction – all the spectator ions
are cancelled out and not shown.
net ionic equation
be sure to include the phases,
Pb (aq) + 2I (aq)
2+ −
PbI2 (s) etc. in net ionic equations!

Net Ionic Equations
Writing Net Ionic Equations

1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong
electrolytes completely dissociated into
cations and anions.
3. Cancel the spectator ions on both sides of
the ionic equation
4. Check that charges and number of atoms
are balanced in the net ionic equation

Net Ionic Equations

example: Write the net ionic equation for the reaction of silver
nitrate with sodium chloride.
molecular AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
ionic Ag+ + NO3− + Na+ + Cl− AgCl (s) + Na+ + NO3−
net ionic Ag+ (aq) + Cl− (aq) AgCl (s)

example: Write the net ionic equation for the reaction of CuS
with acetic acid
molecular CuS (s) + 2 HC2H3O2 (aq) Cu(C2H3O2)2 (aq) + H2S (g)

ionic CuS (s) + 2 H+ + 2 C2H3O2− Cu2+ + 2 C2H3O2− + H2S (g)

net ionic CuS (s) + 2 H+ (aq) Cu2+ (aq) + H2S (g)

Net Ionic Equations
Note that if both reactants and products exist as
solvated ions in solution, then NO REACTION has
occurred – you began with a mix of hydrated ions,
and you ended with the same mix of hydrated ions…

example

KNO3 (aq) + NaCl (aq) KCl (aq) + NaNO3 (aq)

K+ + NO3− + Na+ + Cl− K+ + Cl− + Na+ + NO3−

They are ALL spectator ions! We write:

KNO3 (aq) + NaCl (aq) N.R.

Acid-Base
Reactions

+
H O
H
H

H3O+ = hydronium ion

Acid-Base Reactions
A very common type of double displacement reaction involves
the neutralization of an acid with a base.

Arrhenius and Brnsted-Lowry Definitions
There are several “definitions” of acids or bases, from a
chemical standpoint.
The two most important definitions are those given by
Svante Arrhenius in the late 19th century, and by J.N.
Brnsted and Thomas Lowry, who independently developed
similar chemical descriptions of acids and bases in the 20th
century.

Acids and Bases

An Arrhenius base is a substance that dissociates
to produce OH- ions in water

H2O
NaOH (s) Na+ (aq) + O
H OH¯ (aq)
2

i.e., Arrhenius bases are metal hydroxides that are soluble in water.

An Arrhenius acid is a substance that ionizes to
produce H+ ions in water

H2O
HCl (g) H+ (aq) + Cl¯ (aq)

Acids and Bases

An H+ ion is essentially a bare proton – this is an extremely
reactive species!

H+ ions will instantly bond to a water molecule to form the
polyatomic cation, H3O+, called the “hydronium ion.”

So, actually, an Arrhenius acid is a substance that produces
H3O+ ions in water.

Acids and Bases

Brønsted-Lowry made use of the fact that H+ ions are
essentially just a proton in their definition of acids
and bases:

A Brønsted-Lowry acid is a proton (H+) donor
A Brønsted-Lowry base is a proton (H+) acceptor

B-L acid and base is a somewhat more “general”
definition, since it does not require the presence of
water as a solvent. However, one can certainly
have an aqueous B-L acid or base!

Acids and Bases

Consider the reaction between NH3 and H2O:

base acid acid base

In the forward direction, water acts as the proton donor and
NH3 the acceptor…

… in the reverse direction, NH4+ is the proton donor and OH-
is the acceptor.

Acids and Bases
Identify each of the following species as a Brønsted acid,
base, or both. (a) HI, (b) OH- (c) HPO42-

HI (g) + H2O H3O+ (aq) + I− (aq) acid

OH− (aq) + H+ (aq) H 2O base

HPO42− + H2O H3O+ (aq) + PO43−(aq) acid

HPO42− + H3O+ H2PO4−(aq) + H2O base

note that HPO42- can act as both an acid or a base! Such
substances are said to be amphoteric.

Acids and Bases

Acids with only one Monoprotic acids
ionizable H+ are said to be HCl H+ + Cl−
monoprotic acids
HCN H+ + CN−

Acids with 2 ionizable H+ are Diprotic acids
said to be diprotic acids H2SO4 H+ + HSO4−
HSO4− H+ + SO42−

Acids with 3 ionizable H+ are Triprotic acids
said to be triprotic acids. H3PO4 H+ + H2PO4−
H2PO4− H+ + HPO42−
HPO42− H+ + PO43−

Acids and Bases
Neutralization Reaction
A neutralization reaction is a special type of double
displacement reaction in which an acid reacts with a
hydroxide ion (base) to produce an ionic “salt” and water.

acid + base salt + water

HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl− + Na+ + OH− Na+ + Cl− + H2O
net ionic = H+ + OH− H2O
note that the net ionic equation for all neutralization
reactions will be H+ + OH¯ H2O !

Acids and Bases

Neutralization Reaction Examples:

2 HC2H3O2 + Ca(OH)2 2 H2O + Ca(C2H3O2)2

HCN (aq) + KOH H2O + KCN

H2SO4 + 2 NaOH 2 H2O + Na2SO4

Acids and Bases

Not all reactions involving acids and bases are
neutralization reactions. For example, ammonia (NH3), a
Brnsted-Lowry base, can react with acids to form aqueous
ammonium salts.

HCl (aq) + NH3 NH4Cl (aq)

Although this IS an acid base reaction, technically it is NOT
a neutralization reaction: The aqueous NH4Cl formed can
react with the water present in the solution to produce H 3O+ :

NH4+ + H2O NH3 + H3O+  sol’n is still acidic!

Other Reactions with Acids

Acids produce gases when they react with certain salts
containing carbonate, bicarbonate, sulfite and sulfide ions.
examples:
Na2CO3 + 2 HCl (aq) 2 NaCl (aq) + H2O + CO2 (g)
K2SO3 + 2 HBr (aq) 2 KBr (aq) + H2O + SO2 (g)
PbS + 2 HI (aq) PbI2 (s) + H2S (g)
Note that the reaction with PbS and HI is a simple
double displacement reaction. The others involve
both double displacement AND decomposition!

Basic and Acidic Oxides
Basic oxides are metal oxides.
The name “basic oxide” come from the fact that metal oxides
react with water in a synthesis reaction to form
hydroxides:

eg: H2O + Na2O 2 NaOH

Acidic oxides are non-metal oxides.
The name “acidic oxide” comes from the fact that non-metal
oxides react with water in a synthesis reaction to form
oxoacids.

eg: H2O + SO3 H2SO4

The reaction of water with SO3 is how acid rain forms.
Sulfur trioxide is a common pollutant from burning coal. The
SO3 reacts with water droplets in the air to form sulfuric acid.

H2O + SO3 H2SO4

Basic oxides can react with acidic oxides in a
type of acid-base reaction:

Na2O + SO3 Na2SO4
base acid salt

Some oxides are somewhat acidic and somewhat basic.
They are said to be amphoteric oxides. Aluminum oxide is
a good example:
Al2O3 + 6 HCl 2 AlCl3 + 3 H2O
Al2O3 + 2 KOH 2 KAlO2 + H2O

Amphoteric
oxides are
oxides of
metals at the
far right end
on the
periodic table

basic oxides amphoteric oxides acidic oxides

Aqueous acids react with basic oxides.
Not surprisingly, since metal oxides are basic, they react
with acids in a double displacement reaction that is
essentially a neutralization reaction:

2 HC2H3O2 (aq) + CaO (s) Ca(C2H3O2)2 (aq) + H2O (l )

The reaction occurs because the metal oxide first reacts
with water to form hydroxides, which in turn reacts with
the acid in a neutralization reaction:
i. CaO + H2O Ca(OH)2
ii. Ca(OH)2 + 2 HC2H3O2 Ca(C2H3O2) + 2 H2O
net: CaO (s) + 2 HC2H3O2 (aq) Ca(C2H3O2)2 (aq) + H2O (l )

Aqueous bases react with acidic oxides.
Similarly, since non-metal oxides are acidic, they react with
bases in a double displacement reaction that is essentially a
neutralization reaction:

2 NaOH (aq) + N2O5 (g) 2 NaNO3 (aq) + H2O (l )

The reaction occurs because the non-metal oxide first
reacts with water to form an oxoacid, which in turn reacts
with the hydroxide ion in a neutralization reaction:

i. N2O5 + H2O 2 HNO3
ii. 2 HNO3 + 2 NaOH 2 NaNO3 + 2 H2O
net: N2O5 (g) + 2 NaOH (aq) 2 NaNO3 (aq) + H2O (l )

Titrations
In a titration, a solution of accurately known concentration is
gradually added to another solution of unknown concentration
until the chemical reaction between the two solutions is
complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color (this is called the
end point). Indicators must be carefully chosen so that the
end point occurs at the equivalence point in the titration.

Slowly add base
to unknown acid
UNTIL
the indicator
changes color

example: A 24.00 ml sample of an unknown acid is titrated to
a phenolphthalein end point, which required 31.46 ml of a
0.104 M solution of NaOH. What is the molar concentration
of the unknown acid?

i. At the end point (equivalence point) the moles of acid =
moles of added base:

ii. MAVA = moles acid = moles base = MBVB

MBVB 0.102 M x 31.46 ml
iii. MA = = = 0.134 M
VA 24.00 ml

What volume of a 1.420 M NaOH solution is required to
titrate 25.00 mL of a 4.50 M H2SO4 solution to the eq. point?

Write the chemical equation:

H2SO4 + 2NaOH 2H2O + Na2SO4

MAVA = moles acid* (H+) = moles base (OH-) = MBVB
Note that H2SO4 is diprotic, so moles H+ = 2 x moles H2SO4
M mol M
volume acid moles acid moles base volume base
acid ratio base

4.50 mol H2SO4 2 mol NaOH 1000 ml soln
25.00 mL x x x = 158 mL
1000 mL soln 1 mol H2SO4 1.420 mol NaOH

Oxidation–Reduction
Reactions

In precipitation reactions, there is a transfer of
ions between two compounds.

In acid-base reactions, there is a transfer of
protons between acids and bases.

In the next grouping of reactions, called redox
reactions, there is a transfer of electrons
between reacting species.

Redox Reactions

The term redox is a shortened form of oxidation-
reduction.
Oxidation is a process in which one atom loses or
donates electrons to another.

Reduction is a process in which one atom gains or
accepts electrons from another.

In any redox reaction, one species is ALWAYS
oxidized and one is ALWAYS reduced. You cannot
have oxidation without reduction also occurring.

Redox Reactions

Mnemonic Aid: to help remember these two
definitions, use this aid:

LEO says GER

LEO = Loss of Electrons is Oxidation;
GER = Gain of Electrons is Reduction

Redox Reactions

Redox reactions are a very important class of
reactions. Everything from burning fossil fuels to the
action of household bleach on stains is a redox
reaction. In addition, most metals and nonmetals
are obtained from their ores by redox chemistry.

In fact, nearly every class of reaction we have
already looked at, with the exception of double
displacement reactions, is a form of a redox reaction.

Redox Reactions

Redox reactions often involve ions, but despite the
fact that we speak of electrons being transferred,
redox reactions do not always involve ions.

In all cases, however, this loss or gain of electrons
describes the difference in the electron density
around a bonded atom in a compound, compared to
the electron density in that atom’s elemental, un-
bonded state.

Redox Reactions
We assign an oxidation number to an atom to
indicate the relative electron density in its current
state, compared to that in its elemental state.

The oxidation number of an atom is simply the
“charge” the atom would have in a molecule (or an
ionic compound) if electrons were completely
transferred.

Note carefully that this is usually only a “pretend”
complete electron transfer, and has meaning only as
a book-keeping tool in accounting for shifts in the
electron density around bonded atoms!

Redox Reactions

For ionic compounds, cations are assigned positive
oxidation numbers because they really have lost
electrons compared to the electrons present in their
elemental state.
Similarly, anions really have gained electrons,
compared to their elemental state, and are assigned
negative oxidation numbers.

Redox Reactions

For non-ionic compounds, however, a negative
oxidation number simply means the atom has a
greater electron density in its bonded state than the
atom has in its elemental, un-bonded state.

A positive oxidation number means the atom has less
electron density around it in its bonded state than
when in its elemental, un-bonded state.

Rules for Assigning Oxidation Numbers
1. The oxidation state of any neutral element in its naturally
occurring state is zero.
Na, Be, K, Cl2, H2, O2, P4 = 0
2. The oxidation number of any cation or anion composed of
just one atom is that ion’s actual charge.
Li+ = +1; Fe3+= +3; O2- = -2

3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
4. The oxidation number of hydrogen is +1 except when it is
bonded to metals in binary compounds. In these cases,
its oxidation number is –1.

Rules for Assigning Oxidation Numbers continued

5. The oxidation number of F is always -1.

6. Oxidation numbers do not have to be integers. Oxidation
number of oxygen in the superoxide ion, O2-, is -½.

7. When you begin, assign the most electronegative element
present the charge it would have if it were an anion. If
oxygen is present, it has the higher priority and will always
be –2.

8. The sum of the oxidation numbers of all the atoms in
a molecule or ion is equal to the charge on the
molecule or ion.

Redox Reactions
Example: Consider the compound PCl3:

Cl is the more electronegative element, so assign
each Cl an oxidation # equal to its charge as an
anion (= –1).

PCl3 is a neutral molecule, so the sum of the
oxidation numbers for P and Cl must add to 0.
Let x = oxidation # of P: x + 3(–1) = 0  x = +3

PCl3: P = +3, Cl = -1

Redox Reactions
example: Consider HClO4.
Since oxygen is present, but not as a peroxide ion, the
oxygen is assigned an oxidation # of –2. The H is not bonded
to a metal, so it must have an oxidation # of +1. Let x =
oxidation number of Cl, and note that the sum of the oxidation
#’s must equal zero since HClO4 has a net charge of zero:

0 = +1 + x + 4(–2) solving for x gives x = +7 = Cl.

HClO4: H = +1, Cl = +7, O = -2

Note carefully that Cl is NOT a +7 cation in HClO4 !! The +7 oxidation
state simply tells us the electron density around Cl in HClO 4 is significantly
lower than the electron density around Cl in its elemental state.

Redox Reactions

Example: Consider Cr2O72–
Oxidation number for oxygen is: O = -2
The sum of the charges in the compound = the
charge of the polyatomic ion = -2
Let x = oxidation number of Cr. Solving for x
gives:

-2 = 2x + 7(–2) or x = +6

Thus: O = -2 and Cr = +6

Common oxidation numbers of
elements in their compounds

Redox Reactions
Oxidizing and Reducing Agents

If an element’s oxidation number has increased in a reaction,
this means the element has lost electrons and has been
oxidized.

If an element’s oxidation number has decreased, this means
it has gained electrons in the reaction and has been
reduced.

Example: 2Mg + O2 2MgO

2Mg0 2Mg2+ Mg has been oxidized

O02 2O2- O has been reduced

Redox Reactions

The element that was oxidized “donated” its electrons to the
element that was reduced (gained electrons). Thus, the
species that contains the element being oxidized is said to
be the reducing agent.

The element that was reduced “stole” electrons from the
element that was oxidized (lost electrons). Thus, the
species that contains the element being reduced is said to
be the oxidizing agent.
The more readily an element is oxidized, the better it is as a
reducing agent. Conversely, the more readily an element is
reduced, the better it is as an oxidizing agent.

Example: Identify the oxidizing agents and reducing agents
in the following reactions:

1. Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)

Zn Zn2+ Zn is oxidized Zn is the reducing agent

Cu2+ Cu Cu2+ is reduced CuSO4 is the oxidizing agent

2. 2 K (s) + 2 H2O (aq) 2 KOH (aq) + H2 (g)

K K+ K is oxidized K is the reducing agent
H+ Ho H+ is reduced H2O is the oxidizing agent

Redox Reactions

Disproportionation Reaction
One element is simultaneously oxidized and reduced.
0 +1 -1
Cl2 + 2OH− ClO− + Cl− + H2O

One chlorine atom is being oxidized to a +1 oxidation
number, and at the same time, the other Cl atom is
being reduced to a -1 oxidation number.

Note that the O and H are neither oxidized nor reduced
in this reaction.

Redox and Reaction Classification

Almost every reaction that is not a double
displacement reaction is a redox reaction.
0 0 2+ 2−
Synthesis: Fe + S FeS
1+ 2− 0 0
Decomposition: 2 H 2O 2 H 2 + O2
4− 1+ 0 4+ 2− 1+ 2−
Combustion: CH4 + 2 O2 CO2 + 2 H2O

Single 0 2+ 1− 2+ 1− 0
Displacement: Mg + NiCl2 MgCl2 + Ni

Predicting the products of redox reactions can
be a little tricky.
We will focus primarily on single displacement
reactions

In single displacement reactions, an element
displaces another element in a compound. This
means that one element must be oxidized, and the
other reduced, in order to accomplish this.

Some elements are better oxidizing agents than
others, and so not every element is able to oxidize
another. We may be able to write an equation, but
the reaction may not actually occur if we try it:

Zn (s) + CuCl2 (aq) ZnCl2 (aq) + Cu (s)
This reaction does occur; the Cu2+ ion can oxidize Zn

2 Ag (s) + CuCl2 (aq) 2 AgCl (s) + Cu (s)

This reaction does NOT occur; the Cu2+ ion cannot oxidize Ag

So…How can you tell if a given reaction will occur?

To determine if a single Activity Series of Metals
displacement reaction
can occur, we need to strong reducing
know the relative agents
strengths of elements as
oxidizing agents. This
table of values is called
an activity series.
The table at right lists
common metals in
order of decreasing
strength as reducing
poor reducing
agents – or increasing
agents
strength as oxidizing
agents.

Single Displacement Rxns
The Activity Series for Metals
Hydrogen Displacement Reaction

reactive metals above Mg can
react with cold water

Ca + 2H2O Ca(OH)2 + H2

all metals above Cu will
react with acids
Ni + 2 HCl NiCl2 + H2

metals below Cu do not react
even with acids
Au + HCl N.R.

Single Displacement Rxns
The Activity Series for Metals

Metal Displacement Reaction

a metal can displace any metal
that lies below it in the table

Mg + CdS Cd + MgS

no reaction occurs if you try to
displace a metal with a metal
that lies above it in the table

Cu + NiCl2 N.R.

Activity Series of Halogens
Halogens, unlike metals, are poor reducing agents.
However, they make good oxidizing agents.

Halogen Displacement Reaction

strongest
oxidizing agent
F2 A halogen can displace any
halogen below it in the
Cl2 activity series

Br2 Cl2 + 2KBr 2KCl + Br2
weakest I2
oxidizing agent I2 + 2KBr 2KI + Br2

Predicting Products of Reactions
In order to predict the products of a reaction, keep in mind
the following points:
1. Precipitation reactions always involve reacting two ionic
compounds (exchange cations).
2. Neutralization reactions always involve reacting an acid
with a metal hydroxide (exchange H+ and metal cation)

3. Synthesis reactions – put all the elements together –
does the formula look like a compound you recognize?
Try changing the subscripts by multiplying or dividing
them by an integer – now does the formula look familiar?
Remember the synthesis reactions of acidic and basic
oxides with water, and with each other!

Predicting Products of Reactions continued
4. Decomposition reactions – typically, there will be only
one compound as the only reactant. Break it down into
two compounds or two elements, etc.
5. Combustion reactions – look for hydrocarbons reacting
with elemental O2 to form CO2 + H2O
6. Single displacement reactions – almost always involve
a metal replacing another metal in a compound, or a
metal replacing H in water or acids. Also, halogens can
displace other halogens.
Be sure to use the activity series to
determine if the rxn will occur at all.
Also remember that alkali and alkaline earth metals
displace H from water to form hydroxides, not oxides!

example: predict if a product will form. If so, complete and
balance the reaction.

1. Na + Fe(NO3)3 2 NaNO3 + Fe

2. Ni + CdBr2 N.R.

3. Pb + H2O N.R.

4. Ba + 2 H2O Ba(OH)2 + H2

5. 2 KCl + F2 2 KF + Cl2

The Behavior
Of Gases
Chapter 5


The Kinetic
Molecular Theory
of Gases

Kinetic Molecular Theory of Gases

1. A gas is composed of molecules that are widely
separated from each other. The volume occupied
by the molecules is but a tiny fraction of the total
volume of the gas, to the extent that the gas
molecules can be considered to be point masses;
that is, they possess mass but have negligible
volume.
2. Gas molecules are in constant motion in random
directions, and they frequently collide with one
another.

Kinetic Molecular Theory of Gases continued

3. Collisions among molecules are perfectly elastic,
that is, there is no net gain or loss of kinetic
energy during these collisions.
4. There are neither attractive nor repulsive forces
between gas molecules.

5. The average kinetic energy of the molecules is
proportional to the temperature of the gas in
kelvins. Any two gases at the same temperature
will have the same average kinetic energy.


The average kinetic energy (KE) of a particle
depends on its mass, m, and its average velocity, u:

KE = ½ mu2

This means the higher the temperature, the faster a
particle moves. At a given temperature, low mass
particles move faster than higher mass particles.

* a bar over a variable just means, “the average value.”


The kinetic molecular theory of gases can be used to
explain several properties and behaviors of gases:

1. gases can be compressed
2. gases have low densities
3. gases are fluids
4. gases can undergo diffusion and effusion

Properties of Gases

Compressibility
There is a lot of empty space between the particles
in the gas state. This means that gases are very
compressible. A liter of gas can be compressed
down to about 1 milliliter.

Properties of Gases
Low Density
Because the particles of a gas are so spread out,
the density of gases is usually about 1000 x less
than solids and liquids.
We use the units of grams per LITER instead of
g/mL for the density of gases.

Fluidity
Since there are no attractive forces between gas
particles, they can “slide” past one another freely.

Properties of Gases
Effusion and Diffusion
Diffusion is the spontaneous mixing of particles
due to their random motion. Gas particles move
randomly and have no attractive forces between
themselves that would limit their ability to mix.
Effusion is the process whereby particles under
pressure escape through small openings. Gas
particles are very tiny, and can pass through small
holes at a rate that depends on their velocities.

Properties of Gases

When discussing gases, we must be aware of how
the temperature, the volume, the number of
particles, and the pressure of the gas all affect each
other.
The way each of these affects the other is the
subject of a set of gas laws, which were investigated
in countless experiments over several centuries.
But first, we need to get a better understanding of
this new unit, pressure.

Properties of Gases
Pressure
Pressure is defined as the
force pushing on one
square meter of a surface:
10 miles 0.2 atm
P = Force/Area.
Normal atmospheric
pressure is caused by the
4 miles 0.5 atm
weight of a column of air
pressing down on an
Sea level 1 atm
object.

Pressure
We measure pressure with a barometer.
vacuum
Evangelista Torricelli invented the
barometer in 1643. weight of
mercury
The air pushes down on a bowl column

of mercury, which in turn pushes
the mercury up a column.
The height of the column
depends on the air pressure –
higher pressures can support a
taller column of mercury. mercury

Units of Pressure
SI Unit = pascal (Pa) = N/m2

1 atmosphere = 101.325 kPa

1 atmosphere = 760 mm Hg

1 torr = 1 mm Hg

The conditions 0 oC and 1 atm of pressure is called
standard temperature and pressure (STP).

Manometers Used to Measure Gas Pressures

Open to the
atmosphere

Pressure

Pressure in a container is caused by the force
with which particles strike the walls of a container.
The pressure increases with increased frequency of
collisions, since more particles striking the wall
means more force.

Pressure also increases if the speed of the particles
increases – faster speed = greater force of impact
on the walls of the container.

The Gas Laws
Boyle’s law

At constant temperature, the pressure of a
gas in a confined space will increase when
the volume decreases -- and the pressure
will decrease when the volume increases.

P P P

volume decreases volume increases

pressure increases pressure decreases

Boyle’s law
We say that pressure and volume are
inversely proportional to each other.
k
P= , where k = a constant
V

P V P V

Pressure goes up, Pressure goes down,
volume goes down volume goes up

Boyle’s law
Graphically, P vs V produces a curved line as
shown below:

Boyle’s law
Explanation
At constant temperature, the average kinetic energy of gas
particles (and therefore the average speed) is constant.
If the volume increases, the distance the particles must
travel between collisions with the wall increases, which
means the number of collisions/sec decreases (it takes
longer for particles to reach the walls of the container).
This means the pressure decreases.

If you decrease the volume, then the number of
collisions/second increases and the pressure goes up.

http://highered.mcgraw-hill.com/classware/ala.do?isbn=0072980605&alaid=ala_729140&showSelfStudyTree=true#

The Gas Laws
Charles’ Law
At a fixed pressure, the volume of a gas will
increase if the temperature increases, and
the volume will decrease if the temperature
decreases.
P
P
P
temp decreases temp increases

volume decreases volume increases

Charles’ Law
We say that temperature and volume are
directly proportional to each other.

V = kT , where k = a constant

V T V T

Volume goes up, Volume goes down,
temperature goes up temperature goes down

Charles’ Law
A graph of Volume vs Temperature would look
like the graph below.

The Kelvin scale was
developed based on this
graph. When the
volume of a gas is
reduced to zero, the
temperature was found
to be -273.15 oC, which
was defined as Absolute
Zero on the Kelvin scale.

Charles’ Law
Explanation:
As the temperature increases the particles increase
in kinetic energy, moving faster and faster. They
then hit the walls of the container with a greater
force, causing the walls to expand.

As the temperature decreases the particles
decrease in kinetic energy and hit the walls of the
container with less force. The external air pressure
causes the walls of the container to contract.


Gay-Lussac & Charles’ Law

At constant volume, the pressure of a gas
will increase with increasing temperature
and decrease with decreasing temperature.
P
P

P
temp decreases temp increases

pressure decreases pressure increases

We say that temperature and pressure
are directly proportional to each other.

P = kT , where k = a constant

P T P T

Pressure goes up, Pressure goes down,
temperature goes up temperature goes down

Explanation:
As temperature increases, the average kinetic
energy increases and therefore the speed
increases. The particles travel the same distance
in a shorter period of time, which increases the
number of collisions/sec, and the pressure goes up.

As temperature decreases the average kinetic
energy decreases therefore speed decreases.
Particles travel the same distance in longer periods
of time resulting in fewer collisions/sec and a
decrease in pressure.

Avogadro’s Law
Avogadro’s Law
At a fixed pressure and temperature, the volume of
a gas is directly proportional to the number of gas
particles (i.e., the number of moles) present.

A B

Gas “A” has twice as many particles as gas “B” so gas “A”
has twice the volume of gas “B,” when both gases are at
the same temperature and pressure.

Avogadro’s Law

Explanation:
If the temperature and pressure of two gases are
the same, then the particles in both gases are
moving at the same speed, and colliding with the
container walls with the same frequency.

If one container’s volume is greater than the other
container, then there must be more particles to
increase the frequency of collisions in order to
maintain an equal pressure with the smaller volume
container.

The Ideal Gas Equation

Boyle’s law: V α 1 (at constant n and T)
P
Charles’ law: V α T (at constant n and P) T in kelvins!

Avogadro’s law: V α n (at constant P and T)

nT nT
Vα or V = R x where R = constant of
P P proportionality

This equation is usually rearranged to:

PV = nRT R is the gas constant

Solving Gas Law Problems

To see how changing one variable affects the others, we look
at the ratio of the ideal gas equations “before” and “after” the
changes:

P1V1 nRT1
=
P2V2 nRT2

We then cancel out those variables (if any) being held
constant, enter the values for the other variables, and solve
for the desired variable’s value.
Note:
The temperature MUST be in Kelvins when using the gas
law equations!

Gas Law Problems
Example: A sample of chlorine gas occupies a volume of 946
mL at a pressure of 726 mm Hg. What is the pressure of the
gas (in mm Hg) if the volume is reduced at a constant
temperature to 154 mL?

P1V1 nRT1 We note that the temperature is held
i. = constant, and presumably, no gas enters
P2V2 nRT2 or escapes, so n is also constant. And, of
course, R is also constant.
P1V1
ii. = 1
P2V2

iii. P1V1 726 mm Hg x 946 ml
= P2 = = 4460 mm Hg
V2 154 ml

Gas Law Problems
Example: Argon is an inert gas used in lightbulbs
to retard the vaporization of the filament. A
certain lightbulb containing argon at 1.20 atm and
18 oC is heated to 85 oC at constant volume.
What is the final pressure of argon in the lightbulb
(in atm)?

P1V1 nRT1
= n, V and R are constant
P2V2 nRT2
P1 = 1.20 atm P2 = ?
P1 T1
= T1 = 291 C
18 o K T2 = 358 C
85 o K
P2 T2

T2
P2 = P1 x = 1.20 atm x 358 K = 1.48 atm
T1 291 K

Gas Law Problems
Example: A sample of carbon monoxide gas occupies 3.20 L
at 125 oC and 740 torr. At what temperature, in oC, will the
gas occupy a volume of 1.54 L if the pressure is increased to
800 torr?
P1V1 nRT1 We assume that no gas enters or escapes,
i. = so n is constant. And, of course, R is also
P2V2 nRT2 constant. All other variables are changing

P1V1 T1
ii. = Solve for T2. Watch your algebra!
P2V2 T2

T1P2V2 398.15oC x 800 torr x 1.54 L
125 K
iii. T2 = =
P1V1 740 torr x 3.20 L

iv. T2 = 207.1 K = − 66.0 oC

Experiments show that at 1.0 atm of pressure and a
temperature of 0oC (STP):

1 mole of an ideal gas occupies 22.414 L
We use this information to determine the value for R, the
ideal gas constant:

PV = nRT

PV (1 atm)(22.414L)
R= =
nT (1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Note that the value for R depends on the units used
for pressure, as follows:

Pressure units Gas Constant
atmosphere 0.08206 L·atm/mol·K
kilopascals 8.314 L·kPa/mol·K
torr (mm Hg) 62.36 L·torr/mol·K

Rather than memorize three different R-values, simply
always convert the pressure to atmospheres first, then
use 0.08206 L·atm/mol·K for all problems.


Example: What is the volume (in liters) occupied by 49.8 g of
HCl (g) at STP?

i. at STP, T = 0 0C = 273.15 K, and P = 1.00 atm

ii. n = 49.8 g x 1 mol HCl = 1.37 mol
36.45 g HCl
iii. PV = nRT
1.37 mol x 0.08206 L•atm x 273.15 K
iv. V = nRT = mol•K
P 1.00 atm

= 30.6 L

Density of Gases at various T and P
m
i. d =
V
ii. but V = nRT/P from the ideal gas equation…

m mxP m x P
iii. d = = =
nRT/P nRT n RT

But m/n = mass/mole which is the molecular weight (MW)
of the gas. We can write:

P
d = MW x
RT

Finding the Molecular Weight of an
Unknown Gas
We can easily determine the density of an unknown gas
experimentally.
More importantly, knowing the density, we can rearrange the
equation we just developed to solve for the molecular weight
of the unknown gas

Molar Mass (MW) of a Gaseous Substance

dRT
MW = d is the density of the gas in g/L
P

Molar Mass of Unknown Gas
Example: A 2.10-L vessel contains 4.65 g of a gas at 1.00
atm and 27.0 oC. What is the molar mass of the gas?

dRT
i. MW =
P
m 4.65 g g
ii. d = = = 2.21
V 2.10 L L

g
2.21 x 0.08206 L•atm x 300.15 K
L mol•K
iii. MW =
1 atm

MW = 54.4 g/mol

PV = nRT

Gas Stoichiometry

Avogadro’s Law
We can make use of Avogadro’s Law in solving gas
stoichiometry problems – we can measure volume
ratios to determine mole ratios, if all the gases have
the same temperature and pressure.

Avogadro’s Law
Example: Ammonia burns in oxygen to form nitric oxide
(NO) and water vapor. How many liters of O2 are required to
completely react with 2.4 liters of ammonia at the same
temperature and pressure?
rxn: 4NH3 + 5O2 4NO + 6H2O

Mole ratios: 4 mole NH3 5 mole O2

Recall: at constant T and P, n  V

 4 volumes NH3 5 volumes O2

 2.4 L NH3 x 5 L O2 = 3.0 L O2 required
4 L NH3

Gas Stoichiometry
We can also solve gas stoichiometry problems in
a way very similar to the stoichiometry problems
we have been doing – that is, convert to moles,
use the mole:mole ratio from the balanced
equation, etc.

The ideal gas equation is simply yet another
way of finding moles! Or converting moles to
pressure, or moles to volumes, or…

PV
n =
RT

Example: What volume of N2 gas at STP conditions is
required to react completely with 24.00 liters of H2 at 0.974
atm and 24.5 oC? Rxn: N2 + 3 H2 2 NH3

i. As always, convert to moles first, then use the mole:mole
ratio, then convert to the desired unit.

ii. nH2 = PV = 0.974 atm x 24.00 L = 0.957 mol H2
RT 0.08206 L atm x 297.65 K
mol K
1 mol N2
iii. From bal eqn: 0.957 mol H2 x = 0.3190 mol N2
3 mol H2
L atm
nRT
iv. V = = 0.3190 mol x 0.08206 mol K x 273.15K
P 1.00 atm
= 7.15 L

Gas Stoichiometry
Example: What is the volume of CO2 produced at 37 oC and
1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

1 mol C6H12O6 6 mol CO2
i. 5.60 g C6H12O6 x x = 0.187 mol CO2
180 g C6H12O6 1 mol C6H12O6

L•atm
0.187 mol x 0.08206 x 310.15 K
nRT mol•K
ii. V = = = 4.76 L
P 1.00 atm

Dalton’s Law of Partial Pressures
The total pressure of a gas in a container is equal to the sum
of the “partial pressures” of each component gas in the
container.

P1 P2 Ptotal = P1 + P2

+

V and T are constant

Collecting a Gas over Water
A common method of collecting a gas is to collect it “over
water” using a pneumatic trough.

The volume of the water displaced is equal to the volume of
the gas collected, and the temperature of the water is equal
to the temperature of the gas.

The gas collected will be
“contaminated,” however, by a
small amount of water vapor
since the gas was collected
over water. We use the Law
of Partial Pressures to find the
pressure of the “dry” gas.
2KClO3 (s) 2KCl (s) + 3O2 (g)

Collecting a Gas over Water continued
After collecting the gas, the bottle is raised or lowered until
the level of water in the collection bottle is even with the
water in the pneumatic trough.

At this point, the atmospheric pressure equals the total
pressure of the gas + water vapor in the collection bottle.

We now can find the pressure atm = gas + H2O

of the “dry” gas collected:

Pgas + Pwater = Ptotal = Patm

Collecting a Gas over Water continued

Example: A sample of KClO3 was decomposed in the lab to produce
182 ml of oxygen gas: 2 KClO3 2 KCl + 3 O2. The oxygen was
collected over water. Atmospheric pressure in the room at the time
was 731.0 torr. The temperature of the water (= temp of the gas) was
20.0oC. According to the table on page 196, the vapor pressure of
water at this temperature is 17.54 torr. How many moles of oxygen
were collected?
PV
i. n =
RT
ii. Pgas = Patm – Pwater = 731.0 – 17.54 = 713.46 torr

1 atm
iii. Pgas = 713.46 torr x = 0.9388 atm
760 torr
0.9388 atm x 0.182 L
iv. n = = 7.10 x 10-3 moles
L atm
0.08206 mol K x 293.15 K

Mole Fraction
The mole fraction (X) of a mixture is the ratio of the
number of moles of one component to the total number of
moles of particles in the mixture:
nA
mole fraction (XA ) = nA + nB + …

Consider a mixture of two gases, A and B. Then
(nA + nB) x RT
i. PT = multiply by XA gives:
V
(nA + nB) x RT nA nART
ii. PT = x = = PA
(nA + nB) V
V

 PA = XAPT

Mole Fraction

Example: A sample of natural gas contains 8.24 moles of
CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the
total pressure of the gases is 1.37 atm, what is the partial
pressure of propane (C3H8)?

0.116
i. Xpropane = = 0.0132
8.24 + 0.421 + 0.116
ii. Pi = Xi PT

iii. Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

Mole Fraction
Example: Recall the sample problem in which KClO3 was
decomposed in the lab to produce 182 ml of oxygen gas. The
oxygen was collected over water. Atmospheric pressure in
the room at the time was 731.0 torr. The temperature of the
water (= temp of the gas) was 20.0oC. According to the table
on page 196, the vapor pressure of water at this temperature
is 17.54 torr. What was the mole fraction of oxygen in the
collection bottle?

i. Pgas = Patm – Pwater = 731.0 – 17.54 = 713.46 torr

ii. Pgas = PT Xgas
Pgas 713.46 torr
iii. Xgas = = = 0.9760
PT 731.0 torr

More Gas
Mathematics
urms = √ 3RT
MW

Distribution of Molecular Speeds
James Maxwell (1831-1879) analyzed the proportion of
gas particles in a sample moving at a given speed and
obtained a graph similar to the one below, called a
Maxwell Distribution Curve.
The peak of each graph gives
the most probable velocity of
the particles – that is, the Speed Distribution for
speed of the largest number of Nitrogen at Different
Temperatures
particles.
Note that the most probable
speed increases with increasing
temperature, and as the
temperature increases, the
peaks shift and flatten out.

Root Mean Square Speed
It can be shown that the total kinetic energy of one mole of
gas is equal to ( 3 )RT
2

Since the average KE of one molecule is ½ mu2, we can
say that the total KE of a gas sample is given by:
i. KEtotal = 3 RT = N (½ mu2) ; m = mass of one molecule
2 A

ii. but NAm = # molecules x molecule = grams = MW
mole
grams
mole

iii.  3RT = ½ MW u2
2

iv. 
3RT
MW
= u2, OR urms = √ 3RT
MW

Root Mean Square Speed continued

The distribution of speeds of three different gases
at the same temperature

Note that the speed
is inversely
proportional to the
molar mass of the
molecule.

urms = √ 3RT
MW

Gas Diffusion and Effusion
Gas diffusion is the gradual mixing of molecules of
one gas with molecules of another by virtue of their
kinetic properties.

The rate at which a gas diffuses depends on its
velocity. Particles moving at a higher rate of speed
will diffuse faster than slow moving particles.

Recall that the velocity of a gas is inversely
proportional to it’s molar mass:

urms = √ 3RT
MW

Graham’s Law of Diffusion
If we compare the rate at which two gases diffuse
at the same temperature, we obtain the relation:

√ √
3RT 1
urms A =
MWA rateA MWA
=

√ √
3RT rateB 1
urms B =
MWB MWB

√
rateA MWB
This reduces to =
rateB MWA

Graham’s Law of Diffusion
Example: Compare the rate at which NH3 and HCl diffuse at
the same temperature.

√ √
rateHCl MWNH3 17 g/mol
= = = 0.687
rateNH3 MWHCl 36 g/mol

so HCl diffuses 0.687
times as fast as NH3… NH4Cl

put another way, NH3 NH3 HCl
diffuses 1/0.687, that is,
17 g/mol 36 g/mol
NH3 diffuses 1.46 times
faster than HCl.

Gas effusion is the is the process by which gas under
pressure escapes from one compartment of a container to
another by passing through a small opening.

As with diffusion, the rate of effusion is inversely proportional
to the molar mass of the gas.

√
rA MWB
=
r2 MWA

Example: Nickel forms a gaseous compound of the formula
Ni(CO)x. What is the value of x, that is, what is the actual
formula of the Ni compound, given that under the same
conditions, methane (CH4) effuses 3.26 times faster than the
compound?
Let A = CH4 and B = Ni(CO)x

√ √
rateA 3.26 MWB MWB
i. = = =
rateB 1 MWA 16.04 g/mol

ii. MWB = 3.262 x 16.04 = 170.5

iii. 170.5 = Ni + x(CO) = 58.70 + x(28.01)

iv. x = 170.5 – 58.70 = 3.99 ~ 4  formula is Ni(CO)4
28.01

Deviations from Ideal Behavior
Ideal gases have no attractive or repulsive forces
acting between molecules, and ideal gas molecules
have essentially zero volume.

Real world gases do have attractive and repulsive
forces and their molecules do occupy some space
(albeit only a small space).

Effects of Intermolecular Attractions
When intermolecular attractions come into play, the
particles begin to “stick” together. This effectively
reduces the total number of particles colliding with
the container walls, which lowers the pressure.

In addition, the attractive forces
reduces the speed of the particles
impacting the walls, which reduces
the force at impact, which again
lowers the pressure.

THUS, the measured pressure of a REAL gas will
always be slightly LESS than the ideal pressure that
gas should have.

A corrective factor must be made in the gas
equation for the pressure of a real world gas:
an2
Ptrue = Pmeasured +
V2

a is a constant that depends on the individual molecule.
The likelihood of forming attractions increases with
increasing particle density (moles/liter = n/V) which
explains the second term.

Effects of Particle Volume
Since a gas molecule does, in fact, occupy some
volume, the total volume available for any one
molecule will be LESS than the total volume of the
gas container.

measured volume true volume

At sufficiently high concentrations, and for sufficiently small
volumes, this can have an impact on the observed behavior
of the gas.

For real gases, the TRUE volume the gas occupies
is always slightly LESS than the container’s volume.

A corrective term must be subtracted from the
measured volume of an ideal gas:

Vtrue = Vmeasured – nb

n = # moles of the gas and b = a constant that is different
for each molecule (essentially, it gives the volume per
molecule)

Van der Waals Equation
The Dutch physicist J.D. Van der Waals
was the first to take into account the
effects of these attractive forces and
volume corrections. The “corrected”
form of the ideal gas equation is thus:

Van der Waals equation
nonideal gas
an2 (V – nb) = nRT
(P + V 2 )
corrected corrected
pressure volume

What good is the Ideal Gas Equation?

At sufficiently high temperatures (T ≥ 0 oC), the
particles have enough kinetic energy to overcome
the weak attractions that exist between gas phase
molecules, to the point that these attractions can
be ignored.

At sufficiently low pressures (< 5 atm) the particle
density is low enough that the particles are rarely
close enough for attractive forces to form, and so
this effect can also be ignored.

What good is the Ideal Gas Equation?

Thus, gases at temperatures at or above 0 oC and
with pressures below 5 atm do, in fact, behave as
ideal gases, that is, the corrective terms are
negligible (insignificant).

Most gases that we work with under ordinary lab
conditions do fall within these limitations, so most
gases behave ideally!

Energy Changes in Reactions

Energy is the capacity to do work.

To do work is to apply a force (F) to an object and
cause the object to move through some distance (d):

W = Fd

units for work (energy): Nm = Joule (J)

Energy

Other units for (thermal) energy:

calorie (c) = (heat) energy required to raise the
temperature of 1.00 gram of water by 1ºC.

1 calorie = 4.184 joules

Food Calorie (C) –capital “C” is not the same as a
little “c” calorie. There are 1000 calories in a food
Calorie (that is, a food Calorie is a kilocalorie.)

Energy
Energy is always conserved in any process. It is possible,
however, to convert energy from one form to another.
o Radiant energy comes from the sun and is earth’s
primary energy source
o Thermal energy is the energy associated with the random
motion of atoms and molecules ( a form of kinetic energy)
o Chemical energy is the energy stored within the bonds of
chemical substances
o Nuclear energy is the energy stored within the collection
of neutrons and protons in the atom
o Potential energy is the energy available by virtue of an
object’s position

Some Definitions
Thermochemistry is the study of heat change in
chemical reactions.
Heat is the transfer of thermal energy between
two bodies that are at different temperatures.
Temperature is a measure of the average kinetic
energy of the particles that make up a substance.
Note, however, that higher temperature does
NOT always mean greater thermal energy.
example: A 90ºC cup of coffee is at a higher temperature,
but a bathtub full of 40ºC water has more thermal energy,
since there are more particles in the bathtub of water!

Thermodynamics
“Thermodynamics is a funny subject. The first time
you go through the subject, you don’t understand it
at all. The second time you go through it, you think
you understand it, except for one or two small points.
The third time you go through it, you know you don’t
understand it, but by that time, you are so used to
the subject that it doesn’t bother you anymore.”

-Arnold Sommerfield (1868-1951) Well known German
physicist and textbook author; response as to why he had
never written a book on thermodynamics.

Thermodynamics

Thermodynamics is the scientific study of the
interconversion of heat and other kinds of energy.

In order to observe and study energy changes and
conversions, we must first define two terms: system
and surroundings.

The system is the specific part of the universe
that is of interest in the study.

The surroundings is everything else in the universe
outside the system.

Thermodynamics
There are three types of systems, based on
whether mass, energy, both or neither can be
exchanged with the surroundings.

Three
systems:

What can be open closed isolated
exchanged with mass & energy nothing
surroundings? energy only

Laws of Thermodynamics
There are 3 fundamental Laws of Thermodynamics.
We will study and use the first two laws in this
chapter:

1. The First Law of Thermodynamics deals with
conservation of heat energy
2. The Second Law of Thermodynamics is a
complex law dealing with ways that energy can be
distributed within a system and its surroundings.

First Law of Thermodynamics – (heat) energy
can be converted from one form to another, but
cannot be created or destroyed. This is just the law
of conservation of energy, applied to heat energy
conversions
Thus, any energy lost by the system, must be
transferred to the surroundings, and vice versa.
Mathematically:
ΔEsystem + ΔEsurroundings = 0

or ΔEsystem = –ΔEsurroundings


The Second Law of Thermodynamics deals with
energy transformations and the distribution of
energy within a system. There are many ways the
Law can be stated, depending on what aspect of
these energy distributions you want to emphasize.
For our current purposes, the most important form
of the second law states that heat energy always
flows from the object at a higher temperature, to
the object at the lower temperature.

* In college, a friend who was a physics major had to use a text titled, “An
Introduction to the Second Law” – which was about 1,000 pages in length!

The Second Law of Thermodynamics

The reason why heat must flow from hot objects to
cooler objects relies on the definition of temperature :

A fast moving particle can collide with a slow moving
particle. During the collision, the fast moving particle
transfers kinetic energy to the slower moving
particle.
As a result, the slower particle speeds up (and its
temperature increases) and by conservation of
energy, the faster particle must slow down ( so its
temperature decreases).

The internal energy of a system (E) is the total of the
kinetic plus potential energies of the particles making
up that system. There are two ways that this internal
energy can change:

1. Heat energy can be transferred to the system
from the surroundings, or heat can be transferred
from the system to the surroundings

2. Work can be done on the system by the
surroundings, or work can be done by the system
on the surroundings.


Mathematically, if
q = heat energy absorbed or given off
w = work done on or by the system.
Then the change in the internal energy (ΔE) of a
system is the sum of the heat energy changes
plus the work done on or by the system:

ΔEsystem = q + w

We assign (+/-) signs to q and w depending on whether
energy is leaving the system or entering the system.

Sign Conventions in Thermochemistry
 Energy is transferred INTO the system
We assign positive values to q and w when they represent
a transfer of energy into the system from the surroundings,
increasing the internal energy of the system.

 Energy is transferred OUT OF the system
We assign negative values to q and w when they represent
a transfer of energy out of the system to the surroundings,
decreasing the internal energy of the system.

Summary of Sign Conventions
in Thermochemistry

(energy leaves the system)
(energy enters the system)
(energy enters the system)
(energy leaves the system)

Esystem = q + w

where q = heat energy absorbed or given off, and
w = work done on or by the system.

Example: A beaker containing water is brought to a boil by
absorbing 44.0 kJ of heat energy, while at the same time,
the increased pressure of the water vapor as it expands
does 11.5 kJ of work on the air above the beaker. What is
ΔE of the system?

i. We define our system as the water in the beaker. All
else becomes the surroundings.

ii. The water absorbs heat energy, so q is (+) 44 kJ.

iii. The expanding water vapor does work on the
surroundings, so we assign w = – 11.5 kJ

iv. Thus, ΔE = q + w = (+44 kJ) + (–11.5 kJ) = +32.5 kJ

State Functions
It is important to recognize that there are literally an infinite
number of combinations of heat and work that can lead to the
same value for E for the system.

For example: We could bring about a change of 100 J in the
system by allowing it to absorb 125 J of heat energy, and
then do 25 J of work on the surroundings – or by absorbing
50 J of heat energy and having 50 J of work be done on it
by the surroundings, or…
We say that E of the system is a state function.

State functions are properties with a unique value, which
is determined only by the current state of the system,
regardless of how that condition was achieved.

State Functions
Example: Potential energy is a state function. It does not
matter what path is taken when you determine the change in
gravitational potential energy -- all that matters is the change
in altitude, h:

E = Efinal - Einitial

At the summit, the potential
energy of hiker 1 = hiker 2,
even though they took
different paths to get there.

Other examples of state ∆P = Pfinal − Pinitial
functions include pressure, ∆V = Vfinal − Vinitial
volume and temperature. ∆T = Tfinal − Tinitial

State Functions

Heat and work, on the other hand, are not state
functions. These two functions can only describe
how the system is changing, rather than a current
state or condition.

We cannot say a system “has” work or “has heat” –
only that the system is doing work, or exchanging
heat.

For example, if your car stalls and you need to push it off the
street, the path you follow in pushing the car will determine
the amount of work you do on the car (W = Fd). The longer
the distance you push the car, the more work you have to do.

w =Fd more work
using this path!
w = Fd

note that the net change in the car’s position is the same, so
we can conclude that displacement is a state function – but the
net work done in moving the car is different for each path --
thus, work is not a state function.

State Functions

Even though the different quantities of work and
heat applied or produced do depend on how the
internal energy change occurs, the change in the
internal energy itself, (the sum of q + w ), does not.

Work and heat are not state functions, but
their sum, q + w =  E, is a state function.

To measure the internal energy changes of the
system, we must be able to measure heat changes
and the work done on or by the system.

Measuring heat and work done on a system

We measure the heat exchanged
between the system and the
surroundings by measuring
temperature changes in the
system.

Measuring heat

An exothermic process is any
process that gives off heat –
i.e., thermal energy is transferred
to the surroundings from the
system. The temperature of the
surroundings increases

eg: burning methane in a Bunsen burner, and condensing
steam are both exothermic processes.

CH4 (g) + 2 O2 (g) CO2 (g) + 2H2O (g) + energy

H2O (g) H2O (l) + energy

Measuring heat

An endothermic process is any
process that absorbs heat – i.e.,
thermal energy is transferred to the
system from the surroundings. The
temperature of the surroundings
decreases
eg: decomposing HgO by heating and melting ice are both
endothermic processes.

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

Energy Changes in Reactions

We can depict the energy changes in a reaction using
a reaction coordinate diagram.

CH4 + 2 O2 E released 2 Hg + O2

Energy
Energy

CO2 + 2 H2O 2 HgO E absorbed

reactants products reactants products

An Exothermic Rxn An Endothermic Rxn

Measuring P-V Work
Recall: work = force applied through a distance, or:
W = Fd. This is called mechanical work.

One type of mechanical work is the expansion and
compression of a gas. We call this “P-V work”
because it depends on pressure and volume
changes. We can derive a mathematical expression
for this type of work as follows:

i. Pressure = area = F2
force
d
ii. Volume = d3
iii. and P x V = F2 x d3 = F x d = Work
d

P-V Work

We will only consider cases in which a gas expands or is
compressed under constant pressure conditions (if the
pressure is variable, you have to use differential calculus to
solve the problem...)

Under constant pressure conditions, the pressure of the gas
is in equilibrium with, and must be equal to the external
pressure (usually, this is atmospheric pressure).

Because it is easier to measure, you are typically given the
“external” pressure acting on the system in work problems.
But remember, the external pressure = pressure of the system
In order to do work, however, regardless of the pressure,
the volume of the gas must change…

P-V Work
If the gas expands, its change in volume, V, will be positive.
An expanding gas exerts a force on the surroundings and
therefore must do work against the external pressure; that is,
work is done ON the surroundings.
If the system does work on the surroundings, then energy is
leaving the system, and the internal energy of the system will
be reduced. We say that (+) work is being done on the
surroundings so negative work is done on the system.
since V >0 we write:
 w = –PV
A gas expanding (+V)
AGAINST the external
+ΔV
pressure, P, of the
surroundings

*we often say that work done BY the system is (-) work.

P-V Work
If the gas is compressed, its change in volume, V, will be
negative. And if the gas is compressed, the surroundings
must have done work on the system.

If the surroundings does work on the system, then energy is
entering the system and the internal energy must increase.
This means we assign the work done on the system as it is
compressed to be positive work.
Since V < 0, we write:
 w = –PV
Gas is compressed (−ΔV)
as the surroundings does
– ΔV work against the pressure
of the gas.

P-V Work

Thus, the formula for the P-V work done as a
gas expands or is compressed that reflects the
+/- sign convention is:

w = –PV

P = pressure exerted on the system
V = change in volume of the system

Calculations with w = -PV
The units for work are the Nm or joule. We need a
way to convert pressure x volume to joules.
It can be shown (see appendix 2) dimensionally, that
if we measure volume in liters and pressure in
kilopascals, the product unit (L-kPa) is equivalent to
the unit: Nm = joule.
Recall that there are 101.325 kPa in 1.00 atm. If a
L-kPa = J, then the atm kPa pressure conversion
is also a conversion to obtain work in joules:

PV = L-atm x 101.325 J = Joules
L-atm

P-V Work
example: An 18-L sample of a gas is compressed to 12-L
under a constant atmospheric pressure of 1.087 atm. What is
the work done on the system?

i. W = –PV = –1.087 atm x (12L – 18L) = + 6.522 L-atm

ii. + 6.522 L-atm x 101.325 J = + 661 J of work was
L-atm
done ON the gas.

=

w = –P∆V
initial final

Example: A sample of nitrogen gas expands in volume from
1.6 L to 5.4 L at constant temperature. What is the work done
on the system, in joules, if the gas expands (a) against a
vacuum and (b) against a constant pressure of 3.7 atm?
w = –P ∆ V
(a) ∆V = 5.4 L – 1.6 L = 3.8 L ; P = 0 atm
w = – (0 atm x 3.8 L) = 0 L•atm = 0 joules

(b) ∆V = 5.4 L – 1.6 L = 3.8 L ; P = 3.7 atm
w = – (3.7 atm x 3.8 L) = –14.06 L•atm
101.3 J = –1424 J = work done on the
w = -14.06 L•atm x
1L•atm
system (gas)
*we could also say that +1424 J of work was done on the surroundings.

Heat of Reaction
Although there are many ways in which work can be done
on or by the system (the mechanical work done in stirring a
mixture, for example), PV work is perhaps the most
common type, especially in an open container. If we focus
on just the PV work ad the heat energy changes, we obtain
a more useful equation with which we can measure the
change in the internal energy of a system (E) during any
process:
Esystem = q + w

 Esystem = q + (–PV)

Chemistry in Action: Making Snow
∆E = q + (−PΔV)
The compressed air + water vapor
expands so fast, that no heat is
exchanged with the surroundings:
 q = 0 and E = −PV
Since the compressed air + water
vapor expands (ΔV >0), it does
work on the surroundings,
lowering the internal energy of the
system. ∆E < 0

Since KE is part of the internal energy,
if E < 0, then KE <0 as well, which ∆T < 0 = SNOW!
means that the temperature drops!

Heat of Reaction

Most reactions are carried out under constant
pressure conditions (the beaker or test tube is open
to the atmosphere, etc.) However, reactions can
also be carried out under constant volume conditions
(within a sealed container).

It is relatively easy to measure heat energy changes
during a reaction. Solving the eqn ΔE = q + (−PΔV)
for q we obtain, what we shall see, is a much more
useful form:
q = ΔE + PΔV

Heat of Reaction

If a chemical reaction is run at a constant volume,
then V = 0 and no P-V work is done. This means
that:

q = E + PV 0

 qv = E the “v” subscript is used to remind us
that this is under constant volume
conditions.

Under constant volume conditions, the net change
in the internal energy of a system (ΔE) is equal to the
heat energy (q) absorbed or given off by the system.

Heat of Reaction
Under constant pressure conditions, we write:

qp = E + PV the “p” subscript reminds us that
this is at constant pressure

Since the majority of reactions are carried out under constant
pressure conditions (for example, in an open beaker), we use
this equation to define a new function whose change gives
the heat of reaction, qp .

This new thermodynamic function is called the enthalpy (H)
of the system:

H = E + PV

Enthalpy
For any process under constant pressure conditions,
the change in the enthalpy (H) gives the heat
energy absorbed or given off in the reaction:

H = E + PV

*Note that under constant pressure conditions, the change in
enthalpy, H, depends only on the change in internal energy,
E and/or the change in volume, V. These are both state
functions, which means that H is also a state function!

Comparing E and H
From the equation: H = E + PV, we can make the
following statement:

If V is zero (no work done; no expansion or compression,
etc.) or if V is at least very small, then:

H ≈ E

Thus, by measuring the enthalpy (heat energy) changes,
which is relatively easy to do, we are able to determine the
changes in the internal energy of the system (which would
otherwise be very difficult, if not impossible, to do)!

Enthalpy of Reaction
We define the enthalpy of reaction, Hrxn, as the
difference between the enthalpies of the products
and reactants:

∆ Hrxn = H (products) – H (reactants)

Hrxn can be (+) or (-), depending on the process.

If Hrxn is (+) the process is endothermic
If Hrxn is (-) the process is exothermic.

Enthalpy of Reaction
Thermochemical equations are equations that
show enthalpy changes as well as mass relationships.
example: 6.01 kJ are absorbed for every 1 mole of
ice that melts at 00C and 1 atm.
Note that since heat energy is absorbed, the
reaction is endothermic  ΔH > 0.

thermochemical equation:

H2O (s) H2O (l) ∆H = + 6.01 kJ/mol

Thermochemical Equations
example: 890.4 kJ are released for every 1 mole
of methane that is combusted at 25°C and 1 atm.

Since heat energy is given off in the reaction, the
reaction is exothermic and ΔH < 0.

thermochemical equation:

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)

∆Hrxn = –890.4 kJ/mol


In general, we often call the ΔH of a reaction the
“heat of reaction .” Specific types of reactions
are given their own specific names.

• ΔH for vaporizing a liquid is called the “heat of
vaporization”
• ΔH for a combustion reaction is called the
“heat of combustion.”
• ΔH for a neutralization reaction is called the
“heat of neutralization,” etc.

Writing Thermochemical Equations

• The physical states of all reactants and
products must be specified in thermochemical
equations.

H2O (s) H2O (l) ∆H = 6.01 kJ/mol

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)

∆H = -890.4 kJ/mol


• If you reverse the reaction, the sign of ∆ H also
reverses

N2O4 (g) 2 NO2 (g) ∆H = +58.04 kJ/mol

2 NO2 (g) N2O4 (g) ∆H = –58.04 kJ/mol

• If you multiply both sides of the equation by a
factor n, then ∆ H must change by that same
factor, n:
2 H2O (s) 2 H2O (l) ∆H = 6.01 x 2
= 12.0 kJ/mol

Note the units for ΔH are kJ/mol. The “per mole”
refers to the coefficients (moles) of a particular
species present in the equation as written. :

Hence, in the thermochemical equation:

2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 5 H2O (l)
∆H = −5757 kJ/mol

The –5757 kJ/mol means that 5757 kJ of heat
energy are released for every 2 moles of C4H10, or
for every 13 mole of O2 that reacts, and for every 8
moles of CO2 or for every 5 moles of H2O that form.


Often, thermochemical equations are written with
fractional coefficients so that the species of
interest in the reaction is unimolar


2 Al2O3 (s) 2 Al (s) + 3 O2 (g) H rxn = 3352 kJ/mol

 3
Al2O3 (s) Al (s) + 2 O2 (g) H rxn = 1676 kJ/mol

Note that the magnitude of ΔH changes whenever the
coefficients in the balanced equation changes.
Again, the “per mole” in kJ/mol refers to the molar
ratios of the equation as written.


We see that the amount of heat energy absorbed or
released is fixed for a given reaction – that is,  Hrxn
is stoichiometrically equivalent to the specific
number of moles of reactants and products for
that particular thermochemical equation.

Think of it this way:
Think of “heat” as being a “compound,” and the
magnitude of the heat energy as the “coefficient” for
that “compound.” Then treat the ΔH value as you
would any other species in the reaction when solving
stoichiometry problems.


As always, convert the starting material to moles,
then use the mole:mole ratios from the balanced
equation. Now, however, one of the conversion
factors is kJ of heat energy per mole(s) of reactant
or product:

mol:mol ratio H / mol Joules
Moles of Moles of
of Heat
Reactant Product
Energy

Example: How much heat is given off when 266 g of white
phosphorus (P4) burns in air, given that:

P4 (s) + 5O2 (g) P4O10 (s) ∆H = -3013 kJ/mol

We see that -3013 kJ of heat energy are given off for every
mole of P4 that reacts; thus we need to determine the number
of moles of P4, and then use the ratio: -3013 kJ released per
mole of P4 to determine how much heat is released:
1 mol P4
266 g P4 x x -3013 kJ = – 6470 kJ
123.9 g P4 1 mol P4


Example: Al2O3 undergoes thermal decomposition as follows:

 3
Al2O3 (s) Al (s) + 2 O2 (g) H rxn = 1676 kJ/mol

How many grams of O2 will be formed if 1000 kJ of heat is
absorbed by excess Al2O3?

1000 kJ x 3/2 mole O2 x 32.00 g
= 28.65 g of O2
+1676 kJ mol O2

*note that we can solve this problem using the fractional
coefficient 3/2. You do not have to convert all the coefficients
to integer values.

example: Ammonia is formed by reacting nitrogen and
hydrogen at elevated temperature and pressures:

N2 (g) + 3 H2 (g) 2 NH3 (g) ΔHrxn = -92.6 kJ/mol

What is the enthalpy change when 2.00L of H2 gas at 600K
and 4.00 atm are reacted with excess nitrogen gas?
i. nH = PV/RT = (4.00 atm x 2.00 L) / (0.08206 x 600 K)
2

ii. moles H2 = 0.1625 moles

–92.6 kJ
iii. ΔH = 0.1625 mole H2 x = – 5.02 kJ
3 mol H2
alternately, we could say that +5.02 kJ
of heat energy is released

AP Only
Measuring E of reactions with gases
We start with the equation for H, and solve for E:
H = E + PV E = H – PV

but PV = (PV) = (nRT) from ideal gas equation

and (nRT) = n (RT) at a fixed temperature

E = H – nRT if n = 0 then E = H

where n = moles product gases – moles reactant gases

note that nRT has units of L-atm when R = 0.08206 L-atm/mol K. Thus,
we must multiply this by 101.3 to convert to Joules. Note, too, that
0.08206 x 101.3 = 8.314 kPa-L/mol K. If we use 8.314 for R, then nRT
equals joules directly.

Measuring E of reactions with gases

If the reaction results in a net increase in the number of moles
of a gas, then some of the internal energy goes into doing
work on the surroundings as the system expands.

If the reaction results in a net decrease in
the number of moles of a gas, then the
system is compressed and work is done on
the system by the surroundings.

If there is no net change in the number
of moles of gas, then the system
neither expands nor is it compressed.
Thus, V = 0 and no work is done.

AP Only
Example: Find E for the reaction if 1.30 mol of Na react with excess
water at 298 K, given that:

2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) ∆H = -367.5 kJ/mol

i. E = ∆H - nRT -- we must find n and H:
1 mol H2
ii. 1.30 mol Na x = 0.650 mol H2 produced
2 mol Na
iii. n = mol product gas – mol reactant gas = (0.650 – 0) = 0.650 mol

-367.5 kJ
iv. H = x 0.650 mol H2 = - 238.9 kJ
1 mol H2
= 1610 J ~ 1.61 kJ
v.  E = (-238.9 kJ) - [ (0.650)(8.314 kPa-L/mol K) x 298 K ]

vi.  E = - 238.9 kJ – 1.61 kJ = - 240.5 kJ

note that H ≈ E

Calorimetry

measurement of heat energy changes

Heat Capacity and Specific Heat Capacity

There are three things that determine the amount of
heat energy that is lost or gained by an object:

1. The temperature change of the object

2. What the object is made of (its composition).

3. The quantity (mass) of matter present

The heat capacity (HC) of a substance is the
amount of heat (q) required to raise the temperature
of a substance by one degree Celsius.

q
HC = units for HC = J/oC
t

The heat capacity varies depending on what the sample is
made up of (metal vs wood, for example), and the mass of
sample used. Thus, it would be helpful if we had a term that
takes into account the nature and mass of the substance
being heated (or cooled).

The specific heat capacity (c) – usually shortened to
just “specific heat” -- is the amount of heat energy (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.

q = heat energy (joules)
q m = mass (grams)
c= t = temp change (ºC)
m x t
units for c = J/goC

Note that specific heat is an intensive property. Every
substance has a unique specific heat value.


Note that the specific heat of
metals tends to be quite small –
typically less than 1 J/goC.
Adding a small amount of heat
energy will cause a relatively large
increase in temperature – that is,
the metal gets HOT quickly.
Metals are good
conductors of heat.

On the other hand, metals
cool off quickly as well: a small
loss of heat energy = large
change in temperature.

The specific heat of water, on the
other hand, is by comparison very
large -- 4.184 J/goC. This means
that water can absorb a lot of heat
energy, yet its temperature will
not increase very much. It can
also lose a lot of heat energy,
without cooling down. Water is
an insulator.
Note that the energy unit we call
the calorie is actually another unit
for the specific heat of water:
1 calorie = heat energy/goC
1 calorie = 4.184 J

Lake Effect Weather and Specific Heat
Because of water’s high specific heat, large bodies of water like
Lake Michigan can absorb heat all summer without having any
significant increase in temperature.
In the summer, the hot air comes over Lake Michigan. Since
the lake is cooler than the air, heat energy is transferred from
the air to the water. This cools the air, but barely warms the
water at all. The cool air now blows onshore. The hot sand
then radiates away heat to the cooler air. Thus it is always
cooler near the beach in the summer.
cooler air
heat

cools land
heat hot land

Lake Effect Weather and Specific Heat
In the winter, the water has lost a lot of heat energy, but has
not cooled down that much – water stays warmer than the air
above it. When a cold air mass comes over the water, heat
energy is transferred from the water to the air, warming the
air.
Warm air can hold more moisture than cold air. The warm,
moist air now moves over the land. The land is colder than
the air and so the air transfers heat to the land. The cooler
air cannot hold its moisture, and so we get lake effect snow
storms! warm, heat
cold air
moist air

heat cold land

Solving Specific Heat Problems
To determine the amount of heat energy absorbed or
given off by a substance, start with the definition of
specific heat, and solve for q:

q = cmt where ∆t = tfinal - tinitial

Example: How much heat is given off when an 869 g iron
bar cools from 94oC to 5oC?

specific heat of Fe = 0.444 J/g • 0C

∆t = tfinal – tinitial = 5oC – 94oC = -89oC

q = c m∆t = 869 g x 0.444 J/g •oC x (–89oC) = -34,000 J

Calorimeters
Calorimeters are devices used to determine heat
energy changes for a given process or reaction.
There are two types:
1. Constant-volume calorimeter (also called a bomb
calorimeter) used to measure heat energy
changes when a substance undergoes
combustion.
2. Constant-pressure calorimeter (often nothing
more than two nested styrofoam coffee cups!).
This type is used for measuring heat changes for
reactions in solution.

The constant-volume (bomb) calorimeter
Bomb calorimeters are used to find heats of
combustion.
The bomb calorimeter
uses an electric current to
“burn” a substance of
known mass under about
30 atm of O2 gas. The
heat given off during the
combustion is transferred
to a water jacket. The
heat absorbed by the
water is measured using a
thermometer.

The Bomb Calorimeter

Bomb calorimeters have precisely known heat
capacities, so that very precise enthalpy changes
can be measured.
A bomb calorimeter holds a known, fixed amount
of water. As a result, the heat capacity of a bomb
calorimeter typically includes both the “bomb” and
the water in the jacket as part of the “calorimeter.”

HCcal = HCbomb + HCwater

The Constant-Volume Bomb Calorimeter
qsys = qcal + qrxn
assuming no heat enters or leaves the bomb calorimeter,
we can say that qsys = 0 !
heat given off in the reaction =
 - qrxn = + qcal
heat absorbed by the calorimeter
 - qrxn = +(qwater + qbomb) qcal = heat absorbed by the
bomb + heat absorbed by the
qwater = cm∆t water jacket…
▪ dimensionally, HC = J/oC so qbomb = HCbomb x t
▪ finally, note that Δt of the bomb = Δt of the water, so:

 − qrxn = qcal = [ HCcal Δt ] = + (cmΔtwater + HCΔtbomb )

The Bomb Calorimeter
Example: A 1.922 g sample of methanol (CH3OH) is
combusted in a bomb calorimeter which has a heat capacity
of 10.4 kJ/ºC. The temperature of the water rose by 4.20 ºC.
What is the molar heat of combustion of methanol?
convert HC to
i. −qrxn = + qcal joules/ºC !

ii. −qrxn = HC x Δt = 10,400 J/ºC x (+4.20 ºC) = 43,680 J

iii.  qrxn = −43.68 kJ for a 1.922 g sample of methanol.

32.04 g
iv.  q per mole = −43.68 kJ x = −728 kJ/mol
1.922 g mol CH3OH

see sample problem 6.6
on page 242

Constant Pressure Calorimetry
Reactions under constant pressure
conditions are much easier to work
with. We can use a simple “coffee-
cup” calorimeter to measure energy
changes for many non-combustion
reactions.
In this type of calorimeter, the heat
energy given off by a substance or a
reaction is absorbed by the water in
the calorimeter. We measure the
temperature changes of the water to
determine the heat given off. We
can usually assume no heat is
exchanged with the surroundings.

Constant Pressure Calorimetry

Determining the specific
heat of a metal

qsys = qwater + qcal + qmetal
qsys = 0

- qmetal = + (qwater + qcal)
hot metal

where qwater = [ cm∆t ] H O
2
water
and qcal = HCcal x ∆tcal

-qmetal
finally: c (metal) =
mtmetal
No heat enters or leaves!

Example: Determining sp heat of a metal (see 6.7 pg 243)
A lead pellet (m = 26.47 g) was heated to 89.98oC and then
placed in a coffee-cup calorimeter containing 100.00 g of
water at a temperature of 22.50oC. The final temperature of
the water was 23.17oC. What is the specific heat of Pb?
Assume HC of the calorimeter was 0 J

i. q lost by the lead pellet = + q gained by the water
watch your signs!!
ii.  - cmtmetal = +cmtwater
cmtwater qwater
iii.  cmetal = = 0.67oC = 2 sig fig
-mtmetal
= 4.184 J/goC x 100.00 g x (23.17 – 22.50oC)
-26.47g x (23.17 – 89.98oC)

= 280.3 J/1768.5 goC = 0.158 ~ 0.16 J/goC

Example: To determine the specific heat of titanium, a
student placed a 32.872 gram sample of Ti in a beaker of
boiling water for several minutes, and then transferred the
metal to a calorimeter. The calorimeter contained 98.86 ml
of water at an initial temperature of 20.10 oC . The final
temperature of the system was 23.04 oC. The heat capacity
of the calorimeter was determined to be 18.9 J/ oC . What was
the reported value for the specific heat of Ti?

i. First we note that to for the metal was 100.0oC = boiling pt
of water
ii. Next we note that the mass of water = 98.86 grams since
the density of water is 1.00 g/ml
iii. Since we are given HC for the calorimeter, we use the
“expanded” form: –qmetal = + (qwater + qcal)

Solution:

i. = - qmetal = qwater + qcal

ii.  -cmtmetal = +cmtwater + HCtcal

iii.  -c(32.872 g)(23.04 – 100.00oC) =
+4.184 J/goC x 98.86 g x (23.04 – 20.10oC)
+ 18.9 J/oC (23.04 – 20.10oC)

iv.  -c(-2529.8 goC)= 1216.07 + 55.57J

1216.07 + 55.57 J
v.  c = = 0.503 J/goC
-(-2529.8 goC)

Give a possible source of experimental error to explain why
the student’s value for the specific heat of Ti might be less
than the accepted value?
+qwater + HCtcal
c=
mtmetal

If sp heat is too small, then the water absorbed LESS heat
energy than was lost by the hot metal. Thus, some heat
energy was lost to the surroundings when the metal was
transferred to the calorimeter.
What would explain why the student’s value for the sp heat
of Ti might be greater than the accepted value?

This is the opposite case – some excess heat energy from
the surroundings was transferred to the water.

If the accepted value for the specific heat of Ti is 0.550 J/g oC,
determine the Δ% between the student’s value and the
accepted value.

% = experimental - accepted
accepted

0.503 -0.550 J/goC
% = = -8.5%
0.550 J/goC

since the percent difference was negative, the sp heat value
is too small and heat energy was lost to the surroundings
during the experiment.

Measuring Hrxn in aqueous solutions
We can also use a coffee-cup calorimeter to measure Hrxn for
reactions that take place in aqueous solutions.
Example: 100.0 ml of 0.500M NaOH was placed in a coffee-
cup calorimeter. To this, 100.0 ml of 0.500 M HCl was added.
The temperature of the solution rose from 22.50oC to a final
temperature of 25.86oC. Find Hrxn per mole of acid that
reacts.

i. we assume that sp heat of the aqueous sol’n = 4.184 J/g oC
ii. we assume that the density of the sol’n is ≈ 1.00 g/ml so
that the mass of the sol’n ≈ 100 + 100 = 200 g

iii. we assume V ≈ 0 for the sol’n so that H ≈ qrxn

Solution:

i. –qrxn = + qsol’n + q cal assume a perfect insulator
0

ii. –qrxn = cmtsoln

iii. – qrxn = 4.184 J/goC x 200.0g x (25.86 – 22.50oC)

iv. – qrxn = 2812 J ≈ 2.81 kJ

v. thus + qrxn = 2.81kJ

vii. moles of HCl = MAVA = 0.500M x 0.100L = 0.0500 mol

vii.  Hrxn = qrxn = -2.81 kJ/0.0500 mol = -56.2 kJ/mol of HCl

Hess’s Law
and
Standard
Enthalpy of
Formation
(Hf )
o

Hess’s Law of Heat Summation

You can determine Hrxn for any reaction that can be
written as the sum of two or more other reactions for which
thermodynamic data is known.
Recall that enthalpy is a state function, It doesn’t matter how
you get there, it only matters where you start and end!

Hess’s Law of Heat Summation states that, the enthalpy
change for an overall process or reaction is the sum of the
enthalpy changes of its individual steps.

Hrxn = H1 + H2 + H3 + …

Hess’s Law of Heat Summation

This means that, for any reaction that can be written
as a sum of 2 or more steps, we can find Hrxn if we
know the H values for each step; or we can find H
for any given step, if we know the enthalpy changes
for the other step(s) and the overall Hrxn.

Since nearly every reaction can be written as the sum
of two or more reactions, we can calculate the
enthalpy change for nearly any reaction we might
want to look at!

Hess’s Law
Consider the formation of carbon monoxide from its elements:

C (graphite) + ½ O2 (g) CO (g)

Although we can write this reaction on paper, in the real
world, you cannot synthesize CO in this way. But we can
still determine Hrxn by combining the enthalpies of reactions
that do occur in such a way that the net reaction is the
desired reaction given above:
1. The following two reactions will occur and go to completion:

C (graphite) + O2 (g) CO2 (g) Hrxn = −393.5 kJ/mol
CO (g) + ½ O2 (g) CO2 (g) Hrxn = −283.0 kJ/mol

I will have to give you the reaction steps – you
cannot do this on your own – yet!

2. If we reverse the second reaction and add it to the first
reaction, we obtain the desired reaction for the formation of
CO. Remember when you reverse a reaction, you must
reverse the sign of its enthalpy change!
½
C (graphite) + O2 (g) CO2 (g) Hrxn = −393.5 kJ/mol

CO2 (g) ½ O2 (g) + CO (g) Hrxn = +283.0 kJ/mol

C (graphite) + ½ O2 (g) CO (g) H rxn = −110.5 kJ/mol

Note that identical substances on opposite sides of the “ “ will cancel
out, just like identical numbers or variables cancel when they appear on
opposite sides of the “=“ sign in math problems.

Hess’s Law
Example: CO and NO are toxic gases found in car exhaust.
One way of reducing these emissions is to convert them to
less toxic gases. Determine the H for the reaction:

CO(g) + NO (g) CO2 (g) + ½ N2 (g) H = ??

Given the following thermodynamic data:
rxn A: CO (g) + ½ O2 (g) CO2 (g) HA = -283.0 kJ/mol
rxn B: N2 (g) + O2 (g) 2 NO (g) HB = + 180.6 kJ/mol

1. We will need to use rxn A as written. Rxn B must be
reversed to obtain N2 as the desired product.
2. Next, note that we only want ½ N2 as a product, so we
must multiply rxn B by ½ as well.

Hess’s Law

rxn A: CO (g) + ½ O2 (g) CO2 (g) HA = −283.0 kJ/ml

rxn B: ½ [2 NO (g) N2 (g) + O2 (g) ] HB = ½ (−180.6 kJ/mol)
reverse the sign of ΔH when you
reverse the reaction direction
net
rxn CO(g) + NO (g) CO2 (g) + ½ N2 (g)

net Hrxn :
ΔHA = − 283.0 kJ/mol
+ ΔHB = ½ (−180.6) kJ/mol

Hrxn = −373.3 kJ/mol

Chemistry in Action: Bombardier Beetle Defense
The bombardier beetle ejects a hot chemical spray of quinone
(C6H4O2). Given the following thermodynamic data, determine
Hºrxn for the reaction that produces quinone shown below:
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l)

1. C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g); ∆Hº = 177 kJ/mol

2. H2 (g) + ½ O2 (g) H2O (l); ∆Hº = -286 kJ/mol

3. H2O2 (aq) H2O (l) + ½O2 (g); ∆Hº = -94.6 kJ/mol

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l)

∆Hºrxn = ΔH1 + ΔH2 + ΔH3
= 177 kJ + (– 286 kJ) + (– 9.4 kJ)
= −204 kJ/mol

Standard Enthalpies of Formation

Recall that Hrxn = H (products) – H (reactants)
Unfortunately, there is no way to measure the
absolute enthalpy, H, of a substance. We can only
measure enthalpies relative to an arbitrary
reference point.
It is like measuring the height of a mountain – we
always arbitrarily measure it relative to sea level.

The reference point for measuring enthalpy changes
is called the standard enthalpy of formation (Hºf ).

Standard Enthalpies of Formation

The standard enthalpy of formation of any element
in its most stable form is zero.
∆Hfo (O2) = 0 ∆Hfo (C, graphite) = 0

but…∆Hfo (O3) = 142 kJ/mol ∆Hfo (C, diamond) = 1.90 kJ/mol

The standard enthalpy of formation (∆Hfo) for a
compound is the heat change that results when one
mole of that compound is formed from its elements at
a pressure of 1 atm.
There is a table of standard enthalpies of formation for many
inorganic compounds on page 247 in your textbook. A more
extensive list in given in Appendix 3 at the back of the text.

Standard Enthalpy of Reaction (Hºrxn )
The standard enthalpy of reaction (∆Hºrxn ) is the
enthalpy of a reaction in which reactants and
products are in their standard states at 1 atm.
aA + bB cC + dD

∆Hº = [ c∆Hºf (C) + d∆Hºf (D) ] - [ a∆Hfº (A) + b∆Hfº (B) ]
rxn

∆Hrxn = Σ n∆Hºf (products) - Σ m∆Hºf (reactants)
º

In order to use this equation, we must know Hºf values for each species.
If not already known, these must be calculated first.

Example: Benzene (C6H6) burns in air to produce carbon
dioxide and liquid water. How much heat is released per mole
of benzene combusted (i.e., determine Hºrxn) ? The standard
enthalpy of formation of benzene is 49.04 kJ/mol. See also
table 6.4.
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

∆Hrxn = Σ n∆Hºf (products) - Σ m∆Hfº (reactants)
º

0
∆Hº rxn = [ 12∆Hºf (CO2) + 6∆Hºf (H2O) ] - [ 2∆Hºf (C6H6) + 15Hºf (O2)]

∆Hrxn = [ 12(–393.5) + 6(–187.6) ] – [ 2(49.04) ] = -5946 kJ
º

this is for 2
-5946 kJ
 Hºrxn / mol = = - 2973 kJ/mol C6H6 mol C6H6
2 mol C6H6

Example: It is found, experimentally, that the production of
NH3 from its component elements releases 92.6 kJ,
according to the equation below. Calculate Hfº of NH3.
N2 (g) + 3 H 2 (g) 2 NH3 (g) Hrxnº = -92.6 kJ

Horxn = 2 Hfº(NH3, g) – [Hfº(N2, g) + 3 Hfº(H2 , g)]
But Hfº(N2, g) = 0 and Hfº(H2, g) = 0
Hºrxn = -92.6 kJ = 2 Hfº(NH3, g) – (0 + 0)
Hfº(NH3, g) = -92.6 kJ / 2 = -46.3 kJ/mol NH3

Hess’s Law of Heat Summation Revisited

To produce a table that listed the heats of formation for
every compound known would be impossible. Luckily, we
can use Hess’s Law to determine Hºf for virtually any
compound from even a limited list of enthalpy of formation
values.

Equipped with a means of finding Hºf values for virtually
any compound, we can also then calculate Hºrxn values for
virtually any reaction – even reactions that are not possible
to perform!

Example: Calculate Hºf of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) ∆Hºrxn = −393.5 kJ/mol
S(rhombic) + O2 (g) SO2 (g) ∆Hºrxn = −296.1 kJ/mol
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) ∆Hºrxn = −1072 kJ/mol

1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l) ΔHºf = ?
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g) CO2 (g) ∆Hrxn = −393.5 kJ/mol
º

2S(rhombic) + 2O2 (g) 2SO2 (g) ∆Hºrxn = −296.1x 2 kJ/mol
+ CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) ∆Hrxn = +1072 kJ/mol
º

C(graphite) + 2S(rhombic) CS2 (l) ΔHºf = ?
Hºf = Hºrxn = −393.5 + 2(−296.1) + 1072 = + 86.3 kJ

The enthalpy of solution (∆ Hsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
∆Hrxn = Hproducts − Hreactants

∆Hsoln = Hsoln − Hcomponents

Which substance(s) could be
used for melting ice?

Which substance(s) could be
used for a cold pack?

Dissolving ionic solutes
There are two enthalpy steps involved in dissolving a solid
ionic solute in water:

1. all the ions in the solute must be separated. The
energy for this step is called the lattice energy, U.
This is always an endothermic step.

2. the solute must be hydrated by the solvent water
molecules. The energy for this step is called the
enthalpy of hydration, ΔHhyd. This is always an
exothermic step.

The enthalpy of solution is the sum of these two steps

The Solution
Process for
NaCl

∆Hsoln = U + ΔHhyd = +788 – 784 = +4 kJ/mol

Chapter 7

Quantum Theory & the Electronic
Structure of Atoms

Electromagnetic Waves & Energy
Recall that around the turn of the century (20th
century that is), scientists were beginning to explore
the nature of the atom.
In a flurry of discoveries over a relatively brief time,
our understanding of matter changed from Dalton’s
“tiny billiard ball-atoms” to Thompson’s plum-
pudding model, and finally in 1919, Rutherford
showed that the atom was composed of a very tiny,
dense, positively charged “nucleus” around which a
swarm of negatively charged “electrons” were
located.


The initial assumption was that these electrons
followed circular orbits, not unlike the planets in orbit
around the sun.
The electrostatic forces between the (+) charged
nucleus and the (–)charged electrons -- rather than
gravity – was what kept the electrons in orbit.


Unfortunately, there was a major problem with this
“planetary” model.

According to the laws of physics, a charged particle
in orbital motion should radiate away energy. And
by conservation of energy, this loss of energy would
result in a decrease in the kinetic energy – and
hence, the speed of the electron.

As the electron slowed down, it would be pulled
ever closer to the nucleus.

A quick calculation showed that the time it would
take for an electron to radiate away sufficient energy
to spiral into and collide with the nucleus was less
than one microsecond.

uch!
O
Clearly, this does not
happen in the real world…


Strangely enough, the solution to this conundrum
came out of other scientist’s investigation into the
nature of light...We were on the verge of a
fundamental overhaul of all our macroscopic
assumptions about matter and light.


“We were about to discover that matter is not
as “solid” and “particle-like” as we would like
to think; and that light is not as “insubstantial”
and “wave-like” as we thought.

“On the atomic scale of things, the difference
between matter and electromagnetic waves
begins to blur rather alarmingly...”

-Michael Munowitz
Principles of Chemistry

Wave-Particle Duality
Light and matter -- on a sub-atomic scale – exhibits
both wave-like and particle-like behaviors, which we
call a wave-particle duality.
Because of this duality, we have come to refer to
light as a “wave-packet” – a little bundle of energy
that generally behaves as a wave, but can interact
with matter on a sub-atomic scale as if it were a
particle.

To better understand this duality, we need to review
just what it is that makes something a wave or a
particle. We will start with waves.

Properties of Waves
1. Waves are vibrations that transfer energy. These
vibrations can be in the form of the troughs and crests of
ocean waves, or high-low pressure changes in the air
which we call sound waves, or even oscillating electric and
magnetic fields. Waves are thus a series of of maximas
and minimas.

2. Waves travel through a medium by displacing the medium:
the medium either vibrates parallel with the passing waves,
called longitudinal waves, or the medium vibrates
perpendicular to the direction of the passing waves, which
is called a transverse wave.

Properties of Waves
A Transverse Wave

A direction of
wave
propagation



3. Wavelength,  (lambda) is the distance over which the
wave pattern repeats (the distance between two crests, for
example).

4. Amplitude (A). The distance between the “rest” position of
a wave and the point of maximum displacement .

Properties of Waves
5. How often the wave pattern repeats is called the frequency
of the wave, and is given the symbol, v (nu). The units for
frequency are waves per second, or simply sec-1, which is
called a hertz (Hz).

v = 4 waves/sec = 4 Hz v = 2 waves/sec = 2 Hz

note that higher frequency means shorter wavelengths!

Properties of Waves

6. How fast the wave moves forward (or propagates) is the
speed of the wave. Speed (u) has units of distance/time.
Looking at the units for frequency and wavelength, we find
that the speed of a wave is given by:

meters wave meters
u =  units:
wave x sec
=
sec

Note that the wavelength = u

and the frequency of a wave
are inversely proportional to
each other. = u


Properties of Waves

Example: Ultrasound waves are high frequency sound waves
with a wavelength of about 1.2 cm. If the speed of sound in
air is 340 m/s, what is the frequency of these sound waves?

u 340 m/s
= = = 2.83 x 104 Hz
 0.012 m

Example: What is the wavelength of a radio wave with a
frequency of 102 MHz? (u = 3 x 108 m/s.)

u 3 x 108 m/s
= = = 2.94 m
 1.02 x 108 Hz

Properties of Waves
7. The energy of a light wave, as we shall see later, is
directly proportional to its frequency.

8. The intensity of a wave is the power output (energy
delivered per second) of the wave per square meter of
the medium through which the wave is moving.

Wave Interactions
When two particles collide, they bounce off each other;
when two waves collide, they undergo interference.

constructive interference occurs when the two waves
meet in phase and their amplitudes add to yield a wave with
an amplitude of A+B at the point of interference. The energy
of the wave at that point is (A+B)2.

destructive interference occurs when the two waves meet
out of phase and their amplitudes subtract to yield a wave
with an amplitude of A – B at the point of interference. The
energy of the wave at that point is (A – B)2

Destructive Constructive
Interference Interference

before

during

after

Properties of Waves

Note that it is possible for two waves to interfere
destructively to the point that the net amplitude at the point
of interference is zero (said to be complete destructive
interference).

If the amplitude is zero, then the energy of the wave is
also zero at that point. Such a point is called a node.

On the other hand, if the amplitudes undergo complete
constructive interference, then the wave amplitude (and
energy) has a maximum at the point of interference, which is
called an anti-node.

Properties of Waves

Standing Waves
If a wave is limited to a particular path length, such as when a
guitar string is plucked, the waves undergo constructive and
destructive interference to produce what is called a standing
wave – it appears as if the wave is stationary:
nodes


A standing wave must be some integer multiple of
2

Properties of Waves
10. Waves come in trains – for a wave, the transfer of energy
is not a “one shot” affair, as it is with a particle; rather,
waves “pump” energy in a cumulative fashion, over time.
(For example, You can start a fire using the sun’s rays
and a magnifying glass to focus them, but it takes a few
minutes for the tinder to absorb enough heat energy to
reach its combustion temperature.)

11. Waves can undergo diffraction
– that is, they bend when they
encounter a barrier.

A diffraction pattern is built
up when the diffracted waves
interfere with each other.

Light As Electromagnetic Waves:
If light is a wave, then how does it travel through the vacuum
of space -- there is no medium to vibrate!

In 1873, James Clerk Maxwell proposed that light is a type of
wave which he called electromagnetic radiation (or EM
radiation) for short.
As we shall see, EM radiation is emitted by oscillating
charged particles (mainly electrons, but certain high energy
EM radiation is produced by the vibration of protons.)

Moving charged particles also produce a magnetic field.
Thus, EM radiation has both an electric and a magnetic
component.

Electromagnetic Waves
Maxwell showed that electromagnetic waves are traveling
magnetic and electric fields that oscillate at right angles to
each other. Since these fields vibrate perpendicular to both
each other and the direction of wave propagation, EM
waves are transverse waves.

EM radiation is thus its own medium – and only EM radiation
can travel through a vacuum.

Electromagnetic Waves
What we call “visible light” is only a tiny fraction of a much
broader range of waves called the electromagnetic
spectrum, which includes not only visible light waves, but
radio waves, microwaves, infrared, ultraviolet, x-rays, and
gamma rays.

The regions of the spectrum are defined somewhat loosely,
and depend on the wavelength and/or frequency of the
wave.

Every type of electromagnetic wave travels at the same
speed in a vacuum, namely, the “speed of light,” to which we
give the symbol, c. The speed of light is 2.997924 x 108 m/s
(which is rounded to 3.00 x 108 m/s for our purposes).

see page 271

The Electromagnetic Spectrum

Particle-Like Behaviors
Light also exhibits several distinctly particle-like behaviors.

One of the most obvious is the fact that light casts distinct
shadows, and the image of a window cast onto the floor as
sunlight passes through it has well-defined edges,
meaning light does not appear to diffract as it encounters
large barriers, nor as it passes through large openings.

The Photoelectric Effect
One of the most intriguing, and clearly particle-like
behaviors of light is something called the
photoelectric effect.

This is the process by which light, striking a
metal, causes the metal to emit an electron
(the emitted e– is called a photoelectron and
the process is called photoemission).

However, there are some strict limitations on
just when such photoemission can occur.

If light was, in fact, a wave, then even low energy light waves
should be able to slowly pump an electron up to a high
enough energy state that the electron is ionized, or removed
from the atom, just as light focused by a magnifying lens can
slowly increase the kinetic energy of the tinder to cause it to
burst into flame. However, this is not what happens...

Instead, if the frequency of light waves used is too low
(and hence, the wavelength is too long) no photoemission
occurs at all, regardless of the intensity of the light or how
long the light shines on the metal.

On the other hand, if light of the correct minimum
frequency is used, then photoemission occurs instantly,
even if low intensity light is used.


Furthermore, if light had the correct frequency to cause
photoemission, then it was found that increasing the intensity
of the light merely increased the number of electrons emitted,
but did not increase the kinetic energy of the emitted
electrons.

The photoelectric effect was explained by Albert Einstein.
He recognized that light was apparently transferring
momentum to the electrons, which is clearly a particle-like
behavior, but the amount of momentum somehow depended
on frequency, which is a wave-like property…
Einstein’s explanation relied heavily on work done by Max
Planck…

Quantized Energy Changes
About 1900, Planck was studying the radiation of heat energy
from high temperature objects, called “black-body radiation.”

He discovered that heat did not radiate in a continuous fashion
as “classical” physics predicted. Instead, Planck found that:

The energies of the vibrating molecules
that make up the heated substance were
quantized, meaning, the energy could
only change by small, discrete amounts of
energy. These allowed energy states
were called quantum states.

Max Planck


“Classical” physics said that an object’s energy
should change as it absorbs or emits heat energy in
a smooth, continuous fashion, like a ball rolling up or
down a hill. Instead, Planck showed that these
energy changes were more like a stair-step, that is,
they were quantized.

1 quantum

continuous changes quantized changes

The molecules of the heated object, in turn, could only emit
energy in discrete units of light energy called quanta (now
called photons ). They do so by “jumping” from one quantum
state to another. The minimum energy change between these
quantum states was found to be proportional to the frequency
of the emitted photon:

E = h

where ν = frequency (Hz) and
h = Planck’s constant = 6.626x 10-34 Js

Thus, a quantum = minimum energy that
an atom can absorb or emit.

The Photoelectric Effect Explained
Albert Einstein received his only
Nobel prize in Physics for
explaining the photoelectric
effect. Einstein applied Planck’s
idea that energy is quantized, but
said this quantization was also
true for electromagnetic energy,
not just molecular energies.

Einstein argued that light acted as a particle (i.e., a photon)
with an energy that depended on its frequency, according to
Planck’s idea: E = h. Einstein called the energy required to
remove an electron from the metal its work function, Φ,
(phi).
*your text calls Φ the “binding energy.”

To produce photoemission, a single photon strikes the
electron. If the photon’s energy is less than Φ, then the
electron cannot be “kicked loose” from the surface of the
metal (the photon has insufficient momentum).

If that one photon’s energy is equal to Φ, then the electron
would absorb that energy and photoemission occurs.

If the photon’s energy was greater than Φ, then the “excess”
energy imparted to that one electron would simply increase its
kinetic energy – a single photon would not be able to knock
two electrons loose.

If the intensity of the light used is increased, then the number
of photons striking the metal is increased, but the energy of
each photon is still hv, so each photoelectron would have the
same kinetic energy.

Putting it all together, Einstein said the kinetic energy of an
emitted photon was equal to the photon’s energy minus the
work function required to emit the electron:

Kinetic Energy of a Photoelectron:

KE = hv – Φ

The unit for energy is the joule. The energy for a typical
photon of visible light is only about 10-17 J.
This is such a small number that scientists often use a unit
called the electron volt (eV) instead of the joule for these
small energy changes.

The electron volt is the energy an electron gains being
accelerated across a 1-Volt potential. It’s relation to the joule
is given by:

1 eV = 1.602 X 10-19 J

Example: When cesium metal is illuminated with light of
wavelength 300 nm, the photoelectrons emitted have a
maximum kinetic energy of 2.23 eV. What is the work function
for Cs?

i. KE = hv - Φ

ii.  Φ = hv - KE

1.602 x 10-19 J
iii. KE = 2.23 eV x = 3.57 x 10-19 J
1 eV
c 3.0 X 108 m/s
iv. v = = = 1.00 x 1015 Hz
 3.00 x 10-7 meter

v.  Φ = (6.626 x10-34 Js x 1.00 x 1015 Hz) – 3.57 x 10-19 J

= 3.06 x 10-19 J or 1.91 eV

Emission Spectra and
Bohr’s Model of the
Hydrogen Atom

Line Emission Spectra
It was found that when a substance in the gas phase was
energized (by heating or passing a current through it) the
substance gave off a unique spectrum of EM radiation.

But rather than being a continuous spread of energies or
wavelengths, these spectra were composed of individual
wavelengths, or lines, when dispersed by a prism and cast
upon a screen. Emitted light

Line emission
spectrum for
Hydrogen

Line Absorption Spectra
It was also found that when all wavelengths of light (given off
by a so-called “black body radiator” such as an incandescent
lightbulb, or the sun) were passed through a gas, certain
wavelengths were absorbed, and the rest passed through.

gas

black-body prism
radiator
emits all  black lines = those 
that were absorbed

Line Spectra

Each element produces a unique emission or absorption
spectrum. In fact, He was first discovered by examining the
absorption spectrum of light coming from the sun!
Furthermore, the line emission and line absorption spectra
for a given element were “negative copies” of each other.

emission spectrum

absorption spectrum

Niels Bohr
In 1913, Niels Bohr, a Danish physicist, proposed an
explanation for these line spectra, and in so doing, presented
a new model for the atom.

The Bohr model of the atom worked
well in explaining and predicting the
spectrum of hydrogen and other
one-electron atoms, such as He+,
but his model could not explain the
behavior of multi-electron atoms.
Still, his model served as an
important foundation for our current
understanding of the atom.

The Bohr Model of the Atom
1. Atoms consist of a positively charged nucleus around which
the electrons orbit like planets orbiting the sun.

2. The potential energy of the electron depends on the radius
of the orbit in which an electron is found. The greater the
radius, the greater the potential energy of the electron.

3. There are only certain allowed energy states that an
electron can have (and hence, only certain allowed orbits).
This is due to Planck’s restriction that energy can only
change in small discrete (or quantum) steps.

*This same “quantum restriction” also prevents electrons from
spiraling into the nucleus, as classical physics would require
an orbiting electron to do !

1. The atom has only certain allowable energy
levels or “shells.” Each shell has a fixed radius.

2. The energy of the shell depends on its distance
from the nucleus – larger shells have higher
energy.

The lowest energy shell is called the “ground state.”
Higher energy shells are said to be “excited states.”


1 2 3 4 5 6 increasing
energy

first shell is the
ground state (n = 1)

(Shells are designated with the letter “n”)

Energy vs Shell Size in the Bohr Model
Note that, although the energy of the shell depends on the
radius, it does not increase in direct proportion to the
increased shell size. Rather, the energy states get closer
and closer together as the shell size increases.

n=6
6 5 4 3 2 1 n=5
n=4
n=3 shell
number
n=2
Energy

n=1

3. An electron moves to a higher energy shell by
absorbing energy.
energy
4. An electron moves to a lower energy
energy shell by giving off
energy in the form of light.

An electron can move
from one shell to
another, but it cannot
occupy the space
between shells.


6. This means that the energy absorbed or given off
must be exactly equal to the difference in energy
(ΔE) between the initial and final shells.
+ΔE

E E
Photon given off
-ΔE
absorbed

The Bohr Model
Now, recall:
The energy of a photon depends on its frequency:
E = h. The frequency, in turn, determines the
wavelength: λ = c/ .
This means that the electron can absorb or emit only
those photons which have the correct energy (i.e.,
specific wavelength and frequency) which exactly
corresponds to the energy difference between two
shells.
This explains the line emission and absorption
spectrum of an atom!
http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/pem1s3_1.swf

The Bohr Model

In the case of an absorption spectrum, only those
photons of light with an energy exactly equal to an
allowed transition from one shell to a higher energy
shell are absorbed by the atom’s electrons. The
other photons are not absorbed and pass through.
As a result, most of the photons pass through the
gas sample, but those that are absorbed are now
“missing” and leave a black band in the spectrum.
This produces the observed emission line.

The Bohr Model

ΔE3-1
ΔE3-1
ΔE3-2
3
2
1

E3-2

In the absorption spectrum, only certain wavelengths
have the correct energy to be absorbed.

The Bohr Model

In the case of an emission spectrum, the applied
voltage excites all the electrons in an atom to the
highest possible shell.
The electrons then relax to lower shells. Any drop
in shells is a drop in energy, so the electrons can
drop in multiple steps by moving from shell to shell
to shell until they arrive at the ground state, or they
can drop to the ground state in one step.
Each transition releases a photon with an energy
exactly equal to the ΔE between shells. This
produces the observed emission line.

The Bohr Model

4 E3-4
3 E1-3
2 E1-4
1

Excited electrons relax to lower shells and give off photons
with an energy equal to E between the shells. This creates
the emission line spectrum.

The Bohr Model
Bohr’s model was able to predict the energy and wavelengths
of the emission lines for hydrogen.

These emission lines
were named by their
discoverers. Each
nf = 4
grouping of lines fell in a
different region of the EM nf = 3
spectrum
nf = 2
The Balmer series, in
visible
which e- relax from region
excited states to the
second shell (nf = 2) is the
only group with lines in
the visible region. nf = 1

Mathematical Description of Bohr’s Model

The energy of an electron in a specific shell, n is given by:
n (principal quantum number) = 1,2,3,…
2
En = −RH Z RH (Rydberg constant) = 2.18 x 10-18J
n2
Z = nuclear charge (atomic number)

Note the (−) sign in the equation. We define the reference as
being a “free electron” no longer attached to the atom ( n = ∞),
An electron with n < ∞ is at a lower potential state, and hence
E is negative.

This equation only holds for a one-electron atom.
For this reason, the Z term often seems to be “left out” (as it
is in your text), since the only neutral atom with one electron
is hydrogen, for which Z = 1. Bohr’s equation for the energy
of an electron in hydrogen is then:

En = −RH Z2 12 1
= −RH 2 = −RH 
n2 n n2

So the Z term is there – it just happens to be “1.”
However, this equation also holds for He+ or Li2+ ions, (Z= 2
and Z=3, respectively) since these, too, are one-electron
atoms, and you must include the Z2 term for the energy of
the electron in these ions!

Mathematical Description of Bohr’s Model
The energy of a photon absorbed or emitted during an
allowed electron transition between two shells, nfinal and ninitial
is given by:
–(–) = +
ΔE = Ef – Ei

1 1
= −Z2RH
n2
− −Z2RH 2
n = +
Z2RH 12 − Z2RH 1
f i n n2
i f

1 1
 ΔE = h = Z RH
+ 2 −
n2 n2
i f

Example: Calculate the wavelength (in nm) of a photon
emitted by a He+ atom when its electron drops from the n = 5
state to the n = 3 state.

12 1
Ephoton = ΔE = Z RH( 2 )
n n2
i f

Ephoton = (2)2 x 2.18 x 10-18 J x (1/25 - 1/9)

Ephoton = ΔE = −6.20 x 10-19 J note that ΔE is negative since e- is
relaxing to a lower energy shell…
Ephoton = h x c / λ
but always use a +ΔE when determining the
λ = h x c / Ephoton wavelength using this equation…
λ = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/ (+6.20 x 10-19J)
λ = 3.208 x 10-7 m or ~ 321 nm

Louis deBroglie
We have seen that photons of electromagnetic radiation can
posses both distinctly wave-like properties and particle-like
properties.
In 1924, the French physicist Louis deBroglie made the
rather daring proposal (as part of his doctoral thesis) that
matter could also exhibit this wave-particle duality.

If one made this rather large leap of
faith assumption, then a lot of the
paradoxes and problems with the
nature of the electrons in atoms began
to make some sense…

The Wave Nature of an Electron
We saw that waves can undergo destructive and constructive
interference in such as way as to produce standing waves. In
standing waves, an integral number of half-waves can “fit” in
a given length, L, such that: L = ½ n λ , where n = 1,2,3,…

n=1 n=2 n=3

What would happen if this standing wave were to form a
circle, such that L = circumference of the circle?


For such a circular standing wave to persist, a whole number
of wavelengths would have to fit into the circumference, such
that:
circumference = 2πr = nλ, where n = 1,2,3,4, etc.
That is, there are only certain “allowed” circumferences for
a given wavelength – or put another way, there are only
certain allowed wavelengths for a given circumference.

Not
Allowed
allowed

Furthermore, the circumference can only change in quantized
steps of nλ. This is exactly what Bohr’s model required!

Standing Waves
In this simplified version of de Broglie’s theory of the atom, the
waves are shown only in circular paths around the nucleus. In an
actual atom, the standing waves make up spherical and ellipsoidal
shells rather than flat, circular ones.


Properties of electron matter-waves
1. There are only certain allowed orbital radii for the
electron to occupy (limited by the requirements that the
circumference be an integer multiple of the wavelength of
the electron’s matter-wave.)
2. The energy of the electron is quantized, (which follows
from the fact that the circumference can only change in
integer multiples of λ, and E = hc/λ. )
3. Being a standing wave, the electron is actually not
moving – it is in a stationary state – and since it is not
moving, it will not radiate energy and thus will not spiral
into the nucleus, as would happen otherwise according to
the laws of classical physics.

DeBroglie was able to derive an expression for the
wavelength of such an electron “matter-wave”:

=
h
mv

In physics, the quantity mv is called momentum and is
usually given the symbol, p. Hence, deBroglie’s expression
is often given as:
=
h
p

The wavelength of an electron matter-wave is thus inversely
proportional to its momentum; the constant of proportionality
is simply Planck’s constant, h.

Louis deBroglie
So why haven’t we ever seen evidence of this wave-like
characteristic of macroscopic matter?

Well...consider, for example, the matter-wave of a 0.17 kg
baseball after it has been thrown at 30 m/s (~70 mph)...
From deBroglie’s matter-wave equation, we have:

h 6 .6 2 6 × 1 0 − 3 4 J sec
λ = = = 1 .3 × 1 0 -3 4 m
mv 0 .1 7 k g × 3 0 m / s

This is an exceedingly small wavelength! For comparison,
this is 1/10,000,000,000,000,000,000 th (that’s 19 zeroes!)
the radius of a proton. Thus, the wavelength for ordinary,
macroscopic objects is so small that we cannot even
measure it, more or less notice it!

Louis deBroglie
On the other hand, consider the wavelength of an electron’s
matter-wave :

Example: The mass of an electron is 9.11 × 10– 31 kg. A
typical value for the velocity of an electron about the nucleus
of an atom is 1.0 × 106 m/s. What would be the wavelength
of the electron?
−34
h 6 .6 2 6 × 1 0 J sec −10
λ = = −31 6
= 7 .3 × 1 0 m
m v 9 .1 1 × 1 0 k g × 1 .0 × 1 0 m / s

This wavelength is about the same as the diameter of a an
atom of oxygen, i.e., the wavelength of the electron is about
the same size as an atom itself – still a small number, but at
least it is measurable!

Werner Heisenberg and the Uncertainty Principle
Just as things started to go well, a new problem arose…

To locate something, we need to “bounce” something off the
object, e.g., airport air traffic control towers “bounce” radar
beams (radio waves) off of airplanes to locate and guide
them.
But note that when any two objects collide, a transfer of
momentum always occurs.
One object ends up “losing” some momentum, and the other
object “gains” an equal amount of momentum (thus the total
amount of momentum does not change – momentum, like
energy, must be conserved).

Heisenberg’s Uncertainty Principle
Ideally, the object we are trying to locate should not be
affected by the particle we are “bouncing” off it.
It would not make sense to locate a moving airplane by
throwing, for example, a dump truck at it! The dump truck
would transfer sufficient momentum to the plane to
dramatically alter the plane’s momentum (hence its position,
as well)...
So we locate the plane using something which cannot
transfer a significant amount of momentum to it – say, a
photon of electromagnetic radiation in the radio region of the
spectrum – i.e., radar!


In 1927, Werner Heisenberg realized that we
run into a real problem when we are trying to
locate very tiny objects, like an electron.
It seems that the momentum of something
as insignificant as a photon of light is
sufficient to alter the position of the electron
(as is evidenced in the photoelectric effect,
among other things)!

The very act of trying to find the electron would move it
as the photon transferred momentum to the electron!


To avoid this problem we must use very low momentum
photons. But, as deBroglie showed us, to reduce the
momentum, we must increase the wavelength of the photons
used, since p =h/λ
Unfortunately, this presents yet another problem:

To locate an object precisely, the wavelength of light used
must be smaller than the object being located. Not a
problem for finding an airplane, but to locate a tiny electron
means a very short wavelength is needed.
But a shorter wavelength means we have greater momentum.

There is a trade-off here: we cannot have both low momentum
and short wavelength.

Heisenberg reasoned that, if we let the minimum uncertainty
in the position, Δx, of the electron correspond to the
wavelength of light (λ) used to locate it, and Δp be the
minimum uncertainty in the momentum of the photon that
would be transferred to the electron, then:
Since p = h /λ, according to deBroglie’s equation, the
combined minimum uncertainty is given by:

xp > ( ) x (h/ ) or xp > h*

This is called Heisenberg’s Uncertainty Principle. This
means that we cannot simultaneously determine both the
momentum and position of an electron with a high degree of
precision.
*actually, Heisenberg’s equation is xp > h/4π


So...how bad is this, really?
Suppose the speed of an electron in an atom is on the order
of 106 m/s. It’s momentum would be:
mv = (9.11 x 10– 31 kg)(106 m/s) = 9.11 x 10– 25 kgm/s
The uncertainty in the position of that electron can be
estimated using Heisenberg’s equation: ΔxΔp ≥ h/4π
Δx ≥ h / 4πΔp
x ≥ 6.626 x 10– 34 Js / 1.114 x 10– 23 kgm/s = 5.79 x 10– 11 m.

This is ~ 1½ times larger than the diameter of a hydrogen
atom! Wow! The uncertainty in the position of the electron in
a hydrogen atom is larger than the whole atom! This is not
good news...

To use the Bohr model of the atom to predict the allowed
energy states of the electrons we must know both the position
and momentum of the electron. Heisenberg just showed us
we cannot do this! How do we get around this problem? We
need to find a way of understanding the energy of the electron
without having to know its position.

The solution came when scientists took a closer look at what
a matter-wave meant.

The Development of the
Quantum Mechanical
(or Wave Mechanical)
Model of the Atom

The Quantum Wave Mechanical Model
In 1926, Erwin Schrödinger developed a model of
the atom, called the Quantum Wave Mechanical
Model of the Atom, by plotting the
position and energy of an electron
based on its wave-like properties.
Instead of plotting the electron’s
position to find its energy, he plotted
the wave’s amplitude (intensity)
to find its position.

Erwin Schrödinger

Schrödinger’s Wave Equation (ψ)
The intensity of a light wave is proportional to the square of
its amplitude: I = kA2

High intensity means a brighter light. Think of the
interference pattern when light is passed through two slits:
we get bright bands (constructive interference) separated by
dark bands (destructive interference).

The brightest band is always the
central one because that’s where
the greatest degree of constructive
interference occurs.
From a particle viewpoint, we could
say that this is where the greatest
concentration of photons occurs.


This means we can relate the amplitude2 (intensity)
with the particle concentration. We could also say
there is a greater likelihood, if you will, of finding a
photon at the point of greatest intensity.

Schrödinger reasoned that if an electron behaved as
a matter-wave, we can develop an equation that
describes the energy of the electron in terms of its
amplitude.

This equation is called a wave function symbolized
as ψ (psi).

Think of ψ as a “pointer” that gives us the amplitude of a
wave at any given point in space, such that ψ(x,y,z) gives
the amplitude at the 3-dimensional Cartesian coordinates,
x,y, and z. z

re-



y


x

Since ψ is a function of the amplitude of the matter-wave, then
ψ2 is a measure of the intensity of the matter-wave, that is, a
measure of the probability of finding an electron at that point.

Quantum Wave Mechanical Model

As an analogy: Imagine a beehive with only one
bee. If you take a time exposure photo of the one
bee as it buzzes around, you might get an image
like this:

If I ask you where the bee is right now – you could
not say for certain. But you could be about 90%
certain that the bee was most likely somewhere
inside the circle shown around the bee hive…

When the probability (using ψ2 ) of finding an electron
vs distance from the nucleus is plotted , we obtain
“shells” just like Bohr’s model.


The shells can be
sub-divided into
regions of highest
probability called
orbitals.

All of the orbitals in all of the
shells combined is called the
electron cloud.

Finding an e- in an atom is a little like guessing where a
person might be in a house at any given time.
The electron cloud is the broadest, or most general
probability region of the atom. This is like saying that at
sometime during the day, we know the person will be at
home.
The shell is a more specific probability region. This is like
asking, “Which floor of the house is the person most likely
on, between 3:00 and 5:00 pm?

The orbital is a very specific probability region. This is like
asking, “Which room of the house is the person most l likely
in, at exactly 1 am?” Note that the person might be in the
kitchen getting a late night snack, but the most likely location
of the person at that time is in bed.


Schrödinger’s actual equation was:

∂ 2ψ ∂ 2ψ ∂ 2ψ 8π 2 m
+ + + (E − V )ψ = 0
∂ x ∂ y ∂ z h2

When Schrödinger’s equation is solved, it turns out that there
are 3 terms in the solution. We call these terms “quantum
numbers,” and we use these numbers to more fully describe
the electron’s energy state. (A 4th quantum number, unrelated
to Schrödinger’s numbers, popped up in later studies.)
Keep in mind as we discuss these quantum numbers that they
are mathematical solutions to a “probability density plot,” and
have no physical reality to them.


Thus, the atom is no longer viewed as a nucleus
with orbiting particles called electrons. We now
understand the atom as being a nucleus surrounded
by a diffuse “cloud” of electrons in which the
electrons act more as a “probability wave” than as a
particle.

Electron Configurations

The work of Louis DeBroglie and Erwin Schrödinger in the
1920's showed that the electron could be treated as a wave,
specifically a standing wave, rather than as a particle. The
electron wave must have a whole number of wavelengths
“encircling” the nucleus, or else destructive interference could
occur, which effectively destroys the electron-wave...and thus
the electron!

To increase the number of waves by whole wavelengths
means that there are only certain allowed radial distances the
electron wave can be from the nucleus and have the waves
“fit” without interference --- these are the same restrictions as
those arbitrarily imposed by Bohr on his model to create the
shells. b

Thus, the quantum model, like the Bohr model, also
has shells. However, these “shells” are not of exact
dimensions, as was the case with the Bohr model –
they are merely “probability regions.”

We can describe the matter-wave by its wave
function (ψ) , which is a three dimensional pattern
of “ups” and “downs” and nodes analogous to the
standing wave on a vibrating string. It is a function of
the amplitude of the “electron wave,” and thus is
partly determined by the energy of the electron.

The square of the wave function, ψ2 , gives the probability
distribution mapping of the electron in space (as an analogy,
the square of the amplitude of a light wave equals the
intensity of that wave, which essentially is where there is the
greatest density of photons).
The “point by point tabulation” of the most probable location
of an electron of a specified energy is called an orbital. Note
that the orbital is simply a “mapping” of the most likely
position for the electron – like mapping the most likely
location of a bee in a time lapsed photo of a bee-hive. An
orbital is a mathematical construct, and has no physical
reality to it.

To describe the location of any given point, we need
three dimensions. Using polar coordinates, these
dimensions are radial distance, r, longitudinal angle,
Φ, and latitudinal angle, θ (see drawing below).

There are three quantum numbers,
which are essentially the solutions for z

the three dimensions, r, θ, and Φ for r
θ
an electron of a specified energy 

state: y

Φ

x

The Quantum Numbers
1. Principal quantum number (n) determines the
radial value, r, for an electron. The potential
energy of the e– increases with increasing
distance, r, from the (+) nucleus. The “direction”
of r (north, south, east, etc.) is immaterial – the
potential energy depends only on how far away the
electron is from the nucleus.

Thus, the principle quantum number gives the
shell number for the electron, and has values
n = 1,2,3,4... This is by far the most important
factor in determining the energy of the electron.

The Quantum Numbers

2. Angular (azimuthal) quantum number ( l )
determines the longitudinal angle, θ. This angle
depends on the electron’s angular momentum.

Angular momentum is a bit abstract – you are most
familiar with its conservation in rotating systems: it is
conservation of angular momentum that prevents
gyroscopes and bicycles from tipping over when the
gyroscope or tires are rotating fast, but when the
gyroscope or bicycle tires are not rotating, there is
no angular momentum and they fall over.

The Quantum Numbers

The values for l depend on the values for the
principle quantum number, n, as follows:

l = 0…(n -1)
Thus
 when n = 1 l = 0…(1-1) = 0, that is, when n = 1, l = 0
 when n = 2, l = 0, … (2-1) = 1, so when n = 1,
l has two possible values: 0 and 1
 when n = 3, l = 0, 1, … (3-1) = 2, so when n = 1,
l has three possible values: 0, 1 and 2

…etc.

The Quantum Numbers

The angular quantum number defines the subshell of n,
and determines the SHAPE of the orbital.

s-subshell is made up of s-orbitals which are
spherical shape

p-subshell is made up of p-orbitals which are
"peanut" shaped

d-subshell is made up of d-orbitals which are
"double peanut" or "donut peanut" shaped

f-subshell is made up of f-orbitals which are
"flower" shaped

The Quantum Numbers

Remember, the “shape” of these orbitals is just the
probability distribution mapping of the most likely
location of an electron with the specified angular
momentum (l-value) – the orbital has no other
physical “reality” to it beyond this. There are no little
spheres or peanuts, etc. inside the atom!

electron
distribution a p-orbital
plot

The Quantum Numbers

The energy of the electron's shell puts limits on the
angular momentum of the electron, hence, as we
noted earlier, different shells can contain different
types of subshells (orbital types):

shell (n-value) l -values orbital types
1 0 s
2 0, 1 s, p
3 0, 1, 2 s, p, d
4 0, 1, 2, 3 s, p, d, f

The Quantum Numbers
3. Magnetic quantum number (m l ): a charged
particle in motion creates a magnetic field. These
magnetic interactions between the electron and
the nucleus adds a second “directional”
component, Φ, to the orbital’s angular momentum.
What this means is that the magnetic quantum
number gives the spatial orientation of each
orbital. Different orbitals have different numbers of
possible orientations in space.

Thus, each subshell is made up of one or more
orbitals, and each orbital is oriented differently
than the others.

The Quantum Numbers
The magnetic quantum number (and hence, the
number of orientations possible for each orbital
type) depends on the angular number, l, and can
have the following values:

ml = (– l…0..+ l )

orbital ml values # of
(l-value) for orbital type orientations
s=0 0 1
p=1 -1, 0, +1 3
d=2 -2, -1, 0, +1, +2 5
f=3 -3, -2, -1, 0, +1, +2, +3 7

The Quantum Numbers
The three orientations of p-orbitals

The five orientations of d-orbitals

The Quantum Numbers
4. Spin quantum number (ms ): In addition to
locating the electron based on its energy, a fourth
quantum number was required to account for a
property known as electron spin.

A moving electron creates a magnetic field. The
random motion of the electron as it moves around
the nucleus produces random magnetic fields that
tend to cancel each other out. However, another
way an electron can move is to spin -- although the
e– is not actually spinning like a top, it produces a
consistent magnetic field as if it were.

The Quantum Numbers

N if the e- spins clockwise, it produces
a magnetic field pointing up (we call
this spin up).

if the e- spins counter clockwise, it
produces a magnetic field pointing
down (we call this spin down). N

The Quantum Numbers

The spin quantum number can have one of two
values: it is either + ½ or – ½ , depending on
whether the electron’s magnetic field has spin “up”
or spin “down.”

From a chemical point of view, the spin of the
electron makes no difference in how that atom
behaves. However, what is important for our
purposes is that, in a given orbital, no two electrons
can have the same spin.

The Quantum Numbers

We know two electrons in a given orbital will repel
each other due to their like charges. But if the
electrons are also repelling because their magnetic
fields both point in the same direction, that is just too
much repulsion to handle.
Instead, the electrons will pair up with their spins
opposite (we say they are “spin paired”). This
means that, magnetically, the two electrons are
actually attracting each other, which helps off-set the
electron-electron repulsion and stabilizes the
electrons in the orbital.

The Quantum Numbers

This means that an orbital can have a maximum
of two electrons, and those two electrons must be
spin-paired – that is, their spin quantum numbers
cannot both be the same value.

This is called the Pauli Exclusion Principle, which
states that no two electrons in an atom can have
the exact same four quantum numbers. At least
one (typically the spin) must be different.

eg: only one electron can be in the first shell (n = 1),
in the px-orbital, (ml = -1) of the p-subshell, (l = 1),
with its spin up (s = + ½).

The Quantum Numbers
Thus, since there are at most two electrons per
orbital, we can determine the maximum number of
electrons per subshell or shell.

1st shell:
1 s-subshell x 1 s-orbital x 2 e- per orbital = 2 e-

2nd shell:
1 p-subshell x 3 p-orbitals x 2 e- per orbital = 6 e-
maximum number of electrons = 8 e-

The Quantum Numbers

3rd shell:
1 d-subshell x 5 d-orbitals x 2 e- per orbital = 10 e-

4th shell:
1 d-subshell x 5 d-orbitals x 2 e- per orbital = 10 e-
1 f-subshell x 7 f-orbitals x 2 e- per orbital = 14 e-


The Quantum Numbers

Mathematically, the maximum number of
electrons in any given shell follows the formula:

# of electrons = 2n2

Thus
n = 1 holds 2 x (1)2 = 2 e-

n = 2 holds 2 x (2)2 = 8 e-
n = 3 holds 2 x (3)2 = 18 e-

n = 4 holds 2 x (4)2 = 32 e-

Periodic Relationships
Among the Elements

Chapter 8

The Search for Patterns Among the Elements
In the latter part of the 19th century, chemists began
trying to organize the elements into some kind of
pattern.

John Newlands (1838-1898) noted that, if the
elements were arranged in order of increasing atomic
mass, every eighth element had similar properties.
He called this the Law of Octaves. However, this law
only worked for the first 20 elements, and then broke
down. Most chemists did not pay much attention to
it.

Dmitri Mendeleev (1834-1907)
The first chemist to
develop a meaningful
pattern to the elements
was Dmitri Mendeleev.

Mendeleev

1. First, he arranged the elements in order of
increasing atomic mass.

2. Elements that had similar properties were placed in
the same column or group.

3. The properties were more important than the
mass – this resulted in several “blank” spaces.
Mendeleev realized that these blanks represented
missing elements.

Mendeleev
For example…
Arsenic (As) followed zinc (Zn) by mass – but
arsenic belonged in the group with phosphorus (P)
based on its properties -- this left two blanks below
Al and Si… VIIIA
2
13 14 15 16
VIA
17
He
IIIA IVA VA VIIA 4.003
5 6 7 8 9 10
B C N O F Ne
10.81 12.011 14.007 15.999 18.998 20.179
12
13 14 15 16 17 18
IIB Al Si P S Cl Ar
26.982 28.086 30.974 32.06 35.453 39.948
30 31 32 33 34 35 36
Zn ? Ge As Se Br
Ga ? Kr
65.38 69.72 72.59 74.922 78.96 79.904 83.80
48 49 50 51 52 53 55
Cd In Sn Sb Te I Xe
112.41 114.82 118.69 121.75 127.60 126.90 131.30
80 81 82 83 84 85 86

Mendeleev

4. Mendeleev noticed trends in the properties of
the elements in his table.

For example: he noticed that the reactivity of some
elements, like Na, Li and K, located on the left side
of his table, increased as you moved down the
column in which they were located.
He also noted that all of these elements formed
compounds in a one to one ratio with chlorine:
LiCl, NaCl, KCl, RbCl, etc.

Mendeleev

On the other hand, oxygen was more reactive than
sulfur, which was more reactive than selenium,
etc. – for these elements, located at the right side
of the table, reactivity seemed to increase as you
moved up the column.
In addition, these elements all formed compounds
with Na in a two to one ratio: Na2O, Na2S, Na2Se,
etc.

Mendeleev

He also noted that the density and melting points
of the elements generally increased as you
moved down a column, and that they increased,
and then decreased again as you moved across
a period. these elements have the
highest densities and melting
points within their rows

and melting points
increasing density

Mendeleev

5. Based on the observed trends, Mendeleev was
able to predict the properties of the missing
elements below Al and Si in his table.

Once we knew what the properties of these missing
elements might be, it was easier to search for and
find the elements. As a result, these two elements,
(which we now know as gallium and germanium)
were discovered within a short time.

Mendeleev

Predicted and actual properties of the missing
element eka-silicon (now called germanium)
based on Mendeleev’s periodic table trends:

property predicted actual
atomic mass 72 amu 72.6 amu
density 5.5 g/cm3 5.6 g/cm3
formula with oxygen XO2 GeO2

When the Elements
Were Discovered

Modern Periodic Table

Mendeleev believed the properties of the elements
were a periodic function of their atomic mass.
There were some elements that were “glitches” to
this hypothesis.
For example, based on its properties, iodine clearly
goes in the same group as F, Cl and Br -- but
iodine’s mass of 126.9 amu is LESS than the
atomic mass of Te (127.6 amu). When ordered by
mass, iodine should follow Te, rather than precede
it. This would place I in the same group as S and
O, whose properties are not at all similar to iodine.

Modern Periodic Table

In 1912, Henry Moseley proposed that the elements
be listed in order of increasing atomic number rather
than increasing atomic mass. When he did this, the
glitch with iodine and tellurium disappeared, since
iodine’s atomic number is greater than the atomic
number of tellurium.

The Modern Periodic Law states that the chemical
and physical properties of the elements are a
periodic function of their atomic number.

Organization
of the Periodic
Table

Recall we learned earlier that the periodic table is
“broken down” into periods (rows) and groups
(columns):

group
period

Organization of Periodic Table
As we have seen, the periodic table can be divided
into s, p, d and f “blocks” based on which subshell
the highest energy electron in the atom occupies:

Periodic Classification of the Elements
Depending on the type of subshell being filled, we
can categorize the elements into several sub-
groups:
Representative elements (also called the main
group elements) are elements in Group 1A through
7A. These elements have incompletely filled s or p
subshells in their valence (outermost) shell.

The representative elements can be further broken
down into families, each with similar but unique
valence electron configurations, as follows:

Classification of the Elements

Families of Representative Elements

1A = alkali metals: ns1 electron configuration
2A = alkaline earths: ns2 electron configuration
3A = boron family: ns2 np1 electron configuration
4A = carbon family: ns2 np2 electron configuration
5A = pnictides: ns2 np3 electron configuration
6A = chalcogens: ns2 np4 electron configuration
7A = halogens: ns2 np5 electron configuration
note that for representative elements, the group A number gives
the number of valence electrons for atoms within that family.


The noble gases are actually not representative
elements; these elements have a completely filled
s and p-subshell (ns2np6). Recall that this is the
most stable electron arrangement for any atom.

The transition metals are the elements in groups
3-11 and are characterized by having incompletely
filled d-subshells, or readily form cations with
incompletely filled d-subshells. Note that Zn, Cd
and Hg are technically not transition metals,
although many texts label them as such.


The inner-transition metals or the rare earths as
they are often called, are those elements in which
the f-subshell is being filled.

There are two rows:
1. The lanthanides, which are the elements
following lanthanum (La)

2. The actinides, which are the elements following
actinium (Ac).

Periodic Trends
It turns out that the electron configuration (shell
and subshell) of an atom, especially its valence shell
configuration, is the most important factor in
determining many of the physical, and most of the
chemical properties of a given atom.
We can identify certain trends that arise which also
relate to the electron configuration of the atoms.

So, before we look at the specific characteristics of
specific atoms or families, lets look at some of the
general trends which we find within the periodic table
as a whole.

Periodic Trends

For most trends, there are two main factors
that must be examined:

Within a family, the trends are dependent on
which shell the valence electrons are in.

Across a period, the trends depend on the
number of electrons in that shell, and on the
number of protons in the nucleus for that
particular shell.

Isoelectronic series
Atoms and ions that have the same number of
electrons (hence, the same ground state electron
configuration) are said to be isoelectronic.

e.g.: Na+, Al3+, O2- and N3- all have a total of 10
electrons – that is, they all have the same electron
configuration as [Ne].
Example: what would be an isoelectronic series for elements
# 19 – 25?

K+ Ca2+ Sc3+ Ti4+ V5+ Cr6+ Mn7+ = [Ar]

Periodic Trends
Effective Nuclear Charge (Zeff)
Electrons in interior filled shells (which occupy the space
between the nucleus and the valence electron) act to
“screen” the nucleus from the valence electrons.

Each interior or “screening” (-) electron partially cancels out
or reduces the effective (+) charge acting on the valence
electron, which weakens the attractive forces acting on that
electron.

In addition, electrons in the same shell can repel each other,
which further limits the net attraction a given electron
experiences towards the nucleus.

Effective Nuclear Charge
Effective nuclear charge (Zeff) is the “net” positive
charge felt by a valence electron, and is given by:

Zeff = Z - σ Where Z = atomic number (nuclear charge)
and σ = shielding constant.

Electrons in the same shell do not shield each other
nearly as effectively as those in interior shells, so to
a very good approximation, the effective nuclear
charge can be given as:

Zeff ≈ Z – (number of inner shell screening electrons)

For example:
Consider sodium (Na). Na has 11 protons and 11 electrons.
Of the 11 electrons, 1 is a valence electron, and the remaining
10 are screening electrons in interior filled shells,

Valence
electron
Zeff = Z – interior e-
Screening Zeff = (11-10) = +1
electrons
(2 + 8 = 10)


Finding the effective nuclear charge of an element

Element Z screen e- Zeff
Note: even though Ca has
Li 3 2 +1
8 more protons than Mg,
they have the same Zeff
Mg 12 10 +2
(+2). The valence
electrons in both atoms
S 16 10 +6 “feel” as if only 2 protons
are pulling on them.
Ca 20 18 +2

Al3+ 13 2 +11 pay careful attention to Zeff
of cations and anions!
Se2- 34 28 +6

Trends in Zeff : Within a family
Consider the trend within the Group 1A family

H (Z = 1) has 1 valence electron, with no inner shell
screening electrons, so it has a Zeff of (1-0) = +1

Li (Z = 3) has 1 valence electron, with 2 inner shell
screening electrons, so it has a Zeff = (3-2) = +1

Na (Z = 11) has 1 valence electrons, with 10 inner
shell screening electrons, so it’s Zeff = (11 -10) = +1


We see that the effective nuclear charge stays
constant within a family.

Even though we are adding protons, we are also
adding interior shell screening electrons at the
same time, and so the Zeff does not change.

Trends in Zeff : Across a period

Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17)
[Ne]3s1 [Ne]3s23p1 [Ne]3s23p3 [Ne]3s23p5

As you move across a period, the number of protons
(Z) increases steadily. The added electrons are
entering the same shell, so the number of interior
shell screening electrons does not change.

Trends in Zeff : Across a period

Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17)

Zeff = (11-10) Zeff = (13-10) Zeff = (15-10) Zeff = (17-10)
= +1 = +3 = +5 = +7

Since Zeff = Z – core electrons, we see that the
effective nuclear charge steadily increases from left
to right across a period.

Trends in Zeff across the transition metals

For transition elements, the electron configuration is:
ns2(n-1)dx . All the added electrons as you move
across the period are entering the (n-1) d-subshell –
that is, they are entering an interior shell and they
therefore act as screening electrons.
As a result, although the number of protons
increases across the period, the Zeff stays nearly
constant, just as it does down a family. Thus Zeff will
be ≈ +2 for most transition metals, and +1 for
Groups 6 and 11 (why?)

Atomic Radius
Defining the size of an atom is difficult, because the
electron cloud is diffuse. We use the following
definitions for atomic radius:

atomic radius = ½ the distance
between the nuclei in:
a) two adjacent metal atoms, or
b) two adjacent atoms in a diatomic
molecule

Size Trends
Trends in the radius (size) of the elements
The radius of the elements ranges in size from 31
picometers (He) to about 280 picometers (Fr). A picometer
is 10-12 m.
Note that, like Zeff ,
there are two distinct
trends in the size of
the atoms:

1. Within a family
2. Across a period

Size Trends

Size trend within a family valence
shell #
The size of the atoms gets
larger as you move down a H 1

family
The valence electrons are Li 2
entering successively larger
shells – larger shell means
larger atomic radius.
Na 3
recall that Zeff ≈ constant within a
family, so it has no significant
effect on size trend within a family.

Size Trends

Size trend across a Period
As we move across a period from left to right, the electrons
are entering the same shell, and the number of protons is
increasing. We have already seen that this causes the Zeff
to increase. This increase in Zeff pulls more strongly on the
electrons. The electron cloud is pulled inward, and the size
of the atom gets smaller as you move across the period.

Na (Z = 11) Al (Z = 13) P (Z =15) Cl (Z = 17)
Zeff = +1 Zeff = +3 Zeff = +5 Zeff = +7

General Size Trends for
Representative Elements
Increasing size
Increasing size

Size Trends

So far, we have only looked at the “general” trends
in size. There are some “adjustments” in this
general trend that we need to include.
The “good news” is the fact that the quantum
model of the atom predicts both the general trend
AND these adjustments.
This is one piece of physical evidence that the
quantum model is the correct model.

Size Trends

Electron-electron repulsion effects
Although the atoms tend to get smaller due to the
increasing Zeff as you move from left to right across
a period, the rate at which they get smaller tends to
level off.
As you move from left to right, the number of
valence electrons increases. This causes an
increase in electron-electron repulsion, which tends
to limit how much the electron cloud can be pulled
inwards, despite the increasing Zeff.

Size Trends
Thus, the trend in decreasing size as you move
across a period “levels off” as you move farther to
the right.

increasing
electron-electron
radius

repulsion

atomic number

Size Trends

Size trends across the Transition Metals

As you move across the transition metals (and the
inner-transition metals, as well) we have already
seen that Zeff ≈ constant.
Thus, since all the added electrons are entering
the same (n-1) shell as you go across a given the
period, the size of the elements across a given
period among the transition metals is roughly
constant.

Size Trends

Putting it all together, we obtain the
following graph of atomic radius vs
atomic number…

Size Trends

The Atomic Radii of Elements 1-86

Note the jump in size as you move to a new shell…

…and the roughly
…and the fairly steady constant size across
decrease in size moving the transition metals
across a period, which
levels off at the far right.

*note: unfortunately, the graph in your textbook has several mis-
plotted points, so it does not look like the graph above…but it should.

Size of Ions
Size trends of ions
If we compare the size of ions with the neutral atom
from which the ion was formed, we find that the…

+ --

Cation is always smaller than the atom from which
it is formed.
Anion is always larger than the atom from which it
is formed.

Size of Ions
Explanation: Size of the cation
A cation is formed when an atom loses valence
electrons. When an atom loses its valence electrons,
it tends to lose the entire valence shell – so the
cation is much smaller than the original atom.

Note also that the
When this 3 shell
rd
Zeff acting on the
e- is removed… +
remaining valence
shell electrons is
now (11-2) = +9,
…only the 2 inner so the 2nd shell is
shells remain pulled inward,
making the cation
Na : [Ne]3s1 even smaller.
Na+ : [Ne] 3s0

Size of Ions
Explanation: size of the anion
An anion is formed when an atom gains electrons in its
valence shell. But the atom gains electrons without gaining
any protons – there is no increase in Zeff to off-set the
electron-electron repulsion of the extra electrons in the
valence shell. This causes the electron cloud to “swell” and as
a result, the anion becomes larger than the original atom.

This electron is
added to the
valence shell
Cl
Cl—

Radii of some common cations and anions

neutral
atoms

anions

cations
neutral
atoms

Ionization Energy
Ionization Energy (IE)
Ionization energy is the energy required to remove
a valence electron from an atom (in the gas
phase, to insure that there are no interactions with
neighboring atoms that might influence the IE.).
Multiple electron atoms will have a different IE for
each electron removed:
1st IE: energy + X (g) → X+ (g) + e-
2nd IE: energy + X+ (g) → X2+ (g) + e-
3rd IE: energy + X2+ (g) → X3+ (g) + e-

Trends in Ionization Energies
To understand the trends in ionization energies
among the elements, we must recognize that the
strength of the attraction between the nucleus
and the electron is what determines the energy
required to remove that electron.

The force of attraction for an electron depends on
the Zeff and the distance of the electron from the
nucleus.
Zeff
Force of attraction = k 2
r
*k is just a constant of proportionality

Ionization Energy

The greater the Zeff , and the closer the electron is
to the nucleus, the stronger the attraction.

The stronger the attraction, the more energy will be
required to pull the electron loose.

this is my Unnhh…! Let
electron! go already!!

high Zeff IE
e-
close to
nucleus

Ionization Energy

1st Ionization Energy within a Family

1. Within a family, the Zeff remains constant
2. As you move down a family, the valence
electrons are farther from the nucleus (larger
shells), which weakens the attraction.

The 1st Ionization energy DECREASES
as you go down a family

Ionization Energy

1st Ionization Energy Across a Period

1. Across a period, the Zeff steadily increases
from left to right.
2. Across a period, the size of the atoms gets
steadily smaller as you move from left to right.
3. Both of these factors increases the attraction
for the valence electrons.

1st Ionization energy INCREASES as you
move from left to right across a period.

General Trend in 1st Ionization Energy

Increasing IE
Increasing IE

Variation of Ionization Energy with Atomic Number

Note that there are several
irregularities, but the general trend
shows IE increasing across a period
and decreasing down a family

Irregularities in the General Trends for 1st IE
It turns out that the “irregularities” in the first
ionization energy trends are actually predicted by
the quantum model!

Consider the apparent “glitch” in the 2A family:
Na = 495.9 kJ/mol
Mg = 738.1 kJ/mol ??
Al = 577.9 kJ/mol

The general trend would have the IE of Al greater
than that of Mg, but in fact, Mg is greater than Al…

Irregularities in IE Trends
The explanation for this apparent glitch is as follows:

1. The electron being removed from group 2A
atom is an ns2 electron – meaning, you are
breaking up that very stable filled sub-shell
arrangement, so it takes a little more energy
than expected.
2. In addition, the electron being removed from a
group 3A atom is an ns2np1 electron. It turns out
that s-orbitals have greater electron density near
the nucleus than do p-orbitals (we say the s-
orbital electron is is able to “penetrate” better.)


As a result, s-orbital electrons can partially screen
a p-orbital electrons, even though they are in
the same shell. Thus, it takes less energy to
remove the partially screened p-orbital electron.

The combination of slightly higher IE for Mg and
slightly lower IE for Al makes a marked “glitch”
in the trend.


A second “hiccup” in the general trend of increasing
IE across a period occurs among members of the
Group 5A elements.

Consider the apparent “glitch” in the first IE of P:
Si = 786.3 kJ/mol
P = 1012 kJ/mol ??
S = 999.5 kJ/mol

The general trend would have the IE of S greater
than that of P, but in fact, P is greater than S…


The explanation for this apparent glitch is as
follows:
1. In group 5A atoms there are 3 p-orbital electrons
– one each in the px, py and pz orbital, following
Hund’s rule.

group 5A

Removing an electron from a group 5A atom
involves breaking up the special stability of exactly
½ filled sub-shells, so its IE is higher than expected.


2. On the other hand, group 6A atoms have one set
of spin-paired electrons in one p-orbital:

Group 6A

The spin-paired electron in group 6A has a lower
than expected IE, since removing one of these
electrons would eliminate the electron-electron
repulsion effects on the remaining electron, lowering
its potential energy state.


Again, the higher than expected IE for group 5A
elements, coupled with lower than expected IE for
group 6A elements makes a marked “spike” in the
graph of IE.

Irregularities across the Transition Metals

Finally, we note that the IE is relatively constant
across the transition metal elements, reflecting the
fact that these atoms have relatively constant size
and Zeff.


However, the IE of group 13 (3A) is much lower
than that of group 12 for the same reason that the
IE of group 3A is lower than that of group 2A – the
electron being removed is coming from a higher
energy p-subshell.
The IE of group 12 is also higher than expected,
because you are breaking up the special stability of
a filled s,d subshell combination. This produces the
marked “glitch” between groups 12 and 13.

Irregularities in 1st Ionization Energy Trends

Multiple Ionization Energies
Another important trend appears when we examine multiple
ionization energies ( i.e., 1st, 2nd, 3rd, etc.) for atoms in a given
family

As you remove an electron, the amount of electron-electron
repulsion in the atom’s remaining valence shell electrons is
reduced. This, in turn, reduces the shielding effect, σ, and
increases the attraction for whatever valence electrons
remain. As a result, successive ionization energies are
higher:
IE1 < IE2 < IE3 < …


Removing an electron may leave the atom with a filled or
half-filled subshell. Removing the next electron will require
a somewhat larger than expected jump in energy, due to
the special stability of such filled and exactly half-filled
subshells.

Example: consider the successive IE for oxygen:
O O+ O2+ O3+ O4+ O5+

[He]2s22p4 [He]2s22p3 [He]2s22p2 [He]2s22p1 [He]2s2 [He]2s1

1,314 3,390 5,300 7,470 11,000 13,000

increase: +2,076 +1,910 +2,170 +3,530 +2,000

If removing an electron leaves the atom with a noble gas
core (s2p6 valence configuration) then removing the next
electron will require a much larger jump in energy, since a
noble gas core electron configuration is the most stable
possible configuration.
Compare the first four ionization energies for Na and Al:
2nd IE
1 IE
st
(kJ/mol) 3rd IE (kJ/mol) 4th IE (kJ/mol)
(kJ/mol)

Na [Ne]2s1 [Ne] [He]2s22p5 [He]2s22p4

496 4,560 6,900 9,540

Al [Ne]2s22p1 [Ne]2s2 [Ne]2s1 [Ne]

578 1,820 2,750 11,600

Electron Affinity
Electron Affinity (EA)
Electron affinity is a measure of the tendency of a
single neutral atom (in the gas phase) to gain an
electron in its valence shell.
The force of attraction for an “added electron” will
also depend on the Zeff pulling on it, and how far
from the nucleus the added electron will be (i.e.,
which shell the electron goes in)..
Zeff
Force of attraction = k 2
r
*k is just a constant of proportionality

Electron Affinity
Electron Affinity
Just as was the case for ionization energy, the
greater the Zeff , and the closer the electron is to the
nucleus, the stronger the attraction.

The stronger the attraction, the more likely an atom
will gain the extra electron and form an anion.
I’ll take you!
Sob…no
one wants
me… high Zeff
high Zeff IE
e-
close to
nucleus close to
nucleus

Measuring Electron Affinity
Experimentally, EA is measured by first adding an
electron to the neutral, isolated atom – in the gas
phase -- then measuring how much energy it takes
to “pull” the electron off again.
That is, EA is the ionization energy of an anion.
e.g., F− (g) F (g) + e− H = +328 kJ/mol

As a result, EA has a positive H. The larger the +H
for the process, the more stable the anion, which
means it has a high electron affinity.

Electron Affinity
Some atoms do not readily accept an electron, so it is difficult
to measure their electron affinity.
These atoms are assigned a (-) EA, which cannot be precisely
determined (what is the energy required to remove an electron
from an atom that will not gain an electron?)

EA values for
some
representative
elements

Electron Affinity
Electron Affinity Down a Family

1. Within a family, the Zeff remains constant
2. As you move down a family, the added electron
would enter a valence shell farther from the
nucleus (larger shells), which weakens the
attraction.

Electron Affinity DECREASES as you
go down a family

Electron Affinity
Electron Affinity Across a Period
1. Across a period, the Zeff steadily increases
from left to right.
2. Across a period, the atom gets steadily smaller
as you move from left to right, so the added
electron would be ever closer to the nucleus.
3. Both of these factors increase the attraction for
an added electron.

Electron Affinity INCREASES as you
move from left to right across a period.

Electron Affinity
(General Trend)

Increasing EA
Increasing EA

Irregularities in EA Trends

There are more irregularities in EA trends than for IE
trends. Some of these irregularities can, however,
be explained.

If an atom gains a stable filled or half-filled subshell
when it gains an electron, it will require more energy
to remove it, and hence it’s EA is greater as well.
Thus, the EA of Group 1A, 4A and 7A are higher
than might otherwise be expected.


If an atom already has a stable filled or half-filled
subshell, then it’s tendency to gain another electron
is very low.
As a result, the EA of Group 2A, 5A and 8A
elements tends to be lower than expected.
Coupled with the higher than expected EA values
for Group 1A and 7A, we can account for the
“glitches” in the general trend, so that the EA of
carbon is greater than that for nitrogen, even
though nitrogen is a smaller atom with a higher Zeff.

Electron Affinity for the Group 8A Family
The EA values for the noble gases deserves a
special note. As we might expect, since the noble
gases have a filled s and p subshell, their tendency
to gain another electron is quite low.
If we examine the Zeff acting on any electron that
might be gained, we see a secondary reason for
the low EA value for noble gases…


Since the valence shell is already filled for all noble gases,
the added electron would have to enter the next larger shell.
The Zeff acting on the added electron would be zero because
all the other electrons in the atom would now act as
screening electrons:

added electron must
go to a new shell

all electrons are
now screening e-
Neon (Z=10)
Zeff = 10 - 10 =
filled shell!
0;  EA < 0

Irregularities in Electron Affinity Trends

Glitches:
EA of 2A, 5A
and 8A are
much lower
than expected

EA of 1A, 4A
and 7A are
much higher
than expected

Be N Ne

Finally, we note that…

The observed trends in the radius of the atoms, the
first ionization energies, multiple ionization energies,
and the electron affinities for the atoms are all in
agreement, including the “glitches,” with what the
quantum mechanical model of the atom would
predict.
This is strong evidence in support of our quantum
mechanical model of the atom.

Variation in the
Physical & Chemical
Properties of the
Representative
Elements

General Trends in Physical Properties
Before we examine the chemical properties of
individual groups in detail, there are some general
trends in the physical properties of the elements
which we can point out.

The periodic table is broken down into three major
groups: metals, non-metals and metalloids.
Note that metals are on the left; non-metals are on
the right.

Classification of the
Elements

Classification of
Elements
General Properties of Metals
 malleable: can be hammered into thin sheets
 ductile: can be pulled into long, thin wires
 lustrous: shiny (has a metallic “sheen.”)
 good conductors: allow heat and electricity to
easily pass through them
 tend to lose e- due to their low IE, they tend to
lose electrons in reactions to form cations

Metals

The density & melting point of metals tends to
increase as you go down a family (atomic mass
increases faster than atomic radius).
Across a period, these values tend to increase to a
maximum among the d-block transition metals, and
then slowly decrease again. (Transition metals are
relatively small and have a close-packed structure.)
Li is the lightest metal known (0.53 g/cm3), and
osmium has the highest density of any element
(22.6 g/cm3). Tungsten has the highest melting point
of any known element (3,410 ºC); gallium will melt
in your hand.

Metals

beryllium magnesium

sodium copper

gallium gold

Classification of
Elements
General Properties of Non-metals
 brittle (break into fragments when stressed)
 dull (not lustrous)
 poor conductors (that is, good insulators)
 gain e- to form anions (have high EA)

Non-Metals
The density & melting points of non-metals is
somewhat more varied than that for metals.
However, like metals, these values tend to increase
moving down a family.

For example, fluorine
and chlorine are gases,
bromine is the only
liquid non-metal, and
iodine is a solid (which
readily sublimes).

Classification of
Elements
General Properties of Metalloids
$ properties are intermediate between those of
metals and nonmetals
Si

$ many are semi-conductors

As Sb Ge

Trends in the Metallic
Properties of the
Elements

increasingly metallic in properties…(or less non-metallic)

Classification of Elements

The general trend is for elements to become more
and more metallic as you move from right to left
across a period.

Across the 3rd period, if we begin at the far right
and move to the left, we have the non-metals
argon (Ar), chlorine (Cl) sulfur (S) and phosphorus
(P); moving farther left we reach the metalloid
silicon (Si), then a true metal, aluminum (Al) and
continuing, we have the metals magnesium (Mg)
and sodium (Na).

Trends in the Metallic
Properties of the

increasingly metallic
Elements


The general trend is for elements in the same family
to become more and more metallic as you move
downward.

Within the Group 4A family, moving downward, we
see the first element is carbon (C), a non-metal,
followed by silicon (Si) and germanium (Ge), both
metalloids, and finally, there are two metals, tin
and lead (Sn and Pb).


Note that the trend in increased metallic properties
follows the same general trend as that for
decreasing IE. Because one of the most important
chemical properties of metals is that they tend to
lose electrons in reactions, it follows that the lower
the IE, the more metallic the element!

Across a period, the most reactive metals are
those at the far left (Group 1A).

In a family, the most reactive metals are at the
bottom of the family.

Note that the trend in increased non-metallic
properties follows the same general trend as that
for increasing EA. Because one of the most
important chemical properties of non-metals is that
they tend to gain electrons in reactions, it follows
that the greater the EA, the more non-metallic the
element! As a result:

Across a period, the most reactive non-metals
are those at the far right (halogens);

Within a family, the most reactive non-metals
are at the top of the family.

A Brief Guided
Tour of the
Elements

Group Properties
Since the electron configuration of an atom –
especially its valence shell configuration -- is the
most important factor in determining most of the
chemical properties of a given atom, it is not
surprising that:
Elements with similar valence electron
configurations tend to have similar chemical
properties.

Thus, members of the same group tend to have
similar properties. Chlorine reacts very similarly to
how bromine reacts; sodium behaves much the
same as potassium, etc.

Group Properties

Note also that all the transition metal elements have
similar valence electron configurations – nearly every
one has an ns2 (n-1)dx configuration.
Thus, as a group, many transition elements are fairly
similar in their chemical behaviors.
However, the presence of an incomplete d-subshell
does affect the properties of the transition metals, so
they are different from those of, say Group 2A metals,
which also have a valence ns2 configuration, but do
not have any (n-1) d-subshell electrons.

Group Properties
There are exceptions to this “similar e-
configuration means similar properties” group rule.

For example, carbon, silicon, lead and tin all have an
s2p2 valence electron configuration, yet carbon is a
non-metal, silicon is a metalloid, and lead is a metal!

The first element in a given family also tends to be
somewhat different from the rest of the elements in
the family. This is due to the fact that elements at
the top of a family tend to be smaller, with higher IE
and EA compared to other elements in the family.

Group Properties
Finally, similarities exist between pairs of elements
in different adjacent groups and periods, known
as the diagonal rule.
Specifically, the first three members of the second
period exhibit many similarities to those elements
located diagonally below them in the periodic
table.
The reason for this has to do
with the fact that “diagonal
elements” have similar charge
densities (charge per unit
volume); this makes them
react similarly with anions.

Now let us examine specific groups

Hydrogen: Its Own Special Group
Hydrogen is unique among the elements. It has only one
electron, which it loses quite readily. For this reason, we
place it in the same vertical row as the alkali metals, which
also only have one electron -- but H is not a metal.
e.g., HCl + H2O → H3O+ Cl-
H can also gain one electron, like the halogens, to form the
hydride ion, H‾. For this reason, some periodic tables place
H at the top of both groups 1A and 7A.
e.g., 2 NaH + 2 H2O → 2 NaOH + H2
Probably the most important compound containing H is water
2 H2 (g) + O2 (g) → 2 H2O (l)

Hydrogen

Hydrogen is highly flammable. Because it is also lighter
than air, it was used in making dirigibles (blimps) for many
years.
A spectacular accident
occurred in New Jersey in
1937 when a spark ignited
the hydrogen in the
Hindenburg dirigible as it
was “docking,” killing many
people aboard. Today, we
use helium instead, which
is not flammable!

Group 1A Elements: The Alkali Metals
The alkali metals are all metals with an ns1valence
electron configuration (n >1).
Increasing reactivity

The Alkali Metals

 soft, low density metals
 lose 1 electron to form +1 ions in reactions
 very reactive; not found in nature in elemental state. Alkali
metals are found only in compounds, such as NaCl and KI.
 react with water to form alkaline (basic) solutions, (that is,
they form hydroxides); hence the name:

e.g., 2 Na + 2 H2O → 2 H2 + NaOH
 all react with oxygen to form oxides; all but Li can also
form peroxides:
e.g., Na + O2 → Na2O2

Group 1A Elements: Alkali Metals

8.6

Group 2A Elements: The Alkaline Earths
The alkaline earths are all metals with an ns2 valence
electron configuration.

The Alkaline Earths
 lose 2 electrons to form +2 ions in reactions
 relatively soft, low density metals (but harder and more
dense than the group 1A metals)
 very reactive, but not as reactive as alkali metals. Alkaline
earths are still too reactive to be found in nature in elemental
state (they, too, are only found in compounds).
 Ca, Ba and Sr react with water (Mg reacts with steam) to
form alkaline (basic) solutions (that is, hydroxides) – hence
the name. Be does not react, however, even with steam.
 Sr-90, a radioactive isotope of Sr, was released during the
Chernobyl accident. This toxic isotope, which is chemically
very similar to Ca, is readily taken up by bones, and led to a
sharp increase in cases of leukemia.

Group 2A Elements : Alkaline Earths

The Transition Metals
The transition metals are the elements that make up the
“B” groups, or groups #3-11 using the newer numbering
system.

3 4 5 6 7 8 9 10 11 12
Sc Ti V Cr Mn Fe Co Ni Cu


 tend to be harder, with higher melting points than the Group
1A and 2A metals (their smaller size allows them to pack
more tightly together).
 most form colorful compounds (most compounds with
representative elements are white)
 not as reactive as 1A and 2A metals; many found in
elemental state, such as gold and silver.
 like Al, most react with oxygen and form a protective oxide
layer that prevents further oxidation (e.g., chrome plating).


 most can form +2 or +3 ions (often both). The +3 charge
state is most stable at the left side, and the +2 state is more
common on the right side.

Charge states in red
are the most common.

Note that the oxidation
state climbs to a
maximum at Mn, and
then drops back
down.

Some Properties & Trends among
the 1st Row Transition Metals

Transition Metals

transition metal compounds in aqueous solutions

Ti3+ Cr3+ Mn2+ Fe3+ Co2+ Ni2+ Cu2+

A Comparison of Group A and Group B metals
Consider the electron configurations of Groups 1A and
Groups 2A with their B group counterparts (groups 11,12):

Group number electron configuration
group 1A metals [noble] ns1

group 1B (11) metals [noble] (n-1)d10ns1

group 2A metals [noble] ns2

group 2B (12) metals [noble] (n-1)d10ns2

Note that the A groups and B groups differ only in the
presence or absence of a filled d-subshell of electrons.

Compare A & B Metals

The d-subshell electrons are not as effective at screening the
s-orbital electrons. As a result, the 1B (11)and 2B (12)
metals have higher IE than their A-group counterparts, which
makes them less reactive.
As a result, although alkali metals are never found in nature
in their elemental state, the 1B elements of copper, silver and
gold are very commonly found in their elemental state;
indeed, they are sufficiently inert to be used in making coins
(they are often called the coinage metals for this reason.)
Similarly, the alkaline earths are too reactive to be found in
their elemental states, but Zn, Cd and Hg (group 2B) are less
reactive and can be found in their elemental states.

The Rare Earths
Elements in the bottom two rows on the periodic table are
called the rare earth (metals). The rare earths are broken
down into two groups…

The Rare Earths

 The Lanthanides: Elements # 58-71 immediately following
lanthanum (La).

 The Actinides: Elements #90-103 immediately following
actinium (Ac).

 Elements with atomic numbers greater than 92 are called
the trans-uranium elements. None of these elements are
naturally occurring – they are all man-made.

Group 3A Elements: The Boron Family
 These elements all have an ns2np1 configuration. Boron
is a metalloid. The rest of the elements in this family are
true metals.

Group 3A Elements

 Aluminum is the most abundant metal in the earth’s crust.
It is usually found in the ore, bauxite (Al2O3). About 200
years ago, it was so expensive to extract pure Al from
bauxite that aluminum was worth more than gold! We
have since found cheaper methods of extracting the Al.
 Al is actually quite reactive, but quickly forms a thin coating
of Al2O3 when exposed to air. This protects the Al from
further corrosion, making it a good, lightweight, structural
metal.
 In addition to forming ionic compounds, these metals
can also form molecular compounds.

Group 3A Elements: The Boron Family

8.6

Group 4A Elements: The Carbon Family
 These elements all have a ns2np2 electron configuration.
Carbon is a non-metal, silicon (Si) and germanium (Ge) are
metalloids; tin (Sn) and lead (Pb) are true metals.

Group 4A Elements

 Carbon comes in several allotropes. Recall that an
allotrope is one of two or more distinct molecular forms of
an element, each having unique properties. The allotropes
of carbon are: graphite, diamond, and buckminster
fullerene (also called “buckyballs” for short!)

graphite and diamond buckminster fullerene (C60) – its
structure looks like a soccer ball

Group 4A Elements

 Carbon is also unique among the elements in its ability to
bond to itself to form long chains. This process is called
catenation. This allows carbon compounds to have a wide
range of structures, etc., which enables them to have
biological roles. The study of carbon based compounds is
called organic chemistry.

 Silicon is the second most abundant element in the earth’s
crust (SiO2 is the chemical formula for sand). Si is also
important in the semi-conductor industry, and is an
important ingredient in making computer chips.

 The most stable oxidation state for both C and Si is 4+.
(Recall that this does NOT mean they form +4 ions!)

Group 4A Elements
 Lead is a toxic metal. It was once used in paints and also
as an “anti-knock agent” in gasoline. Exposing young
children to lead can lead to mental retardation and
neurological problems, so its use has been cut back
drastically in the past 20 years. In ancient Rome, the
wealthy used lead pipes to carry water to their homes.
Some historians believe the effects of lead poisoning on the
people of Rome contributed to the decline and fall of the
Roman Empire!

 Elements at the top of the family are most stable in the +4
oxidation state, while lead, at the bottom of the family, is
most stable in the +2 oxidation state (although the Pb 4+ also
exists.)

Group 4A Elements: The Carbon Family

Group 5A Elements: the Pnictides
These elements all have an ns2np3 electron configuration.
Elements in the nitrogen family include non-metals,
metalloids (As and Sb) and one true metal (bismuth).

The Pnictides

 Nitrogen in its diatomic form (N2) is a chemically inert gas
(will not react) which makes up about 70% of our air.
Nitrous oxide (N2O), or laughing gas, is used by many
dentists as an analgesic. Ammonium nitrate (NH4NO3) and
other nitrates (TNT) are explosives. Ammonia (NH3) is a
common cleaning agent.

 Phosphorus has two allotropes: white (yellow) phosphorus
exists as P4 molecules and must be stored under water
because it burns in air with an intense, hot flame. Red
phosphorus, the other allotrope, is less reactive.

The Pnictides

The oxoacids of P and N are formed by reacting the oxides
with water:
N2O5 + H2O → 2 HNO3

P4O10 + 6 H2O → 4 H3PO4

 Arsenic is used as a pesticide. There is evidence that
Napoleon died of arsenic poisoning; it was first believed he
was poisoned by a rival, but we now believe the arsenic
actually came from his wallpaper, which used arsenic in its
dyes!

Group 5A Elements:
the pnictide family

8.6

Group 6A Elements: The Chalcogens
These elements all have an ns2np4 valence electron
configuration. Most of the members of this family are non-
metals, except for Te and Po. Oxygen is the most reactive
element in the family.

reactivity increases

The Chalcogens

 Oxygen is a diatomic element. It is the most abundant
element in the earth’s crust – mainly as silicate rocks (SiO2)
-- and makes up about 20% of the air we breathe. Ozone
(O3) is an allotrope of oxygen that is present in the upper
atmosphere that screens out harmful UV rays. The ozone
layer now has “holes” in it, caused partly by pollutants.
 Most oxides of non-metals form acids in water (the word
oxygen means “acid former”). SO3 is a common pollutant
from burning coal that reacts with water in the air to form
sulfuric acid (H2SO4) – or acid rain:
SO3 + H2O → H2SO4

The Chalcogens

 Sulfur comes in several allotropic forms. Crown sulfur
(S8), the most common allotrope of sulfur, is used in
making matches; it is also added to latex rubber in a
process called vulcanization, which makes the rubber less
“tacky” and harder (the inventor of this process used the
vulcanized rubber to make tires – his name was Charles
Goodyear.)

Group 6A Elements: the chalcogen family

Group 7A Elements: the Halogens
These elements all have a ns2np5 valence electron
configuration. The word halogen means “salt former.” All the
halogens are diatomic elements. Bromine is a liquid, iodine
and astatine are solids, and fluorine and chlorine are gases.


The Halogens

 Fluorine is the most reactive of all the non-metals. It is a
pale yellow-green gas at room temperature. F is used in
making non-stick teflon, and compounds containing F-
ions are used to fight tooth decay.
 Chlorine is a pale green gas that is also quite reactive.
Chlorine bleach is actually the compound, NaClO
(sodium hypochlorite). Never mix bleach with other
household cleaning agents such as toilet bowl cleaners
and ammonia – they react with bleach to produce toxic
fumes!

The Halogens
 Organic compounds containing Cl and F are called
chlorofluorocarbons or CFC’s. These compounds, used in
making refrigerants, have been shown to destroy the
ozone in the earth’s upper atmosphere.
 Bromine is a corrosive, reddish brown liquid. Compounds
containing bromine or chlorine are often used as
antibacterial agents in swimming pools and hot tubs.
 Iodine is a purple solid that sublimes very easily. Tincture
of iodine (iodine dissolved in alcohol) used to be a common
antiseptic to put on cuts, etc.
 All the halogens react with water to form binary halic acids:
e.g., Cl2 + H2O → 2 HCl (aq)

Group 7A Elements: the Halogens

Group 8A Elements: The Noble Gases
The noble gases all have a filled s2p6 valence shell (except
He which is filled with only a 1s2 configuration). Most are
chemically inert (they will not react with anything). As their
name implies, these elements are all gases at room
temperature -- they have some of the lowest boiling points
of any element (e.g., He boils at -269 oC!)

The Noble Gases

 Helium has the distinction of being first discovered in the
spectrum of the sun before it was discovered here on
earth. The name “helium” comes from “Helios,” the Greek
god of the sun.
Neon is often used in lighted signs.

Properties of Oxides Across a Period
There is one last comparison we will make of the properties of
the representative elements will help distinguish the behavior
of metals from that of non-metals.

We will examine the properties of the oxides of the 3rd
period: Na2O, MgO, Al2O3, SiO2, P4O10, SO3, and Cl2O7

Oxides Across a Period

Oxygen tends to form -2 ions, especially when bonded to
metals with low IE, such as Na, Mg and Al.
These compounds form ionic bonds (as attested to by their
very high melting points and extensive 3-dimensional lattice
structures.)

As the ionization energy of the elements increases across a
period, the ionic character of the oxides decreases, while the
molecular nature of the oxides increases. Most molecular
oxides exist as small, discrete molecules, except for SiO2, or
quartz, which has a three dimensional structure very similar
to that of diamond.


Trends in Ionic vs Molecular Nature of Oxides

ionic oxides molecular oxides


Acid-Base Properties of Oxides

Basic oxides are metal oxides that react with water to form
bases (hydroxides) and/or react with acids.
e.g., Na2O + H2O → 2 NaOH (aq)
e.g., MgO + 2 HCl (aq) → MgCl2 + H2O

Acidic oxides are non-metal oxides that react with water to
form acids, and/or react with bases to form salts and water.
e.g., P4O10 + 6 H2O → 4 H3PO4 (aq)

e.g., SO3 + 2 NaOH → Na2SO4 + H2O

Amphoteric oxides can display both acidic and basic
properties.

Al2O3 will react with acids to form salts and water, which is
what we would expect of basic oxides:
Al2O3 + 6 HCl (aq) → 2 AlCl3 + 3 H2O

Al2O3 will also react with bases, which is what we would
expect of an acidic oxide…
Al2O3 + 2 NaOH + 3 H2O → 2 NaAl(OH)4 (aq)


Trends in Basic/Acidic Nature of Oxides

basic oxides amphoteric oxides acidic oxides

Oxides Within a Family

Finally, note also that because the metallic character of the
elements increases as you move down a family,
oxides at the top of the family tend to be more acidic than
those at the bottom of the family
oxides at the bottom of the family are more basic than
those at the top.

AP Inorganic Chemistry

Chemical Bonding:
Basic Concepts
Chapter 9


Why do atoms form bonds?
We now know that the most energetically stable
electron arrangement for an atom is to have a filled
valence shell, or to have an s2p6 valence electron
configuration (the so-called octet rule) – that is, to
be isolectronic with a noble gas.

Gilbert Lewis suggested that atoms obtain this
stable valence electron configuration by sharing,
gaining or losing electrons, which results in the
formation of chemical bonds.

Types of Chemical Bonds
1. ionic bond: electrostatic attraction between cations and
anions, which are created by the transfer of electrons.

2. covalent bond: a bond formed by the equal sharing
of valence electrons between the nuclei of two atoms.

3. polar covalent bond: a bond formed by the unequal
sharing of valence electrons between two atoms.

The type of bond formed depends on the ionization
energy and electronegativity of the atoms involved

Chemical Bonding
Electronegativity
Electronegativity (EN) = a measure, developed by
Linus Pauling, of the tendency for an atom to
attract or pull electrons towards itself in a chemical
bond.

Note the difference between EN and EA:
Electron affinity refers to an isolated, gas phase
atom’s attraction for electrons
Electronegativity involves shifts in electron density
between atoms that are bonded to each other.

Electronegativity

The electronegativity of an atom is a relative value
that can only be measured in relation to the EN of
other elements. Thus, EN is a calculated value,
not an experimentally determined value.

In general, however, we can say that the factors
that increase EA also increase EN, so we may
predict periodic trends in EN.

In general, elements with high EA and high IE also
have high electronegativities.

The Electronegativities of Common Elements

The trend for EN is very similar to the trend for EA:
EN increases up a family, and L→R across a period.
*note the EN values for some of the common elements

Electronegativity

Variation of Electronegativity with Atomic Number

Note that there are fewer glitches in EN compared to EA!

Bond Type and EN
By comparing the difference in electronegativity, E,
between two atoms, we can estimate the degree to
which electrons shared or transferred between the
atoms.
ΔEN = 3.1

ΔEN = 1.0

ΔEN = 0.4

K Na Li Al B H C N O F
0.8 0.9 1.0 1.5 2.0 2.1 2.5 3.0 3.5 4.0

Bond Type and EN

A large ΔEN between the two atoms (ΔEN ≥ 1.7),
indicates the electrons will be shifted or transferred
to the more electronegative atom to a significant
extent (greater ionic character to the bond).

A smaller ΔEN between the two atoms indicates
that each atom’s pull on the electrons is more equal,
which means the electrons will be shared between
them (greater covalent character to the bond).
The smaller the ΔEN, the more equally the electrons
are shared.

Bond Type and ΔEN
Approximate ΔEN values and bond type:
ΔEN 0.3 1.7 3

covalent polar covalent ionic
% ionic 5% 50% 100%
character

Covalent Polar Covalent Ionic
roughly equal unequal sharing or complete
sharing of e- partial transfer of e- transfer of e-

The Ionic
Bond

EN ≥ 2
+

Ionic Bonding
The Ionic Bond

If an atom with a low IE collides and reacts with an
atom which has a high EA, the first atom can transfer
electrons to the second atom.

The atom that gives up the electron becomes a
positively charged cation, and the atom that gains
the electron becomes a negatively charged anion.

An ionic bond is the “electrostatic” attraction formed
between the oppositely charged cation and anion.

Ionic Bonding
This electron transfer most often occurs between
fairly reactive metals (very low IE) such as group 1A
or 2A metals, and reactive non-metals (high EN)
such as the halogens and oxygen.

+

ΔEN ≥ 1.7

metal (low IE) non-metal (high EN) this creates oppositely
loses electron gains electron charged cations and anions

Ionic Bonding
Please note: you may have learned in a previous science
class that “an ionic bond is a transfer of electrons.” This is
incorrect. An ionic bond is simply the electrostatic attraction
that occurs between oppositely charged ions.
The cations and anions involved in an ionic bond, as often as
not, were already ions before the reaction. But we can break
apart and recombine these ions to form new ionic bonds.
For example, consider the reaction between Ag+ and Cl− to
form the ionic compound, AgCl in the reaction:
AgNO3 + NaCl → AgCl + NaNO3
No electron transfer occurs between Ag and Cl – the Ag+ and
Cl‾ ions were already ions before the reaction began! But the
bond in AgCl is an ionic bond.

Ionic Bonding

Note that compounds
involving alkali metals
and halogens always
form ionic bonds
between them --
Bond is at least 50%
ionic, and ΔEN ≥ 1.7
For this class, we will
generally assume that
any compound
involving a metal and
a nonmetal will be
primarily ionic in its
characteristics.

Lewis Dot Notation
To keep track of the valence electrons involved in bonding,
we use Lewis dot symbols which consists of the element’s
symbol, with one dot for each valence electron.
Thus, alkali metals have 1 dot, chalcogens have 6 dots, etc.

Lewis Dot Symbols
We can represent the ionic bond that forms between
two atoms using Lewis dot symbols, as follows:

Li + F Li F 1s22s22p6 = [Ne]

1s22s1 1s22s22p5 1s2 = [He]

We often use a curved arrow to depict shifts
in the e- during reactions. The arrow always
points in the direction of e- movement.

Ionic Bonding

Arrangement of ions in ionic compounds
Ionic compounds do not form individual “molecules,”
composed of just two or three atoms.
Instead, ionic compounds are composed of literally
billions of cations and anions electrostatically
attracted to each other.
The empirical formula for an ionic compound is just
the simplest ratio of the cations to the anions in
compound.

Ionic Bonding

The regular, repeating arrangement of ions in an
ionic compound is called the crystal lattice.

expanded view of the
actual crystal lattice
crystal lattice of NaCl
structure of NaCl

Ionic Bonding
Factors that Determine the Structure of an
Ionic Compound’s Crystal Lattice
1. The charge of the ions: this determines the
ratio of cations to anions
2. The size of the ions: larger ions will not fit in the
same arrangement as smaller ions will.
3. Ions are arranged within the lattice in such a
way as to maximize the separation of like
charged ions, and minimize the separation of
un-like charged ions. (Like charges repel and
opposite charges attract.)

Ionic Bonding
Lattice Energy
The same factors that determine the structure of
the lattice also determine the stability of a solid
ionic compound, which is measured by its lattice
energy.
Lattice energy = energy required to completely
separate 1 mole of a solid ionic compound into its
gaseous ions.
Thus, this is the same amount of energy released
when those ions come together to form the solid
compound.

Lattice Energy

Knowing the structure and formula of the compound,
we can determine it’s potential energy, and hence its
lattice energy using Coulomb’s Law:

Q+ Q- +/- Q = charges of the ions
E=k
r r = distance between the ions

Note that the lattice energy will be greatest when r is small
and the charges are large and opposite in sign.

The Born-Haber Cycle
The lattice energy of an ionic compound cannot be
measured directly. Instead, we measure it indirectly
by assuming the formation of an ionic compound
takes place in a series of steps, involving the
ionization energy of the cations and the electron
affinities of the anions,
and then use Hess’s Law!

This process was developed by Max Born and Fritz
Haber, hence, it is referred to as the Born-Haber cycle.

example: determine the lattice energy for LiF

1. convert: Li (s) → Li (g) ΔHº1 = 155 kJ/mol
2. dissociate F2: ½ F2 (g) → F (g) ΔHº2 = 75 kJ/mol
3. ionize Li: Li (g) → Li+ (g) + e– ΔHº3 = 520 kJ/mol

4. add e- to F: F (g) + e– → F‾ (g) ΔHº4 = -328 kJ/mol
add lattice
5. energy Li+ (g) + F– (g) → LiF (s) ΔH°5 = U = ??

6. when rxns 1-5 are added, we obtain the net reaction:
Li (s) + ½ F2 (g) → LiF (s)
But this is the rxn for the enthalpy of formation for LiF.
From the Appendix we find that, for LiF: ΔH°f = –594 kJ/mol

By Hess’s Law, we have:
ΔHº1 + ΔHº2 + ΔHº3 + ΔHº4 + ΔHº5 = ΔH°f

Solving for ΔHº5 = lattice energy (U) , we obtain:
ΔH5 = U = ΔH°f − (ΔHº1 + ΔHº2 + ΔHº3 + ΔHº4 )
U = −594.1 −[(155 + 75 + 520 + (−328)] kJ/mol
U = −594 − (422) = −1016 kJ/mol LiF

That is, −1016 kJ of heat energy is released in forming 1.0
mole of LiF from its component ions (or +1016 kJ of energy
are needed to separate 1 mole of LiF into its gaseous ions).


Lattice Energy =

∆Hoverall = ∆H°1 + ∆H°2 + ∆H°3 + ∆H°4 + ∆H°5

Lattice Energy

Forming a cation, even if forming the cation results
in an s2p6 electron configuration, requires energy.
Indeed, it requires more energy for each electron
removed, so an Al3+ ion requires more energy to
form than an Al+ ion, even though Al3+ has a noble
gas configuration. So where does the energy come
from to form cations?

It is the huge release of the lattice energy that
supplies the energy required to remove the
electrons from all the cations.

Lattice Energy
If the lattice energy is not sufficient to “off-set” the energy
needed to form the cations and anions from the neutral
atoms, then no electron transfer occurs, and no ionic bonding
is possible between those two atoms.
−
+
+ − ions
− +
formed
neutral Na
atoms & Cl2
molecules

lattice
Energy

Na + Cl2 energy

net release of combined ions
in lattice
energy
Na+-Cl-

reaction progression

Lattice Energy

There is a “trade-off” between increasing the
charge of the metal cations (which increases the
lattice energy) and the greater ionization energy
required to remove multiple electrons.

If, for example, Mg obtains a noble gas electron
configuration by losing 2 e- to form Mg2+, this gain in
stability does not outweigh the energy required to
remove the two electrons.

Lattice Energy

However, the increase in the lattice energy
released by forming ionic attractions with +2 ions
instead of +1 ions is more than enough to
compensate for the extra energy needed to form the
doubly charged Mg2+ ion.
Thus, Mg reacts with Cl‾ ions to form MgCl2, rather
than MgCl:

Mg Mg2+ ΔH = +2188 kJ/mol
Mg2+ + 2 Cl‾ MgCl2 (s) ΔHlat = -2527 kJ/mol

net energy change ΔHnet = -339 kJ/mol

Lattice Energy

By the same token, Na does not form Na2+ ions in
its compounds – the gain in lattice energy is not
sufficient to make up the difference for the huge
increase in ionization energy required to remove
the second electron from the stable [Ne] electron
configuration of the Na+ ion.

What is true about the cations also holds for the
anions: there is a trade off in the (-) EA of forming
higher charged anions (eg, F3−) vs the lattice energy
released when such higher charged ions combine to
form compounds.

Lattice Energy

The lattice energy vs energy required to form the
ions also helps explain why elements with higher
EN do not form cations (and hence, do not form
ionic bonds), even though such a cation could form
a stable ionic bond with an anion.
It is simply because the energy required to form a
cation from an element with high EN is greater than
the lattice energy that could be released if such a
compound were to form.

Lattice Energy
The energy needed to form P+ and Se− ions is greater
than the lattice energy that would be released if PSe
formed. As a result, P+ and Se− ions do not form, and
the ionic compound PSe does not exist.
−
+
+ − ions
− +
formed
neutral P
atoms & Se2 energy
molecules required lattice
energy
Energy

PSe
combined ions
in lattice
P + Se2 this much
energy is still
needed!

reaction progression

Lattice Energy

Thus, there is always a balance between the
IE or EA needed to form the highest charge
that can reasonably be obtained on the atoms,
and the increased lattice energy released
when these higher charged ions come
together to form a solid ionic compound.

Ionic Bonding

Properties of Ionic Compounds
1. Definite crystalline structure
2. High melting points
3. Brittle as solids
4. Most dissolve in water
5. Conduct electricity only when in the molten state
or when dissolved in water.

Ionic Bonding
Definite Crystalline Structure:
This is explained by the factors already discussed concerning
the crystal lattice structure: charge of ions, size of ions, and
attraction/repulsion of ions.

Ionic Bonding
High Melting Points:
To melt a substance requires that the particles of the
substance be separated, which requires energy to accomplish.
In ionic compounds, each particle is attracted to multiple other
particles of opposite charge in a three-dimensional lattice
network.
Melting ionic compounds requires breaking multiple ion-ion
attractions, which requires more energy = higher melting pt.

this cation is held in place by
+ attractions to three anions, all
three of which must be overcome
in order to melt the compound.

Ionic Bonding
Brittle as Solids:
If you “hammer” on an ionic crystal, it shatters. Any attempt
to distort the crystal lattice causes the “layers” to shift, which
causes like-charged ions to become aligned. The resulting
repulsive forces shatters the crystal.

+ +

+ + + +
+ + + + + +

+ + +

force applied to crystal like charges crystal shatters due to
become aligned repulsive forces between
like-charged ions

Ionic Bonding
Dissolve in Water:
As we shall see later, water is a polar molecule. Polar
molecules are molecules which have a small degree of
charge separation, so that one end of the molecule is
slightly negatively charged, and the other end is slightly
positively charged.

red = region of higher blue = region of
electron density,
which gives a partial H reduced electron
density, which gives a
negative charge to the partial positive charge
oxygen atom in this O to the hydrogen atoms
area of the water in this area of the water
molecule H molecule

water molecule

Ionic Bonding
Dissolve in Water: continued
The partial positive charge on the hydrogen atoms in water
are attracted to the anions in an ionic compound; the partial
negative charge on the oxygen atom in water are attracted to
the cations in an ionic compound.
The water molecule is able to literally “pluck apart” the ionic
compound, and carry the ions off into the solution.

+ = water = anion ( ) = cation (+)
+

Ionic Bonding
Ionic Compounds Conduct Electricity when
Molten or Dissolved in Water

When a voltage is applied, a current will flow
only if there are charged particles present that
are free to move. These charge carriers can be
electrons or ions.
AA
We refer to these as
“mobile charge carriers.”

Ionic Bonding

When an ionic compound is in the solid phase,
the ions are “locked” in place and cannot move.
Thus, ionic solids are poor conductors.

Ionic Bonding

However, when an ionic compound
is dissolved in water or in the
molten (liquid) phase, the ions e− e−

are separated and held less tightly.
They are thus free to move,
which allows a current to flow.
− +
−
+ +
+
− −

Ionic Bonding

Electrolytes are compounds that, when dissolved
in water, produce a solution that conducts
electricity. All ionic compounds are electrolytes.
Compounds can be classified as strong or weak
electrolytes, depending on the extent to which they
dissociate into ions in solution. Strong electrolytes
dissociate 100% into ions.
Non-Electrolytes are compounds that, when
dissolved in water, produce a solution that does
NOT conduct electricity. Most molecular, covalently
bonded compounds are non-electrolytes

Covalent
Bonds
N
N

- - - -
O C O O C O O C O

O O O
- -

Covalent Bonding
The Covalent Bond
Covalent bonds involve nonmetals with relatively
high, but similar electronegativities. These atoms
compete roughly equally for the electrons, and thus
must obtain a stable octet of electrons by sharing
electrons between them, rather than transferring
electrons from one atom to another.

This sharing of electrons by overlapping orbitals
between two atoms is called a covalent bond.

Covalent Bonding

As the orbitals of two atoms
overlap, electrons are
attracted to the nuclei of both
atoms. This “traps” the
electrons in a region between
the two atoms, which holds the
atoms together in the covalent
bond. The trapped electrons
are “shared” by both atoms in
the molecule formed.

RED = electron rich area
overlap = covalent bond
BLUE = electron poor area

Covalent Bonding

Pure covalent bonds occur only in diatomic
nonmetal elements, such as Cl2 or N2, which have
identical electronegativities (ΔEN = 0).

Polar covalent bonds form when there is a slight
to moderate difference in EN between the two
bonded atoms ( 0.30 < ΔEN < 1.7).

Polar Covalent Bonds

If an ionic bond is a complete transfer of electrons,
and a pure covalent bond is an equal sharing of
electrons, then we can say a polar covalent
bond is an un-equal sharing of electrons, or a
“partial transfer” of electrons.

In polar covalent bonds, the shared electrons are
shifted towards the atom with the greater
electronegativity. This electron shift can produce a
charge separation within the molecule as a whole.


For example: F is more EN than H. As a result, the
electrons are shifted away from H and towards F.
We represent this shift with an arrow, pointing
towards the more electronegative atom, with a
cross-bar at the more electropositive end of the
bond.

H F = δ+ δ-
H — F

The region with increased e- density is symbolized
δ- and the region with reduced e- density is
symbolized δ+ .
Molecules like HF and H2O which have a net
charge separation are said to be polar molecules.
δ
O
H H
δ + δ+

We will investigate the nature of polar molecules in more detail
when we discuss intermolecular forces in Chapter 11

Covalent Bonds

Properties of Covalently Bonded Compounds
Because molecules can exist as individual, isolated
units, many of the properties of molecular
compounds depend on the nature of the attractive
forces that exist between two or more molecules.
Note carefully that these attractive forces are NOT
covalent bonds.
We will investigate the nature of these so-called
“intermolecular” attractive forces in more detail in
the next chapter…

Covalent Bonds

Properties of Covalently Bonded Compounds

1. Molecular compounds have lower melting points
than ionic compounds.
2. Molecular compounds are poor conductors of
electricity in any phase.
3. Molecular compounds have a wide range of
solubility in water – some dissolve easily in
water, some do not dissolve at all.

Covalent Bonds
Low Melting Points
The bonding within a molecule is a covalent bond
between two atoms – however, there are only very
weak attractions between separate molecules. As
a result, it does not take a lot of energy to separate
two molecules (melting).

weak or no attractions Molecules are easily
between molecules… separated = low melt pt.

Covalent Bonds
Poor Conductors of Electricity
Recall that, in order to conduct a current of
electricity, you must have charged particles that
are some distance apart to create an electric field.

Molecules are made up of neutral atoms. By
sharing the electrons, neither atom has gained or
lost electrons to become ions, even when the
molecules are separated.
Thus, there are no charged particles with which to
create an electric field, and so they are non-
conductors.

Covalent Bonds
Solubility in Water
Water is a polar molecule, in which there is a charge
separation within the molecule so that one end is
partially positive and other end is partially negative.
Polar molecules can dissolve in water because they
form attractions to water molecules, much like ionic
compounds.
Non-polar molecules do not dissolve in water
because they cannot form attractions to the polar
water molecules.

Some “Special Case” Types of Covalent Bonding
Some molecular compounds do not exhibit the properties
typical of most molecules. Some of these special
molecules have extremely high melting and boiling points;
some have an actual electric charge to them.

The two types of special molecules are:
1. network solids
2. polyatomic ions

Some “Special Cases”
Network Solids
Network solids (also called covalent crystals) are
a type of crystal in which every atom is covalently
bonded to its neighbors in a vast, complex
interconnected “network.”
Quartz crystals (SiO2) and
diamonds are examples of these
network solids.

Network Solids

Because every atom is covalently bonded to its neighbor,
network solids can be thought of as giant “macro-
molecules.” In order to melt network solids, you must
break multiple covalent bonds, much as we have to
overcome multiple ion-ion attractions in ionic solids.
As a result, network solids tend to have extremely high
melting points, and tend to be very hard.

diamond:
melting point = 3550 oC
diamond is the hardest
substance known

Polyatomic Ions

Polyatomic ions are groups of atoms covalently bonded
together, which have a net charge.
Most polyatomic ions are anions, in which the group of
atoms picks up one or more electrons to fill the valence
needs of one of the atoms in the group.

cyanide anion (CN-)

nitrate anion (NO3-) sulfate anion (SO42-)

F N F

F

Lewis
Structures
- - -
O C O - O C O
O C O

O O
O -
-

Lewis Structures
We are frequently interested in which atoms are
bonded to which in a molecule. To show this, we
use Lewis structures, which were developed by
Gilbert Lewis.
In the Lewis structure (or structural formula, as it
is sometimes called), we indicate all the valence
electrons around each atom, and which electrons
are shared in order to form the covalent bonds
between two atoms.

Lewis Dot Symbols for the
Representative Elements
and Noble Gases

9.1

Lewis structure terminology
Lewis proposed that a single covalent bond requires
the sharing of a pair of electrons between two
atoms.
A double bond occurs whenever two atoms share
two pair of electrons between them.
A triple bond occurs when atoms share three pair
of electrons between them.
Unshared electron pairs on an atom are called
lone pair electrons.

RULES FOR DETERMINING THE LEWIS STRUCTURE
OF MOLECULES

1. Determine the TOTAL number of valence electrons
supplied by ALL the atoms in the compound. If the
compound has a net charge, add or remove e- from the
total count accordingly.

2. Determine which element is the central element. This
will be the element that is the LEAST electronegative
element of those present in the compound, with the
exception of hydrogen. Hydrogen can NEVER be the
central element, since H can never form more than one
bond.

3. Arrange the other atoms around the central atom.

Lewis Structure Rules

Example: Consider the molecule CHF3
CHF3 = 1C + 1 H + 3 F
= 1(4) + 1(1) + 3(7) = 26 valence electrons
The least electronegative element is H, but H cannot
be the central atom; carbon has the next lowest EN
so the central atom will be carbon. Arranging the
other elements around C we write:
F
the arrangement of the outer atoms is arbitrary –
typically, if the molecule has 3 atoms we place them in
H C F a straight line, for 4 atoms we use a T-shape, and for 5
atoms we use a + sign pattern as shown here.
F


4. Form a single bond between the central element and all
the other elements in the compound. Each single bond
requires 2 electrons. Subtract the number of e- used in
making the single bonds from the total number of
electrons available, which you determined in step #1.

F It requires 8 electrons to form the 4
:

H : C: F single bonds needed to attach the
outside elements to the carbon atom
:

F This leaves: (26 — 8) = 18 electrons


5. IF the number of electrons needed to make the single
bonds is LESS than the total number of electrons
available, then place electrons, IN PAIRS, around the
outside elements until their valence requirements are
met. IF the number of e- needed for this equals the
number of e- available, you are finished

add electrons in pairs around each F
: :

: F: atom until each has a total of 8
electrons; none are added to H because
: :

H : C : F: it already has 2 electrons.
: :

: F: This used an additional 18 electrons,
which is all that were left – we are done.
This is the Lewis structure for CHF3.


We often show single bonds as a single line, rather
than as 2 dots. We still show all the lone pairs as
dots, however.
: :

:
:F: :F:
: :
: :

: :
H : C:F : H C F:
: :

:F: :F:
:
:

Example: consider the molecule OF2
total number of valence e- = 2(7) + 1(6) = 20 e-
Arranging the elements with O as the central atom and
forming the 2 bonds required to attach the F’s to the O
requires 4 electrons. This leaves (20 – 4) = 16 e-
:

:
: F : O :F :
:

:
Next we place 6 e- around each F to fill their valence needs.
This uses 12 e- . So far we have used 4 + 12 = 16 electrons.
This leaves us with 4 e- left over: (20 – 16) = 4

6. IF there are any electrons left over after filling the valence
needs of the outer atoms, add these extra electrons to the
central atom, in pairs.

If the valence requirements of all atoms has been met at
this point, you are finished.
:
:
:

:
:
:
: F : O :F : :F O F :
:
:
:

:
:
:
When we add the 4 remaining electrons to the central
oxygen atom (in pairs), we have now met the valence
needs for every atom in the compound. We are done.

Example: consider the molecule CH2O
total number of valence e- = 1(4) + 2(1) + 1(6) = 12 e-
Arranging the elements with C as the central atom and forming the 3
bonds required to attach the H’s and the O to the C requires 6 electrons.
This leaves (12 – 6) = 6 e-

:
:O :

:
H :C:H

The 2 H’s now have 2 e- each, so their valences are filled. The remaining
e- are placed on the oxygen atom. At this point, the oxygen’s valence
needs are met.

However, the carbon only has 6 electrons in its valence,
and it needs 8 – and we are out of electrons!


7. If there are insufficient electrons to meet the valence
requirements of every element, you must share 2 or more
pairs of electrons with the central atom.

shift this pair of electrons down
:
:O : so that there are now 4 electrons
:

being shared between carbon
H :C:H and oxygen…
:

:O Now every element has a filled
octet (or a filled shell, which only
::

H :C:H requires 2 e- in the case of H)


When an atom shares 4 electrons (2 pairs) we say
the bond is a double bond. We represent a
double bond as 2 lines.

:
:

:O O:
::

H :C:H H C H

Double bonds are stronger and
shorter than single bonds.


Example: Consider HCN

The compound contains 1 + 4 + 5 = 10 e-. We form the bonds
needed to attach the H and N to the central C atom. Next we
place the remaining 6 e- around the nitrogen atom.

: :
H:C : N:
To meet the valence needs of the carbon, we will have to shift
TWO pairs of electrons from the nitrogen, so that C and N
share 3 electron pairs.
: :

H:C : N: H : C : :: N :


When an atom shares 6 electrons (3 pairs) we say
the bond is a triple bond. We represent a triple
bond as 3 lines.

H : C : :: N : OR H C N:

Triple bonds are stronger and shorter
than double bonds.

Example: Write the Lewis structure of nitrogen trifluoride (NF 3).
Step 1 – N is less electronegative than F, put N in center

Step 2 – Count valence electrons N = 5 and F = 7
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on outside F atoms. This uses 24 e-
Step 4 - Add remaining two electrons to central atom.
Check, do all atoms have filled octets?

F N F N
F F
F F

Resonance
When drawing out Lewis structures, you may encounter a
situation where a double bond is needed, but there are two or
more different atoms that could donate the extra electrons.

For example, consider the molecule O3 (ozone). When you
draw out its Lewis structure, you reach this structure after
putting in all the available electrons:
The question becomes, which of the two
O O O oxygens will donate a second e- pair with
the central atom with which the required
double bond can be made? Does it make
any difference?


The answer turns out to be – BOTH of the oxygens
will donate electrons. If only one did, that would
imply there is something unique about that particular
O—O bond, when in fact, all O—O bonds should be
identical.
I’m an oxygen,
they’re an oxygen,
wouldn’t you like
to be an oxygen,
too?


The problem is, how do we show that both oxygens
are simultaneously donating electrons? This would
require that the oxygen atoms share 3 electrons –
that is, we need to show there are two “1½ bonds”
to the central atom…

O O O O O O
?? ??
this would imply that two orbitals on the central oxygen have
3 electrons in them – that is, it does not really have an s 2p6 e-
configuration, and violates both Pauli’s and Hund’s rules, too!

We rectify this by drawing both possible Lewis structures, in
which the central oxygen forms a double bond with one
oxygen, and then the other. We use a double arrow between
the two to show that the true structure is a blending of the
two possible Lewis structures:

O O O O O O

The double bond is NOT switching back and forth
between the left and right hand oxygens!

This “double” Lewis structure is called a
resonance structure.


When your Lewis structure calls for resonance structures,
you must show all possible resonance structures to receive
full credit!
Shown below are two compounds – one requires
resonance and the other does not. Can you tell which one
requires resonance?

NOT resonance – the
Cl P O Cl P O atoms are not the same!

identical atoms – this
O S O O S O requires resonance

Exceptions to the Octet Rule

1. The Expanded Octet only for central atoms in 3rd shell
and larger; (uses available empty d-orbitals to form the bonds)

F
S = 6e- F F S has 12 e-: 4 are in
SF6 6F = 42e- S the 3 d-subshell, the
rest are in 3s and 3p
48e- F F subshells.
F

2. Odd-Electron Molecules
N only has 7 valence e-.
N = 5e- Odd electron molecules
NO O = 6e- N O always have multiple
11e- bonds to central atom.

Exceptions to the Octet Rule

3. The Incomplete Octet
Be = 2e-
BeH2 2 H = 2e- H Be H
4e-
only 2 valence e- on Be

B = 3e-
BF3 3 F = 21 e- F B F
24e-
F

only 6 valence e- on B

Formal Charge
When setting up the Lewis structure for CH2O (this
is formaldehyde), we find that there are actually two
possible skeletal structures:
H
H C O H C O
H

We use a concept called “formal charge” to
predict which of several possible Lewis structures
is the most likely structure for that compound.

Formal Charge

An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
formal charge total number
total number 1 total number
on an atom in
a Lewis
structure
of valence
= electrons in
the free atom
- of nonbonding
electrons
+ 2 ( of bonding
electrons )
The sum of the formal charges of the
atoms in a molecule or ion must equal
the charge on the molecule or ion.

Formal Charge
Example: What is the formal charge on the possible CH2O
structure shown below?
-1 +1
H C O H
on an atom in
a Lewis
structure
of valence
= electrons in
the free atom
- of nonbonding
electrons
+ 2 ( of bonding
electrons )
formal charge
on Carbon
= 4 -[ 2 + ½ x 6 ] = -1

formal charge
Σ charge = 0 
on Oxygen
= 6 - [ 2 + ½ x 6 ] = +1

Formal Charge
Example: What is the formal charge on the possible CH2O
structure shown below?
H 0 0
C O
H
on an atom in
a Lewis
structure
of valence
= electrons in
the free atom
- of nonbonding
electrons
+ 2 ( of bonding
electrons )
formal charge
on Carbon
=4- [0 +½x8]=0
Σ charge = 0 
formal charge
on Oxygen
=6- [4+½x4]=0

Formal Charge & Stability of Lewis Structures
1. For neutral molecules, a Lewis structure in which there
are no formal charges is preferable to one in which
formal charges are present.
2. Lewis structures with large formal charges are less
plausible than those with small formal charges.
3. Among Lewis structures having similar distributions of
formal charges, the most plausible structure is the one in
which negative formal charges are placed on the more
electronegative atoms.

Formal Charge
We see that for CH2O, the second structure yielded the
smallest valued formal charges (zero) on each atom. Thus,
this is the most likely structure for CH2O.
0 0
-1 +1 H
C O
H C O H H

example: Use formal charges to determine the correct
Lewis structure for PClO.
-2 +2 0 0 0 0

possible Lewis P Cl O Cl P O
structures and their
formal charges: -1 +2 -1 +1 0 -1

P Cl O Cl P O

There is evidence that the correct Lewis structure
for H2SO4 involves an expanded octet around S.
Compare the Lewis structures with and without an
expanded octet.
-1
0
O O
0 0 0 0 0 0 0 0
0 +2
H O S O H H O S O H

O O
0 -1

with expanded octet, all without expanded octet,
formal charges are 0 S has +2 formal charge

Metallic Bonding
Metallic bonding is the type of bonds that form
between two or more metal atoms.
Metals have low IE and they also have low EN.

This means that the electrons
are only loosely held. When
the energy levels of many
adjacent metal atoms overlap,
it creates a “common” energy
level called a molecular orbital
which the valence e- of any
one of the atoms can occupy. molecular orbital

Metallic Bonding
Note that the electrons are NOT being shared between two
nuclei as in covalent bonds – rather, the electrons are free to
“associate” with any metal atom, and can and do move from
atom to atom.

We say the electrons are delocalized.

Metallic Bonding

A metallic bond can be thought of as a group of
cations in a “sea” of delocalized electrons, which
are mutually attracted to and associated with the
metal cations.
delocalized
electrons cations

+ + +
+ +
The (-) electrons act as +
+ +
the “glue” that holds the +
+ +
(+) cations together. + +
+
+ +

Metallic Bonding
Properties of metals
1. Metals are good conductors of heat & electricity
2. Metals are malleable and ductile
3. Metals are lustrous

Our model of the metallic
bond can be used to
explain why metals have
these properties.

Metallic Bonding
Metals are Good Conductors
Recall that to be a conductor of electricity requires
that there be charges that are free to move.
The delocalized electrons shared

+
by all the cations in metallic bonds − +

+

+

+
are not locked in place between

+
+
− +
two specific cations, as they are

+
+

+
+
+
with covalent bonds, and so they − +

+
+ +
+
are free to move in the presence of

+
− +
an applied voltage. This makes
metals very good conductors in the
solid phase.
6V

Band Theory of Conductivity see page 876-877 in text

The band theory is a model to explain the ability of
metals to conduct electricity, It gets its name from
the idea that delocalized electrons freely move
through “bands” formed by overlapping molecular
orbital energy levels.
Because of their small size, the atoms in metals are
closely packed together, and their outer shells can
interact with each other, creating “molecular
orbitals” which are very closely spaced in energy –
so close that we refer to them as bands rather than
shells.

Band Theory
The closely spaced filled energy shells make up the
valence band, and the higher energy, unoccupied
“molecular orbitals” formed by the overlap of empty
p-orbitals make up the conduction band.
Each band is actually made up of
individual energy lines so close
Conduction and together they essentially merge

valence bands for
Mg. Note that the
conduction band
involves the
overlap of empty
3p orbitals.

Band Theory
To conduct a current, electrons in the valence band (which
are held to the atom) must be promoted to the conduction
band where the electrons are now delocalized and free to
move.
The energy gap (called the forbidden zone) between the
valence band and the conduction band varies widely from
metals to non-metals, and explains why non-metals do not
conduct (large forbidden zone), but metals (very small
forbidden zone) conduct easily.

Metallic Bonding

Heat is a measure of the transfer of kinetic energy
between particles. The mobile e- are able to
collide with other e- and cations, transferring kinetic
energy as they do so. Thus, metals conduct heat
very well.

+ +
+
+ + +
+
+ + +

Metallic Bonding
Metals are Malleable and Ductile
Malleability and ductility requires that the bonding between
atoms be very “flexible” so that, even when the substance is
distorted, the atoms are still bonded.

When a force is applied, the cations are
free to move – as long as there are some
force
e- between the cations, the atoms are still
bonded together…

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

Metallic Bonding
Metals are Lustrous
An object that appears lustrous or shiny is reflecting or
emitting multiple photons of light with very similar, but slightly
different, energies.
The delocalized e- in metals move between shells that are
close together in energy levels. This e- shift produces
photons of similar energy = “lustrous” quality of the metal.

Larger shells are
close together in
energy states, so the
photons emitted are
also similar in energy
– this makes it shiny.

Bonds & Energy
As the orbitals of two atoms begin to overlap, the extra
attraction of the electrons towards two nuclei lowers the
potential energy of the atoms – as the atoms approach closer
and closer, the shared e- are also closer to both nuclei and
the potential energy of the atoms/electrons is lowered more.

However, if the atoms
approach too closely,
the repulsion of the two
(+) nuclei becomes a
factor, increasing the
energy state of the
bonded atoms quite
dramatically. The bond
is now destabilized.

Bonds & Energy

There is a point of maximum overlap where the
attraction of the e- for the two nuclei is greatest, and
the repulsion of the nuclei is not too great, which
yields a minimum potential energy for the two atoms.

The bond energy is
the ΔE associated
with the point of
ΔE lowest potential
energy in forming
the bond.
stable bond

Bonds & Energy

Chemical bonds form because the potential energy
of the bonded atoms is lower than the potential
energy of the atoms in the free or unbonded state.
THUS:

Forming bonds is always an exothermic process
– that is, ΔH < 0

Breaking bonds is always an endothermic process
– that is, ΔH > 0

Bond Enthalpy (energy) in Binary Compounds
The enthalpy change required to break a bond in one
mole of gaseous molecules is the bond energy.

molecule Bond Energy

H H H2 (g) H (g) + H (g) ∆H° = 436.4 kJ

H Cl HCl (g) H (g) + Cl (g) ∆H° = 431.9 kJ

O O O2 (g) O (g) + O (g) ∆H° = 498.7 kJ

N N N2 (g) N (g) + N (g) ∆H° = 941.4 kJ

Bond Energy
The bond energies for multiple bonds is always
greater than that for single bonds:

Single bond < Double bond < Triple bond

C C ΔHº = 347 kJ/mol


Bond Enthalpies for Polyatomic Molecules
In compounds that contain more than one bond,
the bond energies can differ for different bonds,
even if the bonds involve the same two kinds of
atoms.
example: the bond energy to break the first O-H
bond in water is greater than the energy needed
to break the second O-H bond:

H2O (g) H (g) + OH (g) ΔH = 502 kJ/mol

OH (g) H (g) + O (g) ΔH = 427 kJ/mol

Bond Enthalpies

The structure of the molecule is different after the
first O-H bond is broken, so it is not too surprising
that the enthalpy change in breaking the second
O-H bond is also different.

Note that this also means that the bond energy for
the O-H bond in water is different than the bond
energy for the O-H bond in, say, methanol (CH3OH)
or hydrogen peroxide (H2O2) etc.
For polyatomic molecules, we usually only refer to
the average bond energy for a particular bond.

Average O-H bond energy in water:

H2O (g) H (g) + OH (g) ∆Hº = 502 kJ/mol

OH (g) H (g) + O (g) ∆Hº = 427 kJ/mol

502 + 427
Average OH bond energy = = 464 kJ/mol
2

see page
386

ΔH values in
red are for
diatomic
molecules

Bond Energies (BE) and Enthalpy changes
in reactions

Since all reactions involve making and breaking
chemical bonds, we can analyze the enthalpies of
reactions by examining the enthalpy changes in
breaking the bonds of the reactant molecules and
forming the bonds among the product molecules.

This technique only gives an approximation of the
actual enthalpy of reaction, since we are only using
average bond energies, and seems to work best
when ΔHrxn > 100 kJ/mol.

Bond Energies and Enthalpy changes in Rxns

Mathematically, we imagine a reaction proceeding
by breaking all the bonds in the reactants and then
using the gaseous atoms to form all the bonds in
the products. Then:

ΔHº = total energy input – total energy released
ΔHº = ΣBE(reactants) – ΣBE(products)

Where
ΣBE (react) = bond energy in breaking reactants’ bonds
ΣBE (prod) = bond energy in forming product bonds

Bond Energies and Enthalpy changes in Rxns

Net
Net ΔH
ΔH

endothermic reaction exothermic reaction

Example: Use bond energies to calculate the enthalpy
change for:
H2 (g) + F2 (g) 2HF (g)

∆Hº = ΣBE(reactants) – ΣBE(products)
Type of bonds Number of Bond energy Energy change
broken bonds broken (kJ/mol) (kJ)
H H 1 436.4 436.4

F F 1 156.9 156.9

Type of bonds Number of Bond energy Energy change
formed bonds formed (kJ/mol) (kJ)
H F 2 568.2 1136.4

∆Hº = ( 436.4 + 156.9 ) – (2 x 568.2 ) = -543.1 kJ

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