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Contents
1
Stoichiometric relationships
Introduction
of
The
matter
mole
Reacting
to
the
and
particulate
chemical
change
12
and
Condensation
Atomic structure (AHL)
Electrons
in
atoms
291
volumes
heavy
13
polymers
Environmental
1
concept
masses
12
nature
528
impact—
metals
534
The periodic table—the
B
20
Biochemistry
transition metals (AHL)
2
Atomic structure
The
nuclear
Electron
atom
First-row
d-block
Coloured
complexes
elements
50
14
to
Proteins
enzymes
biochemistry
539
301
and
547
319
37
conguration
Introduction
Lipids
565
Carbohydrates
580
Chemical bonding and
Vitamins
590
structure (AHL)
Biochemistry
3
and
the
Periodicity
Further
aspects
of
covalent
environment
Periodic
table
bonding
and
structure
329
Proteins
Periodic
trends
and
enzymes
606
75
Hybridization
345
Nucleic
acids
Biological
4
597
67
619
pigments
629
Chemical bonding and
15
Energetics/thermochemistry
Stereochemistry
in
biomolecules
641
structure
(AHL)
Ionic
bonding
and
structure
93
Energy
Covalent
bonding
cycles
Entropy
Covalent
357
C
Energy
97
structures
and
spontaneity
364
Energy
Fossil
Intermolecular
forces
bonding
657
Nuclear
Chemical kinetics (AHL)
fusion
and
ssion
665
133
Rate
expression
and
Solar
reaction
mechanism
5
653
fuels
122
16
Metallic
sources
104
375
energy
674
Environmental
impact—global
Energetics/thermochemistry
Activation
Measuring
energy
changes
energy
warming
384
679
139
Electrochemistry,
Hess’s
Law
148
batteries
1
7
Bond
enthalpies
and
Nuclear
equilibrium
law
fusion
Chemical kinetics
solar
18
theory
and
687
ssion
and
reaction
cells
710
acids
and
bases
395
161
D
involving
Medicinal chemistry
acids
Pharmaceutical
and
bases
drug
pH
Equilibrium
products
and
397
Equilibrium
curves
action
717
403
179
Aspirin
and
penicillin
725
Opiates
19
8
pH
Electrochemical
of
Properties
acids
of
and
acids
bases
pH
and
bases
scale
and
acids
and
bases
deposition
stomach
737
impact
745
of
some
Organic chemistry (AHL)
medications
of
organic
reactions
751
437
Taxol—a
200
routes
chiral
auxiliary
448
case
204
Stereoisomerism
study
758
451
Nuclear
Drug
9
the
medications
Environmental
195
Synthetic
Acid
of
413
197
weak
regulation
Anti-viral
Types
Strong
cells
191
20
The
732
Redox processes (AHL)
Acids and bases
Theories
702
dye-sensitized
Acids and bases (AHL)
Calculations
7
cells
nuclear
rates
Lewis
of
and
389
Photovoltaic
Collision
fuel
Equilibrium (AHL)
152
The
6
rechargeable
medicine
detection
765
and
analysis
775
Redox processes
2
1
Oxidation
and
reduction
Measurement and analysis
209
(AHL)
Electrochemical
cells
Internal Assessment
226
Spectroscopic
organic
10
of
compounds
(with
461
thanks
assistance
to
with
Mark
this
Headlee
chapter)
for
his
785
Organic chemistry
Fundamentals
of
organic
A
chemistry
Materials
Functional
Index
235
Materials
group
chemistry
science
introduction
791
471
248
Metals
and
plasma
11
identication
inductively
(ICP)
coupled
spectroscopy
475
Measurement and data
Catalysts
484
processing
Liquid
Uncertainties
and
errors
crystals
489
in
Polymers
measurement
and
results
494
261
Nanotechnology
Graphical
techniques
Environmental
Spectroscopic
identication
impact—plastics
compounds
509
of
Superconducting
organic
501
272
metals
and
X-ray
277
crystallography
516
iii
Course book denition
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knowledgeable
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rights.
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have
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work,
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work
verbatim
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acknowledged
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treated
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books
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journals
acknowledged.
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all
assignments,
written
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oral,
own
completed
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used
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photographs,
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must
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requirements.
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student.
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advantage
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material
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v
Using your IB Chemistry
Online Resources
What is Kerboodle?
Kerboodle
is
subscription
able
to
to
an
to
access
guide
you
online
IB
a
learning
Chemistry
huge
bank
through
this
platform.
Kerboodle
of
resources,
If
your
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school
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Resources
assessments,
and
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you
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presentations
course.
What is in your Kerboodle Online Resources?
There
are
three
planning,
main
resources,
areas
and
for
students
on
the
IB
Chemistry
Kerboodle:
assessment.
Resources
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hundreds
Kerboodle
develop
your
Watch
of
Online.
skills
videos
science
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perform
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answer
questions.
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Find
at
galleries
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answer
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of
by
images
looking
questions,
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do
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book
and
recommended
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close
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resources
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Plan
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about
and
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vi
for
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on
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work
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own.
conclusions
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Click
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assessment
examinations.
style
practice
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tests:
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year.
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test.
Don't forget!
You can also nd extra
resources on our free website
www.oxfordsecondary.co.uk/ib-chemistry
Here you can nd answers to questions in the book .
vii
Introduction
This
book
is
a
companion
Baccalaureate
Chemistry
is
one
Programme.
with
is
the
often
the
Diploma
It
is
of
called
the
from
a
prerequisite
and
environmental
being
Programme
knowledge
IB
With
full
the
structure
Subject
Topics
both
1
SL
higher
-
11
and
Each
as
study
and
of
the
a
two
are
its
wide
International
closely
skills.
own
as
underpin
right,
A
of
years
chemistry
of
an
the
nature
study
of
additional
IB
of
numeracy
Diploma
to
develop
nature
scientic
and
is
biological
problem-solving,
encouraged
of
both
systems.
medicine,
range
study
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biological
skills,
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Diploma
academic
critical-thinking,
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level
of
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such
understanding
issues
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engineering.
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skills,
students
develop
on
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principles
in
analytical
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of
combines
and
disciplines
scientic
book
in
detail
courses.
optional
Materials,
of
science
literacy
global
scientic
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perspective.
follows
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Topics
Topics
topics
core
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A,
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B,
the
chemistry
cover
number
of
material
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chemistry
topics
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live
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chemical
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act
the
subjects
materials
such
and
will
material).
option
analytical
of
explain
HL
The
Chemistry:
of
other
Course,
on
to
the
programme
in
Guide.
level
options.
All
them
as
worthy
skills.
learners
prepare
in
investigational
communication
focus
understanding
The
skills,
chemistry
its
and
which
involves
Chemistry
of
in
many
and
Chemistry
science
science,
sciences,
literacy
Chemistry
better
IT
of
science
laboratory
subject
for
transferable
scientic
inquiry.
a
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of
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chemistry
pivotal
environment
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students
experimental
acquisition
physical
generic,
the
an
for
Programme.
of
D
common
environmental
cover
the
Energy
that
the
the
major
and
strands
–
is
common
AHL
content
domains
Medicinal
and
of
in
the
Applied
Chemistry.
quantitative
perspectives
to
(additional
aspects,
integrated
organic
linkages.
in
the
book
include
the
following
elements:
Understandings
The
in
specics
detail.
of
the
Concepts
content
are
requirements
presented
in
ways
for
each
that
sub-topic
promote
are
covered
enduring
understanding.
Applications and skills
These
a
sections
specic
method
help
approach
involving
you
illustrative
key
or
to
develop
example,
by
your
often
considering
laboratory
understanding
following
a
a
by
considering
step-by-step
particular
chemical
working
experiment,
techniques.
Nature of science
Here
you
issues,
can
endeavour.
chemical
world.
viii
explore
theories,
This
is
research
NOS
the
methods
hypotheses
done
that
underpins
and
using
led
to
each
of
laws
science
that
carefully
paradigm
topic
and
are
selected
shifts
presented
some
of
associated
in
examples,
our
and
the
with
knowledge
scientic
including
understanding
throughout
the
of
book
the
there
are
a
wide
your
chemical
NOS
is
style
questions
an
range
of
NOS
understanding
assessable
are
based
and
component
integrated
questions
draw
of
the
on
and
your
programme
throughout
the
exercises
scientic
and
to
challenge
perspectives.
sample
NOS
book.
Theory of K nowledge
These shor t sections have headings that are equivocal 'knowledge questions'.
The text that follows often details one possible answer to the knowledge question.
We encourage you to draw on these examples of knowledge issues in your TOK
essays. Of course, much of the material elsewhere in the book , par ticularly in the
NOS sections, can be used to prompt TOK discussions.
TOK provides a space for you to engage in stimulating wider discussions about
questions such as whether there should be ethical constraints on the pursuit of
scientic knowledge. It also provides an oppor tunity for you to reect on scientic
methodologies, and how these compare to the methodologies of other areas of
knowledge. TOK is not formally assessed in the IB Chemistry programme, but it
plays a pivotal role in the teaching of IB science.
Activities and quick questions
A variety of shor t topics or challenging questions are included with a focus
on active learning. We encourage you to research these topics or problems
yourselves using information readily available in textbooks or from the Internet.
The aim is to promote an independent approach to learning.
End -of-topic questions
At
the
end
of
each
(multiple-choice,
problems
Answers
and
can
topic
you
data-base
hypothesis
be
found
at
will
nd
exercises,
style
a
wide
range
extended
of
questions
response,
NOS
style
questions).
www.oxfordsecondary.co.uk/ib-chemistry
ix
Meet the authors
Sergey
Bylikin
in
1998
and,
in
Chemistry.
University,
UK.
since
2007
Gary
since
Head
Hong
in
Gary
and
is
where
he
to
at
Athlone
been
IB
David
Chemistry
in
USA,
project
of
the
this
appreciation
Brian
an
earlier
devotion,
for
through
age,
x
to
my
and
wife
have
to
IB
had
Kong.
in
and
the
in
take
of
up
he
1998
is
science
in
IB
School
coordinator
with
the
in
IBO,
India,
curriculum
course.
to
from
the
position
UK
at
Gary
and
a
University
and
UAE
to
senior
involved
in
Ireland,
University,
associate
Ireland
currently
was
CAS
Chemistry
DP
the
Chemistry
workshops
Chemistry
back
and
of
the
Kong.
posts
a
in
with
review.
taught
that
IB
latest
Medical
International
involvement
Chemistry
moved
has
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Hong
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years
IB
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Chemistry
of
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cum
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Chemistry
support
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Australia.
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chemistry
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latest
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Russian
at
State
Russian
curriculum.
writing,
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professor
Canada
Emirates
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Northwest
David
has
team
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boards
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curriculum.
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patience.
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to:
parents,
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Murphy
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Head
Chemistry.
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role
University
George
Arab
a
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United
Institute
curriculum
A
the
Tarcy
the
State
textbooks.
Europe
graduated
College
the
of
up
leadership
2010,
Chemistry
assistant
the
the
King
Following
Whitworth
in
In
member
at
in
2000,
across
Japan.
became
Inorganic
of
in
the
took
Switzerland
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teaching
of
latest
a
he
several
various
Science.
Cork.
moved
has
held
Murphy
College
he
graduate
PhD
was
involved
Australia,
and
currently
Brian
a
he
of
a
received
point
author
workshops
Kong
review
2009,
was
has
of
attending
is
and
later,
which
an
Horner,
1986
career
and
is
awarded
year
Until
after
Sergey
was
one
and
for
all
(RIP)
her
family;
support
professional
support,
career
for
as
their
invaluable
her
who
of
love,
Gary
their
expertise;
well
as
the
support,
comments.
understanding
instilled
in
me
Horner
children's
David
many
advice,
an
-
my
Tarcy
unremitting
parents
happiness,
friends
and
to
and
appreciation
-
to
and
my
Tina
of
Dennis
sister
friendship;
Sergey
and
Susan
Walton,
professional
support
(míle
internationalisation
my
Myrtle
for
her
brothers
colleagues
Bylikin
-
to
I
for
from
their
eternal
Gary
have
and
met
Natasha
for
her
1
S T O I C H I O M E T R I C
R E L AT I O N S H I P S
Introduction
There
is
within
and
a
a
broad
wide
community
variety
approaching
of
their
of
people
scientic
inquiry
working
disciplines
with
reactions
terminology
and
equations,
Chemistry
can
be
chemists,
regarded
as
science,
and
mathematics
the
science.
In
this
chapter
we
begin
to
many
of
the
foundations
on
to
of
which
of
of
chemistry
matter
to
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the
based.
IUPAC
the
nomenclature
of
organic
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representations
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gas
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Its
is
of
denitions
in
of
particles,
universal
mass
and
the
understanding
the
the
relative
quantitative
amounts
of
method
of
reactants
the
products
in
developed.
a
particular
Treatment
of
chemical
the
gas
reaction
laws
and
the
inorganic
application
compounds
concept
organization
is
of
mole
an
and
classication
the
importance.
the
a
examining
understanding
the
lay
and
down
discusses
chemistry.
language
volume
of
chapter
the
relation
central
of
reasoning
fundamental
processes.
this
language
common
For
methodology,
by
comprehensive
of
of
volumetric
analysis
complete
this
chemical
introductory
chapter.
1.1 Itotio to t a ti at at
o att a ia a
Understandings
Applications and skills
➔
Atoms of dierent elements combine in xed
➔
Deduction of chemical equations when
ratios to form compounds, which have dierent
reactants and products are specied.
proper ties from their component elements.
➔
➔
Application of the state symbols (s), (l), (g),
Mixtures contain more than one element and/
and (aq) in equations.
or compound that are not chemically bonded
➔
Explanation
of
obser v a ble
changes
in
together and so retain their individual proper ties.
➔
physical
proper ties
changes
of
and
tem pera tu re
du rin g
Mixtures are either homogeneous or
state.
heterogeneous.
Nature of science
➔
Making quantitative measurements with replicates to ensure reliability – denite and multiple
propor tions.
1
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
The atomic theory
A
universally
today
is
that
However,
the
this
matter
has
seventeenth
was
of
accepted
all
a
widely
within
it
called
was
revealed
the
To
been
was
that
said
released
magnesium
of
so.
explain
phlogiston,
of
in
to
atoms.
scientists
During
breaking
theory
conditions
the
a
Over
process
be
Some
found
burning
gains
The
burning.
rather
the
mass
when
phlogiston
Scientists
use
instruments,
to
obtain
it
burns
in
oxygen,
to
feed
for
in
do
all
not
available
made
appetite
stand
those
and
the
knowledge.
have
been
understanding.
test
that
to
ground-
primitive
for
hypotheses
precision
are
the
not
often
relatively
and
renewed
theories
was
who
their
theories
theories
best
today
past,
are
of
time.
simple
and
facts.
than
The
loses
the
discoveries
with
account
metals
used
in
time,
tested
re-like
during
fact
commonly
science
phlogiston
proposed
investigations
that
of
composed
always
belief.
substances,
Quantitative
is
century
held
combustion
element
not
axiom
atomic
theory
states
that
all
matter
is
contradicting
composed
of
atoms.
These
created
destroyed,
atoms
cannot
be
theory.
a
wide
and
range
of
advanced
evidence
experimentation.
through
Much
of
computing
the
reactions.
properties
power
observation
or
chemical
methodologies,
of
matter
arrangement
and
of
and
are
Physical
depend
these
rearranged
and
on
during
chemical
the
bonding
and
atoms.
technology
TOK
States of matter
Antoine
often
referred
modern
to
Lavoisier
as
is
well
Lavoisier
phosphorus
they
gained
His
that
the
Matter
contribution
In
through
when
were
is
of
documented.
mass.
contradicted
“father
discovered
experimentation
and
the
chemistry”.
science
1772
to
(1743–1794)
sulfur
is
everywhere.
surrounds
form
of
planet
seeks
and
matter
and
to
us,
the
we
which
entire
expand
our
We
can
we
are
see
made
and
know
universe
is
are
up
of
touch
there,
made
understanding
of
matter,
many
though
up
of
matter
we
forms
we
and
see
belief
results
that
particles –
atoms,
would
be
lost
during
combustion
molecules,
as
phlogiston
was
released.
or ions
Could
phlogiston
mass?
from
Empirical
Lavoisier ’s
eventually
of
data
the
His
a
negative
derived
experiments
accepted
community.
some
have
by
work
first
the
was
scientific
contained
examples
of
particles are
quantitative
law
of
chemistry
conser vation
experiments
present-day
may
of
appear
in
the
mass.
standards
ground-breaking
and
they
occupies a
MATTER
motion
His
simple
but
their
in constant
by
day.
The discovery of oxygen by Joseph
Priestly and Carl Scheele
the phlogiston theory.
invalidated
This is an
example of a paradigm shift. The
has a mass
dominant paradigm or belief is
replaced by a new paradigm. Is this
how scientic knowledge progresses?
Figure 1 The characteristics of matter
2
volume in
space
were
it.
itsproperties.
made up of
mass
it
Air
is
a
Our
chemistry
combusted
These
it,
matter.
cannot
matter
and
consume
of
1 . 1
The
I n T r O d u c T I O n
properties
of
T O
the
T h e
three
p A r T I c u l A T e
states
of
n A T u r e
matter
are
Soi
xed
volume
●
xed
●
xed
shape
●
no
of
●
cannot
be
attractive
a
xed
the
forces
hold
c h A n g e
below.
gas
–
be
takes
it
the
shape
●
no
xed
volume
●
no
xed
shape
occupies
occupy
compressed
●
can
●
forces
be
the
–
space
expands
to
available
compressed
forces
between
particles
are
between
particles
are
between
the
close-packed
particles
shape
container
cannot
particles
than
in
solids
taken
as
zero
in
arrangement
●
●
c h e m I c A l
volume
weaker
particles
summarized
A n d
compressed
●
●
m A T T e r
liqi
●
●
O f
vibrate
in
particles
vibrate,
rotate,
●
and
particles
vibrate,
rotate,
and
xed
translate
(move
around)
translate
faster
than
in
a
liquid
positions
The
way
the
particles
of
matter
move
depends
on
the
temperature.
As
SI (Système International) units are a
the
temperature
increases
the
average
kinetic
energy
of
the
particles
set of standard units that are used in
increases
–
the
particles
in
a
solid
vibrate
more.
The
particles
in
liquids
science throughout the world. This will be
and
gases
also
vibrate,
rotate,
and
translate
more.
discussed in great detail in sub-topic 1.2.
Temperature
When describing room temperature, we
There
are
a
number
of
different
temperature
scales.
The
most
commonly
might say ‘25 degrees Celsius (25 °C)’ or
used
are
the
Fahrenheit,
Celsius,
and
Kelvin
scales.
All
three
are
named
‘298 kelvin (298 K)’ (to the nearest kelvin).
in
honour
of
the
scientist
who
developed
them.
Note that we use just the word kelvin, not
The
SI
unit
for
temperature
energetics
calculations
(see
Absolute
zero
on
is
the
topic
kelvin
(K).
The
Kelvin
scale
is
used
in
degrees kelvin. The boiling point of water
is 100 °C or 373 K, and the melting point of
5).
water is 0 °C or 273 K.
this
is
stops.
even
You
273
At
in
can
is
°C).
It
zero
is
the
temperatures
solid
the
Kelvin
temperature
greater
than
scale,
at
0
K
which
absolute
(on
all
the
Celsius
movement
zero,
all
scale
of
particles
particles
vibrate,
matter.
convert
temperatures
from
the
Celsius
scale
to
the
the
Kelvin
evaporation
scale
using
the
algorithm:
100
(K)
=
temperature
(°C)
+
C8 /erutarepmet
temperature
273.15
Changes of state
If
you
heat
a
block
of
ice
in
a
beaker
it
will
melt
to
form
liquid
water.
condensation
water
melting
If
0
you
continue
heating
the
water,
it
will
boil
to
form
water
vapour.
Figure2
freezing
ice
shows
during
a
heating
these
temperature
curve
changes
and
the
for
of
water
state.
kinetic
–
it
We
energy
shows
shall
of
how
look
at
particles
its
temperature
the
changes
relationship
during
these
between
changes
of
state.
Figure 2 The heating curve for water
3
steam
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Atiity
1
What happens to the particles during
Explain why the
changes of state?
temperature of a boiling
liquid does not increase
2
●
As
a
sample
despite energy being
in
constantly applied.
until
Deduce which would be
●
the
The
solid
it
ice
reaches
ice
begins
more painful, scalding your
Figure
skin with water vapour or
melting
boiling water.
lattice,
2
is
at
to
–10
begin
the
shows
melt
the
(263
point
and
All
a
is
of
to
is
heated,
more.
of
The
water
at
solid–liquid
no
the
attractive
molecules
K)
vibrate
there
occurring.
the
°C
to
melting
that
breaking
allowing
3
of
lattice
change
energy
forces
move
0°C
(273
molecules
being
freely.
used
the
The
increases
K).
is
temperature
between
more
water
equilibrium
in
is
the
temperature
to
set
up.
while
disrupt
molecules
level
of
the
and
disorder
Explain why you might feel
increases.
(The
nature
of
the
forces
between
molecules
is
discussed
cold and shiver when you get
in
sub-topic
4.4.)
out of the water at the beach
on a very hot, windy day.
●
Once
all
the
molecules
until
the
it
ice
has
vibrate
reaches
water
the
starts
melted,
more
to
further
and
boiling
move
point
heating
faster.
of
makes
The
water
at
the
water
temperature
100
°C
(373
rises
K),
and
boil.
Freeze-drying is a food
●
At
100
°C
a
liquid–gas
equilibrium
is
established
as
the
waterboils.
preservation technique
Again
the
temperature
does
not
change
as
energy
isrequired
to
that uses the process of
overcome
the
attractive
forces
between
the
moleculesin
the
liquid
sbiatio. Foods that require
water
in
order
to
free
water
molecules
from
theliquid
toform
a
dehydration are rst frozen and
gas.
(Equilibrium
The
curve
is
covered
in
sub-topic
7.1.)
then subjected to a reduced
pressure. The frozen water
●
in
then sublimes directly to water
temperature
vapour, eectively dehydrating
converted
gure
2
remains
to
steam,
shows
that
at
°C.
100
the
while
Once
temperature
the
all
water
the
will
is
liquid
increase
boiling
water
above
its
has
100
been
°C.
the food. The process has
●
Melting
and
boiling
are
endothermic
processes.
Energy
must
widespread applications in
be
transferred
to
the
water
of
state.
from
the
surroundings
to
bring
about
areas outside the food industry
these
changes
The
potential
energy
(stored
energy)
of
the
such as pharmaceuticals
molecules
increases
–
they
vibrate
more
and
move
faster.
(vaccines), document recovery
for water-damaged books, and
●
Cooling
brings
about
condensation
scientic research laboratories.
freezing
●
of
is
transferred
changes
they
●
of
less
Vaporization
during
to
a
and
potential
move
change
boiling,
or
by
form
to
heating
liquid
water,
–
the
and
the
solid.
exothermic
surroundings
The
the
form
are
processes
to
from
energy
the
of
processes.
water
the
Energy
during
molecules
these
decreases
–
slower.
of
state
from
liquid
evaporation
at
to
gas
which
temperatures
may
below
u
b
boiling
the
solid
point.
In
sublimation
matter
changes
state
directly
from
e
d
m
a
il
the
gas
phase
without
becoming
a
liquid.
Deposition
is
the
it
to
reverse
process
of
sublimation
–
changing
directly
from
a
gas
to
a
solid.
n
o
it
m
o
n
le
p
it
o
n
is
g
g
n
iz
e
e
r
f
s
happen
is
reverse
vapour
freezing
the
state.
vibrate
water
and
to
the
water
liquid
Condensation
solid
of
vaporization
Elements and compounds
An
condensation
in
element
a
xed
contains
ratio
to
rearrangements
liquid
atoms
form
of
the
of
only
compounds
particles
of
one
type.
matter
are
of
the
of
elements
molecules
or
fundamental
combine
ions.
These
cornerstone
gas
of
chemistry,
represented
in
formulae
and
balanced
Figure 3 Changes of state for water
(Atoms
4
Atoms
composed
are
covered
in
detail
in
sub-topic
2.1.)
chemical
equations.
1 . 1
I n T r O d u c T I O n
Chemists
the
study
many
created
in
these
important
The
and
group
oxidation
Sodium
abundant
p A r T I c u l A T e
elements
chemical
reactions,
and
and
and
n A T u r e
compounds
physical
how
they
O f
react
with
properties
can
be
m A T T e r
of
used
one
the
in
A n d
c h e m I c A l
c h A n g e
another,
substances
many
sodium
chloride,
NaCl,
is
made
up
of
the
elements
chlorine.
1
in
is
T h e
applications.
compound
sodium
The
how
different
T O
alkali
air
metal
and
stored
violently
under
element
sodium
on
oil
the
is
a
reacts
to
soft
with
prevent
planet,
metal
water,
these
(2.26 %
that
undergoes
creating
reactions.
by
It
rapid
alkaline
is
the
solutions.
sixth
most
mass).
Figure 5 The structure of sodium chloride. It consists of a crystalline
Figure 4 Elemental sodium is a reactive alkali metal
The
halogen
chlorine
is
a
gas
at
room
lattice of sodium ions (purple) and chloride ions (green)
temperature.
Chlorine,
Cl
,
is
2
highly
The
the
highly
ionic
table
and
irritating
to
the
reactive
and
uses
of
constituent
daily
chloride
and
sodium
compound
consumed
sodium
skin,
elements
crystalline
salt
eyes,
the
and
sodium
in
are
the
respiratory
chlorine
chloride,
food
very
upper
we
to
commonly
eat.
different
combine
tract.
The
from
form
called
properties
those
of
its
elements.
Mixtures
A
pure
substance
chemical
include
and
the
is
matter
physical
elements
that
properties
nitrogen,
has
are
N
a
constant
distinct
and
composition.
and
argon,
Ar
consistent.
and
Its
Examples
compounds
such
as
2
water,
H
O,
table
salt,
NaCl,
and
glucose,
C
2
Pure
6
substances
example,
sea
substances
such
as
H
can
water
can
be
ltration,
physically
contains
separated
fractional
combine
mainly
from
to
12
form
sodium
the
O
6
a
mixture.
chloride
mixture
distillation,
.
by
and
physical
For
water.
Pure
techniques
orchromatography.
The
5
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
elements
or
compounds
that
make
up
a
mixture
are
not
chemically
post’s aw o ostat oositio
bound
together.
(1806) stated that compounds have
Homogeneous
mixtures
have
both
uniform
composition
and
uniform
distinct properties and the same
properties
throughout
the
mixture.
Examples
include
salt
water
or
a
elemental composition by mass.
metal
alloy
such
composition
Examples
as
and
hence
include
prawn
soup)
Figure
9
or
brass.
foods
Irish
and
their
properties
such
stew
summarizes
compounds,
Heterogeneous
the
as
(a
tom
vary
yum
mixture
of
classication
mixtures
throughout
goong
cubed
of
have
(Thai
meat
matter
non-uniform
the
hot
and
into
a
mixture.
and
sour
vegetables).
elements,
mixtures.
matter – any substance that occupies
mixture – a combination
pure substance – has a
of two or more pure
denite and constant
substances that retain their
composition
individual properties
homogeneous mixture –
heterogeneous mixture –
has both uniform
has
element – made up of
non-uniform
compound – made up
atoms that each have
of a combination of
composition
composition and
the same atomic
atoms or ions in a xed
and properties
varying properties,
number, eg lead, Pb,
ratio and having dierent
eg salad dressing,
mercury, Hg,
throughout,
eg salt water,
paint, garden soil
properties from the
bromine, Br
constituent elements, eg
Figure 6 Chlorine reacts vigorously
metal alloys
water, H
O, carbon dioxide,
2
with sodium metal
CO
, sodium chloride, NaCl
2
Figure 9 Elements, compounds, and mix tures
The language of chemistry
Chemistry
enables
the
has
wider
symbols
chloride, NaCl(s). It has very dierent properties
and
from those of its constituent elements
a
the
to
process
Chemical
symbols
and
in
The
International
an
nomenclature
is
to
provide
language
country
of
or
are
a
for
of
of
and
st u de n t s ,
with
t ha t
e l em e nt s
each
in
a l lo w i ng
bo rd er s
an d
oth e r.
re qu i re s
an d
d is p la yed
in
expressing
both
Pure
no
ba la n c e d
a nd
t h e ir
eq u at i on
und er s t an d in g
the
Applied
a
inorganic
naming
to
which
that
of
elements
inorganic
of
the
Chemistry
system
of
no
( IUPAC)
resulting
translation
from
The IUPAC Gold Book (http://goldbook .iupac.org/index.html) is IUPAC’s
compendium of chemical terminology.
is
IUPAC’s
another.
investigate industrial dyes by separating them
present
standardized
compounds.
compounds,
require
are
compounds.
us rso
6
of
C he m ic a l
Figure 8 Paper chromatography is used to
into their pure constituent components
and
c i t iz en s
t r a ns l at i on .
c om po un d s
a
and
monitors
and
words
language
organic
and
and
organic
consistency
culture’s
in
develops
both
symbols
l a ng ua ge
as
t ra n sc e n ds
e xa mined .
way
Union
that
a
for
i nf or ma ti on,
b e i ng
th a t
l e ctur er s ,
communic at e
are
proportions,
organization
a nd
a no the r
of
chemical
which
to
s y mb ol s
one
wealth
l anguag e
te a che r s,
equa ti o ns
of
relationship
unlocks
uni v e r sa l
communi ty
Knowledge
Figure 7 T
able salt is the compound sodium
a
scientists,
in
role
a
one
1 . 1
I n T r O d u c T I O n
T O
T h e
p A r T I c u l A T e
n A T u r e
O f
m A T T e r
A n d
c h e m I c A l
na o
TOK
c h A n g e
foa
na o
oyatoi io
foa
oyatoi io
Language is a crucial component in the communication of
+
ammonium ion
NH
3
phosphate(V) ion
PO
4
4
knowledge and meaning. Does the language of chemistry with
carbonate ion
2
3
phosphonate ion
CO
its equations, symbols, and units promote or restrict universal
PO
3
hydrogencarbonate
understanding? What role does linguistic determinism play?
3
2
sulfate(VI) ion
HCO
SO
3
4
ion
For example, the concept of equilibrium is often initially
hydroxide ion
2
sulfate(IV) ion
OH
misinterpreted. Preconceived ideas focus on a 50:50 balance
SO
3
between reactants and products. It requires an understanding
nitrate(V) ion
2
ethanedioate ion
NO
C
3
O
2
4
that equilibrium means that both the forward and reverse
nitrate(III) ion
2
peroxide ion
NO
O
2
2
reactions are occurring at the same rate before we can see that
an equilibrium reaction might favour the formation of products
T
able 1 Common polyatomic ions
or reactions, or that such a reaction could be non-spontaneous.
na o ai
foa
hydrochloric acid
HCl
Common combinations of elements: Background
nitric(V) acid
HNO
3
to writing equations
phosphoric(V) acid
H
PO
3
An
ion
is
positively
a
charged
species.
Anions
are
negatively
charged
and
cations
4
are
sulfuric(VI) acid
charged.
H
SO
2
There
the
are
a
numb e r
substances
yo u
of
common
wi l l
s tudy
p o ly a t om i c
a nd
w o rk
io ns
wit h .
t ha t
You
ex is t
ne e d
to
in
m a ny
be
of
ethanoic acid
CH
4
COOH
3
fa m i li a r
T
able 2 Common acids
with
the
names
a nd
fo r mula e
of
the s e
io n s,
s h own
in
t a bl e s
1
to
3.
na o aio
foa
sulde ion
ability
to
write
equations
is
essential
to
chemistry
and
requires
-ide
S
Writing and balancing equations
An
nai sx
2
2
sulfate(VI) ion
-ate
SO
4
a
full
understanding
of
the
language
of
equations.
At
the
most
2
sulfate(IV) ion
-ate
SO
3
fundamental
hand
side
level,
along
formulae
with
their
for
state
the
reactants
symbols
(s),
are
(l),
put
(g),
on
the
(aq),
left-
and
those
T
able 3 Naming anions. The prex identies the
for
the
products
on
the
right-hand
side.
The
arrow
represents
a
element present and the sux the type of ion
boundary
deduced
matter
between
by
of
reactants
referring
the
to
element
the
or
and
products.
solubilities
compound
of
at
a
State
ionic
given
symbols
salts
and
can
the
be
state
(eg element or polyatomic ion)
of
temperature.
Worked example
A
reaction
The
may
process
equation
ionic
and
of
starts
be
described
transforming
with
covalent
the
in
terms
these
will
starting
words
construction
formulae
of
be
of
into
a
materials
balanced
chemical
discussed
in
and
chemical
formulae.
depth
in
products.
Writing
topic
4.
Magnesium
form
a
Write equations for the following chemical reactions, including state symbols. Refer
to the working method on the next page on balancing equations if you need to.
oxygen
oxide.
equation
represent
to
state
known
Write
a
this
to
as
chemical
change,
symbols.
Solution
The
reactants
magnesium,
are
a
temperature,
1
in
powder
magnesium
including
Qik qstios
burns
white
the
solid
and
metal
at
the
room
diatomic
Zinc metal reacts with hydrochloric acid to form the salt zinc chloride. Hydrogen
molecule,
oxygen,
which
is
a
gas is evolved.
gas.
2
Hydrogen gas and oxygen gas react together to form water.
3
At a high temperature, calcium carbonate decomposes into calcium oxide and
The
magnesium,
which
carbon dioxide.
product
is
a
the
oxide
magnesium
solid
2Mg(s)
is
+
of
oxide
substance.
O
(g)
→
2MgO(s)
2
7
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Working method: how to balance
Step
1:
Balance
the
metal
Ca
rst.
chemical equations
It
The
examples
Figure
10
below
reminds
involve
you
that
reactions
metals
of
are
is
balanced.
metals.
below
and
Ca(s)
+
H
O(l)
→
Ca(OH)
2
to
the
left
of
the
metalloids
Remember
that
change
coefcient
to
balance
in
an
the
periodic
equation
of
a
formula
(add
a
front
of
the
formula).
You
do
not
change
it
(g)
2
2:
occurs
Balance
in
only
O
one
next,
formula
on
the
each
formula
H
number
as
in
+
you
Step
the
(aq)
2
table.
side.
(H
occurs
in
both
products.)
itself.
Multiply
O
H
by
2
to
balance
O.
2
Step
1:
First
balance
each
in
side
front
of
of
the
the
the
metallic
element
equation–
symbol
on
add
one
a
on
number
side
if
Ca(s)
+
2H
O(l)
→ Ca(OH)
2
necessary
of
atoms
so
of
that
this
there
is
element
the
on
same
each
2:
Balance
any
elements
that
occur
You
in
has
only
one
formula
on
the
reactant
can
ions
side.
remain
they
can
be
Sometimes
unchanged
balanced
been
3:
Balance
the
remaining
now
reactions
easily
at
this
elements
if
balanced
5
6
7
8
9
C
N
O
F
Boron
Carbon
Nitrogen
Oxygen
Fluorine
13
14
15
16
17
Al
Si
P
S
Cl
Aluminium
Silicon
Phosphorus
Sulfur
Chlorine
30
31
32
33
34
35
Zn
Ga
Ge
As
Se
Br
equation
Always
check
to
which
make
often
sure.
is
now
Potassium
hydroxide
neutralize
the
produce
two
Zinc
Gallium
Germanium
Arsenic
Selenium
Bromine
48
49
50
51
52
53
Cd
In
Sn
Sb
Te
I
Cadmium
Indium
Tin
Antimony
Tellurium
Iodine
H
SO
2
the
Balance
80
81
82
83
84
85
Hg
Tl
Pb
Bi
Po
At
Mercury
Thallium
Lead
Bismuth
Polonium
Astatine
(aq)
+
a
overall.
soluble
acid
base
sulfuric
hydrogen
Balance
1:
is
diprotic
the
ions
following
K
by
KOH(aq)
→
when
Step
2:
two
Both
O
can
Diprotic
they
equation.
KOH
on
side.
K
SO
2
in
that
acid.
doubling
reactant
4
occur
116
balanced
Example 2
Step
115
hydrogen
2,
and
dissociate.
114
step
stage.
acids
113
by
3:
that
necessary.
B
112
see
polyatomic
in
The
Step
(g)
2
and
happens.
products
+ H
side.
Step
Step
(aq)
2
number
and
compounds
(aq)
+
H
4
O(l)
2
H
on
both
117
sides
of
the
unchanged
equation.
in
the
The
reaction
sulfate
and
is
ion
is
balanced,
so
metals
the
coefcient
for
H
SO
2
will
stay
the
same.
4
Figure 10 Metals are below and to the left of the metalloids in
There
are
4
H
atoms
on
the
reactant
the periodic table
side,
so
multiply
H
O
by
2.
2
Example 1
The
alkaline
earth
metal
calcium
reacts
with
SO
H
2
water
to
produce
an
alkaline
solution.
Balance
H
SO
2
following
+
2KOH
(aq)
→
K
SO
2
(aq)
+
2KOH(aq)
4
→
K
SO
2
equation.
The
8
(aq)
4
(aq)
+
H
4
O(l)
2
the
equation
is
now
balanced.
(aq)
4
+
2H
O(l)
2
1 . 1
I n T r O d u c T I O n
T O
T h e
p A r T I c u l A T e
n A T u r e
O f
m A T T e r
A n d
c h e m I c A l
So tys o atio
The names and symbols of
the elements can be found in
cobiatio or sytsis reactions involve the combination of two or more
section 5 of the Data booklet
substances to produce a single product:
C(s) + O
(g) → CO
2
c h A n g e
(g)
2
doositio reactions involve a single reactant being broken down into two or
more products:
CaCO
(s) → CaO(s) + CO
3
(g)
2
Si at reactions occur when one element replaces another in a
compound. An example of this type of reaction is a redox reaction (topic 9):
Mg(s) + 2HCl(aq) → MgCl
(aq) + H
2
(g)
2
dob at reactions occur between ions in solution to form insoluble
substances and weak or non-electrolytes, also termed tatsis reactions:
HCl(aq) + NaOH(aq) → NaCl(aq) + H
O(l)
2
This example is an acid-base reaction discussed fur ther in topic 8.
Some applications and reactions of butane
Fuels and refrigerants
Butane,
C
H
4
create
the
variety
of
is
in
a
the
same
that
other
hydrocarbons
petroleum
gas
(LPG).
such
This
is
as
propane
used
in
a
to
wide
applications.
different
CFCs
with
liqueed
Methylpropane
have
mixed
10
fuel
(also
called
chemical
way.
were
isobutane)
formula
but
Methylpropane
previously
used
is
for
is
an
their
used
this
isomer
atoms
as
a
of
are
butane.
Isomers
arranged
refrigerant,
structurally
replacing
the
purpose.
cfcs a t iat o
si a tooy
Ozone
occurs
Ozone
lters
Without
many
this
forms
naturally
out
most
in
of
protection
of
life,
the
the
the
causing
stratosphere,
harmful
the
ultraviolet
ultraviolet
skin
in
radiation
cancer
in
upper
rays
from
would
humans
atmopshere.
and
be
the
harmful
other
The process of refrigeration
sun.
to
problems.
involves the energy changes of
a condensation–evaporation
cycle using volatile liquids.
H
Chlorouorocarbons (CFCs)
H
H
were traditionally used
O
in refrigerators and air-
H
C
C
O
conditioning units. They cause
O
depletion of the ozone layer
H
C
in the atmosphere, which
H
C
H
H
protects us from the harmful
eects of ultraviolet radiation
H
in sunlight.
H
CFCs are now banned in many
Figure 11 Ozone, O
Figure 12 Methylpropane is used as a refrigerant
countries, and non-halogenated
3
hydrocarbons such as propane
CFCs
undergo
reactions
with
the
ozone
in
the
stratosphere,
causing
it
are more commonly used
to
break
down.
The
‘ozone
hole’
is
a
thinning
of
the
ozone
layer
that
instead. There is more about this
appears
over
the
polar
regions
of
the
Earth
each
spring.
The
use
of
in sub-topic 5.3.
CFCs
been
has
caused
replaced
by
this
depletion
of
the
ozone
layer,
so
they
have
now
methylpropane.
9
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
The combustion of
hydrocarbons, C
x
H
produces
y
carbon dioxide and water.
Since 1997, taxis in Hong Kong
have been powered by liqueed
petroleum gas (LPG). Today
there are over 18 000 LPG
taxis and 500 LPG light buses
operating there. LPG, consisting
of butane and/or propane,
undergoes combustion to
release energy to power the
vehicle. The reaction produces
carbon dioxide and water
(sub-topic 10.2). LPG burns
much more cleanly than petrol
or diesel.
Figure 13 The ozone ‘hole’ was rst noticed in the 1970s and is monitored by scientists
worldwide
Balancing the equation for the combustion of butane
The
combustion
C
H
4
Step
(g)
O
10
1:
atoms
+
of
butane
(g)
→
is
CO
2
There
rst
by
an
(g)
exothermic
+
H
2
are
no
O(l)
2
metal
multiplying
reaction.
atoms
CO
by
to
balance,
so
balance
the
carbon
4.
2
C
H
4
(g)
Step
2:
leave
this
the
+
O
10
(g)
Oxygen
right,
→
4CO
2
until
so
(g)
+
H
2
is
found
last.
in
two
Hydrogen
multiply
H
O
O(l)
2
by
compounds
has
10
atoms
on
on
the
the
product
left
and
side
2
so
atoms
on
5.
2
C
H
4
Step
To
(g)
3:
H
Figure 14 Rush hour in Hong Kong
10
+
6.5O
(g)
are
not
by
+
5H
now
(g)
contain
6.5
→
13
oxygen
molecules
4CO
(g)
+
of
5H
2
used
(see
O(l)
2
2
in
15).
are
an
odd
number.
required.
O(l)
2
balanced
topic
atoms,
oxygen
We
equations,
therefore
except
when
multiply
the
calculating
whole
2.
(g)
+
13O
10
complex
balancing
4CO
equation
enthalpy
4
The
products
the
(g)
equation
→
2
10
Fractions
2C
(g)
2
The
H
4
lattice
O
10
balance
C
+
(g)
→
8CO
2
coefcients
equations
on
(g)
+
10H
2
in
this
page
8
example
is
O(l)
2
more
show
efcient
why
than
the
method
just
trial
of
and
error.
1 . 1
I n T r O d u c T I O n
T O
T h e
p A r T I c u l A T e
n A T u r e
O f
m A T T e r
A n d
c h e m I c A l
c h A n g e
The atom economy
The
global
demand
population,
and
dwindling
need
to
must
be
fewer
the
To
for
nite
conserve
less
toxic
and
resources
resources.
increasingly
and
goods
developing
services
have
led
Synthetic
efcient
along
economies,
to
emissions.
to
a
an
raw
Sustainable
and
increasing
levels
heightened
reactions
preserve
with
increasing
of
awareness
industrial
materials
world
pollution,
and
development
is
of
the
processes
produce
the
way
of
future.
this
Trost
of
rapidly
end
of
the
efciency
atoms
atom
Stanford
in
of
the
economy
University
chemical
reactants
reactions
with
the
Molecular
percentage
was
developed
Stanford,
by
CA,
of
mass
atoms
of
Professor
This
comparing
molecular
mass
by
USA.
looks
the
of
Barry
at
the
molecular
useful
useful
level
mass
of
compounds.
products
____
=
atom
The
atom
which
the
×
economy
economy
we
will
amount
economy
Molecular
of
of
is
discuss
important
later
reactants
100%
=
would
in
mass
in
this
the
atoms
of
that
in
In
an
of
ideal
products
no
atoms
100%
reactants
discussion
book.
amounts
suggest
of
Green
produced.
are
Chemistry,
chemical
process
So
an
atom
wasted.
Atiity
a)
Suggest why even if a chemical reaction has a yield close to 100%, the atom
economy may be poor. Carry out some research into this aspect.
b)
Discuss some other ways a chemical process may be evaluated other than
the atom economy, eg energy consumption etc.
)
Deduce the percentage atom economy for the nucleophilic substitution
reaction:
CH
(CH
3
)
2
OH + NaBr + H
3
SO
2
→ CH
4
(CH
3
)
2
Br + H
3
O + NaHSO
2
4
Qik qstios
Identify the type of reaction and then copy and balance
4
Al(s) + O
(g) → Al
2
2
O
(s)
3
the equation, using the smallest possible whole number
5
KClO
(s) → KCl(s) + O
3
(g)
2
coecients.
6
1
SO
(g) + H
3
O(l) → H
2
SO
2
7
2
NCl
(g) → N
3
2
(g) + Cl
CH
4
(g) + O
2
H
(g) + O
8
2
(g) → CO
(g) + H
2
O(g)
2
Ni(OH)
(s) + HCl(aq) → NiCl
2
(g)
(aq) + H
2
O(l)
2
2
8
3
C
3
(aq)
4
(g) → CO
2
(g) + H
AgNO
(aq) + Cu(s) → Cu(NO
3
O(g)
3
)
(aq) + Ag(s)
2
2
9
Ca(OH)
(s) → CaO(s) + H
2
O(l)
2
11
1
S TO I c h I O m e T r I c
r e l AT I O n S h I p S
1.2 T o o t
Understandings
Applications and skills
➔
The mole is a xed number of par ticles and
Calculation of the molar masses of atoms, ions,
➔
refers to the amount, n, of substance.
molecules and formula units.
➔
Masses of atoms are compared on a scale
Solution of problems involving the
➔
12
relative to
C and are expressed as relative
relationships between the number of par ticles,
atomic mass (A ) and relative formula/
r
the amount of substance in moles and the
molecular mass (M ).
r
mass in grams.
1
➔
Molar mass (M) has the units g mol
➔
The empirical formula and molecular formula
Interconversion of the percentage composition
➔
by mass and the empirical formula.
of a compound give the simplest ratio and the
Determination of the molecular formula of
➔
actual number of atoms present in a molecule
a compound from its empirical formula and
respectively.
molar mass.
Obtaining and using experimental data for
➔
deriving empirical formulas from reactions
involving mass changes.
Nature of science
➔
Concepts – the concept of the mole developed from the related concept of ‘equivalent mass’ in the early
19th century.
SI: the international system of measurement
Throughout
history
measurement.
so
an
internationally
measurements
Units
of
million
tonnes
do
and
desire
the
seven
from
for
12
the
a
of
allows
language
e ss e ntia l
US
in
do ll a rs,
p r eci ous
mi ll io n
me tal s
use s
a
of
all
the
ran g e
(p pm)
of
and
us
our
to
w a l ks
of
to
another,
understand
of
li fe .
in
m e a su r es
a g en c i e s,
pa r t ic u l at e
Th e
i n du s t ri es
me a su r ed
of
forms
culture
culture.
re s ourc e s
a re
standard
international
a
that
system
Système
base
these
in
units
different
country
us e
o un c e s ,
inc l ud in g
a mo n gs t
m a t t e r.
y i el d
ot h e rs ,
W h ic h
un i t s
use?
development
cultures–
are
the
of
one
e nv i r o nme nta l pr ot e c t io n
partspe r
chemists
The
(MT ) ,
of
developed
from
set
manuf a cturi ng
hectare,
about
sp e a k s
have
vary
agreed
measureme nt
agricultural
talk
may
regardless
 nancialworld
per
societies
These
units
seven
of
base
International
the
SI
units.
set
transcends
system.
of
all
units
d’Unités
All
led
languages
other
(SI).
units
to
the
and
Ta b l e
are
1
shows
derived
1 . 2
T h e
m O l e
c O n c e p T
Accuracy and SI units
Continual
in
the
improvements
measurement
physical
constants
Bureau
of
initials
in
applications
of
Aeronautics
and
are
all
precision
have
changed
and
Space
in
the
units
the
from
over
instrumentation
that
time.
use
school
Administration
the
The
(known
correct
the
of
meant
Measures
monitors
science,
equivalent
in
SI
have
Weights
French)
of
as
of
values
units,
laboratory
(NASA),
some
International
BIPM
SI
used
of
to
SI
from
so
the
units
its
that
US
are
in
all
National
used
and
cases.
po ty
uit
Sybo
mass
kilogram
kg
temperature
kelvin
K
time
second
s
amount
mole
mol
electric current
ampère
A
luminosity
candela
cd
length
metre
m
Figure 1 A platinum–iridium cylinder
at the National Institute of Standards
and Technology, Gaithersburg, MD, USA,
represents the standard 1 kg mass
Sty tis
T
able 1 The seven base units of the SI system
Physical constants and unit
Table
2
shows
chemistry,
used
to
two
along
convert
quantities
with
SI
their
units
to
that
units.
a
are
used
Table
suitable
3
size
throughout
is
a
for
list
the
of
the
study
standard
application
of
conversions are available in
prexes
you
section 2 of the Data booklet.
are
The value of Avogadro’s
measuring.
constant (L or N
) will be
A
provided in Paper 1 questions,
moa o o a ia as at 273 K a
Aoao’s ostat (N
and may be referred to in the
)
Data booklet when completing
A
10 0 kpa
both Papers 2 and 3.
23
1
2
mol
6.02 × 10
2.27 × 10
3
m
1
mol
3
(= 22.7 dm
1
mol
)
T
able 2 Useful physical constants and unit conversions
px
Abbiatio
Sa
nano
n
micro
µ
milli
m
centi
c
deci
d
standard
–
kilo
k
10
mega
M
10
giga
G
10
Amount of substance: The mole
Chemists
to
need
control
processes
involved
and
such
in
companies
precise
All
of
understand
as
use
processing
amounts
of
of
of
constituent
aspects
reaction.
smelting
food
and
medicines
reacting
substances
all
the
electrolytic
synthesizing
chemical
their
to
make
are
atoms,
which
a
chemical
From
of
up
of
vary
drugs,
is
of
to
ability
crucial
the
and
in
order
6
10
industrial
industries
pharmaceutical
the
elements
in
reaction
large-scale
aluminium
beverages,
and
substances
made
of
9
10
to
3
10
2
10
measure
1
10
importance.
that
number
are
of
composed
1
3
protons,
6
neutrons,
and
electrons
(topic
2).
Chemists
use
a
system
to
measure
9
equal
are,
SI
amounts
which
unit,
of
allows
symbol
different
them
mol,
to
elements
calculate
dened
as
a
regardless
reacting
xed
of
how
big
quantities.
amount,
n,
of
a
their
The
atoms
mole
substance.
is
an
This
T
able 3 Useful prexes, their
abbreviations and scales
13
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
denition
can
be
applied
to
atoms,
molecules,
formula
units
of
ionic
Stoiioty uses the
compounds,
and
electrons
in
the
process
of
electrolysis.
quantitative relationships
This
xed
amount
is
a
number
of
particles
called
Avogadro’s
constant
between amounts of
23
(symbol
L
or
N
reactants and products in
)
and
it
has
a
value
of
6.02
×
1
10
mol
.
Avogadro’s
A
constant
enables
us
to
make
comparisons
between
chemical
species.
A
a chemical reaction. These
mole
of
any
chemical
species
always
contains
an
identical
number
of
relationships depend on the
representative
units.
law of conservation of mass
and denite propor tions. They
allow chemists to calculate
Relative atomic mass, relative formula mass,
the propor tions of reactants to
mix, and to work out expected
yields, from the ratios of
and molar mass
Isotopes
reactants and products
of
according to the balanced
topic
chemical equation.
2.1).
relative
occurs
The
a
rati
Atoi
are
protons
a
sample
of
atom
an
of
of
atom
of
atoms
of
is
the
nucleus
Isotopes
masses
of
atoms
the
abundance
in
single
A
in
an
of
each
isotope
equals
weighted
is
a
with
12
average
of
the
mass
measure
same
neutrons
of
number
(see
numbers.
the
sub-
The
percentage
that
4).
one
units.
of
have
different
(table
compared
that
numbers
have
element
carbon-12
a
element
different
element
the
are
same
but
the
another
The
on
a
relative
atomic
masses
scale
in
atomic
of
its
which
mass
isotopes
r
Isoto
aba
ass
and
in
their
relative
carbon
having
abundances.
an
A
of
The
12.01.
existence
The
of
relative
different
isotopes
molecular
mass
results
or
r
35
Cl
35.0
75%
relative
formula
mass
M
for
a
molecule
or
formula
unit
is
determined
r
37
Cl
37.0
25%
by
combining
the
A
values
of
the
individual
atoms
or
ions.
r
have
rati atoi ass A
no
units
as
A
and
M
r
they
are
both
r
ratios.
35.5

The
molar
mass
is
dened
as
the
mass
of
one
mole
of
a
substance.
It
T
able 4 The relative atomic mass of
–1
has
the
unit
of
grams
per
mole,
g
mol
(gure
2).
chlorine is the weighted average of
the atomic masses of its isotopes
and their relative abundance
Mg
24.31 g
58.44 g
23
6.02 ×
10
H
NaCl
18.02 g
23
6.02 ×
O
2
10
atoms
formula units
of Mg
of NaCl
23
6.02 ×
10
molecules
of H
O
2
Figure 2 The molar mass of a substance contains Avogadro’s number of representative
par ticles (the par ticles may be atoms, molecules, or ions)
TOK
Scientic discoveries are the product of many dierent ways of knowing
(WOK). To construct knowledge and understanding, scientists can use
intuition, imagination, reasoning, and even emotion, as well as detailed
investigation and analysis of large volumes of data that either suppor t or
disprove observations and hypotheses. Sometimes it can just be a matter
of serendipity. The scale of Avogadro’s constant (602 000 000 000 000 000
000 000) passes beyond the boundaries of our experience on Ear th. The
population of the planet is dwarfed by this number. How does this experience
limit our ability to be intuitive?
14
1 . 2
T h e
m O l e
c O n c e p T
13
Worked examples: A
and M
r
r
Al
Example 1
26.98
State
the
relative
atomic
mass
A
of
aluminium.
r
Figure 3 The element aluminium as
represented in the periodic table
Solution
Figure
A
3
shows
(Al)
=
the
periodic
table
entry
for
aluminium.
26.98
r
Example 2
Calculate
the
molar
mass
M
of
sulfuric
acid,
H
r
SO
2
4
nati iis a its
An ix or ow is a mathematical
Solution
notation that shows that a quantity
Table
5
shows
the
data
needed
to
answer
this
question.
or physical unit is repeatedly
multiplied by itself:
et
rati atoi
nb o
cobi
atos
ass/
2
m × m = m
ass A

A ati ix shows a reciprocal:
hydrogen
1.01
2
2.02
sulfur
32.07
1
32.07
oxygen
16.00
4
64.00
1
_
1
= x
x
1
_
3
dm
=
3
dm
cotatio (oaity): units
T
able 5
3
may be written as mol dm
, M, or
1
M
(H
r
SO
2
)
=
(2
×
1.01)
+
(1
×
32.07)
+
(4
×
16.00)
mol L
(US).
4
1
M
(H
r
SO
2
)
=
98.09
g
etay o taizatio: units
mol
4
1
are kJ mol
Example 3
Iitia at o atio: units are
3
Calculate
M
of
copper(II)
sulfate
pentahydrate,
CuSO
r
5H
4
mol dm
O.
1
s
2
Solution
Many
transition
molecules
metal
bonded
to
complexes
the
central
(sub-topic
metal
ion.
13.1)
The
contain
formula
water
CuSO
5H
4
shows
that
5
mol
et
of
water
combines
with
rati atoi
ass A
1
mol
of
copper(II)
O
2
sulfate.
nb o
cobi
atos
ass/
Sty tis

•
copper
63.55
1
63.55
sulfur
32.07
1
32.07
When adding and subtracting
numbers, always express
the nal answer to the same
oxygen
16.00
oxygen
16.00
4
number of decimal places as
64.00
the least precise value used.
5 × 1 = 5
80.00
•
hydrogen
1.01
5 × 2 = 10
When dividing or multiplying,
10.10
always express the answer to
T
able 6 Calculating the molar mass of copper(II) sulfate pentahydrate
the same number of signicant
gures as the least precise
–1
M
(CuSO
r
5H
4
O)
=
249.72
g
mol
value used.
2
15
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Qik qstio
piay staas
Calculate the molar mass of the
A iay staa is any substance of very high purity and large molar mass,
following substances and ions.
which when dissolved in a known volume of solvent creates a primary standard
solution.
a)
Mg(NO
)
3
2
Primary standard solutions are used in acid–base titrations to improve the
b)
Na
CO
2
3
accuracy of the nal calculation. The concentration of a primary standard can be
)
Fe
(SO
2
)
)
4
determined accurately.
3
S
8
)
Zn(OH)
Mole calculations
2
)
Ca(HCO
All
)
3
chemists,
industries,
)
whe the r
in
the
s ci e nt i c
c om m u n i t y,
m an u fac t u r in g
2
or
rese a r ch
facil i ti es ,
wor k
e ve r y
da y
wi t h
r ea c t i n g
I
2
quantities
)
MgSO
7H
4
O
[Al(H
O)
2
stoichiometric
P
s ubs ta nce s
calcul a ti ons .
The
a nd
so
n ee d
r el a t i on sh i p
to
pe r for m
be t wee n
t he
amou n t
(in
mol),
number
of
p a r ti cl e s,
a nd
the
mass
of
the
s a mp le
is
s u m m ar i z ed
]
6
in
j)
chem i ca l
2
3+
i)
of
gure
4.
O
2
5
×
Avogadro’s constant, L
×
molar mass
÷
molar mass
number of
particles
÷
Avogadro’s
constant, L
Figure 4 The relationship between amount, mass, and number of par ticles
Worked examples: mole calculations
➔
1.50
mol
of
glucose
contains
9
mol
of
C
atoms.
Example 1
number
Calculate
the
amount
(in
mol)
of
carbon
of
atoms
=
n(CO
)
in
a
sample
of
1.50
×
10
amount
(in
mol)
n
×
dioxide,
Avogadro's
23
constant,
L
molecules.
2
23
=
9
mol
=
5.42
×
6.02
×
10
1
mol
Solution
24
number
of
×
10
C
atoms
particles
___
amount
(in
mol)
n
=
Avogadro’s
constant,
L
Sty ti
Rearranging
and
substituting
values:
The answer is recorded to 3 signicant gures, as this is
the precision of the data given by the examiner (1.50 mol).
23
1.50 × 10
__
n(CO
)
=
2
23
6.02
=
×
0.249
1
10
mol
Example 3
mol
Calculate
Example 2
Calculate
in
1.50
the
mol
the
amount
(in
mol)
of
water
molecules
22
in
number
of
of
glucose,
carbon
C
H
6
atoms
contained
3.01
×
10
ethanedioic
formula
acid,
H
C
2
units
O
2
of
2H
4
hydrated
O.
2
O
12
6
Solution
Solution
➔
1
➔
molecule
carbon,
12
of
glucose
atoms
of
contains
hydrogen,
6
atoms
and
6
of
of
every
16
1
formula
1
mol
of
a
substance
of
particles.
oxygen.
number
➔
1
mol
of
glucose
contains
6
mol
of
C
unit
there
are
2
molecules
water.
atoms
➔
of
For
atoms.
contains
Avogadro’s
1 . 2
Therefore,
T h e
m O l e
c O n c e p T
Example 6
number
of
particles
___
amount
(in
mol)
n
=
Calculate
Avogadro’s
constant,
the
number
of
chlorine
atoms
in
a
L
6.00
mg
sample
of
the
anti-cancer
drug
cisplatin,
22
3.01 × 10
__
C
n(H
2
O
2
2H
4
O)
=
O)
=
2
×
0.0500
Pt(NH
mol
)
3
Cl
2
2
23
6.02
n(H
cis-diamminedichloroplatinum(II),
=
2
0.0500
×
mol
10
=
0.100
Solution
mol
2
uits
➔
First
➔
Next
the
Amount of substance n has the units mol
convert
nd
the
the
molar
mass
amount
in
in
mg
to
mol
g.
by
calculating
mass.
m
___
n =
➔
Finally
remember
that
there
are
2
mol
of
molar mass
chlorine
atoms
in
every
mol
of
cisplatin.
1
Mass m has the units g; molar mass has the units g mol
3
6.00
mg
n[Pt(NH
=
)
3
6.00
Cl
2
×
10
g
]
2
3
Example 4
6.00
×
10
g
____
=
Calculate
dioxide,
the
amount
(in
mol)
in
8.80
g
of
195.08
carbon
+
2(14.01)
+
6(1.01)
+
2(35.45)
CO
5
2
=
Solution
2.00
×
10
mol
5
n(Cl)
=
2
×
2.00
×
10
5
mol
=
4.00
×
10
mol
m
__
n(CO
)
5
=
number
2
molar
of
atoms
(Cl)
=
4.00
×
mol
10
×
mass
23
8.80
6.02
×
2.41
×
10
1
mol
g
___
=
19
=
1
12.01
=
0.200
+
2(16.00)
g
10
mol
mol
H
H
Cl
Example 5
Calculate
the
mass
in
g
of
0.0120
mol
of
sulfuric
N
acid,
H
SO
2
4
Pt
H
H
Solution
Calculate
the
molar
mass
of
H
SO
2
into
the
and
substitute
N
4
equation:
Cl
H
H
mass
(g)
=
n(H
SO
2
=
)
×
M
4
0.0120
(H
r
mol
×
SO
2
)
4
[2(1.01)
+
32.07
+
–1
4
(16.00)]
g
mol
Figure 5 The anti-cancer drug cisplatin
=
1.18
g
Qik qstios
1
Calculate the amount (in mol) in each of the following masses:
a)
8.09 g of aluminium
b)
9.8 g of sulfuric acid
)
25.0 g of calcium carbonate
)
279.94 g of iron(III) sulfate.
17
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
2
Calculate the mass (in grams) in each of the following:
a)
0.150 mol of nitrogen, N
2
b)
1.20 mol of sulfur dioxide, SO
)
0.710 mol of calcium phosphate, Ca
2
(PO
3
)
0.600 mol of ethanoic acid, C
H
2
3
)
4
2
O
4
2
Calculate the number of par ticles present in the following:
a)
2.00 mol of vanadium, V
b)
0.200 mol of sodium chlorate(VII), NaClO
4
)
72.99 g of iron(III) chloride, FeCl
)
4.60 g of nitrogen(IV) oxide.
3
Experimental empirical and molecular formula
determination
The
term
“empirical”
observation
and/or
laboratories
involved
or
food
production
compound
in
describes
information
investigation,
in
will
medical
often
processes
that
using
research
carry
may
out
be
that
is
scientic
and
through
Chemical
development,
analyses
either
derived
methods.
of
the
qualitative
manufacturing,
composition
or
of
quantitative
a
in
nature.
Qualitative
present
in
a
analysis
Quantitative
of
elements
The
of
or
elements
formula
in
Therefore
formula
(in
one
the
formula.
Table
for
ionic
the
within
the
the
shows
compounds
compound,
the
of
Sbsta
work
each
actual
or
some
to
is
one
is
a
of
simplest
present
atoms
of
the
is
the
relative
a
compound.
amount
ratio
as
of
(in
empirical
the
formula
formula
is
the
represents
same
the
moa oa
C
water
as
H
of
ions
eiia oa
CH
3
O
H
H
C
C
6
HO
2
H
4
O
2
O
2
O
8
H
12
C
2
O
6
ions in its empirical formula. It is used in some
18
formula
ratio
6
H
hydrogen peroxide
countries to enhance the health of teeth
empirical
the
simplest
2
glucose
of
molecular
6).
2
Figure 6 Sodium uoride, NaF has a 1:1 ratio of
mol)
respectively.
the
the
ratio
The
examples.
ethane
butanoic acid
masses
composition.
compound,
same
are
substance.
whole-number
in
or
the
the
exact
whole-number
formula
of
determine
the
mole
elements
purity
their
element
number
which
the
out
compound
of
unit
(gure
verify
chemists
to
formula
the
determining
also
empirical
since
structure
a
mol)
molecular
Sometimes
7
them
structural
formula.
For
is
on
could
enables
allows
amount
molecular
It
analysis
which
empirical
atoms
focuses
compound.
T
able 7 Some examples of molecular and empirical formulae
H
2
O
4
CH
2
O
1 . 2
T h e
m O l e
c O n c e p T
Worked examples: percentage composition by mass
You
the
can
use
molar
your
mass
percentage
by
understanding
of
a
compound
mass
of
of
to
elements
how
to
calculate
in
a
calculate
the
Sometimes
ratio
to
multiplication
whole
is
needed
to
convert
the
numbers:
compound.
example
1
1:1.25
Multiply
each
4(1):4(1.25)
side
≈
by4:
4:5
Example 1
example
Calculate
the
percentage
by
mass
of
sulfur
2
1:1.33
Multiply
each
3(1):3(1.33)
sulfuric
acid,
H
side
by3:
in
≈
3:4
SO
2
4
Sty ti
Solution
Empirical formulae are based on experimental
A
(S)
r
_
%
sulfur
=
×
M
(H
r
SO
100%
data; those for example 2 would likely have been
)
2
4
determined by a combustion reaction. The value
32.07
___
=
×
2(1.01)
×
(32.07)
×
100%
of 3.97 rather than 4 for hydrogen comes from
4(16.00)
experimental error.
=
If
you
but
have
you
mass
the
32.69%
of
a
compound
know
the
of
unknown
percentage
elements
empirical
molecular
the
present,
formula
and,
in
formula
composition
you
can
some
Example 3
by
calculate
cases,
Upon
analysis,
mass
formula.
0.25
of
g
194.13
of
oxygen.
Example 2
the
compound
that
empirical
formula
of
an
hydrogen
by
75 %
carbon
and
mol
hydrogen,
of
an
was
acid
8.0
g
the
found
of
sulfur,
empirical
8.0
_
=
mass.
=
1.0
=
n(C)
to
determine
the
ratio
of
the
=
1
=
4
=
1
0.25
_
=
0.25
1.01
to
g
and
0.25
0.25
_
is
16.0
1.0
_
=
Solution
step
molar
0.25
16.0
_
n(H)
and
formula
0.25
32.07
16.00
rst
a
contain
0.25
_
=
25%
n(O)
The
with
to
formula.
organic
n(S)
contains
sample
g
Determine
molecular
Determine
a
–1
the
0.25
of
n(H):
Therefore
the
empirical
formula
is
HSO
.
4
%
composition
__
relative
amount
of
substance
=
molar
To
mass
calculate
the
molecular
formula,
calculate
the
75
_
n(C)
=
=
6.24
empirical
formula
empirical
formulae
mass
and
determine
how
many
12.01
make
up
the
molar
mass.
25
_
n(H)
=
=
24.75
molar mass
___
1.01
empirical
Now
the
take
the
divisor
to
smallest
quotient
determine
the
(6.24).
lowest
Use
this
formula
mass
as
194.13
___
whole-number
=
ratio
of
the
194.13
_
=
1.01
+
32.07
+
=
2
97.08
4(16.00)
elements:
6.24
_
carbon
=
1
6.24
24.75
_
O
percentage
S
2
=
2
.
This
formula
of
compound
is
the
acid
called
is
2(HSO
)
or
peroxodisulfuric
8
3.97
acid
6.24
the
molecular
H
4
hydrogen
Because
The
composition
(gure
7).
O
is
O
O
experimentally
determined
it
is
acceptable
H
to
O
S
round
to
the
nearest
whole
number
if
the
number
S
is
close
to
a
whole
number.
Therefore
the
simplest
O
H
O
O
whole-number
ratio
of
carbon
to
hydrogen
is
O
1:4
Figure 7 Molecular model of peroxodisulfuric acid
and
the
empirical
formula
is
CH
.
4
19
1
S TO I c h I O m e T r I c
r e l AT I O n S h I p S
1.3 rati asss a os
Understandings
Applications and skills
➔
Reactants can be either limiting or excess.
➔
The experimental yield can be dierent from
➔
quantities, limiting and excess reactants,
the theoretical yield.
➔
Solution of problems relating to reacting
theoretical, experimental, and percentage yields.
Avogadro’s law enables the mole ratio of
➔
reacting gases to be determined from volumes
Calculation of reacting volumes of gases using
Avogadro’s law.
of the gases.
➔
➔
The molar volume of an ideal gas is a constant
graphs involving the relationship between
at specied temperature and pressure.
➔
temperature, pressure, and volume for a xed
mass of an ideal gas.
The molar concentration of a solution is
determined by the amount of solute and the
➔
volume of solution.
➔
Solution of problems and analysis of
Solution of problems relating to the ideal gas
equation.
A standard solution is one of known
➔
concentration.
Explanation of the deviation of real gases from
ideal behaviour at low temperature and high
pressure.
➔
Obtaining and using experimental values to
calculate the molar mass of a gas from the ideal
gas equation.
➔
Solution of problems involving molar
concentration, amount of solute, and volume of
solution.
➔
Use of the experimental method of titration to
calculate the concentration of a solution by
reference to a standard solution.
Nature of science
➔
Making careful obser vations and obtaining evidence for scientic theories – Avogadro’s initial
hypothesis.
Stoichiometry
A
balanced
reactants
and
also
chemical
and
the
equations
relative
also
include
(see
topic
of
examining
the
understanding
chemical
20
to
of
relative
this
reactions,
the
success
is
provides
their
amounts
may
the
linked
are,
reaction
of
of
equation
products
of
reactants
specic
5).
in
and
of
the
protability
is
reactants
data
the
and
processes
percentage
of
the
about
their
products.
quantitative
industrial
particularly
and
symbols,
Stoichiometry
amounts
vital
information
chemical
on
what
state
the
of
matter,
Chemical
the
enthalpy
quantitative
products.
where
yield,
organization.
An
the
is
method
efciency
directly
1 . 3
From
a
balanced
chemical
equation
the
coefcients
r e A c T I n g
can
be
m A S S e S
A n d
v O l u m e S
interpreted
TOK
as
the
ratio
equation
of
for
the
the
amount,
reaction
in
mol,
used
for
of
reactants
the
and
products.
manufacture
of
This
ammonia
is
in
the
the
When comparing the eight
Haber
process
(see
topic
7):
areas of knowledge (AOK),
N
(g)
+
3H
2
(g)
⇋
2NH
2
(g)
∆H
=
−92.22
Mathematics involves
kJ
3
knowledge and understanding
It
shows
that
one
molecule
of
nitrogen
gas
and
three
molecules
of
of the highest certainty. The
hydrogen
gas
combine
in
an
exothermic
reaction
to
produce
two
Nature of Science (NOS)
molecules
of
ammonia.
However,
when
setting
up
a
reaction
the
reactants
informs us that experimental
may
not
always
be
mixed
in
this
ratio
–
their
amounts
may
vary
from
the
data is often quantitative
exact
stoichiometric
amounts
shown
in
the
balanced
chemical
equation.
and mathematical analysis
is required to enable precise
descriptions, predictions
The limiting reagent
and, eventually, laws to be
Experimental
designers
of
industrial
processes
use
the
concept
of
a
developed. Mathematics
limiting
reagent
as
a
means
of
controlling
the
amount
of
products
is an integral part of
obtained.
The
limiting
reagent,
often
the
more
expensive
reactant,
will
scientic endeavours. The
be
completely
consumed
during
the
reaction.
The
remaining
reactants
use of numbers and an
are
present
in
amounts
that
exceed
those
required
to
react
with
the
understanding of the mole
limiting
reagent.
They
are
said
to
be
in
excess.
concept have helped develop
It
is
the
Using
reagent
measured,
specic
here
limiting
is
calculated
amounts
that
identical
to
that
of
the
determines
amounts
products
to
of
be
the
amount
limiting
obtained.
the
experimental
or
the
theoretical
predicted
or
the
actual
yield
of
yield
of
reagent
The
formed.
This
is
science. “Why is mathematics
made
achieved
products.
Chemistry into a physical
enables
assumption
products
of
products
so eective in describing the
is
natural world?”
rarely
the
IB Diploma Chemistry Syllabus
case.
as
Much
this
effort
equates
to
is
focused
increased
on
improving
prots
and
the
yield
efcient
use
of
industrial
of
raw
processes,
materials.
Worked example: determining the limiting reagent
In
the
manufacture
of
phosphoric
m
_
acid,
n(O
)
=
2
molten
and
elemental
then
phosphorus
hydrated
according
is
to
M
oxidized
the
following
100.0
g
__
=
chemical
=
2(16.00)
P
(l)
+
5O
4
(g)
+
6H
2
O(l)
→
4H
2
PO
3
24.77
g
of
phosphorus
reacts
with
g
mol
mol
(aq)
P
4
(l)
+
4
If
3.125
1
equation:
100.0
g
5O
(g)
+
6H
2
O(l)
→
4H
2
PO
3
(aq)
4
of
______________________________
oxygen
and
excess
water,
determine
the
limiting
1
M(g
reagent,
the
amount
in
mol
of
phosphoric(V)
mol
)
123.88
32.00
acid
______________________________
produced
in
g,
of
(the
theoretical
phosphoric
yield)
and
the
mass,
acid.
m/g
24.77
100.0
excess
______________________________
Solution
n /mol
0.200
3.125
excess
0
i
The
amount
in
mol
of
phosphorus
and
oxygen
______________________________
is
determined
using
the
working
method
from
n /mol
sub-topic1.2:
______________________________
f
To
m
_
n(P
)
determine
the
amount
of
oxygen
that
will
=
4
M
react
24.77
with
the
phosphorus
we
can
use
a
cross-
g
__
=
multiplication
=
0.2000
technique:
mol
1
4(30.97)
g
mol
21
1
S TO I C H I O M E T R I C
P
:
R E L AT I O N S H I P S
O
4
P
2
1
:
5
0.200
:
α
(s)
+
5O
4
(g)
+
6H
2
→
O(l)
4H
2
PO
3
(aq)
4
_________________________________
1
M(g mol
)
123.88
32.00
_________________________________
1
×
α
=
0.2000
×
α
=
0.2000
×
5
m/g
24.77
100.0
excess
5
_
_________________________________
1
n /mol
0.2000
3.125
excess
0
i
α
=
1.000
mol
_________________________________
Therefore
0.2000
mol
of
phosphorus
requires
n /mol
0.0
2.125
excess
0.8000
f
1.000
mol
of
oxygen
to
completely
react.
There
is
_________________________________
3.125
and
mol
of
oxygen
phosphorus
is
available
the
limiting
so
this
is
in
reagent.
excess
All
the
The
mass
of
phosphoric
acid,
H
PO
3
phosphorus
will
be
consumed
in
the
reaction
produced
be
determined
by
multiplying
n
by
M
f
3.125
1.000
=
2.125
mol
of
oxygen
will
the
reaction
comes
to
:
r
remain
m
after
can
4
and
=
M
×
n
=
[3(1.01)
completion.
1
The
limiting
reagent
dictates
the
amount
×
phosphoric
to
acid
determine
times
the
produced.
the
amount
amount
produced
in
compared
mol
The
of
of
with
mole
product,
ratio
in
the
amount
is
acid
of
30.97
+
4(16.00)]
g
mol
0.8000
mol
=
78.40
g
used
mol.
phosphoric
+
of
Four
will
be
phosphorus:
This
value
represents
phosphoric
achieved
in
acid.
the
theoretical
Theoretical
yields
yield
are
of
rarely
practice.
Qik qstios
1
Butane lighters work by the release and combustion
CO
)
Calculate the mass, in g, of K
)
Calculate the mass, in g, of O
2
of pressurized butane:
produced.
3
produced.
2
2C
H
4
(g) + 13O
10
(g) → 8CO
2
(g) + 10H
2
O(l)
4
A solution of 155 g of potassium iodide, KI is added to
2
a solution of 175 g of nitric acid, HNO
Determine the limiting reagent in the following
. The acid acts
3
as an oxidizing agent.
reactions:
a)
20 molecules of C
H
4
6KI(aq) + 8HNO
and 100 molecules of O
10
(aq) → 6KNO
3
+ 3I
b)
10 molecules of C
H
4
)
0.20 mol of C
H
4
)
8.72 g of C
4
(s) + 4H
2
and 91 molecules of O
10
O(I)
2
2
a)
Deduce which reagent is in excess.
b)
Determine how many grams of this reactant will
and 2.6 mol of O
10
H
(aq) + 2NO(g)
3
2
2
and 28.8 g of O
10
2
remain unreacted.
2
Two aqueous solutions, one containing 5.3 g of sodium
)
Determine how many grams of nitrogen
carbonate and the other 7.0 g of calcium chloride, are
monoxide, NO will be produced.
mixed together. A precipitation reaction occurs:
5
Na
2
CO
(aq) + CaCl
3
(aq) → 2NaCl(aq) + CaCO
2
Chlorine gas is produced by the reaction of
(s)
3
hydrochloric acid, and the oxidizing agent
Determine the limiting reagent and the mass, in g, of
manganese(IV) oxide, MnO
:
2
precipitate formed (the theoretical yield).
MnO
(s) + 4HCl(aq) → MnCl
2
3
(aq) + Cl
2
(g) + 2H
2
O(l)
2
The oxygen required in a submarine can be produced
At 273.15 K and 100 kPa, 58.34 g of HCl reacts with
by a chemical reaction. Potassium superoxide, KO
2
3
0.35 mol of MnO
reacts with carbon dioxide, CO
to produce oxygen and
to produce 7.056 dm
of chlorine gas.
2
2
potassium carbonate, K
2
.
Write the balanced chemical equation for this reaction.
b)
28.44 g of KO
reacts with 22.00 g CO
2
a)
Deduce the limiting reagent.
b)
Calculate the theoretical yield of chlorine.
3
a)
the limiting reagent.
22
CO
2
. Deduce
1 . 3
r e A c T I n g
m A S S e S
A n d
v O l u m e S
Theoretical and experimental yields
The
balanced
what
is
chemical
theoretically
carried
out
under
equation
possible
ideal
●
represents
when
conditions.
a
reaction
It
allows
changes
is
amount
of
products
to
be
reaction
calculated
–
reverse
of
in
industry
reactions
and
work
maximize
to
maximize
prots.
the
the
yield
reactions
conditions
and
However,
especially
in
processes,
many
products.
factors
result
in
a
These
factors
could
of
in
loss
●
impurity
products
from
reaction
of
side-reactions
due
to
the
impurities.
calculate
made
the
percentage
between
the
yield
theoretical
a
comparison
yield
and
the
reduced
amount
produced
in
the
process
–
the
include:
experimental
●
of
products
under
actual
of
consuming
largeis
yield
as
systems
existence
presence
To
experimental
scale
such
yield.
●
Scientists
conditions,
pressure
the
equilibrium
theoretical
and
the
●
expected
in
temperature
yield:
vessels
experimental
yield
__
of
%
reactants
yield
=
×
theoretical
100%
yield
Worked example: determining theoretical yield
Respirators
concern
are
for
being
used
workplace
environmental
increasingly
safety
pollution.
and
rising
Iodine(V)
with
Step
levels
oxide,
I
of
2:
5
I
O
2
with
carbon
monoxide,
CO
and
can
be
remove
I
O
2
this
(s)
+
5CO(g)
I
g
with
of
I
(g)
from
+
5CO
of
:
CO
:
5
0.3000
(g)
:
α
1
×
α
=
0.3000
×
α
=
0.3000
×
α
=
1.500
5
5
33.6
5
_
g
1
mol
carbon
The
reaction
of
0.3000
mol
of
I
O
2
and
requires
1.50mol
5
given
of
an
limiting
5
Calculate
dioxide
the
2
theoretical
yield
determine
air:
O
2
CO.
gas
2
reacts
the
→
5
100.0
of
poisonous
ratios,
used
1
to
mole
O
2
reacts
Using
reagent.
CO
for
completion.
However,
only
1.20
mol
of
CO
experimental
is
yield,
0.900
in
mol,
mol
available;
therefore
this
is
the
limiting
reagent.
of
CO
The
,
ratio
of
limiting
reagent
CO
to
product
CO
2
2
calculate
is
the
percentage
5:5
or
1:1.
The
number
of
mol
of
CO
Figure 1 A chemist wearing a
yield.
that
is
2
theoretically
possible
It
that
is
therefore
1.2
mol.
respirator for safety
was
found
0.90
mol
or
39.61
g
of
CO
was
2
Solution
produced.
Step
1:
Calculate
the
initial
amount
in
mol
and
determine
the
limiting
is
the
experimental
yield.
of
To
reactants
This
determine
the
percentage
yield
of
CO
we
rst
2
reagent:
need
to
calculate
the
theoretical
yield
of
CO
:
2
m
_
n(I
O
2
)
=
m = M × n
5
M
1
100.0
g
=
[12.01
=
52.8
+
2(16.00)]
g
×
mol
1.20
mol
_
_
_
=
1
2(126.90)
=
0.2996
+
5(16.00)
g
mol
mol
g
Then:
m
_
n(CO)
experimental
=
yield
__
%
M
yield
×
=
theoretical
33.6
100%
yield
g
_
_
_
=
1
12.01
+
16.00
g
39.61
mol
g
_
=
×
52.8
=
100%
=
75.0%
g
1.20mol
23
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Qik qstios
1
Acetylsalicylic acid, also known as aspirin, C
H
9
synthesized by reacting salicylic acid, C
H
7
acetic anhydride, C
H
4
C
H
7
O
6
(s) + C
3
H
4
O
6
O
6
6
O
8
O
is
3
3
9
H
8
O
(s) + C
4
H
2
3
O
4
(l)
(g)
2
produced.
2
3
can be produced in the following
Sulfur trioxide, SO
3
H
CH
+
O(l) + CO
2
If a 1.68 g sample of sodium hydrogen carbonate is
O
O
O
(s) + H
3
heated, calculate the mass, in g, of sodium carbonate
(l) → C
H
CO
2
:
3
3
O
(s) → Na
2NaHCO
4
with
two-step reaction:
O
O
+ H
O
C
3
OH
CH
OH
4FeS
O
3
(s) + 11O
2
(g) + O
2SO
CH
(g) → 2Fe
2
2
O
2
(g) → 2SO
2
(s) + 8SO
3
(g)
2
(g)
3
3
30.0 g of iron disulde (pyrite), FeS
reacts in the
2
a)
Calculate the theoretical yield, in g, of aspirin
presence of excess oxygen to completion.
when 3.0 g of salicylic acid is reacted with 4.0 g of
a)
Calculate the theoretical yield, in g, of sulfur
acetic anhydride.
trioxide.
b)
If the experimental yield of aspirin is 3.7 g,
b)
If an experimental yield of 28.0 g of sulfur
calculate the percentage yield.
trioxide is achieved, deduce the percentage
2
yield.
The thermal decomposition of sodium hydrogen
results in a 73.8% yield of sodium
carbonate, NaHCO
3
carbonate, Na
CO
2
:
3
Avogadro’s law and the molar volume of a gas
The
kinetic
behaviour
number
to
hold.
1
theory
of
of
gases
are
distances.
2
a
Gaseous
gases
or
postulates
made
Most
up
of
particles
is
a
model
microscopic
postulates
These
Gases
of
at
assumptions
to
The
that
explain
theory
must
be
is
and
predict
based
true
for
upon
the
the
a
theory
are:
of
very
the
are
used
level.
small
volume
particles,
occupied
constantly
separated
by
moving
in
a
gas
is
by
large
empty
straight
lines,
space.
but
random
directions.
3
Gaseous
walls
4
SI
unit
of
pressure
particles
the
Gaseous
Under
The
of
particles
conditions
is
obeys
these
Many
theory.
undergo
container.
of
No
exert
no
standard
postulates
and
elastic
loss
of
force
collisions
kinetic
of
attraction
temperature
the
with
energy
equations
and
that
each
other
and
the
occurs.
on
other
pressure,
follow
gases.
an
from
ideal
the
gas
kinetic
–2
the
pascal
other
are
units
of
N
m
.
forces
pressu re
commonly
different
the
(Pa),
used
countries,
atmosphere
high
in
inclu d ing
gas
At
of
degree
of
Hg),
torr,
of
separation
per
and
the
and
low
gas
they
pressure,
molecules
act
in
a
the
is
way
signicance
minimized
that
–
adheres
of
any
there
to
the
is
a
ideal
at
high
pressure
and
low
temperature
the
particles
of
a
gas
mercury
bar,
more
slowly
and
the
distances
between
the
particles
decrease.
an d
Intermolecular
pounds
between
model.
move
(mm
temperature
(atm ) ,
However,
millimetres
high
attraction
square
attractions
(sub-topic
4.4)
become
signicant
and
in ch
eventually
the
gas
can
liquefy.
These
responses
to
changing
conditions
5
(psi).
The
bar
(10
Pa)
is
now
mean
widely
used
as
a
that
behaviour
unit ,
as
it
is
very
atmospheric
clos e
pressure,
of
can
real
depart
from
ideal
gas
behaviour
and
exhibit
the
gases.
to
1
a tm .
The
early
terms
and
24
gases
convenient
by
postulates
scientists
Joseph
Louis
of
the
such
as
kinetic
Robert
Gay-Lussac.
theory
Boyle,
were
Edme
explained
Mariotte,
in
quantitative
Jacques
Charles,
1 . 3
In
1806,
Gay-Lus s a c
volumes
ratio
of
of
reacting
whole
p r opo se d
gases
a nd
tha t
the
the
r e A c T I n g
re l at i on s h ip
pro duc t s
c o ul d
bet we en
be
m A S S e S
A n d
v O l u m e S
the
e xpr e ss e d
as
a
Sty tis
numb e r s .
Physical
There
and
are
examine
the
many
Avogadro’s
gaseous
behaviour
property
of
particles
Imagine
gases
a
of
taking
and
same
(0
is
are
s imp le
inate
K)
(gure
as
a
and
a pp ly.
the
force
2).
equal
balloon.
an d
th e ir
the
gas
used
i mpo rt a n t
exerted
by
a
conversions
la w s
be h a vi ou r
m o de ls
An
pressure
These
to
the
Under
particular
the
the
and
to
e xp la i n
gas
as
Data
volume
found
p hys i c al
of
in
can
be
booklet.
an
fo u nd
The
ideal
section
u nit
in
molar
gas
is
2.
its
and
different
will
and
STP
Cl
2
1
of
have
pressure
pressure,
O
2
2
1
16.05 g mol
of
conditions
balloons
temperature
N
1
mass
same
the
temperature
4
4.00 g mol
molar
(100kPa)
CH
1
2.02 g mol
T he
to
standard
He
2
r e a c t io ns
unde rs t a n d
numerically
°C/273
H
to
and
s ur fa ce .
to
known
us
pressure ,
a
mass
g as - phase
qua nti ta ti vel y.
are
i ts
each
volume
conditions
s y s te m s
w i th
a
using
e na bl e
gases
gas
collide
temperature
the
imp o r ta nt
la w
constants
1
28.02 g mol
1
32.00 g mol
70.90 g mol
Figure 2 The molar volume of any gas is identical at a given temperature and pressure
Figure 3 Amedeo Avogadro
(1
776–1856) proposed in 1811
3
At
STP
the
balloo ns
w i ll
hav e
id ent ic a l
v ol u m e s
of
22. 7
1
dm
mol
.
that equal volumes of any gas
This
is
the
molar
volume
of
an
id ea l
g as
and
it
is
constant
at
a
at the same temperature and
given
temperature
a nd
p re s sur e .
Eac h
bal lo on
c on t a in s
1
mol
of
the
pressure contain the same number
23
gas
so
it
contains
6.02
×
10
atoms
or
molecu l e s
of
the
g a s.
T his
of molecules
relationship
measured
number
at
The
3
H
8
the
s a me
law
as
Avogadro’s
te mpe r a tur e
simplies
coefcients
correspond
C
kno wn
law :
a nd
equal
pr es s ur e
volumes
c o n t ai n
of
the
a ny
gas
s a me
ofmolecul e s .
Avogadro’s
gases.
is
(g)
to
the
+
of
ratio
stoichiometric
a
of
balanced
volumes
calculations
chemical
of
the
5O
2
involving
equation
gases
(gure
reacting
involving
gases
4).
(g)
2
+
4H
O(l)
2
Figure 4 Volumes of gases obey Avogadro’s law
25
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Worked examples: Avogadro’s law
Qik qstio
Ammonium carbonate
Example 1
decomposes readily when
3
Calculate
n(O
heated:
)
found
in
a
6.73
dm
sample
of
oxygen
gas
at
STP
.
2
3
(NH
)
4
2
CO
(s) → 2NH
3
1
(g)
mol
O
(g) + H
2
occupies
22.7
dm
at
STP
2
3
+ CO
O(l)
2
Solution
Determine the volume, in
3
3
dm
6.73 dm
_
, of the individual gases
n(O
)
=
=
2
0.296
mol
3
22.7
produced on decomposition
dm
of 2.50 mol of ammonium
carbonate.
Example 2
The
hydrogenation
gas,
H
The
product
of
ethyne,
C
H
2
in
the
presence
of
a
involves
reaction
with
hydrogen
2
nely
divided
nickel
catalyst
at
150
°C.
2
C
H
2
(g)
is
+
ethane,
2H
2
(g)
C
→
H
:
2
6
C
H
2
2
(g)
6
3
When
100
3
cm
of
C
H
2
volume
and
reacts
with
250
cm
of
H
2
,
determine
the
2
composition
of
gases
in
the
reaction
vessel.
Solution
According
to
2molecules
Avogadro’s
of
law,
hydrogen,
for
every
1molecule
1molecule
of
ethane
of
will
ethyne
be
and
formed.
3
Looking
at
the
volumes
reveals
that
only
200
cm
of
the
hydrogen
is
3
required,
of
gases
and
that
contains
100
both
cm
of
ethane
ethane
C
and
H
2
(g)
will
be
formed.
unreacted
+
2H
2
(g)
→
The
nal
mixture
hydrogen:
C
2
H
2
(g)
6
3
initial
volume,
V /cm
100
250
0
0
50
100
i
3
nal
volume,
V /cm
f
3
After
reaction
there
will
be
3
50cm
150
cm
of
gases
in
the
vessel
comprising
3
of
H
and
100
cm
of
C
2
H
2
6
The gas laws
The
gas
mass
of
laws
gas
are
in
a
series
changing
of
relationships
conditions
of
that
predict
temperature,
the
behaviour
pressure,
and
of
a
xed
volume.
3
You
at
have
STP)
seen
is
that
Avogadro’s
independent
of
the
law
states
that
composition
of
the
the
molar
volume
(22.7
dm
gas.
Boyle’s law
Robert
Boyle
constant,
Gases
(1627–1691)
inverse
contained
collisions
The
an
with
in
the
relationship
discovered
relationship
smaller
surface
between
volumes
of
the
∝
or
V
p
1
V
26
=
1
V
p
2
2
when
between
will
have
container,
pressure
1
_
p
that
exists
p
and
so
the
temperature
pressure
an
increased
exert
volume
and
V
a
number
higher
can
be
remains
volume.
of
pressure.
expressed
as:
1 . 3
where
V
and
p
1
the
nal
represent
the
initial
volume
and
r e A c T I n g
pressure
and
m A S S e S
V
1
and
and
pressure,
v O l u m e S
p
2
volume
A n d
2
respectively.
aP/P ,erusserp
aP/P ,erusserp
3
–3
volume, V/dm
1/V / dm
Figure 5 Boyle’s law: the pressure of a gas is inversely propor tional to the volume at constant temperature
Worked example: Boyle’s law
A
helium-lled
weather
balloon
is
designed
to
rise
to
altitudes
as
high
3
as
37
000
of
101
m.
kPa
is
atmospheric
A
balloon
released
pressure
with
and
is
a
volume
rises
68
kPa.
to
an
of
5.50
altitude
Calculate
dm
of
the
and
3500
new
a
m
pressure
where
the
volume,
3
in
dm
.
remain
It
is
assumed
that
the
temperature
and
amount,
in
mol,
constant.
Solution
First
make
p
=
a
101
summary
of
the
data:
kPa
1
3
V
=
5.50
dm
1
p
=
68
kPa
2
3
V
=
α
dm
2
Making
V
the
subject
of
the
expression:
2
p
1
_
V
=
V
2
×
1
p
2
101 kPa
_
3
=
5.50
dm
×
68
kPa
3
=
8.17
dm
Charles’s law
Jacques
the
Charles
(1746–1823)
temperature
mass
of
gas
at
a
and
volume
constant
investigated
of
a
gas.
pressure,
the
He
the
relationship
discovered
volume
V
of
that
the
between
for
gas
is
a
xed
directly
27
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
proportional
Absot zo
can
be
to
the
expressed
absolute
temperature
T
in
kelvin.
This
relationship
as:
We saw in sub-topic 1.1
V
V
1
2
_
V
that absolute zero is zero
∝
T
_
or
=
T
T
1
2
on the kelvin scale, 0 K
(
273.15 °C). The idea of
When
an
inated
negative temperatures and
(boiling
point
the existence of a minimum
The
possible temperature had
frequency
been widely investigated
balloon
by the scientic community
room
gaseous
is
balloon
196
°C),
particles
and
then
placed
and
removed
the
into
average
collide
energy
temperature
is
the
it
with
the
begins
from
the
balloon
a
energy
internal
to
deate
liquid
will
container
kinetic
wall
–
of
the
nitrogen
of
of
liquid
the
the
balloon
volume
and
nitrogen
particles
decreases.
with
reduces.
allowed
to
If
less
the
return
to
reinate.
before Lord Kelvin’s time
(1824–1907). Kelvin stated
that absolute zero is the
temperature at which molecular
motion ceases. According to
Charles’s law, if the temperature
of a system was to double from
10 K to 20 K, the average kinetic
energy of the par ticles would
double and the volume would
correspondingly double.
Figure 6 Reducing the temperature reduces the average kinetic energy of the
par ticles of a gas, and the volume reduces
V ,emulov
Worked example: Charles’s law
3
A
glass
gas
ice-cold
syringe
water
contains
over
the
76.4
outside
cm
of
the
of
a
gas
gas
at
27.0
syringe,
°C.
the
After
running
temperature
3
of
the
gas
reduces
to
18.0
°C.
Calculate
temperature, T (K)
occupied
by
the
gas.
Figure 7 Charles’s law: the volume of a
gas is directly proportional to absolute
Solution
temperature at constant pressure
3
V
=
76.4
cm
=
27.0
+
1
T
273.15
=
300.15
K
273.15
=
291.15
K
1
3
V
=
α
cm
=
18.0
2
T
+
2
V
V
1
2
_
_
=
T
T
1
2
V
×
1
T
2
_
=
V
2
T
1
3
=
28
74.1
cm
the
new
volume,
in
cm
,
1 . 3
r e A c T I n g
m A S S e S
A n d
v O l u m e S
Gay-Lussac’s law
Collaboration
Having
established
proportional
is
directly
laws
volume
proportional
remaining
constant
to
gas
at
to
relationship
stating
that
constant
temperature
temperature
involves
pressure
at
pressure
is
and
constant
and
inversely
that
The
volume
pressure,
temperature,
is
the
scientic
highly
Evidence
at
to
volume.
collaborative.
that
(1778–1850)
work
with
ideal
gases
led
him
to
that
when
the
volume
of
a
gas
is
constant,
the
the
gas
is
directly
proportional
to
its
absolute
temperature.
can
be
expressed
by
investigate
_
or
8
K),
no
kinetic
pressure.
walls
of
increased
As
that
energy
the
the
general
laws.
of
when
the
temperature
container
with
the
ideal
temperature
gas
particles
increases,
increased
the
force
reaches
is
zero
particles
and
absolute
and
it
collide
frequency,
zero
exerts
with
P erusserp
the
the
developing
2
demonstrates
the
of
T
1
(0
to
=
T
Figure
new
2
_
1
T
scientists
understanding
as:
p
p
∝
other
new
possibility
p
often
and
The
and
relationship
is
tested,
pressure
develop
of
fundamental
the
utilized
understanding
is
understanding
challenged,
Gay-Lussac’s
community
causing
pressure.
The combined gas law
absolute
The
three
gas
laws,
Charles’s
law,
Boyle’s
law,
and
Gay-Lussac’s
law,
are
zero, 0 K
combined
of
gas,
in
the
one
law
called
relationship
the
combined
between
gas
temperature,
law.
For
pressure,
a
xed
and
amount
volume
is:
temperature T
p
V
p
1
2
Figure 8 Gay-Lussac’s law: the pressure of
_
1
_
V
2
=
T
T
a gas is directly propor tional to absolute
2
1
temperature at constant volume
The ideal gas equation
The
ideal
volume,
gas
established
both
of
a
that
pressure
gas
these
equation
temperature,
and
pressure
and
the
describes
and
and
volume
amount
the
of
a
relationship
amount,
volume
have
gas
a
in
are
direct
mol,
between
of
gas
inversely
the
ideal
Having
proportional
relationship
particles,
pressure,
particles.
with
gas
the
and
that
temperature
equation
combines
interrelationships:
pV
=
nRT
T as ostat a t its o t ia as qatio
–1
R is called the as ostat and it has a value of 8.31 J K
–1
mol
. This value is
provided in section 2 of the Data booklet
The inclusion of R in the ideal gas equation requires the following units: p (Pa),
3
V (m
–3
), and T (K). Note that 1 Pa = 1 J m
; this allows you to see how the units in
the ideal gas equation are balanced:
–3
p(J m
3
1 dm
3
) × V(m
–1
) = n(mol) × R(J K
–3
= 1 × 10
–1
mol
) × T (K)
3
m
29
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
TOK
Worked example: using the ideal gas
The ideal gas equation is a
equation to calculate volume
model which is the product
3
Calculate
the
volume,
in
m
,
of
a
balloon
lled
with
0.400
mol
of
of a number of assumptions
hydrogen
gas
at
a
temperature
of
22.90
°C
and
a
pressure
of
1.20
Pa.
about the ideal behaviour
of gases. These have been
Solution
discussed earlier in the
–1
topic. Scientic models
Convert
all
data
to
SI
units
to
enable
the
use
of
R
as
8.31
J
K
–1
mol
are developed to explain
p
=
1.20
Pa
observed behaviour. In the
3
V
=
α
m
n
=
0.400
R
=
8.31
T
=
22.90
development of models what
role do imagination, sensory
mol
perception, intuition, or the
1
J
K
1
mol
acquisition of knowledge in
the absence of reason play?
+
273.15
=
296.05
K
nRT
_
V
=
p
ra ass deviate from
1
1
0.400 mol × 8.31 J K
mol
× 296.05 K
____
ideal behaviour at very low
=
1.20
Pa
temperature and high pressure.
3
Under these conditions
=
820
m
the forces between the gas
par ticles become signicant,
and the gas gets closer to the
Worked examples: determining the molar
point where it will condense
from gas to liquid.
mass of a substance
An
organic
hydrogen,
compound
and
A
oxygen
containing
was
only
the
elements
carbon,
analysed.
Example 1: Empirical formula
A
was
being
found
to
oxygen.
contain
54.5%
Determine
the
C
and
9.1 %
empirical
H
by
formula
mass,
of
the
the
remainder
compound.
[3]
Solution
54.5
_
n(C)
=
4.54
_
=
4.54
12.01
9.1
_
n(H)
=
9.0
=
4
=
1
2.28
_
=
2.28
2.28
16.00
empirical
≈
2.28
36.4
_
The
2
9.0
_
=
1.01
n(O)
≈
2.28
formula
is
C
H
2
O.
4
Example 2: Relative molecular mass
3
A
0.230
at95
30
°C
g
sample
and
102
of
A
kPa.
when
vaporized
Determine
the
had
a
relative
volume
of
molecular
0.0785
mass
of
dm
A.
[3]
1 . 3
r e A c T I n g
m A S S e S
A n d
v O l u m e S
Solution
pV
=
nRT
n
=
m/M
pV
=
mRT
_
M
1
0.230
mRT
_
M =
g
×
8.31
J
1
K
mol
×
368
K
____
=
1
=
3
pV
102
×
3
10
Pa
×
0.0785
×
10
87.9
g
mol
3
m
Example 3: Molecular formula
Figure 9 A homogeneous mix ture
is characterized by a constant
Determine
the
molecular
formula
of
A
using
your
answers
composition throughout
from
parts
(a)
and
(b).
[1]
Solution
molar mass
___
87.8
___
=
empirical
formula
2(12.01)
mass
+
4(1.01)
+
(16.00)
87.8
_
≈ 2
=
44.06
molecular
formula
=
C
H
4
IB,
Nov
O
8
2
2005
Concentration
In
a
typical
solution
solutions
A
a
of
solution
solvent.
the
laboratory
rather
known
is
a
The
solvent
than
is
the
in
majority
the
gaseous
reactions
phase.
carried
Chemists
out
need
are
to
in
make
up
concentrations.
homogenous
solute
water
of
is
the
mixture
usually
a
solution
of
solid,
is
a
solute
but
could
described
as
that
be
an
has
a
been
liquid
aqueous
or
dissolved
gas.
in
When
To make up solutions of known
solution
concentration, volumes must
be measured accurately.
The
molar
concentration
of
substance
of
a
solution
is
dened
as
the
amount
(in
mol)
Apparatus used to do this
3
a
dissolved
in
1
dm
3
of
solvent.
1
dm
=
1litre
(1
L).
include burettes, pipettes and
volumetric asks.
amount of substance n/mol
___
3
concentration
c/mol
dm
=
3
volume
of
solution
V/dm
Par ts per million (ppm) is not an SI unit but is often used
uits o otatio
for very dilute concentrations such as when measuring
Units of concentration include:
pollutants (see sub-topic 9.1).
–3
●
mass per unit volume, g dm
●
mol per unit volume, mol dm
–3
Concentration in mol dm
may also be referred to as oaity,
–3
and square brackets are sometimes used to denote molar
–2
6
●
parts per million (ppm): one par t in 1 × 10
par ts.
concentration, for example [MgCl
] = 4.87 × 10
–3
mol dm
2
3
1 ppm = 1 mg dm
31
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Worked examples: concentration calculations
Example 1: Molarity of solution
Solution
3
Calculate
the
concentration,
in
mol
dm
,
of
a
n(C
H
8
solution
formed
when
0.475g
of
O
5
K)
=
V
×
[C
4
H
8
O
5
K]
4
magnesium
3
1 dm
_
3
chloride,
MgCl
is
completely
dissolved
in
water
=
to
250
cm
×
–3
×
1.25
mol
dm
3
2
1000
cm
3
make
a
solution
with
a
volume
of
100cm
=
0.313
mol
Solution
m
=
H
n(C
8
First
calculate
n(MgCl
O
5
K)
×
M
4
):
2
=
0.475
m
_
n(MgCl
)
0.313
=
mol
×
[8(12.01)
+
g
5(1.01)
+
–1
_
_
_
4(16.00)
+
39.10]
g
mol
=
2
1
M
24.31
+
2(35.45)
g
mol
=
63.9
g
3
=
4.99
×
mol
10
Example 4: Concentration of standard
3
Convert
the
volume
in
3
cm
to
dm
:
solution
3
1 dm
_
3
100
cm
×
A
3
=
0.1
standard
solution
is
prepared
by
dissolving
dm
3
1000
3
5.30g
cm
of
sodium
carbonate,
Na
CO
2
in
250
cm
of
3
3
distilled
Calculate
the
concentration
of
the
sample
n
_
]
in
a
volumetric
ask.
A
10.0
cm
solution:
3
[MgCl
water
of
this
pipetteand
solution
diluted
is
with
removed
water
to
by
the
bulb
nal
volume
4.99 × 10
mol
__
=
=
3
2
3
of
V
0.1
0.100dm
.
Calculate
the
concentration,
in
dm
–3
mol
2
=
4.99
×
10
dm
,
of
the
diluted
solution.
3
mol
dm
Solution
Example 2: Concentration of ions
3
First
calculate
n(Na
CO
2
)
in
a
10.0
cm
sample
of
3
3
Determine
chloride
the
ions
concentration,
in
example
1
in
mol
dm
of
the
the
standard
solution:
above.
3
10.0 cm
_
m
_
n(Na
CO
2
)
=
×
3
3
M
250
Solution
cm
5.30
g
____
When
solid
MgCl
is
dissolved
in
water,
the
=
1
2
2(22.99)
constituent
ions
are
+
12.01
+
3(16.00)
g
mol
liberated:
3
10.0 cm
_
2+
MgCl
(s)
→
Mg
(aq)
+
2Cl
(aq)
×
2
3
250
cm
3
2(4.99
n
_
[Cl
]
=
×
10
mol)
__
=
=
0.00200
mol
3
V
0.1
dm
Finally
2
=
9.98
×
10
calculate
the
concentration
3
mol
dm
–3
solution
in
mol
dm
:
Example 3: Mass of solute
n
_
[Na
CO
2
Calculate
the
phthalate,
C
mass,
H
8
O
5
K
in
(a
g,
of
potassium
primary
]
=
3
V
standard)
3
32
3
of
a
dm
in
=
cm
0.100
4
3
250
0.00200 mol
__
=
3
hydrogen
1.25
mol
dm
solution.
0.0200
mol
dm
of
the
diluted
1 . 3
r e A c T I n g
m A S S e S
A n d
v O l u m e S
Titrations
Quantitative
analysis
includes
a
range
of
laboratory
techniques
used
An
to
determine
the
amount
or
concentration
of
an
analyte.
The
results
aayt
that
expressed
as
numerical
values
with
is
involving
of
analysis
two
known
a
is
until
the
monitored
quantitative
A
concentration
concentration
progress
is
solutions.
titration
which
is
chemical
through
being
a
substance
analysed
by
a
units.
given
Volumetric
is
are
technique
involves
added
to
reaction
colour
a
is
a
used
solution
of
complete.
changes
by
standard
using
analytical
proced u re.
chemists
solution
unknown
The
reaction
indicators
(topic8).
A staa sotio or iay
Qik qstios
3
1
Calculate the mass, in g, of H
SO
2
required to prepare 500 cm
3
sotio is prepared using a
of a 2.0 mol dm
4
volumetric ask. Solvent is
solution of sulfuric acid.
added to a high purity sample
2
is to be used in the laboratory during an
A solution of aluminium bromide, AlBr
3
until the level of the solution
3
electrolyte investigation. Calculate the total number of ions present in 2.5 dm
of a
reaches the mark on the ask.
3
1.6 mol dm
solution of AlBr
3
Worked example: acid–alkali titration
Sty ti
When solving quantitative
calculation
problems involving
3
Calculate
the
volume,
in
dm
3
,
of
0.390
mol
dm
potassium
concentrations and volumes
3
hydroxide,
KOH
solution
that
will
neutralize
25.0
cm
of
0.350
mol
of solutions, the focus is on
–3
dm
sulfuric
acid,
H
SO
2
2KOH(aq)
+
H
SO
2
the amount, in mol, of the
4
(aq)
→
K
4
SO
2
(aq)
+
2H
4
substances reacting and their
O(l)
2
relationship as shown by the
Solution
mole ratios in the balanced
chemical equation.
Step
1:
Calculate
the
amount,
in
mol,
of
H
SO
2
n(H
SO
2
)
=
c
×
=
0.350
=
8.75
:
4
V
4
–3
mol
dm
3
×
0.0250
dm
–3
×
10
mol
–3
Step
2:
The
mole
ratio
of
acid:alkali
is
1:2.
Therefore
–3
of
acid
reacts
with
2(8.75
×
10
8.75
×
10
mol
–2
mol
)
=
1.75
×
10
mol
of
KOH.
not
Step
3:
Calculate
the
volume
of
KOH:
In topic 9, we will introduce a
n
V
general, simple-to-use formula.
=
c
This formula can also be used for
2
the type of volumetric chemistry
1.75 × 10
mol
__
V(KOH)
=
3
=
0.0449
dm
3
0.390
mol
dm
question shown above.
33
1
S TO I C H I O M E T R I C
R E L AT I O N S H I P S
Questions
1
Epsom
used
salts
as
(magnesium
bath
salts.
sulfate)
However,
the
are
commonly
f)
Determine
anhydrous
Fe(NH
)
4
form
the
of
the
water
salt
of
is
a
drying
hydration
of
agent.
To
Epsom
determine
salts,
a
2.50
value
of
the
evaporating
Bunsen
were
salt
dish
burner
was
and
gently
ame
observed.
placed
until
Table
8
in
a
shows
over
further
the
changes
May
3
The
mol,
of
hence
the
2
x.
equation
results.
CO
mass of evaporating basin
24.10
+
of
3H
2
[2]
that
for
is
→
reaction
CH
OH
be
occurring
in
the
is:
+
H
3
the
can
a
methanol
2
What
xH
in
and
2008
synthesis
mass/
4
2
O
a
dsitio
mass of evaporating basin + MgSO
4
amount,
·xH
porcelain
heated
no
2
of
)
g
IB,
sample
the
(SO
O
2
maximum
formed
amount
from
2
mol
of
of
methanol
carbon
26.60
O
2
dioxide
mass of evaporating basin after heating
and
3
mol
of
hydrogen?
25.32
A.
1
mol
B.
2
mol
C.
3
mol
D.
5
mol
IB,
May
T
able 8
a)
Calculate
the
evaporated
b)
Calculate
of
H
mass,
from
the
in
the
g,
of
water
sample.
amount
amount,
in
[1]
mol,
2006
O.
2
c)
Calculate
the
mass,
in
g,
d)
Calculate
the
amount,
of
MgSO
4
in
mol,
of
4
MgSO
Calcium
carbonate
decomposes
on
heating
as
4
e)
Calculate
MgSO
:
the
ratio
amount
of
of
amount
H
4
value
O
and
shown
below.
CaCO
→
of
deduce
the
2
of
When
f)
State
CaO
the
formula
of
the
hydrated
2
g
of
50
g
of
The
value
of
x
in
Fe(NH
)
4
found
mol
of
in
the
was
)
4
the
xH
2
in
added.
water
The
O
can
oxide
carbonate
yield
A.
7%
B.
25%
C.
50%
D.
75%
IB,
November
of
are
formed.
are
decomposed,
calcium
What
is
the
oxide?
2
amount
compound.
wasdissolved
(aq)
(SO
2
determining
sulfate
sample
BaCl
by
calcium
calcium
percentage
be
CO
salt.
7
2
+
3
x
A
in
0.982
and
g
excess
precipitate
[1]
of
2
BaSO
was
separated
and
dried
and
found
to
2006
4
weigh
a)
1.17
g.
Calculate
the
amount,
in
mol,
of
BaSO
4
5
Ethyne,
C
H
2
in
the
1.17
g
of
precipitate.
the
b)
Calculate
the
amount,
in
,
reacts
with
oxygen
according
mol,
of
sulfate
in
equation
below.
What
volume
3
(in
dm
)
of
0.982
g
sample
of
Fe(NH
)
4
(SO
2
)
4
·xH
2
with
0.40
dm
of
C
H
2
2C
Calculate
0.982
g
the
amount,
sample
of
in
Fe(NH
mol,
)
4
d)
Determine
present
Fe(NH
)
4
(i)
e)
in
(SO
2
iron
Use
the
the
the
your
g
·xH
2
in
g
g,
in
of
the
iron
in
·xH
O.
2
(g)
2
A.
0.40
B.
0.80
+
5O
(g)
→
2
from
of
(d)
water
Fe(NH
)
+
2H
O(g)
2
[1]
2
following
sulfate.
to
)
4
[3]
determine
present
(SO
2
(g)
2
C.
1.0
D.
2.0
IB,
November
of
(iii)
part
4CO
the
2
of
H
2
[1]
O:
mol
sample
)
4
sample
4
34
(SO
2
ammonium
answer
amount
0.982
)
4
(ii)
mass,
0.982
of
?
2
O.
[1]
c)
oxygen
3
reacts
2
the
to
2
[2]
in
·xH
2
the
O.
2
[2]
2007
Q u e S T I O n S
6
A
xed
mass
of
an
ideal
gas
has
a
volume
of
10
A
toxic
gas,
A,
consists
of
53.8 %
nitrogen
3
800cm
under
certain
conditions.
The
pressure
and
46.2%
carbon
by
mass.
At
273
K
and
5
(in
kPa)
What
with
is
and
the
other
temperature
volume
of
conditions
(in
the
K)
gas
are
after
remaining
both
these
the
doubled.
1.01
changes
Determine
same?
×
3
10
Calculate
Pa,
1.048
the
the
200
cm
B.
800
cm
C.
1600
cm
D.
3200
cm
IB,
May
2005
3
IB,
specimen
An
oxide
of
empirical
molar
anddetermine
3
A.
g
its
A
occupies
formula
mass
of
the
molecular
paper
of
462
cm
.
A.
compound
structure.
[3]
2009
3
3
[1]
11
of
of
hydrogen
Assuming
complete
reaction,
what
volume
of
were
After
gas
apparatus
7
copper
hydrogen.
was
had
was
reduced
heating,
the
maintained
cooled.
The
in
a
stream
stream
until
of
the
following
results
obtained.
–3
0.200
mol
dm
potassium
hydroxide
solution,
3
KOH(aq)
is
required
to
neutralize
25.0
–3
of
0.200
mol
dm
aqueous
sulfuric
Mass
of
Mass
of
SO
2
dish
dish
and
=
13.80
contents
g
before
heating
=
acid,
21.75
H
empty
cm
g
(aq)?
4
Mass
of
dish
and
contents
after
heating
and
3
A.
12.5
cm
B.
25.0
cm
C.
50.0
cm
D.
75.0
cm
IB,
May
leaving
to
cool
=
20.15
g
3
a)
Explain
why
the
stream
of
hydrogen
gas
was
3
maintained
until
the
apparatus
cooled.
[1]
3
[1]
b)
2007
Calculate
the
of
using
copper
complete
c)
8
Copper
metal
may
be
produced
by
the
Write
an
empirical
the
reduction
equation
formula
data
of
for
of
above,
the
the
oxide
assuming
oxide.
the
[3]
reaction
that
reaction
occurred.
of
copper(I)
oxide
and
copper(I)
d)
according
to
the
below
equation.
State
two
O
+
Cu
2
A
→
6Cu
+
kg
of
of
IB,
2
10.0
kg
copper(I)
of
copper(I)
sulde
was
oxide
reaction
heated
until
Determine
as
that
it
would
was
be
observed
heated.
[2]
November
2004
no
0.502
g
of
an
alkali
metal
sulfate
is
dissolved
occurred.
in
a)
tube
and
12
further
the
SO
2
mixture
5.00
S
changes
[1]
inside
2Cu
[1]
sulde
the
limiting
reagent
in
water
and
excess
barium
chloride
solution,
this
BaCl
(aq)
is
added
to
precipitate
all
the
2
reaction,
showing
your
working.
[3]
sulfate
ions
as
barium
sulfate,
BaSO
(s).
The
4
b)
Calculate
that
of
the
could
maximum
be
obtained
mass
from
of
copper
these
precipitate
masses
reactants.
0.672
May
Calculate
organic
of
compound
carbon,
remainder
24.1%
being
A
contains
by
mass
of
62.0 %
the
amount
Determine
nitrogen,
the
c)
Determine
metal
hydrogen.
the
percentage
by
mass
hydrogen
and
the
empirical
formula
(in
mol)
of
barium
Dene
c)
The
the
term
relative
molecular
molecular
mass
of
A
of
A.
mass.
is
the
November
molecular
formula
(in
mol)
of
the
present.
[1]
the
Deduce
the
molar
and
mass
state
identity
of
its
of
the
alkali
units.
the
alkali
[2]
your
metal,
workings.
[2]
[3]
Write
an
equation
for
the
precipitation
[2]
including
state
symbols.
[2]
116.
IB,
Determine
amount
sulfate
sulfate
reaction,
relative
[2]
of
e)
b)
the
metal
showing
IB,
weighs
by
d)
a)
and
formed.
Determine
alkali
mass
dried
2006
b)
An
and
g.
sulfate
9
ltered
[2]
a)
IB,
is
of
A.
May
2007
[1]
2006
35
1
S TO I C H I O M E T R I C
13
Aspirin,
the
one
world,
equation
of
the
can
be
given
R E L AT I O N S H I P S
most
widely
prepared
used
drugs
according
to
in
B.
the
A
student
times,
below.
and
OH
of
out
three
obtained
Mass
OCOCH
carried
with
the
brass
=
this
following
0.456
g
±
brass
three
nails,
results.
0.001
g
+
O
H
CH
COOH
3
C
COOH
H
Tit
1
CH
O
3
Iitia o o
aspirin
excess
solid
aspirin
ethanoic
salicylic
O
2
acid
anhydride.
0.00
0.00
28.50
28.60
28.40
28.50
28.60
28.40
0.10 0 o 
2–
some
0.00
–3
ethanoic acid
S
reacted
with
3
COOH
ethanoic anhydride
student
2
C
C
3
salicylic acid
A
small
3
O
A.
experiment
identical
3
(± 0.05 
)
3
fia o o
Impure
–3
was
obtained
by
ltering
0.10 0 o 
the
2–
reaction
mixture.
Pure
aspirin
was
O
S
obtained
2
by
recrystallization.
Table
9
shows
the
3
(± 0.05 
)
3
data
vo a o
recorded
by
the
student.
–3
0.10 0 o 
2–
O
S
Mass of salicylic acid used
3.15 ± 0.02 g
Mass of pure aspirin obtained
2.50 ± 0.02 g
2
3
(± 0.10 
)
3
Aa o
28.50
a o 0.10 0 o
–3
2–
S

Table 9
O
2
3
3
(± 0.10 
i)
Determine
salicylic
the
acid,
amount,
C
H
6
ii)
Calculate
the
in
mol,
(OH)COOH,
used.
[2]
Table 10
4
theoretical
yield,
in
g,
i)
ofaspirin,
C
H
6
(OCOCH
4
)COOH.
Calculate
the
[2]
Determine
the
percentage
yield
of
S
O
ii)
the
gures
Calculate
number
associated
of
the
aspirin
with
obtained,
percentage
this
the
mass
Calculate
and
Another
iv)
Calculate
uncertainty
obtained
150%.
an
The
repeated
the
14
May
Brass
uses.
checked
found
no
is
a
the
percentage
of
samples
of
analysis
are
is
carried
copper
brass.
[1]
alloy
out
present
The
shown
to
in
reactions
with
many
determine
three
the
identical
involved
in
this
below.
+
Step
1:
Cu(s)
+
2HNO
(aq)
+
2H
(aq)
→
3
2+
Cu
(aq)
+
2NO
(g)
+
2H
2
O(l)
2
2+
Step
2:
4I
(aq)
+
2Cu
(aq)
→
2CuI(s)
+
I
(aq)
2
2
Step 3: I
(aq) + 2S
2
36
O
2
of
3
mass
of
[1]
copper
the
percentage
in
[1]
by
mass
in
the
brass.
[1]
The
manufacturers
2
(aq) → 2I
of
brass
claim
contains
that
the
44.2 %
mass.
Determine
the
copper
percentage
the
result.
copper-containing
analysis
the
copper
in
errors.
2009
An
mol,
brass.
error
IB,
in
brass.
yield
IB,
Commenton
the
experiment
experimental
teacher
and
in
[2]
student
calculations
amount,
associated
mass.
by
of
the
of
sample
and
[2]
calculate
v)
v)
in
3.
3
present
iii)
of
with
step
signicant
the
pure
amount,
in
[1]
copper
State
added
of
pureaspirin.
iv)
average
2
mol,
3
2
iii)
)
of
(aq) + S
O
4
6
(aq)
May
2010
the
result.
[1]
2
AT O M I C
ST R U CT U R E
Introduction
Australian-born
British
Lawrence
(1890–1971)
1915
Bragg
Nobel
Prize
in
physicist
Physics
with
A
William
shared
his
wrong
than
the
no
theory
theory
is
at
always
so
much
father,
William
Sir
William
crystal
the
structures
development
William
on
of
Henry
Bragg
record,
only
1969
is
using
of
the
having
25.
Bragg
In
a
Bragg,
for
their
X-rays,
X-ray
analysis
which
led
received
Nobel
the
tape-recorded
to
prize
at
the
Chemistry
science”
laureate
interview
Lawrence
Bragg
of
crystallography.
youngest
better
all.
age
in
said:
theory.
terms
is
sometimes
and
at
Every
of
the
chemical
atoms.
In
described
centre
this
various
theories
current
understanding
and
of
reaction
topic
models
of
as
the
chemistry
we
can
that
the
be
shall
“central
lies
atomic
explained
examine
have
led
structure
of
to
in
the
our
the
atom.
2.1 T  to
Understandings
Applications and skills
➔
Atoms contain a positively charged dense
A
➔
Use
of
the
nuclear
sy m bol
notation
X
to
Z
nucleus composed of protons and neutrons
deduce
the
number
of
proto ns ,
neu trons,
an d
(nucleons).
electrons
➔
in
atoms
and
ions .
Negatively charged electrons occupy the space
➔
Calcu lations
involving
non- integer
relative
outside the nucleus.
atomic
➔
The mass spectrometer is used to determine
from
masses
given
and
data,
a bu nd a nce
inclu d ing
of
ma ss
is otope s
s pectra.
the relative atomic mass of an element from its
isotopic composition.
Nature of science
➔
Evidence and improvements in instrumentation – alpha par ticles were used in the development of the
nuclear model of the atom that was rst proposed by Rutherford.
➔
Paradigm shifts – the subatomic par ticle theory of matter represents a paradigm shift in science that
occurred in the late 1800s.
37
2
ATO M I C
S T R U C T U R E
Background to atomic theory
Two
Greek
philosophers,
Democritus
stated
wascomposed
of
Leucippus
around
440
indivisible
BC
contain
and
that
particles
by
matter
However,
no
concrete
scientic
The
law
of
given
to
support
accepted
to
any
this
hypothesis
great
degree
by
and
the
at
so
it
1808
the
developed
supported
formed
an
the
schoolteacher
atomic
model
experimental
origin
of
of
matter
data.
atomic
John
This
theory
of
destroyed.
of
a
modern
this
a
not
only
science.
We
shall
that
see
scientic
model
over
was
progressively
called
comprise
●
Postulate
very
Matter
total
mass
cannot
of
be
matter
the
is
start
equal
of
the
to
the
total
reaction.
an
to
be
accepted
explanation
but
correctly
outcomes
should
be
able
of
to
it
should
known
the
of
future
predict
experiments.
to
deduce
was
Dalton
used
his
theory
another
law:
underpins
in
this
The
law
of
rened
multiple
proportions :
If
two
topic
X
and
Y
combine
in
different
ways
and
form
more
than
one
compound,
the
time.
the
matter
summarized
theory
provide
masses
Dalton
matter:
reaction
before
observations
to
replaced
chemical
matter
elements
how
of
The
Dalton
model
that
For
●
much
oxygen
thetime.
English
by
parts
was
Dalton’s atomic theory
In
3
scientic
mass
community
conservation
or
following
not
and
evidence
created
was
sulfur
termed
●
atomos.
1part
mass.
as
1:
small
indivisible
“atoms”.
building
Dalton’s
blocks
theory
that
can
of
be
Y
of
can
whole
X
be
that
combine
expressed
as
with
a
a
ratio
xed
of
mass
small
numbers.
follows.
All
matter
particles
(materials)
called
consists
of
Example: The law of multiple
atoms
propor tions
●
Postulate
one
2:
type
An
element
consists
of
atoms
of
only.
Chemists
as
a
an
●
Postulate
3:
Compounds
consist
of
atoms
example
than
one
element
and
are
formed
atoms
in
whole-number
of
of
the
the
time
did
amount
law
of
not
of
use
the
mole
substance.
multiple
As
proportions,
measuring
the
mass
of
carbon
and
oxygen
by
in
combining
Dalton’s
of
consider
more
in
measure
forming
the
two
compounds
carbon
monoxide,
ratios.
CO(g),
and
carbon
dioxide,
CO
(g).
An
experiment
2
●
Postulate
4:
In
a
chemical
reaction
atoms
are
might
not
The
created
simple
“laws
wereknown
1800s
and
or
to
The
by
law
a
that
3
g
of
carbon
combines
with
destroyed.
of
chemical
the
Dalton’s
scientic
theory
combination”
community
explains
a
in
the
4
g
of
oxygen
3
g
of
carbon
to
form
have
number
to
form
carbon
combines
carbon
combined
with
dioxide.
in
monoxide,
8
g
Carbon
different
ratios
of
oxygen
and
to
whereas
oxygen
give
different
compounds:
ofthese.
●
measure
of
denite
French
proportions:
scientist,
This
Joseph
was
Proust,
proposed
in
CO(g):
C:O
ratio
=
3:4
CO
C:O
ratio
=
3:8
1799.
(g):
2
The
law
states
that
a
compound
always
has
The
the
same
proportion
of
elements
by
mass.
ratio
with
example,
if
you
measure
the
mass
of
of
sulfur
the
same
in
sulfur
trioxide,
SO
it
will
masses
mass
of
of
oxygen
carbon
to
that
form
combine
the
two
and
compounds
oxygen
the
For
is
1:2
(a
simple
ratio
of
whole
numbers).
always
3
Stdy tip
In science, a w can be considered a summary of several observations.
38
2 . 1
T h e
n u c l e a r
TOK
a T O m
Qik qstio
Deduce the ratio of the mass
John Dalton was a brilliant scientist. He never married and said: “My head
of oxygen per gram of sulfur in
is too full of triangles, chemical proper ties, and electrical experiments to
the compounds sulfur dioxide,
think much of marriage!” He was an multidisciplinary scientist, who worked
SO
(g), and sulfur trioxide,
2
in the disciplines of physics, mathematics, biology, and philosophy, as well
SO
(g).
3
as chemistry. Do you think philosophy still has a place in modern scientic
thinking? Debate this question in class and consider why scientists should
always try to embrace an interdisciplinary approach in their thinking.
Dalton was colour blind and saw himself as being dressed in grey clothes.
His only known pastime was bowling. Could the wooden balls on the bowling
green have inuenced his theories of the atom? How impor tant is the work–life
balance for the scientic practitioner or indeed for society as a whole?
atoms of element X
atoms of element Y
compound consisting
of elements X and Y
Figure 1 Schematic showing some of the principles of Dalton’s
theory. Examine each of the four postulates and discuss each one
in relation to the three representations shown here
Thomson’s “plum-pudding” model of the atom
Although
Dalton’s
theory
did
“What
is
not
the
1808
answer
atom
postulates
one
had
merit,
fundamental
composed
his
question:
the
atom
eaten
with
of?”
on
was
negatively
embedded
It
was
almost
another
100
years
before
One
to
of
physicist
at
the
gather
the
rst
J.J.
evidence
leaders
Thomson
Cavendish
Cambridge,
UK.
to
in
eld
this
was
(1856–1940),
laboratory
In
answer
the
1906
at
the
Thomson
in
a
to
a
day
charged
positive
plum
in
the
pudding
UK
particles
region
and
(like
(the
(a
dessert
Ireland),
raisins)
“pudding”)
of
scientists
the
began
similar
Christmas
question.
the
who
English
worked
University
won
atom.
the
+
of
Nobel
+
Prize
in
Physics
for
the
discovery
of
the
+
electron.
+
+
Thomson
worked
on
cathode
rays,
which
he
“pudding” of positive
+
suggested
consist
of
very
small
negatively
+
charged
+
charge spread over
negatively
particles
called
electrons.
+
+
the entire sphere
charged par ticles
The term “electron” was originally proposed by the Irish
scientist George Johnstone Stoney in 1891.
Figure 2 Thomson’s “plum-pudding” model of the atom. In
the analogy, raisins represent negatively charged par ticles
embedded in a pudding of positive charge. Overall there is a
balance between the positive and negative charges since the
Thomson
proposed
what
is
now
termed
the
atom is electrically neutral
“plum-pudding”
model
of
the
atom
–
he
said
that
39
2
aTO m I c
S T r u c T u r e
Rutherford’s gold foil experiment
Most
gold
Thomson’s
model
raised
a
number
of
of
the
foil
matter
is
electrically
neutral,
of
negatively
charged
particles
in
that
very
they
must
particles.
The
also
contain
search
for
for
a
New
more
detailed
Zealand
model
physicist
of
Ernest
the
sli g ht l y
some
p a r ti cle s
we re
as
de  e c t e d
back
angle s
a nd
towar d s
so me
the
e ve n
so ur ce .
bo un c e d
Th e s e
the
had
colli d e d
hea d- on
wi th
wha t
we
particles
know
to
be
t he
nucleus
in
the
gold
atom
atom
( gure
led
th ro u gh
positively
these
now
and
But
large
particles
charged
we nt
de  e cte d
atoms
straight
implies
were
the
by
presence
p a r ti cle s
some
questions.
expected.
Because
alpha
and
4) .
Ruther f o r d
d e scr ibe d
thi s
r es ult
by
Rutherford
commenting:
(1871–1937)
foil
and
experiment
experiment
model
by
tested
placing
evacuated
co-workers
in
1909.
to
Thomson’s
a
thin
chamber
gold
and
conduct
Published
in
the
1911,
gold
It
this
“plum-pudding”
metal
foil
bombarding
it
in
was
as
15inch
itcame
an
straight
particles
( gure3).
Alpha
( α)
as
shell
back
if
at
you
a
and
had
piece
hit
red
of
a
paper
and
you!
with
Rutherford
alpha
incredible,
artillery
based
his
explanation
on
the
fact
that
particles
the
gold
foil
consists
of
thousands
of
gold
atoms.
2+
are
high-energy,
positively
charged
He
ions
When
emitted
from
elements
naturally
such
as
occurring
the
beam
of
positively
charged
alpha
radioactive
particles
bombarded
particles
passed
the
foil
the
majority
of
the
radium.
atom
consists
through
mainly
of
unde ected,
empty
space.
since
the
However,
some alpha par ticles are deected
at
(scattered) at large angles
the
core
of
the
atom
lies
a
dense
region
of
most alpha par ticles
positive
charge
called
the
nucleus.
When
an
are undeected
alpha
gold
particle
atom
it
when
it
hit
initial
path.
came
close
de ected
the
to
the
through
nucleus
it
nucleus
a
large
re ected
of
a
angle,
back
and
along
its
par ticle deected
thin gold foil
beam of par ticles
+
+
+
deected
par ticle
+
lead block containing
circular zinc sulde
a source of radioactive
uorescent screen
beam of alpha
+
+
deected
alpha par ticles
par ticles
par ticle
+
Figure 3 Rutherford’s experiment. The zinc sulde uorescent
screen was used to detect alpha par ticles that had passed
+
through or been deected by the gold foil
+
+
The
results
of
Rutherford’s
ground-breaking
at
the
experiment
time.
Based
were
on
Thomson’s
+
model
alpha
of
the
atom,
particles
Rutherford
would
have
expected
sufcient
that
energy
+
the
to
+
pass
directly
mass
that
that
the
their
40
made
alpha
direction
would
his
through
involve
results
up
the
the
particles
on
were
gold
a
astonishing.
He
the
of
reected par ticle
predicted
decelerate
through
minor
distribution
atoms.
would
going
only
uniform
gold
deection.
and
gold atoms
that
foil
However,
Figure 4 Rutherford’s model, which explains his ndings in
the gold foil experiment
2 . 1
T h e
n u c l e a r
a T O m
T s i si
The scale of the atom
The word “nucleus” means the
Rutherford’s
work
has
formed
the
basis
of
much
of
our
thinking
on
central and most important part
the
structure
of
the
atom.
Rutherford
is
rumoured
to
have
said
to
of an object. The word is used
hisstudents:
in both chemistry (the nucleus
All
science
is
either
physics
or
stamp-collecting!
of an atom) and biology (the
nucleus of a cell).
The
is
vast
hard
space
to
in
fully
rugby-playing
and
on
looking
the
centre
represent
The
relative
diameter
the
of
an
nucleus.
sub-topic
the
the
volume
The
of
the
centre
stand.
open
is
a
pitch
space
in
occupies
a
to
the
the
atom
of
row
at
you
and
4
the
scale
is
a
great
of
6)
seats
the
would
nucleus.
vast
of
nucleus
(gure
placed
and
times
the
top
gure
volume
100000
idea
the
the
is
in
the
Zealand
were
and
of
stadium
grape
model
tiny
New
Park
grape
the
atomic
size
from
electron
approximately
return
Eden
small
the
tiny
native
at
the
the
between
between
nucleus
shall
of
If
Rutherford’s
atom
We
being
distance
of
to
Rutherford’s
Imagine
distance
representation
unrealistic.
at
of
eld,
the
compared
appreciate.
tier
the
atom
nation.
down
upper
of
the
our
is
atom
the
of
simple
obviously
and
the
diameter
the
atom
of
in
Figure 5 The 100 New Zealand dollar note,
issued in 1999, shows a picture of Lord
2.2.
Rutherford, reecting his immense contribution
Atoms
themselves
are
extremely
small.
The
diameter
The
unit
of
most
atoms
to science. Do any bank notes in your own
10
is
in
the
range
1
×
10
10
to
5
×
10
m.
used
to
describe
the
country have pictures of famous scientists?
dimensions
of
atoms
is
the
picometre,
pm:
12
1
In
is
pm
=
X-ray
the
10
m
crystallography
angstrom,
a
symbol
commonly
used
unit
for
atomic
dimensions
Å:
10
1
For
Å
=
10
m
example,
the
atomic
radius
of
the
uorine
atom
is
quoted
in
12
section
toÅ
9
we
given
of
the
can
Data
use
booklet
as
60
dimensional
×
10
m
analysis,
(60
using
pm).
the
To
convert
conversion
this
factors
above:
12
10
m
_
60
pm
×
1 Å
_
1
×
=
0.60
Å
=
6.0
×
10
Å
10
1
pm
10
m
Figure 6 Eden Park, Auckland, New Zealand
Can we see atoms and are they real?
All
the
models
However,
be
seen.
working
for
In
1981
at
IBM
tunnelling
generates
This
The
for
have
two
in
discussed
people
physicists,
Zurich,
microscope
allowed
scientists
Prize
in
the
ground-breaking
are
Gerd
ability
in
an
images
to
1986
assumed
only
and
invented
electron
of
atoms
when
Heinrich
the
at
the
to
real.
can
Rohrer,
scanning
that
atomic
individual
awarded
are
they
microscope
surfaces
observe
was
that
“real”
Binnig
Switzerland
(STM) ,
Physics
have
objects
three-dimensional
Nobel
their
we
many
atoms
Binnig
level.
directly.
and
Rohrer
work.
41
2
ATO M I C
S T R U C T U R E
TOK
us so
A simulation of Rutherford’s
The American theoretical physicist Richard Feynman (1918–1988) said:
gold foil experiment has been
If ... all of scientic knowledge were to be destroyed, and only one
developed by PhET at the
sentence passed on to the next generation... I believe it is that all things
University of Colarado, Boulder,
are made of atoms.
USA and is available on their
Are the models and theories that scientists create accurate descriptions of the
website.
natural world, or are they primarily useful interpretations for the prediction,
http://phet.colorado.edu/
explanation, and control of the natural world?
No
subatomic
do
we
use
to
par ticles
interpret
ca n
be
ind ire ct
directly
obs er v ed.
evidence,
ga i ned
Which
ways
throu g h
t he
of
know ing
use
of
technology?
Subatomic particles and descriptions of the atom
After
Rutherford’s
followed
in
having
much
a
Atoms
the
consist
more
of
●
the
proton
●
the
neutron
●
the
electron.
Section
4
of
the
coulombs,
C,
very
and
small
these
of
masses
experiment
period
Data
types
booklet
each
of
1909
picture
of
a
number
1935,
of
the
subatomic
gives
these
atomic
(table
in
approximately
detailed
three
the
to
the
mass
unit,
experiments
structure
of
in
the
scientists
atom.
particle:
mass,
subatomic
of
culminating
kg,
and
the
particles.
in
The
masses
amu,
is
a
charge
convenient
in
given
unit
are
for
1).
–24
1
amu
=
1.660539
Sbtoi
×
10
g
c
mss/
+1
∼ 1
lotio
p ti
proton
neutron
0
electron
nucleus
nucleus
∼ 1
1
__
1
outside the nucleus in the
1836
electron cloud
T
able 1 A comparison of the subatomic par ticles
The
neutron
1932
(gure
Chadwick’s
which
was
discovered
by
British
physicist
James
Chadwick
in
7).
discovery
beryllium,
of
Be,
the
neutron
placed
in
a
was
vacuum
based
on
chamber
an
experiment
was
in
bombarded
with
2+
alpha
particles,
to
emit
to
prove
He
neutrons
,
emitted
and
categorically
based
that
from
on
the
polonium.
Chadwick’s
particles
were
Figure 7 British physicist Sir James Chadwick
gamma
rays
as
had
been
previously
(1891–197
4), who was awarded the Nobel Prize
4
in Physics in 1935 for discovering the neutron
α
2
42
9
+
Be
4
12
→
C
6
1
+
n
0
The
mass
thought:
in
beryllium
was
calculations
fact
neutrons
he
found
was
and
able
not
2 . 1
The
discovery
jigsaw
the
of
the
neutron
puzzle
of
atomic
existence
of
the
Chadwick’s
was
structure.
neutron
but
at
the
time
the
Rutherford
had
no
last
always
conclusive
piece
of
T h e
n u c l e a r
a T O m
the
postulated
evidence
until
discovery.
The atomic number, Z
The
●
atoms
The
an
of
each
atomic
atom
of
element
number
an
have
is
the
element.
an
individual
number
Different
of
atomic
protons
elements
in
have
number,
the
Z:
nucleus
different
of
atomic
numbers.
For
a
neutral
protons,
●
Z
for
for
atom
the
number
of
electrons
equals
the
number
of
example:
oxygen,
O,
is
8.
Therefore
the
oxygen
atom
has
8
protons
and
8electrons.
●
Z
for
29
copper,
Cu,
is
29.
Copper
atoms
have
29
protons
and
electrons.
The mass number, A
The
mass
both
●
of
the
protons
The
mass
neutrons
For
●
●
atom
and
is
number,
in
concentrated
in
the
nucleus,
which
contains
neutrons.
the
A,
is
nucleus
the
of
number
an
of
protons
+
the
number
of
atom.
example:
Z
for
9
electrons.
A
uorine,
for
F
,
uorine-19
is
9.
Therefore
is
19.
uorine
Therefore
has
uorine-19
9
protons
has
19
9
and
=
10
neutrons.
The nuclear symbol
The
for
nuclear
a
symbol
particular
includes
element
X
and
both
is
A
and
Z
For
example,
hydrogen
has
three
isotopes:
represented
3
H
(tritium)
1
like
this:
1
proton,
1
electron,
2
neutrons
1
neutron
0
neutrons
A
X
2
H
Z
(deuterium)
1
1
proton,
1
electron,
Isotopes
1
H
As
you
forms
saw
of
atomic
in
the
sub-topic
same
number,
Z,
1.2,
element
but
isotopes
that
different
have
mass
are
the
different
same
numbers,
1
they
have
different
numbers
of
neutrons
1
electron,
nature
most
elements
exist
as
mixtures
of
isotopes.
in
For
their
proton,
A,
In
because
(hydrogen)
1
example,
boron
contains
the
two
naturally
nuclei.
occurring
19.9%)
isotopes
and
boron-10
boron-11
(natural
(natural
abundance
abundance
80.1%).
43
2
ATO M I C
S T R U C T U R E
Stdy tip
Isotop it: n y d  wpos
An easy way to remember the
Uranium found in nature consists of three isotopes with the relative abundances
A
X
order of the nuclear symbol
and atomic compositions shown in table 2.
Z
is “A to Z”, after the rst and last
Isotop
rtiv
nb o
nb o
nb o
bd
potos
tos
tos
U
0.0055%
92 protons
92 electrons
142 neutrons
U
0.7200%
92 protons
92 electrons
143 neutrons
U
99.2745%
92 protons
92 electrons
146 neutrons
letters of the alphabet.
234
235
238
T
able 2 Isotopes of uranium
Uranium-235
(split ting)
has
be
Stdy tips
a
much
used
the
higher
id
uranium
●
is
with
to
into
in
nuclear
release
of
a
abundance
increase
enriched
the
reactors
large
of
U-238
propor tion
uranium
and
where
amount
than
of
of
it
u n d e r g o e s  i ss i o 
energy.
U-235
U-235.
depleted
so
The
Natural
uranium
uranium
ore
separation
uranium
is
the
of
may
natural
p h y s i ca l
Isotopes are often written
process
of
i s o to p 
s  p   ti o .
with just their mass number
37
Cl may be
A. For example,
Because they are the same element (same Z) isotopes have the same chemical
17
37
written as chlorine-37,
●
Cl,
proper ties but they show dierent physical proper ties due to their dierent mass
or Cl-37.
numbers,
The atomic number, Z, can be
The dierence in mass between U-235 and U-238 can be used to enrich a fuel
obtained directly from the
with U-235. In some nuclear reactors natural uranium is used as the fuel but
periodic table (sub-topic 3.1;
uranium used for nuclear weapons needs to be of higher grade and is usually
section 6 of the Data booklet).
enriched.
rdiotiv isotops can
occur naturally or can be
A
ativity
1
In class, discuss the pros and cons of nuclear energy and debate the issue of
ar tically produced. Carbon-14
countries developing nuclear weapons programmes.
is an example of a radioisotope
2
)
Deduce the number of protons, electrons, and neutrons in the isotopes
that occurs naturally.
37
35
Cl and
17
Cl.
17
37
b)
Deduce the number of protons, electrons, and neutrons in the ion,
Cl
17
us so
Radioisotopes
The WebElements website,
developed by Professor Mark
Winter at the University of
Sheeld, UK , contains lots
of information about the
elements. It includes a link to
isotopes, showing the naturally
As
well
as
boron-10
radioisotopes
boron-12,
for
and
boron-11,
boron-13.
diagnostics,
pharmaceutical
archaeological
and
(radioactive
boron
isotopes).
Radioisotopes
treatment,
research,
and
and
are
research,
as
also
has
Examples
as
“chemical
used
a
are
in
tracers
clocks”
number
of
boron-8,
nuclear
in
boron-9,
medicine
biochemical
in
geological
and
and
dating.
occurring isotopes and
radioisotopes of the various
Iodine radioisotopes as medical tracers
elements of the periodic table.
The
http://www.webelements.com/
into
thyroid
the
control
44
gland
in
the
bloodstream.
the
body’s
neck
These
growth
releases
thyroxine
hormones
and
or
metabolism.
and
chemical
An
triiodothyronine
messengers
overactive
thyroid
2 . 1
gland
the
produces
anxiety,
Iodine
goitre
is
and
also
iodine-131
example
of
in
in
and
cancer
the
In
image
rays
to
is
and
such
the
gland)
gland.
in
In
as
is
gland
levels
of
loss.
(short-
treatment
whether
contrast,
a T O m
accelerates
high
weight
n u c l e a r
radioisotope
the
patient
thyroid
this
high-energy
determine
a
and
The
are
used
hospital,
of
hormones
symptoms
which
camera.
brain
to
thyroid
( γ)
Iodine-131
gamma
and
two
thyroid
diagnostics,
an
a
these
leading
the
gamma
normally.
using
of
body
photons.
functioning
prostate
the
(swelling
emits
wavelength)
cancer
excess
of
concentrated
iodine-131
is
an
metabolism
T h e
the
given
can
be
of
thyroid
thyroid
gland
radioactive
obtained,
iodine-125
is
for
used
to
treat
tumours.
Figure 8 A single-photon emission computed
Positron
emission
images
tracer
tomography
(PET)
scanners
give
three-dimensional
tomography scanner can be used to detect the
of
concentration
in
the
body,
and
can
be
used
to
detect
gamma rays from iodine131
cancers
(see
sub-topic
tomography
emitted
from
D.8).
(SPECT)
Single-photon
imaging
can
be
emission
computed
used
to
detect
used
to
treat
the
gamma
rays
iodine-131.
Cobalt-60 in radiotherapy
us so
Cobalt-60
also
emits
gamma
rays
and
is
cancer.
The Nobel Prize in Chemistry is
awarded annually by the Royal
Carbon-14 in cosmic, geological, and archaeological dating
Swedish Academy of Sciences,
Radioisotopes
cosmic,
are
often
geological,
and
used
as
radioactive
archaeological
clocks
matter.
for
The
the
dating
American
of
Stockholm, Sweden.
scientist
The Nobel Prize website
Professor
Willard
Libby
won
the
Nobel
Prize
in
Chemistry
in
1960
for
gives information about the
his
method
that
uses
carbon-14
for
age
determination
in
archaeology,
various Nobel Prize winners in
geology,
geophysics,
and
other
branches
of
science.
chemistry and other elds of
Nitrogen
is
present
nitrogen-14.
The
in
the
Earth’s
atmosphere
is
atmosphere
constantly
as
the
isotope
bombarded
by
science and medicine.
highly
Chemistry was deemed the
penetrating
cosmic
rays
from
outer
space
and
this
neutron
most impor tant science for the
bombardment
causes
radioactive
carbon-14
to
form,
along
with
work of Alfred Nobel.
hydrogen,
according
to
the
nuclear
equation:
●
14
1
N
+
7
This
14
n
→
neutron
H
Chemistry Nobel laureate?
1
bombardment
atmosphere,
gas
consists
The
+
6
the
of
half-life,
What is the average age of a
1
C
0
as
it
78%
t
is
is
of
results
in
continuously
the
the
Earth’s
time
it
air
takes
a
constant
formed
by
for
supply
from
of
carbon-14
nitrogen-14.
in
Nitrogen
volume.
an
●
When will this year ’s Nobel
Prize in Chemistry be
announced?
amount
of
radioactive
isotope
1/2
to
decrease
decay
process
Carbon-14
carbon
other
their
to
is
can
dioxide
5730
be
carbon
continually
When
a
the
is
initial
to
form
photosynthesis
in
their
ratio
bodies.
and
between
essentially
exchanged
living
its
The
or
carbon
and
half-life
they
exhale
carbon-12
the
dies
with
at
dioxide.
assimilate
Animals
constant
with
organism
atmosphere
value.
for
the
carbon-14
years.
compounds,
the
atmosphere
of
oxidized
for
compounds
organisms
with
one-half
consume
carbon
and
any
other
its
in
plants,
dioxide.
time,
the
carbon
is
plants
carbon
carbon-14
given
atmosphere
however,
Living
the
taking
In
found
since
all
in
longer
of
in
living
the
carbon
processes
no
absorb
into
is
life.
exchanged
organisms.
45
2
ATO M I C
S T R U C T U R E
The
carbon-14
emitting
beta
14
→
N
+
result
is
tha t
the r e
to
carb o n- 1 2
carbon-14
in
the
of
measured.
artefacts
and
undergo
in
the
decay
to
form
nitrogen,
process:
e
carbon-14
be
then
-1
7
net
may
(electrons)
0
14
C
6
The
isotope
particles
body
of
Scie nti s ts
such
as
in
is
a
g ra dual
d ec re a se
or g a ni sm ’s
pl a nt
ca n
w o o d,
a
the
use
or
a ni m a l
this
to
p a py ru s ,
the
The
that
met h o d
p ai ntings,
in
bod y.
ra tio
of
am ou n t
wa s
on c e
d et er m i n e
a nc i e n t
of
l iv i n g
the
ca n
age
man u s c ri pt s ,
scrolls.
The Shroud of Turin
The
Shroud
of
one
used
wrap
shows
the
many
in
at
from
on
the
is
the
of
that
Rome
the
Technology,
dating
Turin
image
believe
Vatican
based
to
a
it
a
person
cloth
the
of
who
All
originated
Christ
of
three
of
the
Swiss
Arizona,
results
and
In
to
ad,
and
laboratories
out
the
the
the
Institute
carry
that
1390
be
cloth
1988
analytical
Federal
USA,
to
The
traumatized
Jesus.
conrmed
1260
people
death.
independent
UK,
between
many
his
physically
crucixion
three
Oxford,
by
after
appears
the
University
Shroud.
believed
Jesus
commissioned
and
the
of
represents
University
cloth
linen
body
of
carbon-14
samples
suggesting
taken
that
the
Figure 9 The Shroud of Turin
Shroud
was
not
and
debate
and
historians
the
about
to
burial
the
this
cloth
Shroud
of
Jesus.
continues
Nevertheless,
amongst
the
controversy
scientists,
theologians,
day.
us so
In
July
2013
Giulio
Fanti
and
co-workers
from
the
University
of
Padua,
An app (application) has
Italy,
published
research
in
the
journal
Vibrational
Spectroscopy
which
been developed by the
shows
a
two-way
relationship
between
age
and
a
spectral
property
Diocese of Turin in Italy and
of
ancient
ax
textiles.
The
media
reported
their
ndings
worldwide,
the International Centre of
claiming
that
the
results
dated
the
Shroud
of
Turin
between
300
bc
and
Sindonology (scientic study
400
ad,
which
could
date
from
the
time
of
Christ.
of the Shroud). The app is
named Shroud 2.0 and using
this you can explore the
ativity
various images, scientic and
In class, consider and debate the aspects of hypothesis, theory, technology, and
theological interpretations.
analytical evidence surrounding the Shroud of Turin.
Relative atomic mass
1
____
The
mass
of
the
electron
is
negligible
amu
(
).
The
mass
of
the
atom
1836
is
concentrated
the
in
mass
of
a
section4
of
the
relative
the
●
single
atomic
unied
mass
are
The
atomic
as
is
booklet,
masses.
unied
nucleus
atom
Data
dened
dened
the
The
mass
tiny,
and
the
as
it
seen
is
atomic
unit
protons
in
more
mass
46
amu
or
mass
one-twelfth
1
u
table
according
neutrons.
1
of
this
convenient
unit
This
unit
of
is
unit
the
is
mass
used
to
a
to
(more
IUPAC)
non-SI
of
a
=
1.6605402
×
10
kg.
unit
to
However,
sub-topic
use
a
correctly
and
of
carbon-12
express
27
1
and
and
system
of
termed
relative
atomic
asfollows:
atomic
ground-state.
in
masses
mass
atom
of
and
in
atomic
is
its
particles:
2 . 1
●
The
relative
atomic
mass,
A
,
is
the
ratio
of
the
average
mass
T h e
of
n u c l e a r
a T O m
the
r
atom
to
the
unied
atomic
mass
Stdy tip
unit.
rtiv toi ss is a ratio
As
mentioned
in
sub-topic
1.2,
the
average
mass
of
the
atom
is
a
weighted
so it does not have units.
average
of
the
atomic
masses
of
its
isotopes
and
their
relative
abundances.
The mass spectrometer
The
mass
atomic
spectrometer
mass
of
an
is
element.
an
It
instrument
can
also
used
show
its
to
determine
isotopic
the
relative
composition.
detector
(stage 5)
lighest par ticles
positive ions are
(deected most)
accelerated in the electric
eld (stage 3)
heating lament to vaporize
magnet (stage 4)
sample (stage 1)
inlet to inject
heaviest par ticles
sample
(deected least)
N
electron beam to
ionize sample (stage 2)
S
Figure 10 Schematic diagram of a mass spectrometer
There
●
are
ve
Stage
1
where
stages
in
this
process:
(vaporization):
it
is
heated
and
The
sample
vaporized,
is
injected
producing
into
the
gaseous
instrument
atoms
or
molecules.
●
Stage
2
(ionization):
energy
electrons,
X(g)
e
The
gaseous
generating
atoms
positively
are
bombarded
charged
by
high-
species:
100
+
●
Stage
3
→
X
(g)
+
(acceleration):
charged
plates
Stage
(deection):
eld
the
2e
4
and
80.1 (area under peak)
The
positive
accelerated
perpendicular
mass-to-charge
The
to
the
positive
their
ratio
in
path.
(the
ions
attracted
electric
ions
The
m/z
are
are
ratio).
The
negatively
eld.
deected
degree
to
of
by
a
magnetic
deection
species
with
depends
the
on
ecnadnuba evitaleR
●
+
50
19.9 (area under peak)
smallest
0
mass,
m,
and
the
highest
charge,
z,
will
be
deected
the
most.
0
Particles
with
no
charge
are
not
deected
in
the
magnetic
2
4
6
8
10
eld.
m/z
●
Stage
ratio.
The
(detection):
ions
instrument
charge
The
5
The
are
mass
isotope)
can
spectrum
the
m/z
the
mass
so
a
detects
and
deection
therefore
or
relative
detector
counter
adjusted
The
is
The
the
be
detected.
versus
indicates
hit
an
that
will
plot
only
then
of
number,
abundance
of
the
species
electrical
positive
depend
relative
A.
The
of
a
particular
signal
is
ions
of
only
respective
of
a
on
abundance
height
m/z
generated.
Figure 11 Mass spectrum of boron. The two
peaks correspond to two isotopes
single
the
(of
each
mass.
each
peak
isotope.
47
12
2
ATO M I C
S T R U C T U R E
Stdy tips
Worked examples: calculations involving
In the periodic table of
non-integer relative atomic masses and
elements in section 6 of the
Data booklet, the atomic
abundances of isotopes
number Z is given above the
symbol for each element. The
Example 1
number below the symbol
Boron
has
two
naturally
occurring
isotopes
with
the
natural
represents the relative atomic
abundances
mass, A
shown
in
table
3.
(gure 12).
r
Isotop
nt bd/%
10
5
Z
B
19.9
B
80.1
11
B
Table 3 Isotopes of boron
10.81
A
r
Calculate
Figure 12 Periodic table entry
the
relative
atomic
mass
of
boron.
Solution
for boron
The
●
of
Don’t confuse the nuclear
relative
the
atomic
isotopes
and
mass
their
is
the
weighted
relative
average
of
the
atomic
masses
abundance:
11
symbol, eg
B for boron-11,
5
80.1
_
19.9
_
relative
atomic
mass
=
10
(
with the representation
×
)
+
(
11
×
)
=
10.8
100
100
given in the periodic table.
The nuclear symbol refers
Example 2
to a par ticular isotope or
Rubidium
has
a
relative
atomic
mass
of
85.47
and
consists
of
two
naturally
id, and shows both
85
occurring
isotopes,
87
Rb
(u
=
84.91)
and
Rb
(u
=
86.91).
Calculate
the
atomic number Z and mass
percentage
composition
of
these
isotopes
in
a
naturally
occurring
sample
number A, with A shown
of
rubidium.
above Z. In the periodic
table however, the relative
atomic mass A
Solution
is given
r
along with Z, with Z above A .
Note
that
in
this
example
exact
u
values
are
given
correct
to
two
r
decimal
●
places
so
you
need
to
use
this
information
in
your
answer.
Never round the relative
In
Example
1
no
such
precise
information
was
given.
atomic mass when
answering a question.
Always use the data given
85
Take
a
sample
of
100
atoms.
Let
x
=
number
of
Rb
87
(100
x)
=
number
of
Rb
atoms
in
the
sample.
in section 6 of the Data
84.91x
+
86.91(100
x)
___
booklet and express values
A
=
85.47
=
r
100
to two decimal places,
eg A (H) = 1.01.
cross-multiplying:
r
84.91x
+
86.91(100
84.91x
+
8691
solve
by
making
x
x)
86.91x
the
2.00x
=
-144
x
=
72.00
=
=
subject
8547
8547
of
the
expression:
85
The
48
sample
contains
72.00 %
Rb
87
and
28.00%
Rb.
atoms
and
2 . 1
T h e
n u c l e a r
a T O m
45
Example 3
Deduce
the
spectrum
relative
in
gure
atomic
13
and
mass
of
identify
the
X
element
from
the
X
from
periodic
its
mass
40
table.
35
●
The
●
In
of
mass
spectrum
theory
atoms
taken
peak
as
the
of
area
each
an
shows
under
two
each
isotope.
In
approximation
heights
are
●
The
naturally
●
The
total
X-69
=
occurring
isotopes,
peak
is
X-69
27
the
isotopes
the
relative
units
X-71.
proportional
calculations
of
and
and
sum
=
41
to
the
height
numbers
X-71
must
peak
to
of
number
can
be
atoms.
The
units.
ecnadnuba evitaler
Solution
30
25
20
15
100 %
10
height
of
both
peaks
is
68
units.
To
deduce
the
relative
5
atomic
mass
of
X
we
need
to
determine
the
relative
abundance
of
69
each
7
1
isotope:
0
0
20
40
60
80
100
27
_
X-69:
(
)
×
100
=
40%
m/z
68
Figure 13 Mass spectrum of X showing the
41
_
X-71:
(
)
×
100
=
relative abundances of its naturally occurring
60%
68
isotopes
●
The
relative
procedure
atomic
from
mass
worked
of
X
can
example
now
be
determined
40
_
relative
atomic
mass
=
(
69
×
●
From
Ga
(Z
the
=
periodic
31),
table
which
is
in
70.2
(or
section
quoted
as
the
60
_
+
)
(
71
×
100
=
using
1:
70
6
)
100
of
correct
the
having
A
to
Data
=
2
SF)
booklet,
69.74.
X
must
be
The
value
of
this
calculation
r
70.2
if
from
you
the
use
this
calculation
peak
calculations
gures
were
heights
will
be
expressed
is
closest
instead
2
to
SF
2
at
of
to
this
peak
best,
so
value.
areas,
this
is
the
the
In
precision
reason
of
why
all
SF
.
49
2
aTO m I c
S T r u c T u r e
2.2 eto otio
Understandings
Applications and skills
➔
Emission spectra are produced when photons
➔
Description of the relationship between colour,
are emitted from atoms as excited electrons
wavelength, frequency, and energy across the
return to a lower energy level.
electromagnetic spectrum.
➔
The line emission spectrum of hydrogen
➔
Distinction between a continuous spectrum
provides evidence for the existence of
and a line spectrum.
electrons in discrete energy levels, which
➔
Description of the emission spectrum of the
converge at higher energies.
hydrogen atom, including the relationships
➔
The main energy level or shell is given an
between the lines and energy transitions to the
integer number, n, and can hold a maximum
rst, second, and third energy levels.
2
number of electrons, 2n
➔
➔
Recognition of the shape of an s atomic orbital
A more detailed model of the atom describes
and the p
, p
x
, and p
y
atomic orbitals.
z
the division of the main energy level into s,
➔
Application of the Aufbau principle, Hund’s
p, d, and f sublevels of successively higher
rule, and the Pauli exclusion principle to write
energies.
electron congurations for atoms and ions up
➔
Sublevels contain a xed number of orbitals,
to Z = 36.
regions of space where there is a high
probability of nding an electron.
➔
Each orbital has a dened energy state for
a given electron conguration and chemical
environment and can hold two electrons of
opposite spin.
Nature of science
➔
Developments in scientic research follow improvements in apparatus – the use of electricity and
magnetism in Thomson’s cathode rays.
➔
Theories being superseded – quantum mechanics is among the most current models of the atom.
➔
Use theories to explain natural phenomena – line spectra explained by the Bohr model of the atom.
The electromagnetic spectrum
What
visions
Samuel
the
The
Beckett
Nobel
Prize
light.
in
Visible
light
include
light,
is
radio
dark
of
one
Literature
that
structure
Visible
the
(1906–1989),
developments
electronic
50
in
of
the
of
Irish
in
led
atom
light
type
waves,
have
the
light!
we
novelist,
poet,
and
playwright
who
won
1969
to
much
have
see,
is
of
come
full
of
our
infrared
experiments
scientic
electromagnetic
microwaves,
understanding
from
the
intrigue.
radiation .
radiation
of
involving
Other
(IR),
examples
ultraviolet
2 . 2
radiation
(UV),
spectrum
various
X-rays,
(EMS)
types
of
is
a
and
gamma
spectrum
of
electromagnetic
rays.
The
e l e c T r O n
c O n f I g u r a T I O n
electromagnetic
wavelengths
that
comprise
Stdy tip
the
radiation.
The wavelengths of the dierent
types of waves in the EMS are
The
energy,
E,
of
electromagnetic
radiation
is
inversely
proportional
to
given in section 3 of the Data
the
wavelength,
λ:
booklet
1
_
E
∝
λ
High-energy
radiations
wavelengths,
and
such
as
low-energy
gamma
rays
radiations
and
such
as
X-rays
radio
have
small
waves
have
long
wavelengths.
Wavelength
is
related
to
the
frequency
of
the
radiation,
ν,
by
the
expression:
c
=
νλ
8
where
c
The
unit
for
SI
is
the
of
frequency
speed
of
energy
the
light
is
hertz,
the
(3.00
joule,
×
J;
10
-1
m
for
s
).
wavelength
the
metre,
m;
and
Hz.
absoptio, issio d otios spt
A white-hot metal object such as an incandescent light bulb lament emits the
full range of wavelengths, producing a otios spt including all the
colours of the rainbow from red to violet.
If a pure gaseous element such as hydrogen is subjected to an electrical
discharge the gas will glow – it emits radiation. The resultant issio spt
Figure 1 The aurora borealis in Lapland,
Sweden. The aurora borealis (or Nor thern
consists of a series of lines against a dark background.
Lights) is a display of coloured light visible in
If a cloud of a cold gas is placed between a hot metal and a detector, an
the night sky at high latitudes. It occurs when
bsoptio spt is observed. This consists of a pattern of dark lines against
charged and energetic par ticles from the sun
a coloured background. The gaseous atoms absorb cer tain wavelengths of light
from the continuous spectrum.
are drawn by the Ear th’s magnetic eld to the
polar regions. Hundreds of kilometres up they
collide with gaseous molecules and atoms,
Absorption and emission spectra are widely used in astronomy to analyse light
causing them to emit light
from stars.
Emission spectra and Bohr ’s theory of the
hydrogen atom
In
the
1600s
passed
colours
of
all
Isaac
a
generating
merging
continuous
400
to
a
prism
(1642–1727)
the
continuous
and
into
so
the
spectrum
700
Newton
glass
wavelengths
colour
from
Sir
through
is
as
with
a
no
rainbow.
showed
light
spectrum.
appears
next
a
visible
is
This
The
if
The
series
familiar
wavelengths
of
sunlight
into
spectrum
continuous
gaps.
that
separated
contains
of
is
different
light
colours,
example
visible
each
of
light
a
range
nm.
Figure 2 White light as perceived by the
Many
sources
of
radiation
produce
a
line
spectrum
rather
than
a
human eye consists of many colours or
continuous
spectrum.
If
a
pure
gaseous
element
is
subjected
to
a
high
wavelengths of light. Shown here is the
voltage
colour
under
of
reduced
light.
For
pressure,
example,
the
sodium
gas
will
emits
emit
yellow
a
certain
light.
If
characteristic
this
light
is
continuous spectrum of white light emitted by
an incandescent light bulb lament
51
2
ATO M I C
S T R U C T U R E
passed
through
consists
each
a
prism,
black
element
identify
emission
to
a
the
has
the
to
its
a
resultant
spectrum
own
of
spectrum
a
small
example,
sodium
nm
is
not
number
continuous
of
coloured
but
lines
wavelength.
characteristic
For
589.0
(gure
with
characteristic
element.
wavelengths
background
the
background
corresponding
Each
to
of
line
in
spectrum
the
two
distinct
and
589.6
visible
yellow
nm,
can
which
region
lines,
be
can
of
the
be
used
line
corresponding
seen
on
a
black
3).
Flame tests
Figure 3 Line emission spectrum of sodium.
Flame
The spectrum looks like a single bright yellow
The
tests
colour
are
of
often
the
used
ame
in
the
varies
laboratory
for
different
to
identify
elements
certain
and
can
metals.
be
used
line but at high resolution it is possible to see
to
identify
unknown
of
electrons
substances.
The
colours
are
due
to
the
excitation
two lines very close together corresponding to
in
the
metals
by
the
heat
of
the
ame.
As
the
electrons
the wavelengths 589.0 nm and 589.6 nm
lose
the
energy
they
have
just
gained,
they
emit
photons
oflight.
Quantization of energy
aoy
The
precise
lines
in
the
line
emission
of
an
element
have
specic
You might think of a line
wavelengths.
Each
characteristic
wavelength
corresponds
to
a
discrete
emission spectrum as being
amount
of
energy.
This
is
the
basis
of
quantization,
the
idea
that
analogous to a barcode. Every
electromagnetic
radiation
comes
in
discrete
“parcels”
or
quanta.
A
product in a shop has its own
photon
is
a
quantum
of
radiation,
and
the
wavelength,
λ,
and
energy,
unique barcode which gives
E,
of
a
photon
are
related
by
the
equation:
it an identity, and the same
hc
_
E
is true of the line emission
=
hν
=
λ
spectra of the elements. Each
where:
line emission spectrum is
34
dierent and is characteristic
h
=
Planck’s
constant
=
6.63
×
10
J
s
of a specic element.
ν
=
frequency
of
the
radiation
8
c
=
speed
This
E
is
equation
the
1913
proposed
the
way
of
s
section
the
vice
1
of
the
greater
Data
the
booklet.
energy
of
It
the
shows
that
photon,
the
versa.
Neils
Bohr
hydrogen
based
(1885–1962)
atom.
on
Bohr
classical
examined
proposed
a
the
line
theoretical
mechanics.
His
model
following:
at
atom
its
electron
that
consists
centre,
is
balanced
velocity
in
suggested
energy
its
a
the
two
by
a
positively
which
circular
sun.
charged
negatively
path
or
Although
oppositely
the
a
acceleration
of
orbit,
there
charged
particle
is
particle
similar
an
species,
the
called
charged
inherent
this
electron
to
force
moving
of
at
orbit.
that
of
in
orbit
the
of
around
moves
planets
attraction
the
m
λ:
spectrum
between
Bohr
in
to
and
the
1
10
physicist
the
hydrogen
the
×
found
attraction
it:
52
for
proton
high
●
be
Danish
the
called
the
can
3.00
proportional
spectrum
explanation
The
=
wavelength,
the
emission
●
light
inversely
smaller
In
of
the
each
orbit
electron
has
a
denite
orbiting
the
energy
positively
associated
charged
with
centre
2 . 2
in
a
particular
electron
in
a
orbit
is
xed
particular
or
orbit
quantized.
is
given
by
The
the
energy
e l e c T r O n
of
c O n f I g u r a T I O n
the
aoy
expression:
Think about standing on the
E
=
1
_
–R
(
H
bottom step of a ight of stairs.
)
2
n
You could jump to the second
where:
step, or you could jump higher
18
R
=
Rydberg
constant
=
2.18
×
10
to the third or four th step.
J
H
Suppose you jump from the
n
=
principal
quantum
number,
with
positive
integer
values
1,
2,
3,
4,
...
rst step to the fth step. You
depending
on
the
orbit
or
energy
level
the
electron
occupies
stay there for a few seconds
●
When
an
electron
subjecting
level
and
it
to
in
an
stays
in
its
ground-state
electrical
this
is
discharge),
excited-state
excited
it
for
moves
a
(for
to
fraction
a
example,
higher
of
a
by
energy
and then jump back down. You
might jump down to the rst
step, or jump two steps down
second.
to the third step, or jump three
●
When
the
electron
falls
back
down
from
the
excited-state
to
a
lower
steps down to the second step.
energy
level
it
emits
a
photon,
a
discrete
amount
of
energy.
This
This is analogous to the way
photon
corresponds
to
a
particular
wavelength
λ,
depending
on
the
excited electrons can jump
energy
difference
between
the
two
energy
levels
(gure
4).
from a higher energy level to a
lower one.
n > 1
You always jump to a step,
not to some place between
steps. This shows the idea of
quantization – each step is
analogous to an energy level,
n = 1
which has a denite, discrete
ground-state
excited-state
e
falls back down to a lower level and
energy. Jumping up steps
energy is emitted as a photon of light of
wavelength, λ, corresponding to the energy
dierence between the two energy levels
requires an amount of energy,
and jumping down steps
releases discrete amounts
Figure 4 Principles of the Bohr model of an atom when an electron is excited. n is the
of energy.
principal quantum number
Note
that
current
excited
an
electron
level:
to
n
in
can
gure
=3,
n
=
4
4,
be
excited
instead
etc.
The
of
to
any
being
electron
energy
excited
can
level
to
also
n
fall
=
higher
2
it
back
than
could
down
its
be
to
any
Wy t tiv si?
lower
energy
level.
The negative sign in the
The
difference
in
energy
between
the
two
energy
levels
can
be
expressed
expression for E is an arbitrary
as
follows,
where
i
represents
the
initial
state
and
f
represents
the
convention. It means that the
nalstate:
energy of the electron in the
ΔE
=
E
E
f
atom is less than its energy
i
if the electron was located an
hc
_
=
hν
=
innite distance away from the
λ
nucleus.
We
can
rearrange
this
expression
noting
that:
Conventions are often used in
E
=
chemistry. Another example
1
_
–R
(
H
2
)
n
ΔE
=
E
of an arbitrary convention is
E
f
=
i
1
_
R
H
(
[
2
n
-
[
)]
1
_
R
f
(
H
2
n
i
always placing the cathode
)]
on the right-hand side in a cell
diagram (see topic 9). Can you
1
_
R
=
[
(
H
2
n
1
_
R
)]
[
H
(
2
n
i
)]
think of any other conventions
f
that we use in chemistry?
ΔE
1
_
R
=
[
(
H
2
n
i
hc
_
1
_
=
2
n
)]
hν
=
λ
f
53
2
ATO M I C
S T R U C T U R E
The
hydrogen
different
of
the
spectrum.
series,
upper
line
colours
which
energy
Colour
emission
(violet,
The
series
comprises
levels
spectrum
blue,
of
lines
back
consists
blue–green,
lines
shown
associated
down
to
the
n
in
2
a
series
red)
in
gure
with
=
of
and
5
is
lines
of
visible
called
electronic
energy
of
the
region
the
Balmer
transitions
level.
violet
blue
blue–green
red
410
434
486
656
λ/nm
n = 6
n = 5
to
Transition
from
to
n = 2
n = 3 to n = 2
n = 4 to n = 2
n = 2
n = 6
Stdy tip
n = 5
You are not required to know the names
n = 4
of the individual series of spectral lines.
However, you are required to know which
n = 3
transition corresponds to which region
n = 2
of the EMS, eg the transition
n = 4 to n = 1 will be seen in the UV
n = 1
region, etc.
outside the atom
n = ∞
n = 5
continuum
inside the atom
n = 4
n = 3
Figure 5 Line emission spectrum of the hydrogen atom. Four lines are seen in the visible
and ultraviolet regions of the spectrum; these make up the Balmer series
n = 2
Other
n
=
3
series
of
energy
infrared
lines
levels
regions
Sis
of
exist
(table
the
n

n = 1
Figure 6 Some transitions to the n = 1 level
corresponding
1).
These
are
to
transitions
observed
in
to
the
the
EMS.
n
rio o emS
i
Lyman
1
2, 3, 4, 5, ...
UV
Balmer
2
3, 4, 5, 6, ...
visible and UV
Paschen
3
4, 5, 6, 7, ...
IR
from higher levels for the Lyman series (in the
UV region) of spectral lines that occur in the
emission spectrum of the hydrogen atom
54
n
=
1
ultraviolet
T
able 1 Dierent series of lines in the hydrogen line emission spectrum
and
and
2 . 2
e l e c T r O n
c O n f I g u r a T I O n
Quantization and atomic structure
The
line
emission
of
electrons
to
converge)
merge,
any
in
at
forming
energy;
outside
the
it
spectrum
discrete
higher
a
is
hydrogen
levels,
energies.
continuum.
no
atom.
of
energy
longer
Such
the
the
electron
get
limit
Beyond
under
an
At
provides
which
this
of
be
this
of
for
the
together
the
referred
the
as
a
are
the
electron
nucleus
to
existence
(they
convergence
continuum
inuence
may
evidence
closer
and
free
is
said
lines
can
have
therefore
electron.
n =
2
n =
3
Models of the atom and electron arrangements
n =
The
Bohr
theory
arrangements.
electrons
in
the
each
atom
electron
shell
or
is
a
basis
for
orbit,
for
arrangement
of
H:
1
arrangement
of
P:
2,
8,
5
electron
arrangement
of
Ca:
2,
8,
8,
arrangements
the
the
Bohr
are
chemical
model
of
a
very
useful
properties
the
atom
gives
electron
the
number
of
example:
electron
predicting
writing
arrangement
electron
Electron
In
of
An
1
of
the
2
tool
an
for
explaining
and
element.
energy
levels
are
often
drawn
as
Figure 7 Electron arrangement for phosphorus
according to the Bohr model
concentric
circles,
as
shown
in
gure
7
for
phosphorus.
Limitations of the Bohr theory
This
of
model
has
●
It
assumes
is
incorrect;
introduce
It
●
It
●
is
Bohr
suggests
were
superseded
and
is
associated
with
a
number
idea
an
positions
orbits
of
an
energy
do
of
the
not
electron
actually
orbits
exist
(we
are
xed.
shall
This
shortly
orbital).
levels
incorrect
that
some
limited
elements
Bohr
●
the
fact
are
circular
or
spherical
in
nature.
incorrect.
2.1
hydrogen.
the
the
scale
atom
fundamental
for
is
the
made
atom
–
up
mainly
theoretical
of
remember
problems
from
empty
space.
pertaining
to
the
the
●
It
to
containing
that
did
more
the
to
not
just
one
explain
than
one
electron
is
element,
the
line
namely
spectra
of
other
electron.
a
subatomic
particle
orbiting
Bohr
made
structure
a
and
signicant
in
contribution
particular,
some
of
to
the
our
understanding
merits
of
his
theory
following:
was
that
It
calculations
model
nucleus.
electronic
are
his
The
suggested
Nevertheless,
●
been
model:
Bohr
●
in
that
also
sub-topic
There
that
the
assumes
This
of
now
misconceptions:
based
on
electrons
the
exist
incorporated
the
fundamental
in
denite,
idea
of
idea
of
discrete
electrons
quantization
energy
moving
–
the
fact
levels.
from
one
energy
level
another.
55
2
ATO M I C
S T R U C T U R E
The quantum mechanical model of the atom
The
Bohr
and
in
replaced
of
theory
by
the
more
quantum
the
provided
particular
electron.
a
sophisticated
mechanics,
Some
Heisenberg’s
rst
approximation
arrangement
of
the
uncertainty
electrons.
mathematical
which
key
of
are
principle
the
described
states
atomic
has
that
structure,
since
theories
incorporates
ideas
of
It
been
from
the
wave-like
eld
nature
of
below.
it
is
impossible
to
TOK
determine
accurately
simultaneously
both
(topic
the
12).
momentum
This
means
and
that
the
the
position
more
we
of
a
know
particle
about
Heisenberg’s uncertainty
the
position
of
an
electron,
the
less
we
know
about
its
momentum,
and
principle states that there
vice
versa.
Although
it
is
not
possible
to
state
precisely
the
location
of
is a theoretical limit to the
an
electron
in
an
atom
and
its
exact
momentum
along
a
trajectory
at
precision with which we can
the
same
time,
we
can
calculate
the
probability
of
nding
an
electron
in
know the momentum and the
a
given
region
of
space
within
the
atom.
position of a particle. What are
the implications of this for the
limits of human knowledge?
One aim of the physical
sciences has been to
give an exact picture of
the material world. One
Schrödinger’s
physicist
mathematical
nature
equation
Erwin
of
the
subsequent
was
Schrödinger
equation
electron.
integrates
This
development
Schrödinger
received
The
to
formulated
the
in
(1887–1961).
the
dual
1926
His
wave-like
ground-breaking
of
the
Nobel
eld
of
Prize
in
by
the
Austrian
sophisticated
work
quantum
Physics
and
led
to
particle
the
mechanics.
with
Paul
birth
In
and
1933
Dirac.
achievement ... has been
solution
Schrödinger’s
equation
generated
a
series
of
to prove that this aim is
mathematical
functions
called
wavefunctions
describing
the
electron
unattainable.
in
Jacob Bronowski
the
can
hydrogen
occupy.
atom
Each
and
associated
wavefunction
is
possible
energy
represented
by
the
states
the
electron
symbol, ψ.
The
2
(1908–1974):
Polish-born British
square
an
of
the
electron
wavefunction,
in
a
region
of
ψ
,
space
represents
at
a
given
the
probability
point
a
of
distance,
r,
nding
from
2
mathematician,
biologist, scientic
historian, inventor,
the
nucleus
equations
the
basic
of
are
the
atom.
very
ψ
is
complex
principles
termed
but
at
underpinning
the
this
the
probability
level
all
we
density.
need
to
The
consider
are
results.
and poet.
The
What are the implications of
wavefunctions
of
electrons
in
an
atom
are
described
by
atomic
orbitals:
this claim for the aspirations
An
●
atomic
orbital
is
a
region
in
space
where
there
is
a
high
of the natural sciences in
probability
of
nding
an
electron.
par ticular and for knowledge
in general?
Any
orbital
types
shape
of
can
atomic
and
hold
a
orbital:
associated
maximum
s,
p,
d,
of
and
f,
two
etc.
electrons.
Each
type
There
has
a
are
several
characteristic
energy.
Stdy tip
aoy
Atomic orbitals have dierent
Imagine that you are a student in an IB chemistry class in Quito in Ecuador, waiting
shapes. For SL you need to be
for your teacher to arrive at 8.00 am. At 8.15 am there is no sign of your teacher and
familiar with the shapes of the
your class decide to go looking for him. You decide rst to dene the most probable
s and p atomic orbitals, while
places the teacher is likely to be. Suggestions from the class include:
for HL you need to know the
shapes of the s, p, and d atomic
The teacher:
orbitals. We shall return to the
●
is possibly in the sta room, the chemistry laboratory, or the library
●
may be in the school principal’s oce or in the school car park
●
could be at his house in Quito
shapes of the d orbitals in topic
13 when we discuss crystal
eld theory.
56
2 . 2
●
could perhaps be at the airpor t
●
might even have gone home to South Africa!
e l e c T r O n
c O n f I g u r a T I O n
If the class went looking for the teacher they would most likely star t looking in
the most probable locations closest to the classroom. But at 8.15 am they do
not know with any degree of cer tainty precisely where the teacher is.
A three-dimensional graph could be drawn with a cluster of dots showing areas
where there is a high probability of nding him. This is the idea of an obit. A
boundary surface could be drawn around this cluster of dots to dene a region
of space where there is a 99% chance of nding the teacher. This might be the
school perimeter, or Quito where he lives.
If you were also asked to measure the distance from the classroom to the
exact location where the teacher is you could not do this at 8.15 am, as you do
Figure 8 An orbital is a three-dimensional
not know his exact location with absolute cer tainty.
graph with a cluster of dots showing the
What aspects of quantum mechanics does this analogy capture?
probability of nding the electron at dierent
distances from the nucleus
The s atomic orbital
An
s
orbital
is
spherically
boundary
surface,
chance
probability
or
symmetrical.
meaning
of
that
nding
The
within
an
sphere
the
represents
sphere
electron
there
(gure
is
a
a
y
99 %
9).
The p atomic orbital
A
p
orbital
is
dumbbell
shaped.
There
are
three
p
atomic
orbitals,
p
,
p
x
and
p
,
all
with
boundary
surfaces
conveying
probable
electron
,
y
density
z
pointing
in
different
directions
along
the
three
respective
Cartesian
x
axes,
x,
y,
and
z
(gure
10).
Energy levels, sublevels, orbitals, and electron spin
The
by
Bohr
n,
model
which
positive
is
integer
increases,
the
introduced
called
the
values
mean
the
idea
principal
1,
2,
3,
position
etc.
of
In
an
of
a
main
quantum
the
energy
quantum
electron
is
level,
number.
described
This
can
mechanical
further
from
the
z
have
model,
as
nucleus.
n
The
Figure 9 The s atomic orbital is spherically
energies
of
the
orbitals
also
increase
as
n
increases.
Each
main
of
electrons
given
energy
level
symmetrical
2
or
shell
can
electron
we
in
have
the
The
hold
capacity
two
energy
maximum
for
n
elements
second,
common
a
=
1
in
is
number
2,
the
for
rst
n
=
row
2
of
is
8,
the
for
n
=
3
periodic
is
by
2n
18.
table,
.
So
That
eight
is
the
why
elements
y
etc.
levels
types:
s,
are
p,
split
d,
up
and
f.
into
sublevels,
Each
sublevel
of
which
contains
a
there
are
number
y
four
of
x
orbitals,
each
of
which
can
hold
a
maximum
of
2
electrons
(table
2).
z
x
nb o obits i
mxi b o
sbv
tos i sbv
x
x
z
z
p
Sbv
y
p
p
z
y
Figure 10 The three p atomic orbitals are
dumbbell shaped, aligned along the x, y, and
s
1
2
p
3
6
d
5
10
f
7
14
z axes
T
able 2 Sublevels of the main energy levels in the quantum mechanical model
57
2
ATO M I C
S T R U C T U R E
For
convenience,
an
“arrow-in-box”
notation
called
an
orbital
us so
diagram
is
used
to
represent
the
electrons
in
these
atomic
orbitals
The “Orbitron” website,
(gure
11).
We
shall
use
orbital
diagrams
to
represent
electron
developed by Professor Mark
congurations.
Winter at the University of
Sheeld, UK is an excellent
s sublevel (one box representing an s orbital)
resource for exploring the
shapes of the various atomic
orbitals. It also provides
information on the associated
sophisticated mathematical
p sublevel (three boxes representing the three p orbitals p
, p
x
, and p
y
)
z
wavefunctions.
http://winter.group.shef.ac.uk/
orbitron/
d sublevel (ve boxes representing the ve d orbitals)
f sublevel (seven boxes representing the seven f orbitals)
Figure 11 Orbital diagrams are used to represent the electron congurations for atoms.
Arrows are drawn in the boxes to represent electrons, a maximum of 2 electrons in each
box (orbital)
Two
electrons
in
the
same
orbital
have
opposite
values
of
the
1
N
S
magnetic
quantum
number,
m
.
The
sign
of
m
s
indicates
electron.
facing
by
in
two
the
A
orientation
pair
of
opposite
arrows
in
of
the
electrons
in
directions
a
box
an
(gure
+
s
magnetic
and
(
eld
orbital
is
-
2
as
)
2
generated
behaves
therefore
spin
1
or
by
two
commonly
the
magnets
represented
12).
Qt bs
In
this
mathematical
four qt
S
N
N
S
represents
qt
number,
model
of
the
electronic
structure
bs. The rst is the piip
the
energy
b, l,
level.
The
describes
the ti
second
the
qt
quantum
sublevel,
and
b, m ,
the
of
the
atom
there
are
qt b, n, which
number,
the
third
atomic
the zit
quantum
orbital.
The
four th
l
quantum
number,
the spi
ti
qt
b, 
,
describes
the
s
magnet analogy
spatial
orientation
examined
of
S
in
the
levels,
IB
the
electron
Chemistry
sublevels,
spin.
Diploma,
atomic
Quantum
but
you
orbitals,
numbers
need
and
to
are
know
electron
not
the
formally
principles
spin.
N
half-arrows representing
electrons
energy
of
You might think of the four quantum numbers as an electronic postal
of opposite spin
address. The country represents the energy level, the province the
in an orbital
sublevels, the town the orbitals, and the street number or postal code the
spin of the electron.
Figure 12 Electron spin is represented by
arrows in orbital diagrams
58
2 . 2
e l e c T r O n
c O n f I g u r a T I O n
Writing electron congurations
We
shall
now
develop
congurations
for
these
atoms
ideas
and
further
by
writing
electron
ions.
3p
3s
ygrene
2p
2s
1s
Figure 13 This is the order of energy levels of the rst few sublevels
There
are
electron
1
The
three
Aufbau
orbital
rst
Up
that
few
to
with
3d
principles
that
must
be
followed
when
representing
congurations.
principle
is
available
energy
Ca
(Z
=
since
that
electrons
Figure
13
ll
shows
the
the
lowest-energy
sublevels
for
the
levels.
20)
the
experimental
level
states
rst.
it
is
Aufbau
data
and
lower
in
principle
the
4s
correlates
level
energy.
The
is
lled
precisely
rst
condensed
before
the
electronic
2
conguration
(Z
=
21),
level
the
the
now
3d
is
for
Ca
two
slightly
lled
is
written
levels
are
higher
rst.
The
in
as
comparable
energy
correctly
3d
in
written
sublevel.
energy
For
than
as
Zn
the
[Ar]3d
(Z
=
3d
30),
Zn
is
best
written
and
as
However,
energy
the
3d
for
level
conguration
Sc
with
the
and
for
Sc
4s
hence
therefore
2
4s
.
the
the
This
4s
trend
level
[Ar]3d
continues
now
condensed
10
for
.
in
than
condensed
1
is
[Ar]4s
is
much
electron
along
the
higher
conguration
2
4s
for
this
reason.
This
is
Stdy tips
consistent
with
experimental
data
which
shows
that
when
•
the
3d-block
elements
are
ionized,
the
electrons
are
For the IB Chemistry
removed
Diploma you need to be
from
the
4s
before
the
3d
levels,
which
makes
sense
since
the
able to deduce the electron
4s
is
higher
in
energy
than
the
3d
for
this
block
of
elements.
congurations for the atoms
The
situation
overall
is
quite
complex
as
in
the
case
of
Sc
the
and ions of the elements up
lling
of
the
last
three
electrons
does
not
continue
in
the
3d
to and including
level,
and
experimental
data
does
not
provide
evidence
for
Z = 36 (Kr).
an
3
[Ar]3d
the
3d
electron
orbitals
electrons
conguration
are
entering
more
the
for
compact
3d
orbitals
Sc.
The
than
will
the
reason
4s
for
orbitals
experience
a
this
and
much
is
that
hence
greater
•
The periodic table showing
atomic number, Z, is provided
in section 6 of the Data booklet
59
2
ATO M I C
S T R U C T U R E
mutual
repulsion.
written
by
and
E.
In
Scerri,
Biochemistry,
an
excellent
Department
at
the
orbitals
article
of
This
Chemistry
University
USA
Chemistry,
7th
explained
as
and
published
November
in
2013,
singly
before
illustrated
in
occupying
gure
them
in
pairs.
14.
of
1
California,
is
Education
the
reason
4
2
3
in
is
2p
feature
can
is
that
relieve
repulsion,
make
follows:
full
although
such
slightly
the
additional
different
use
“The
of
atoms
this
form
unsettling
relevant
s
orbital
Figure 14 Electrons ll each orbital singly before
electron-electron
do
of
not
occupying them in pairs
always
sheltering
There
because
the
situation
is
more
be
than
is
just
that
through
set
of
the
described.
nuclear
the
atoms,
interactions
nucleus
as
One
charge
thing
there
between
well
as
to
increases
and
are
is
the
between
a
we
Pauli
electrons
the
1
full
2
condensed
hold
a
principle
maximum
electron
3
orbital
write
of
states
two
these
electrons
have
opposite
Hund’s
rule
of
maximum
that
an
that
when
lling
of
equal
conguration
“build
the
we
successive
orbitals
described
the
according
to
the
three
above.
periodic
table
to
can
the
be
shown
four
as
sublevels
four
s,
p,
blocks
d,
orbitals
ll
all
15).
the
p block
main-group elements
1
18
1s
1s
2
13
14
15
16
2p
2s
d block
3p
3s
3
4
5
6
7
8
9
10
11
12
4s
3d
4p
5s
4d
5p
6s
5d
6p
7s
6d
4f
f block
5f
Figure 15 The blocks of the periodic table correspond to the sublevels s, p, d, and f
60
use
electrons
multiplicity
electrons
s block
up”
spin.
degenerate
energy),
and
any
(gure
(orbitals
conguration
representation.
electron
corresponding
states
electron
electrons,
The
3
conguration
diagram
table,
principles
and
can
electrons
To
exclusion
can
congurations
and
in
orbital
electron
move
complicated
periodic
The
ways
illustrated:
consider
as
themselves”.
2
three
complicated
1
7
and
f
2 . 2
e l e c T r O n
c O n f I g u r a T I O n
Full electron congurations
Table
3
shows
the
full
electron
congurations
for
some
of
the
rst
36elements.
et
Z
eto otio
Period 1 elements:
1
H
1
1s
He
2
1s
2
Period 2 elements:
2
Li
3
1s
Be
4
1s
B
5
1s
C
6
1s
N
7
1s
O
8
1s
1
2s
2
2
2s
2
2
2s
2
2
2s
2
2
1s
Ne
10
1s
3
2p
2
2s
2
9
2
2p
2s
2
F
1
2p
4
2p
2
2s
2
5
2p
2
2s
6
2p
Period 3 elements: continue with the same lling pattern, for example:
2
Na
11
1s
Mg
12
1s
Al
13
1s
Ar
18
1s
2
2s
2
6
2p
2
1
3s
6
2
2s
2p
3s
2
6
2
2
2s
2
2p
2
2s
3s
6
2p
1
3p
2
3s
6
So xptios:
3p
coi d opp
Period 4 elements:
After Z = 30 the 4p sublevel is lled:
2
K
19
1s
Ca
20
1s
Sc
21
1s
Ni
28
1s
Zn
30
Ga
31
1s
Br
35
1s
Kr
36
1s
2
2s
2
Two of the rst 36 elements have
6
2p
2
2
3s
6
6
3p
2
1
electron congurations that dier
4s
6
from what you may predict. These
2
2s
2p
3s
3p
4s
2
6
2
6
1
two elements are Cr (Z = 24) and
2
2s
2p
3s
3p
3d
2
4s
Cu (Z = 29):
2
2
2s
6
2p
2
3s
6
3p
8
3d
2
4s
2
2
1s
2
6
2
6
2s
2p
3s
3p
2
6
2
6
10
3d
Cr
1s
Cu
1s
2
2
2s
2p
3s
3p
10
3d
2
2s
6
2p
2
3s
6
3p
5
3d
1
4s
2
4s
2
4s
2
2s
6
2p
2
3s
6
3p
10
3d
1
4s
1
4p
In these two elements electrons
2
2
2s
6
2p
2
3s
6
3p
10
3d
2
4s
5
4p
go into the 3d orbitals before
2
2
2s
6
2p
2
3s
6
3p
10
3d
2
4s
6
4p
completely lling the 4s orbital.
Chromium has a half-lled 3d
T
able 3 Full electron congurations for some of the rst 36 elements
sublevel of 5 electrons and copper
has a completely lled 3d sublevel
Condensed electron conguration
of 10 electrons. Half-lled and
You
can
see
above
that
full
electron
congurations
become
quite
completely lled 3d sublevels
lengthy
and
cumbersome
with
increasing
atomic
number.
An
element’s
reduce the overall potential
chemistry
is
dictated
by
its
outer
valence
electrons
(as
opposed
to
the
energy of an atom, so the electron
inner
core
electrons),
and
a
more
convenient
way
of
representing
5
congurations 3d
electron
congurations
is
as
the
condensed
electron
1
4s
conguration :
10
and 3d
4
are more stable than 3d
9
[nearest
noble
gas
core]
+
valence
electrons
3d
1
4s
2
4s
and
2
4s
, respectively.
61
2
ATO M I C
For
S T R U C T U R E
example:
He
[He]
O
[He]2s
Ne
[He]2s
P
[Ne]3s
2
4
2p
2
6
2p
2
or
simply
[Ne]
3
3p
Orbital diagrams
Orbital
arrows
boxes
joined
Orbital
two
make
the
steps
types
of
orbital
1s
2
2s
to
of
show
the
the
core
energy
diagrams
arrows-in-boxes
boxes
their
all
gas
the
and
show
noble
showing
2
F
may
nearest
use
electrons
together
diagrams
beyond
have
diagrams
representing
energy
orbitals
as
levels
that
in
or
can
notation
representing
in
Degenerate
gures
orbitals
11
and
are
13,
with
represented
as
in
the
full
condensed
may
be
used
electron
electron
represented
to
represent
on
conguration,
conguration.
one
line.
For
or
just
Orbital
the
orbitals
diagrams
example,
gure
may
16
shows
uorine:
5
2p
1
4
2
5
3
1
5
or
2p
4
2
5
3
[He]
2
5
2s
2p
ygrene
2
2s
2
1s
Figure 16 Orbital diagrams showing the electron conguration for uorine
The
condensed
diagrams
for
version
the
5
Cr
[Ar]3d
is
more
elements
convenient
chromium,
and
cobalt,
will
and
be
used
bromine
in
are
this
1
2
3
4
5
4
5
[Ar]
1
5
4s
3d
7
Co
[Ar]3d
2
4s
1
6
2
7
3
[Ar]
2
7
4s
3d
10
Br
[Ar]3d
4s
2
5
4p
1
4
2
5
[Ar]
2
4s
62
10
3d
5
4p
book.
For
represented
4s
1
by
equivalence.
the
be
described
orbitals.
3
as
example,
follows:
the
orbital
2 . 2
e l e c T r O n
c O n f I g u r a T I O n
Worked examples: electron congurations
Example 1
Deduce
the
Example 2
full
2+
Mg
electron
congurations
for
Mg,
Deduce
the
condensed
O,
and
2
O
S,
S
2+
,
Fe,
Fe
congurations
of
+
,
Cu,
2
From
table
2
Mg
1s
2
2s
●
3:
6
2p
S
[Ne]3s
Cu
2
4
3p
2-
3s
For
the
S
anion
2
To
and
Solution
Solution
●
electron
2
,
write
the
electron
conguration
for
the
2
S
[Ne]3s
we
add
2
electrons:
6
3p
or
simply
[Ar]
2+
Mg
cation,
2
electrons
must
be
removed.
●
These
are
principal
taken
from
quantum
the
orbital
number
n;
of
in
The
electron
conguration
this
case,
be
deduced
Fe
(Z
=
26)
n;
in
can
Fe
[Ar]3d
as:
the
6
2
3s
for
highest
orbital:
2
4s
2+
2+
2
Mg
1s
2
2s
6
For
2p
the
Fe
removed
●
From
table
2
1s
2
2s
the
2
electrons
orbital
of
are
highest
3:
this
O
cation
from
case,
the
4s
orbital:
4
2p
2+
6
Fe
[Ar]3d
2
To
write
the
electron
conguration
for
the
O
●
anion,
to
the
two
electrons
same
must
principles
as
be
added
The
copper
one
of
electron
conguration
is
according
the
two
exceptions
that
you
must
before:
remember:
2
2
O
1s
2
2s
6
2p
10
Cu
Notice
that
the
electron
congurations
[Ar]3d
1
4s
for
+
To
2+
the
species
Mg
form
the
O
are
identical:
the
ion,
again
the
electron
is
they
removed
contain
Cu
2
and
same
number
of
electrons
from
the
orbital
of
highest
n;
in
this
and
case,
the
4s
level:
+
are
said
to
be
isoelectronic .
Na
,
F
,
and
+
2+
Ne
are
also
However,
number
table
isoelectronic
each
of
of
these
protons
with
species
(atomic
2
Mg
and
has
a
number
O
.
Cu
10
[Ar]3d
different
Z
4):
Spis
atoi b, Z
nb o
(b o potos)
tos
2
O
F
Ne
8
10
9
10
10
10
11
10
12
10
+
Na
2+
Mg
T
able 4 Isoelectronic species
63
2
ATO M I C
S T R U C T U R E
Stdy tip
Example 3
Warning: Do not be tempted
2+
Deduce
the
orbital
diagrams
for
Ni,
Ni
,
and
Se.
to rearrange the conguration
2+
for Fe
5
to [Ar]3d
1
4s
as for
Solution
5
chromium. The 3d
10
and 3d
First
write
the
condensed
electron
conguration
for
the
species.
congurations for chromium
Then
draw
the
orbital
diagram,
remembering
that
two
electrons
in
and copper apply only to
the
same
orbital
have
opposite
spin
quantum
numbers:
neutral atoms, not to ions.
8
Ni
[Ar]3d
2
4s
1
6
2
7
3
8
4
5
4
5
[Ar]
2
8
4s
3d
2+
Ni
8
[Ar]3d
1
6
2
7
3
8
[Ar]
0
8
4s
3d
2+
Notice
that
electrons
For
in
in
the
the
orbital
4s
diagram
orbital
–
the
for
box
the
Ni
should
cation
be
left
there
are
no
blank.
selenium:
10
Se
[Ar]3d
2
4s
4
4p
1
4
2
3
[Ar]
2
10
4s
4
3d
4p
Experimental evidence for electron
congurations
Direct
be
of
evidence
found
from
magnetism,
paramagnetic
can
be
eld.
and
In
can
number
scientists
64
by
contrast,
be
a
a
to
in
conguration
measurements.
at
least
magnetic
eld.
greater
the
diamagnetic
by
a
determine
the
in
and
can
types
diamagnetism.
greater
of
element
different
electron
the
number
attraction
has
all
its
the
past
and
in
a
A
hence
of
magnetic
electrons
paired
eld.
research
improvements
The
force
an
are
unpaired
material
magnetic
scientic
one
for
There
paramagnetism
has
the
repelled
of
electron
including
electrons,
Developments
a
the
material
attracted
unpaired
of
magnetic
over
instrumentation
number
of
unpaired
50
years
which
have
have
electrons
in
led
to
allowed
an
atom.
Q u e S T I O n S
Questions
1
What
is
the
neutrons
number
in
of
protons,
electrons,
and
5
What
with
boron-11?
A.
5
protons,
5
electrons,
and
11
B.
5
protons,
5
electrons,
and
10.81
C.
5
protons,
5
electrons,
and
6
D.
11
is
the
the
relative
mass
atomic
spectrum
mass
shown
of
in
an
element
gure
17?
neutrons
100
neutrons
neutrons
80
What
is
the
11
number
34
neutrons
electrons,
in
of
and
5
protons,
neutrons
electrons,
ecnadnuba %
2
protons,
and
2
S
?
60
40
16
A.
18
protons,
18
neutrons
16
16
protons,
34
neutrons
16
protons,
18
neutrons
16
protons,
18
neutrons
electrons
and
20
B.
18
electrons
and
0
22
C.
18
electrons
23
24
25
and
26
27
28
29
30
mass/charge
Figure 1
7
D.
3
Which
16
statements
35
chlorine,
They
about
and
the
isotopes
of
A.
24
B
25
C.
26
D.
27
IB,
May
37
Cl
and
17
I.
electrons
Cl,
are
correct?
17
have
the
same
chemical
[1]
2009
properties.
II.
They
have
the
same
atomic
number.
III.
They
have
the
same
physical
6
Which
is
correct
electromagnetic
for
the
following
regions
of
the
spectrum?
properties.
A.
I
and
II
B.
I
and
III
C.
II
and
D.
I,
II
utviot (uV)
only
III
high
shor t
low
low
energy
wavelength
energy
frequency
A.
only
only
high
low
low
long
energy
frequency
energy
wavelength
B.
IB,
and
May
Id (Ir)
III
[1]
2011
high
shor t
high
long
frequency
wavelength
energy
wavelength
C.
63
4
A
sample
of
element
X
contains
69 %
of
X
high
long
low
low
frequency
wavelength
frequency
energy
D.
65
and
31%
of
X.
What
is
the
relative
atomic
[1]
mass
of
X
in
this
sample?
IB,
A.
63.0
B.
63.6
C.
65.0
7
In
May
the
2009
emission
electronic
D.
69.0
IB,
May
spectrum
transition
of
would
hydrogen,
produce
a
which
line
[1]
in
the
visible
region
of
the
electromagnetic
2010
spectrum?
A.
n
=
2
→
n
=
1
B
n
=
3
→
n
=
2
C.
n
=
2
→
n
=
3
D.
n
=
∞
IB,
May
→
n
=
1
[1]
2011
65
2
ATO M I C
8
Which
S T R U C T U R E
describes
spectrumof
the
visible
b)
emission
spectrum
c)
A.
A
series
Distinguish
hydrogen?
of
lines
converging
at
The
thinning
between
and
of
a
the
line
a
continuous
spectrum.
ozone
layer
[1]
increases
longer
the
amount
the
Earth’s
of
UV-B
radiation
that
reaches
wavelength
B.
A
series
of
regularly
C.
A
series
of
lines
converging
spaced
at
lower
D.
A
series
of
lines
converging
at
higher
May
(table
5).
lines
energy
frequency
IB,
surface
Typ o ditio
Wvt / 
UV-A
320–380
UV-B
290–320
[1]
2010
T
able 5
9
What
is
the
order
of
increasing
energy
of
the
Based
orbitals
within
a
single
energy
on
explain
A.
d
B.
s
<
C.
p
D.
f
IB,
May
<
s
<
f
p
<
d
s
<
f
<
<
d
<
<
<
p
why
dangerous
p
<
the
information
in
table
5
level?
IB,
f
UV-B
than
Specimen
rays
are
more
UV-A.
[3]
Paper
d
<
s
[1]
13
a)
Deduce
the
full
and
Mn
electron
conguration
2+
2009
for
b)
Mn
Deduce
the
condensed
electron
2+
conguration
10
What
is
the
condensed
for
Cu
electron
2+
3+
congurationfor
2
A.
[Ar]4s
c)
?
[Ar]4s
[Ar]3d
D.
[Ar]4s
14
series
are
often
of
IB,
label
an
atom.
of
In
lines
its
energy
your
in
the
emission
labelling
May
level
diagram
diagram
show
for
spheres.
model.
Developments
in
scientic
improvements
in
apparatus.
ultraviolet
and
each
spectrum
are
series.
research
follow
Discuss
how
statement
with
regard
to
the
use
of
visible
and
magnetism
in
Thomson’s
produced,
with
cathode
rays.
[4]
2010
the
following
radiation
in
wavelength
In
many
textbooks
order
types
of
of
electromagnetic
widely
increasing
(shortest
[Ar]4s
I.
Yellow
II.
Red
.
This
the
electronic
vanadium
is
accepted
However,
rst).
for
is
written
as
3
3d
electronic
common
by
suggest
the
why
practice
chemical
this
conguration
for
way
and
community.
of
writing
vanadium
the
may
light
be
at
odds
with
experimental
evidence.
light
You
III.
Infrared
radiation
IV
.
Ultraviolet
might
like
to
read
radiation
the
following
article:
http://www.rsc.org/eic/2013/11/aufbau-
electron-conguration
[1]
your
66
As.
this
a
2
List
as
of
as
conguration
a)
and
the
16
12
use
representation
experiments
clearly
drawn
onthe
electricity
regions
Co
5
this
the
for
3d
15
hydrogen
Atoms
Comment
1
and
diagrams
4
3d
C.
Draw
orbital
3d
6
11
Draw
7
2
B.
Co
answer.
to
guide
you
in
3
P E R I O D I C I T Y
Introduction
Science
some
of
is
resulted
vast
full
the
of
factual
greatest
from
amounts
scientists
of
information.
scientic
data
being
and
However,
discoveries
able
deduce
to
have
interpret
clear
important
table
from
it.
In
1869
the
Russian
Mendeleev
arranged
in
recognized
order
that
according
if
to
(relative
atomic
mass),
a
the
denite
be
led
seen
in
the
ultimately
properties
(after
some
of
the
a
table
topic
theory)
to
the
of
the
core
periodic
of
physical
it
became
properties
clear
of
that
the
the
elements
function
we
shall
of
Z,
the
examine
atomic
the
number.
nature
periodic
table,
establish
what
of
information
elements.
renement
development
the
pattern
be
extracted
the
from
it,
and
explore
how
of
repeated
the
developed
and
periodic
this
can
This
at
atomic
the
could
chemists,
lies
elements
their
In
weight
to
which
chemist
are
were
available
chemistry.
chemical
Dmitri
tool
elements,
patterns
As
emerging
of
(periodic)
patterns
can
be
linked
to
the
most
properties
of
the
elements.
3.1 P a
Understandings
Applications and skills
➔
The periodic table is arranged into four blocks
➔
Deduction of the electron conguration of
associated with the four sublevels - s, p, d,
an atom from the element ’s position on the
and f.
periodic table, and vice versa.
➔
The periodic table consists of groups (ver tical
columns) and periods (horizontal rows).
➔
Nature of science
The period number (n) is the outer energy level
that is occupied by electrons.
➔
➔
➔
Obtain evidence for scientic theories by
The number of the principal energy level and
making and testing predictions based on
the number of the valence electrons in an
them – scientists organize subjects based
atom can be deduced from its position on the
on structure and function; the periodic
periodic table.
table is a key example of this. Early models
The periodic table shows the positions of
metals, non-metals and metalloids.
of the periodic table from Mendeleev, and
later Moseley, allowed for the prediction of
proper ties of elements that had not yet been
discovered.
67
3
P E R I O D I C I T Y
The development of the periodic table
Evidence
for
predictions
theories.
based
table
then
Scientists
on
of
scientic
and
structure
elements
theories
testing
often
and
is
a
try
is
obtained
them
to
classify
function,
good
against
and
example
by
their
the
of
making
proposed
subject
periodic
scale.
In
when
development
over
a
from
of
number
different
each
of
of
the
years
periodic
and
countries
others’
work
has
and
took
involved
on
key
the
scientists
modern
the
the
it
scientists
table,
be
to
It
the
of
the
German
the
chemist
(1780–1849)
elements
calcium,
in
similar
and
(relative
atomic
of
that
mass
strontium
of
mean
calcium
of
elements
other
the
and
as
sum
of
barium.
and
the
a
triads–
triad.
one
or
same
more
iodine,
and
and
called
law
weights
this
also
there
and
and
of
This
triads .
trio
community
at
the
attention
to
elements
this
into
law
triads
a
few
elements.
discovery
time
and
was
H o w e v e r,
suggesting
between
atomic
there
did
apply
elements
in
the
was
an
the
of
the
each
music.
Thus
nitrogen
in
and
seven
elements;
fourteen;
between
between
fourteen;
and
and
bismuth,
lastly,
fourteen
relationship
term
The
Law
of
I
propose
to
Octaves.
Newlands,
News,
‘a
12
letter
(18th
to
the
editor’,
August
1865).
of
of
octaves
known
applied
to
elements.
only
He
a
tried
at
principle
the
to
time).
the
known
However,
elements
they
did
not
pay
neatly
t
this
type
of
pattern:
highly
reactive
classication
such
as
lithium,
sodium,
and
caesium
became
potassium,
to
an
grouped
such
as
silver
and
with
copper.
very
One
Newlands
had
was
to
place
two
elements
properties
stepping
periodic
metals
inherent
the
in
one
box
of
a
periodic
table,
to
take
stone
account
development
to
extremities
D ö b e r e i n e r ’s
and
important
idea
this
60
together,
of
of
the
was
not
limited
was
weight
of
some
the
idea
link
are
peculiar
number
unreactive
hypothesis
by
s u l f u r,
rubidium,
just
stand
as
between
arsenic,
Chemical
metals
of
Members
antimony,
J.A.R.
all
much
or
bromine,
Surprisingly
that
numbers
seven
to
horizontal
of
of
(about
scientic
belonging
same
the
by
relation
antimony
This
recognized
chlorine,
involving
tellurium.
their
barium
to
the
the
that
differ
octaves
phosphorus
limited
selenium,
on
of
t o d a y ’s
atomic
Döbereiner
another
elements
seen
group,
phosphorus
Newlands’s
and
order
transpositions,
approximately
classied
involving
This
atomic
using
was
the
He
be
in
slight
that
seven.
provisionally
the
weight.
the
pattern)
that
also.
terminology)
few
appear
also
the
between
weight
of
Johann
discovered
strontium,
properties
idea
repeated
octaves
arranged
a
elements
nitrogen
arsenic
had
are
with
group
will
one
the
1817
atomic
of
this
a
below.
Döbereiner
Döbereiner
of
is,
development
summarized
other
In
(that
law
observed
same
line.
as
the
elements
will
the
foundations
ideas.
contributed
periodic
order
as
equivalents
place
multiples
of
in
known
analogous
Four
published
elements
this.
table
building
Newlands
of
arranged
became
If
The
1865
periodicity
table
of
this.
Newlands
presented
his
law
to
the
of
Chemical
Society
in
England
but
his
ideas
were
elements.
not
accepted.
Society
of
His
this
presentation
work
in
1866
to
was
the
Chemical
not
published.
Newlands
As
In
1864
the
English
chemist
John
a
his
(1837–1898)
arranged
to
be
of
the
in
discovered
order
evidence
of
of
a
that
atomic
pattern
when
weight,
with
the
elements
repeated
in
there
seven
elements,
such
octaves
that
each
of
appeared
or
below
analogy
is
similar
it.
of
repeated
This
an
at
to
the
term
octave
eighth
was
in
intervals
music
of
–
eight
element
based
the
on
1869,
four
rst
at
a
and
sugar
returned
plant.
years
mooted,
after
Newlands’s
Russian
ideas
chemist
(1834–1907)
discovered,
Dmitri
like
above
on
same
the
the
had
Newlands,
that
in
order
atomic
if
the
of
their
elements
were
arranged
the
of
weight,
a
repeated
pattern
note
properties
could
be
identied.
This
was
musical
termed
68
ridiculed
consisting
element
named
felt
chemist
Mendeleev
Mendeleev
properties
chief
properties
were
of
Newlands
position
were
In
elements
result
Newlands
the
periodic
law.
The
main
difference
to
3 . 1
between
that
Newlands’s
Mendeleev
elements
only
In
very
elements
1869
table
of
and
Mendeleev’s
considered
carefully
that
had
Mendeleev
the
and
grouped
similar
published
work
properties
was
of
the
rst
It
soon
improved
the
table
over
time
atomic
gaps
for
undiscovered
fell
into
the
elements,
correct
so
the
approach
existence
elements.
to
be
J.W
.
version
iodine
placed
in
some
of
the
table
link
of
weight
elements
threes.
in
(1817)–
atomic
different
groups
of
–
did
law.
of
For
of
was
scientic
elements
problematic.
Henry
the
elements
atomic
Moseley
in
number,
weight.
This
table
elements.
the
despite
is
the
basis
atomic
appeared
to
pattern
octaves
seven
be
table
summarizes
over
Dimitri
(1865)
were
order
weight
1
various
scientists
the
Z,
for
the
In
in
1913
order
the
(1887–1915)
periodic
instead
of
table
in
atomic
the
modern
of
there
evidence
with
the
repeated
consisting
in
The
periodic
Mendeleev
–
in
order
weight
pattern
a
were
of
law
the
arranged
atomic
repeated
of
properties
of
the
some
of
periodic
their
was
Henry
Moseley
(1887–1915)
when
elements
developed
time.
periodic
(1869)
contributions
who
(1834–1907)
octaves
in
of
Figure
had
elements
arranged
a
weight
the
the
obey
Newlands
properties
▲
not
126.90)
when
of
to
arranging
predict
tellurium,
Law
atomic
Using
(1837–1898)
triads
and
to
periodic
weight
after
higher
undiscovered
J.A.R.
Döbereiner
between
able
of
elements
the
(atomic
(1780–1849)
Law
was
properties
However,
Mendeleev’s
example,
Mendeleev
and
a
that
group.
order
this
apparent
that
physicist
arranged
element
had
and
British
each
tellurium
became
community
of
left
that
(127.60).
Moseley
periodic
elements.
Mendeleev
fact
weight
t A b l e
together
properties.
his
the
P e r i o d i c
found.
The
law
modern
(1913)
elements
in
order
atomic
(Z),
–
periodic
when
were
of
the
arranged
increasing
numbers
their
recurred
properties
periodically.
of
elements.
Figure 1 Scientists who contributed to the development of the periodic table of elements
Avy
In modern science do you think that theoretical research has a much greater
chance of acceptance by the scientic community if it is suppor ted by empirical
evidence? Discuss this in class.
69
3
P E R I O D I C I T Y
toK
A hyphss is a proposal that tries to explain particular
What role did inductive and deductive reasoning play
phenomena. A hy results from testing a hypothesis and
in the development of the periodic table? What role
may subsequently replace the hypothesis. A hypothesis
do inductive and deductive reasoning have in science
can therefore be considered a tentative explanation
in general?
that can be tested through investigation and exploration
inductive reasoning (“bottom-up” approach):
whereas a theory is an established array of ideas or
4. theory
concepts which may then be used to make predictions.
3. hypothesis
In science there are two ways of arriving at a particular
2. pattern
conclusion – nuv asnng and uv
1. obser vation
asnng (gure 2). Inductive reasoning is a “bottom-up”
approach whereas deductive reasoning may be described
deductive reasoning (“top-down” approach):
as a “top-down” approach. With inductive reasoning
1. theory
denite measurements and observations can lead
scientists to establish the existence of possible trends
2. hypothesis
or a pattern. From such a pattern a hypothesis can be
3. pattern
formulated that can ultimately lead to a theory based on
4. obser vation
certain conclusions. In deductive reasoning, the starting
▲
Figure 2 Inductive and deductive reasoning
point involves the theories themselves. These are tested
based on experimental (empirical) work.
The periodic table today
In
the
modern
increasing
physical
called
have
periodic
atomic
properties
groups.
their
The
own
table
number,
placed
elements
with
are
(table
are
elements
underneath
groups
names
the
Z,
each
numbered
arranged
having
other
from
1
in
to
in
order
similar
vertical
18;
of
chemical
and
columns
certain
groups
1).
Gup num
rmmn nam
1
alkali metals
2
alkaline ear th metals
15
pnictogens
16
chalcogens
17
halogens
18
noble gases
Usfu su
Much information on each
▲
T
able 1 Names of groups recommended by IUPAC in the periodic table of elements
element can be found on
the “ WebElements” periodic
table website. This resource
was compiled by Professor
Mark Winter at the University
The
current
gure
For
3.
periodic
Each
example,
there
are
group
the
known
table
is
consists
of
characterized
noble
gases
compounds
in
118
by
group
a
18
containing
elements
number
are
very
noble
and
of
is
shown
distinct
unreactive
gases,
such
in
properties.
as
(though
XeF
).
4
of Sheeld, UK .
http://www.webelements.com/
70
Helium,
the
has
many
not
typically
lightest
industrial
form
of
the
noble
applications
chemical
gases,
is
because
compounds
used
it
is
with
for
lling
balloons
non-ammable
any
elements.
and
and
does
3 . 1
P e r i o d i c
t A b l e
1
18
1
1
2
H
He
2
13
14
15
16
1
7
5
6
7
8
9
10
B
C
N
O
F
Ne
20.180
1.008
4.0026
3
2
4
Li
Be
6.94
9.0122
10.81
12.011
14.007
15.999
18.998
11
12
13
14
15
16
1
7
Na
Mg
22.990
24.305
3
Metalloids
3
4
6
7
8
9
10
11
18
Al
Si
P
S
Cl
Ar
26.982
28.085
30.974
32.06
35.45
39.948
12
19
20
2
1
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.098
40.078
44.956
47
.867
50.942
51.996
54.938
55.845
58.933
58.693
63.546
65.38
69.723
72.63
74.922
78.96
79.904
83.798
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
4
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.468
87
.62
88.906
91.224
92.906
95.96
[97
.91]
101.07
102.91
106.42
107
.87
112.41
114.82
118.7
1
12
1.76
127
.60
126.90
131.29
5
55
56
Cs
Ba
132.91
87
6
7
7
1
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Lu
Hf
T
a
W
Re
Os
Ir
Pt
Au
Hg
Ti
Pb
Bi
Po
At
Rn
137
.33
1
74.97
1
78.49
180.95
183.84
186.2
1
190.23
192.22
195.08
196.97
200.59
204.38
207
.2
208.98
[208.98]
[209.99]
[222.02]
88
103
104
105
106
107
108
109
110
111
112
113
114
115
116
11
7
*
Fr
Ra
[223.02]
[226.03]
118
Lr
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Uut
Fl
Uup
Lv
Uus
Uuo
[262.11]
[265.12]
[268.13]
[27
1.13]
[270]
[277
.15]
[276.15]
[281.16]
[280.16]
[285.1
7]
[284.18]
[289.19]
[288.19]
[293]
[294]
[294]
**
57
*lanthanoids
58
59
60
6
1
62
63
64
65
66
67
68
69
70
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
138.91
140.12
140.91
144.24
[144.91]
150.36
151.96
157
.25
158.93
162.50
164.93
167
.26
168.93
1
73.05
89
90
91
92
93
94
95
96
97
98
99
100
101
102
Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
[227
.03]
232.04
231.04
238.03
[237
.05]
[244.06]
[243.06]
[247
.07]
[247
.07]
[251.08]
[252.08]
[257
.10]
[258.10]
[259.10]
**actinoids
▲
5
Figure 3 The modern periodic table of elements
The
horizontal
periods.
number,
period.
energy
The
n,
of
For
rows
of
period
the
highest
example,
levels
elements
number
with
n
1,
equal
occupied
calcium
=
numbered
is
2,
and
Z
=
from
the
energy
(Ca),
3,
to
level
20,
is
1
to
7
principal
in
in
the
are
termed
quantum
elements
period
4
so
has
of
the
four
4.
Metals, non-metals, and metalloids
The
periodic
these
are
this
line
and
the
table
is
separated
are
the
also
by
a
metals
non-metals
lie
split
broadly
stepped
(excluding
to
the
into
diagonal
metals
line.
and
The
non-metallic
non-metals;
elements
hydrogen
to
the
which
is
left
a
of
gas)
right.
Metals:
●
are
good
●
are
malleable
●
are
ductile
conductors
of
heat
and
electricity
Quk qusn
(capable
of
being
hammered
into
thin
sheets)
Suggest  w reasons why
(capable
of
being
drawn
into
wires)
authorities in Sweden banned
the use of mercury dental
●
have
lustre
(they
are
shiny).
llings since 2008.
Mercury,
Hg,
solutions
formed
Hg
can
be
sub-topic
Z
used
=
80,
in
as
a
is
a
this
liquid
way
lling
and
are
for
can
called
teeth.
dissolve
many
amalgams;
We
shall
for
discuss
other
metals.
example,
metals
Ag
further
The
Sn
in
4.5.
Non-metals
Non-metals
metals
gain
metals
lose
are
poor
conductors
electrons
electrons
in
of
chemical
(they
are
heat
and
reactions
electricity.
(they
are
Typically
reduced),
non-
whereas
oxidized).
Metalloids
Some
of
the
elements
close
metallicand
non-metallic
silicon,
=
Si,
Z
14,
to
the
stepped
properties.
germanium,
Ge,
Z
The
=
diagonal
elements
32,
line
have
boron,
arsenic,
As,
Z
B,
=
both
Z
=
33,
5,
antimony,
71
3
P E R I O D I C I T Y
Sb,
Z
the
=
51,
tellurium,
metalloids.
semiconductors,
dependent
material
Z
=
52,
to
their
as
in
astatine,
such
as
which
computers
At,
silicon
intermediate,
conductivity
such
and
metalloids
due
electrical
science,
Te,
Some
has
and
Z
and
highly
=
85
called
are
temperature-
widespread
smart
are
germanium
applications
in
phones.
Main group, transition elements, and s, p, d, and f blocks
The
●
periodic
the
the
The
on
of
be
properties
of
position
transition
discussed
The
in
periodic
sublevels.
also
group
1
into
two
(excluding
broad
H),
sections:
group
2,
and
the
in
periodic
elements.
is
groups
main-group
the
detail
table
The
elements:
in
split
The
this
chemistry
13.
into
four
of
elements
table;
topic
occupancy
3–11.
blocks
electrons
for
can
is
of
often
less
the
based
each
be
true
predicted
for
the
transition
on
the
sublevel
s,
is
p,
based
properties
elements
d,
and
shown
in
will
f
table2.
Maxmum num f
Num f am
ns n suv
as n ah suv
s
2
1
p
6
3
d
10
5
f
14
7
T
able 2 Occupancy of electrons for each sublevel, and the related number of atomic orbitals
group 1 (excluding H), group 2, and groups 13–18
tansn mns
groups 3–11 (the f-block elements are sometimes described as the nn ansn mns)
s-k mns
groups 1 and 2 and He
p-k mns
groups 13–18 (excluding He)
groups 3–12 (including Z
and Z
f-k mns
=
=
57 (La) and Z
90 (Th) to Z
elements from Z
=
lanhans
Ans
72
divided
Man-gup mns
-k mns
▲
further
elements:
Suv
▲
be
13–18
transition
their
the
can
main-group
groups
●
table
T
able 3 The elements in the blocks of the periodic table
=
=
89 (Ac), but excluding Z
=
58 (Ce) to Z
=
71 (Lu)
103 (Lr), which are classied as f-block elements
58 (Ce) to Z
=
elements from Z
elements from Z
71 (Lu) and from Z
=
=
57 (La) to Z
89 (Ac) to Z
=
=
=
90 (Th) to Z
71 (Lu)
103 (Lr)
=
103 (Lr)
3 . 1
P e r i o d i c
main-group elements
s-block
t A b l e
p-block
1
18
1s
1s
13
2
2s
14
15
16
1
7
2p
d-block
3s
3p
3
4
5
6
7
8
9
10
11
12
4s
3d
4p
5s
4d
5p
6s
5d
6p
7s
6d
4f
f-block
5f
▲
Figure 4 The four blocks of the periodic table corresponding to the s, p, d, and f sublevels
The
number
from
the
calcium
so
has
7
dropped
of
group
is
in
valence
number
group
valence
from
2,
so
electrons
of
has
electrons
the
group
the
2
s-
(outer-shell
and
p-block
valence
(note
that
number
in
electrons)
elements.
electrons.
for
the
order
to
Fluorine
p-block
nd
the
can
For
is
be
found
example,
in
elements,
number
group
the
of
1
17,
is
valence
electrons).
73
3
P E R I O D I C I T Y
Electron congurations and the periodic table
Sub-topic
2.2
showed
●
full
●
condensed
●
orbital
For
electron
that
the
electron
conguration
of
an
element
can
be
expressed
in
three
ways:
conguration
electron
conguration
diagram.
example,
for
uorine,
F
,
Z
=
9:
2
●
full
●
condensed
●
orbital
electron
conguration:
1s
2
2s
5
2p
2
electron
conguration:
[He]2s
5
2p
diagram:
[He]
2
2
2s
2
2p
Figure
4
can
periodic
be
table
a
2p
y
powerful
can
1
2p
x
be
used
z
tool
to
when
deduce
writing
the
electron
electron
congurations:
conguration,
as
the
the
position
following
of
an
worked
element
in
example
the
shows.
Worked example: deduction of the electron conguration from the element’s
position in the periodic table
1
Consider
the
element
number
selenium,
protons
State
the
b)
State
in
c)
State
the
number
of
valence
electrons
d)
State
the
number
of
protons
and
e)
Deduce
the
full
f)
Deduce
the
condensed
g)
Draw
group
of
the
and
has
a)
which
of
which
the
electrons
periodic
table
in
chemical
in
an
atom
selenium
an
symbol
atom
of
Se.
Se.
belongs.
of
Se.
2
the
electron
orbital
electrons
conguration
electron
diagram
for
of
in
the
anion,
Se
Se.
conguration
of
Se.
Se.
Solution
a)
Z
=
34,
b)
Se
c)
Group
d)
For
is
in
so
Se
has
group
16
16
34
protons
(the
elements
and
34
electrons
(atoms
are
neutral).
chalcogens).
have
6
valence
electrons.
2
Se
the
number
two
negative
charges
e)
The
full
f)
The
condensed
g)
The
orbital
of
it
protons
has
equals
gained
Z
two
for
Se,
2
electron
conguration
for
Se
is
namely
electrons,
1s
2s
2
so
6
2p
it
2
3s
34.
has
6
3p
diagram
for
conguration
Se
is
given
for
Se
is
[Ar]3d
However,
total
10
3d
10
electron
a
2
4s
2
4s
4
4p
4
4p
below:
[Ar]
2
4s
10
3d
2
4p
74
1
4p
x
1
4p
y
z
of
since
36
it
is
an
electrons.
anion
carrying
3 . 2
P e r i o d i c
t r e N d S
3.2 P ns
Understanding
Applications and skills
Ver tical and horizontal trends in the periodic
➔
Prediction and explanation of the metallic and
➔
table exist for atomic radius, ionic radius,
non-metallic behaviour of an element based on
ionization energy, electron anity, and
its position in the periodic table.
electronegativity.
Discussion of the similarities and dierences in
➔
Trends in metallic and non-metallic behaviour
➔
the proper ties of elements in the same group,
are due to the trends above.
with reference to alkali metals (group 1) and
Oxides change from basic through amphoteric
➔
halogens (group 17).
to acidic across a period.
Construction of equations to explain the pH
➔
changes for reactions of Na
O, MgO, P
2
O
4
, and
10
the oxides of nitrogen and sulfur with water.
Nature of science
Looking for patterns – the position of an element in the periodic table allows scientists to make accurate
➔
predictions of its physical and chemical proper ties. This gives scientists the ability to synthesize new
substances based on the expected reactivity of elements.
Trends in physical and chemical properties
Electron
congurations
explained
through
understand
as
atomic
and
radius,
quantum
aspects
this
of
chemical
topic,
in
patterns
turn
are
with
mirrored
in
mechanics,
a
and
and
such
reactions
new
these
heart
The
the
show
its
periodic
trends
across
position
allows
about
and
of
properties
table
predictions
the
elements
groups.
periodic
properties
and
–
at
chemical
down
afnity,
understanding
time,
lie
elements
us
described
trends,
chemical
Patterns
be
help
electron
better
same
can
properties
properties,
the
patterns
which
atomic
provide
At
2),
energy,
These
reactions.
peppered
of
ionization
electronegativity.
in
are
many
(topic
of
to
behaviour
therefore
their
periods
an
scientists
in
make
of
atomic
and
element
in
facilitate
table
in
the
accurate
chemical
the
synthesis
of
compounds.
properties.
Atomic radius
The
radius
to
point
of
a
circle,
R
,
is
the
distance
from
the
centre
of
the
circle
c
a
on
the
circumference.
It
is
easily
measured
and
has
the
core
a
denitevalue.
In
is
the
the
Bohr
model
nucleus
model
it
of
while
would
the
the
appear
hydrogen
single
that
atom
electron
the
radius
(sub-topic
lies
of
in
the
a
2.2)
xed
atom,
orbit.
R
,
can
of
Based
also
be
the
on
atom
this
measured,
e
as
according
to
Bohr
the
electron
is
in
a
xed
position
within
a
dened
orbit.
75
3
P E R I O D I C I T Y
However,
the
atom
as
is
orbitals,
nding
xed,
we
which
an
so
described
highly
cannot
measure
need
to
often
that
are
the
move
surface
probability
of
way
is
consider
to
is,
of
consider
of
the
an
overcoming
an
X
space
that
that
in
of
there
When
the
looking
quantum
on
as
spheres
orbital)
in
that
problem
non-metallic
molecule.
the
nding
is
same
way
atoms
we
as
we
are
know
boundaries.
a
of
not
models,
where
represents
of
of
atomic
probability
atomic
xed
fact
chemically
The
in
model
in
electron
mechanics
with
in
high
the
at
Bohr
located
a
model
region
and
atoms
is
of
atom
Based
the
fact
position
Bohr
electron
diatomic
are
simplistic
atomic
this
know
where
the
radius
circle.
the
now
electrons
represented
(i.e.
nding
two
a
we
the
spheres.
be
boundary
of
means
from
as
2
and
measure
radius
cannot
One
This
away
represented
atoms
topic
regions
electron.
we
in
simplistic
The
99 %
space.
the
radius
bonded
distance
of
an
atom
together,
between
the
that
two
2
nuclei
of
the
X
atoms
is
given
by
d,
and
the
bonding
atomic
radius ,
R
,
b
is
dened
as:
1
R
=
d
b
This
2
is
shown
atomic
radius
in
is
gure
1,
using
sometimes
the
termed
example
the
of
iodine.
covalent
The
bonding
radius.
1
For
metals
the
bonding
atomic
radius
is
d’
where
d‘
now
represents
the
2
distance
of
the
An
between
two
atoms
adjacent
to
each
other
in
the
crystal
lattice
metal.
alternative
atomic
radius
is
the
non-bonding
atomic
radius ,
R
.
nb
d = 2R
b
▲
Figure 1 The iodine diatomic
molecule, I
Consider
a
with
another
one
densities.
. The bonding atomic
group
Argon
of
gaseous
there
does
is
not
argon
very
form
atoms.
little
a
When
penetration
diatomic
two
of
species.
argon
their
If
atoms
electron
argon
is
collide
cloud
frozen
in
2
radius, R
, for iodine is 136 pm (d
=
the
solid
not
be
phase
the
atoms
would
touch
each
other
(topic
1)
but
would
b
12
272 pm), where 1 pm
=
10
m
chemically
bonded.
In
this
case
the
distance
between
the
argon
toK
●
We saw in sub-topic 3.1 that Mendeleev examined the proper ties of
elements in minute detail and grouped elements with similar proper ties
Usfu su
together.
The “Periodic Table of Videos”
hence
website, developed by
The predictive power of Mendeleev’s periodic table illustrates the “risk
CBE and co-workers at the
taking” nature of science. What is the distinction between scientic and
University of Nottingham in the
pseudoscientic claims?
UK provides videos for all 118
a research professor and is also
a pioneer in the eld of gn
hmsy which is discussed
at several points in
the IB
Chemistry Diploma programme.
http://www.periodicvideos.com/
76
elements fell into their correct groups. Mendeleev was therefore
able to predict the proper ties of yet undiscovered elements at the time.
Professor Mar tyn Poliako,
elements. Professor Poliako is
When Mendeleev published his rst periodic table of elements
in 1869 he left gaps in the table for as yet undiscovered elements, and
●
The periodic table is an excellent example of classication in science. It
classies elements in several ways – metals, non-metals, and metalloids;
main-group and transition elements; groups and periods; elements
with acidic, basic, and amphoteric oxides; and s, p, d, and f sublevels.
How do classication and categorization help and hinder the pursuit of
knowledge? For example, scandium will be discussed fur ther in topic 13.
Why is it incorrect to classify scandium as a non-transition element?
3 . 2
atoms
could
be
measured
and
hence
R
could
be
found
(gure
2).
P e r i o d i c
t r e N d S
The
nb
non-bonding
atomic
radius
is
often
termed
the
van
der
Waals’
radius.
d = 2R
nb
▲
Figure 2 Atoms of argon in the solid phase. The atoms are touching but not chemically
bonded. The non-bonding atomic radius of argon R
is 188 pm (d
=
376 pm)
nb
Section
of
the
the
9
mean
over
of
the
a
range
radius
approximate
For
The
bonding
wide
atomic
from
Data
elements.
their
is
atomic
of
atomic
for
provides
term
radius
elements
always
bond
example,
booklet
general
length
for
from
the
two
covalent
is
used
atomic
to
Note
that
non-bonding
elements
can
the
atomic
also
be
radii
represent
experimental
compounds.
than
between
the
radius”
obtained
and
smaller
data
“atomic
data
bonding
radius.
The
estimated
radii.
the
interhalogen
compound
BrF:
Quk qusn
atomic
radius
of
bromine
atomic
radius
of
uorine
=
117
pm
Predict the bond lengths in:
bond
Compare
length
this
of
with
Br
the
F
=
=
60
177
pm
a)
iodine monobromide, IBr
)
trichloromethane
pm
experimental
bond
length
of
Br
F
in
the
(chloroform), CHCl
gas
3
phase
(176
nm).
ev nua hag an snng 
In an atom the negatively charged electrons are
the
attracted to the positively charged nucleus. A valence
electrons:
charge, S,
that
is
shielded
or
screened
by
the
core
or outer-shell electron is also repelled by the other
Z
=
Z
S
e
electrons in the atom. The  ns in the inner
where Z
=
actual nuclear charge (atomic number) and
non-valence energy levels of the atom reduce the
S
=
snng  shng nsan
positive nuclear charge experienced by a valence
electron. This eect of reducing the nuclear charge
Z
experienced by an electron is termed snng or
these rules in advanced textbooks on inorganic chemistry,
can be worked out using Sa ’s us. You can read about
e
but you are not required to calculate Z
shng.
using Slater’s rules
e
as part of the IB Chemistry Diploma programme. You do need
The
net
charge
experienced
by
an
electron
is
t er med
to understand the principle of screening and for our purposes
the
 ff   v
n ua
hag ,
.
Z
This
is
the
nuclear
eff
you can consider S as a parameter related to the number of
charge,
Z,
(representing
the
atom ic
nu mber)
m inu s
core electrons in an atom.
77
3
P E R I O D I C I T Y
Worked example: estimating nuclear charge
Estimate
the
effective
nuclear
charge
(For
experienced
comparison,
using
Slater’s
rules
Z
for
eff
by
the
valence
electron
in
the
alkali
metal
potassium
potassium.
need
to
is
be
calculated
aware
assumptions,
of
as
the
2.2.)
As
chemists
limitations
equations,
and
of
we
many
of
our
rules.
Solution
Potassium,
2
1s
K
2
2s
6
2p
has
a
3s
core
does
of
(it
19
charge.
3p
4s
of
is
the
and
19
in
protons
18
Z
=
conguration
19.
electrons
group
1).
(gure
experience
The
electron
1
electrons
not
the
has
6
total
electron
18
K
2
the
that
core
and
This
3).
full
one
The
electrons
there
valence
force
provide
valence
means
of
the
are
19+
19+ nuclear charge
18
18
electron
attraction
nuclear
partially
cancel
core electrons
this
4s valence electron
positive
charge
approximately
and
the
effective
nuclear
charge
is
1:
▲
Z
≈
Z
S
=
19
18
=
Figure 3 Shielding of the outer valence electron in the
1
eff
potassium atom
Periodic trends in atomic radius
Across
a
period
the
increasing
the
period.
from
left
effective
to
right,
nuclear
atomic
charge,
radii
Z
,
decrease.
going
This
from
left
is
to
because
right
of
across
eff
nucleus,
Down
a
period
group
away
The
the
radii
reason
level
of
the
from
from
nuclear
trends
atomic
the
the
period.
are
of
for
top
to
bottom,
nucleus.
charge,
Z,
summarized
the
transition
this
is
that
the
are
atomic
enter
This
in
a
electrons
of
gures
number
number,
added
4
do
of
they
radii
new
has
because
elements
quantum
electrons
(outer-shell)
closer
to
the
radius.
electrons
the
principal
As
valence
atomic
outer-shell
increasing
These
pulls
reducing
the
further
This
a
increase.
energy
greater
shielding
and
not
5.
n,
by
Figure
electrons
in
remains
enter
effect
change
the
5
1)
are
new
located
the
core
shows
greatly
the
each
so
than
the
electrons.
that
across
outermost
almost
(n
In
level
constant
rather
a
the
period.
energy
across
than
the
th
n
energy
level.
So
the
number
of
valence
electrons
and
hence
Z
remain
eff
essentially
constant,
resulting
in
little
variation
in
atomic
radius.
puorg a nwod esaercni iidar cimota
atomic radii decrease across a period
▲
Figure 4 Trends in atomic radii. Some people think of these shapes as snowmen – going down
a group the snowman is standing upright, while across a period the snowman is sleeping!
78
3 . 2
P e r i o d i c
t r e N d S
2
0
0
K
300
L
i
N
a
R
b
2
3
8
C
s
9
M
C
a
g
S
2
0
6
S
c
Y
1
9
4
T
i
L
a
R
a
Z
2
0
1
)mp( suidar
A
H
r
B
a
200
3
2
9
B
e
2
4
2
F
r
1
4
4
V
r
1
5
6
c
N
H
C
r
b
f
M
o
T
a
F
e
T
c
W
R
u
8
4
C
o
R
e
R
h
P
d
O
s
C
P
7
5
Z
A
g
Ir
S
n
7
1
N
P
S
n
4
P
O
A
S
T
i
S
S
6
2
A
B
r
T
e
r
K
r
I
P
o
F
C
l
e
B
i
0
1
6
0
1
0
0
s
b
b
6
4
1
0
4
G
e
t
1
4
C
i
G
a
d
In
A
u
B
A
l
C
u
100
X
e
A
t
2
R
n
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
7
18
▲
Figure 5 Values of atomic radii of elements in pm. These data can be found in section 9 of
the Data booklet
Periodic trends in ionic radius
The
radii
they
The
are
of
cations
formed
radii
of
in
and
the
cations
anions
vary
following
are
smaller
from
the
parent
atoms
from
which
way.
than
those
of
their
parent
atoms;
for
ins
example,
the
atomic
radius
of
K
is
200
pm
while
the
ionic
radius
of
+
K
is
138
electrons
attracted
The
radii
pm.
in
to
of
The
the
the
reason
cation
so
for
the
this
is
that
valence
there
are
electrons
more
are
protons
more
than
anions:
nucleus.
anions
are
An n is a charged species.
Ions are either cations or
strongly
larger
than
those
of
their
parent
atoms;
•
for
A an is positively
+
example,
F
is
133
greater
the
pm.
atomic
This
repulsion
is
radius
of
because
between
F
is
the
the
60
pm
extra
valence
while
electron
the
in
electrons.
ionic
the
radius
anion
charged, such as Na
of
results
in
•
2+
, Mg
An ann is negatively
2
charged, such as Cl
, O
79
3
P E R I O D I C I T Y
Values
for
ionic
radii
are
also
given
in
section
9
of
the
Data
booklet
Suy p
An easy way to remember the
uorine
potassium
dierence in size of ionic radii
is as follows:
Aol
America onine
Anion larger
+
K
F
K
F
(It follows from this that
atomic radius of potassium (K) = 200 pm
atomic radius of uorine (F) = 60 pm
cations are smaller.)
+
ionic radius of K
= 138 pm
ionic radius of F
= 133 pm
▲ Figure 6 Atomic and ionic radii for potassium and uorine
Ionization energy
The
an
ionization
electron
The
rst
energy,
from
a
IE,
neutral
ionization
is
the
minimum
gaseous
energy,
IE
,
atom
of
a
in
energy
its
gaseous
required
to
remove
ground-state.
atom
relates
to
the
process:
1
+
X(g)
The
→
X
second
(g)
+
e
ionization
energy
relates
to
the
removal
of
a
further
electron
+
from
the
ion
X
(g),
and
the
third
ionization
energy
is
associated
with
the
2+
removal
of
another
electron
from
X
(g).
Values
of
ionization
energy
are
1
quoted
The
kJ
values
section
as
in
8
there
of
of
is
mol
rst
the
an
(per
mole
ionization
Data
input
booklet.
of
of
atoms).
energies
for
Ionization
energy
in
order
the
elements
energy
to
values
remove
an
are
are
provided
always
in
positive,
electron.
Periodic trends in ionization energy
Ionization
left
1
to
As
energies
right
the
vary
ionization
effective
across
energy
nuclear
the
periodic
values
charge,
table.
increase
Z
,
for
Across
the
increases
a
period
following
from
left
to
from
reasons:
right
across
eff
Suy p
a
period
the
Trends in ionization energy
valence
attraction
makes
across a period and down a
the
it
electrons
between
more
difcult
the
to
are
pulled
electrons
remove
and
an
closer
the
to
the
nucleus
electron
from
nucleus,
so
increases.
the
This
atom.
group are the pps to the
2
Atomic
radii
decrease
across
a
period
–
because
the
distance
between
trends in atomic radius. The
the
valence
electrons
and
the
nucleus
decreases,
it
becomes
more
snowman diagram (gure 4)
difcult
to
remove
an
electron
from
the
atom.
and its opposite (gure 7) will
Going
down
a
group
from
top
to
bottom
ionization
energy
values
help you remember both, and
decrease
for
the
following
reasons:
you need to know the reasons
underlying these trends.
1
Atomic
radii
electron
80
increase
from
the
down
atom.
a
group,
making
it
easier
to
remove
an
3 . 2
P e r i o d i c
t r e N d S
puorg a nwod esaerced seigrene noitazinoi
ionization energies increase across a period
▲
Figure 7 Trends in ionization energy are the opposite of the trends in atomic radius
2372
He
2081
Ne
2500
1681
F
1312
H
1402
1520
N
1314
Ar
O
2000
1086
1251
C
1351
Cl
Kr
900
1012
1
Be
1000
1140
801
P
S
Br
B
I
E
1
lom Jk/
1500
11
70
787
944
941
738
Xe
Si
As
Mg
Li
590
1008
I
869
Sb
Ge
Al
496
Na
Se
944
762
578
1037
Te
579
Rn
920
709
Ca
1000
Ga
At
812
Sn
419
558
549
K
In
Sr
7
16
Po
703
Pb
Bi
589
403
Ti
503
y
g
r
e
n
e
Rb
Ba
376
500
Cs
atomic number Z
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
7
18
increasing ionization energy
▲
Figure 8 Trends in rst ionization energy, IE
, for groups 1
2 and 13
18 of the periodic
1
table. IE
values increase across a period and decrease down a group
1
2
The
shielding
nuclear
and
If
a
outer
graph
shown
in
increase
is
not
topic
effect
charge,
of
electrons
rst
gure
across
smooth
the
in
the
ionization
9,
a
of
core
weakening
the
a
energy
but
versus
trend
decrease
period.
increases
attractive
force
faster
between
than
the
the
nucleus
atom.
general
period
across
electrons
the
The
is
atomic
that
down
spikes
rst
a
and
number
ionization
group,
dips
is
though
will
be
plotted,
energy
the
as
values
graph
explained
in
12.
81
3
P E R I O D I C I T Y
2500
He
Ne
2000
1
Ar
Kr
Xe
Rn
I
E
1
lom Jk/
1500
Hg
1000
Zn
Ga
Al
500
Li
Cd
Tl
ln
Na
K
10
Rb
20
30
Cs
40
50
Fr
60
70
80
90
100
Z
▲
Figure 9 Plot of rst ionization energy, IE
, versus atomic number, Z. Notice the general
1
trend that IE
increases across a period but decreases down a group, though the graph is
1
not smooth across a period
Electron anity
According
to
IUPAC,
the
electron
afnity,
E
,
is
the
energy
required
to
ea
detach
This
is
X
A
an
the
(g)
more
the
electron
energy
→
X(g)
common
energy
from
the
associated
+
singly
with
mol
of
and
released
neutral
X(g)
+
Electron
e
→
equivalent
(E
E
afnity
example,
for
atoms
X
the
negative
ion
in
the
gas
phase.
process:
e
initial
1
charged
denition
)
when
1
is
that
mol
of
the
electron
electrons
is
afnity
attached
is
to
nal
or
molecules
in
the
gas
in
section
phase:
(g)
values
are
provided
8
of
the
Data
booklet.
For
uorine:
1
F(g)
+
e
→
F
(g)
E
=
-328
kJ
mol
ea
The
negative
process
is
sign
indicates
exothermic
endothermic
(in
process).
that
energy
contrast
The
more
to
is
released
ionization
negative
the
during
energies
E
value,
this
process:
which
the
relate
greater
the
to
is
an
the
ea
attraction
of
the
ion
for
the
electron.
However,
gure
10
shows
that
the
E
ea
values
for
some
elements,
for
example
group
1
1
H
60
3
Na
53
4
K
48
5
Rb
47
noble
gases,
are
positive.
He
2
Li
the
18
13
14
15
16
1
7
73
2
18
>0
Be
>0
Mg
B
27
Al
>0
42
Ca
Ga
2
Sr
5
41
In
29
C
N
122
>0
Si
P
134
72
Ge
As
119
78
Sn
Sb
107
101
O
F
141
Ne
328
S
Cl
200
Ar
349
Se
Br
195
I
190
>0
Kr
325
Te
>0
>0
Xe
295
>0
1
▲
Figure 10 Electron anities E
, in kJ mol
, for a selection of main-group elements.
ea
Notice that some of the elements have positive E
values. The group 18 elements have
ea
theoretical, calculated values
82
3 . 2
P e r i o d i c
t r e N d S
Periodic trends in electron anity
Trends in electron anity across a period
Trends
as
the
in
electron
trends
across
a
afnity
observed
period
from
in
for
left
the
periodic
atomic
to
right
radius
E
table
and
values
are
not
as
ionization
become
well
highlighted
energy.
more
In
negative
general,
(with
ea
some
The
exceptions).
group
17
elements,
the
halogens,
have
the
most
negative
E
values:
ea
1
for
example,
E
(Cl)
=
-349
kJ
mol
.
This
is
expected
since
on
gaining
Quk qusn
ea
an
electron
these
elements
attain
the
stable
noble
gas
conguration.
If
Suggest why the E
values for the
ea
you
look
across
period
4
(n
=
4
energy
level)
in
gure
10
you
can
see
group 2 elements are more positive
1
that
from
left
to
right
E
becomes
more
negative
from
48
kJ
mol
forK
ea
than expected.
1
to
325
kJ
energies,
mol
there
for
are
Br.
However,
examples
of
within
elements
each
that
period,
do
not
as
for
follow
ionization
this
trend.
For
1
example,
arsenic,
As,
has
E
78
kJ
mol
while
you
might
expect
this
to
ea
1
lie
E
between
value
119
for
As
kJ
can
1
mol
be
for
Ge
explained
and
by
195
kJ
mol
examining
its
for
Se.
electron
The
higher
conguration
ea
10
[Ar]3d
1
2
4s
4p
1
4p
x
already
applies
y
contains
for
1
4p
:
an
electron
is
added
it
will
enter
a
4p
orbital
that
z
one
other
if
electron,
members
of
causing
group
repulsion.
15,
in
A
similar
particular
for
argument
nitrogen
where
the
Positive E
values
ea
E
value
is
positive.
ea
A
positive
afnity
value
for
suggests
electron
that
the
anion
Trends in electron anity down a group
is
In
the
case
of
the
group
1
alkali
metals,
values
of
E
generally
become
not
stable,
formed
in
so
the
it
cannot
gas
phase.
be
For
ea
less
negative
going
down
the
group
(table
1).
However,
for
the
last
three
example,
E
for
krypton
is
ea
or
four
elements
there
is
little
difference
between
E
1
values.
positive
(41
kJ
mol
),
so
Kr
ea
does
not
exist.
Interestingly,
3
1
Gup 1 mn
E
the
N
anion
the
solid
is
well
known
in
/kJ m
a
Li
60
sodium
state
(for
nitride,
example,
Na
N),
in
despite
3
Na
the
53
fact
that
E
for
nitrogen
ea
1
is
K
positive
(20
kJ
mol
).
48
3
In
Rb
47
Cs
46
by
crystals,
the
lattice
(sub-topic
provides
Fr
T
able 1 Electron anity values for the group 1 elements
The
do
patterns
not
show
ionization
is
stabilized
enthalpy
15.1),
which
sufcient
energy
47
to
▲
N
of
electron
the
same
energy,
and
afnity
clear
vary
trends
by
electronegativity
repulsion
group,
down
a
overcome
so
group
electron
as
(discussed
do
afnity
atomic
in
the
the
electron
nitride
anion.
values
radius,
next).
Electronegativity
Electronegativity,
atom
In
has
1932
for
the
the
American
electronegativity
attract
scales
electrons
but
which
the
has
symbol
shared
and
to
one
the
χ,
pair
scientist
dened
itself”.
used
symbol
in
χ
.
is
of
it
Linus
as
There
this
8
as
in
the
a
Pauling
“the
are
section
On
dened
electrons
a
of
proposed
power
number
the
scale
relative
covalent
of
an
of
Data
the
atom
the
is
concept
in
different
booklet
uorine,
attraction
that
an
bond.
molecule
to
electronegativity
the
most
a
of
Pauling
scale,
electronegative
p
element
in
the
periodic
table,
has
a
value
of
electronegativity
of
4.0.
83
3
P E R I O D I C I T Y
Periodic trends in electronegativity
As
shown
period
the
same
Across
the
in
and
a
gure
down
a
reasons
period
effective
11,
electronegativities
group
(see
from
nuclear
that
pages
left
to
charge
mirror
show
those
for
periodic
trends
ionization
across
energies,
a
for
80–81).
right
and
electronegativity
atomic
radii
both
values
increase
increase
because
across
a
period.
puorg a nwod esaerced seitivitagenortcele
electronegativities increase across a period
▲
Figure 11 Trends in electronegativity are the same as those in ionization energy and the
opposite to the trends in atomic radius
Down
a
group
because
from
atomic
increases,
its
top
radii
effect
to
bottom
increase
is
shielded
and
by
electronegativity
although
the
core
the
values
nuclear
decrease
charge,
Z,
electrons.
H
B
e
N
a
K
4
M
g
S
c
T
i
R
b
r
Y
Z
r
X
p
B
a
F
r
L
a
N
b
f
A
N
i
R
h
T
a
R
e
F
C
o
W
R
a
O
F
e
T
c
R
u
H
N
C
B
M
o
C
s
2
V
C
r
S
C
u
A
l
P
d
O
s
c
Ir
P
0
Z
A
g
t
A
u
C
n
G
a
In
A
s
i
P
S
e
S
C
l
B
r
S
n
H
g
3
S
d
1
2
G
e
S
b
T
e
I
T
i
4
P
b
B
i
5
P
o
6
A
t
7
8
9
10
11
12
13
14
15
16
1
7
▲
Figure 12 Electronegativity values, χ
, increase across a period from left to right and decrease down a group from top
p
to bottom. Fluorine is the most electronegative element in the table with a χ
p
84
value of 4.0 on the Pauling scale
3 . 2
P e r i o d i c
t r e N d S
Science and peace
Pauling
Nobel
in
was
the
Prizes,
1962
for
as
his
rst
he
person
also
to
won
opposition
to
win
the
two
Nobel
weapons
Pauling
unshared
Peace
of
also
vitamin
Prize
the
mass
is
destruction.
C
common
given
in
Pauling’s
●
Do
you
know
of
any
other
scientists
who
peace
through
their
scientic
role
peace
in
can
scientists
world
play
today?
in
the
Discuss
promotion
this
taking
may
be
structure
of
the
Data
correct?
large
doses
effective
of
ascorbic
booklet.)
Carry
out
of
against
acid
Was
this
aspect,
using
the
some
library,
the
work?
in
literature,
and
an
online
search.
Discuss
of
your
the
(The
35
suggestion
into
scientic
What
cold.
section
that
acid)
have
research
promoted
suggested
(ascorbic
ndings
in
class.
class.
Periodic trends in metallic and non-metallic
character
As
described
classied
topic
in
into
sub-topic
metals,
3.1,
the
elements
non-metals,
and
in
the
metalloids
periodic
(see
table
gure
3
can
in
be
sub-
3.1).
Metallic
shown
character
in
gure
decreases
across
a
period
and
increases
down
a
group,
as
13.
puorg a nwod sesaercni retcarahc cillatem
metallic character decreases across a period
▲
Figure 13 Trends in metallic character in the periodic table
As
well
as
the
properties
metals
also
have
to
electrons
low
of
metals
ionization
described
energy
previously
values
–
they
in
sub-topic
have
a
3.1,
tendency
1
lose
oxidized.
We
during
shall
a
chemical
explore
redox
reaction,
processes
that
is,
they
further
in
tend
topic
to
18
be
9.
1
2
13
14
15
+
The
properties
addition,
of
non-metals
non-metals
show
were
highly
also
described
negative
in
electron
sub-topic
afnities
3.1;
–
in
they
have
2
Li
N
to
gain
electrons
during
a
chemical
reaction,
that
is,
they
tend
3
Na
4
K
reduced.
5
Rb
6
Cs
2+
Mg
14
metals.
always
In
shows
For
1+,
topic
13
different
the
and
we
stable
the
charges
cations
for
of
the
shall
see
of
the
some
alkali
alkaline
that
the
common
metals
earth
in
metals
transition
ions
group
in
of
1
group
metals
metals
the
2
form
it
a
and
charge
is
non-
is
always
number
F
3+
Al
3
P
2
S
Cl
2+
Ca
+
Figure
O
to
+
be
1
7
2
a
+
tendency
16
3
+
2
Se
2+
Sr
Br
2
Te
I
2+
Ba
2 +.
of
▲
Figure 14 The charges of some common ions
ions.
85
3
P E R I O D I C I T Y
Trends in the proper ties of metal and non-metal oxides
Suy p
You can work out the formulas
of the main-group metal
oxides and hydroxides
An
oxide
We
make
deduce
oxide
is
the
ion
formed
use
of
from
the
chemical
to
be
2
,
the
charge
combination
on
formula
for
the
of
a
metal
metal
of
an
cation
oxide,
element
as
with
shown
taking
in
the
oxygen.
gure
charge
15
on
to
the
example:
using the periodic table.
+
●
2
Na
combines
with
O
to
form
Na
2+
of the non-metals are less
O
2
The corresponding oxides
●
Ca
●
Al
2
combines
with
O
to
form
CaO
straightforward and you
3+
2
combines
with
O
to
form
Al
O
2
3
should memorize these
for the elements carbon,
Metal
nitrogen, sulfur, phosphorus,
hydroxides:
oxides
are
basic:
they
react
with
water
to
form
metal
and halogens along with the
CaO(s)
+
H
O(l)
→
Ca(OH)
2
(aq)
2
corresponding acids formed.
Na
O(s)
+
H
2
In
contrast,
form
oxides
acidic
CO
(g)
+
H
(l)
+
H
(g)
O
4
of
the
O(l)
⇋
non-metals
H
+
O(l)
H
→
H
are
acidic:
they
react
with
water
to
O(l)
⇋
+
SO
6H
10
H
(aq)
SO
→
carbonic
acid
sulfuric
acid
4
2
O(l)
(aq)
3
2
2
(s)
CO
2
2
2
P
2NaOH(aq)
2
3
SO
→
solutions:
2
SO
O(l)
2
(aq)
sulfurous
acid
3
4H
2
PO
3
(aq)
phosphoric
acid
4
Namng xanns an as
In naming oxoanions the following rules are useful:
Students often struggle with the names of the oxoanions
•
If only one oxoanion is formed, the ending is “-ate”.
•
If two oxoanions are formed, the one with the smaller
and their corresponding oxoacids. Table 2 summarizes
some of these names.
number of oxygens ends in “-ite” and the one with the
greater number of oxygens ends in “-ate”.
Fmua f xann
Nn-sysma nam
If
•
2
there
are
four
oxoanions ,
the
one
w ith
the
carbonate
CO
3
smallest
number
of
o xy gens
ends
in
“ -ite”
and
is
2
C
ethanedioate (oxalate)
O
2
prefixed
4
ends
in
by
“hypo”;
“-ate”,
and
the
the
next
one
ends
w ith
in
the
“ -ite”;
mos t
the
third
oxyg en s
nitrite
NO
2
is
nitrate
NO
prefixed
oxoanions
3
this
2
by
of
system
“per ”
and
chlorine,
(table
end s
in
bromine,
“ -ate”.
and
T he
four
iodine
follow
3) .
sulte
SO
3
2
sulfate
SO
Fmua
Nn-
Fmua
Nn-
f
sysma
f xa
sysma
xann
nam
4
3
phosphite
PO
3
nam
3
phosphate
PO
4
ClO
ClO
hypochlorite
HClO
hypochlorous
hypochlorite
acid
ClO
chlorite
ClO
2
chlorite
chlorate
ClO
3
chlorate
ClO
4
86
perchlorate
HClO
4
hydroxide
▲
▲
chloric acid
perchlorate
4
OH
HClO
3
3
ClO
chlorous acid
2
2
ClO
HClO
T
able 2 The non-systematic names of some oxoanions
T
able 3 The oxoanions and acids of chlorine
perchloric acid
3 . 2
P e r i o d i c
t r e N d S
Some interesting oxides
●
Silicon
dioxide,
SiO
,
does
not
dissolve
in
water.
However,
it
is
2
classied
as
hydroxide,
an
acidic
NaOH
to
oxide
form
because
sodium
it
can
react
silicate,
Na
with
SiO
2
SiO
(s)
+
2NaOH(aq)
→
Na
2
●
Aluminium
oxide,
Al
O
2
means
SiO
2
it
can
react
is
(aq)
+
H
3
classied
sodium
(aq)
and
water:
3
O(l)
2
as
an
amphoteric
oxide.
This
3
both
as
an
acid
and
as
a
base.
See
topic
8
for
more
information.
Acting
Al
as
O
2
an
(s)
acid:
+
2NaOH(aq)
+
3H
3
O(l)
→
2NaAl(OH)
2
sodium
Acting
as
O
Al
2
a
(aq)
4
aluminate
base:
(s)
+
6HCl(aq)
→
2AlCl
3
(aq)
+
3H
3
O(l)
2
aluminium
chloride
Amph an amphp xs
The terms amphoteric and amphiprotic are often mixed
up.
A par ticular type of amphoteric species is described
•
Amphiprotic species are described fur ther in sub-
as amphp. These are species that are either
+
topic 8.1.
proton (H
) donors or proton acceptors. Examples
include self-ionizing solvents (such as water, H
According to the IUPAC Gold Book, a chemical species
•
O and
2
methanol, CH
that behaves both as an acid and as a base is termed
OH), amino acids, and proteins.
3
amph. Aluminium oxide is classied as an
amphoteric oxide.
Table
that
4
shows
there
across
the
is
a
how
the
trend
period
oxides
from
from
Fmua f x
left
Na
of
basic
to
some
period
through
▲
elements
to
vary.
acidic
It
shows
oxides
right.
O(s)
MgO(s)
Al
2
Nau f x
3
amphoteric
O
2
as
as
(s)
SiO
3
(s)
P
2
amph
O
4
a
(s)
SO
10
(l) and SO
3
(g)
2
a
a
T
able 4 Trend in the proper ties of the oxides of some period 3 elements
Chemical properties within a group: Group 1, the alkali metals
+
The
group
1
potassium,
metals
K,
are
lithium,
rubidium,
Rb,
Li,
sodium,
caesium,
Cs,
Na
Na,
1e
→
Na
and
1
[Ne]3s
francium,
that
Fr
(see
hydrogen
metals
–
it
is
a
is
sub-topic
not
a
3.1,
member
non-metal
and
gure
of
a
the
3).
alkali
On
group
1
metals
are
descending
and
gas.
of
The
the
the
having
one
valence
group
ionization
alkali
1
the
energy
metals
with
atomic
radius
decreases.
water
The
therefore
increases
reactions
become
characterized
more
by
[Ne]
Note
electron;
they
vigorous
further
down
the
group.
Less
energy
therefore
is
required
to
remove
the
valence
electron
from
+
form
the
ion
M
in
ionic
compounds
by
1
potassium,
K
(IE
=
419
kJ
mol
)
than
from
1
losing
this
electron
(they
are
oxidized,
1
sodium,
Na
(IE
=
496
kJ
mol
),
for
example.
1
topic
9).
For
example:
87
3
P E R I O D I C I T Y
alkaline
Reaction with water
solution
liberated
The
a
group
metal
1
metals
hydroxide,
react
with
MOH(aq),
water
to
which
in
this
(table
5).
Hydrogen
gas
Li
2M(s)
an
+
2H
O(l)
→
2MOH(aq)
+
H
+
2H
(g)
2
ran wh wa
2Li(s)
also
form
gives
2
Gup 1 ma
is
reaction:
dspn
O(l) → 2LiOH(aq)
+
H
2
Lithium reacts slowly and oats on the water (due
(g)
2
to its low density). Bubbling is observed.
Na
2Na(s)
+
2H
O(l) → 2NaOH(aq)
+
H
2
Sodium reacts vigorously. Heat is evolved and the
(g)
2
sodium melts to form a ball of molten metal which
whizzes around on the surface of the water.
K
2K(s)
+
2H
O(l) → 2KOH(aq)
+
H
2
Potassium reacts more vigorously than sodium: the
(g)
2
reaction is violent. It evolves enough heat to ignite
the hydrogen, so bursts into ames instantly. A
characteristic lilac-coloured ame is observed.
Rb
2Rb(s)
+
2H
O(l) → 2RbOH(aq)
+
H
2
Both rubidium and caesium react explosively with
(g)
2
water.
Cs
2Cs(s)
+
2H
O(l) → 2CsOH(aq)
+
H
2
▲
(g)
2
T
able 5 Reactions of the alkali metals with water become progressively more violent as you descend the group
Only two elements in the
Chemical properties within a group:
periodic table exist as liquids:
bromine, Br
Group 17, the halogens
and mercury, Hg.
2
The
group
chlorine,
gure
3).
electrons,
noble
17
Cl,
gas
elements,
bromine,
Their
giving
Cl
+
2
The
them
a
group
e
halogens,
iodine,
chemistry
conguration
[Ne]3s
the
Br,
is
17
tendency
(they
are
→
to
the
non-metals
astatine,
gain
by
an
reduced,
At
their
(see
seven
electron
topic
9).
to
For
uorine,
F
,
sub-topic3.1,
valence
attain
the
example:
Cl
2
[Ne]3s
elements
are
and
characterized
5
3p
I,
exist
6
3p
as
or
simply
diatomic
[Ar]
molecules
X
.
Fluorine
and
2
chlorine
room
The
gases,
halogens
structure
the
halogens
highly
group
with
atomic
an
the
most
radius
electron.
in
liquid,
me ta l
of
co v a l ent
tho ug h
decrease
a
compo und s
the
b o ndi ng
form
is
iodine
and
astatine
are
solids
at
pressure.
i o ni c
w i th
and
reactive,
this
and
form
combining
the
for
bromine
temperature
anion
88
are
the
i onic
co mpou n ds .
r e acti ve
do wn
t op ic
ha logen
be i n g
g r ou p
w it h
4
for
Wi th
Ha lo ge n s
d ec r e as e s
d es cendi ng
the
m e t a ls ,
(s e e
co m p ou n ds ).
r e activi ty
r e a ctivi ty
incr e a s e s
wit h
ca tion
in
g ro up
m a ki n g
it
X
de t a il s
is
of
n on - m e t a ls
g en e r al
g oi ng
 uo ri n e .
the
the
down
The
that
l es s
are
the
r ea s on
the
e a sy
to
ga i n
3 . 2
P e r i o d i c
t r e N d S
Reaction between halogens and alkali metals
The
halogens,
X
,
react
with
the
alkali
the
ionic
metals,
M(s)
to
form
ionic
alkali
2
metal
halide
salts,
MX(s).
In
compound,
MX(s),
the
cation
is
+
M
and
the
2M(s)
anion
+
X
is
(g)
X
→
:
2MX(s)
2
For
example:
2Na(s)
+
Cl
(g)
→
2NaCl(s)
2
Reactions between halogens and halides
A
solution
of
a
more
reactive
halogen,
X
(aq),
will
react
with
a
2
solution
of
summary
In
table
an
of
6
complete
ions,
these
the
aqueous
reactions
bromide,
of
and
(aq),
are
is
of
given
can
chlorine
aqueous
the
formed
also
is
Br
as
be
potassium
(aq)
a
reactive
ionic
to
halogen.
A
6.
equations.
written.
a
For
is
A
example,
colourless
chloride
colour
(gure
less
table
added
yellow/orange
bromine,
in
by
represented
equation
solution
colourless,
formation
X
reactions
balanced
potassium
is
halide
formed,
observed
is
when
solution
due
of
which
to
the
15):
2
▲
Figure 15 Gaseous chlorine, Cl
(g), is bubbled
2
Cl
(aq)
+
2KBr(aq)
→
2KCl(aq)
+
2
Br
(aq)
2
colourless
through a solution of potassium bromide, which
yellow/orange
is initially colourless. On reaction, aqueous
bromine is displaced from the potassium
bromide solution and the yellow/orange colour
of Br
(aq) is obser ved
2
Suy p
You can think of this displacement reaction as being a competition between the
chlorine and the bromine for an extra electron. Remember that the atomic radius
increases down a group (gure 4). The atomic radius of chlorine (100 pm) is
smaller than that of bromine (117 pm) so chlorine has a stronger attraction for
a valence electron than does bromine. Therefore chlorine forms the chloride
anion, Cl
more readily than bromine forms the bromide anion, Br
. Going down
group 17 the xzng ay, that is, the ability to gain an electron, decreases.
X
(aq)
c
(aq)
b
(aq)
i
(aq)
2
c
(aq)
no reaction
2
Cl
(aq)
+
2Br
(aq) → 2Cl
(aq)
2
+
Br
(aq)
Cl
2
observation: yellow/orange solution due to
formation of Br
(aq)
+
2I
(aq) → 2Cl
(aq)
(aq)
no reaction
no reaction
2
+
I
(aq)
2
to formation of I
2
b
(aq)
2
observation: dark red/brown solution due
(aq)
2
Br
(aq)
+
2I
(aq) → 2Br
(aq)
2
+
I
(aq)
2
observation: dark red/brown solution due
to formation of I
(aq)
2
i
(aq)
no reaction
no reaction
no reaction
2
▲
T
able 6 Reactions between halogens X
(aq) and halides X
(aq)
2
89
3
P E R I O D I C I T Y
Suy ps
1
The order of oxidizing ability for the group 17 elements follows the order of
electronegativity:
F
χ
:
>
Cl
4.0
>
Br
3.2
>
I
3.0
2.7
p
oxidizing ability:
F
>
Cl
2
2
>
Br
2
>
I
2
2
Be careful with the term s van when describing a chemical reaction. An
observation is something that you directly witness, such as bubbles of a gas,
the colour of a solution, or a precipitate forming. The formation of a gas is not
in itself an observation.
Worked example: explaining pH changes
Construct
the
pH
(see
a
balanced
changes
sub-topic
for
equation,
the
including
reaction
of
state
nitrogen
symbols,
dioxide
to
with
explain
water
8.1).
Solution
●
Nitrogen
NO
is
a
reacts
non-metal
with
water
and
to
therefore
form
a
1 :1
may
form
mixture
of
an
acidic
nitrous
oxide.
acid,
HNO
2
and
,
2
nitric
acid,
HNO
.
Nitrous
acid
is
a
weak
acid
and
nitric
acid
is
3
a
●
strong
We
acid.
next
2NO
+
write
H
2
●
Finally,
2NO
O
we
(g)
Because
less
90
→
balanced
HNO
2
+
2
●
the
+
include
H
O(l)
the
→
a
3
state
HNO
(see
symbols:
(aq)
+
HNO
2
mixture
7
of
topic
equation:
HNO
2
2
than
chemical
acids
8).
(aq)
3
is
formed
the
pH
of
the
solution
will
be
Q U e S t i o N S
Questions
1
What
can
A.
2
is
the
maximum
occupy
a
d
number
of
electrons
that
6
C.
6
D.
10
Which
of
the
following
elements
can
3
as
Cl
B.
K
C.
Br
D.
F
IB,
May
[1]
2010
What
happens
when
sodium
is
added
to
water?
I.
A
gas
II.
The
is
evolved.
temperature
of
the
water
increases.
Si
III.
A
A.
I
and
clear,
II
colourless
B.
I
and
III
C.
II
and
D.
I,
II,
IB,
November
solution
is
formed.
Te
I
and
II
B
I
and
III
C
II
and
D
I,
II,
How
2
B.
6
and
valence
electrons
does
Which
earth
and
III
[1]
2009
selenium
8
Which
II.
34
only
III
P
oxides
to
produce
an
acidic
solution
when
water?
O
4
D.
III
only
I.
16
only
only
III
many
C.
only
only
added
A.
II.
radius?
Al
A
I.
largest
metalloids?
contain?
4
the
be
7
classied
III.
has
+
5
II.
ion
A.
2
B.
I.
Which
sublevel?
10
MgO
III.
SO
A.
I
and
II
B.
I
and
III
C.
II
and
D.
I,
II,
IB,
May
3
of
the
following
elements
are
only
only
alkaline
III
only
metals?
and
III
[1]
Rb
2010
Sr
III.
Ba
A.
I
and
II
9
only
Which
17
B.
I
C.
II
and
and
III
D.
I,
II,
statement
about
the
elements
in
group
correct?
only
III
and
is
A.
only
Br
will
oxidize
Cl
2
B.
III
F
has
the
least
tendency
to
be
reduced.
2
C.
Cl
D.
I
will
oxidize
I
2
5
Which
period
property
generally
decreases
IB,
Atomic
B.
Electronegativity
C.
Atomic
D.
First
IB,
May
a
stronger
2
across
3?
A.
is
May
oxidizing
agent
than
F
.
[1]
2
2011
number
radius
ionization
energy
[1]
2011
91
3
P E R I O D I C I T Y
10
How
many
of
the
following
oxides
12
are
Na
O,
MgO,
Al
2
O
2
A.
None
B.
1
C.
2
D.
4
Describe
,
SiO
3
periodic
table
explain
is
what
bubbled
you
through
will
a
see
if
solution
of:
a)
potassium
iodide
[2]
b)
potassium
uoride.
[1]
IB,
May
The
shows
the
a)
relationship
2010
alkali
periodic
The
gas
2
13
11
and
chlorine
amphoteric?
metals
table
State
the
of
are
found
in
group
1
of
the
elements.
full
electron
conguration
+
between
electron
properties
of
arrangement
elements
and
is
a
and
of
the
valuable
tool
predictions
in
and
Identify
the
elements
in
property
the
Describe
used
periodic
to
arrange
table.
the
c)
[1]
State
rst
Outline
two
what
K
you
reasons
ionization
and
energies
down
b)
ion,
understand
by
the
chemistry.
term
a)
its
for
b)
making
K
explain
of
the
group
energy
how
alkali
the
rst
metals
ionization
vary
going
1.
why
+
electronegativity
period
one
3
in
reason
assigned
the
increases
periodic
why
noble
across
table
gases
electronegativity
d)
and
are
smaller
not
values.
parent
May
2010
why
than
atom,
Suggest
an
why
alkali
working
92
the
the
ionic
radius
atomic
of
radius
K
of
is
the
K.
[3]
e)
IB,
Explain
you
metal
in
the
should
with
never
your
touch
ngers
laboratory.
when
C H E M I C A L
B O N D I N G
A N D
4
S T R U C T U R E
Introduction
At
the
very
heart
understanding
structural
chemical
that
this
can
atoms
oppositely
together
in
topic
of
chemistry
chemical
arrangements
bond
holds
holds
of
the
we
be
in
together
case
of
an
in
a
ions
and
as
the
the
covalent
“glue”
we
different
shall
electron
or
used
species)
we
In
types
of
to
ionic,
covalent,
as
in
how
pair
a
also
For
look
the
at
and
metallic
between
covalent
simple
repulsion
polarity
and
structure
compounds.
see
determine
shall
such
–
differences
A
compound.
three
at
the
molecule,
(charged
ionic
explore
bonding
our
compounds.
considered
charged
shall
lies
bonding
model,
(VSEPR)
shape
some
of
key
–
look
and
compounds
valence
theory,
a
and
ionic
shell
can
molecule,
chemical
intermolecular
be
and
principles,
forces.
4.1 Ioni ondin nd stt
Understandings
Applications and skills
➔
Positive ions (cations) form by metals losing
➔
Deduction of the formula and name of an ionic
valence electrons.
compound from its component ions, including
➔
Negative ions (anions) form by non-metals
polyatomic ions.
gaining electrons.
➔
➔
The number of electrons lost or gained is
determined by the electron conguration of the
Explanation of the physical proper ties of ionic
compounds (volatility, electrical conductivity,
and solubility) in terms of their structure.
atom.
➔
The ionic bond is due to electrostatic attraction
between oppositely charged ions.
➔
Under normal conditions, ionic compounds are
usually solids with lattice structures.
Nature of science
➔
Use theories to explain natural phenomena –
melting points of ionic compounds can be used
molten ionic compounds conduct electricity but
to explain obser vations.
solid ionic compounds do not. The solubility and
93
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Ionic bonding
Dnition of n ioni
Ions
are
formed
when
one
or
more
electrons
are
transferred
from
one
ond
atom
An ioni ond refers to the
to
another.
formation
of
a
The
noble
driving
gas
force
electron
for
this
electron
transfer
is
usually
the
conguration.
electrostatic attraction
For
example,
the
electron
conguration
of
sodium,
Na
is:
experienced between the
1
[Ne]3s
electric charges of a
tion (positive ion) and
where
[Ne]
is
the
noble
gas
core.
A
sodium
an nion (negative ion).
atom
can
lose
its
one
valence
+
(outer-shell)
electron
to
form
the
Na
cation,
[Ne].
That
is:
+
Na
We
e
say
The
→
that
Na
sodium
electron
If
a
oxidized
conguration
2
[Ne]3s
is
of
in
chlorine,
Cl
We
+
say
Cl
(it
loses
an
electron).
is:
3p
atom
gains
an
electron
2
gas
process
5
chlorine
noble
this
conguration,
e
→
that
[Ne]3s
to
form
the
or
[Ar].
That
Cl
anion
it
will
adopt
a
6
3p
is:
Cl
chlor i ne
is
reduced
in
this
process
(it
g a i ns
an
electron).
Hence,
the
formation
Ionic
but
electron
of
the
compounds
note
that
between
a
that
ionic
are
the
by
sodium
is
compound
sodium
chloride,
lost
generally
strict
cation
is
and
formed
denition
an
anion
between
involves
(for
gained
by
chlorine
metals
electrostatic
example,
the
in
the
NaCl.
and
non-metals,
attraction
compound
ammonium
+
chloride,
NH
Cl,
which
consists
of
the
ammonium
cation,
NH
4
chloride
Let
us
anion,
take
Cl
,
is
another
Magnesium
,
and
the
4
is
a
ionic,
but
example
group
2
of
does
an
alkaline
not
ionic
earth
contain
a
metal).
compound,
metal,
and
magnesium
so
has
two
oxide.
valence
electrons:
2
[Ne]3s
2+
A
magnesium
also
adopts
atom
the
can
[Ne]
lose
noble
these
gas
two
core.
electrons
That
forming
Mg
,
which
is:
2+
Mg
–
2e
Magnesium
Oxygen
is
electrons.
→
is
in
oxidized
group
The
2
[He]2s
Mg
16,
electron
in
the
this
process.
chalcogen
conguration
group,
of
and
oxygen
so
has
six
valence
is:
4
2p
2
An
oxygen
adopts
a
atom
noble
can
gas
gain
That
or
[He]2s
electrons
conguration:
2
[Ne]
two
6
2p
is:
2
O
+
Oxygen
94
2e
is
→
O
reduced
in
this
process.
to
form
the
O
anion,
which
4 . 1
Hence,
by
the
oxygen
oxide,
two
in
el e ctr o ns
the
that
f o r ma tio n
of
are
los t
the
by
ion i c
I O N I c
ma gn e s iu m
com p ou n d
a re
b O N D I N g
a N D
S T r u c T u r e
ga i ne d
ma g ne s iu m
Stdy tips
MgO.
•
Under
normal
lattice-type
of
positive
conditions,
structures
and
negative
ionic
that
compounds
consist
ions
of
(gure
are
typically
solids,
three-dimensional
and
repeating
have
You should know the names
of the various ions, their
units
formulas and charges,
1).
including some oxonions
(oxygen-containing anions,
2
+
such as NO
sodium ion Na
, SO
3
, etc.)
4
chloride ion Cl
(table 1).
+
•
+
It is incorrect to use the term
“molecule” when referring to
+
ionic compounds. We never
+
say “molecules of sodium
chloride”, but instead
+
“sodium chloride formula
units” to indicate that ions
+
+
are involved in the lattice
+
structure.
+
+
+
+
Figure 1 Lattice structure of sodium chloride, which consists of sodium cations, Na
chloride anions, Cl
+
Na
1
2
(102 × 10
, and
. From the ionic radii given in section 9 of the Data booklet you can see that
1
2
m = 102 pm) is smaller than Cl
(181 × 10
m = 181 pm)
TOK
Ion
●
Nm
General rules in
chemistry (such as the
+
ammonium
NH
4
octet rule) often have
exceptions. How many
hydroxide
OH
exceptions have to exist
nitrate
NO
for a rule to cease to
3
being useful?
hydrogencarbonate
HCO
3
●
2
carbonate
CO
What evidence do
scientists have for the
3
existence of ions? What
2
sulfate
SO
is the dierence between
4
direct and indirect
3
phosphate
PO
4
evidence? Topic 9 may
▲
help you when reecting
T
able 1 Names of various ions
on this point.
The octet rule
The
octet
and
can
rule
be
a
has
useful
its
own
starting
place
in
point
the
in
discussion
trying
to
of
chemical
understand
how
bonding
chemical
Qik qstion
bonds
(that
are
is,
undergo
electrons
rst
two
basis
of
formed.
in
rule
states
oxidation),
order
processes
covalent
The
to
acquire
are
the
bonding,
gain
a
that
electrons
noble
basis
of
which
elements
gas
ionic
we
tend
(reduction),
core
electron
bonding.
shall
to
discuss
The
in
lose
or
electrons
Can you think of an example
share
conguration.
third
process
sub-topic
is
The
the
in which the octet rule is not
obeyed?
4.2.
95
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Worked example: deduction of the formula and name of an ionic
compound
Deduce
and/or
the
formula
polyatomic
a)
magnesium
b)
aluminium
c)
sodium
and
and
and
and
name
of
the
ionic
compounds
formed
between
the
following
pairs
of
elements
species:
uorine
d)
oxygen
calcium
e)
and
ammonium
nitrate
and
phosphate.
oxygen
Solution
comintion
)
Fom
Nm
2+
magnesium and uorine
Mg is in group 2, so forms Mg
magnesium uoride
;
;
F is in group 17, so forms F
fom is MF
2
)
3+
aluminium and oxygen
Al is in group 3, so forms Al
aluminium oxide
;
2
;
O is in group 16, so forms O
O
fom is a
2
)
3
+
sodium and oxygen
Na is in group 1, so forms Na
sodium oxide
;
2
;
O is in group 16, so forms O
O
fom is N
2
d)
2+
calcium and nitrate
Ca is in group 2, so forms Ca
calcium nitrate
;
;
nitrate is NO
3
)
fom is c(NO
3
)
2
+
ammonium and phosphate
ammonium is NH
ammonium phosphate
;
4
3
;
phosphate is PO
4
)
fom is (NH
4
▲
(PO
3
)
4
T
able 2 Formulas and names of some ionic compounds from their component ions. In naming ionic binary compounds, ab,
consisting of a metal and a non-metal, the ending will be-ide
Physical properties of ionic compounds
Q
Melting and boiling points
,
and
inversely
proportional
to
the
square
2
2
of
Ionic
compounds
have
high
melting
the
distance
high
boiling
points
because
of
the
of
attraction
,
as
given
law
of
electrostatics
from
by
physics:
Q
1
forces
r
strong
Q
electrostatic
them,
points
Coulomb’s
and
between
between
2
_
the
F
∝
2
r
ions
in
their
melting
lattice
point
point
is
there
must
1413
of
structures.
NaCl
°C.
In
is
801
order
to
For
°C
example,
and
melt
its
an
the
boiling
ionic
solid
Hence,
in
charges
the
case
of
correspond
magnesium
to
2+
for
the
oxide,
the
2+
apart
the
be
a
large
input
electrostatic
of
energy
to
break
cation,
these
forces.
Mg
two
,
two
magnesium
2
and
2
charges
for
are
the
oxide
greater
anion,
than
those
O
of
.
As
1+
+
and
The
electrostatic
force
of
attraction,
F,
is
1
in
melting
proportional
to
the
interacting
charges,
Q
and
1
96
the
case
of
the
Na
and
Cl
ions,
the
directly
point
for
MgO
is
higher,
that
is
2852
°C.
4 . 2
such
Volatility
as
c O v a l e N T
hexane.
The
b O N D I N g
molecule
of
water
is
polar
+
and
Volatility
refers
to
the
tendency
of
a
has
vaporize.
For
ionic
compounds
the
of
attraction
are
strong,
and
so
itself,
δ
2
on
H
and
δ
partial
charges
are
attracted
to
the
on
ions
in
O.
the
electrostatic
lattice
forces
charges
substance
These
to
partial
the
(for
example,
in
the
case
of
sodium
chloride,
volatility
+
the
of
such
compounds
is
very
δ
on
each
H
in
the
water
molecule
is
attracted
low.
to
a
the
negatively
result
charged
individual
ions
chloride
are
pulled
anion,
out
of
Cl
the
).
As
lattice
Electrical conductivity
and
For
an
occupy
ions
ionic
xed
are
solid
compound
not
ionic
positions
free
to
contrast,
to
move
in
and
the
in
do
molten
conduct
the
the
move
compounds
In
in
in
solid
state
lattice.
the
not
Hence
solid
state,
conduct
state,
the
the
ions
the
so
electricity.
ions
are
become
case
of
a
between
the
the
solvent
remain
surrounded
non-polar
ions
of
the
molecules,
within
the
by
solvent,
water
there
ionic
so
the
molecules.
is
no
compound
cations
In
the
attraction
and
and
anions
lattice.
free
electricity.
uss of ioni iqids
Ionic liquids are ecient solvents and electrolytes,
Solubility
used in electric power sources and green industrial
Ionic
as
compounds
water,
but
do
dissolve
not
in
dissolve
polar
in
solvents
non-polar
such
processes.
solvents
4.2 c ont ondin
Understandings
Applications and skills
➔
A covalent bond is formed by the electrostatic
➔
Deduction of the polar nature of a covalent
attraction between a shared pair of electrons
bond from electronegativity values.
and the positively charged nuclei.
➔
Single, double, and triple covalent bonds
involve one, two, and three shared pairs of
electrons, respectively.
➔
Bond length decreases and bond strength
increases as the number of shared electrons
increases.
➔
Bond polarity results from the dierence in
electronegativities of the bonded atoms.
Nature of science
➔
Looking for trends and discrepancies –
➔
Use theories to explain natural phenomena – Lewis
compounds that contain non-metals have
introduced a class of compounds which share
dierent proper ties from compounds that
electrons. Pauling used the idea of electronegativity
contain non-metals and metals.
to explain unequal sharing of electrons.
97
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
cont ondin
Dnition of  ont ond
In
A ont ond is formed by the
electrostatic attraction between a shared
pair of electrons and the positively
charged nuclei. According to IUPAC (the
ionic
order
bonding
to
attain
chemical
other
in
a
bond
order
bonding
is
In
to
we
saw
noble
exists,
to
atoms
a
can
electron
however,
attain
covalent
how
gas
in
noble
which
gas
bonding,
either
atoms
electron
and
lose
or
conguration.
it
gain
A
share
occurs
type
electrons
conguration.
usually
electrons
second
with
This
between
in
of
each
type
of
non-metals.
International Union of Pure and Applied
order
look
at
this
type
of
Lewis
of
bonding
in
detail,
it
is
useful
rst
to
Chemistry), a covalent bond is a region of
introduce
the
idea
a
symbol,
which
is
a
simple
and
convenient
relatively high electron density between
method
of
representing
the
valence
(outer
shell)
electrons
of
an
element.
nuclei that arises at least partly from
In
sub-topic
4.3
we
shall
develop
this
further
into
what
we
term
the
the sharing of electrons and gives rise
Lewis
(electron
dot)
structure
of
a
compound,
based
on
a
system
to an attractive force and characteristic
devised
by
the
US
chemist,
Gilbert
N.
Lewis
(1875–1946).
internuclear distance.
In
a
Lewis
number
of
element.
N
Let
us
symbol
dots
Some
consider
species,
F
,
O
2
,
representation,
(or
crosses),
examples
the
N
2
,
are
presence
and
each
which
given
of
element
represent
in
gure
covalent
is
the
surrounded
valence
by
a
electrons
of
the
1.
bonding
in
four
different
HF
.
2
Fluorine, F
2
●
Cl
Fluorine
is
acquiring
electron
●
The
in
group
one
more
17,
conguration
Lewis
symbol
so
has
electron,
for
with
seven
valence
uorine
a
would
complete
uorine
octet
electrons.
attain
of
a
Hence
noble
by
gas
electrons.
is:
F
B
●
If
two
uorine
atoms
share
one
electron
each
with
each
other,
each
Figure 1 Lewis symbols of three
uorine
atom
gains
one
more
electron
to
attain
a
complete
octet
of
elements. Nitrogen has ve valence
electrons,
which
results
in
the
uorine
atoms.
This
formation
of
a
covalent
bond
between
electrons, chlorine has seven
the
two
covalent
bond
is
a
single
bond
and
the
valence electrons, and boron has
shared
pair
can
be
represented
by
a
line:
three valence electrons
+
Stdy tip
Remember, to deduce the nm of
F
F
n tons of an element you can
use the op nm from the periodic
●
Note
that
in
this
Lewis
structure
of
F
there
are
a
total
of
six
non-
2
table of elements. For example, sodium
(s-block) is in group 1, so has one valence
bonding
pairs
bonding
pair
of
of
electrons
(often
called
lone
pairs)
and
one
electrons
electron; calcium (also s-block) is in group
2, so has two valence electrons. For the
Oxygen, O
2
p-block elements you simply drop the ‘1’
●
Oxygen
is
in
group
16,
so
has
six
valence
electrons.
Hence
by
in the group number to nd the number of
acquiring
two
more
electrons,
oxygen
would
attain
a
noble
gas
valence electrons: silicon (p-block) is in
electron
conguration
with
a
complete
octet
of
electrons.
group 14, so has four valence electrons.
Fluorine (also p-block) is in group 17, so
has seven valence electrons, and so on.
●
If
two
oxygen
electron
covalent
double
98
atoms
each
conguration
bond
bond
between
and
the
share
can
be
the
two
two
electrons
achieved
two
and
oxygen
shared
pairs
with
results
atoms.
can
be
each
in
This
the
other,
this
formation
covalent
represented
bond
by
two
of
is
a
a
lines.
4 . 2
c O v a l e N T
b O N D I N g
+
O
●
Note
that
in
this
Lewis
O
structure
of
O
there
are
a
total
of
four
non-
2
bonding
pairs
of
pairs
of
electrons
(the
lone
pairs)
and
two
bonding
electrons
Nitrogen, N
2
●
Nitrogen
is
acquiring
electron
●
If
two
group
three
15,
more
atoms
bond
bond
can
between
and
the
three
a
Note
that
valence
nitrogen
complete
three
two
would
octet
and
nitrogen
shared
electrons.
pairs
can
be
a
by
noble
gas
electrons.
with
results
atoms.
Hence
achieve
of
electrons
achieved
in
each
the
This
other,
this
formation
covalent
represented
by
of
a
bond
is
a
three
lines:
N
N
●
ve
share
be
the
+
N
has
with
each
conguration
covalent
so
electrons
conguration
nitrogen
electron
triple
in
in
this
Lewis
N
structure
of
N
there
are
a
total
of
two
non-
2
bonding
pairs
of
pairs
of
electrons
(the
lone
pairs)
and
three
bonding
electrons
Hydrogen uoride, HF
●
Fluorine
is
acquiring
in
group
one
conguration
1,
so
has
electron,
●
Note
is
just
more
with
one
a
in
has
octet
electron.
does
nature,
seven
attain
not
and
valence
uorine
complete
would
hydrogen
historical
so
valence
hydrogen
that
17,
electron,
would
of
electrons.
Hence
the
key
by
noble
acquire
the
electrons.
attain
an
to
Hence
noble
Hydrogen
acquiring
gas
just
(the
octet
remember
by
gas
is
electron
in
one
conguration
octet
point
a
of
group
more
helium.
rule
here
for
In the Lewis structure of
hydrogen
is
the
formation
of
a
noble
gas
electron
conguration).
a molecule, the electrons
●
The
Lewis
symbols
for
hydrogen
and
uorine
involved in the covalent bond
are:
are indistinguishable.
x
H
F
atiity
For
convenience
electrons
for
the
in
two
each
we
of
use
the
different
two
symbols
Lewis
(a
symbols
cross
to
and
signify
a
dot)
for
different
the
electrons
Using a similar approach to
that of the examples here,
elements.
deduce the Lewis structures
●
To
achieve
noble
gas
congurations,
uorine
and
hydrogen
can
of the molecules carbon
each
share
one
electron
with
each
other,
forming
a
covalent
bond.
dioxide, CO
This
covalent
bond
is
a
single
bond
and
the
shared
pair
can
be
, and water, H
2
O,
2
showing the steps involved in
represented
by
a
line.
the formation of the covalent
bonds in each case.
x
H
+
H
H
x
F
99
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
●
Note
that
in
this
non-bonding
pair
of
Lewis
pairs
structure
of
of
electrons
HF
there
(the
lone
are
a
total
pairs)
of
and
three
one
bonding
electrons
bond stnth nd ond nth
The
examples
above
triplecovalent
bond
describe
bonds.
These
molecules
bonds
with
differ
in
single,
both
double,
bond
and
strength
and
length.
Bond strength
anoy
The
trend
in
bond
strength
is:
You can think of ond stnth
≡
>
=
>
−
in terms of windows – a
That
is,
a
triple
bond
is
stronger
than
a
double
bond,
which
in
turn
is
window that is triple-glazed is
stronger
than
a
single
bond.
stronger than a window that is
double-glazed, which in turn is
stronger than a window with a
The
bond
Bond
enthalpies
enthalpies
in
will
section
be
11
discussed
of
in
the
Data
booklet
sub-topic
show
this
(table
1).
5.3.
single pane.
Bond length
This
is
−
A
a
the
>
=
single
triple
opposite
>
to
bond
strength:
≡
bond
bond.
illustrate
trend
this
is
longer
The
than
covalent
(table
a
double
bond
bond,
lengths
in
which
section
in
turn
10
of
is
longer
the
Data
than
booklet
1).
bond nthpy
cont ond nth / pm
bond
1
(t 298 K) / kJ mo
C≡C
C=C
C
▲
C
839
120
614
134
346
154
T
able 1 Bond strengths (enthalpies) and bond lengths
compison of ont onds nd ioni onds
We
now
bonds.
H
F
understand
Table
2
the
inherent
summarizes
some
difference
of
these
between
ionic
and
covalent
differences.
etontiity
We
saw
the
single
in
the
case
of
both
uorine,
F
,
and
hydrogen
uoride,
HF
,
that
2
+
δ
-
δ
Figure 2 Dipole moment represented
each
covalent
molecule.
atoms
in
F
,
In
there
bond
the
is
is
case
an
made
of
equal
up
of
identical
sharing
a
shared
atoms,
of
the
pair
such
of
as
electrons
electrons
the
in
two
the
for
uorine
shared
pair
2
by a vector in the polar molecule, HF
between
shared
pair
atoms.
In
two
100
the
two
is
unequally
fact,
atomic
atoms.
you
This
shared
might
partners
for
is
think
the
not
the
case,
between
of
this
shared
the
as
pair!
a
In
however,
hydrogen
in
and
“tug-of-war”
reality,
HF
,
and
uorine
between
uorine
the
has
a
the
much
4 . 2
c O v a l e N T
Ioni ondin
b O N D I N g
cont ondin
Usually formed between non-metals.
Formed between a cation (usually metal) and an anion (usually
+
non-metal). Some cations (such as NH
) can be comprised of
4
) can contain metals.
non-metals and some anions (such as MnO
4
Formed from atoms either losing electrons (process of oxidation)
Formed from atoms sharing electrons with each
or gaining electrons (process of reduction) in order to attain a
other in order to attain a noble gas electron
noble gas electron conguration.
conguration.
Electrostatic attraction between oppositely charged ions, that is, a
Electrostatic attraction between a shared pair of
cation (positive ion) and an anion (negative ion).
electrons and the positively charged nuclei.
Ionic compounds have lattice structures.
Covalent compounds consist of molecules.*
Ionic compounds have higher melting points and boiling points.
Covalent compounds have lower melting points and
boiling points.
Ionic compounds have low volatilities.
Covalent compounds may be volatile.
Ionic compounds tend to be soluble in water.
Covalent compounds typically are insoluble in water.
Ionic compounds conduct electricity because ions are free to
Covalent compounds do not conduct electricity
move in the molten state. They do not conduct electricity when
because no ions are present to carry the charge.
solid, however, as the ions are not free to move.
T
able 2 Dierences between ionic and covalent bonding
▲
*We shall discuss covalent network structures that involve lattices later.
greater
to
a
attraction
what
we
partial
for
describe
negative
the
as
a
shared
polar
charge,
δ
,
pair
than
covalent
and
one
hydrogen
bond,
atom
does
with
adopting
and
one
a
this
atom
partial
leads
adopting
positive
+
charge,
the
δ
.
shared
negative
In
this
pair
case,
of
charge,
since
electrons
δ
,
and
uorine
in
the
hydrogen
has
a
covalent
then
greater
bond,
adopts
pulling
it
the
power
acquires
partial
the
for
partial
positive
+
charge,
dipole
δ
.
This
separation
moment,
symbol
of
μ
charge
(gure
can
be
represented
vectorially
by
a
Tnds in tontiitis
2).
●
If
the
two
atoms
involved
in
the
formation
of
the
covalent
bond
are
Going from left to right
across a period, χ
values
P
identical,
covalent
the
bond
bond
is
is
said
to
non-polar
be
and
a
pure
has
no
covalent
dipole
bond;
moment.
that
is,
Hence,
the
the
F
increase.
F
Reasons:
bond
in
F
is
a
non-polar
covalent
bond
2
i) decreasing atomic radii
The
US
chemist
Linus
Pauling
(1901–1994)
introduced
the
idea
of
ii) increasing nuclear charge.
electronegativity
(χ
)
as
the
relative
attraction
that
an
atom
of
an
P
element
has
for
the
shared
pair
of
electrons
in
a
covalent
●
bond.
Going down a group, χ
P
values decrease.
Pauling
in
devised
section
8
of
a
the
electronegative
scale
of
Data
electronegativity
booklet.
element
in
the
On
the
values,
Pauling
periodic
table
which
scale,
with
a
can
uorine
value
of
be
is
χ
found
the
most
=4.0
i) increasing atomic radii
P
(sub-topic
3.2).
There
certain
Reasons:
ii) primary screening
have
a
are
already
trends
seen
for
in
the
electronegativity
ionization
values
energies
across
that
a
mirror
period
what
and
we
down
(shielding) eect of inner
electrons.
group.
101
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
One
Stdy tips
●
of
the
main
uses
estimate,
based
on
a
ionic,
pure
of
electronegativity
electronegativity
values
value
is
that
differences,
we
Δχ
can
,
whether
P
bond
is
covalent
(non-polar),
or
polar
covalent.
This
Polar covalent bonds
estimation
is
based
on
the
rules
in
table
3
which
you
need
to
have an unsymmetrical
remember.
distribution of electron
density and are
bond typ
represented by par tial
Δχ
P
+
charges, δ
and δ
ionic
Δχ
> 1.8
P
●
For ionic bonds, we use full
charges to represent the
pure covalent (non-polar)
Δχ
= 0
P
ions, eg + and
in the case
of NaF. These charges are
polar covalent
0 < Δχ
≤ 1.8
P
not par tial and hence we do
not use the δ sign.
▲
T
able 3 Rules to estimate whether a bond is ionic, pure covalent (non-polar),
or polar covalent
For
example,
χ
F
2
(F)
=
from
Δχ
4.0,
P
χ
=
0,
so
F
(F)
=
4.0
has
a
pure
covalent
2
bond
and
χ
P
NaF
3:
P
(non-polar)
HF
table
(H)
=
2.2,
Δχ
hence
P
polar
covalent
χ
=
(F)
4.0
=
1.8,
so
HF
has
a
highly
P
bond
and
χ
P
(Na)
=
0.9,
hence
Δχ
P
=
3.1,
so
NaF
has
ionic
P
bonding.
chmisty in th kithn
Miows are part of the electromagnetic spectrum (EMS),
as seen from section 3 of the Data booklet. The reason why
microwaves are so ecient at heating food relates to the
interaction of the microwave radiation of wavelength, λ,
(in the range 1 mm to 1 m, corresponding to frequencies, ν, in
the range 300 GHz to 300 MHz) with the molecules of water,
H
O, present in food in the liquid state. Water consists of two
-
2
O
-
-
-
H bonds, which are both polar, and water itself is a polar
molecule (this is explained later). Hence the oxygen side of
the water molecule is negatively charged and the hydrogen
2
2
δ
2
δ
2
δ
δ
side of the molecule is positively charged, resulting in a net
+
δ
+
O
+
δ
+
O
δ
+
δ
+
O
δ
+
δ
O
δ
+
δ
dipole moment:
2
δ
+
δ
O
+
δ
+
δ
O
+
δ
+
δ
O
2
δ
In food, polar molecules of water absorb microwave
radiation and constantly change their orientation (“ip”),
▲
2
δ
+
δ
δ+
O
+
δ
2
δ
Figure 3 Interaction of electrical charges seen on a sinusoidal
wave representation of microwaves (showing their oscillating
capacity) interacting with the water molecules present in
aligning themselves with the alternating electromagnetic
food and causing them to ip
eld created by microwaves (gure 3). The ipping and
rotation of water molecules results in the dissipation of
heat energy and increases the temperature of the food.
102
4 . 2
c O v a l e N T
b O N D I N g
Worked examples: compounds
Example 1
Deduce
Solution
which
of
the
following
compounds
are
In
order
to
deduce
this
we
need
to
work
out
Δχ
,
P
molecular:
●
the
electronegativity
the
various
binary
difference
for
compounds,
each
and
bond
in
remember
SO
2
that
Δχ
=
0
is
indicative
of
a
pure
covalent
P
●
PCl
(non-polar)
3
bond,
0
Δχ
<
≤
1.8
is
indicative
of
P
a
●
Na
polar
covalent
bond,
and
Δχ
O
>
1.8
is
indicative
P
2
of
●
NH
an
ionic
bond.
NO
4
3
Δχ
H
2
=
0,
so
H
has
a
pure
covalent
(non-
2
P
polar)
bond.
χ
=
Solution
HCl
SO
and
PCl
2
are
molecules
as
they
contain
only
(Cl)
and
no
ions.
In
Na
O,
Δχ
2
is
(H)
=
2.2,
hence
=
2.5,
so
=
ionic
(since
>1.8).
Although
1.0,
it
consists
with
NH
NO
4
is,
in
HCl
has
fact,
ionic
chlorine
a
polar
having
a
covalent
partial
of
negative
non-metals,
so
P
P
bond,
all
χ
and
P
Δχ
non-metals
it
3.2
P
3
charge,
δ
,
and
hydrogen
having
a
because
3
+
+
it
consists
of
an
ammonium
cation,
NH
,
partial
positive
χ
=
charge,
δ
and
4
a
nitrate
oxoanion,
NO
.
Remember
that
ionic
3
KBr
(Br)
3.0
and
χ
P
compounds
have
lattice
structures.
(K)
=
0.8,
hence
P
Δχ
=
2.2,
so
KBr
has
ionic
bonding
with
P
potassium
Example 2
Deduce
which
having
of
the
bonds
in
the
following
CO
χ
(O)
a
=
having
1
3.4
and
χ
P
binary
compounds
are
ionic,
pure
covalent
(non-
a
1+
charge
and
bromine
charge.
(C)
=
2.6,
hence
P
Δχ
=
0.8,
so
CO
has
a
polar
covalent
bond,
P
polar),
or
polar
covalent:
with
and
●
oxygen
carbon
having
having
a
a
partial
partial
negative
positive
charge,
charge.
H
2
●
HCl
●
KBr
●
CO
Qik qstion
Explain why, when you heat a refrigerated bowl of soup in a microwave cooker,
the soup close to the outside of the bowl can appear warm but near the centre of
the bowl the soup can often be cold.
103
4
c H e M I c a l
b O N D I N g
a N D
S T r u c T u r e
4.3 c ont stts
Understandings
Applications and skills
➔
Lewis (electron dot) structures show all the
➔
Deduction of Lewis (electron dot) structure of
valence electrons in a covalently bonded
molecules and ions showing all valence electrons
species.
for up to four electron pairs on each atom.
➔
The “octet rule” refers to the tendency of atoms
➔
The use of VSEPR theory to predict the electron
to gain a valence shell with a total of eight
domain geometry and the molecular geometry
electrons.
for species with two, three, and four electron
➔
Some atoms, like Be and B, might form
domains.
stable compounds with incomplete octets of
➔
Prediction of bond angles from molecular
electrons.
geometry and presence of non-bonding pairs of
➔
Resonance structures occur when there is more
electrons.
than one possible position for a double bond in
➔
Prediction of molecular polarity from bond
a molecule.
polarity and molecular geometry.
➔
Shapes of species are determined by the
➔
Deduction of resonance structures, examples
repulsion of electron pairs according to the
2
include but are not limited to C
H
6
, CO
6
and O
3
3
valence shell electron pair repulsion (VSEPR)
➔
Explanation of the proper ties of covalent
theory.
network (giant covalent) compounds in terms
➔
Carbon and silicon form covalent network (giant
of their structures.
covalent) structures.
Nature of science
➔
Scientists use models as representations of the real world – the development of the model of molecular shape
(VSEPR) to explain observable properties.
Nature of science
But
the
a
what
and,
on
scholar
must
knowledge
he
as
says
for
will
what
ultimately
only
he
lives
be
that
soon
is
be
true,
seeing
long
content
what
it
is
Scientists
with
false
world
in
of
exposed
he
can
built
if
on
in
Ronald
Nobel
104
Coase
(Recipient
Economic
(1910–2013)
laureate
until
his
was
death
the
on
2
shape
certain
of
the
Nobel
Sciences
oldest
in
a
representations
has
been
Every
assumptions
for
a
model,
scientist
its
theory
used
to
model
–
is
one
to
as
a
of
the
explain
in
of
science
the
and
is
major
appreciate
limitations,
real
model
the
whether
Prize
it
will
is
one
withstand
the
test
of
time.
VSEPR
theory
1991).
living
September
of
as
VSEPR
properties.
considerations
enough.
Coase
models
example,
molecular
validity
Ronald
use
for
observable
count
accepted,
–
2013.
such
limitations.
model,
although
not
without
its
4 . 3
c O v a l e N T
S T r u c T u r e S
lwis (ton dot) stts
Earlier
which
in
this
shows
represented
idea
of
of
the
by
Lewis
by
two
by
a
can
dots,
we
i ntr od uced
numb er
eithe r
bo nd
be
by
of
do ts
(electron
covalent
electrons
For
top i c
the
in
or
two
cr o ss e s
in
a
( or
a
of
Fr om
struc tu res ,
mol ecul e.
r e p r e s ente d
i de a
a
e le ct r on s
cro ss e s .
d ot )
a
the
v al e nce
In
a
this
an
we
based
Le w i s
nu m be r
Lewis
of
of
de ve l ope d
on
the
of
a
ea c h
wa ys
dot
the
forma ti o n
s t ru c t u r e,
d iffe r en t
com bi n a t ion
symbol ,
e l em e n t
a nd
–
pa ir
of
e ith e r
a
c ro ss ) ,
or
line.
example,
phosphine,
some
PH
,
of
the
might
ways
be
in
which
represented
the
are
Lewis
shown
structure
in
gure
of
1(a).
3
(a)
x x
-
P
P
P
H
H
H
H
H
Cl
(b)
H
Cl
H
C
Cl
H
H
Cl
Figure 1 (a) Two dots, two crosses (or a combination of the two), or a line can be used to
represent each pair of electrons in a Lewis (electron dot) structure. (b) Lewis (electron
dot) structure of CCl
. Remember the bond angles shown in a Lewis structure do not
4
necessarily represent the actual bond angles in the molecular geometry
In
●
such
a
representation
bonding
double,
●
pairs
or
non-bonding
which
In
the
are
Lewis
electrons
Similarly,
of
triple
of
the
one
of
of
to
distinguish
(showing
the
between:
covalent
bond
as
single,
and
electrons ,
electrons
lone
Lewis
important
electrons
structure
and
is
bonds)
pairs
pairs
it
not
often
involved
phosphine
there
called
the
in
bonding.
are
the
three
lone
bonding
pairs,
pairs
of
pair.
structures
of
carbon
dioxide,
CO
,
and
carbon
2
monoxide,
shown
in
CO,
which
gure
contain
multiple
bonds,
can
be
represented
as
2.
Figure 2 Lewis structures of CO
In
CO
,
each
double
bond
represents
two
bonding
electron
pairs,
and
in
2
CO,
the
Lewis
bond
in
triple
structures
(single,
molecules.
actual
Lewis
bond
represents
help
double,
us
or
H o w e v e r,
shapes
structure
of
of
three
understand
triple
Lewis
molecules,
a
bonding
molecule
the
bonds)
and
hence
may
be
different
and
structures
electron
the
tell
the
pairs.
types
existence
us
nothing
of
of
covalent
lone
about
representation
drawn
with
a
of
pairs
the
the
geometrical
105
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
arrangement
For
example,
that
the
differs
Lewis
completely
structure
of
from
its
carbon
real
shape
in
tetrachloride,
space.
CCl
,
4
is
typically
existence
represented
of
90 °
tetrachloride
angles.
very
We
C
Cl
molecule
shall
useful
valence
Cl
as
model
shell
for
figure1b,
bond
is
shortly
in
angles.
tetrahedral
see
how
predicting
electron
pair
to
which
In
fact
with
the
this
molecular
suggest
shape
109.5 °
deduce
repulsion
might
Cl
C
shape
of
Cl
(VSEPR)
the
carbon
bond
based
geometries,
the
on
called
a
the
theory
Lewis (electron dot) structures of cations and anions
Stdy tip: us of
and ionic compounds
sq kts in
Lewis
structures
can
be
written
not
only
for
neutral
molecules
but
also
lwis (ton dot)
for
stts
an
cations
anion
charged
and
anions.
there
ions,
is
an
which
In
a
compound
electrostatic
forms
the
containing
attraction
ionic
bond.
both
between
a
the
However,
cation
and
oppositely
the
bonding
The chemical formula of
within
the
cation
and
anion
separately
may
be
covalent
in
nature;
for
ammonium nitrate is often
example,
written as NH
NO
4
in
ammonium
nitrate,
NH
, but in
NO
4
(gure
3(a))
the
bonding
in
3
3
+
[NH
reality it is made up of a cation,
]
and
in
[NO
4
]
is
covalent,
even
though
the
bonding
between
3
the
cation
NH
Cl
and
the
anion
is
ionic.
In
the
case
of
ammonium
chloride,
ammonium, and an oxoanion,
nitrate. When you write Lewis
(gure
3(b))
the
Lewis
structure
of
the
chloride
anion
can
be
4
represented
with
the
chlorine
surrounded
by
eight
dots
to
represent
structures of cations or anions,
the
eight
valence
electrons
present
in
the
anion.
including oxoanions, you
should always include square
(a)
H
(b)
+
H
brackets and the charge in the
O
representation.
H
N
H
H
N
H
N
H
H
covalent bonds
covalent bonds
covalent bonds in the cation
in the cation
in the anion
Cl
ionic bonds between the cation and
the anion − so the compound overall is ionic
H
H
N
H
Cl
H
ionic bonds between the cation and the
anion − so the compound overall is ionic
Figure 3 (a) Lewis structure of ammonium nitrate. (b) Lewis structure of ammonium
chloride
vn sh ton pi psion (vSePr)
thoy
Much
of
of
the
structure
and
as
chemists
dimensions.
106
core
and
understanding
bonding.
we
need
Every
to
of
chemistry
molecule
have
the
involves
has
ability
to
a
discussions
particular
always
shape
think
in
three
4 . 3
a)
c O v a l e N T
S T r u c T u r e S
b)
CH
3
O
HO
H
C
3
N
CH
3
Figure 4 (a) 2D representation of the drug tramadol, whose molecular formula is C
H
16
NO
25
.
2
Tramadol is a centrally acting synthetic opioid analgesic used in treating severe pain.
(b) Three-dimensional molecular space-lling model of tramadol. The atoms are
represented as spheres and are colour coded: carbon (grey), hydrogen (white), nitrogen
(blue), and oxygen (red)
As
mentioned
previously,
representations
shell
the
electron
shapes
The
basis
charged
apart
In
as
to
balloons
is
as
of
two-dimensional
about
theory
shape.
can
be
Valence
used
to
deduce
pairs
of
–
since
electrons
electrons
repel
one
are
negatively
another
to
be
as
far
space.
the
examine
two
aligning
are
nothing
(VSEPR)
follows
particles,
maximum
pair–electron
Then
case
us
molecules.
theory
in
structures
tell
repulsion
determine
electron
the
pair
this
possible
together.
In
of
Lewis
ultimately
covalent
subatomic
order
this
of
and
pair
the
balloons
at
180°
spatial
a
to
angle
repulsion
shape
linear
each
that
try
the
a
be
Think
is
of
achieved
number
balloons
geometry
other.
can
tying
of
from
balloons
ultimately
obtained,
dividing
with
a
adopt.
the
circle
two
up
into
360
___
halves:
=
180°
(gure
5).
2
Figure 5 Two balloons tied together showing a linear arrangement in space
In
the
case
of
three
balloons,
a
trigonal
planar
arrangement
is
generated,
360
___
similar
to
taking
a
circle
and
slicing
it
into
three
segments:
=
120°.
3
Hence
the
balloons
each
other
Now
consider
you
give
might
a
what
(the
to
on
tying
angle
happens:
might
help
the
in
taking
a
to
the
appreciate
of
be
and
two
109.5 °
of
four
on
means
–
one
at)
Thinking
dividing
the
space
apart
360 °
balloons
the
try
it!
plane
(gure
in
by
of
two
4,
90 °.
balloons
This
at
120 °
dimensions,
which
This
would
is
not
maximize
shape
to
6).
creates
their
a
7).
tetrahedron
the
lie
together.
circle
any
(gure
to
chemistry
three-dimensional
geometry
repulsion
in
balloons
between
imagine
you
themselves
planar
four
arrangement
tetrahedral
You
term
visualize
bond
spatial
arrange
sitting
in
the
three-dimensionality
electron
pairs
(gure
environment
of
this
of
geometry
a
cube
based
8).
107
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
y
x
z
Figure 7 Four balloons tied together
Figure 6 Three balloons tied together showing
showing a tetrahedral arrangement
Figure 8 A tetrahedron ts into a cube
a trigonal planar arrangement in space
in space
in three-dimensional space
The
basic
molecular
in
table
of
electrons
might
1,
like
on
the
is
to
geometries
basis
of
described
imagine
Nm of
as
two,
as
can
therefore
three,
or
occupying
being
a
eld
four
an
of
be
summarized,
pairs
of
electron
electron
Mo omty
as
electrons.
domain,
shown
Each
which
pair
you
density.
bond n
exmps of
ton
mos o ions
domins
hin this shp
linear
AB
2
two
180°
BeCl
, CO
2
2
trigonal planar
AB
3
three
120°
BF
, [NO
3
]
3
Figure 9 The Scottish scientist, engineer, and
inventor, Alexander Graham Bell (1847–1922)
sitting in his tetrahedral chair. Most famous
tetrahedral
for his invention of the telephone, Bell was
also fascinated by the theory of engineering
AB
4
structures and ight. He championed the
four
109.5°
+
CH
cause of tthd stts, frameworks
, [NH
4
]
, [ClO
4
]
4
based on a series of interlocked tetrahedra. He
is seen here watching trials of his kite designs
▲
T
able 1 Molecular geometries based on two, three, and four electron domains
The
set
of
three
molecular
geometries,
AB
(linear),
AB
extended
to
2
planar),
and
AB
(tetrahedral),
can
also
be
(trigonal
3
generate
4
additional
than
108
the
shapes
number
for
of
species
electron
that
have
domains
fewer
bonding
present.
In
pairs
such
of
cases,
electrons
the
electron
4 . 3
domains
not
occupied
non-bonding
additional
pairs
of
by
the
bonding
electrons
molecular
(lone
geometries
are
pairs
of
pairs).
electrons
In
such
generated:
AB
are
cases,
E
c O v a l e N T
lled
S T r u c T u r e S
by
three
(V-shaped
or
bent),
2
AB
E
(trigonal
pyramidal),
and
AB
3
2
represents
We
●
can
the
a
H
The
electrons
domain
domains
this
of
distinguish
molecular
illustrate
O.
pair
electron
the
To
lone
therefore
electron
●
E
let
us
bent),
where
E
2
(table
geometry
geometry
of
or
2).
between:
predicted
idea
number
(V-shaped
from
(based
VSEPR
(which
take
electron
the
gives
on
total
theory);
the
example
domains
the
of
of
and
shape
the
predicted
number
of
the
water
from
molecule).
molecule,
VSEPR
theory
is
2
four
(we
electron
formula
shall
learn
domain
we
presence
of
occupied
by
molecular
see
how
geometry
that
two
to
there
bonding
two
lone
geometry,
deduce
is
only
pairs
pairs
of
two
(not
electrons.
on
an
AB
E
This
However,
O
four).
2
(table
shortly).
tetrahedral.
are
based
this
H
This
from
bonds,
The
implies
is
two
that
the
which
other
structure,
means
the
chemical
suggests
domains
that
the
the
are
actual
V-shaped
or
bent
2
2).
Nm of
eton
Mo
ton
domin
omty
domins
omty
bond n
exmps of
mos o ions
hin this shp
boon noy fo
mo shp
V-shaped
trigonal planar
three
<120°
AB
E
SO
, [NO
2
(bent)
]
2
Returning to the balloon analogy,
2
you can see this in action if you
tetrahedral
again take four balloons and tie
trigonal
2
four
<109.5°
AB
E
NH
, [SO
3
pyramidal
]
+
, [H
3
O]
3
them together. This time have
3
two of the balloons blue and
tetrahedral
two of the balloons yellow, the
V-shaped
four
<109.5°
AB
E
2
H
O, [NH
2
(bent)
]
2
latter representing lone pairs of
2
electrons. Make the two yellow
▲
T
able 2 Geometries involving lone pairs based on three and four electron domains
balloons bigger than the two
blue balloons (the text opposite
explains why). To emphasize the
Bond angles in molecular geometries:
fact that the lone pairs are nonLone
pairs
of
electrons
affect
the
bond
angles
in
a
molecule.
Lone
pairs
bonding pairs of electrons take a
occupy
more
space
than
bonding
pairs,
so
they
decrease
the
bond
angle
black marker and mark two dots
between
bonding
pairs.
The
degree
of
electron
pair–electron
pair
on each yellow balloon. You still
repulsion
follows
this
order:
have four electron domains, so
LP|LP
>
LP|BP
>
the electron domain geometry
BP|BP
is designated as tetrahedral, but
where
LP
represents
lone
pairs
of
electrons
and
BP
represents
bonding
now it is made up of two bonding
pairs
of
electrons.
electron pairs and two non-
Table
3
illustrates
decreases
the
how
bond
repulsion
between
lone
pairs
of
electrons
bonding electron pairs.
angles.
109
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Mo
Nm of
Mo omty
bond n
ton domins
H
C
H
H
four
CH
109.5°
4
H
tetrahedral
AB
4
N
H
H
H
NH
four
107°
3
trigonal pyramidal
AB
E
3
O
H
O
four
104.5°
2
V-shaped
AB
E
2
2
T
able 3 Eect of lone pairs on bond angles
▲
Interpreting the VSEPR model
Using
the
model
of
VSEPR
theory
it
is
not
possible
hydrogen
sulde,
S,
H
is
V-shaped,
based
on
an
2
to
predict
exact
bond
angles
when
lone
pairs
are
AB
E
2
present.
will
the
be
All
angle
However,
students
to
determined
is
to
bond
(104.5°)
be
and
account.
make
state
angles
and
then
molecular
geometries
a
mistaken
that
than
with
the
the
basic
repulsions
the
for
angle
from
shape.
should
mistake
that
be
many
experimentally
ammonia
assume
have
(107 °)
by
feasible.
bonds
ethene,
trigonal
C=C
and
all
V-shaped
be
interpretation
of
bond
the
angles.
For
example,
phosphine,
PH
,
the
92.1°.
so
Two
H
S
the
H
H
bond
C
predicted
to
arrangement
H
is
be
bond
angle
angles
exact
factors
is
occupy
bond
121°,
120°
about
in
angle
even
based
each
play
and
more
example,
is
role
are
just
molecule
117°
a
a
multiple
though
on
affected
space,
the
is
are
predictions
that
differences
also
For
angle
The
making
other
latter
pairs).
bond
and
both
trigonal
of
the
H
would
planar
carbon:
VSEPR
H
model.
(the
lone
all
these
at
factors,
electronegativity
like
and
lower
many
not
but
2
much
that
geometries
also
bond
predicted
common
learn
molecular
is
less
LP|LP
A
pyramidal
This
is
associated
LP|BP
into
water
can
expected
bond
taken
you
structure,
also
12
1°
H
has
3
C
an
AB
E
structure
and
is
trigonal
pyramidal,
11
7°
C
but
3
H
its
H
P
H
bond
angle
drops
to
93.5 °.
H
Likewise,
Working method to deduce both Lewis (electron dot) structures and
electron domain and molecular geometries
We
can
theory
combine
in
a
following
structures
method
and
geometries:
110
Lewis
structures
simple-to-use
can
be
electron
and
working
used
to
domain
VSEPR
method.
deduce
and
The
Lewis
molecular
1
Draw
a
central
ball-and-stick
atom.
electrons
about
draw
in
bond
the
Each
the
in
at
this
any
identifying
represents
covalent
angles
sticks
diagram,
stick
bond.
stage
a
Don’t
–
direction
you
to
pair
the
of
worry
can
commence
4 . 3
the
process.
the
negative
In
the
case
of
oxoanions,
localize
E
AB
2
atoms;
the
charges
on
remaining
any
terminal
bonds
should
oxygen
other
into
anions
double
(not
bonds.
In
oxoanions)
(V-shaped)
–
all
with
associated
bond-
considerations.
be
the
and
S T r u c T u r e S
2
angle
●
converted
c O v a l e N T
case
Oxoanions.
of
cations
Worked examples
use
square
brackets
and
place
the
charge
Example 1: Carbon tetrachloride, CCl
outside
these.
For
central
4
●
2
the
number
in
atom,
the
deduce
periodic
table
from
the
its
group
number
A
ball-and-stick
diagram
for
CCl
:
4
of
Cl
valence
3
From
electrons.
the
number
of
sticks,
count
the
number
Cl
of
single
sigma
bonds,
(σ)
which
we
shall
designate
C
Cl
as
bonds.
Cl
4
Add
one
electron
for
each
negative
charge
●
(but
not
oxygen
for
localized
atoms
electron
for
in
a
charges
oxoanions
positive
already
in
step
charge.
assigned
1 ).
Delete
Subtract
C
one
one
pi
(π)
four
valence
electrons
(it
is
in
group
14);
four
σ
bonds;
for
so
each
has
to
the
total
number
of
valence
electrons
is
eight;
bond.
8
_
=
5
Combining
by
two
which
to
steps
2,
obtain
equals
3
the
the
and
4,
divide
number
number
of
of
this
number
electron
electron
4
so
there
are
four
electron
domains.
2
pairs,
Thus
domains.
(AB
the
electron
domain
geometry
is
tetrahedral
).
4
6
Based
on
deduce
the
the
number
electron
of
electron
domain
domains,
●
geometry.
There
are
are
four
present
tetrahedral
7
Determine
if
the
applicable,
number
and
of
deduce
lone
the
pairs
–
C
the
and
Cl
bonds
molecular
the
bond
so
no
lone
geometry
angle
will
is
pairs
therefore
be
109.5 °
present,
molecular
Cl
geometry.
Then
draw
an
exact
representation
109.5°
of
the
structure,
complete
with
predicted
C
Cl
bond
angles,
electron-pair
LP|LP
>
taking
into
account
the
order
Cl
repulsion:
LP|BP
>
BP|BP
●
Finally
each
8
Finally,
draw
Cl
of
a
Lewis
representation
excluding
the
octets
hydrogen
on
all
(which
terminal
will
need
Cl
to
in
complete
order
to
the
octets
generate
on
the
by
Lewis
completing
you
terminal
structure.
atoms,
already
have
Cl
attained
a
noble
gas
electron
conguration
of
109.5°
two).
Remember
to
include
square
brackets
C
Cl
for
any
cation
or
Cl
anion.
Cl
9
Draw
page
any
115)
resonance
where
structures
(explained
on
applicable.
+
Example 2: Ammonium cation, [NH
]
4
Let
us
put
are
three
this
working
method
to
the
test.
There
+
●
Ball-and-stick
diagram
for
[NH
]
:
4
types
of
to
work
out:
●
Basic
shapes
structure
that
you
are
required
+
H
–
AB
(linear),
AB
2
planar),
and
AB
(trigonal
3
(tetrahedral).
4
H
●
Species
with
lone
pairs
of
electrons
–
AB
E
2
(V-shaped),
AB
E
3
(trigonal
pyramidal),
and
H
111
4
C H E M I C A L
●
N
has
four
ve
σ
B O N D I N G
valence
A N D
electrons
S T R U C T U R E
(as
it
is
in
group
15);
the
lone
the
difference
positive
in
the
bond
electronegativity
angle
is
but
likely
also
to
play
total
role.
charge;
●
so
inuencing
bonds;
a
one
pair
number
of
valence
electrons
=
8;
Finally,
each
8
_
you
need
terminal
F
to
in
complete
order
to
the
octets
generate
on
the
Lewis
structure:
=
4
so
4
electron
domains.
2
N
Electron
domain
geometry
is
tetrahedral
(AB
).
F
4
F
<109.5°
●
There
are
are
four
present
tetrahedral
–
N
H
the
and
bonds
so
molecular
the
bond
no
lone
geometry
angle
will
F
pairs
is
therefore
be
109.5 °
Example 4: Sulfur diuoride, SF
2
+
●
Ball-and-stick
diagram
for
SF
H
:
2
109.5°
F
N
H
H
H
●
The
above
structure
●
structure
as
is
hydrogen
also
is
a
valid
by
has
two
Lewis
surrounded
S
which
is
the
maximum
number
electrons
(group
16);
bonds;
the
total
number
of
valence
electrons
of
is
electrons
σ
valence
two
so
electrons,
six
eight;
permissible.
8
_
=
4,
so
four
electron
domains.
2
Example 3: Nitrogen triuoride, NF
Electron
3
domain
geometry
is
tetrahedral
(AB
E
2
●
Ball-and-stick
diagram
for
NF
).
2
:
3
●
There
are
must
only
be
two
two
S
F
lone
pairs
covalent
present,
bonds.
as
Hence
there
the
F
molecular
angle
geometry
will
be
less
is
V-shaped
than
and
109.5°
due
the
to
bond
the
F
presence
●
N
has
ve
valence
electrons
(as
it
is
in
group
much
three
σ
bonds;
LP|LP
more
is
which
so
the
total
number
of
valence
electrons
of
the
two
lone
pairs,
which
occupy
15);
space.
greater
is
The
than
greater
than
repulsion
that
that
between
between
a
between
a
LP |BP
a
BP |BP
,
so
is
the
bond
angle
is
reduced
signicantly
from
its
eight;
predicted
value
of
109.5 °
8
_
=
4,
so
there
are
four
electron
domains.
2
S
Electron
domain
geometry
is
tetrahedral
(AB
E).
F
3
●
There
are
must
only
three
molecular
bond
presence
than
–
one
N
F
geometry
angle
space)
be
will
of
the
the
that
lone
covalent
is
be
trigonal
less
lone
than
pair
repulsion
between
pair
present,
bonds.
as
Hence
pyramidal
109.5°
(which
between
there
and
due
to
the
is
from
role
more
the
greater
F
experimentally
angle,
the
occupies
LP |BP
The
the
<109.5°
which
the
of
model,
the
LP|LP
Finally
you
need
terminal
F
F
Lewis
structure:
F
S
determined
F
N
F
bond
F
angle,
the
112
which
model,
is
cannot
102.2°,
of
S
showing
in
the
F
bond
precisely
signicant
operation
uorine
will
(also
have
an
BP|BP
.
N
experimentally
98°,
F
determined
repulsion
electronegativity
each
The
is
be
inuence).
●
F
determined
cannot
be
determined
suggesting
precisely
that
not
only
from
is
<109.5°
F
to
in
complete
order
to
the
octets
generate
on
the
4 . 3
Example 5: Nitrite oxoanion, [NO
This
is
an
oxoanion,
so
in
the
S T r u c T u r e S
]
2
●
c O v a l e N T
O xonions
ball-and-stick
In the case of oxoanions, we do not add an additional
diagram
we
rst
localize
the
one
negative
electron here for the negative charge on the nitrite
charge
on
any
one
of
the
two
oxygen
atoms.
oxoanion because this has already been accounted for
Since
oxygen
has
a
valency
of
two,
this
means
in the rst step. This is a very impor tant point and you
the
other
nitrogen-to-oxygen
bond
must
be
a
need to note this dierence for oxoanions.
doublebond.
●
There
must
be
one
lone
pair
present,
as
there
O
are
only
two
σbonds.
Hence
V-shaped
Note
that
in
the
case
of
an
oxoanion,
we
with
square
brackets
with
the
due
outside,
but
we
then
the
molecular
bond
angle
geometry
will
be
is
to
the
presence
of
the
less
one
than
lone
pair,
negative
which
charge
and
the
covalent
rst
120°
begin
nitrogen-to-oxygen
localize
the
occupies
much
more
space.
charge
N
on
any
one
remove
the
stepwise
of
the
two
square
working
oxygen
brackets
atoms
until
and
later
in
O
our
method.
The
●
N
has
ve
two
σ
one
pi
valence
O
<120°
electrons
(as
it
is
in
experimentally
which
group15);
cannot
model,
is
Finally
you
be
determined
determined
bond
angle,
precisely
from
the
115°
bonds;
●
( π)
bond,
which
counts
as
1(it
each
important
to
remember
this;
see
the
need
to
complete
the
octets
on
is
terminal
O
in
order
to
generate
the
box
Lewis
structure:
below);
N
so
the
total
number
of
valence
electrons
is
six;
O
O
<120°
6
_
=
3
so
there
are
three
electron
domains.
3
Note
Electron
domain
geometry
is
trigonal
planar
(AB
that
for
the
oxygen
containing
the
E).
2
double
bond,
addition
oxygen
of
completing
two
with
lone
the
the
pairs,
single
octet
entails
whereas
bond
in
for
this
the
the
structure,
How to hnd π onds in vSePr thoy
completion
of
the
octet
requires
the
addition
π bonding involves o-axis bonding, as we will explain
of
in topic 14. Hence, as the shape of a molecule
three
lone
pairs.
The
actual
structure
of
is
nitrite
is
a
combination
of
two
contributing
controlled by the σ bonding framework along the
Lewis
structures.
This
is
resonance,
which
we
internuclear axis, in counting the valence electrons we
shall
return
to
shortly.
Contributing
resonance
subtract 1 for each π bond present. For SL students,
structures
are
represented
by
a
double-headed
you do not need to go into the reason behind this (the
arrow.
explanation is given at HL) but you do need to know
N
N
the method involved.
O
<120°
O
<120°
You can think of it like this – the shape is controlled
by the geometrical arrangement along the
The
two
nitrogen-to-oxygen
bond
lengths
in
internuclear axis, where the σ bonding framework
nitrite
are
equivalent
and
intermediate
in
length
lies. A double bond is described as a (σ + π) bond
between
a
single
and
a
double
bond.
The
two
and a triple bond is described as a (σ + 2π) bond. To
contributing
resonance
structures
therefore
could
reduce these back to σ bonds, you simply subtract
be
combined,
each
represented
with
square
out the π components.
brackets
and
the
negative
charge
placed
outside.
113
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
S
N
N
O
O
O
<120°
<109.5°
<120°
Note
that
for
the
oxygen
containing
the
2
Example 6: Sulte oxoanion, [SO
]
double
3
●
This
is
stick
also
an
oxoanion,
diagram
charges
on
oxygen
other
bond
we
two
has
a
rst
of
the
valency
order
to
in
three
of
sulfur-to-oxygen
in
so
localize
this
ball-and-
two
oxygen
two,
bond
satisfy
the
the
this
negative
atoms.
means
must
be
valency
a
for
Since
the
the
addition
the
two
bonds
of
octet
the
of
two
pairs.
in
lone
this
The
octet
the
three
involves
whereas
atoms
structure,
actual
of
the
pairs,
oxygen
requires
combination
oxygen.
completing
terminal
single
lone
double
bond,
with
for
the
completion
addition
structure
of
of
three
sulte
contributing
is
a
resonance
structures:
2
2
S
O
S
has
six
valence
electrons
(as
it
is
in
group
σ
<109.5°
π
so
is
bond
the
total
(
three
length
1);
number
of
valence
equivalent
between
a
and
bond
lengths
single
intermediate
in
and
a
double
in
bond.
Inompt nd xpndd otts
8
_
4,
sulfur-to-oxygen
are
electrons
eight;
=
<109.5°
bonds;
sulte
one
O
16);
The
three
O
O
O
<109.5°
●
S
O
O
O
2
S
O
O
2
S
O
O
so
there
are
four
electron
In most Lewis structures, the central atom will be
domains.
2
surrounded by an octet of electrons. However, in some
Electron
domain
geometry
is
tetrahedral
(AB
E).
3
●
There
are
must
only
be
three
one
lone
pair
present,
sulfur-to-oxygen
as
species, the central atom will have less than an octet
there
of valence electrons: these are inompt otts (for
example, the linear molecule beryllium chloride, BeCl
covalent
,
2
σbonds.
Hence
pyramidal
109.5°
and
the
the
because
of
molecular
bond
the
geometry
angle
will
presence
of
be
is
trigonal
less
the
than
one
lone
which has the central beryllium atom surrounded by
only four electrons, or the trigonal planar molecule
, in which the central boron atom
boron trichloride, BCl
3
pair,
which
occupies
much
more
is surrounded by only six electrons):
space.
Cl
S
120°
180°
B
O
Cl
<109.5°
The
experimentally
which
●
model,
is
Finally
you
each
be
determined
determined
bond
angle,
precisely
from
106°
Cl
Cl
O
structure.
to
in
complete
order
to
the
octets
generate
on
the
Cl
In other species, an xpndd ott is possible. This is
the
discussed in topic 14. In such cases alternative Lewis
structures involving octets may be used.
need
terminal
Lewis
114
cannot
Be
4 . 3
c O v a l e N T
S T r u c T u r e S
rsonn stts
Dnitions
As
we
Lewis
saw
in
the
structures
arrangements
of
case
to
of
have
the
the
nitrite
identical
oxoanion,
sometimes
arrangements
of
atoms
it
is
but
possible
for
different
●
rsonn involves using two or
more Lewis structures to represent
electrons.
a par ticular molecule or ion. A
The
individual
Lewis
structures
that
contribute
to
the
overall
structure
resonance structure is one of two or
are
called
resonance
forms.
The
actual
electronic
structure
of
the
more alternative Lewis structures
species
is
called
a
resonance
hybrid
of
these
resonance
forms.
In
order
for a molecule or ion that cannot
to
represent
this
idea
of
resonance,
the
contributing
resonance
forms
be described fully with one Lewis
are
linked
via
a
double-headed
arrow.
structure alone.
One
of
the
best
known
examples
of
resonance
is
the
molecule
benzene,
●
C
H
6
(gure
In doiztion electrons are
10).
6
shared by more than two atoms in a
molecule or ion as opposed to being
localized between a pair of atoms.
(a)
(b)
Figure 10 (a) Two Kekulé structures of benzene showing resonance. (b) Representation of
benzene showing the delocalized nature of its π electrons
C
The
two
resonance
forms
represented
here
are
termed
the
non-polar molecule
Kekulé
section
140
structures
10
pm,
of
the
of
Data
intermediate
benzene .
booklet,
In
each
between
a
benzene,
as
can
carbon-to-carbon
carbon-to-carbon
be
seen
bond
from
length
double
is
bond
F
(134
pm)
structure
where
and
of
the
further
in
a
carbon-to-carbon
benzene,
circle
topic
therefore,
represents
the
is
single
often
bond
(154
drawn
as
delocalization
pm).
in
The
Figure
(which
we
10(b),
shall
discuss
B
10).
F
In
topic
14
we
shall
discuss
resonance
in
more
detail,
together
with
F
π
non-polar molecule
electrons
and
bond
order.
Mo poity
Earlier
We
in
now
this
focus
topic
on
we
O
discussed
molecular
the
idea
polarity ,
of
that
bond
is,
H
polarity .
whether
the
H
molecule
polar molecule
itself
is
distinct
may
polar
from
have
or
non-polar
the
polar
polarity
bonds.
In
(gure
of
11).
individual
order
to
The
polarity
bonds;
deduce
the
a
of
molecules
non-polar
molecular
is
molecule
polarity
we
N
can
follow
a
simple
three-step
working
method
described
in
the
H
box
H
below.
H
Wokin mthod to dd mo poity
polar molecule
1
Using VSEPR theory, deduce the molecular geometry.
2
For each bond present, using electronegativity dierences, Δχ
Figure 11 Examples of non-polar and polar
molecules. The net dipole moment, μ, of a
, deduce the
P
polar molecule can be represented vectorially.
bond polarity for each bond present and draw the associated dipole moments;
The dipole moment represents the non-
these are best represented as vectors.
symmetrical distribution of charge in a polar
3
Using vector addition, sum all the dipole moments present to establish
molecule (compared with a symmetrical
whether there is a net dipole moment, μ, for the molecule. If so, the molecule
distribution of charge in a non-polar
is polar.
molecule). In the vector the head of the vector
+
represents δ
and the tail represents δ
115
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Worked example: deducing molecular polarity
›
Deduce
the
molecular
polarities
of
the
following:
›
ν
+
›
ν
1
=
SF
b)
CO
.
This
results
in
a
net
dipole
net
moment,
a)
ν
2
μ;
the
molecule
is
polar.
2
2
ν
S
ν
1
µ
ν
2
ν
1
2
ν
net
Solution
F
F
a)
ν
SF
net
2
As
seen
on
VSEPR
of
SF
in
is
an
earlier
theory,
worked
the
V-shaped.
example
molecular
From
section
<109.5°
geometry
8
of
the
b)
CO
2
Data
2
booklet:
Using
χ
(S)
=
2.6
and
χ
P
(F)
=
4.0
to
be
VSEPR
a
linear
theory,
carbon
molecule.
dioxide
From
is
section
found
8
of
the
P
Data
Hence,
sulfur
uorine
and
the
is
S
more
F
electronegative
bond
is
polar
with
booklet:
than
the
χ
(C)
=
2.6
and
χ
P
following
dipole
(O)
=
3.4
P
moment:
Hence,
each
C=O
bond
(Δχ
=
0.8)
is
polar.
P
The
S
two
opposite
F
other
+
that
out,
is,
though
deduce
need
to
in
are
equal
direction
resulting
and
in
no
in
magnitude
hence
net
cancel
dipole
but
each
moment,
-
δ
δ
To
vectors
the
sum
molecular
the
two
S
F
polarity,
vectors.
μ
it
=
0.
has
The
two
molecule
polar
is
non-polar
(even
bonds).
we
The
SF
2
molecule
vectors
(see
is
V-shaped
using
Study
so
we
add
the
theparallelogram
tip
two
C
law
below):
µ
●
Stdy tip
=
0
A tug-of-war is a model that can be used to consider
vectors on the same line (axis).
In deducing molecular polarities based on molecular
geometries you need to nd the vector sum of the
individual dipole moments. The parallelogram law is a
useful method.
›
●
The parallelogram law. If you have two vectors
v
and
1
›
v
, and both vectors star t from the same point, the
2
›
sum of the two vectors,
v, can be found by completing
the parallelogram. The diagonal will give the resultant
(the vector sum).
→
ν
1
Figure 12 Students in Montserrat in a tug-of-war. Both teams
→
are pulling along the same axis
ν
More polar bonds (resulting from a greater dierence in
→
ν
2
electronegativity, Δχ
) win the tug-of-war, provided the
P
pull is along the same axis.
that is:
›
›
v
=
v
1
116
›
+
v
2
4 . 3
c O v a l e N T
S T r u c T u r e S
aotops
Dnition of otops
Allotropes
of
the
same
element
can
vary
in
both
physical
and
chemical
As described by IUPAC,
properties.
otops are dierent
Carbon
is
one
and
forms
of
the
most
fascinating
elements
in
the
periodic
table,
structural modications of the
life
on
Earth
are
based
on
carbon.
Carbon
has
a
number
of
same element.
allotropes:
graphite,
diamond,
graphene,
and
C
fullerene.
60
Covalent network solids
●
Graphite,
diamond,
network
solids .
atoms
held
are
dimensional
well
silicon
by
of
a
are
examples
network
covalent
structure
example
dioxide,
graphene
covalent
together
lattice
known
and
A
(in
solid
bonds
large
covalent
is
of
one
in
a
which
giant
networks
network
covalent
in
the
three-
orchains).
solid
is
quartz,
Another
which
is
SiO
2
●
In
contrast,
C
fullerene
is
molecular.
60
Graphite
Graphite
is
an
are
layers
are
connected
called
example
the
pencils
form
geometry,
of
hexagonal
by
London
andin
in
of
of
and
weak
forces,
(the
covalent
network
consisting
intermolecular
leading
so-called
graphite).
is
a
rings
the
‘lead’
Each
covalently
to
in
of
our
of
In
These
attraction,
is
other
as
not
adopts
three
graphite
atoms.
graphite
pencils
atom
to
solid.
carbon
forces
use
carbon
bonded
of
a
a
which
are
lubricant
lead
but
trigonal
each
there
layers
carbon
planar
carbon
atoms
Pop tis of ont
at
a
bond
angle
of
120 °.
The
coordination
number
of
each
carbon
nt wok soids
is
three
within
which
can
be
solids,
in
the
thestructure.
sheets,
allowsthe
used
as
a
graphite
the
Although
London
layers
to
lubricant
is
a
good
the
forces
slide
past
(gure
between
each
13).
conductor
covalent
of
bonds
the
other,
Unlike
layers
and
other
electricity
are
as
strong
are
thus
covalent
it
has
weak,
●
graphite
Melting points. Covalent
network solids have high
network
melting points (typically
delocalized
greater than 1000 °C and
πelectrons.
much higher than the
melting points of molecular
substances).
●
Electrical conductivity.
Covalent network solids are
poor electrical conductors
(though graphite and
graphene are clear
exceptions – electrical
conductivity is one of the
characteristics that makes
graphene remarkable).
●
Solubility. They are typically
insoluble in common solvents.
●
Hardness. Generally, covalent
network solids are hard,
Figure 13 Graphite is a covalent network solid that consists of hexagonal layers of carbon
atoms, which can slide past each other. The layers are connected by weak intermolecular
though in graphite the layers
can slide past one another.
forces of attraction (London forces)
117
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Diamond
Diamond
is
diamond,
atoms
in
(gure
four.
a
are
and
of
is
This
the
and
means
dental
points
single
σ
a
C
in
lattice
C
to
C
structure
four
of
of
cutting
109.5°
diamond
because
arrangement
of
carbon
angle
within
known
heavy-duty
other
bond
carbon
substances
of
tetrahedra.
tools
is
this
such
For
as
drills.
of
diamond
graphite,
covalent
diamond
the
each
structural
used
Unlike
of
In
bonded
with
number
hardest
often
boiling
the
solid.
covalently
interlocking
tools,
in
is
arrangement,
respectively).
localized
freely.
one
diamond
melting
network
atom
coordination
is
polishing
4827°C,
covalent
bonded
reason
The
The
Diamond
saws,
a
carbon
tetrahedral
14).
covalently
this
also
each
does
in
are
bonds,
not
very
diamond
and
conduct
high
the
(3550
and
valence
therefore
electrons
cannot
move
electricity.
Figure 14 Diamond is an allotrope of carbon
with a covalent network lattice structure.
Strong
covalent
bonds
in
diamond
make
it
is
insoluble
in
all
common
Large crystals of diamond are mined for use
solvents.
as gemstones. Small crystals are used as an
industrial abrasive. High-quality crystals of
diamond are found in South Africa, Russia,
Brazil, and Sierra Leone.
Diamonds are forever?
It
Intntion
pspti
has
ever
been
said
that
considered
Unfortunately
if
“diamonds
diamonds
not!
thermodynamically
Under
are
last
ambient
unstable
and
a
girl’s
best
friend”,
but
have
you
forever?
conditions,
eventually
diamond
turns
into
is
another
allotrope
Throughout history diamonds
of
carbon,
graphite.
However,
at
room
temperature
this
process
is
have often been a potential
extremely
slow,
so
diamond
is
said
to
be
kinetically
stable.
At
1000
°C
source of signicant global
the
conversion
of
diamond
into
graphite
accelerates
and
at
1700
°C
it
conict. The term “blood
completes
within
seconds.
When
we
talk
about
stability
in
chemistry
we
diamond” has been coined
need
to
consider
both
thermodynamic
stability
and
kinetic
stability
to describe diamonds mined
in regions of conict and
subsequently sold to fund such
conicts. What responsibilities
Graphene – the super material!
do nations and governments
Graphene
is
materials,
but
not
it
only
one
Graphene
is
covalent
of
the
thinnest
and
strongest
of
known
have in the import of products
is
also
the
rst
two-dimensional
crystal
ever
discovered.
such as gemstones and precious
a
network
solid,
but
differs
from
graphite
in
that
it
metals?
consists
of
a
single
(gure
15),
carbon
atom
and
coordination
is
is
planar
only
one
covalently
number
sheet
of
of
atom
in
bonded
each
carbon
atoms
thickness.
to
three
carbon
in
arranged
As
other
in
graphite,
carbon
graphene
is
hexagonally
each
atoms
three.
The
so
the
carbon
usf so
atoms
are
Look at the history of the
lattice
is
discovery of graphene and
structure.
densely
actually
packed
planar,
in
a
honeycomb
which
makes
it
crystalline
remarkable
lattice,
as
a
but
the
crystalline
current research developments
The
experimental
evidence
for
the
existence
of
graphene
was
obtained
in using this material at the
in
2004
by
the
Russian
scientists
Andre
Geim
and
Konstantin
Novoselov,
University of Manchester, UK ,
who
won
the
Nobel
Prize
in
Physics
in
2010
for
their
ground-breaking
the university where Geim and
experiments
at
the
University
of
Manchester
in
the
UK.
Novoselov did their research to
win the Nobel prize in Physics
Graphene
in 2010, http://www.graphene.
more
efcient
manchester.ac.uk/story/
three
million
When
118
is
an
excellent
than
sheets
graphite
is
thermal
copper.
of
A
graphene,
prised
apart
and
piece
it
of
with
electrical
graphite
one
becomes
conductor,
1
mm
stacked
thick
on
essentially
top
300
times
consists
of
of
another.
graphene.
If
a
4 . 3
graphene
sheet
When
this,
which
looks
Graphene
electrical
in
like
is
rolled
a
a
●
used
high
devices
After
ball
a
into
as
in
especially
the
the
aerospace
nanotube
it
(sub-topic
becomes
a
A.6).
fullerene,
below).
exibility,
“new
because
and
its
Some
research
composite
industry
of
superb
transparency.
silicon”.
following
graphene–plastic
the
sphere
(discussed
material,
lie
carbon
a
S T r u c T u r e S
of
materials
because
of
the
future
areas:
to
their
replace
low
density
strength
due
displays
to
the
(LCD)
and
exibility,
exible
touch-screens
transparency,
and
for
electrical
mobile
conductivity
graphene.
the
discovery
materials
BN
forms
up
described
of
in
it
strength,
graphene
liquid-crystal
of
soccer
been
of
development
metals
up,
folded
remarkable
has
applications
and
is
conductivity,
Graphene
●
is
turn,
c O v a l e N T
and
have
of
graphene
emerged,
molybdenum
in
which
2004
include
disulde,
MoS
.
a
whole
the
BN
class
single
is
an
of
layers
two-dimensional
of
excellent
boron
nitride,
lubricant
and
can
Figure 15 Graphene
2
be
used
in
ceramic
a
vacuum
materials.
so
it
MoS
is
is
important
also
a
very
in
space
good
research
and
is
also
used
in
lubricant.
2
atiity
C
fullerene
60
In
gphn nnoions - th
1985
a
new
form
of
carbon
allotrope
called
fullerene,
with
carbon
ft ndmk in th d of
atoms
arranged
in
closed
shells,
was
discovered
by
Robert
F
.
Curl
Jr
phn!
(working
at
Rice
University
at
Sussex
University
at
Rice
in
the
USA),
Sir
Harold
W
.
Kroto
(working
Find out about graphene
in
the
UK),
and
Richard
E.
Smalley
(also
working
nanoribbons (GNRs) by
University).
In
1996
these
scientists
were
awarded
the
Nobel
accessing the chemical
Prize
in
Chemistry
for
their
discovery
of
fullerenes.
The
number
of
literature or online and why are
carbon
atoms
in
the
shell
was
found
to
vary,
which
led
to
the
discovery
material scientists so excited
of
several
new
carbon
structures.
Fullerenes
were
found
to
form
when
about their future development.
vaporized
of
C
and
carbon
C
70
being
condensed
were
initially
in
an
atmosphere
synthesized,
with
of
an
more
inert
C
60
gas.
than
C
60
formed.
The
structure
of
each
C
molecule
was
Clusters
clusters
70
found
to
consist
of
60
a
truncated
spherically
icosahedral
symmetrical
cage,
C
which
has
molecule
the
was
shape
unique
of
in
a
soccer
nature
ball.
at
its
The
time
of
60
discovery.
In
the
C
polyhedron
cage
there
are
20
hexagonal
surfaces
and
60
12pentagonal
to
three
others
arrangement
The
has
is
geodesic
Fuller
surfaces,
for
this
the
so
that
not
a
each
carbon
coordination
(gure
d e s i gned
196 7
of
the
planar
dom e
shape
and
Mo ntre a l
s o ccer
by
is
number
covalently
is
three,
bonded
but
the
16).
the
US
Wor l d
ba ll
atom
a nd
ar c h i t ec t
E xh i bi t io n
he nc e
C
R.
in
has
B u c km in s t er
Ca n a da
been
named
60
buckminsterfull e re n e .
referred
to
as
These
spherical
ful l e r e ne s
a re
some ti me s
Figure 16 Molecular structure of c
buckyballs
60
fn (kminstfn),
C
fullerene
is
not
a
covalent
network
solid,
and
so
is
different
from
showing 60 carbon atoms arranged
60
graphite,
diamond,
and
graphene.
C
is
composed
of
individual
60
molecules
between
with
the
strong
covalent
molecules.
bonds,
but
with
weak
London
forces
in a spherical structure that consists
of interlinking hexagonal and
pentagonal rings, like a soccer ball
119
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Fullerenes
in
some
they
and
are
black
non-polar
form
coloured
varies
from
solids
that
solvents
solutions;
red
to
do
(for
not
the
brown
dissolve
example,
to
colour
in
water,
benzene).
depends
magenta.
C
,
In
on
but
the
unlike
can
organic
dissolve
solvents
solvent
both
graphite
60
and
graphene,
delocalized
one
C
does
not
electrons,
molecule
to
conduct
the
the
electricity.
electrons
next
do
because
not
of
Although
have
the
the
it
does
contain
capacity
symmetrical
to
move
nature
of
60
Many
with
60
new
compounds
atoms
inclusion
other
or
small
of
the
Fullerenes
the
enzymes
Carbon
of
in
have
active
and
lie
can
unusual
also
fullerenes
molecules
complexes
applications
because
forms
from
C
of
thereby
nanotubes
enclosed
potentially
the
areas
of
ability
the
used
synthesized
the
as
properties
inside
the
subsequently,
fullerene
gene
superconductivity
t
human
inhibit
are
to
been
within
be
electromagnetic
the
site
have
of
and
and
cage.
drug
ferromagnetism
inclusion
hydrophobic
immunodeciency
Such
carriers;
virus
complexes.
cavity
(HIV)
that
protease
them.
tube-shaped
molecules,
closely
related
to
C
,
60
and
have
copper
many
new
electrical
and
are
100
electronic
times
that
incorporated
approximately
stronger
applications,
compounds
when
conductivity
allow
within
than
replacing
unstable
the
steel.
ten
They
silicon,
The
are
and
substances
tubes.
times
to
in
better
also
the
of
used
in
synthesis
become
diameter
than
such
of
stabilized
tubes
is
9
extremely
and
small,
carbon
chemical
in
the
nanotubes
research
in
order
have
of
10
mm,
become
materials
a
on
the
vibrant
science,
with
nano
and
scale.
dynamic
numerous
Buckyballs
area
of
applications.
Serendipitous discoveries
The
accidental
discovery
of
the
importance
in
the
library
discovered.
famous
in
and
What
of
of
buckminsterfullerene
serendipity
online
other
to
in
nd
chemistry.
out
examples
how
of
is
Carry
a
classic
out
example
some
research
buckminsterfullerene
serendipitous
discoveries
was
are
chemistry?
Siion dioxid, SiO
(q tz)
2
Silicon
dioxide,
SiO
,
often
called
silica,
is
found
in
its
amorphous
form
2
(that
is,
a
solid
crystalline
with
form
it
is
three-dimensional
no
ordered
called
structure)
quartz.
covalent
as
Quartz
network
is
solid.
It
sand.
In
another
consists
its
most
common
example
of
arrays
of
a
of
SiO
4
tetrahedra
arranged
covalently
to
four
covalently
to
two
is
bent
on
because
each
(1710°C)
strong
of
oxygen
and
a
covalent
in
a
lattice
oxygen
silicon
the
high
atoms
atoms.
presence
atom.
(gure
Silicon
boiling
and
The
of
each
Si
two
dioxide
point
17).
O
Each
silicon
oxygen
Si
atom
(2230
both
°C)
is
geometrical
non-bonding
has
atom
a
due
pairs
high
to
is
bonded
arrangement
of
electrons
melting
the
bonded
point
existence
of
bonds.
Figure 1
7 Structure of quar tz, which is a
Both
crystalline form of silicon dioxide, SiO
crystalline
and
amorphous
dioxide
are
insoluble
in
water
and
solid
.
2
Crystals of quar tz are used in optical and
crystalline
does
not
conduct
electricity
(since
there
are
no
delocalized
2
scientic instruments and in electronics, such
electrons
in as quar tz watches
electricity
120
SiO
present)
orheat.
however
as
Note
electrons
that
are
molten
free
to
silicon
move
in
dioxide
the
can
molten
conduct
state.
4 . 3
c O v a l e N T
S T r u c T u r e S
coodint ont ondin
The term oodintion ond
We
have
just
considered
covalent
network
solids.
Another
type
of
is often used (based on
covalent
bonding
covalent
bond,
is
called
coordinate
covalent
bonding .
In
a
typical
IUPAC recommendations)
the
shared
pair
of
electrons
originate
from
both
atoms
to designate a coordinate
that
form
pair
and
the
bond;
one
atom
contributes
one
electron
to
the
shared
covalent bond.
the
covalent
the
A
two
second
bonding,
atoms;
number
of
atom
the
this
shared
atom
species
contributes
pair
donates
have
of
the
second
electrons
both
coordinate
electron.
comes
electrons
covalent
to
from
the
In
coordinate
only
shared
bonding.
one
of
pair.
Examples
include:
+
●
[NH
]
4
+
●
[H
O]
3
●
CO
●
Al
Cl
2
●
6
transition
metal
complexes
(discussed
in
topic
13).
+
Ammonium cation, [NH
]
4
+
When
ammonia,
NH
,
reacts
with
an
acid,
H
,
the
lone
pair
on
the
3
+
nitrogen
in
NH
combines
with
the
proton,
H
,
to
form
the
ammonium
3
+
cation,
[NH
]
:
4
+
H
N:
+
+
H
→
[NH
3
]
4
+
H
109.5°
N
H
H
H
+
Hydronium cation, [H
O]
3
+
O
H
H
H
<109.5°
The coordinate covalent bond
is represented by an arrow
Carbon monoxide, CO
to signify the origin of the
electrons in the bond. Once
C
O
formed, however, all the bonds
are equivalent (whether
Dimer of aluminium chloride, Al
Cl
2
coordinate covalent or normal
6
covalent). Previously, the
Cl
Cl
Al
was used for this type of
Cl
Cl
bond but, based on IUPAC
recommendations, this term is
Al
Cl
term dative covalent bonding
Cl
now largely obsolete.
121
4
c H e M I c a l
b O N D I N g
a N D
S T r u c T u r e
In
the
solid
state
aluminium
chloride
is
ionic.
AlCl
Cl
is
six-coordinate
involving
3
an
ionic
lattice
(but
with
signicant
covalent
characteristics).
At
atmospheric
Cl
pressure
it
sublimes
at
180
°C.
On
increasing
the
pressure
it
melts.
On
melting
Al
Cl
Al
at
192.4
°C
it
forms
the
dimer,
Al
Cl
2
structure,
aluminium
three-dimensional
is
tetravalent
structure
of
the
,
which
has
coordination
bonding.
In
this
6
with
a
dimer
coordination
is
shown
in
number
gure
18.
of
four.
The
The
bridging
Cl
chlorines
are
on
a
different
plane
compared
to
the
terminal
chlorines.
Cl
Al
Cl
Cl
2
this
Figure 18 The structure of the Al
Cl
2
molecules
predominate
in
the
gaseous
state
up
to
400
°C.
Above
6
temperature,
it
dissociates
to
form
molecules
of
AlCl
dimer in
with
a
trigonal
3
6
the gaseous phase
planar
geometry
(120°
bond
angles).
Qik qstions
rpsnttions of stts
1
Apar t from aluminium
For tetrahedral structures it is common to use wd-nd-dsh nottion to show
chloride, identify th other
the various planes:
substances that sublime
●
a wedge indicates that the bond is in front of the dening plane
●
a dash indicates that the bond is behind the dening plane
●
a solid line indicates that the bond lies on the dening plane.
readily.
2
Deduce whether both solid
and molten aluminium
chloride conducts
For example, the tetrahedral structure of methane, CH
, can be represented as
4
electricity.
follows using this notation:
H
C
H
H
H
4.4 Intmo fo s
Understandings
Applications and skills
➔
Intermolecular forces include London
➔
Deduction of the types of intermolecular force
(dispersion) forces, dipole–dipole forces, and
present in substances, based on their structure
hydrogen bonding.
and chemical formula.
➔
The relative strengths of these interactions are
➔
Explanation of the physical proper ties of
London (dispersion) forces < dipole–dipole
covalent compounds (volatility, electrical
forces < hydrogen bonds.
conductivity, and solubility) in terms of their
structure and intermolecular forces.
Nature of science
➔
Obtain evidence for scientic theories by making and testing predictions based on them – London (dispersion)
forces and hydrogen bonding can be used to explain special interactions. For example, molecular covalent
compounds can exist in the liquid and solid states. To explain this, there must be attractive forces between
their particles that are signicantly greater than those that could be attributed to gravity.
122
4 . 4
I N T e r M O l e c u l a r
F O r c e S
Theories on intermolecular forces
In
sub-topics
4.2
intramolecular
and
4.3
forces
we
of
saw
that
there
attraction
ionic
are
that
compound
whereas
hold
the
sodium
melting
chloride,
point
of
NaCl(s),
water,
H
is
O(s),
801
°C,
which
is
2
the
atoms
covalent
together
bonding.
molecular
within
Such
geometries,
a
molecule,
resulting
intramolecular
physical
forces
properties,
in
a
covalent
compound,
is
much
lower,
at
0
°C.
affect
and
chmisty in th kithn
reactivities
could
be
of
compounds.
described
as
Intramolecular
“bonding”
forces
forces
of
attraction.
The next time you are having a drink with cubes of ice
in it, take one cube of ice and try to break it with your
Another
type
of
attraction,
intermolecular
ngers. As you will discover this is vir tually impossible
forces,
are
interactions
between
molecules
within
and the reason for this can be associated with the
a
compound
(gure
1).
Intermolecular
forces
are
intermolecular forces of attraction in the vast network of
largely
responsible
for
the
bulk
properties
of
water molecules present in the ice.
matter,
melting
that
is,
point
its
and
physical
properties
such
as
boilingpoint.
One
of
the
relation
+
-
δ
+
δ
to
assumptions
the
made
in
topic
kinetic–molecular
6
in
theory
of
gases
-
δ
δ
is
that
and
that
collisions
another
gaseous
adhere
polar
every
liquid
covalent bond
to
are
one
species
molecules
of
condensed
a
forces
This
can
of
covalent
phase
gaseous
elastic.
molecules
is
be
temperature.
intermolecular
intermolecular
or
another.
some
one
completely
atoms
gaseous
at
between
do
This
not
incorrect
converted
It
is
the
attraction
and
suggests
stick
or
because
into
a
existence
that
compound
(liquid
particle
to
solid).
of
enable
exist
in
Figure
the
2
force of attraction
demonstrates
an
example
of
this.
Figure 1 Dierence between intramolecular and intermolecular
forces of attraction for hydrogen uoride. The intramolecular
forces result in covalent bonding
As seen in topic 1, the par ticles in a solid or liquid are
tightly packed together – that is why we use the term
Intermolecular
forces
of
attraction
are
much
weaker
ondnsd phs
than
covalent
enthalpy
ΔH
bonds.
change
(the
For
of
enthalpy
example,
the
vaporization
change
standard
of
associated
water,
with
the
vap
conversion
boiling
of
point
one
at
mole
of
standard
pure
liquid
pressure,
into
100
kPa)
1
kJ
to
gas
is
at
its
44.02
1
mol
(at
break
298
the
molecule
from
a
K)
two
of
O
water
section
11
whereas
H
polar
(see
of
the
the
926
kJ
covalent
bond
Data
mol
is
bonds
enthalpies
booklet:
2
×
463
required
in
a
for
=
O
H
926
kJ
1
mol
are
).
Since
the
relatively
compound
this
example,
weak,
are
reason,
N
intermolecular
not
many
(g),
molecules
strongly
covalent
O
2
(for
the
held
(g),
CO
2
example,
H
forces
of
of
a
covalent
together
compounds
(g)
attraction
and
and,
are
CO(g))
for
gases
or
(for
liquids
2
O(l)).
In
contrast
ionic
compounds
Figure 2 Sperm bank shipping containers being lled with
liquid nitrogen to keep the sperm frozen. We think of nitrogen
2
have
very
between
are
the
solids
melting
strong
at
electrostatic
ions,
room
points.
meaning
forces
that
temperature
For
example,
of
ionic
and
the
attraction
as being in the gas phase, but all gases can be conver ted into
liquids at some temperature because of the intermolecular
compounds
have
melting
forces of attraction between the molecules. The boiling point of
high
point
of
the
liquid nitrogen is –195.8 °C at atmospheric pressure
123
4
C H E M I C A L
In
order
to
properties,
and
take
The
in
of
be
need
to
account
all
question,
types
can
understand
we
into
attraction
B O N D I N G
three
A N D
S T R U C T U R E
these
bulk
widen
our
intermolecular
phases,
therefore,
intermolecular
is
solid,
what
forces
the
forces
of
and
theories.
of
As
that
stated
the
science
we
obtain
evidence
for
by
making
and
testing
predictions
The
The relative strengths of these
can
that
can
(dispersion)
types
be
exist
the
in
explain
and
special
molecular
liquid
explanation
forces
signicantly
forces
intermolecular
to
how
the
attractive
are
of
used
example,
above,
of
two
and
is
covalent
solid
related
between
greater
states.
to
their
than
those
scientic
that
theories
For
existence
particles
In
attraction
compounds
present?
London
bonding,
interactions.
gas.
various
attraction
the
hydrogen
forces
liquid,
are
of
on
physical
discussion
main
discuss
could
be
attributed
to
gravity.
based
three
types
of
intermolecular
forces
of
attraction
that
we
shall
are:
interactions in general are:
●
London
london fos < dipo–dipo
forces
(also
dipole-induced
called
dipole
dispersion
forces
or
instantaneous
induced
forces)
fos < hydon onds
●
dipole–dipole
●
hydrogen
forces
However, we shall also consider
bonding.
examples where London forces
are stronger than dipole–dipole
forces, but the above order is
what occurs often!
Collectively
induced
by
the
dipole
rst
two
forces)
intermolecular
are
termed
van
forces
der
(as
well
Waals
as
dipole-
forces ,
as
specied
IUPAC.
London (dispersion) forces + dipole–dipole forces + dipole-induced dipole =
The
can
strengths
vary
of
intermolecular
signicantly,
considerably
weaker
but
it
than
forces
must
ionic
be
or
of
attraction
emphasized
covalent
n d Ws fos
between
that
these
molecules
forces
are
bonds.
london fos
London
(dispersion)
recognized
London
by
the
all
molecules.
Such
German–American
forces
theoretical
physicist
(1900–1954),
London,
the
capital
dispersion
forces
forces.
origin
The
happens
to
a
of
or
of
exist
hence
the
in
the
name
United
instantaneous
the
non-polar
latter
(which
Kingdom!)
term
molecule
has
can
such
be
as
were
Fritz
nothing
London
induced
forces
forces
to
rst
Wolfgang
do
are
with
also
dipole–induced
understood
diatomic
if
we
called
dipole
consider
hydrogen,
H
,
what
when
it
is
2
approached
Although
the
another
moment),
nuclei
surrounded
respect
clouds
to
molecule,
However,
one
part
another
the
is
if
of
an
each
by
a
you
the
the
were
In
this
on
of
is
non-polar
molecule
located
symmetrical,
to
take
case,
a
moment
adjacent
a
might
as
random
have
termed
hydrogen
in
shown
the
in
electrons.
no
net
charged
The
distribution
cloud
at
with
throughout
a
given
instant
electron
moment
is
instantaneous
molecules.
is
the
gure3.
more
dipole
an
there
positively
average
snapshot
slightly
temporary
is
The
is,
of
charged
position.
are
(that
consists
negatively
change
electrons
molecule
dipole
inuence
cloud
constantly
time
molecule.
molecule
hydrogen
spherically
part.
temporary
hydrogen
hydrogen
dipole
electron
124
by
of
density
time
than
generated.
dipole
and
This
has
4 . 4
H
H
H
I N T e r M O l e c u l a r
F O r c e S
H
Figure 3 The average distribution over time of the electrons in any H
molecule is spherically
2
symmetrical
Therefore,
molecule
nucleus
of
of
higher
one
hydrogen
has
the
At
each
the
molecule
acquired
rst
electron
molecule.
within
if
that
a
hydrogen
density
same
molecule
in
molecule
the
time,
will
now
short-lived
the
repel
will
electron
two
approaches
be
other
second
dipole,
attracted
cloud
regions
each
a
instantaneous
of
of
on
the
to
the
second
electron
hydrogen
the
region
hydrogen
density
approach,
as
both
are
negativelycharged.
As
described
electron
by
cloud
by
the
be
dispersed,
IUPAC,
of
a
proximity
minimized.
of
the
polarizability
molecular
a
charged
repulsion
Therefore,
the
probability
of
results
in
non-spherical
orbital
is
is
a
nding
pulled
generated
out
that
entity
its
results
is
between
orbital
regions
of
of
distribution
of
electrostatic
the
space
the
of
distortion
eld
(such
mobile
electron
effectively
symmetrically
in
ease
electric
Because
(region
can
the
an
particle).
electrons)
of
by
electron
spherical
can
is
shape,
cloud;
shape.
attractions
there
its
A
the
that
caused
electrons
density
where
change
of
as
a
can
be
high
which
that
is,
the
temporary
between
the
dipole
partial
+
positive
charge,
charge,
δ
,
of
δ
the
,
of
one
hydrogen
neighbouring
molecule
hydrogen
and
the
molecule
partial
(gure
4).
+
δ
+
δ
H
negative
δ
δ
H
H
H
London force
Figure 4 At a given instant in time, a temporary dipole, the instantaneous dipole, is established
This
interaction
such
an
is
the
arrangement
different
pattern
of
basis
is
of
only
induced
a
London
temporary
dipoles
+
force.
–
may
in
δ
next
however,
instant
of
that
time
a
+
δ
δ
Note,
emerge.
+
δ
the
δ
δ
δ
+
+
δ
δ
δ
δ
+
δ
δ
+
+
δ
δ
δ
δ
δ
+
+
δ
δ
δ
+
δ
δ
+
+
δ
δ
δ
δ
+
δ
+
δ
δ
+
δ
δ
δ
δ
+
δ
δ
+
δ
+
δ
+
δ
+
δ
Figure 5 Dierent arrangements of the interactions of the London forces of attraction
between molecules, which result from interactions between an instantaneous dipole on
one molecule and an induced dipole on an adjacent molecule
125
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
What
affects
There
are
the
magnitude
three
●
number
●
size
●
shapes
of
London
forces?
factors:
electrons
(volume)
of
of
of
the
electron
cloud
molecules.
Number of electrons
The
the
greater
electrons
can
the
valence
be
number
electrons
to
the
nucleus
polarized
For
example,
Ne,
and
booklet
more
Kr,
electrons,
the
will
the
nucleus.
be
larger
The
reduced
the
distance
attraction
and
hence
of
the
the
between
valence
electron
cloud
easily.
consider
krypton,
(table
of
and
the
boiling
using
the
points
of
the
information
in
two
noble
section
7
gases
of
the
neon,
Data
1).
No s
boiin point / °c
Nm of tons
10
Ne (Z = 10)
246.0
36
Kr (Z = 36)
153.4
T
able 1 Boiling points and number of electrons for neon and krypton
6
London
forces
decrease
rapidly
with
increasing
distance,
r
,
based
on
the
relationship:
1
_
V
∝
6
r
where
V
Hence,
n
=
are
2
is
in
the
the
and
Hence
the
krypton
the
in
of
level,
much
means
great
case
energy
located
This
potential
neon
but
in
further
krypton
higher
the
of
associated
eight
case
from
the
the
in
that
outer
of
the
in
can
are
are
the
eight
the
n
electrons
cloud
interactions.
electrons
nucleus
krypton
of
with
krypton
outer
electron
forces
than
the
the
attraction
London
is
energy
be
to
=
located
outer
4
the
energy
nucleus
polarized
stronger,
so
the
in
the
electrons
more
level.
is
not
as
easily.
boiling
point
of
neon.
Size (volume) of the electron cloud
As the number of dispersed
The
magnitude
of
the
London
forces
will
also
depend
on
the
size
of
the
electrons can be linked to the
electron
cloud,
that
is
its
volume
in
space.
In
a
large
electron
cloud,
the
molecular mass the greater the
attraction
of
electrons
to
the
nucleus
will
not
be
as
great
as
in
a
smaller
molecular mass, the greater
electron
cloud,
and
hence
the
electrons
in
a
large
electron
cloud
can
be
the number of London forces
polarized
more
easily.
present.
For
example,
propane,
CH
consider
CH
3
number
126
of
CH
2
carbon
results
in
octane
than
stronger
for
the
and
boiling
octane,
points
CH
3
atoms
3
in
London
propane.
octane
forces
of
(CH
is
and
the
)
2
two
CH
6
greater
hence
alkanes
(table
2).
The
3
than
a
in
higher
propane,
boiling
which
point
for
4 . 4
akn
H
3
octane (C
)
42.0
8
H
8
F O r c e S
Sp-in mod
boiin point / °c
propane (C
I N T e r M O l e c u l a r
)
125
18
T
able 2 Boiling points of two alkanes
Shapes of molecules
The
molecular
shape
is
the
third
factor
that
inuences
the
magnitude
the
two
Isom
of
London
of
C
H
5
,
forces.
pentane,
Let
us
CH
12
compare
(CH
3
)
2
CH
3
,
the
boiling
and
points
of
2,2-dimethylpropane,
boiin point / °c
isomers
(CH
3
)
3
C
4
pentane
36.1
(table3).
2,2-dimethylpropane
Both
isomers
contain
the
same
number
of
electrons,
but
the
9.5
boiling
T
able 3 Boiling points of two isomers of pentane
point
of
2,2-dimethylpropane
point
of
pentane.
of
pentane’s
across
the
full
interaction
pentane.
the
In
of
length
the
of
of
contrast,
is
reason
allows
because
molecules
shape
The
shape
the
the
the
for
considerably
for
this
is
that
molecules
molecule;
better
to
that
contact
lower
the
is,
(gure
smaller
there
between
because
than
the
boiling
straight-chain
interact
2,2-dimethylpropane,
considerably
molecule
is
is
the
the
of
with
a
each
large
area
molecules
contact
the
nature
other
area
almost
of
of
for
soccer-ball
6).
In pentane, there is a large contact area across the entire molecule for
In 2,2-dimethylpropane, there is a much smaller contact area for
adjacent molecules to interact.
adjacent molecules to interact.
Figure 6 Space-lling models of pentane and 2,2-dimethylpropane showing areas of
contact between adjacent molecules for London forces of attraction
127
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
As
the
contact
London
which
area
forces
results
is
much
between
in
a
the
higher
larger
between
molecules
boiling
will
point
for
molecules
have
a
of
pentane,
greater
the
magnitude,
pentane.
Final points on London forces – a warning
on misinterpretation!
●
It
must
be
stressed
molecules
are
molecules.
almost
small.
is
to
zero,
A
molecule
of
has
electrons
magnitude
●
London
in
He)
H
),
but
they
This
greater
to
is
exist
with
a
when
useful
are
explaining
it
means
resulting
in
mass
a
greater
forces
have
remember
an
is
very
London
molecular
but
between
molecules
molecules
marker,
mass
between
attraction
between
greater
polarized,
that
if
number
increase
in
the
forces.
attractive
occur
attraction
individual
molecular
be
of
attraction
of
molecules
able
are
also
forces
gravitational
misinterpretation
London
forces
and
masses
forces.
a
are
of
the
that
London
to
gravitational
since
alone
London
attributed
fact,
common
state
greater
a
In
that
not
forces
between
between
between
non-polar
polar
atoms
(for
molecules
molecules
as
well
example
(for
(for
example,
example,
2
HCl).
That
non-polar
is,
every
or
molecule
will
experience
London
forces
(whether
polar).
Dipo–dipo fos
The
second
type
which
exist
dipole
moment,
and
CH
in
CHO.
of
all
μ.
In
intermolecular
polar
molecules
Examples
this
type
of
of
forces
with
such
a
are
dipole–dipole
permanent
molecules
intermolecular
( not
include
force,
forces,
instantaneous)
HF
,
there
is
ICl,
an
HCl,
attraction
3
between
end
Let
of
us
the
positive
another
compare
molar
masses:
halogen,
end
of
one
permanent
the
the
bromine,
boiling
points
interhalogen,
Br
permanent
dipole
(table
of
on
two
iodine
an
dipole
and
adjacent
molecules
that
monochloride,
the
negative
molecule
have
ICl,
similar
and
the
4).
2
Typs of intmo
Isom
boiin point / °c
fos psnt
London forces +
–1
97.4
)
ICl (M = 162.35 g mol
dipole–dipole forces
–1
Br
(M = 159.80 g mol
58.8
)
only London forces
2
T
able 4 Boiling points of ICl and Br
2
Since
ICl
dipole
lead
128
to
is
highly
forces
a
of
higher
polar,
in
attraction
boiling
addition
between
point.
to
the
London
ICl
forces
molecules
it
also
(gure
has
6),
dipole–
which
4 . 4
+
I N T e r M O l e c u l a r
F O r c e S
+
δ
δ
δ
I
δ
CI
I
CI
dipole–dipole force
of attraction
Figure 7 Dipole–dipole force of attraction between permanent dipoles on adjacent molecules
of ICl. Note that iodine has a larger atomic radius compared to chlorine (see section 9 of the
Data booklet), but chlorine is more electronegative (see section 8 of the Data booklet).
Hydon ondin
This
and
third
is
type
one
Hydrogen
O
H,
or
of
of
the
intermolecular
most
bonding
an
N
H
can
bond
force
important
occur
types
between
holds
of
a
special
place
intermolecular
molecules
when
in
chemistry
force.
there
is
Dnition of th
hydon ond
a
H
F
,
an
present.
As recommended by IUPAC
in 2011, a hydon ond
is dened as an attractive
A
typical
hydrogen
bond
may
be
depicted
as:
interaction between a
hydrogen atom from a
X -
- Z
molecule or a molecular
fragment, X–H, in which X is
more electronegative than
H, and an atom or a group
hydrogen bond
of atoms in the same or a
dierent molecule, in which
where
the:
there is evidence of bond
●
hydrogen
●
acceptor
Y
●
Z
in
donor
may
which
hydrogen
a)
water
X
an
H
formation.
atom
or
to
an
Y
is
bonded
bond
is
represented
sometimes
Hydrogen
be
is
anion,
Y,
a
fragment
or
a
molecule
Pure and Applied Chemistry,
83(8), (2011) pp1637-1641
Z
by
the
three
dots
though
dashes
are
used.
bonds
occur,
molecules,
H
for
example,
between
Ky point
O
2
The H
b)
ammonia
molecules,
F, O
H, and N
H bonds
NH
are polar covalent bonds and
3
are not hydrogen bonds.
c)
hydrogen
d)
water
uoride
molecules
molecules,
and
HF
dimethyl
ether
molecules,
(CH
)
3
a)
b)
O
H
H
O
2
c)
d)
H
O
F
H
rpsnttion of
H
N
hydon onds
H
H
F
H
In this book we use dashes
O
H
O
H
H
H
C
3
N
H
to represent hydrogen bonds
CH
3
to distinguish them from lone
H
pairs of electrons.
H
Hydrogen
bonding
structures
of
often
materials.
has
a
large
inuence
on
both
the
properties
and
129
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
exmp of th t of hydon ondin
150
Let us compare the boiling points of some hydrides of
groups 14, 15, 16, and 17.
H
O
2
100
Figure 8 shows a plot of the boiling points of the series
of hydrides versus period number. As you move down
50
a group, the boiling points increase within a par ticular
hydride series, because of an increase in the number of
C°/tniop gniliob
electrons, resulting in a greater number of London forces.
O, HF, and NH
However, in the case of the hydrides H
2
, the
3
boiling points are considerably higher. This is because of
the existence of hydrogen bonding in these compounds.
Methane, CH
HF
H
Te
2
period
0
1
2
3
4
number
6
SbH
3
H
Se
2
H
NH
50
HI
S
2
3
SnH
4
AsH
3
, however, has a lower boiling point as
PH
3
4
HCl
GeH
4
100
expected, because it does not show hydrogen bonding.
SiH
4
The strength of the hydrogen bond depends on the
150
electrostatic attraction between the lone pair of electrons
CH
4
of the electronegative atom and the nucleus of the proton.
Hence the hydrogen bonding in HF is stronger
the hydrogen bonding in H
200
than
O because uorine is more
Figure 8 Boiling points for the series of hydrides (HX, H
2
P
(O) = 3.4]
and XH
P
is
more
random,
compared
water
on
did
Ear th
hydrogen
(ice)
state.
This
molecule
show
be
has
is
molecules,
in
a
gaseous
means
ice
f loats
leads
lat tice
bonds
that
to
on
9).
cavities
In
with
The
in
all
In
the
in
ice,
results
in
a
higher
density
of
liquid
each
water
water
ordered
presence
the
which
ice.
water
the
adjacent
very
to
addition,
phase
water
water.
regular,
(figure
creates
solid
than
bonds
a
state.
the
density
hydrogen
which
the
hydrogen-bonding,
the
lower
why
forms
hydrogen
in
bonding
water
network
not
would
,
3
) from groups 14, 15, 16, and 1
7
4
Hydon ondin nd wt
If
X, XH
2
(F) = 4.0, χ
electronegative than oxygen [χ
of
lat tice.
the
In
Figure 9 Open cavity structure in the lattice structure of ice
contrast ,
in
the
liquid
phase
the
hydrogen
bonding
Hydrogen
bonding
is
also
present
in
biomolecules
such
as
in
the
double
TOK
helix
structure
of
DNA.
(sub-topic
B.8)
Both theoretical and empirical
evidence are used for the
–1
existence of hydrogen bonding.
Typ of intmo fo
rti stnth / kJ mo
The nature of the hydrogen
London forces
bond is a topic of much
weak (1–10) – this can increase with
number of electrons, size (volume) of
discussion and the current
electron cloud, and shape of molecule
IUPAC guidelines contain six
criteria that should be used as
dipole–dipole forces
weak to moderate (3–25)
hydrogen bonds
moderate to strong (10–40)
evidence for the occurrence of
hydrogen bonding.
How does a specialized
vocabulary help and hinder
T
able 5 Comparison of the various relative strengths of intermolecular forces between
molecules
the growth of knowledge?
130
4 . 4
I N T e r M O l e c u l a r
F O r c e S
Worked examples
Example 1
Identify
●
the
intermolecular
forces
in
the
following
substances:
●
He
(CH
)
3
●
CH
(CH
3
●
)
2
●
CH
4
CH
3
O
2
F
3
●
NF
CH
3
CH
3
OH
2
Solution
Sstn
Intmo fos psnt
He
London only
CH
(CH
3
)
2
CH
4
commnt
Non-polar molecule so London only
3
Since F is more electronegative than N [χ
P
(F) = 4.0, χ
(N) =
P
3.0], this trigonal-pyramidal molecule has a net dipole moment
and therefore is polar:
Polar molecule so London +
N
NF
3
F
dipole–dipole
F
F
This molecule contains no H atoms, so no hydrogen bonding is
possible.
Even though the highly electronegative element oxygen is present,
there is no O–H bond so therefore no hydrogen bonding is possible.
Polar molecule so London +
)
(CH
3
O
2
O
dipole–dipole
H
CH
C
3
3
This molecule is tetrahedral; F is more electronegative than H
and C [χ
P
(F) = 4.0, χ
(C) = 2.6, χ
P
(H) = 2.0] so there is a net
P
dipole moment present making the molecule polar:
F
Polar molecule so London +
CH
C
F
3
dipole–dipole
H
H
H
Even though the highly electronegative element uorine
is present, there is no H–F bond so therefore no hydrogen
bonding is possible.
The molecule is polar and an O–H
O
CH
CH
2
3
bond is present, so:
H
London + dipole–dipole +
CH
3
CH
OH
2
hydrogen bonding
O
CH
CH
2
3
H
131
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Example 2
State
and
species
Example 3
explain
can
molecules:
form
which
of
the
hydrogen
ammonia,
NH
,
following
bonds
with
propane,
CH
3
ethanoic
acid,
CH
As
water
CH
3
CH
2
a
general
London
forces
CH
2
does
not
contain
OH,
NH
or
on
hydrogen
basing
the
It
forces
that
cannot
dipole–dipole
on
intermolecular
boiling
point
of
carbon
HF
form
hydrogen
bonds
with
CCl
water
is
76.72
°C,
COOH
have
N–H
and
O–H
bonds,
3
they
the
boiling
4
have
the
ability
to
form
of
uoromethane,
CH
F
is
78.2
°C.
Solution
hydrogen
●
with
whereas
3
CH
3
therefore
bonds
<
answer
the
molecules.
and
of
order:
3
point
NH
strengths
the
bonds
your
fact
tetrachloride,
bonds.
follow
(dispersion)
<
Comment,
forces,
CH
relative
,
3
Solution
3
the
forces
COOH.
3
CH
rule
intermolecular
We
rst
work
out
the
types
of
intermolecular
water:
forces
CCl
:
of
attraction
only
present
London
forces,
in
each
because
compound.
this
is
a
non-
4
H
polar
CH
molecule
F:
London
with
no
forces
net
and
dipole
moment.
dipole–dipole
forces,
3
O
because
this
is
a
polar
molecule.
H
●
On
the
rules,
N
H
point
H
basis
we
of
of
this
would
the
forces
opposite
is
following
that
uoromethane
intermolecular
Infact,
and
expect
the
our
general
boiling
with
its
additional
should
be
much
the
higher.
case!
H
●
H
The
the
of
reason
for
strength
CH
F
,
the
this
of
the
must
be
associated
London
number
of
forces.
valence
In
with
the
case
electrons
is
3
considerably
O
fewer
than
in
CCl
.
In
CCl
4
presence
H
greater
of
more
valence
polarizability
of
4
electrons
the
the
leads
electron
to
cloud.
a
This
CH
3
results
O
C
which
The
O
the
H
in
signicantly
outweigh
key
point
syllabus
challenged
is
here
the
is
that
relative
based
on
stronger
the
and
the
above
every
data
forces,
forces.
order
example
cited
must
in
be
provided!
Qik qstion
atiity
Suggest a second way in which
Researchers have seen hydrogen bonds for the rst time!
ethanoic acid CH
London
dipole–dipole
COOH can
3
Researchers in China recently used tomi fo miosopy (aFM) to produce
hydrogen bond with water.
the rst high quality images of hydrogen bonds that exist between molecules
of 8-hydroxyquinoline. Find out more about this from the chemical literature
or online. What is especially surprising about the atoms involved in this type
hydrogen bond (hint – consider the involvement of carbon!)?
132
4 . 5
M e T a l l I c
b O N D I N g
4.5 Mti ondin
Understandings
Applications and skills
➔
A metallic bond is the electrostatic attraction
Explanation of electrical conductivity and
➔
between a lattice of positive ions and
malleability in metals.
delocalized electrons.
Explanation of trends in melting points of
➔
➔
The strength of a metallic bond depends on the
metals.
charge of the ions and the radius of the metal ion.
Explanation of the proper ties of alloys in terms
➔
➔
Alloys usually contain more than one metal and
of non-directional bonding.
have enhanced proper ties.
Nature of science
➔
Use theories to explain natural phenomena – the properties of metals are dierent from covalent and ionic
substances and this is due to the formation of non-directional bonds with a “sea” of delocalized electrons.
Mti ondin
In
topic
3,
periodic
section
we
table
8
of
delocalized
gure
1,
is
saw
of
the
that
metals
elements.
Data
booklet),
throughout
a
surrounded
regular
by
a
the
giant
“sea”
of
lie
to
Metals
so
the
have
left
low
valence
metal.
lattice
The
that
delocalized
of
the
stepped
ionization
(outer-shell)
structure
consists
of
of
a
line
energies
electrons
metal,
positive
in
the
(see
can
shown
ions
be
in
( cations)
electrons.
Dnition of  mti ond
A mti ond is the electrostatic attraction between a lattice of positive ions
(cations) and delocalized electrons.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Figure 1 Structure of a metal showing an array of positive ions (cations) surrounded by
a “sea” of delocalized electrons
133
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Delocalized
a
metal,
lattice
The
but
electrons
instead
forming
strength
a
of
“sea”
a
●
the
number
●
the
charge
●
the
ionic
are
of
radius
associated
move
the
with
a
throughout
particular
the
entire
nucleus
of
crystalline
electrons.
depends
electrons
metal
of
to
bond
valence
the
not
mobile
metallic
of
of
are
free
on
that
three
can
factors:
become
delocalized
ion
metallic
positive
ion
(cation).
aoys
Dnition of n oy
An
An oy can be dened
alloy
metal
as a metallic material,
or
homogeneous on a
more
and
macroscopic scale,
is
(or
the
mixture
non-metal
metallic
hardness,
elements
that
consists
combined
non-metals
strength,
consisting of two or more
a
metals)
(for
(table
either
an
example,
carbon).
and
with
cast
Alloys
durability
of
two
alloying
iron
have
or
consists
enhanced
which
more
element
differ
metals,
of
the
those
of
metal
properties,
from
or
composed
of
of
a
one
iron
such
the
as
parent
1).
elements so combined
that they cannot be
readily separated by
composition of oy mix ts
mechanical means. Alloys
(mjo mnt: a metal) + (oyin mnt: can be metal or non-metal)
are to be considered as
mixtures for the purpose of
aoy
composition
uss
brass
copper and zinc
door handles, window
classication.
United Nations (2011)
ttings, screws
4th Edition.
steel
Iron, carbon, and other
bridges, buildings
metals such as tungsten
dental amalgam
mercury, silver, and tin
used by dentist for
teeth llings
T
able 1 Examples of alloys
Explanation of electrical conductivity and malleability in metals
Electrical conductivity
Metals
the
is
applied
move
carry
in
are
a
to
the
the
electrical
wiring
can
metal,
the
electric
metal
conductors
delocalized
through
an
through
134
good
mobile
a
high
electricity
and
presence
movement
Hence
degree
in
an
copper
of
because
a
of
of
purity.
can
Metals
of
or
a
are
solid
other
metals
impurities
(cations)
electrons
in
in
electrical
to
a
malleable.
to
be
shape
hence
increase
used
of
potential
electrons
structure
The
the
When
mobile
resulting
resistance.
needs
the
metallic
metal,
of
electrons.
current.
restrict
Malleability
have
without
this
can
dened
bonds
or
past
of
within
direction
non-directional
the
the
(they
as
is
one
they
that
The
the
another,
shape
lattice
are
is
the
hammered
breaking.
property
slide
rearrangement
metallic
Malleability
pounded
often
act
in
of
do
ability
into
a
positive
which
the
not
why
ions
leads
solid.
The
have
described
every
sheet
reason
any
as
direction
4 . 5
about
is
the
applied
over
one
metallic
xed
by
immobile
pounding,
another
bonding
but
cations).
the
there
(gure
Thus
cations
is
no
may
if
pressure
slide
disruption
The
melting
strength
to
the
2).
M e T a l l I c
positive
of
ions
electrons.
than
point
the
the
within
The
a
metal
attractive
the
melting
melting
following
of
b O N D I N g
“sea”
point
point
depends
forces
of
that
of
of
on
the
hold
the
delocalized
calcium
potassium
is
for
higher
the
reasons:
applied force under pressure
●
Calcium
atom,
has
delocalized
+
+
+
+
+
two
whereas
electron
ions
+
+
+
+
●
and
1+
+
+
+
+
atom.
per
one
Therefore,
between
delocalized
the
the
electrons
positive
will
be
+
●
From
a
2+
Therefore
the
electrons
+
calcium.
forms
ion.
between
+
attraction
the
in
Calcium
a
+
only
+
greater
+
per
electrons
has
+
electrostatic
+
delocalized
potassium
ion
the
positive
will
also
section
9
ions
be
of
whereas
and
greater
the
potassium
electrostatic
Data
the
in
forms
attraction
delocalized
calcium.
booklet
+
we
see
that
+
the
size
of
the
ionic
radius
of
K
is
138
pm,
2+
whereas
the
size
of
the
Ca
ionic
radius
is
structure of metal after being pounded into a sheet
smaller
Figure 2 Metallic bonding remains intact even after a metal is
at
100pm,
delocalized
which
electrons
hammered into a sheet or other object without breaking. This
will
implies
be
that
more
the
strongly
2+
attracted
to
the
Ca
ion.
illustrates the proper ty of malleability
This
Aluminium
this
foil,
property
of
often
used
to
malleability
wrap
for
the
food,
shows
variation
observed
metals.
metal
As
increases
aluminium.
going
expntion of tnds in mtin
in
melting
descending
we
saw
in
will
a
attractive
the
group,
decrease
forces
points
group
topic
(remember
down
points
the
on
and
with
3,
can
1,
the
radius
diagram)
the
melting
decreasing
(table
be
alkali
ionic
snowman
hence
also
the
strength
of
3).
points of mts
Metallic
bonds
are
very
strong
and
Mt
therefore
Ioni dis of
Mtin point / °c
+
metals
often
have
Forexample,
3414
occur
Table
°C,
metal,
as
potassium
calcium
some
38.8
compares
(section
melting
melting
although
such
2
the
high
°C
the
of
low
for
the
the
of
tungsten
melting
is
points
also
/ pm
lithium (Li)
76
180.5
sodium (Na)
102
97.8
mercury.
melting
and
7
point
M
points.
points
alkaline
Data
of
the
earth
alkali
potassium (K)
138
63.5
rubidium (Rb)
152
39.3
caesium (Cs)
167
28.5
metal
booklet).
T
able 3 Melting points for the group 1 metals
Mt
Mtin point / °c
When
comparing
metals,
potassium (K)
the
number
melting
of
points
delocalized
of
the
valence
alkali
electrons
63.5
per
atom
does
calcium (Ca)
the
not
(one)
and
change
the
within
charge
the
on
the
group,
so
cation
the
(1+)
only
factor
842
thatinuences
T
able 2 Melting points of potassium and calcium
the
melting
point
comparisonis
the
+
size
of
the
ionic
radius
ofM
135
4
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
Even
expntion of th poptis of
adding
element
oys in tms of non-dition
ondin
an
alloy
As
seen
the
Alloys
can
have
a
number
of
improved
to
the
parent
metallic
greater
greater
strength
to
enhanced
magnetic
ductility
for
a
metal
example
into
a
to
(a
mechanical
deform
being
as
those
the
of
alloying
the
the
metallic
non-directional
can
be
modied
if
(cations)
different
network
then
becomes
property
to
the
slide
the
properties
parent
bonds
shape
of
metal.
within
of
by
force
a
because
of
can
slide
atoms
positive
are
past
one
the
present
ions
is
another.
the
disturbed
more
difcult
for
the
and
positive
able
to
under
tensile
metal
stretch
the
past
each
(gure
other
3).
This
and
is
change
why
the
alloys
shape
are
that
generally,
allows
3,
an
properties
of
greater
with
of
change
corrosion
ions
●
gure
are
ions
However,
resistance
amount
element:
it
●
in
metal
regular
●
small
properties
positive
●
a
dramatically
compared
lattice
pure
compared
can
much
stronger
than
pure
metals.
stress,
metal
wire).
applied force
under pressure
applied force
under pressure
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
structure of metal after being pounded into a sheet
Figure 3 The presence of dierent atoms in an alloy disturbs the regular lattice and hinders the movement
of positive ions past one another
Mts – n intntion dimnsion
In this sub-topic we explored some key aspects of metals. The availability
of metal resources and the means to extract them varies greatly in dierent
countries, and is a factor in determining national wealth. As technologies develop,
the demands for dierent metals change and careful strategies are needed to
manage the supply of these nite resources.
Discuss this in class in the context of your own country and the regional economy.
136
Q u e S T I O N S
Qstions
1
What
is
the
name
of
K
SO
2
?
7
What
the
A.
potassium
calcium
C.
potassium
D.
calcium
the
electr o n
molecular
d o ma in
geo me tr y,
a nd
g e ome try,
the
Cl – P– Cl
sulte
bond
B.
is
4
angle
for
the
mo l ecul e
phos phor u s
sulte
trichloride,
PCl
?
3
sulfate
eton
sulfate
Mo
c–P–c ond
domin
omty
n /
°
omty
2
What
is
the
A.
MgO
B.
MgO
formula
of
magnesium
oxide?
A.
tetrahedral
B.
tetrahedral
tetrahedral
109.5
trigonal
109.5
2
pyramidal
C.
MnO
D.
MnO
trigonal
tetrahedral
C.
100.3
2
pyramidal
3
What
is
the
A.
NaNO
B.
NaNO
C.
Na
formula
of
sodium
nitrate?
trigonal
trigonal
pyramidal
pyramidal
100.3
D.
2
3
N
8
3
D.
4
A.
is
NaCN
Which
of
Na
Which
the
following
species
is
molecular?
O
following
KBr
C.
NH
graphite
B.
graphene
C.
C
D.
diamond
9
What
are
the
intermolecular
forces
present
4
in
the
molecule
CH
F
2
Which
molecule
oxygen
bond
A.
CH
B.
(CH
C.
(CH
CH
3
has
the
shortest
carbon-to-
A.
London
forces
B.
London
forces
C.
London
forces,
3
2
and
dipole–dipole
forces
OH
O
hydrogen
2
)
?
length?
2
)
3
dipole–dipole
forces,
and
bonding
CO
2
D.
only
hydrogen
bonding
CO
10
6
carbon
3
O
2
D.
of
NO
4
5
allotropes
60
B.
N
the
A.
2
D.
of
molecular?
The
electronegativities,
χ
,
for
four
elements
Which
statement
best
describes
metallic
are
P
bonding?
given
in
table
emnt
4.
H
c
O
A.
Electrostatic
B.
Electrostatic
charged
2.2
χ
2.6
3.4
attractions
between
oppositely
between
a
c
ions.
attractions
lattice
of
3.2
P
positive
▲
ions
and
delocalized
electrons.
T
able 4
C.
Which
bond
is
the
most
negative
polar?
D.
A.
C
Electrostatic
attractions
ions
Electrostatic
and
between
delocalized
attractions
a
lattice
of
protons.
between
protons
H
and
B.
O
H
C.
H
Cl
D.
C
O
IB,
electrons.
May
[1]
2009
137
4
C H E M I C A L
11
Consider
B O N D I N G
the
following
A N D
S T R U C T U R E
species
15
Compare
and
(diamond,
BF
Cl
NCl
2
contrast
graphite,
the
allotropes
graphene,
and
)
carbon
in
60
OF
3
of
C
2
termsof:
For
a)
b)
each
species
●
structure
●
bonding
●
intermolecular
●
melting
●
electrical
deduce:
(i)
its
electron
(ii)
its
molecular
(iii)
its
bond
(iv)
its
molecular
draw
an
domain
geometry
forces
geometry
points
angle(s)
polarity
appropriate
Lewis
(electron
conductivity.
dot)
structure.
16
In
chemistry
can
12
Consider
the
following
]
[ClO
3
both
lead
to
terminology
certain
]
[BF
3
]
COF
4
Suggest
is
why
incorrect
the
term
based
on
each
“macromolecular”
IUPAC
for
covalent
network
species
solids
a)
models
2
recommendations
For
and
assumptions.
species:
a)
[NO
often
such
as
graphite.
deduce:
b)
The
O–Cl–O
bond
angle
in
OCl
is
110.9°.
2
(i)
its
(ii)
electron
its
domain
molecular
geometry
Discuss
with
geometry
the
(iii)
b)
its
draw
bond
an
whether
predictions
model
of
this
of
bond
bond
VSEPR
angle
angles
agrees
based
on
theory.
angle(s)
appropriate
Lewis
(electron
dot)
17
structure.
Suggest
work
why
for
the
VSEPR
theory
majority
of
does
not
transition
metal
2
complexes,
such
as
[FeCl
]
,
but
4
13
Deduce
the
intermolecular
forces
present
in
few
complexes,
such
as
[MnO
]
4
each
of
●
Ar
●
CH
the
CH
3
●
CH
following
CH
2
species:
OH
2
Cl
3
●
CH
CH
3
14
Deduce
form
●
which
CH
CH
●
NH
●
C
3
H
2
4
PH
3
138
3
of
the
following
bonds
OCH
2
CH
2
hydrogen
3
●
OCH
2
CH
2
3
with
species
water
may
molecules:
does
for
a
5
E N E R G ET I C S
AND
T H E R M O C H E M I ST RY
Introduction
Chemistry
involves
elements
of
energy
a
is
the
study
periodic
use
models,
and
scientic
with
reactions
to
of
Conservation
principle
empirical
terminology
associated
of
table.
fundamental
The
changes
of
the
of
data,
central
of
examine
science.
mathematics
explain
chemical
the
the
energy
reactions
is
to
the
the
chemistry
function
Hess’s
the
nature
and
energy.
enthalpy,
Law
of
science.
relationship
and
of
We
a
will
greater
bond
In
this
exists
investigate
gain
applications
that
topic
introduce
the
we
between
the
state
applications
understanding
of
of
enthalpies.
5.1 M   
Understandings
Applications and skills
➔
Heat is a form of energy.
➔
Temperature is a measure of the average kinetic
➔
using
Total energy is conser ved in chemical
reactions.
➔
➔
A
q
=
the
of
a
hea t
pure
cha nge
when
s u bs tance
is
the
changed
mc ∆T
calorimetry
reaction
Chemical reactions that involve transfer of heat
of
temperature
energy of the par ticles.
➔
Calculation
experiment
should
be
f or
covered
an
and
enthalpy
the
of
resu lts
evaluated.
between the system and the surroundings are
described as endothermic or exothermic.
➔
The enthalpy change (ΔH) for chemical
1
reactions is indicated in kJ mol
➔
.
ΔH values are usually expressed under
standard conditions, known as ΔH
, including
standard states.
Nature of science
➔
Fundamental principle – conser vation of energy is a fundamental principle of science.
➔
Making careful obser vations – measurable energy transfers between systems and surroundings.
139
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
What is thermodynamics?
The
utilization
Agricultural,
consume
of
energy
industrial,
vast
amounts
is
central
and
of
to
our
domestic
energy
will
lives.
activities
remain
law
all
is
the
study
of
how
it
is
interconverted.
The
can
only
states
that
rst
energy
law
can
from
one
form
to
be
of
neither
law
created
converted
is
often
energy
and
nor
between
called
states
the
that
destroyed;
different
it
another
that
for
a
given
system
we
forms.
can
of
for
and
quantify
all
the
energy
changes.
be
This
converted
be
means
account
thermodynamics
This
energy
This
and
constant.
conservation
energy
daily.
can
Thermodynamics
of
and
is
one
of
the
most
fundamental
principles
that
ofscience.
the
total
amount
of
energy
for
a
given
system
Chemical potential energy, heat, and entropy
In
a
chemical
energy
the
is
Heat,
of
q,
cooler
the
is
a
to
as
conduction,
When
as
heat
of
a
total
the
of
reacting
ions,
and
energy
result
thermal
of
and
transferred
to
is
conserved.
bonds
of
mixture
is
is
a
It
an
function
transferred
can
be
object,
and
potential
products,
of
the
while
kinetic
present.
temperature
radiation.
Chemical
reactants
molecules
that
the
energy.
convection,
is
energy
chemical
the
atoms,
form
body,
referred
system +
in
temperature
energy
universe =
reaction
stored
from
transferred
Heat
the
a
gradient.
has
result
by
the
is
warmer
Heat
is
the
processes
ability
an
body
to
a
sometimes
to
do
increase
of
work.
in
the
surroundings
average
kinetic
temperature
At
absolute
or
reaches
its
kelvin)
is
matter.
As
a
zero,
theoretically
surroundings
energy
change
0
stops
K
(
minimum
its
in
particles
the
°C),
all
entropy
possible
to
and
therefore
an
increase
in
its
phase.
273.15
and
proportional
the
of
the
temperature
value.
motion
S
(see
The
average
the
absolute
kinetic
increases,
of
the
particles
sub-topic
15.2)
a
temperature
energy
kinetic
of
of
the
energy
system
(in
particles
of
the
of
particles
increases.
system
(contents
of ask)
Chemical energy
When
examining
the
energy
changes
involved
in
a
chemical
reaction,
Figure 1 The universe is the combination of the
system and its surroundings
we
divide
chemical
think
of
the
universe
reaction
the
is
system
into
taking
as
two
being
solvents.
The
reaction,
thermometers
parts:
place,
surroundings
all
and
the
include
the
its
system
reactants,
the
in
which
surroundings.
products,
apparatus
that
the
You
and
can
any
contains
the
In an op tm the transfer
or
other
measuring
devices,
the
laboratory,
and
of matter and energy is possible
everything
external
to
the
reacting
substances.
across its boundary (eg matter
can be added to a beaker, and
When
energy can be transferred
rearranged
to
through its sides). A od
are
and
tm allows no transfer of
Energy
matter, though energy may be
endothermic
transferred across the boundary.
energy
In an otd tm, matter
released
can neither enter nor exit, but
exothermic
can only move around inside.
and
a
broken
is
the
changes
140
chemical
reaction
create
new
required
and
it
can
is
an
This
energy
reaction.
for
bonds
transfer
important
the
are
is
are
type
of
the
reactants
are
in
the
reactants
form
the
products.
bond
the
made:
energy
of
to
bonds:
termed
of
bonds
made
each
of
part
atoms
Chemical
chemical
quantied
The
place,
bonds
the
chemical
process.
system
a
break
be
new
takes
products.
chemical
to
process.
when
in
new
breaking
bond
bond.
bond
is
an
dissociation
Energy
making
is
is
an
between
the
surroundings
understanding
the
energy
5 . 1
M e a s u r i n g
e n e r g y
c h a n g e s
Enthalpy and thermochemistry
A omt is any apparatus
Enthalpy
is
an
example
of
a
state
function.
For
a
state
function
any
used to measure the amount
change
in
value
is
independent
of
the
pathway
between
the
initial
and
of heat being exchanged
nal
measurements.
Other
examples
of
state
function
include
volume,
with the surroundings. In an
temperature
and
pressure.
exothermic reaction heat is
For
example,
pool
early
in
afternoon
a
simple
and
the
(nal
uctuations
is
taking
value),
one.
temperature
morning
that
cooling
the
may
does
have
However,
which
(the
has
initial
not
tell
does
occurred
the
water
value)
the
occurred
it
of
and
whole
give
any
throughout
a
then
story
throughout
not
in
the
again
of
in
any
day.
indication
the
generated which is transferred
swimming
temperature
The
of
to the surroundings. In an
the
calculation
the
heating
endothermic reaction heat
is consumed. In the school
laboratory, experiments focus
on the change in temperature
day.
of the reaction solvent, which
∆T
=
T
T
(nal)
in most cases is water.
(initial)
Thermochemistry
chemical
is
reactions.
dened
reaction.
as
The
commonly
unit
of
the
is
heat
term
used
the
At
of
“change
in
by
∆H
changes
a
is
the
the
closed
enthalpy”
describing
change
heat
pressure,
transferred
when
enthalpy
study
constant
or
that
change
system
“heat
occur
in
during
enthalpy
during
of
a
reaction”
thermodynamics
of
a
∆H
chemical
is
reaction.
The
exothermic reaction
kJ.

ygrene laitnetop
Endothermic and exothermic reactions
A
chemical
to
the
reacti o n
in
surrounding s
is
which
he a t
d ene d
as
is
an
t r an s fe r re d
e xoth e r m i c
from
t he
r e act i on
sy s t em
wi t h
a
a
reactants
∆H
products
negative
their
∆H.
In
contrast,
surrounding s
positive
are
chemi ca l
de ned
as
r e a ctio ns
tha t
e n dot h e r m i c
a bso rb
r e ac t ion s
he a t
wi t h
fro m
a
reaction pathway
∆H
Figure 2 In an exothermic reaction the
enthalpy of the products is lower than that
The
calculations
involved
in
investigating
the
energetics
of
the
reaction
of the reactants. The products are described
between
zinc
and
aqueous
copper(II)
sulfate,
CuSO
are
described
4
as being energetically more stable than the
later
in
the
topic.
Measured
quantities
of
copper(II)
sulfate
solution
reactants. (We shall study the activation
and
zinc
are
mixed
in
a
calorimeter.
The
mixture
is
stirred
and
the
energy of a reaction in topic 6)
change
in
temperature
measured
using
a
thermometer
or
data-logging
equipment.
endothermic reaction
Zn(s)
+
CuSO
(aq)
→
Cu(s)
+
ZnSO
4
result
involving
of
calculating
the
exothermic
displacement
reaction
Ammonium
ΔT
nitrate,
shows
of
(gure
NH
this
single
copper(II)
ion
replacement
by
the
metal
reaction
zinc
is
an
2).
NO
4
the
that
is
an
important
component
of
fertilizers.
3
ygrene laitnetop
The
(aq)
4
products

a
∆H
reactants
When
the
nitrate
solid
ions,
dissolves
the
reaction
in
water
requires
to
form
heat
to
aqueous
proceed.
ammonium
This
heat
is
and
absorbed
reaction pathway
by
the
system
from
the
surroundings,
resulting
in
a
decrease
in
the
Figure 3 In an endothermic reaction the
temperature
of
the
surroundings
as
recorded
by
a
thermometer.
The
enthalpy of the products is greater than that
apparatus
of
an
containing
endothermic
the
reaction
reaction
feels
(gure
3).
cold
to
touch.
This
is
an
example
of the reactants. The products are described
as being energetically less stable than the
reactants
+
NH
NO
4
(s)
3
→
NH
(aq)
4
+
NO
(aq)
3
141
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
Calculating enthalpy changes in aqueous solutions
We
stated
earlier
that
the
change
in
enthalpy ∆H
or
a
copper
1
K.
For
example,
the
1
is
dened
as
the
heat
transferred
by
closed
is
0.385
1
system
∆H
for
heat
a
a
a
chemical
reaction
change.
pure
an
during
we
When
substance
therefore
of
as
the
To
need
calculating
such
understanding
reaction.
the
water,
to
nd
heat
we
physical
calculate
the
change
need
to
quantity,
2.44
of
of
have
a
J
K
heat
capacity,
specic
heat
capacity
of
1
K
.
The
while
lower
substance,
achieved
transferred
that
of
ethanol
is
for
to
the
the
the
the
specic
higher
same
heat
the
amount
capacity
temperature
of
heat
sample.
the
Specic
specic
g
1
g
given
rise
J
heat
capacity
is
an
intensive
property
c
that
does
not
vary
in
magnitude
with
the
size
of
3
the
1
The
units
The
specific
for
specific
hea t
ca pacity
are
kJ
kg
system
being
described.
For
example,
a
10
cm
1
K
.
sample
of
copper
has
the
same
specic
heat
1
heat
cap a city
of
w ater
is
4 . 18
kJ
capacity
kg
as
a
1
tonne
block.
1
K
and
this
can
be
found
in
s ection
2
of
the
Specic
Data
q
of
a
q
The
specic
dened
the
heat
capacity
is
used
to
calculate
the
heat
booklet
as
heat
the
capacity
amount
temperature
of
1
g
of
a
pure
of
heat
of
the
substance
needed
to
substance
1
=
using
the
relationship:
mcΔT
is
raise
by
system
°C
where
m
is
mass
in
kg
and
∆T
is
the
change
in
temperature.
Wokd xmp : t tp 
Example 1
Example 2
When a 1.15 g sample of anhydrous lithium chloride, LiCl
180.0
was added to 25.0 g of water in a coee-cup calorimeter,
iron,
resulting
a temperature rise of 3.80 K was recorded. Calculate the
26.0
°C.
J
of
heat
is
in
trans f erred
a
Calculate
to
tempera ture
the
specif ic
a
1 00 . 0
ris e
hea t
g
from
s a mp le
2 2.0 °C
capa city
of
of
to
iron .
enthalpy change of solution for 1 mol of lithium chloride.
Solution
Solution
∆T = (299
295) K = 4 K .
q = mc∆T
Make c the subject of the equation and solve:
1
= 0.025 kg × 4.18 kJ kg
q
__
1
K
× 3.80 K
c =
m∆T
= 0.397 kJ
0 180 kJ
___
=
Conver t to energy gained for 1 mol of LiCl.
0.100 g × 4 K
0.397kJ / 1.15g LiCl × 42.394 g/mol = 14.6 kJ/mol LiCl
1
=
0.450 kJ K
1
∆H = -q = -14.6 kJ mol
thermometer
co-p omt
Performing reactions in a polystyrene coee cup to measure the enthalpy
change is a convenient experimental procedure. The methodology introduces
glass stirrer
cork stopper
systematic errors that can be analysed and the eect of their directionality
assessed.
stmt o are a consequence of the experimental procedure. Their
eect on empirical data is constant and always in the same direction. With the
two polystyrene cups nested
coee-cup calorimeter, the measured change in enthalpy for a reaction will
together containing reactants
always be lower than the actual value, as heat will be transferred between the
in solution
contents and the surroundings in every experiment.
Figure 4 A coee-cup calorimeter
142
5 . 1
M e a s u r i n g
e n e r g y
c h a n g e s
Investigation to nd the molar enthalpy change for a reaction
Earlier
we
looked
displacement
at
the
reaction
exothermic
between
zinc
●
metal
and
The
copper(II)
specic
solution
is
heat
the
capacity
same
as
of
that
an
of
aqueous
water.
sulfate:
Loss
Zn(s)
+
CuSO
(aq)
→
Cu(s)
+
ZnSO
4
is
(aq)
following
heat
main
from
method
is
used
to
calculate
that
is
enthalpy
change
for
this
reaction
the
this
surroundings
experiment
to
quantify.
The
change
and
in
the
=
mc∆T
∆T
systematic
means
that
will
lower
be
calculated
Experimental method to determine ΔT
calculated
or
The
effect
the
from
directional
Using
an
electronic
balance,
a
graph
error.
will
This
include
loss
CuSO
the
mass
solution.
of
25
Subtract
from
solution
following
to
the
the
coffee-cup
mass
of
the
the
of
a
1.0
mass
mol
of
heat
the
of
the
cylinder
An
transfer
of
the
+
accepted
or
record
a
temperature
the
errors
in
of
curve
the
probe
temperature
of
until
a
every
30
constant
seconds
for
temperature
the
than
making
the
actual
procedure
is
on
important
in
the
value.
the
result
of
considering
up
is
to
3
in
experimental
procedures.
to
data
of
calculating
compensate
is
to
after
this
look
the
back
at
for
the
the
maximum
systematic
cooling
reaction
to
the
is
section
complete,
point
of
and
introduction
and
the
zinc,
as
shown
in
gure
5.
A
more
accurate
the
value
solution
lower
in
method
temperature
solution
calorimeter.
thermometer
software,
q
value,
recorded
dm
of
related
of
actual
calculations
extrapolate
Using
the
errors
improvements
3
cm
4
cylinder
than
temperature
accurately
3
measure
maximum
value
of
subsequent
2
to
in
equation:
q
1
system
error
from
a
the
of
difcult
temperature
molar
the
source
4
one
The
of
the
minutes,
for
ΔT
can
then
be
determined.
or
achieved.
Calculation of molar enthalpy change
3
At
3
minutes,
(between
and
4
1.3
g
and
commence
Continue
to
introduce
5
to
g,
zinc
previously
weighed)
M of opp(ii) ft oto/
28.8
M of z/
1.37
stirring.
take
minutes
1.4
powdered
temperature
after
the
readings
maximum
for
up
c  tmpt T()
39.0
T(t)/°c
temperature
T
able 1 Sample results
has
5
been
Produce
reached.
a
temperature
determine
the
change
versus
in
time
graph
to
temperature.
Taking
molar
q
the
results
enthalpy
=
mc∆T
=
0.0288
=
4.69
in
table
change
as
1
we
can
calculate
the
follows:
Assumptions and errors
1
A
number
this
●
of
assumptions
The
when
heat
released
coffee
loss
coffee
is
cup
to
cup
difcult
also
to
polystyrene
The
acts
the
transferred
be
from
transferred
as
an
reaction
the
has
it
a
insulator
heat
from
quantify
cup,
so
it
the
is
water.
heat
and
It
to
would
be
reached
the
heat
capacity
assumed
temperature
against
representation
of
kJ
kg
1
K
×
39.0
K
kJ
is
of
∆T
a
zero.
an
0
accurate
4.18
is
However,
capacity
the
×
water.
surroundings.
to
maximum
the
to
kg
using
C°/erutarepmet
The
heat
●
made
method:
completely
●
are
the
heat
2
4
6
8
10
evolved
time/min
during
the
reaction.
Figure 5 Determination of the change in temperature
in calorimetry experiments
143
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
1
__
amount
of
CuSO
=
1.37
g
The
×
4
shape
of
the
graph
and
the
change
in
1
65.38
g
mol
temperature
=
0.0210
from
a
lower
to
a
higher
value
lead
mol
to
the
conclusion
that
the
reaction
is
exothermic:
1
4.69 kJ
__
molar
enthalpy
change
ΔH
=
-223
kJ
mol
=
0.0210
mol
1
=
223
kJ
mol
TOK
Tmpt 
In theory of knowledge there
The SI unit of temperature is the kelvin (K). Note that a change in temperature ΔT
are eight specic ways of
calculated from experimental data in Celsius will be identical to the value of ΔT
knowing. These are: language,
calculated in kelvin.
sense perception, emotion,
Throughout the world, the majority of countries use the Celsius scale for the
reasoning, imagination,
everyday description of temperature. As the Celsius and kelvin scales are linked,
faith, intuition, and
you will often see both scales being used in an IB question. The USA uses a
memory. Scientists perform
mixture of metric and imperial units of measurement. For example, the Fahrenheit
experiments and process
scale of temperature is used in the USA.
the raw data to enable us
to draw conclusions. We
compare experimental and
°F
theoretical values. What
100
criteria do we use when
80
making these comparisons?
60
Are our judgments subjective
40
or objective? When analysing
20
and appraising experimental
0
20
limitations and making
theoretical assumptions,
which of the ways of knowing
Figure 6 We use SI units in science, but a mix ture of imperial and metric systems of
are we utilizing?
measurement is used in dierent countries
The tdd tp  of
 to ΔH
is determined
Enthalpy change of formation
298
at temperature 25 °C/298 K and
The
change
in
enthalpy
during
a
reaction
can
be
determined
using
the
pressure 100 kPa with all species
following
equation:
in their standard state. stdd
ΔH
reaction
=
∑(ΔH
odto are denoted by the
symbol
.
products)
–
∑(ΔH
f
ΔH
is
the
standard
reactants)
f
enthalpy
change
of
formation
of
a
substance.
f
This
is
from
the
its
existing
energy
change
constituent
enthalpy
upon
elements
of
the
in
formation
formation
their
data
of
standard
to
1
mol
state.
calculate
the
of
We
a
substance
can
enthalpy
use
of
std tp
reaction.
The
value
and
sign
of
the
calculated
enthalpy
of
formation
Section 12 in the Data booklet
informs
us
about
the
energetics
of
the
reaction.
gives the standard enthalpy of
formation for a large number
For
example,
the
standard
enthalpy
change
of
formation
of common compounds. In
for
methane
is:
1
C(s)
+
2H
(g)
→
CH
2
(g)
∆H
4
=
-74.9
kJ
mol
f
examinations, questions will
provide any other values not
included in the Data booklet
It
is
in
to
note
their
that
the
standard
elements
states.
carbon
Equations
and
for
hydrogen
∆H
must
are
represent
f
the
144
important
represented
formation
of
1
mol
of
a
substance.
In
some
cases,
such
as
the
5 . 1
M e a s u r i n g
e n e r g y
c h a n g e s
1
formation
of
phenol
shown
below,
this
results
in
mol
of
diatomic
2
Qk qto
oxygen
O
appearing
on
the
reactant
side:
2
Write equations to describe
1
_
6C(s)
+
3H
(g)
+
1
O
2
(g)
→
C
2
H
6
OH(s)
∆H
5
=
-165.0
kJ
the standard enthalpy change
mol
f
2
of formation for the following
compounds and state the
Enthalpy change of combustion
enthalpy value by referring to
The
standard
enthalpy
change
of
combustion
∆H
is
the
heat
the Data booklet
c
evolved
The
are
upon
the
enthalpies
values
classied
of
complete
combustion
derived
as
combustion
under
liqueed
found
standard
petroleum
in
of
1
mol
section
conditions.
gas
(LPG)
is
of
13
substance.
of
the
Butane,
highly
Data
one
of
)
propane
b)
chloromethane
)
ethanol
d)
benzoic acid
)
carbon monoxide
booklet
the
gases
ammable:
13
_
C
H
4
(g)
+
O
10
(g)
→
4CO
2
(g)
+
5H
2
O(l)
2
2
1
∆H
=
-2878
kJ
f)
mol
methylamine
c
This
thermochemical
enthalpy
of
equation
combustion
value
can
also
included
in
be
written
the
with
equation.
the
The
negative
1
compod
∆H
/kJ mo
f
enthalpy
change
indicates
an
exothermic
reaction
so
the
value
would
C
H
6
be
included
on
the
product
(l)
+49.0
6
side:
CO
(g)
393.5
2
13
_
C
H
4
(g)
+
1
O
10
(g)
→
4CO
2
(g)
+
5H
2
O(l)
+
2878
kJ
mol
O(l)
H
2
2
285.8
2
T
able 2 Standard enthalpy changes
Working method
of formation.
Benzene,
2C
C
H
6
H
is
6
6
(l)
+
highly
15O
6
∆H
(g)
ammable,
→
12CO
2
reaction
=
(g)
producing
+
6H
2
∑(∆H
[12
×
(
=
(
=
-6535
393.5)
4722
+
products)
6
sooty
ame:
O(l)
2
-
∑(∆H
f
=
a
reactants)
f
×
1714.8
(
285.8)
98.0)
2
×
(+49.0)
15
×
0]
kJ
kJ
kJ
Investigation to nd the enthalpy change of combustion
3
The
enthalpy
change
of
combustion
of
common
2
Accurately
determine
the
mass
of
30
cm
of
3
alcohols
From
a
can
be
determined
homologous
series
in
of
the
laboratory.
alcohols,
water
enthalpy
change
of
combustion
values
can
and
subsequently
Place
the
beaker
with
the
Using
either
temperature
following
spirit
of
in
burners
the
procedure
standard
are
alcohols
butan-1-ol
1
a
and
Determine
using
an
utilizes
school
required,
methanol,
initial
electronic
cm
beaker.
or
metal
spirit
calorimeter
burner
on
a
beneath.
each
Five
containing
ethanol,
mass
of
balance.
temperature
determine
of
the
probe
and
or
record
a
the
initial
water.
equipment
laboratory.
one
propan-1-ol,
pentan-1-ol.
the
a
thermometer,
Experimental method
available
250
analysed.
4
The
a
be
tripod
determined
in
patterns
3
in
contained
the
spirit
burners
5
A
spirit
the
burner
alcohol
period
over
in
of
a)
one
allow
is
is
lit
burnt
which
two
under
to
it
heat
burns
different
each
alcohol
temperature
the
calorimeter
the
can
water.
be
and
The
monitored
ways:
to
change
burn
of
until
30 °C
is
a
achieved
145
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
b)
Qk qto
Write equations to describe the
6
standard enthalpy change of
allow
After
cap,
this
each
time
re-weigh
alcohol
period,
each
to
burn
for
extinguish
one
and
a
period
each
record
the
spirit
of
2
minutes.
burner
change
in
by
mass
replacing
of
the
the
alcohol.
combustion for the following
compounds, and state the
Calculation of enthalpy of combustion
enthalpy value by referring to
aoo
∆m of oo/
∆T/°c
0.348
30.0
M of wt/
section 134 of the Data booklet
methanol
)
octane, C
31.2
H
8
18
T
able 3 Sample results
b)
chloroethane, C
H
2
Cl (hint:
5
q
a corrosive strong acid is
=
mc∆T
=
0.0312
=
3.91
one of the products)
)
cyclohexanol, C
H
d)
methanoic acid, CH
6
1
kg
×
4.18
kJ
kg
1
K
×
12
0.348
g
__
2
amount
of
CH
OH
=
=
3
6
H
12
0.0109
mol
1
32.05
glucose, C
K
kJ
O
2
)
30.0
O
g
mol
O
6
3.91 kJ
__
molar
energy
change
=
0.0109
mol
1
=
359
kJ
mol
As in all investigations, rst
1
ΔH
=
–359
kJ
mol
determine the dependent and
independent variables and the
variables that will be controlled.
Assumptions
●
Heat
loss
to
signicant
●
All
the
the
but
environment
cannot
alcohols
are
be
pure
is
negligible
(in
reality,
it
is
quantied).
and
undergo
complete
combustion.
Dt o can be used to
record temperature changes
accurately and the associated
software to perform data
Obt d t  ott of food
analysis and graphing. The
The world increase in obesity
use of data-logging equipment
Obesity, eating disorders, and unhealthy diet are serious health issues facing many
demonstrates the practical
cultures throughout the world, as many societies become more auent and food is
application of technology in the
readily available. Obesity is generally dened as an excessive accumulation of fat
laboratory.
that can lead to health problems. The bod m dx (BMi) is found by taking a
person’s mass (in kilograms) and dividing it by the square of their height (in metres).
An adult with a BMI above 25 is considered overweight while one with a BMI
greater than 30 is obese. The World Health Organization (WHO) has been
Energetics experiments
monitoring the eect of changes in diet on dierent nations for decades. Their
provide a useful set of
research has found the following:
raw data and involve
●
In 2013 the occurrence of obesity worldwide was more than double the level
experimental procedures
in 1980.
that can be evaluated for
random and systematic errors
●
(topic 11). The identication
In 2008 over 1.4 billion adults worldwide were overweight, with approximately
200 000 000 men and 300 000 000 women being classied as obese.
of the systematic errors
●
65% of the world’s population reside in countries where more people die from
and examination of their
obesity-related causes than from being underweight.
directionality is an essential
●
More than 40 000 000 children under the age of 5 years were overweight in 2010.
●
Obesity is a preventable disease.
aspect of the analysis of
experimental results.
146
5 . 1
M e a s u r i n g
e n e r g y
c h a n g e s
In China, the rapid increase in auence and the globalization of the economy
has seen an unprecedented expansion in the fast-food industry and of nutritional
choices. With these has come a signicant increase in the number of children
who are overweight and obese. Type 2 diabetes is normally associated with
adults, but the rise in the prevalence of the disease amongst children in China
and in other countries is seen as a signicant threat to the wellbeing of future
generations.
Food labelling and determination of energy content
Governments throughout the world have a responsibility to their citizens to
provide leadership, education, and guidance in health and nutrition. Linked
to the globalization of economies and free-trade agreements has been the
standardization of labelling of food products to include an analysis of the contents
including energy content.
To determine their energy content, foods were traditionally placed in a calorimeter
surrounded with water and completely burnt, causing the water to rise in
temperature. This temperature change was then used to calculate the energy
content (sometimes referred to as “caloric value”) of the food. Today the
preferred method of calculation is using the Atwater system. This system relies on
average energy values for proteins, carbohydrates, fats, and alcohol being applied
to foods of a known composition. The National Data Laboratory (NDL) in the USA
holds information on the energy content of over 6000 foods.
Figure 7 Nutritional information displayed on food packaging
147
5
e n e r g e T i c s
a n D
T h e r M O c h e M i s T r y
5.2 h’ w
Understandings
Applications and skills
➔
The enthalpy change for a reaction that is
Application of Hess’s law to calculate enthalpy
➔
carried out in a series of steps is equal to
changes.
the sum of the enthalpy changes for the
Calculation of ∆H reactions using ∆H
➔
data.
f
individual steps.
Determination of the enthalpy change of a
➔
reaction that is the sum of multiple reactions
with known enthalpy changes.
Nature of science
➔
Hypotheses – based on the conser vation of energy and atomic theory, scientists can test the hypothesis
that if the same products are formed from the same initial reactants then the energy change should be
the same regardless of the number of steps.
TOK
Testing hypotheses
In TOK , a primary focus is on
Experimental
questions about knowledge
hypothesis.
which are open ended
atomic
with multiple perspectives
that
and expressed without
change
using subject-specic
and
language. Hess’s law can be
can
evidence
Based
theory,
when
in
the
be
on
enables
the
scientists
products
enthalpy
number
analysed
of
are
are
able
formed
should
be
chemical
and
scientists
principles
used
as
to
of
test
from
to
or
identical
reactions
disprove
of
same
set
regardless
involved.
for
of
of
a
energy
experimentally
the
evidence
prove
conservation
the
and
hypothesis
reactants,
the
route
Quantitative
the
taken
data
thishypothesis.
considered an application
of the law of conservation
of energy. What are the
Overall and net reactions
challenges in applying
If
you
have
travelled
to
New
York,
Tokyo,
London,
Hong
Kong,
Paris,
general principles of a law
Beijing,
Berlin,
or
Seoul,
you
will
have
experienced
the
subways
that
to something as specic as
criss-cross
these
enormous
cities
and
transport
millions
of
people
every
Hess’s law?
day.
In
any
between
often
The
point
same
idea
shows
reactions,
Hess’s
law
Regardless
always
Figure
be
2
can
is
of
can
the
an
the
then
B.
For
which
be
net
is
true
by
that
in
a
the
applied
of
to
–
eld
it
is
than
the
all
each,
chemistry.
summary
result
reaction
and
the
nal
in
of
A
a
to
half
travel
the
of
an
enthalpy
chemical
number
overall
energy
proceeds,
states
in
of
an
equations,
results
the
way
traveller,
conservation
chemical
one
fun
is
route.
of
a
together
nd
more
adventurous
initial
summing
is
fastest
the
of
which
magnitude
be
the
added
providing
there
the
reaction
when
route
shows
network
application
same
and
and
out
which
the
direction
148
A
working
usually
law
transport
the
the
taking
change
enthalpy
are
into
change
the
the
will
same.
account
equation.
for
different
reaction.
law:
system
overall
equation
of
the
Hess’s
reaction.
5 . 2
h e s s ’ s
l a W
A
H
H
y
x
C
H
z
B
Figure 2 Hess’s law
Figure 1 You can take many alternative routes on the Paris Metro
Worked example : calculating enthalphy change
Using
Hess’s
calculate
law,
the
and
the
enthalpy
following
change
ΔH
reversing
information,
for
the
change
reaction:
the
the
chemical
sign
of
the
equation
enthalpy
we
must
value:
4
1
_
1
_
C
+
2H
+
CO
O
2
→
CH
2
OH
∆H
3
(4)
+
2H
2
O
→
CH
2
OH
+
1
O
3
2
2
4
2
∆H
1
_
CH
OH
+
1
=
+676
kJ
1
O
3
→
CO
2
+
2H
2
O
2
2
●
∆H
=
-676
kJ
Because
oxygen
is
found
in
all
three
(1)
1
equations,
should
C
+
O
→
CO
2
∆H
2
=
-394
kJ
be
O
→
H
2
point
hydrogen.
We
of
focus
require
2
mol
of
(2)
1
_
+
2
next
2
hydrogen
H
the
O
∆H
2
=
-242
kJ
3
(3)
can
be
as
a
used
reactant.
in
the
Therefore,
direction
as
equation
written
but
3
2
with
IB,
May,
that
2006
double
the
●
of
at
the
overall
formation
of
equation
4
for
the
enthalpy
From
Now
equations
1–3,
the
reactant
carbon
product
focus
●
For
of
methanol
your
carbon,
should
be
the
require
a
2H
can
and
O
add
∆H
three
species
summing
be
=
This
means
doubled:
-484
kJ
3
the
those
moles.
must
2
the
in
table
equations
common
enthalpy
to
together,
both
values
as
1.
main
rtt
reaction
of
and
methodology.
we
→
value
2
we
shown
the
O
eliminating
methanol.
sides
●
+
2
●
Look
number
enthalpy
2H
Solution
the
that
Podt
C +
uses
→
etp
ΔH
=
–394 kJ
2
1
mol
of
carbon
as
a
reactant.
Carbon
is
a
_
+
O
→
CH
OH + 1
ΔH
3
reactant
used
as
in
equation
2,
so
this
equation
can
= +676 kJ
1
be
2
1
_
written:
2H
+
O
2
→
ΔH
2
2
= –484 kJ
3
2
C
+
O
→
CO
2
∆H
2
=
-394
kJ
2
C + 2H
2
●
For
methanol
we
need
to
use
equation
→
CH
3
OH
ΔH
= –202 kJ
4
1
_
1,
+
O
2
but
we
need
methanol
is
to
a
reverse
product
the
not
equation
a
so
reactant.
that
2
When
T
able 1
149
5
E N E R G E T I C S
The
combination
A N D
of
T H E R M O C H E M I S T R Y
the se
re a ctio ns
can
a l so
1
C
+
2H
+
O
2
be
represented
gure
You
diag r a mma tical l y
as
sh o wn
CH
2
OH
3
2
in
3.
will
notice
that
the
enthalpy
cycle
diagram
1
O
O
2
1
2
O
2
2
shows
the
reversed,
combustion
in
summation
the
of
same
of
methanol
way
equations
it
was
equation
during
the
method.
CO
2
+
2H
O
2
Figure 3 Enthalpy cycle for the formation of methanol
T t of 
With the world population exceeding 7 billion people,
phosphor dust, and other hazardous substances. Guiyu in
and increasing international concern over the world’s
China has become a centre for e-waste disposal and the
resources, recycling of materials has become mainstream
rapid expansion of recycling processes in villages around
in many countries. Recycling developed from a desire to
Guiyu has resulted in heavy metal contamination of the
use raw materials more eciently, to reduce energy use
groundwater and soil, and air pollution from the burning of
in the production of goods, to protect the environment
plastics within the waste.
from excessive pollution, and to utilize waste materials
Despite good intentions, the end result of recycling can
and thereby reduce landll waste disposal. However,
be extremely negative. What is more, the eciency
economic and other pressures to recycle can lead
of recycling processes in energy terms varies widely.
to potentially harmful impacts on the environment.
Many countries and environmental organizations are
For example, electronic or e-waste is an escalating
investigating how we can address the long-term
problem throughout the world. This waste contains
eects of recycling programmes on the environment
heavy metals such as lead and cadmium, highly toxic
and communities.
Figure 4 Technology waste recycling in Guiyu, China
150
5 . 2
h e s s ’ s
l a W
Worked example
Determine
the
enthalpy
of
formation
C
the
enthalpy
of
combustion
H
2
using
of
ethane,
data
In
summary:
in
6
■
section
13
of
the
Data
The
equation
is
■
equation
hydrogen
Write
a
balanced
chemical
equation
for
of
1
combustion
of
for
the
combustion
of
combustion
of
carbon
mol
of
is
tripled.
the
■
formation
the
doubled.
The
Solution
●
for
booklet
The
ethane:
is
∆H
equation
for
the
ethane
reversed.
f
2C(s)
+
3H
(g)
C
2
H
2
(g)
6
2C(s)
+
3H
(g)
→
C
2
●
Write
equations
hydrogen,
and
for
the
ethane
combustion
and
of
determine
H
2
(g)
6
carbon,
∆H
their
=
–84
kJ
c
enthalpy
1....C(s)
values
+
O
from
(g)
→
the
CO
2
∆H
=
Data
booklet:
(g)
Figure
5
can
found
be
shows
how
using
the
an
enthalpy
enthalpy
of
formation
cycle
diagram.
2
-393.5
kJ
c
∆H
1
_
2....H
(g)
+
O
2
(g)
→
H
2
2C(s)
O(l)
+
3H
2
f
2
2
H
6
(g)
2
∆H
=
-286
kJ
c
7
_
3....C
H
2
(g)
+
O
6
(g)→
2CO
2
(g)
+
3H
2
O(l)
2
2
3
7
O
∆H
=
-1561
kJ
c
●
Multiply
2O
or
●
Now
for
reverse
each
the
sign
combine
enthalpy
of
1....(×2):
equation
of
the
the
2C(s)
to
for
=
-787
→
the
156
1
net
(g)
2CO
2
(g) +
3H
2
O(l)
kJ
Figure 5 Alternative method: enthalpy cycle to nd the
c
__
2....(×3):
3H
(g)
=
-858
enthalpy of formation of ethane
+
2
∆H
+
ethane:
2
∆H
3(-286)
2
enthalpy
form
equation
+
2
accordingly.
equations
formation
2
2
2(-394)
change
O
2
→
2
2
2
kJ
c
3....(reversed):
2CO
(g)
2
+
→
2
2
_
(g)
+
2
∆H
=
+1561
kJ
c
std tp
You may nd the summation of equations method easier when working out the
direction of the equations and the mole coecients. During examinations you
may be asked to use the summation of equations method and/or construct an
enthalpy cycle.
Often candidates make simple arithmetical errors when calculating the enthalpy
of reaction. It is advisable to clearly show your full working rather than simply
recording the nal answer. This gives the examiner the oppor tunity to assign par t
marks where applicable.
151
5
e n e r g e T i c s
a n D
T h e r M O c h e M i s T r y
5.3 Bod tp
Understandings
Applications and skills
➔
Bond forming releases energy and bond
➔
Calculation of the enthalpy changes from
breaking requires energy.
known bond enthalpy values and comparison
➔
Average bond enthalpy is the energy needed
of these with experimentally measured values.
to break 1 mol of a bond in a gaseous molecule
➔
Sketching and evaluation of potential energy
averaged over similar compounds.
proles in determining whether reactants or
products are more stable and if the reaction is
exothermic or endothermic.
➔
Discussion of the bond strength in ozone
relative to oxygen in its impor tance to the
atmosphere.
Nature of science
➔
Models and theories – measured energy changes can be explained based on the model of bonds broken
and bonds formed. Since these explanations are based on a model, agreement with empirical data depends
on the sophistication of the model and data obtained can be used to modify theories where appropriate.
Modelling energy changes
Scienti c
certain
directly.
of
the
models
processes
Based
on
processes,
evidence
inform
in
are
d e v el ope d
tha t
a
ca nnot
the or etica l
s uch
suppo r t
of
modi cati o ns
mod el s
the
to
to
be
e x pla i n
models
und e r st a n di n g
can
degree
p r odu c e
the or i es,
the
changes
ob se r ve d
or
the or i es .
the
can
in
of
of
empirical
accuracy
ca n
br e a k i ng
agreeme nt
dependent
E n er g y
reactio ns
bond
data
on
of
the
the
be
unde rs too d
a nd
b ond
b e twe e n
o b ta i ne d
v a l i di ty
the s e
in
of
the
the
u s in g
m a ki ng.
m ode l s
The
an d
l a bor a t or y
mode l
and
is
t he
da ta .
Bond enthalpy
The
breaking
requires
example)
gaseous
Bond
of
energy.
is
covalent
for
hydrogen
The
dened
breaking
value,
the
is
bond
as
the
energy
molecules
an
molecule
enthalpy
into
a
to
standard
process
individual
bond
required
under
endothermic
of
(the
break
hydrogen
H
1
H
mol
bond
of
atoms
in
this
bonds
and
has
a
positive
enthalpy
example:
1
H
(g)
→
2H(g)
∆H
=
+436
kJ
mol
2
Bond
152
enthalpy
and
selected
and
in
is
also
values
table1.
referred
are
These
to
provided
are
as
in
average
bond
dissociation
section
values
in
conditions.
11
and
of
the
are
enthalpy ,
Data
booklet
therefore
only
an
5 . 3
approximation.
They
are
derived
from
experimental
data
B O n D
e n T h a l P y
involving
Bod
the
breaking
of
For
example,
the
same
bond
found
in
a
wide
variety
of
av bod
compounds.
1
tp/kJ mo
series
If
a
as
the
the
would
upon
be
of
signicant
for
each
of
enthalpy
of
time,
at
as
successive
the
in
the
that
the
vary
the
a
the
of
These
bond
in
is
take
are
an
the
account
not
H
O
O
436
one
144
bond
of
O
H
C
H
C
C
463
the
considered
value
498
O=O
changes
accepted
H
enthalpy
Additionally,
not
limitations
enthalpy
changes.
which
environment
does
alkane
dissociation
atoms.
state
the
bonds
steps
bond
chemical
hydrogen
exist.
through
individual
series
time,
gaseous
average
reaction
a
will
of
underwent
removed
forces
and
enthalpy
environment
molecules
intermolecular
be
was
removal
for
bond
methane
atom
different
the
enthalpy
H
chemical
molecule
hydrogen
C
to
to
414
use
346
calculations.
614
C=C
Bond length
Consider
the
molecules
hydrogen
uoride
H
F
,
hydrogen
chloride
C
H
Cl,
hydrogen
discussed
with
how
839
C≡C
bromide
as
increasing
you
H
Br,
move
atomic
and
down
number,
hydrogen
group
Z.
The
17
iodide
the
H
I.
atomic
consequence
of
In
topic
radius
this
is
3
increases
that
358
O
we
804
C=O
bond
615
C=N
length
you
increases,
move
and
down
bond
group
17
strength
(table
decreases,
in
the
hydrogen
halides
as
N
Bod
H
F
H
890
C≡N
2).
Cl
H
Br
H
158
N
I
470
N=N
Bod t/pm
92
128
141
160
945
N≡N
T
able 2 Bond lengths of the hydrogen halides
Bond strength
The
bond
enthalpy
reects
the
strength
of
the
covalent
bond.
As
Cl
Cl
Br
Br
the
from
bond
to
increases
shortening
carbon
single
of
bond
the
of
double
resulting
bond
the
to
triple
in
length.
an
increase
This
homologous
bonds
trend
series
of
the
in
can
number
of
electrostatic
be
seen
alkanes,
in
electrons
forces
the
alkenes,
151
I
in
and
a
carbon–
and
193
we
I
move
242
T
able 1 Average bond enthalpies
at 298 K
alkynes.
Bond polarity
The
polarity
of
a
electronegativity
bond
of
can
the
be
described
bonded
atom
etotvt
H
2.2
F
4.0
Cl
3.2
Br
3.0
atoms
Bod
by
the
(tables
difference
3
and
in
4).
∆(etotvt)
Bod tp/
1
kJ mo
H
F
H
Cl
H
Br
1.8
567
1.0
431
0.8
366
δ
δ+
T
able 3 Electronegativity values
T
able 4 Bond polarity and bond enthalpy at 298 K
H
Fluorine
The
has
polarity
atom.
attract
The
one
the
of
highest
the
bond
is
H
F
said
another,
electronegativity
bond
to
results
have
increasing
in
ionic
the
a
value
partial
of
charge
character .
strength
of
any
The
the
F
element.
on
each
partial
bond
charges
(gure
1).
Figure 1 The polar H
F bond
153
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
Worked example: using bond enthalpies to
nd the enthalpy change of reaction
Using
the
change
form
data
for
from
the
section
11
electrophilic
of
the
addition
Data
of
booklet
hydrogen
nd
the
enthalpy
bromide
to
ethene
to
bromoethane.
C
H
2
(g)
+
HBr
(g)
→
C
4
H
2
∆H
=
∑(BE
∆H
=
[4BE
bonds
+
broken)
BE
C–H
∆H
=
[(4
×
∆H
=
2636
+
+
-
(g)
∑(BE
BE
C=C
414)
Br
5
]
bonds
[5BE
H–Br
614
+
formed)
+
BE
C–H
366]
[(5
×
+
BE
C–C
414)
+
346
]
C–Br
+
285]
1
The
bond
using
∆H
2701
enthalpy
=
-65
kJ
calculated
reaction
=
∑∆H
mol
will
vary
signicantly
(products)
-
∑∆H
f
(
105.7kJ)
when
do
because
liquids
not
take
are
into
bond
account
in
the
the
calculation
f
enthalpies
involved
from
(reactants)
the
are
average
reaction,
bond
intermolecular
values.
Additionally,
enthalpy
forces
within
calculations
the
liquids.
std tp
A frequent error made by candidates is to confuse the dierent equations for the
calculation of a change in enthalpy. For bond enthalpy, think in terms of bond
breaking and formation:
∆H
= ∑(BE bonds broken) -
∑(BE bonds formed)
For enthalpy of formation, think in terms of products and reactants:
∆H
reaction = ∑(∆H
products) -
∑(∆H
f
reactants)
f
Bond enthalpy values and enthalpies
of combustion
Gasoline
used
to
or
witnessed
developed
The
petrol
power
is
produced
various
signicant
of
the
fractional
of
transport.
changes
in
its
economies
enthalpy
by
modes
is
being
combustion
markets
overtaken
of
The
as
by
enthalpy
values
from
octane,
section
11
the
that
C
H
8
bond
distillation
of
of
automotive
demand
of
the
can
petroleum
industry
for
automobiles
developing
Data
be
calculated
using
booklet:
25
_
C
H
8
(g)
+
O
18
(g)
→
8CO
2
(g)
+
9H
2
O(g)
2
2
∆H
=
∑(BE
bonds
broken)
-
∑(BE
bonds
formed)
c
25
_
=
(18BE
+
7BE
C–H
+
BE
C–C
)
(16BE
O=O
+
C=O
18BE
)
O–H
2
25
_
=
(18(414)
+
7(346)
+
(498))
(16(804)
2
1
=
154
16
099
21
198
=
-5099
kJ
mol
+
in
economies.
18
the
and
has
18(463))
5 . 3
In
comparison,
change
of
the
experimentally
combustion
∆H
for
determined
octane
(section
value
13
of
for
the
the
B O n D
e n T h a l P y
enthalpy
Data
booklet)
is
Qk qto
c
1
5470
kJ
mol
1
The
reason
change
takes
for
using
place
However,
octane
difference
bond
in
the
the
and
the
enthalpy
gaseous
is
in
their
when
values,
state,
experimentally
water
that
it
with
assumed
no
derived
standard
is
calculating
the
that
intermolecular
enthalpy
states,
of
forces
liquid.
with hydrogen gas
reaction
combustion
namely
chlorine gas is combined
enthalpy
the
In the chlor-alkali industry
to produce hydrogen
involved.
chloride:
involves
Additionally,
Cl
2
as
mentioned
above,
all
bond
dissociation
enthalpy
values
are
(g) + H
(g) → 2HCl(g)
2
averaged
If the enthalpy change for
across
a
wide
range
of
related
compounds
so
they
represent
only
an
his reaction is
approximation
of
the
true
185 kJ,
value.
calculate the bond enthalpy
for the H–Cl bond.
Ozone
2
Ozone,
O
is
both
created
and
d e s tr oy ed
in
the
st r a t os ph e ri c
l a ye r
Earth’s
Methane and chlorine react
of
to produce chloromethane
3
a tmosphere .
Ul tr av iol e t
(UV )
r ays
fro m
the
s un
ar e
a bso rbe d
and hydrogen chloride.
by
oxygen,
O
,
splitting
the
mol e cul e
into
s ingl e
ox yge n
a t om s .
2
)
These
oxygen
ato ms
ca n
the n
comb in e
w it h
ox yg e n
m ol ec u l es
Write the balanced
to
chemical equation for
form
ozone:
this reaction.
UV
b)
O
(g)
→
O
(g)
+
O
Using bond enthalpy
(g)
2
values from the Data
O
(g)
+
O
(g)
→
O
2
Ozone
UV
(g)
booklet, determine
3
is
very
effective
radiation.
molecular
This
oxygen
at
absorbing
absorption
and
a
harmful
breaks
single
down
oxygen
long-
the
atom.
and
ozone
molecule
Without
the enthalpy change
short-wavelength
the
to
for this reaction.
reform
presence
)
of
ozone
in
the
stratosphere,
life
on
Earth
would
change
forever,
Deduce whether this
as
reaction is exothermic
harmful
UV
radiation
would
damage
cells
in
both
plants
and
animals.
or endothermic.
1
The
bond
dissociation
enthalpy
of
an
oxygen
molecule
is
498
kJ
mol
.
In
d)
comparison,
the
energy
required
to
break
an
oxygen–oxygen
bond
Which are more
within
energetically stable,
1
an
ozone
molecule
is
364
kJ
mol
.
The
consequence
of
this
is
that
an
the reactants or the
ozone
molecule
is
decomposed
by
UV
rays
more
readily
than
an
oxygen
products?
molecule.
reaction
The
with
potential
atom
molecule.
exist
at
a
the
energy
Examination
oxygen
ozone
of
required
prole
this
have
The
photolysis
a
for
energy
this
energy
of
is
from
reaction
that
energy
are
above
the
shown
reveals
combined
this
described
coming
reaction
prole
greater
products
higher
reaction
UV
the
to
an
the
The
2.
oxygen
be
endothermic
radiation.
ingure
than
said
is
molecule
reactant
less
stable,
and
ozone
as
they
energy.
ygrene laitnetop
O
+ O
2
products

a
∆H
O
3
reactants
reaction pathway
Figure 2 Endothermic energy prole for ozone photolysis
155
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
Dpto of t ozo 
Since the early 1980s scientists have been monitoring ozone depletion globally,
par ticularly the giant holes in the ozone layer which have appeared above the
Arctic and Antarctic polar icecaps (gure 3).
Figure 3 ERS-2 satellite map of Antarctic ozone hole in 2010
Chlorouorocarbons (CFCs) are a type of hydrocarbon containing carbon,
hydrogen, and halogen atoms. The compounds themselves are considered to be
non-toxic and have a low level of both ammability and reactivity. First used in
industry in the late 1920s, a CFC called Freon gas became an industry standard
refrigerant in domestic refrigerators throughout the world. Over the following
decades the use of CFCs in commercial air conditioning and the automobile
industry, as an aerosol propellant, and as a solvent saw the demand for this class
of compounds increase rapidly.
Scientific
close
these
of
to
research
the
Ear th’s
compounds
chlorine.
The
destruction
directly
of
the
ClO
in
in
w hen
this
tha t
exposed
chemical
releas ed
the
then
lay er
of
reaction,
de s tru ction
(g) → ClO
3
discovered
un d er w ent
ozone
for
Cl(g) + O
1973
su rf a ce,
chlorine
consumed
responsible
in
of
(g) + O
a
CFCs
rema i ned
ra d iation
in
resulting
ca talytic
atm osp her e.
s mall
large
UV
reaction s
had
the
while
to
in
effect
As
the
the
am ou nts
of
CFCs
qu an tities
of
ozon e.
the
on
h armless
s tratosp her e
rel ea se
the
chlor ine
wer e
is
fou nd
not
to
be
(g)
2
(g) + O
3
(g) → Cl(g) + 2O
(g)
2
Science has made many advances that have improved people’s daily lives and
extended life expectancy and quality. As a consequence the world’s population
is increasing. The use of CFCs has had a massive economic (positive) and
environmental (negative) impact on the world. Whether the blame for the
environmental consequences lies with multinational companies who utilize
technology or the scientists who invent it is the focus of many discussions.
156
Q u e s T i O n s
Questions
1
Consider
the
following
specic
metals
heat
(table
capacity
of
the
5
4).
Use
data
each
1
385
Ag
234
Au
130
Pt
134
these
section
12
enthalpy
of
the
change
chemical
Data
of
booklet
reaction
to
for
reactions:
1
a)
Cu
of
the
K
sp t pt/J k
Mt
from
calculate
H
C
2
b)
(g)
+
H
4
CH
(g)
+
H
4
6
a)
→
C
H
2
O(g)
→
(g)
6
CO(g)
+
3H
2
Dene
T
able 4
(g)
2
the
formation,
(g)
2
term
ΔH
standard
enthalpy
change
of
.
[2]
f
Which
metal
will
temperature
to
a
0.001
initial
A.
show
increase
kg
sample
the
if
50
of
greatest
J
of
each
b)
heat
metal
is
at
(i)
Use
supplied
the
same
information
the
complete
temperature?
C
C
Au
H
4
Ag
D.
Pt
(g)
to
+
May
the
6O
8
table
of
→
to
for
the
but-1-ene
following
(g)
5
change
4CO
2
equation.
(g)
+
4H
2
O(g)
2
[1]
compod
IB,
in
enthalpy
combustion
according
Cu
B.
the
calculate
C
H
4
2007
(g)
CO
8
(g)
H
2
O(g)
2
1
∆H
+1
/kJ mo
394
42
f
[3]
2
When
solid
40
H
O
joules
at
of
16.0
heat
°C
are
the
added
to
a
temperature
sample
of
increases
T
able 5
to
2
(ii)
8.0
°C.
What
is
the
mass
of
the
solid
H
O
Deduce,
giving
a
reason,
whether
sample?
2
1
Specic
heat
capacity
of
H
O(s)
=
2.0
J
g
thereactants
1
or
the
products
are
K
2
morestable.
A.
2.5
g
C.
10
B.
5.0
g
D.
160
(iii)
IB,
Nov
g
Predict,
[1]
giving
enthalpy
based
The
temperature
of
a
2.0
g
sample
of
of
heat
from
25
energy
°C
to
were
30
°C.
added?
How
many
(Specic
1
capacity
of
aluminium
=
0.90
J
g
0.36
C.
9.0
B.
2.3
D.
11
of
with
on
for
the
how
but-2-ene
that
average
of
the
complete
would
but-1-ene
bond
enthalpies.
[1]
May
2007
joules
heat
1
K
.)
7
The
of
A.
reason,
aluminium
IB,
increases
a
change
combustion
2007
compare
3
[2]
g
∆H
values
nitrogen
are
for
the
given
formation
of
two
oxides
below.
1
_
1
N
[1]
(g)
+
O
2
(g)
→
NO
2
(g)
∆H
=
–57
∆H
=
+9
kJ
mol
2
2
IB,
May
1
2003
N
(g)
+
2O
2
Use
these
following
4
What
is
the
energy
temperature
of
20
change
g
of
(in
water
kJ)
when
increases
(g)
→
B.
20
×
10
×
20
×
283
values
to
reaction
(g)
calculate
(in
kJ
mol
4
∆H
for
the
kJ):
the
by
10
°C?
4.18
×
O
2
2NO
(g)
→
N
2
A.
N
2
a)
4.18
b)
O
2
(g)
4
105
–48
c)
+66
d)
+123
20 × 10 × 4.18
__
IB,
C.
November
2007
1000
20 × 283 × 4.18
__
D.
[1]
1000
IB,
November
2003
157
5
E N E R G E T I C S
8
The
standard
combustion
A N D
T H E R M O C H E M I S T R Y
enthalpy
reactions
change
is
given
of
Reaction
three
below
in
II
kJ.
1
_
SO
(g)
+
O
2
(g)
⇋
SO
2
(g)
∆H
=
–92
kJ
3
2
2C
H
2
(g)
+
7O
6
(g)
→
4CO
2
(g)
+
6H
2
O(l)
2
∆H
=
a)
–3120
State
with
2H
(g)
+
O
2
C
(g)
→
2H
2
H
2
(g)
+
O(l)
∆H
=
4
(g)
→
name
reason,
accompanied
2CO
2
(g)
+
2H
2
of
the
term
whether
∆H
.
reaction
State,
I
would
be
–572
2
3O
the
a
by
a
decrease
or
increase
in
temperature.
O(l)
[3]
2
∆H
=
–1411
b)
At
room
temperature
sulfur
trioxide,
SO
,
is
3
Based
the
on
the
standard
following
above
information,
change
in
a
calculate
enthalpy,
∆H
,
for
solid.
∆H
the
Deduce,
value
negative
reaction.
if
with
a
reason,
would
be
SO
instead
(s)
more
of
SO
3
formed
C
H
2
(g)
→
C
6
H
2
(g)
+
H
4
(g)
November
(g)
or
the
less
were
3
reaction
II.
[2]
[4]
2
c)
IB,
in
whether
negative
Deduce
the
∆H
value
of
this
reaction:
[1]
2009
1
_
S(s)
+
1
O
(g)
→
SO
2
(g)
3
2
IB,
9
Approximate
values
of
the
average
November
2005
bond
1
enthalpies,
shown
in
in
kJ
table
mol
,
of
three
substances
are
6.
12
H
carbon
430
H
But-1-ene
the
F
following
and
in
oxygen
water
to
vapour
produce
according
to
equation.
C
H
4
+
6O
8
→
4CO
2
+
4H
2
O
2
565
F
a)
T
able 6
What
burns
155
F
H
gas
dioxide
Use
of
is
the
enthalpy
change,
in
kJ,
for
this
the
∆H
data
for
in
the
Bod
C
table
7
to
calculate
combustion
C
C=C
C
of
the
value
but-1-ene.
H
[3]
O=O
C=O
496
743
O
H
reaction?
av
2HF
→
H
+
2
348
612
412
463
bod
F
2
tp/
A.
+545
IB,
May
B.
+20
C.
20
D.
545
1
kJ mo
2006
T
able 7
b)
10
The
is
reaction
between
ethene
and
hydrogen
gas
State
and
above
is
explain
whether
endothermic
or
the
reaction
exothermic.
[1]
exothermic.
IB,
a)
Write
b)
Deduce
an
equation
the
for
relative
this
reaction.
stabilities
of
Explain,
by
the
reactants
and
2006
[1]
and
13
energies
May
products.
Given
the
following
data:
[2]
1
c)
the
referring
molecules,
why
to
the
the
bonds
reaction
C(s)
in
+
2F
(g)
→
CF
2
(g);
∆H
4
=
–680
kJ
mol
=
+158
kJ
mol
=
+715
kJ
mol
1
is
1
F
(g)
→
2F(g);
∆H
2
exothermic.
2
[2]
1
C(s)
→
C(g);
∆H
3
IB,
November
2007
1
calculate
for
11
Two
reactions
occurring
in
the
sulfuric
Reaction
I
S(s)
(g)
+
O
2
158
acid
⇋
are
SO
shown
(g)
2
below:
∆H
C
F
average
bond
bond.
manufacture
IB,
of
the
the
=
–297
kJ
November
2003
enthalpy
(in
kJ
mol
)
Q u e s T i O n s
14
Methanol
is
made
in
large
quantities
as
c)
it
The
Data
booklet
value
for
the
enthalpy
of
1
is
used
in
in
the
production
of
polymers
combustion
and
methanol
is
726
kJ
mol
fuels.
Suggest
values
The
be
of
enthalpy
of
determined
combustion
theoretically
of
or
methanol
why
this
value
calculated
in
differs
parts
a)
from
and
the
b)
can
experimentally.
(i)
Part
a)
[1]
(ii)
Part
b)
[1]
1
_
CH
OH(l)
+
1
O
3
(g)
→
CO
2
(g)
+
2H
2
O(g)
2
2
a)
Using
of
the
the
IB,
information
Data
booklet,
from
section
determine
enthalpy
of
11
One
important
property
of
a
rocket
fuel
methanol.
the
large
volume
of
gaseous
products
provide
thrust.
Hydrazine,
N
H
2
The
e nthalpy
of
co mb us ti on
of
also
be
determi ne d
e xp er ime nta ll y
used
school
laborator y.
A
b ur ne r
water
in
 gure
4.
a
was
we i g he d
test
tub e
as
a nd
as
a
rocket
fuel.
The
4
combustion
is
represented
by
the
equation
of
below.
containi n g
N
methanol
is
in
hydrazine
a
,
m e t h a no l
often
can
formed
[3]
which
b)
mixture
combustion
is
of
2011
the
15
theoretical
May
us e d
to
i ll us tr a te d
in
he a t
H
2
(g)
+
O
4
(g)
→
N
2
(g)
+
2H
2
O(g)
2
1
∆H
=
–585
kJ
mol
c
a)
Hydrazine
nitrogen
reacts
and
with
uorine
hydrogen
to
uoride,
produce
all
in
the
thermometer
gaseous
state.
State
an
equation
for
the
reaction.
[2]
b)
Draw
the
Lewis
and
nitrogen.
c)
Use
the
structures
for
hydrazine
[2]
test tube with water
stand
section
the
shield
11
of
a)
Based
bond
the
enthalpy
part
d)
average
Data
change
enthalpies
booklet
for
the
to
given
in
determine
reaction
in
above.
on
[3]
your
answers
to
parts
a)
and
c),
burner
suggest
and
whether
uorine
mixture
of
is
a
a
mixture
better
hydrazine
of
hydrazine
rocket
and
fuel
than
a
oxygen.
[2]
Figure 4
IB,
The
data
shown
in
table
8
were
May
2010
collected.
Initial mass of burner and methanol/g
80.557
16
Two
students
from
Final mass of burner and methanol/g
the
were
Data
asked
booklet
to
to
use
information
calculate
a
value
for
the
80.034
enthalpy
Mass of water in test tube/g
20.000
C
H
2
26.4
Final temperature of water/°C
hydrogenation
of
ethene
to
form
ethane.
21.5
Initial temperature of water/°C
of
John
(g)
H
(g)
→
C
2
used
section
+
4
11.
the
average
Marit
H
2
used
(g)
6
bond
the
enthalpies
values
of
from
enthalpies
T
able 8
of
i)
Calculate
the
methanol
ii)
Calculate
by
iii)
the
amount,
in
mol,
of
burned.
the
heat
absorbed,
in
kJ
of
1
mol
the
mole
,
for
enthalpy
of
the
section
b)
change,
Marit
section
value
of
bond
for
12.
the
ethene
enthalpy
obtained
enthalpies
given
of
using
in
11.
[2]
arranged
section
combustion
methanol.
the
theaverage
[3]
1
in
Calculate
from
hydrogenation
kJ,
water.
Determine
a)
[2]
combustion
12
into
the
an
values
energy
she
found
cycle
in
(gure
5).
[2]
159
5
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
∆H
C
H
2
+
(g)
H
4
(hydrogenation)
(g)
C
2
1
1
lom Jk 682
O
1
2
O3
2
lom Jk 1141
2
-
l
o
m
J
k
0
6
5
1
-
2CO
+
(g)
3H
2
O(l)
2
Figure 5
Calculate
of
the
value
hydrogenation
for
of
the
enthalpy
ethene
from
the
energycycle.
c)
Suggest
one
isslightly
[1]
reason
less
why
accurate
John’s
than
answer
Marit’s
answer.
d)
John
[1]
then
enthalpy
to
decided
of
produce
C
H
6
(i)
+
H
(g)
the
average
a
value
of
of
John’s
was
hypothesis
ofaverage
less
accurate
bond
for
forethene.
is
moreaccurate
160
May
2009
was
[1]
between
the
for
for
that
Determine
use
to
of
enthalpies
greater
it
information
enthalpy
(average
and
is
equationshown
IB,
enthalpies
the
difference
enthalpies
same.
(l)
12
cyclohexenethan
was
the
cyclohexene
cyclohexene.
methods
combustion)
bethe
of
H
bond
for
percentage
thesetwo
bond
C
6
hydrogenation
The
→
2
deduce
(ii)
determine
cyclohexane.
(l)
Use
to
hydrogenation
10
H
2
ethene.
it
why
would
the
enthalpies
is
cyclohexene
above,
Deduce
needed
answer.
than
what
to
it
extra
provide
a
[2]
(g)
6
6
C H E M I CA L
K I N ET I C S
Introduction
In
this
topic
kinetics.
the
of
will
proper
razor
also
in
learn
discuss
the
form
a
the
key
science
the
topic
part
a
the
the
of
as
a
exploring
gases
and
guide
the
the
of
to
aspect
the
of
reaction.
of
this
a
In
and
namely
reaction.
outline
molecules
addition
of
collide
we
numerical
branch
reactions,
chemical
the
We
chemical
will
then
importance
of
describe
the
principle
theories.
that
and
chemical
of
theory
developing
graphical
treatment
of
rate
collision
probability
rate
analysis
our
important
by
greater
higher
of
very
chapter
theory
that
orientation
Throughout
will
will
begin
kinetic-molecular
Occam’s
We
we
We
data
physical
with
consider
sufcient
the
obtained
effect
from
of
energy
a
and
catalyst
rate
on
rate.
experiments
chemistry.
6.1 C oo o   o co
Understandings
Applications and skills
➔
Species react as a result of collisions of
➔
Description
of
the
kin etic
theory
in
terms
of
sufficient energy and proper orientation.
the
➔
The rate of reaction is expressed as the change
movement
kinetic
in concentration of a par ticular reactant/
in
energy
of
is
par ticles
w hose
prop or tional
to
average
tempera ture
kelvin.
product per unit time.
➔
➔
Concentration changes in a reaction can be
Analysis
rate
of
graphical
a nd
nu merical
d ata
from
experiments.
followed indirectly by monitoring changes in
➔
Explanation
of
the
eff ects
of
temperatu re,
mass, volume, and colour.
pressure/concentration,
➔
Activation energy (E
) is the minimum energy
rate
a
of
a nd
par ticle
s ize
on
reaction.
that colliding molecules need in order to have
➔
Construction
of
Maxw ell -Boltzmann
successful collisions leading to a reaction.
energy
➔
By decreasing E
, a catalyst increases the rate
the
a
of a chemical reaction, without itself being
probability
factors
permanently chemically changed.
a
➔
distribution
of
affecting
to
succes s f u l
thes e,
accou nt
f or
collisions
inclu ding
and
the
effec t
of
the
resu lts.
energy
profiles
catalyst .
Investigation
of
experimentally
➔
cu r ves
Sketching
with
and
and
rates
and
of
reaction
eva lu a tion
explan a tion
without
of
of
cata ly s ts .
Nature of science
➔
The principle of Occam’s razor is used as a guide to developing a theory - although we cannot directly
see reactions taking place at the molecular level, we can theorize based on the current atomic models.
Collision theory is a good example of this principle.
161
6
C H E M I C A L
K I N E T I C S
Studying reaction rates
When
to
a
chemical
reaction
takes
1
What
occurs
2
What
is
3
Does
4
What
the
the
This
the
of
which
is
given
four
information
involved
sufcient
can
be
questions
are
of
interest
discussed
in
terms
3
This
is
the
of
the
study
This
is
the
energy
or
nothing
One
of
the
whether
much
be
eld
heat
about
main
it
point
in
a
period
of
of
the
in
how
a
take
from
out
a
airbags
in
or
a
reaction
chemical
(and
(topic
is
5),
reaction.
their
states),
a
given
if
reactions
while
for
it
it
to
takes
occur
others
sub-topic
a
the
branch
focus
which
is
of
7.1).
of
this
the
reaction
a
useful.
150
very
years
as
of
tells
us
place.
reaction
There
for
quickly,
such
takes
chemical
be
topic.
study
Thermodynamics
examining
enough
reaction
reactions,
which
slowly
when
cars,
balanced
products
equilibrium
chemical
chemical
chemical
could
1.1).
of
fast
and
following:
the
reactants,
kinetics,
quickly
place
the
chemical
rate
reaction
time?
thermodynamics
ow
carrying
of
of
chemical
Many
ination
the
considerations
will
formed!
rapid
over
called
in
slowly?
involves
(sub-topic
is
reaction?
or
obtained
identies
This
4
rapidly
questions
stoichiometry
chemistry
chemical
happen
2
to
four
reaction?
the
transfer
occur
these
equation
and
extent
energy
Answering
1
during
reaction
potentially
is
place,
chemists:
the
such
rusting
is
not
product
as
take
the
place
years.
Rate of reaction
The
rate
of
reactants
reaction
or
is
products
3
Units:
mol
dm
dened
per
unit
as
1
s
,
the
change
3
mol
in
concentration
of
time.
dm
1
min
,
etc.
Experimental measurements of reaction rates
Δc,
the
change
property
that
Examples
concentration,
change
when
1
change
in
pH
2
change
in
conductivity
3
change
in
mass
4
change
in
colour
In
order
is
to
of
(for
or
acid
base
(for
volume
(for
determine
(or
plotted.
gradient
be
measured
reactants
are
by
monitoring
converted
into
a
products.
reactions)
reactions
(for
reactions
involving
reactions
electrolytes)
involving
involving
solids
transition
or
metals
gases)
or
other
compounds).
concentration
time
can
the
include:
coloured
162
in
will
the
the
The
the
rate
of
property
rate
of
tangential
reaction,
associated
reaction
line
at
is
time
then
t.
at
time
with
t,
a
graph
of
concentration)
determined
from
versus
the
slope
or
6 . 1
Let
us
consider
(limestone),
the
following
CaCO
and
reaction
hydrochloric
C O l l i s i O n
between
acid,
t h e O r y
calcium
a n d
r a t e s
O f
r e a C t i O n
carbonate
HCl:
3
CaCO
(s)
+
2HCl(aq)
→
CaCl
3
In
this
reaction,
volume,
rate
of
set-up
V,of
gaseous
carbon
reaction
for
(aq)
+
CO
2
this
can
carbon
dioxide
be
dioxide
gas
(g)
+
H
2
is
one
produced
determined.
Figure
is
1
O(l)
2
of
the
products.
recorded
shows
over
the
If
time
the
t,
the
experimental
reaction:
clamp
delivery tube
measuring cylinder
carbon dioxide
ask
trough
hydrochloric acid
water
calcium carbonate
Figure 1 Experimental set-up for measuring the rate at which carbon dioxide is produced in
the reaction between calcium carbonate and hydrochloric acid
Table
1
shows
the
t/
data
recorded
during
this
experiment.
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
0.0
19.0
30.0
37.5
45.0
50.0
52.0
53.0
53.0
53.0
53.0
3
V(CO
)/cm
2
T
able 1 Volume, V, of carbon dioxide evolved at time t in the reaction between calcium
carbonate and hydrochloric acid
60
Figure
2
shows
a
plot
of
V
against
t
50
The
rate
of
reaction
can
be
expressed
in
three
ways:
average
mc/)
3
●
rate
instantaneous
●
initial
OC(V
2
●
rate
rate.
40
30
20
10
0
0
Average rate
20
40
60
80
100
t/s
The
average
reactant
or
rate
is
product
a
measure
in
a
given
of
the
time
change
interval,
in
concentration
of
Figure 2 Plot of volume of carbon dioxide gas
t
evolved V(CO
) against time t in the reaction
2
between calcium carbonate and hydrochloric
Mathematically,
average
rate
can
be
expressed
as:
acid
Δc
_
average
rate
=
Δt
where:
Δc
=
change
Δt
=
time
For
gases,
in
concentration
interval
the
over
average
which
reaction
of
reactant
the
change
rate
can
be
or
in
product
concentration
also
expressed
as
was
the
measured
change
in
volume:
ΔV
_
average
rate
=
Δt
where
during
ΔV
the
is
the
change
in
volume
of
the
gas
produced
or
consumed
reaction.
163
6
C H E M I C A L
K I N E T I C S
Hence
in
this
example,
as
the
reaction
53.0
_
average
rate
is
complete
1
=
=
7.57×
10
3
cm
at
70.0
s,
then:
1
s
70.0
Instantaneous rate
The
instantaneous
Δc
_
lim
rate
is
given
by:
dc
_
=
Δt
→
0
dt
Δt
Initial rate
When
t
=
t
,
the
instantaneous
rate
=
the
initial
rate.
0
In
order
drawn
gives
to
to
the
deduce
the
the
curve
initial
at
rate
initial
0
s,
rate
and
(gure
of
the
reaction
slope
or
at
t
=
gradient
0
s,
of
a
tangent
the
is
tangential
line
3).
60
(x
, y
2
) = (20, 50)
2
50
(x
, y
3
) = (50, 50)
3
40
, y
4
2
mc/)
3
(x
) = (20, 39)
4
30
OC(V
20
10
(x
, y
1
) = (0, 0)
1
0
0
20
40
60
100
80
120
t/s
Figure 3 Plot of volume of carbon dioxide evolved versus time showing tangents used to
measure both initial rate at t = 0 s and instantaneous rate at t = 50 s
(x
,
y
1
(x
)
=
(0.0,
0.0)
=
(20.0,
1
,
y
2
)
50.0)
2
Δy
50.0
0.0
_
_
Initial
rate
=
=
So
at
the
start
of
the
3
=
20.0
Δx
2.50
cm
1
s
0.0
reaction,
carbon
dioxide
gas
is
being
generated
at
a
3
rate
In
of
2.50
order
tangent
to
is
tangential
(x
,
y
3
(x
,
4
)
cm
per
deduce
drawn
line
second.
the
to
gives
instantaneous
the
the
curve
at
instantaneous
=
(50.0,50.0)
=
(20.0,39.0)
t,
of
and
rate
reaction,
the
at
slope
this
at
t
or
time
=
50.0
s,
gradient
(gure
y
)
3).
4
Δy
50.0
39.0
__
_
rate
=
=
Δx
1
=
50.0
20.0
3.67
×
10
3
cm
1
s
a
of
3
instantaneous
164
rate
point
the
6 . 1
C O l l i s i O n
t h e O r y
a n d
r a t e s
O f
r e a C t i O n
Rate equation
A
rate
equation
rate
expressed
A
B
+
→
C
+
in
D,
is
a
mathematical
terms
a
rate
d[A]
=
equation
=
In
the
general
=
xA
where
+
yB
x,
y,
→
q,
qC
and
=
+
as
which
the
shows
reaction
follows:
_
=
+
dt
dt
p
pD
are
the
stoichiometric
d[B]
1
_
_
-
=
x
written
in
where:
d[A]
1
_
rate
expression,
example,
d[D]
+
dt
case
be
_
-
dt
can
For
d[C]
_
-
differential
concentration.
d[B]
_
rate
of
-
d[C]
1
_
_
·
=
y
dt
coefcients:
·
=
q
d[D]
1
_
_
+
dt
+
_
·
p
dt
dt
Kinetic–molecular theory of gases
The
the
ideal
gas
equation
(sub-topic
1.3)
is
given
in
by
At
pV
=
kelvin.
a
given
equation
does
not
order
gases,
to
a
of
describes
explain
why
understand
model
Clausius
how
gases
the
devised
the
kinetic
more
simply
gases
act
as
physical
was
called
gases,
in
behave,
they
mo v e me nt
due
to
instant
by
the
as
of
the
of
gases.
The
theory
can
be
the ir
in
of
a to ms
the r mal
ti me ,
than
energy
temperature,
theory
have
tr a v e l l ing
the
temperatur e
kinetic
Rudolf
molecular
will
velocities
In
properties
1857
termed
but
do.
so me
or
ene r g y.
the
wi l l
the
same
o the r
at
of
t he
pa rticl es .
i ncr ea s es
be
g re a te r
the
hig he r.
p a r ti cl es
ave r a g e
of
At
all
ki ne tic
H oweve r,
ave ra g e
a
gi ven
g ase s
wil l
e ne r gy.
kinetic
Hence,
theory
is
nRT
particles
The
The
molecules
expression:
be
the
kinetic
molecular
theory
can
offer
summarized
an
explanation
of
temperature
and
pressure
at
asfollows:
the
1
Gases
consist
which
are
of
a
large
moving
at
number
high
of
particles,
velocities
in
molecular
molecular
equation
level.
theory
can
be
Starting
postulates,
with
the
the
ideal
kinetic
gas
derived.
randomdirections.
2
The
In
size
this
a
gaseo us
ther e
even
though
their
volumes
can
the
be
or
are
the
is
co ns id er e d
at
to
ha ve
or
the
that,
m a s s,
ne gl igib l e.
no rma l
a to m s
compar e d
neg l ig ibl e.
a ss ump ti o n
pa r ti cl e s
since ,
between
large
p a rticle
is
gaseo us
justied
space
very
3
of
model
mole cul es
si ze
Th is
pr e s su r e,
of
th e
is
a t om
molecule.
Collisions
and
between
another
particles
are
collide,
energy
energy
one
gaseous
completely
there
is
is
no
simply
particle
elastic .
net
loss
When
in
transferred.
A
Figure 4 In the kinetic
useful
analogy
here
is
what
happens
molecular theory of gases, a sample
when
of gas is modelled as consisting of a collection of par ticles
two
snooker
balls
collide.
The
total
kinetic
moving at high velocities in random directions. The sizes of
energy
is
equivalent
before
and
after
the
the par ticles are negligible. The par ticles collide with each
collision
of
the
snooker
balls.
other and with the walls of the container containing the
gaseous sample. All the collisions are c. The pressure
4
The
average
kineti c
e ne rg y
of
the
pa r t ic l es
exer ted by the gas results from collisions of its par ticles
is
proportional
to
the
a b so lute
tem pe r at u r e
with the walls of the container
165
6
C H E M I C A L
K I N E T I C S
Collision theory
Occam’s
razor
is
a
principle
fourteenth-century
theologian,
though
was
and
many
known
states
logician
suggest
earlier
to
Franciscan
William
references
much
attributed
English
than
of
of
the
a
Ockham,
that
this.
the
The
quantum
razor
friar,
is
theory
can
a
taking
formulate
models.
that:
as
guide
although
reactions
principle
principle
mechanics.
used
we
place
principle
the
the
is
directly
molecular
based
theory
of
a
Occam’s
development
cannot
at
theories
Collision
The
in
on
level,
current
good
of
see
we
atomic
example
of
this
principle.
“Entities
should
not
be
multiplied
For
unnecessarily.”
a
chemical
reacting
Scientists
have
formulated
the
principle
like
reaction
particles,
a
to
occur
number
of
between
two
conditions
must
this:
befullled:
“When
two
competing
theories
exist
that
1
explain
observed
facts,
both
giving
The
two
that
the
same
two
be
predictions,
theories
used
is
until
the
then
the
optimum
more
simpler
one
evidence
particles
must
collide
with
each
other,
essentially
that
of
there
must
be
physical
contact.
the
2
should
transpires
is,
The
colliding
mutual
to
particles
must
have
the
correct
orientations.
proveotherwise.”
3
The
of
British
the
theoretical
University
of
physicist
Cambridge
Stephen
and
The
author
History
of
Time
of
could
laws
for
that
some
observe
without
of
us
the
that
determines
present
disturbing
universe
mortals.
the
imagine
supernatural
the
principle
all
still
It
are
it.
features
as
of
there
events
being,
state
of
of
to
Occam’s
the
who
the
helps
set
three
must
initiate
conditions
is
that
cannot
on
new
out
lead
the
and
on
form
have
reaction.
the
basis
why
a
rates
of
hydrogen
slowly
principle
is
often
encountered
widely
For
justication
of
example,
the
the
only
reacting
a
taking
particles
formation.
H
In
the
, and
The
at
idea
it
of
has
been
used
uncertainty
in
iodide,
room
as
of
place.
bounce
Let
us
In
off
resulting
gaseous
iodine,
I
,
For
fraction
of
most
one
in
no
reaction
to
form
2
the
reaction
temperature,
collisions
proceeds
since
only
a
very
results
in
a
small
reaction:
a
the
eld
H
(g)
+
I
2
do o cvo
HI,
model
theory.
small
unchanged,
that
in
number
chemistry.
of
model
chemical
molecular
reaction
hydrogen,
a
temperature.
kinetic
to
is
2
observed.”
The
sufcient
the
theory
reactions,
remain
product
between
be
the
chemical
another
cut
depend
collisions,
to
This
understand
collisions
the
and
based
most
models
interest
employ
us
theory.
reactions
could
such
razor
theory
a
universe
much
better
is
completely
However,
not
seems
known
to
said:
collision
“We
particles
energy
of
These
ABrief
reacting
kinetic
Hawking
consider
each
of
(g)
⇋
2HI(g)
2
these
three
conditions
in
the
box
above
separately:
g, E

1
The
rst,
that
physical
contact
is
necessary
for
a
reaction
to
occur,
is
The activation energy is the
relatively
obvious
to
appreciate.
minimum energy that colliding
2
Figure
5
illustrates
the
second
condition.
In
(a)
the
orientation
of
par ticles need for a reaction
the
colliding
particles
is
not
favourable,
so
no
reaction
will
occur.
In
to occur.
(b)
the
particles
effective
have
166
and
sufcient
have
a
a
suitable
result
in
kinetic
energy .
orientation,
chemical
reaction,
so
the
provided
collision
the
can
reacting
be
particles
6 . 1
C O l l i s i O n
t h e O r y
a n d
r a t e s
O f
r e a C t i O n
(a) ineective collision
No reaction occurs, since
+
orientation is not favourable.
(b) eective collision
+
+2
Orientation is favourable, so may result
in reaction if there is sucient kinetic
energy.
x
Figure 5 The possibility of a collision leading to a reaction depends on the
orientation of the par ticles

a
y
A
useful
analogy
somebody
trying
in
understanding
to
push
a
large
activation
rock
over
a
energy
hill
is
to
(gure
imagine
available
6).
energy release
Initially,
there
must
be
a
minimum
input
of
energy
in
order
for
the
person
z
to
shift
the
rock
over
the
hill.
Once
over
the
hill,
the
rock
can
fall
to
point
z
Figure 6 Analogy of activation energy, E
a
Catalysts
The
analogy
(gure
just
described
can
be
conveyed
in
a
potential
energy
prole
7).
(a)
(b)
transition state
transition state

ygrene laitnetop
ygrene laitnetop
a
reactants
∆H

a
products
∆H
reactants
products
reaction coordinate
reaction coordinate
Figure 7 Potential energy prole for (a) an exothermic reaction with ΔH<0; (b) an endothermic reaction with ΔH>0
The
arrangement
transition
state
of
atoms
or
activated
at
the
crest
of
the
energy
prole
is
termed
the
complex
do o  c
Catalysts
may
be
either
homogeneous
or
heterogeneous.
A catalyst is a substance
that increases the rate of a
Homogeneous catalysts
chemical reaction, but is not
A
homogeneous
catalyst
is
in
the
same
physical
phase
or
state
as
the
consumed in the reaction
reactants.
The
destruction
of
gaseous
ozone,
O
,
by
chlorine
atoms
is
an
3
example
of
homogeneous
catalysis,
since
itself. A catalyst provides
chlorine
atoms
(which
act
as
an alternative pathway for
the
catalyst)
have
the
same
state
as
the
gaseous
reactants.
the reaction and lowers the
In
the
stratosphere
(upper
atmosphere),
ozone
in
the
ozone
layer
activation energy, E
a
absorbs
over
protecting
us
95%
of
from
the
this
UV
radiation
harmful
reaching
Earth
from
the
sun,
radiation.
167
6
C H E M I C A L
K I N E T I C S
hν
O
(g)
→
O
3
O
(g)
+
O
→
O
(g)
+
O
(g)
2
Thus,
With
(g)
2
(g)
+
heat
3
there
the
is
a
net
energy
progressive
radiation
can
increased
reach
risk
of
conversion
depletion
the
skin
of
Earth’s
cancers
the
from
ozone
surface.
This
(melanomas)
UV
to
layer
can
and
heat
energy.
(gure
be
8),
more
associated
UV
with
an
cataracts.
Uu rouc:
Ozo ho Wc - images,
data and information for both
the southern and nor thern
hemispheres can be found on
this website hosted by NASA -
Goddard Space Flight Centre:
http://ozonewatch.gsfc.nasa.gov
Figure 8 The largest ozone hole to date seen here shown in purple was
recorded on September 24, 2006 for the Antarctic Hemisphere
Chlorine
(CFC)
atoms
with
UV
refrigerators,
are
produced
light.
and
CFCs
aerosols.
in
the
were
reaction
previously
Freon,
CF
Cl
2
presence
is
of
greater),
UV
and
light,
the
radicals
weaker
are
C
of
a
used
(g),
chlorouorocarbon
in
air
is
one
is
broken
conditioning
example
of
a
units,
CFC.
In
the
2
Cl
produced.
bond
The
chlorine
(the
radicals
C
F
bond
then
strength
attack
ozone.
hν
CF
Cl
2
rst
step:
Cl
(g)
→
CF
2
(g)
Cl
(g)
+
+
O
(g)
→
ClO
(g)
+
O
3
second
step:
ClO·(g)
reaction:
2O
+
O
(g)
→
Cl·(g)
acts
represent
in
gure
the
as
a
catalyst
potential
(g)
and
energy
(g)
+
2O
(g)
2
→
3O
3
Chlorine
(g)
2
3
overall
Cl
2
(g)
2
is
regenerated
prole
for
this
in
the
second
reaction
step.
scheme
We
as
can
shown
9.
transition states
ygrene laitnetop

a
Cl•(g) + 0
(g)
3
+ O•(g)

(uncatalysed)
a
(catalysed)
(catalysed)
∆H
Cl•(g) + 20
(g)
2
(catalysed)
reaction coordinate
Figure 9 Potential energy prole for the catalytic destruction of ozone, showing both the
catalysed and uncatalysed pathways
168
6 . 1
As
can
for
be
the
and,
in
prole
seen
from
catalysed
general,
the
more
the
catalysed
state
Although
ozone
harmful
for
to
may
the
in
of
formed
typical
the
are
in
for
uncatalysed
pathway.
acts
as
UV-a
a
of
as
gaseous
of
is
a
(the
shield
just
one
10.
from
high
atmosphere)
emphysema.
nitrogen
oxides,
r e a C t i O n
energy
gure
radiation,
lower
and
O f
states
shows
in
r a t e s
reaction
potential
shown
UV-b
a n d
transition
particular
protective
and
asthma
t h e O r y
pathways
This
troposphere
such
reaction
two
this
representation
and
the
actually
specic
stratosphere
problems
in
is
high-energy
ozone
respiratory
be
of
there
catalysed
the
effects
concentrations
lead
9,
This
showing
transition
the
gure
pathway.
C O l l i s i O n
can
Ozone
NO
and
NO
,
2
from
in
car
the
exhaust
gases
troposphere
with
also
acts
VOCs
as
a
(volatile
organic
greenhouse
compounds).
Ozone
gas
transition states at crests
ygrene laitnetop

(uncatalysed)
a
reactants

(catalysed)
a
∆H
products
reaction coordinate
Figure 10 Typical potential energy prole showing catalysed and uncatalysed pathways
Heterogeneous catalysts
acv
A
heterogeneous
catalyst
is
in
a
different
phase
or
state
from
the
Can you name ve countries
reactants.
Typically,
the
heterogeneous
catalyst
is
in
the
solid
phase
around the world that still
and
the
reactants
are
in
the
liquid
or
gaseous
states.
An
example
of
a
continue to use leaded petrol?
heterogeneous
system
CO,
of
a
catalyst
car.
nitrogen
In
a
is
car
the
catalytic
engine
monoxide,
NO,
converter
pollutants
and
such
unburned
used
as
in
the
carbon
exhaust
monoxide,
hydrocarbons,
C
H
x
produced.
of
The
carbon
hydrocarbon
atmospheric
car.
The
fuels.
nitrogen
catalytic
substances,
monoxide
Nitrogen
and
oxygen
converter
namely
comes
carbon
from
monoxide
at
converts
dioxide,
is
high
,
incomplete
temperature
these
CO
the
produced
in
substances
water,
H
2
from
O,
into
and
are
combustion
the
the
,
y
reaction
engine
less
of
of
the
harmful
nitrogen,
N
2
:
2
catalyst
2NO(g)
N
(g)
+
O
2
(g)
2
catalyst
2NO
(g)
N
2
(g)
+
2O
2
(g)
2
catalyst
2CO(g)
+
O
(g)
2CO
2
(g)
2
catalyst
CH
CH
3
CH
2
propane
Examples
+
5O
catalysts
and
vanadium(V)
Cr
O
2
if
a
car
poison
(g)
+
4H
O(g)
2
used
oxide,
V
O
.
Leaded
in
and
,
catalytic
converters
transition
copper(II)
metal
oxide,
are
oxides
CuO,
platinum,
such
and
as
chromium(III)
5
petrol
(leaded
gasoline)
is
not
used
in
modern
cars
–
3
tted
the
3CO
2
rhodium,
2
oxide,
(g)
2
fuel
of
palladium,
(g)
3
with
a
catalytic
converter
used
leaded
petrol,
the
lead
would
catalyst.
169
6
C H E M I C A L
K I N E T I C S
Maxwell-Boltzmann energy distribution and temperature
Rates
of
reaction
at
in
the
the
gas
phase
molecular
can
level
be
using
different
approaches:
1
the
collision
2
Maxwell
theory
Boltzmann
energy
distribution
curve
3
temperature
We
have
model.
effects
already
We
shall
on
kinetic
explored
now
the
look
at
energies.
collision
theory
approaches
2
and
3.
ygrene citenik htiw selcit rap fo noitcarf
interpreted
area A

a
for catalysed
reaction
area B
kinetic energy
Maxwell–Boltzmann energy distribution
Figure 11 Maxwell
curve
The
Boltzmann energy distribution cur ve
showing the activation energy for a catalysed reaction. Area
kinetic-molecular
theory
says
that
A shows the fraction of particles in the sample that do not
particles
have sucient energy to react. Area B shows the fraction
of
gas
move
randomly
in
different
directions
at
of particles in the sample that have the minimum energy
high
velocities.
and,
since
However,
these
velocities
differ
required to initiate a reaction with the use of a catalyst
with
other
changes
this
particles
the
of
scenario.
curve.
with
a
of
kinetic
with
of
is
This
is
value
a
needs
of
energy.
of
representation
the
curve
is
is
the
seen
particle
discuss
particle
is
in
the
particles .
by
the
distribution
(that
energy
be
to
described
fraction
energy
can
the
of
the
considered
energy
the
kinetic
As
be
of
gaseous
gaseous
to
colliding
sides
realistic
velocities
plot
the
single
velocities
kinetic
of
a
not
Boltzmann
given
that
It
constantly
individual
What
distribution
Maxwell
of
an
are
and
velocity
constantly.
velocity
distribution
The
particles
of
is
particles
the
probability
occurring)
from
asymmetric.
versus
gure
The
ygrene citenik htiw selcit rap fo noitcarf
container,
the
the

a
area A
the
total
number
area
of
reaction
E
a
for uncatalysed
reaction
area B
area C
kinetic energy
11,
under
Figure 1
2 Maxwell
represents
for catalysed
Boltzmann energy distribution cur ve
gaseous
showing the activation energy for an uncatalysed reaction.
particles
the
in
the
majority
kinetic
time,
energy
some
either
high
sample.
of
of
or
gaseous
near
the
the
At
certain
mean
gaseous
low
a
particles
value.
particles
velocities,
temperature,
will
but
have
At
will
the
a
a
not have sucient energy to react. Area B + area C shows
the fraction of par ticles that have the minimum energy
given
required in order to initiate a reaction using a catalyst. Area
have
majority
Area A shows the fraction of par ticles in the sample that do
will
C shows the fraction of par ticles in the sample that have the
minimum energy required to initiate a reaction without the
have
velocities
and
hence
kinetic
energies
close
to
use of a catalyst
the
170
mean
kinetic
energy.
6 . 1
C O l l i s i O n
t h e O r y
a n d
r a t e s
O f
r e a C t i O n
Temperature eects on kinetic energies
increasing
of
particles
to
overcome
increase.
the
that
As
greater
in
atter
and
Therefore,
frequency
be
T
more
now
activation
increases,
the
proportion
with
a
higher
increasing
collisions
successful
more
at
particles
curve
be
of
an
having
the
Maxwell
becomes
temperature.
increase.
collisions,
which
will
temperature,
will
will
velocity
particles
result,
energy
barrier
there
of
a
proportion
kinetic
mean
so
distribution
broader
of
As
the
energy
the
and
energy.
energy
T,
sufcient
increases
kinetic
Boltzmann
have
the
particles
increase
temperature
since
have
ygrene citenik htiw selcit rap fo noitcarf
With
T
1
T
2

a
the
There
there
sufcient
kinetic energy
will
are
kinetic
Figure 1
3 Maxwell
Boltzmann energy distribution cur ves
for two temperatures, T
> T
2
energy
to
overcome
the
activation
energy
barrier.
. Notice that, at the higher
1
temperature, the energy distribution is broader and the mean
kinetic energy is greater. The proportion of particles that have
This
results
in
an
increase
in
the
rate
of
reaction.
with
an
increase
in
temperature
sucient thermal energy to overcome the activation energy
Typically
of
barrier has increased. The area under both curves is the same as
10°C,
the
reaction
rate
will
double.
this signies the total number of gaseous particles in the sample
tOK
su p
In a potential energy prole,
Temperature can be considered as a measure of the average amount of kinetic
the y-axis is the potential
energy of par ticles, which is the energy due to motion. As the temperature
energy and the x-axis is the
increases, so does the kinetic energy of the par ticles. The lowest temperature
reaction coordinate, which
that can be attained theoretically is bou zo, which is
273.15 °C. This is
represents the progress of
taken as the zero point on the Kelvin scale. At 0 K , all thermal energy has been
the reaction. In a Maxwell
removed from a substance and the motion of par ticles has eectively ceased.
Boltzmann energy distribution
However, if you consider Heisenberg’s uncer tainty principle, all molecular
cur ve, the y-axis is the
movement may not cease entirely at 0 K . Consider why this may be the case.
fraction of par ticles with a
The Kelvin scale gives a natural measure of the kinetic energy of a gas, and
cer tain kinetic energy and the
is independent of physical proper ties, whereas the ar ticial Celsius scale is
x-axis is kinetic energy. Don’t
based on the physical proper ties of water. The Celsius scale is dened about an
arbitrary zero point and hence negative °C values occur. Are physical proper ties
mix up the two axis labels in
these two representations.
such as temperature invented or discovered? Could Anders Celsius, the
Swedish astronomer credited with the Celsius scale, have chosen the melting
and boiling points of another substance other than water to devise the ar ticial
Celsius scale?
171
6
C H E M I C A L
K I N E T I C S
Factors that aect the rate of a chemical reaction
There
are
four
factors
that
can
increase
the
rate
of
4 Decreasing the par ticle size of
a
chemical
reaction:
reactants in the solid phase
1
increasing
the
temperature
reaction
conducted
at
which
the
In
is
a
2
addition
of
a
catalyst
a
increasing
the
concentration
of
the
decreasing
the
particle
size
of
reactants
in
us
the
examine
each
of
these
the
for
1 Increasing the temperature at which
surface
just
discussed
this
factor.
A
the
surface
rate
have
example
involves
of
the
the
above
milk
to
down
0
°C
of
involving
rate
area
up
of
of
into
a
gas
reaction
the
solid
smaller
is
(or
will
increased
pieces.
is
that
of
the
the
reaction
takes
The
place
only
turn
bacterial
if
in
can
the
a
of
milk.
sour
temperature
on
At
solid
a
This
is
liquid
is
of
reactant.
reacting
the
solid
reactant.
used,
and
particles
of
liquid
So,
there
will
will
a
for
be
be
for
example,
greater
For
only
direct
As
the
contact
solid
increase
reaction.
increase.
solid,
divided
an
a
a
have
nely
will
will
available
reaction
If
there
with
in
the
number
of
a
the
result,
grain
can
combust
example,
explosively.
nely
This
has
led
rate
to
a
in
grain
number
of
major
explosions
in
conned
spaces
room
over
reactions.
a
good
temperature
refrigeration
temperature,
due
effect
of
area
divided
slowed
solid
particles
powder
solid
the reaction is conducted
time
surface
the
this
case
surface
individually.
or
of
the
phase.
with
We
the
reaction
solid,
the
in
Let
a
reactants
on
solid
if
breaking
reason
4
and
increase
by
3
heterogeneous
liquid)
factories.
period
process
brought
to
is
just
refrigerator.
acv
Find out some examples from dierent locations
around the world where this has happened.
2 Addition of a catalyst
We
have
also
already
discussed
this
in
detail.
Measuring the rate of a chemical
reaction
3 Increasing the concentration of the
We
reactants
If
in
a
xed
reactant
be
volume
species
corresponding
collisions.
the
concentr a ti on
incr e a s e s ,
incr e a s e
Hence
the r e
in
ther e
the
wil l
wi l l
an
look
to
stated
at
a
number
monitor
the
of
rate
techniques
of
a
that
chemical
can
reaction
previously:
a
fre que nc y
be
as
of
be
shall
used
1 Change in pH (for acid-base
of
inc re a s e
in
the
reactions)
number
of
success f ul
the
of
reactio n
col li si ons ,
a nd
t he r efor e
In
rate
will
incr ea s e.
An
a
reaction
where
+
H
of
this
involves
the
d e s tr ucti on
of
st a t u es
m a de
O
limestone
(calc i um
car bo na te ,
hydronium
cations,
+
(or
Ca C O
).
If
simply
H
)
or
hydroxide
ions,
OH
,
are
3
present
of
either
ex am pl e
as
either
the
reactant
or
product
species,
the
3
the
concentration
of
the
p o ll uta nt
s ulfur
change
probe
SO
,
in
the
atmosphe r e
i ncr e as e s,
in
pH
can
be
monitored
using
a
pH
di ox ide ,
the
ra te
and
meter
of
2
destruction
of
the
l i me stone
statue s
i n c re a se s .
2 Change in conductivity (for reactions
2CaCO
(s)
+
3
2SO
(g)
+
2
O
(g)
→
2CaSO
2
(s)
4
involving electrolytes)
+
2CO
(g)
2
Consider
The
solubility
of
calcium
sulfate
is
greater
the
following
reaction:
than
+
that
of
calcium
carbonate,
leading
to
erosion
of
IO
(aq)
+
5I
(aq)
+
6H
(aq)
→
3I
3
the
In
this
in
an
the
172
(aq)
+
3H
2
O(l)
2
limestone.
reaction
acidic
of
iodate
medium
concentration
of
ions
there
ions
is
with
a
from
net
a
iodide
ions
decrease
total
of
12
in
on
6 . 1
the
reactant
This
be
side
decrease
in
monitored
meter.
If
to
zero
the
using
the
net
on
the
product
concentration
a
of
conductivity
number
of
C O l l i s i O n
ions
t h e O r y
a n d
r a t e s
O f
r e a C t i O n
side.
ions
can
probe
absorbance
and
decreases
photoelectric
light
during
the
reaction,
then
the
total
electrical
cell
source
coloured
lter
conductivity
will
also
decrease
and
vice
versa.
meter
solution
Figure 1
4 Schematic diagram of a colorimeter that records
3 Change in mass or volume (for
absorbance. According to Beer ’s law the absorbance A is directly
proportional to the concentration c, that is A ∝ c or A = cεl,
reactions involving gases)
where l is the path length and ε is the extinction coecient.
We
have
seen
an
example
of
this
physical
method
earlier.
In
is
colorimetry,
passed
light
through
monitored.
The
of
the
a
certain
coloured
colorimeter
or
wavelength
solution
being
spectrophotometer
4 Change in colour (for reactions
then
measures
the
intensity
of
the
transmitted
light.
involving transition metals or other
The
coloured compounds)
absorbance
absorbed
curve
Colorimetry
(gure
14)
is
used
to
by
can
the
be
reactions
thathave
a
colour
in
or
a
coloured
intensity
product
One
can
be
example
product.
corresponding
concentration
of
the
of
this
to
the
a
(aq)
+
5C
4
(aq)
the
+
light
calibration
versus
of
the
standard
coloured
solution
15).
change
or
colorimeter.
ethanedioate
+
O
2
A
absorbance
of
in
reaction:
2
2MnO
change
reactant
using
involves
mixture.
amount
ecnabrosba
manganate(VII)
of
the
monitor
The
coloured
monitored
reaction
of
coloured
(gure
reactant
indicative
plotted
concentration
chemical
is
16H
(aq)
4
purple
2+
→
2Mn
(aq)
+
10CO
(g)
+
8H
2
pale
O(l)
2
pink
3
concentration/mol dm
Figure 15 Sketch of a typical calibration cur ve
Worked examples: rates of reaction
1
b)
M(CO
Example 1
)
=
12.01
+
2
×
16.00
0.25
_
n(CO
In
the
=
44.01
g
mol
2
chemical
reaction
of
calcium
carbonate
)
=
3
=
5.7
×
10
mol
2
44.01
with
hydrochloric
acid,
0.25
g
of
carbon
dioxide
3
5.7 × 10
_
Average
was
a)
generated
Deduce
in
the
60.0
rate
=
s.
balanced
5
=
9.5
×
10
1
mol
s
60.0
chemical
equation
su p
forthe
reaction,
including
state
symbols.
Be careful with g c gu in a question like
-1
b)
Calculate
the
average
rate,
in
mol
s
this. Since division is involved the answer should be
expressed with the smallest number of signi cant
Solution
 gures from the experimental data, which in this
a)
CaCO
(s)
3
+
2HCl(aq)
→
CaCl
(aq)
2
+
CO
(g)
+
case will be two.
2
H
O(l)
2
173
6
C H E M I C A L
K I N E T I C S
Example 2
Figure
16
dioxide
shows
formed
hydrochloric
calcium
how
the
varies
acid
volume
with
solution
carbonate
in
a
time,
is
of
carbon
when
added
to
Then
a
as
the
gradient
excess
dioxide
ask.
of
acid
The
is
proceeds
as
generated
the
slows
with
amount
as
the
time,
of
the
carbon
concentration
decreases.
plot
acid
reaction
decreases
eventually
attens
as
hydrochloric
consumed.
ii)
OC( V
2
)
OC( V
2
)
0
time
0
Figure 16
time
i)
Explain
ii)
Copy
the
shape
of
the
curve.
Figure 1
7
[3]
Decreasing
the
graph
and
sketch
the
rate
youwould
obtain
if
double
the
hydrochloric
acid
solution
of
concentration,
as
in
the
less
gas
instead,
with
all
other
constant
shape
of
from
the
thecurve
original.Explain
is
Outline
one
other
way
in
will
less
this
reaction
which
can
be
Sketch
a
be
the
selected
variable
studied
graph
in
to
There
wouldchange
the
state
between
term
one
activation
reason
calcium
takes
slower.
dioxide
are
a
why
is
volume
time.
halved,
Since
collisions
the
rate
of
will
reaction
there
is
the
same
amount
acid,
will
the
be
same
volume
of
produced.
to
of
possible
measure
the
answers
rate
at
here.
which
as
place
carbonate
at
a
and
reasonably
the
rate
gas
is
the
beginning
of
[2]
graph.
the
produced
quickest,
hydrochloric
acid
is
as
reaction,
the
carbon
contents
versus
time
dioxide
concentration
greatest.
of
given
mass
off.
of
time
(gure
or
a
plot
of
This
ask
mass
18).
ssol ssam
of
plot
at
2010
gradient
a
hydrochloric
fast
temperature.
=
involve
reaction
Figure 18
174
is
hence
Since
decreases
simply
time
is
longer
energy
the
Solution
At
and
number
stnetnoc + ksa fo ssam
room
Rate
frequent
mass
versus
i)
a
[2]
Dene
May
maximum
over
with
+
IB,
same
concentration
method
would
acid
the
evolved
a
the
and
the
(blue)
illustrate
time.
iv)
but
be
hydrochloric
One
how
curve
the
iii)
schoollaboratory.
decreases
the
[4]
carbon
rateof
17
why
different.
of
iii)
steep,
acid
will
the
gure
variables
be
kept
In
example
the
above,isused
concentration
reaction.
half
of
the
of
volume
is
of
the
curve
of
time
loss
6 . 1
Other
the
possibilities
pH
or
the
might
include
C O l l i s i O n
monitoring
t h e O r y
b)
a n d
Draw
a
versus
pressure.
r a t e s
graph
O f
of
r e a C t i O n
total
volume
of
oxygen
time.
3
c)
iv)
Activation
energy
is
the
minimum
energy
Calculate
correct
colliding
particles
need
to
have
in
order
the
average
rate,
in
1
cm
min
,
the
to
one
decimal
place.
for
3
a
reaction
calcium
takes
to
carbonate
place
at
temperature
barrier
occur.
is
a
The
and
between
hydrochloric
reasonably
because
quite
reaction
the
fast
rate
activation
d)
acid
at
Deduce,
oxygen
in
to
s,
how
be
long
it
took
for
40
cm
of
collected.
room
3
e)
Determine
the
initial
rate,
in
f)
Determine
the
instantaneous
at
=
cm
1
min
energy
rate,
in
low.
3
1
cm
g)
min
Explain
,
t
whether
4
min.
the
catalyst
used
is
Example 3
homogeneous
The
data
in
table
decomposition
manganese(IV)
2
of
were
recorded
hydrogen
oxide
for
peroxide
catalyst.
The
using
total
a
Solution
1
H
O
2
oxygen
gas
collected
was
heterogeneous.
volume
a)
of
or
the
measured
(aq)
→
H
2
O(l)
+
O
2
(g)
2
2
at
differenttimes.
b)
70
tm/m
to voum o ox g g
3
coc/cm
0
1
18
2
32
3
42
4
50
5
56
6
61
7
64
8
64
9
64
negyxo fo emulov latot
0
60
50
40
30
20
10
0
0
2
4
6
10
8
12
time/min
Figure 19 Plot of total volume of oxygen given o versus time
3
64 cm
_
c)
Average
rate
=
10
64
T
able 2
d)
2.8
min
e)
(x
y
,
1
)
=
=
2.8
×
(2,50);
3
=
7
9.1
60
(x
1
s
,
2
=
y
)
168
Initial
Deduce
the
the
balanced
reaction,
chemical
including
equationfor
rate
=
(0,0);
2
3
=
=
0
1
min
s
0
50
_
a)
cm
min
25
cm
1
min
2
statesymbols.
175
6
C H E M I C A L
f)
(x
,
3
y
)
=
K I N E T I C S
(6,
64);
(x
3
,
y
4
)
=
(2,
36);
g)
Manganese(IV)
a
Instantaneous
rate
at
t
=
4
different
3
=
2
7cm
phase
to
is
solid,
aqueous
so
is
in
hydrogen
min:
peroxide.
36
64
_
oxide
4
Manganese(IV)
oxide
is
acting
as
1
min
heterogeneous
6
catalyst
in
this
reaction.
su p
There are a number of points that you have to be careful
2
about in graphical questions:
1
Graphs
should have a title which involves a plot of y
versus x (not the other way round!).
176
Both the
x- and y-axes must be labelled and units
included.
3
When
finding
choose
two
the
slope
points
as
of
far
a
tangential
apar t
as
line,
possible.
try
to
a
Q U e s t i O n s
Questions
1
Consider
and
I
the
gaseous
(g)
+
H
2
reaction
between
gaseous
iodine
4
hydrogen.
(g)
⇋
2HI(g)
∆H
=
-9
Equal
masses
of
powdered
carbonatewere
added
of
acid.
hydrochloric
to
calcium
separate
The
calcium
solutions
carbonate
kJ
2
was
Why
do
some
hydrogen
collisions
not
result
in
between
the
iodine
formation
of
and
in
excess.
the
intervals.
product?
The
Which
represent
against
A.
I
and
H
2
molecules
do
not
The
dioxideproduced
the
time
volume
was
of
carbon
measured
at
curves
in
gure
evolution
of
carbon
forthe
acid
regular
20
solutions
best
dioxide
shown
in
have
2
table
sufcient
3?
energy.
system
The
temperature
is
in
equilibrium.
D.
The
activation
of
the
energy
system
for
this
is
too
high.
reaction
is
mc/)g(
The
C.
3
B.
I
II
IB,
low.
May
[1]
2011
3
2
At
25
OC fo emulov
2
very
°C,
200
cm
III
IV
3
of
1.0
mol
dm
nitric
acid
time/s
is
added
to
5.0
experiment
of
is
in
the
of
magnesium
repeated
magnesium
result
g
powder,
same
using
the
which
initial
powder.
same
If
mass
conditions
reaction
the
▲
Figure 20
will
3
rate?
25 cm
3
o
50 cm
3
o
3
2 mo m
Voum
Coco o
25 cm
3
1 mo m
1mo m
hC
hC
hC
A.
I
III
IV
B.
I
IV
III
C.
I
II
III
D.
II
I
III
tmpu / °C
o
3
3
/
o hnO
hnO
3
/mo m
3
3
cm
200
A.
2.0
25
B.
200
1.0
50
C.
100
2.0
25
D.
100
1.0
25
▲
T
able 3
[1]
[1]
IB,
IB,
May
5
3
Which
for
of
rate
A.
s
B.
min
C.
cm
May
2009
2011
the
of
following
reaction?
is
an
appropriate
unit
Hydrochloric
of
calcium
acid
repeated
using
does
change
and
this
the
is
reacted
carbonate;
the
calcium
collision
affect
large
carbonate
the
is
pieces
then
powder.
activation
How
energy
frequency?
acvo g
3
with
reaction
Coo
s
quc
3
D.
mol
dm
1
min
A.
increases
increases
B.
stays constant
increases
C.
increases
stays constant
D.
stays constant
stays constant
[1]
IB,
November
2009
177
6
C H E M I C A L
6
Which
K I N E T I C S
factors
can
affect
the
rate
of
a
chemical
9
a)
A
solution
of
hydrogen
peroxide,
H
O
2
reaction?
added
to
acidied
I.
The
concentration
of
the
a
solution
with
of
sodium
hydrochloric
acid,
,
is
2
iodide,
NaI,
HCl.
The
reactants.
yellow
colour
of
the
iodine,
I
,
can
be
used
2
II.
The
temperature
at
which
the
reaction
to
takes
determine
H
III.
The
the
rate
of
reaction.
place.
physical
state
of
the
O
2
reactants.
(aq)
+
2NaI(aq)
+
2HCl(aq)
2
→
2NaCl(aq)
+
I
(aq)
+
2H
2
A.
I
and
II
B.
I
and
III
The
experiment
II
D.
I,
and
II,
is
repeated
with
III
only
and
of
the
III
to
the
reaction
changes
reason,
i)
its
The
that
effect
conditions.
follow,
on
the
concentration
predict,
rate
of
H
of
O
2
at
7
In
an
ethyl
acid-catalysed
ethanoate,
hydrolysis
the
reaction
in
3.5
from
min,
1.50
concentration
at
a
mol
given
dm
of
the
ii)
ester
The
3
to
0.35
temperature,
mol
T
.
constant
For
each
stating
a
reaction.
is
increased
2
temperature.
[2]
of
3
changes
some
only
changes
C.
O(l)
2
only
a
dm
solution
ne
of
powder
NaI
is
prepared
instead
of
from
from
large
crystals.
Which
[2]
1
of
the
following
statements
are
correct?
b)
Explain
when
I.
The
average
rate
of
the
3
0.33mol
reaction
If
the
rate
of
a
of
reaction
the
increases
system
increases.
[3]
min
IB,
II.
the
temperature
is
1
dm
why
the
reaction
temperature,
is
T
carried
,
the
out
at
reaction
a
November
2009
higher
rate
will
2
begreater.
10
III.
The
products
of
the
reaction
will
Models
the
ethanoic
acid
and
I
and
II
B.
I
and
III
C.
II
I,
II,
and
the
Factors
and
that
reaction
affect
and
Dene
the
rate
particle
the
of
a
size,
chemical
Design
in
concentration
temperature
of
the
an
rate
the
term
rate
of
a
an
of
appropriate
reaction
alkaline
the
Design
an
rate
three
characteristic
properties
hydrolysis
to
measure
reaction
particles
reaction
as
theory.
2011
of
the
ester
methyl
ethanoate
medium.
appropriate
experiment
to
measure
of
reaction
using
that
described
affect
by
the
the
rate
involving
the
the
“clock
reaction
of
reaction”
magnesium
of
with
178
experiment
a
[1]
reactant
May
of
chemical
technique
IB,
theory.
of
the
List
collision
reaction.
reaction.
ii)
Discuss
molecular
III
12
i)
chemistry.
only
include
reactants,
in
kinetic
only
III
(saponication)
8
vital
the
only
the
D.
of
ethanol.
11
and
prove
principles
theory
A.
can
be
of
collision
[3]
dilute
hydrochloric
acid
solution.
7
E Q U I L I B R I U M
Introduction
An
understanding
of
reactions
that
are
equilibrium,
in
examine
the
equilibrium
constant
K
,
c
equilibrium
the
to
and
equilibrium
science
and
how
is
of
to
control
the
fundamental
society.
The
Haber
position
of
importance
process,
used
examine
the
the
effects
the
value
We
will
information
of
of
changing
K
it
conveys
and
experimental
applying
Le
discuss
conditions
Châtelier’s
on
principle.
c
for
an
the
equilibrium
impact
we
large-scale
on
discuss
the
manufacture
system
history
how
that
of
has
the
reactions
of
ammonia,
had
World.
can
be
in
a
profound
In
a
is
this
state
chapter
of
Q,
also
which
products
not
in
a
is
introduce
a
and
state
measure
the
of
reactants
of
term
the
reaction
relative
present
in
a
quotient,
amounts
reaction
of
that
is
equilibrium.
7
.1 E
Understandings
Applications and skills
➔
A state of equilibrium is reached in a closed
➔
The characteristics of chemical and physical
system when the rates of the for ward and
systems in a state of equilibrium.
reverse reactions are equal.
➔
➔
Deduction of the equilibrium constant
The equilibrium law describes how the
expression (K
) from an equation for a
c
equilibrium constant (K
) can be determined for
c
homogeneous reaction.
a par ticular chemical equation.
➔
➔
Determination of the relationship between
The magnitude of the equilibrium constant
dierent equilibrium constants (K
) for the
c
indicates the extent of a reaction at equilibrium
same reaction at the same temperature.
and its temperature dependence.
➔
➔
Application of Le Châtelier ’s principle to
The reaction quotient (Q) measures the relative
predict the qualitative eects of changes of
amount of products and reactants present
temperature, pressure, and concentration on
during a reaction at a par ticular point in time.
the position of equilibrium and on the value of
Q is the equilibrium constant expression with
the equilibrium constant.
non-equilibrium concentrations. The position
of the equilibrium changes with changes in
concentration, pressure, and temperature.
➔
A catalyst has no eect on the position of
equilibrium or the equilibrium constant.
Nature of science
➔
Obtaining evidence for scientic theories –
➔
Common language across dierent disciplines –
isotopic labelling and its use in dening
the term dynamic equilibrium is used in other
equilibrium.
contexts, but not necessarily with the chemistry
denition in mind.
179
7
E Q U I L I B R I U M
Equilibrium reactions in chemistry
Many
state
important
of
occurring
control
being
reaction
synthesized
understanding
Systems
that
example,
with
a
leave
An
the
Water
liquid
evaporation.
energy
in
have
molecules
through
the
and
same
collision
of
and
in
the
enter
time,
and
the
liquid
liquid
the
enter
in
in
that
of
being
enables
the
chemists
desired
relies
everyday
a
storage
product
heavily
liquid
life.
on
in
water
phase
For
container
sufcient
phase
gaseous
the
a
constantly
Industry
have
gaseous
some
yield
in
reactions
reactions.
common
hot
exist
equilibrium
reactions.
equilibrium
are
and
reverse
reactants
maximize
some
and
chemical
equilibrium
equilibrium
you
reversible
forward
products
and
controlling
phase
At
with
are
the
understanding
involving
imagine
lid.
both
conditions
and
are
reactions
with
simultaneously
interconverted.
to
chemical
equilibrium,
the
process
molecules
in
the
sealed
energy
will
of
will
process
lose
of
condensation.
As
evaporation
saturated
particles
with
resulting
evaporation
further
gas
will
change
present
The
continues,
water
in
forward
constantly
described
in
the
of
being
as
a
atmosphere
and
the
condensation
equal
and
concentration
in
the
vapour
the
the
rate
amount
closed
of
gas
and
and
is
change,
between
the
the
at
at
two
the
rate
will
of
be
amount
no
of
equilibrium.
equal
but
becomes
between
rates.
molecules
phases.
This
The
are
is
equilibrium
H
O(l)
⇋
H
2
O(g)
2
liquid phase
▲
system
there
then
occurring
not
closed
collisions
Eventually
present
system
are
does
interconverted
dynamic
increases.
The
reactions
and
the
of
condensation
liquid
system.
reverse
liquid
of
in
chance
gaseous phase
Figure 2 In a dynamic equilibrium the for ward and reverse reactions occur at equal rates
Heterogeneous equilibrium and solubility
▲
Figure 1 Bromine in a closed
A
system will establish a dynamic
equilibrium
equilibrium between the liquid
and gaseous phases. Eventually
saturated
chloride,
The
soluti o n
if
NaCl
ther e
and
concentration
in
is
a
clo se d
e xce s s
wa ter,
of
i ons
sy st e m
s ol id
initia ll y
pr e se nt
w il l
p re s e nt .
the
in
s ol id
the
e st a bli sh
If
you
wil l
a
mi x
beg i n
a que ou s
dyn a m i c
so li d
to
so di um
di ss ol ve.
s ol u t io n
wi l l
the colour stops changing as the
increase.
Some
solution.
At
aq ue o us
io ns
wi ll
r e c o mbin e
and
pr e c ip it a t e
ou t
of
equilibrium concentration of vapour
this
is reached
precipitation.
180
s ta g e
the
ra te
of
dis so lut i on
is
fas t e r
t h an
the
r ate
of
7. 1
Eventually,
when
the
solution
becomes
saturated,
the
rate
of
E q u i l i b r i u m
dissolving
TOK
will
equal
the
rate
of
precipitation.
A
dynamic
equilibrium
has
been
established.
Scientists use specialized
terminology to
precipitation
+
NaCl(s)
⇋
Na
(aq)
+
Cl
facilitate unambiguous
(aq)
communication of meaning
dissolution
and understanding. They
initial
recognize the need for a
equilibrium
common language. For
example, a closed system
is one in which no matter
is exchanged with the
surroundings.
The language of science and
its internationally agreed
terminology and symbols
aid communication and
understanding. However,
when meaning is inexible
Cl
+
(aq)
(aq)
Na
+
+
Na
Cl
does language also play a
(aq)
Na
(aq)
(aq)
(aq)
Cl
(aq)
Cl
+
Na
role in forming boundaries
(aq)
to our experience of new
knowledge?
▲
Figure 3 A saturated solution is in dynamic equilibrium
Chemical systems
When
describing
“reactants”
and
completion,
a
state
of
depend
chemical
producing
equilibrium.
on
chemical
the
and
on
give
new
The
physical
conditions
products),
reactions
“products”
(a
the
the
in
an
substances.
relative
in
presence
In
rates
conditions
change
open
impression
of
system,
that
reality
of
of
a
many
forward
temperature
concentration
the
of
terms
reactions
and
and
proceed
reactions
reverse
exist
in
reactions
pressure,
reactants
to
on
Stdy tp
and
catalyst.
For non-reversible reactions,
Nitrogen(IV)
oxide
or
nitrogen
dioxide,
NO
is
a
toxic
brown
gas
that
the use of a single arrow → in a
2
exists
in
equilibrium
with
colourless
dinitrogen
tetroxide,
N
O
2
chemical equation represents
:
4
the complete conversion of
N
O
2
When
(g)
⇋
2NO
4
a
(g)
2
sample
of
reactants to products. Reversible
colourless
frozen
N
O
2
which
is
then
sealed,
initially
N
O
2
colour
is
is
is
placed
into
a
reactions and systems in
container
4
the
only
substance
present
and
no
equilibrium are designated by
4
the use of a double-headed
observed.
arrow: ⇋
At
room
temperature
N
O
2
to
brown
signals
the
begins
to
decompose
and
a
colour
change
4
presence
of
NO
.
Your
understanding
of
kinetic
decomposition
of
N
2
theory
will
tell
you
that
initially
the
rate
of
O
2
be
greatest,
when
the
concentration
[N
O
2
passes,
sufcient
NO
molecules
will
be
]
is
at
a
maximum.
As
will
4
time
4
present
in
the
closed
system
for
2
successful
collisions
to
reform
N
O
2
The
concentration
of
N
O
2
concentration
of
NO
.
4
decreases
progressively
while
the
4
increases.
The
system
approaches
a
dynamic
2
equilibrium
become
as
the
rates
of
the
forward
and
reverse
reactions
equal.
181
7
E Q U I L I B R I U M
At
equilibrium:
NO
2
noitartnecnoc
N
●
The
forward
●
There
●
There
●
The
is
no
and
reverse
change
in
reactions
the
are
occurring
concentrations
of
at
equal
reactants
rates.
and
products.
O
2
4
is
no
change
equilibrium
in
can
macroscopic
be
properties
approached
from
such
either
as
colour
the
and
forward
density.
or
equilibrium achieved
reverse
direction.
0
time
▲
The
●
time for the reaction N
O
2
(g) ⇋
2NO
4
equilibrium
continue
Figure 4 Change in concentration versus
but
products
(g)
is
no
dynamic.
overall
The
change
forward
in
the
and
reverse
concentration
reactions
of
reactants
and
occurs.
2
approaching and achieving equilibrium
Any
●
changes
pressure,
or
in
the
reaction
concentration
conditions,
of
reactants
such
or
as
temperature,
products,
can
affect
the
rate of for ward
equilibrium,
demonstrating
its
dynamic
nature.
reaction
qck eston
etar
Which statement is always true for a chemical reaction that has reached
equilibrium achieved
equilibrium?
(rates are equal)
rate of reverse
reaction
0
a)
The yield of product(s) is greater than 50%
)
The rate of the forward reaction is greater than the rate of the reverse
time
reaction.
▲
Figure 5 At equilibrium, the rates of the for ward
and reverse reactions remain constant
c)
The amounts of reactants and products do not change.
d)
Both forward and reverse reactions have stopped.
[1]
ib, Nov 20 05
Animated computer
simulations are available in
The equilibrium law
which the user can decide
When
a
reaction
system
has
established
equilibrium,
the
forward
and
how to change the conditions
reverse
reactions
are
occurring
at
equal
rates
and
there
is
no
change
of a reaction to illustrate
in
the
concentration
of
reactants
and
products.
To
make
best
use
of
the concept of dynamic
the
reaction
we
need
to
understand
the
position
of
the
equilibrium ,
equilibrium.
that
is,
whether
manipulate
of
this
reactants
to
or
products
maximize
the
are
yield
of
favoured,
products
and
and
how
the
we
can
protability
industry.
TOK
The
Scientists have a common
terminology and a common
reasoning process, which
involves using deductive
the
law
ratio
molar
of
of
chemical
of
the
coefcients)
their
molar
equilibrium
equilibrium
concentration
to
the
states
products
concentration
coefcients)
constant,
of
K
is
.
a
of
constant.
The
that
a
given
to
reactants
This
subscript
at
(raised
‘c’
the
temperature
power
(raised
constant
indicates
is
to
the
called
that
of
values
for
products
and
reactants
are
being
concentration
used.
through analogies and
For
example:
generalizations. They share
O
mathematics, the language
(g)
+
4HCl(g)
⇋
2
2H
O(g)
+
2Cl
2
[
H
O
[
]
(g)
2
2
2
of science, as a powerful
Cl
2
]
2
__
K
tool. Indeed, some scientic
=
c
4
[
O
]
[
HCl
]
2
explanations exist only in
mathematical form.
The
value
reaction
of
at
a
the
equilibrium
given
constant
temperature.
is
Values
specic
of
K
c
182
for
have
each
no
power
the
c
logic and induction
their
chemical
units.
7. 1
The
magnitude
of
K
tells
you
about
the
position
of
the
E q u i l i b r i u m
equilibrium.
c
If
K
is
a
very
large
number,
K
c
1,
this
indicates
that
at
a
given
c
temperature,
of
>>
K
,
the
products
greater
the
are
favoured
proportion
of
over
reactants.
products
that
The
exists
larger
the
value
compared
with
c
reactants
at
equilibrium.
Conversely,
a
very
small
value
of
K
(
K
c
<<
1
)
c
In a hoogeneos e
indicates
that
the
reaction
is
unfavourable
at
this
given
temperature.
all the reactants and products
are present in one phase. The
most common example is
Writing equilibrium constant expressions
reversible reactions that occur
When
constructing
the
equilibrium
constant
expression
for
a
in the gaseous phase.
homogeneous
reaction,
the
following
need
to
be
considered:
In a heteogeneos
1
For
an
aqueous
reaction
the
concentration
of
the
solvent
water
e the reactants or
does
not
appear
in
the
equilibrium
constant
expression,
as
its
products exist in more than
concentration
does
not
change
during
the
reaction.
one phase, such as gaseous
2
If
the
reaction
esterication
takes
place
discussed
in
in
a
non-aqueous
sub-topic
10.2),
solution
water
(such
must
be
as
the
and solid, liquid and solid.
included
in
The equilibrium constant
the
K
expression
as
any
other
reactant
or
product.
c
Table
1
expressions derived here are
gives
different
some
examples
of
equilibrium
constant
expressions
for homogeneous equilibria.
for
reactions.
Checa eaton
E constant expesson
2+
3+
Fe
]
[ Fe(SCN)
___
2+
(aq) + SCN
(aq) ⇋ Fe(SCN)
(aq)
K
=
c
3+
1
[ Fe
[
C
][ SCN
H
3
]
COOCH
7
]
3
__
CH
OH(aq) + C
3
H
3
COOH(aq) ⇋ C
7
3
H
COOCH
7
(aq) + H
3
O(l)
K
2
=
c
[
CH
OH
][
C
3
H
3
COOH
]
7
2
[
NH
]
3
__
N
2
(g) + 3H
(g) ⇋
2NH
2
(g)
K
3
=
c
3
[
▲
N
2
][
H
2
]
T
able 1 Equilibrium constant expressions for some chemical reactions
TOK
Scientists approach their understanding of the universe
from varying perspectives. Observations made using
our senses examine phenomena at the macroscopic
level. The models and theories developed by the
scientic community focus on our understanding of the
microscopic world. Which of the ways of knowing (WOK)
qck eston
enable us to make the transition from the macroscopic to
Deduce the equilibrium constant expression for each of
the microscopic?
the following homogeneous equilibrium reactions.
macoscopc properties of substances can be identied
3
1
a)
using our senses, and can be directly determined by
measurements. Examples include colour, texture, and
N
2
)
(g) +
2
H
2
ClNO
(g) ⇋ NH
2
(g)
3
(g) + NO(g) ⇋ NO
2
(g) + ClNO(g)
2
density. mcoscopc properties exist at an atomic level
c)
and can be determined only indirectly.
4NH
3
(g) + 5O
2
(g) ⇋ 4NO(g) + 6H
O(g)
2
183
7
E Q U I L I B R I U M
Worked example: deducing equilibrium constant expressions
For
the
homogeneous
0.300 × 0.300
__
equilibrium:
=
2
(3.00)
H
(g)
+
I
2
(g)
⇋
2HI(g)
2
2
=
1.00
×
10
3
the
equilibrium
concentrations
(in
mol
dm
)
are
Note:
as
follows:
1
_
K
=
c(1)
[
H
(g)
=
]
0.300,
[
I
2
Deduce
(g)
]
=
0.300,
[HI(g)]
=
K
3.00
c(2)
2
the
equilibrium
constant
expression,
The
K ,
value
of
K
and
determine
the
value
of
K
for
the
forward
also
depends
on
how
the
chemical
c
c
equilibrium
and
reaction
is
balanced:
c
reverse
1
_
reactions.
1
_
H
(g)
+
I
2
(g)
⇋
HI(g)
2
2
2
Solution
[HI]
_
=
K
1
c(3)
For
the
forward
reaction:
1
2
H
[
]
2
I
[
2
]
2
2
[HI]
3.00
__
_
K
=
K
c(1)
=
c(3)
[
H
I
][
2
1
1
]
2
2
(
0.300
)
2
(
0.300
)
2
(
)
3.00
__
=
=
0.300
×
10.0
0.300
___
2
=
1.00
×
10
=
K
c(3)
For
the
reverse
⇋
H
(g)+
2
I
by
(g)
2
last
][
c(1)
shows
gives
a
that
dividing
new
the
equation
equilibrium
constant
2
I
2
example
throughout
which
[
H
K
reaction:
This
2HI(g)
√
is
equal
to
the
square
root
of
the
original
]
2
_
K
equilibrium
=
c(2)
constant.
2
[HI]
qck estons
1
2
The equilibrium constant for the reaction between
The equilibrium constant for the equilibrium between
4
hydrogen and chlorine gas to produce hydrogen
N
O
2
and NO
4
is 7.7 × 10
at 273 K:
2
33
chloride gas is 2.40 × 10
at 298 K .
N
O
2
H
(g) + Cl
2
(g) ⇋ 2NO
4
(g)
2
(g) ⇋ 2HCl(g)
2
Calculate the equilibrium constant for
1
Calculate the equilibrium constant for the reverse
N
2
O
2
(g) ⇋ NO
4
(g) at 273 K .
2
reaction, namely the decomposition of HCl.
Combining equilibrium constants
Table
and
2
summarizes
equilibrium
the
ways
constants
that
can
be
the
equilibrium
constant
expressions
combined.
Change n eacton eaton
E constant expesson
reverse the reaction
inverse of the expression
E constant
1
_
K
(reverse) =
(forward)
1
_
c
K
or K
c
c
K
c
__
halve the coecients
square root of the expression
√
K
c
2
double the coecients
square the expression
K
c
sum equations
product of the expressions
K
c
▲
T
able 2 The equilibrium constant K
184
K
c1
× K
×
…
c2
for the same reaction at the same temperature can be
c
expressed in a number of ways
=
7. 1
E q u i l i b r i u m
The eect of changing experimental conditions
on the equilibrium constant
When
equilibrium
constant
change
provided
in
is
established,
that
the
experimental
However,
the
value
of
the
conditions
K
position
temperature
remains
can
and
of
affect
constant
the
equilibrium
pressure
the
do
not
equilibrium
unless
the
remains
change.
A
position.
temperature
c
changes
(table
3).
Change n condton
E poston
K
c
concentration of
changes in response to a change in
product or reactant
[reactants] or [products]
no change
pressure
in a reaction with gaseous reactants or
no change
products, the pressure can aect the
equilibrium position
temperature
usually changes: the direction of change
unless
exothermic or endothermic
ΔH = 0
catalyst
▲
changes,
depends on whether the reaction is
no change
no change
T
able 3 The eect of changing conditions on the equilibrium position and the value of K
c
Le Châtelier ’s principle
Le
Châtelier’s
principle
changing
conditions
If
is
At
of
a
change
the
forward
the
system
a
given
made
and
to
will
to
a
reactants
or
a
useful
on
system
reverse
tool
the
that
reactions
for
predicting
equilibrium
is
in
will
equilibrium,
shift
to
the
effect
that
position:
offset
the
this
balance
change
between
and
return
equilibrium.
temperature,
equilibrium
is
have
products
constant
K
table
does
.
The
3
shows
not
that
result
in
equilibrium
changing
a
change
position
the
in
of
concentration
the
the
value
of
reaction
will
c
change
in
response
to
the
change
in
concentration
so
as
to
return
K
c
to
its
original
value.
For
example,
gure
6
illustrates
the
chromate–
▲
dichromate
Figure 6 The chromate–dichromate equilibrium.
equilibrium:
2
Aqueous dichromate Cr
O
2
2
Cr
2
(aq)
+
H
7
O(l)
⇋
2CrO
2
orange
(aq)
+
2H
aqueous chromate CrO
(aq)
yellow
(aq) is yellow. The
4
4
dichromate
(aq) is orange while
2
+
2
O
7
colour of the solution gives an indication of the
chromate
position of equilibrium
If
the
tells
concentration
us
that
change.
the
of
a
reactant
forward
Decreasing
forward
reaction.
mixture
results
the
The
is
reaction
increased,
will
be
concentration
addition
of
Le
Châtelier’s
favoured
of
a
to
product
hydroxide
ions,
principle
counteract
will
OH
,
also
to
this
favour
the
the
reaction
+
to
form
orange
The
water.
in
a
The
dichromate
reverse
increased
reduction
equilibrium
reacts
reaction
or
the
in
is
to
concentration
mixture
form
favoured
concentration
+
H
if
of
becomes
yellow
the
a
a
as
OH
paler
reacts
colour
with
as
H
more
chromate.
concentration
reactant
is
of
a
product
decreased.
is
The
addition
HCl
results
+
of
H
the
ions
in
reverse
observed.
the
form
reaction
In
both
of
concentrated
being
cases,
favoured
the
value
hydrochloric
and
of
a
deeper
equilibrium
acid,
orange
colour
constant
K
in
being
remains
c
unchanged
as
long
as
the
temperature
remains
the
same.
185
7
E Q U I L I B R I U M
Le
Châtelier’s
conditions
an
to
principle
equilibrium
ammonia
N
(g)
uses
+
the
reaction.
the
3H
2
The
allows
maximize
(g)
The
yield
nitrogen
is
⇋
2NH
equilibrium
of
chemists
the
to
desired
process
for
manipulate
product
the
reaction
formed
manufacture
in
of
reaction:
(g)
3
increased
and
Haber
following
2
industrial
amount
by
hydrogen,
using
and
high
concentrations
removing
the
product
of
reactant
ammonia
gases
from
the
mixture.
Cause and eect
“Cause
and
situation
direct
where
develop
or
a
second
They
which
is
this
the
rst
and
(the
test
that
the
collection
either
to
the
effect)
event
(the
and
suggests
factors
to
refers
concept
that
then
used
event
the
between
false.
disprove
of
hypothesis
experimentation
data,
“causation”
understand
causation
true
a
or
consequence
Scientists
or
effect”
a
a
relationship
predict
of
the
not
empirical
modify,
or
of
a
effects
demonstrate
a
changing
change
of
in
helping
changes
reaction
and
However,
explanation
the
tool
concentration
constant.
an
on
useful
and
equilibrium
changes
resultant
hypothesis.
an
provide
between
is
qualitative
pressure,
equilibrium
not
these
by
principle
the
temperature,
does
be
hypothesis
support,
to
Châtelier’s
position
often
may
of
is
cause).
Le
for
equilibrium
in
on
the
the
the
the
value
principle
effects
reaction:
it
of
does
cause-and-effect
relationship
these
and
in
the
conditions
equilibrium
the
system.
The Haber process
Over
the
past
century,
of
agriculture
have
of
the
land
and
Earth’s
the
over
7
employment
and
being
population
billion
food
of
people.
sector
security
many,
including
World
Trade
advances
resulted
in
in
cultivated
the
planet
developing
remains
of
with
is
crops
a
to
major
countries
great
international
science
one-third
increasing
Agriculture
in
the
over
concern
bodies
such
a
process
a
large
one
of
(WTO),
the
the
the
world.
in
fertilizers,
manufacture
to
as
Food
Organization
Organization
(FAO),
and
the
demand
decade
for
demands
for
endeavours
fertilizers.
and
of
food
increases
More
will
and
the
the
not
water
elds
than
onset
with
plastics,
of
the
production
exponentially
nitric
of
50 %
be
able
without
to
meet
HNO
scientic
biotechnology
of
current
and
used
(aq)
on
nitrate
Haber
+
this
will
the
use
of
was
1918
(1868–1934)
awarded
for
his
was
a
German
ammonia,
the
work
Nobel
on
the
Prize
in
NH
from
NO
century
wor l d
w ar,
mi li ta r y
to
H a be r
a dva n c e
NH
its
at
elements.
the
A
Ammon ia
(g)
→
ammo nium
re a c t s
NH
also
NO
4
as
a
(aq)
n i t ra t e ,
which
is
3
fe rtil ize r. Ammo ni u m
und e r g o ex pl os ive
whe n
d e to na te d:
→
N
O(g)
+
2H
2
O(g)
2
is
sometimes
warfare”,
Haber
referred
to
advancing
as
the
the
“father
research
of
into
utilization
rst
world
of
many
war.
He
poisonous
was
not
gases
alone
in
during
making
shortage
beginning
prompted
fo r m
of
of
advances
to
in
science
with
military
the
during
this
time.
Gustav
Hertz
research
and
186
rs t
e x p l o s ive s .
to
day
(s)
applications
twentieth
the
3
signicant
fertilizers
and
Chemistry
synthesis
3
natural
explosives,
chemist
the
of
the
fertilizers.
and
in
in
food
chemical
who
processes
application
used
bres,
Ge r m an
of
acid
decomposition
Haber
Fritz
also
3
to
4
relies
is
industrial
major
World
NH
production
its
on
now
(WHO).
we
food
in
as
ammonia
3
every
well
is
the
with
Global
widespread
As
ammonia
manufacture
and
the
Health
manufacture
pharmaceuticals.
worked
Agriculture
could
Ammonia
most
in
With
Organization
that
scale.
James
Franck
were
physicists
who
devised
7. 1
investigations
the
atom
mechanics
see
the
in
atomic
an
(for
the
Hahn
1944
his
the
for
on
course
Bohr’s
the
in
topic
of
Later
in
life
range
of
awards
and
the
founder
was
the
he
They
1925,
in
for
of
his
on
Prize.
Franck,
Haber’s
and
research
developing
were
serve
to
while
Their
research
illustrate
understanding
ethical
to
their
receive
work,
and
several
discoveries.
is
how
were
who
the
while
the
Nature
of
goes
the
of
on
members
Their
of
questions
results
utilize
(NOS).
scientic
of
intellectual
to
on
stories
Science
advanced
raising
concept
society
all
worked
weapons.
considerations
scientic
He
team
signicantly
of
heavy
Hahn
chemical
Physics
continued
Peace
atomic
in
ssion
nominated
Nobel
of
12).
Prize
the
Hertz,
of
quantum
mechanics,
Physics
Nobel
model
for
quantum
book,
Prize
discovery
nuclei.
impressive
Niels
groundwork
more
Nobel
achievements
as
the
received
for
occasions
support
lay
Physics
awarded
Otto
to
and
E q u i l i b r i u m
about
some
the
of
property
scientic
regarded
age.
The eect of pressure on reactions in the
Stdy tp
gas phase
We
have
effect
of
position
rather
seen
a
When considering the eect
how
change
of
Le
in
Châtelier’s
equilibrium.
than
the
principle
concentration
A
aqueous
system
phase
of
a
that
will
can
involves
be
be
product
affected
used
or
to
predict
reactant
substances
by
changes
on
of changes in pressure on a
the
reaction, you must refer to the
the
in
the
gaseous
in
applied
moles of gaseous reactants or
products in the reaction. For
example:
pressure:
4HCl(g)
+
O
(g)
⇋
2H
2
O(g)
+
2Cl
2
C(s) + H
(g)
O(g) ⇌ H
2
2
(g) + CO(g)
2
In this heterogeneous reaction,
In
this
reaction
there
are
5
moles
of
gas
on
the
reactant
side
and
4
moles
the solid carbon is not included
on
the
product
side.
A
change
in
pressure
applied
to
the
system
will
when considering the eect of
result
in
a
shift
in
the
equilibrium
position.
If
the
pressure
is
increased,
changes in pressure.
Le
Châtelier’s
change.
of
the
the
N
principle
forward
system.
reverse
In
The
In
reaction
Haber
(g)
+
the
3H
are
decrease
Such
a
same
being
4
(g)
side.
⇋
of
A
way,
2NH
gas
a
favoured
decrease
in
will
to
shift
to
reduce
pressure
reverse
the
will
this
pressure
result
in
the
(g)
in
on
high
pressure
change
equilibrium
equilibrium
3
moles
the
the
becomes
favoured.
2
product
that
process:
2
there
the
says
reaction
the
of
the
reactant
pressure
the
K ,
if
side
and
favour
the
only
2
moles
forward
of
gas
reaction,
on
to
system.
equilibrium
constant,
will
the
position
will
temperature
not
affect
remains
the
value
of
constant.
c
Temperature and the equilibrium constant
An
understanding
considering
constant.
N
(g)
the
For
+
this
the
of
thermodynamics
changing
the
of
a
reaction
temperature
on
is
the
required
when
equilibrium
example:
3H
2
In
of
effect
(g)
⇋
2NH
2
reaction
(g)
ΔH
=
–92
kJ
3
energy
can
be
considered
a
product
and
is
released
to
the
surroundings.
187
7
E Q U I L I B R I U M
exothermic reaction
TOK
The work of Fritz Haber

a
had mixed outcomes for
society. While his method of
ygrene laitnetop
ammonia production had a
signicant and long-lasting
eect on increasing food
production for humanity,
reactants
∆H
his work with the German
military in the manufacture
products
of explosives and the use of
chlorine in chemical warfare
brought into question
reaction progress
the role of scientists in
▲
Figure 7 The potential energy prole of an exothermic reaction
society. The outcomes
of scientic endeavours
The
reverse
reaction
is
endothermic
–
it
requires
energy
from
the
often have signicant
surroundings.
ethical implications.
At
equilibrium
the
forward
and
reverse
reactions
occur
at
equal
rates
Should scientists be held
and
there
is
no
net
change
in
energy.
responsible for the way in
which society utilizes their
discoveries?
For
the
exothermic
equilibrium
accordance
move
to
change;
This
in
with
the
the
results
the
left,
reaction,
direction
Le
Châtelier’s
favouring
concentrations
in
a
an
decrease
in
increase
that
in
the
reactants,
nitrogen
the
temperature
consume
principle,
the
of
will
to
and
equilibrium
the
extra
will
equilibrium
minimize
hydrogen
constant
the
.
effect
the
In
position
will
K
shift
energy.
will
of
the
increase.
Conversely,
c
a
decrease
in
equilibrium
the
temperature
to
the
concentration
right,
of
for
the
exothermic
favouring
ammonia,
the
NH
.
reaction
forward
This
will
reaction
results
in
an
shift
and
increase
3
Table
4
summarizes
the
effects
on
the
increasing
in
K
.
c
the
equilibrium
system
of
changing
temperature.
Type of
Change n
eacton
tepeate
E poston
E
constant K
c
increase
moves to the left,
decreases
favouring reactants
exothec
decrease
moves to the right,
increases
favouring products
increase
moves to the right,
increases
favouring products
endothec
decrease
moves to the left, favouring
reactants
▲
188
T
able 4 The eects on the equilibrium system of a change in temperature
decreases
7. 1
E q u i l i b r i u m
The eect of a catalyst on equilibrium reactions
The
addition
lowering
a
catalyst
energy
In
a
of
the
means
to
on
catalyst
that
overcome
reversible
effect
a
activation
both
forward
and
position
of
greater
the
reaction,
the
the
a
energy
lowered
reactions
alternative
reaction
proportion
and
equilibrium
constant
an
In
activation
forward
reverse
the
equilibrium
a
provides
energy.
increase
will
not
of
by
to
an
and
The
equal
and
for
have
has
rates
is
reaction,
sufcient
an
of
amount.
there
a
completion,
become
energy
reactions.
change
goes
reactants
barrier
activation
reverse
pathway
that
no
products.
equal
the
The
effect
on
the
K
c
qck eston
PCl
(g) ⇋ PCl
5
(g) + Cl
3
(g) ∆H
= +92.5 kJ
2
Predict and explain any shift in the equilibrium position when:
a)
the temperature of the system is decreased
)
additional chlorine gas is injected into the system
c)
the pressure applied to the system is increased
d)
a catalyst is added.
Reaction quotient
If
a
system
product
to
has
not
reached
reactants
will
equilibrium,
not
equal
K
.
the
This
ratio
ratio
of
is
concentration
called
the
of
reaction
c
quotient
as
it
Q
moves
favoured
to
and
this
toward
helps
you
equilibrium
establish
to
determine
and
equilibrium
the
(table
the
direction
progress
of
the
of
the
reaction
reaction
that
is
5).
The concentration of products is greater than at equilibrium and the
Q > K
c
reverse reaction is favoured until equilibrium is reached.
The concentration of reactants is greater than at equilibrium and the
Q < K
c
forward reaction is favoured until equilibrium is reached.
The system is at equilibrium and the forward and reverse reactions
Q = K
c
▲
occur at equal rates.
T
able 5 The relationship between the reaction quotient Q and the equilibrium constant K
c
189
7
E Q U I L I B R I U M
Questions
1
The
equation
industry
shown
to
for
a
reversible
convert
methane
reaction
to
used
hydrogen
in
Poston of
is
Vae of e
e
constant
shifts towards the
decreases
below.
A.
CH
(g)
+
H
4
O(g)
⇋
CO(g)
+
3H
2
Δ
=
reactants
(g)
2
+210
kJ
shifts towards the
increases
B.
reactants
Which
statement
is
always
correct
about
shifts towards the
thisreaction
when
equilibrium
has
decreases
been
C.
products
reached?
shifts towards the
A.
The
concentrations
of
methane
and
increases
D.
carbon
products
monoxide
are
equal.
IB,
B.
The
rate
than
C.
The
of
the
the
rate
amount
forward
of
of
the
reaction
reverse
hydrogen
is
is
of
three
times
the
The
equation
The
value
of
ΔH
for
the
reverse
reaction
IB,
May
kJ.
oxides
one
of
reversible
nitrogen
is
reaction
shown
below:
is
N
210
for
methane.
involving
D.
2003
reaction.
4
amount
November
greater
O
2
[1]
(g)
⇋
2NO
4
Experimental
2006
(g)
ΔH
=
+58
kJ
2
represented
data
on
for
the
this
reaction
following
can
graph
be
(gure
8).
1.0
2
Sulfur
dioxide
and
oxygen
react
to
form
sulfur
product
according
to
the
3
trioxide
equilibrium:
0.8
(g)
+
O
2
How
(g)
⇋
2SO
2
are
the
md lom/noitartnecnoc
2SO
(g)
3
amount
of
SO
and
the
value
2
of
the
equilibrium
affected
A.
The
by
an
constant
increase
amount
of
SO
in
for
the
reaction
pressure?
and
the
value
of
0.6
reactant
0.4
0.2
the
3
equilibrium
constant
both
increase.
0
0
B.
The
amount
of
SO
and
the
value
of
2
4
6
8
10
the
3
time/min
equilibrium
constant
both
decrease.
▲
C.
The
amount
of
SO
increases
but
Figure 8
the
3
valueof
the
equilibrium
constant
a)
Write
an
expression
for
the
equilibrium
decreases.
constant,
K
,
for
the
reaction.
Explain
c
D.
The
amount
of
SO
increases
but
the
3
value
of
the
equilibrium
constant
the
signicance
of
the
horizontal
the
lines
graph.
parts
of
does
notchange.
on
the
State
what
can
be
[1]
deduced
about
the
magnitude
of
K
for
the
c
IB,
November
2007
reaction,
b)
3
What
and
will
the
happen
value
of
to
the
the
position
equilibrium
of
equilibrium
temperature
is
increased
in
constant
the
Use
Le
and
explain
Use
+
Cl
Le
(g)
⇋
2BrCl(g)
ΔH
=
+14
kJ
d)
the
the
of
to
predict
increasing
position
of
the
equilibrium. [2]
Châtelier’s
the
effect
principle
of
to
predict
increasing
the
and
pressure
State
position
and
of
explain
equilibrium.
the
effects
of
[2]
a
catalyst
2
onthe
the
forward
position
value
of
K
of
and
IB,
.
November
reverse
equilibrium,
reactions,
and
on
on
the
[6]
c
190
principle
effect
[1]
on
(g)
on
[4]
when
reaction?
2
the
reason.
following
explain
Br
a
Châtelier’s
temperature
c)
the
giving
2005
8
A C I D S
A N D
B A S E S
Introduction
The
theories
applications
and
of
the
of
Nature
scientists
and
of
laid
development
acids
acids
of
bases.
bases
as
bases
the
The
This
of
that
donors
for
work
a
and
the
acids
acceptors,
of
are
acid
The
of
measure
acids
of
examines
properties
ions
the
describes
and
respectively.
their
scientists
de nitions
chapter
theory
of
early
foundations
range
proton
and
work
Science.
the
a
Brønsted–Lowry
and
and
exemplify
of
the
all
general
acids
the
and
a
the
topic.
reactions
bases,
characteristics
explored
threat
the
concentration
deposition,
age
and
to
a
in
of
of
of
scale
as
hydronium
strong
detail.
product
the
and
pH
The
the
and
weak
chemistry
industrial
environment,
concludes
8.1 To o   
Understandings
Applications and skills
+
➔
A Brønsted–Lowry acid is a proton/H
donor
➔
Deduction of the Brønsted–Lowry acid and
+
and a Brønsted–Lowry base is a proton/H
base in a chemical reaction.
acceptor.
➔
➔
Deduction of the conjugate acid or conjugate
Amphiprotic species can act as both
base in a chemical reaction.
Brønsted–Lowry acids and bases.
➔
A pair of species diering by a single proton is
called a conjugate acid–base pair.
Nature of science
➔
Falsication of theories – HCN altering the theory
of
that oxygen was the element which gave a
false.
compound its acidic proper ties allowed for other
acid–base theories to develop.
➔
Theories
theory
of
being
super s ed ed
acidity
derived
–
f rom
➔
a
sour
taste,
but
this
ha s
been
proven
Public understanding of science – outside of the
arena of chemistry, decisions are sometimes
one
the
ea rly
referred to as “acid test” or “litmus test”.
s ensation
191
8
A C I D S
A N D
B A S E S
The role of acids and bases
The process of to
Acids
and
bases
are
familiar
in
our
everyday
lives
and
have
a
signicant
refers to the heating of materials
role
in
chemistry.
For
many
hundreds
of
years
scientists
have
been
to very high temperatures in
investigating
acid-base
reactions
and
developing
a
range
of
denitions
air in order to bring about their
and
theories
that
will
be
discussed
in
this
topic
and
topic
18.
Each
of
thermal decomposition (in the
these
theories
has
its
strengths
and
weaknesses
and
some
of
the
earliest
case of limestone), the removal
ideas
about
acids
and
bases
have
now
been
disproved.
of water from a hydrated
compound (for bauxite), or the
It
removal of a volatile matter from
The
has
long
minerals and ores.
associate
word
a
(ethanoic
acid).
A
derived
been
acid
understood
is
sour
derived
taste
acid),
with
lemon
base
that
from
the
is
that
from
many
juice
soluble
Arabic
acids
the
or
in
word
and
Latin
bases
acidus
substances
grapefruit
water
is
al-qaly
behave
that
(citric
called
as
meaning
are
meaning
acidic
acid),
an
and
alkali.
opposites.
sour
sour
The
calcined
and
such
we
as
milk
word
still
vinegar
(lactic
alkali
is
ashes.
Early theories about acids
Antoine
believe
Lavoisier’s
that
reactions,
source
of
oxygene
that
was
their
had
present
acidic
previously
Lavoisier’s
disproving
However,
acidic
oxygen
integral
in
the
his
in
been
work
that
led
acids
and
He
Greek)
was
the
subsequently
by
theory
even
the
from
Joseph
developed
time
presence
the
by
inuenced
by
differences
explain
The
being
by
and
a
served
the
of
new
laws,
on
and
to
used
to
propose
leads
Modern
trying
Evidence
is
theories
disproved,
theory.
focus
scientic
observation
these
or
bases,
important
and
community
supported,
bases
and
phenomenon.
and
testing
cultures
acids
an
curiosity
generalize
scientic
to
about
over
to
or
existing
even
theories
explain
on
why
are
they
sometimes
the
replaced
globe
to
theories,
theories
of
disproved,
generating
hypotheses.
1.1).
disproved.
around
in
specic
theories
experimentation
acids
Theories
later
develop
in
vocabulary
Early
endeavour
characteristic
the
if
purpose
name
element
(sub-topic
to
of
languages.
the
fundamental
due
meaning
to
was
gave
to
properties
were
him
combustion
discovered
phlogiston
belief
compounds
was
all
to
properties.
(“acid-forming”
Priestly.
of
investigations
oxygen,
in
react
in
the
way
they
do.
the
Arrhenius’s theory of acids and bases
Svante
August
Arrhenius
Chemistry
in
1903
for
an
as
a
substance
alkali,
a
soluble
(1859–1927)
his
work
in
the
was
eld
awarded
of
acids
the
and
Nobel
bases.
Prize
He
in
dened
+
acid
An
of
an
the
acid
and
base
combination
of
that
base,
is
in
produces
well
the
ionizes
known
hydrogen
water
to
hydroxide
as
a
ion
produce
ions,
OH
neutralization
and
the
hydrogen
.
The
H
involving
ion.
+
H
(aq)
+
OH
(aq)
→
H
O(l)
2
An
example
acid
in
the
of
this
type
stomach
of
with
neutralization
aluminium
is
the
hydroxide
reaction
of
contained
hydrochloric
in
an
antacid
tablet:
3HCl(aq)
+
Al(OH)
(s)
→
AlCl
3
Arrhenius’s
Figure 1 The reaction between the
base
vapours of concentrated ammonia
explained,
theory
ammonia
as
and
had
its
limitations.
hydrogen
ammonia
does
chloride
not
and hydrogen chloride solutions
NH
produces solid ammonium chloride,
visible as a white smoke
192
(g)
3
+
HCl(g)
→
(aq)
+
3H
3
NH
Cl(s)
4
O(l)
2
The
gas
contain
reaction
(gure
between
1)
hydroxide
could
ions.
.
combination
reaction
hydroxide
ions,
the
not
weak
be
8 . 1
T h e O r i e s
O f
a c i d s
a n d
b a s e s
TOK
Science in society
Terminology
community,
often
have
general
the
say
serves
is
as
often
life,
is
life
as
the
test
add
of
test”
Public
decisions
of
meaning
by
a
or
specic
“litmus
event
ability
to
is
within
everyday
or
a
to
of
as
understanding
of
scientic
associated
at
nal
science
ndings
vital
hypotheses is obtained in the
teacher
laboratory through observation
in
data
in
and experimentation. One
Science
everyday
assumption in this process
that
is that scientists are able to
uninuenced
is
understanding. Evidence to test
with
a
university.”
and
to develop knowledge and
in
examination
situations
and
variety of methodologies
can
example,
providing
experimentation,
about
is
the
study
perceived
For
Scientists employ a wide
scientic
phenomenon:
in
myriad
the
understanding,
life.
test”
performance
their
process
in
some
credibility
supported
bias.
informed
“acid
student’s
scientic
rigorous,
human
“the
to
meaning
certainty
litmus
used
accepted
communicating
term
the
that
the
an
different
of
a
has
clearly
a
testing
might
that
recreate conditions in the
by
laboratory that accurately
making
represent what is occurring in
issues.
the universe outside. How then
is this methodology used in
chemistry dierent from the
Brønsted–Lowry acids and bases
Scientists
of
often
information
subject
of
the
sometimes
idea
work
and
ideas
research.
discovering
simultaneously.
Thomas
Lowry
Arrhenius’s
Referring
could
be
collaboratively,
that
In
and
This
leads
other
developed
participating
to
a
cases
better
a
the
case
denition
in
work
theorizing
when
of
an
open
understanding
scientists
subsequently
was
methodologies employed in
and
of
the
independently,
about
Johannes
acids
other areas of knowledge?
exchange
the
same
Brønsted
bases
that
and
broadened
theory.
to
a
hydrogen
dened
as
a
ion
as
proton
a
proton,
donor
and
they
a
proposed
base
as
a
that
proton
an
acid
acceptor.
+
H
In
an
aqueous
solution
a
proton
can
be
represented
+
hydrogen
ion,
as
either
O
the
+
H
or
as
the
hydronium
ion,
H
O
.
The
hydronium
O
3
H
ionis
formed
when
a
water
molecule
forms
a
coordinate
bond
H
with
+
H
H
aproton
(gure
2).
For
H
example:
+
+
HCl(aq)
+
H
O(l)
→
H
2
Common
acids
hydrochloric
are
often
acid),
phosphoric
acid).
phosphoric
acid
O
(aq)
+
Cl
Figure 2 The hydronium ion, H
(aq)
referred
diprotic
(such
Hydrochloric
is
a
weak
O
3
3
to
as
as
and
being
sulfuric
sulfuric
monoprotic
acid),
acids
or
(such
triprotic
are
strong
as
(such
acids
as
while
acid.
+
HCl(aq)
→
H
(aq)
+
Cl
(aq)
+
H
SO
2
(aq)
→
2H
2
(aq)
+
SO
4
(aq)
4
+
H
PO
3
(aq)
⇋
3H
3
(aq)
+
PO
4
Ethanoic
(aq)
4
acid,
CH
COOH
is
also
a
weak
acid:
3
+
CH
COOH(aq)
⇋
CH
3
COO
(aq)
+
H
(aq)
3
or
+
CH
COOH(aq)
+
H
3
In
the
and
last
water
reaction,
O(l)
⇋
CH
2
equilibrium
is
the
acting
as
ethanoic
a
ethanoate
COO
+
H
acid
CH
O
(aq)
3
is
acting
Brønsted–Lowry
ion,
(aq)
3
COO
is
as
base.
a
Brønsted–Lowry
Focusing
acting
as
a
on
proton
the
acid
reverse
acceptor
and
3
193
8
A C I D S
A N D
B A S E S
the
hydronium
A to  is assumed to
base
undergo complete dissociation
ion
of
is
the
the
ion
as
a
proton
Brønsted–Lowry
conjugate
acid
donor.
acid
of
The
ethanoate
(ethanoic
another
acid),
ion
while
Brønsted–Lowry
is
the
the
conjugate
hydronium
base,
water.
in water (sub-topic 8.4).
The
conjugate
acid
and
base
differ
from
one
another
by
a
single
proton.
This
For example, in hydrogen
is
termed
a
conjugate
acid–base
pair.
Figure
3
shows
another
example.
chloride, HCl the hydrogen ion
has almost no anity for the
conjugate base 1
chloride ion.
A wk  undergoes only
par tial dissociation in water,
-
-
H
CO
2
establishing an equilibrium,
(aq) + OH
(aq) ⇋ HCO
3
(aq) + H
3
O(l)
2
and a solution of a weak acid is
only a weak electrolyte.
conjugate acid 2
Figure 3 Conjugate acid–base pairs in the neutralization of carbonic acid
(Brønsted–Lowry acid) with a hydroxide ion (Brønsted–Lowry base)
Quk quto
1
Identify
a)
H
the
bases
of
the
SO
2
b)
conjugate
following
d)
C
4
H
6
HNO
acids.
OH
5
e)
OH
f)
H
3
c)
C
H
2
2
OH
5
Identify
O
2
the
conjugate
acids
of
the
following
bases.
2
a)
OH
d)
CO
b)
H
e)
HNO
c)
NH
f)
C
3
O
2
3
3
3
H
2
Identify
the
conjugate
acid–base
pairs
NH
5
2
HCO
(aq)
+
2
in
the
following
equations:
2
S
(aq)
⇋
HS
(aq)
+
CO
3
(aq)
3
2
CH
COOH(aq)
+
HPO
3
(aq)
⇋
CH
4
COO
(aq)
+
H
3
PO
2
(aq)
4
Amphiprotic species
Some
a
substances
part.
These
molecule
Lowry
acid.
N
3
are
donate
It
can
the
base
species
can
Polyprotic
+
H
have
Brønsted–Lowry
said
a
also
species
are
ability
to
depending
to
be
proton
accept
in
a
act
on
as
the
either
amphiprotic.
a
reaction,
proton,
frequently
a
Brønsted–Lowry
reaction
involved
For
thus
acting
in
in
which
example,
acting
as
a
they
as
a
the
in
or
taking
water
Brønsted–
Brønsted–Lowry
reactions
acid
are
which
base.
they
-
CH
COO
behave
amphiprotically
for
example:
2
H
PO
2
(aq)
+
OH
(aq)
⇋
HPO
4
(aq)
+
H
4
O(l)
2
+
H
PO
2
R
Amino
acids
(aq)
+
H
4
O
(aq)
⇋
H
3
acids
(sub-topic
contain
a
PO
3
weakly
B.2)
also
acidic
(aq)
+
4
act
as
carboxyl
H
O(l)
2
amphiprotic
group
and
a
species.
weakly
All
2-amino
basic
amino
zwitterion
group.
Figure 4 A 2-amino acid is amphiprotic
as
an
In
acid
presence
194
the
in
of
ionized
the
a
form
presence
strong
acid
(a
of
it
zwitterion,
a
strong
acts
as
a
gure
base,
base
4)
the
donating
and
accepts
a
a
compound
proton.
proton.
In
acts
the
8 . 2
p r O p e r T i e s
O f
a c i d s
a n d
b a s e s
8.2 po t o   
Understandings
Applications and skills
➔
Most acids have obser vable characteristic
Balancing chemical equations for the reactions
➔
chemical reactions with reactive metals, metal
of acids.
oxides, metal hydroxides, hydrogencarbonates,
Identication of the acid and base needed to
➔
and carbonates.
make dierent salts.
➔
Salt and water are produced in exothermic
Candidates should have experience of acid–
➔
neutralization reactions.
base titrations with dierent indicators.
Nature of science
➔
Obtaining evidence for theories – obser vable proper ties of acids and bases have led to the modication
of acid–base theories.
Acid–base theories
For
scientists,
provide
refute
sense
experimentation
evidence
the
of
that
theories
our
can
we
and
either
have
observation
support
formulated
has
or
to
come
and
make
world.
have
about
been
the
data,
rationalizing
reactions
modi ed
over
of
time
acids
as
and
more
observed
Scientists
quantitative
de ning
Theories
from
bases.
properties
analyse
establishing
discrepancies
of
acids
qualitative
patterns
with
the
and
and
goal
of
relationships.
bases
evidence
Properties of acids and bases
Acids
and
bases
(concentrated
block
domestic
remover,
perform
sodium
and
changing
many
useful
hydroxide)
commercial
iron(III)
functions
dissolves
drains.
oxide
(Fe
.
Ammonia
is
used
as
a
general
daily
and
Phosphoric
O
2
FePO
in
grease
,
rust)
acid
into
life.
oil
is
Caustic
deposits
an
iron(III)
soda
that
effective
can
rust
phosphate,
3
cleaner
while
mild
acids
such
as
4
vinegar
Table
▲
1
are
sometimes
shows
some
put
on
wasp
properties
of
stings,
acids
which
and
are
alkaline.
bases.
a
b
taste sour
taste bitter
pH < 7.0
pH > 7.0
litmus is red
litmus is blue
phenolphthalein is colourless
phenolphthalein is pink
methyl orange is red
methyl orange is yellow
T
able 1 General proper ties of acids and bases and their effects on some common indicators
195
8
A C I D S
A N D
B A S E S
The reactions of acids with metals, bases,
Tt o yo
and carbonates
Putting a lighted splint in
Most
acids
react
with
metals,
metal
oxides,
hydroxides,
hydrogencarbonates,
hydrogen gas results in a
and
carbonates.
combustion reaction, and
a “squeaky pop” is heard.
Hydrogen gas is highly
ammable.
2H
(g) + O
2
All
as
these
reactions
“common
compound
(g) → 2H
2
salt”
produce
but
this
composed
of
a
is
an
salt.
just
Sodium
one
anion
chloride
example
and
of
a
is
referred
salt,
which
to
is
a
cation.
O(g)
2
It
is
important
to
understand
how
a
salt
is
derived
from
an
acid
and
a
+ energy
base.
The
following
reactions
illustrate
the
formation
of
a
wide
variety
ofsalts.
Metals
react
that
with
are
found
acids
to
above
form
acid
+
2HCl(aq)
a
hydrogen
salt
and
metal
+
→
Zn(s)
in
salt
→
the
hydrogen
+
ZnCl
activity
SO
2
(aq)
2CH
(aq)
+
Fe(s)
→
+
H
These
COOH(l)
The
salt
+
Figure 1 Hydrogen gas
with
of
give
the
2Na(s)
off
metal
produced
example,
(aq)
+
H
⇋
2CH
(g)
2
COONa(l)
+
H
3
reactions
reactivity
(g)
4
3
hydrogen
and
the
depends
magnesium
hydrochloric
on
gas
the
is
at
acid
a
(g)
2
strength
chloride
9.1)
2
FeSO
4
(sub-topic
hydrogen
2
H
series
gas:
different
and
from
chloride
rates
according
concentration
which
salt
it
was
of
the
to
the
acid.
produced,
produced
in
a
for
reaction
acid.
explodes upon ignition
The
standard
change
reaction
with
between
conditions.
an
enthalpy
associated
This
exothermic
a
change
the
strong
enthalpy
process
of
neutralization
formation
acid
and
change
(sub-topic
a
has
of
1
mol
strong
a
of
base
negative
is
water
under
value
the
energy
from
the
standard
–
neutralization
5.1):
+
H
(aq)
+
OH
(aq)
→
H
O(l)
2
The
salt
from
produced
the
include
which
base
and
metal
is
a
in
neutralization
an
anion
hydroxides,
weak
from
metal
reactions
the
acid.
oxides,
2HCl(aq)
+
SO
2
+
base
Ca(OH)
(aq)
→
salt
→
CaCl
(aq)
COOH(aq)
dissolves
+
CaO(s)
→
in
a
cation
of
bases
hydroxide,
water
(aq)
+
2H
+
NH
OH(aq)
⇋
CH
not
to
(s)
+
H
directly
an
COONH
with
alkaline
O(l)
2
3
react
create
O(l)
2
4
does
water
+
CaSO
4
oxide
of
examples
ammonium
2
4
3
Calcium
Common
and
2
CH
composed
base:
acid
H
is
(aq)
+
H
4
aqueous
solution
of
O(l)
2
acids.
This
calcium
base
hydroxide,
Figure 2 Zinc reacting with
which
neutralizes
the
acid:
hydrochloric acid
CaO(s)
+
H
O(l)
→
Ca(OH)
2
Many metal oxides act as bases
Calcium
an
hydroxide
alkali.
Many
(aq)
2
is
slightly
other
bases,
soluble
in
such
iron(II)
in aqueous solutions whereas
hydroxide,
most non-metal oxides are
acidic (sub-topic 3.2).
196
are
insoluble
in
water.
as
water.
A
soluble
hydroxide
base
or
is
called
aluminium
is
8 . 3
Metal
carbonates
carbon
acid
dioxide
+
metal
carbon
and
and
hydrogencarbonates
with
acids
to
+
p h
s c a l e
produce
water:
Tt o o ox
carbonate/metal
dioxide
react
T h e
hydrogencarbonate
→
salt
Passing carbon dioxide gas
+
through limewater (calcium
water
hydroxide) results in a cloudy
2HCl(aq)
+
Na
CO
2
HCl(aq)
+
(s)
→
2NaCl(aq)
+
CO
3
NaHCO
(g)
+
H
2
(s)
→
NaCl(aq)
+
CO
3
O(l)
2
(g)
+
H
2
(milky) suspension of insoluble
O(l)
calcium carbonate, CaCO
2
:
3
Ca(OH)
(aq) + CO
2
(g) →
CaCO
2
(s)
3
Acid–base titrations
+ H
O(l)
2
A
titration
involves
with
a
a
a
progress
loggers
into
of
point
the
the
to
1.3)
a
volumetric
between
a
substance
solution
solution
reaction
combined
plotted
colour
reaction
is
standardized
burette
be
(sub-topic
with
produce
change
as
(sub-topic
the
a
being
can
a
pH
pH
(the
be
of
titrant).
monitored
curve.
can
An
be
in
titrant
small
using
used
and
that
concentration
is
delivered
increments.
several
to
acid–base
approaches
technique
unknown
The
analysed,
probe
titration
analysis
techniques.
collect
indicator
reaches
from
The
data
that
undergoes
the
Data
can
a
equivalence
18.3).
Figure 3 Limewater provides
a test for carbon dioxide
The colour changes of dierent indicators can be found in section 22 of the Data
booklet
8.3 T h 
Understandings
Applications and skills
+
➔
➔
pH =
-log [H
+
(aq)] and [H
pH
] =
10
+
➔
Solving problems involving pH, [H
], and [OH
➔
Students should be familiar with the use of a
].
A change of one pH unit represents a 10-fold
+
change in the hydrogen ion concentration [H
].
pH meter and universal indicator.
➔
pH values distinguish between acidic, neutral,
and alkaline solutions.
+
➔
The ionic product constant, K
=
[H
][OH
Nature of science
] =
w
14
10
at 298 K .
➔
Occam’s razor – the pH scale is an attempt to
scale the relative acidity over a wide range of
+
H
concentrations into a very simple number.
197
8
A C I D S
A N D
B A S E S
Occam’s razor
Scientic
theories
can
be
comprehensive
models
works.
from
Derived
of
complex
how
while
and
the
analysis,
and
hypothesis,
they
elegant
and
also
high
capacity
for
gaining
pH
scale
is
a
very
effective
method
of
can
representing
be
a
experimentation,
The
observations,
maintaining
understanding.
universe
a
continuous
range
of
hydrogen
ion
imposing.
+
concentration
The
for
of
principle
the
Occam’s
development
elds
theory
of
of
of
knowledge.
should
remain
razor
is
theories
Its
as
a
in
blueprint
a
number
philosophy
simple
as
is
ease
is
that
and
a
of
]
as
simple
interpretation
the
general
distinguishes
possible
[H
for
basic/alkaline
both
population.
between
numbers
students
The
acids,
that
pH
of
scale
neutral
enable
science
clearly
solutions,
and
solutions.
The pH scale
The
pH
scale
is
a
simple
and
effective
way
of
representing
the
+
concentration
of
hydrogen
ions,
[H
]
in
a
solution.
This
concentration
+
is
often
very
low;
for
example,
in
water
[H
7
]
=
1.0
×
10
3
mol
dm
.
+
Comparing
result
in
values
ratios
that
+
[H
∶
[H
1
Scientists
:
0.000
oven
employ
treatment
and
pH
for
different
difcult
to
substances
to
water
can
comprehend:
of
a
cleaner
a
number
to
ion
of
of
is
and
dened
use
results
valid
by
mathematical
The
concentrations
non-scientists
solution
different
data.
the
for
=
-log
[H
of
a
in
a
following
approaches
logarithmic
simple
scientists
+
pH
relative
001
presentation
hydrogen
accessible
The
are
]
]
water
is
[H
+
]
display
of
visual
(gure
two
scale
to
the
to
scale
that
1).
expressions:
+
(aq)]
or
pH
=
-log
[H
O
(aq)]
3
+
[H
pH
]
=
10
neutral
acidic
pH
0
1
2
alkaline
3
4
5
6
7
gastric
pure
juice
water
8
9
10
11
12
13
14
oven
detergent
cleaner
Figure 1 The pH scale
h
+
[h
]
As
1
the
pH
scale
equivalent
2
7
10
14
a
logarithmic
10-fold
change
scale
in
to
the
base
10,
hydrogen
a
change
ion
of
1
pH
unit
concentration.
A
is
small
in
the
pH
of
a
solution
represents
a
large
change
in
the
hydrogen
5
ion
1 × 10
concentration
(table
1).
Note
that
the
pH
scale
has
no
units.
7
1 × 10
10
1 × 10
Calculating pH
14
1 × 10
T
able 1 pH values and their corresponding
hydrogen ion concentrations
198
a
2
Strong
acids
aqueous
▲
to
1 × 10
change
5
is
1
1 × 10
and
strong
solutions.
bases
Therefore
are
the
assumed
to
completely
concentration
of
a
ionize
strong
in
monoprotic
8 . 3
acid
will
mol
dm
be
the
same
as
the
concentration
of
the
hydrogen
ion.
A
T h e
p h
s c a l e
0.1
TOK
3
+
solution
of
hydrochloric
acid
equates
to
[H
3
]
=
0.1
mol
dm
The language of mathematics
is integral to the development
Ionization of water
of scientic theories based on
The
pH
scale
covers
both
the
acidic
and
alkaline
regions
of
aqueous
the analysis of quantitative
systems.
When
we
to
considering
solutions
involving
strong
and
weak
bases,
data. Chemists use the
need
examine
the
relationship
between
hydrogen
ion
and
language of mathematics as a
hydroxide
ion
concentrations.
Water
undergo
means of communicating their
can
equilibrium
auto-ionization,
expression
(sub-topic
according
to
the
following
ndings. It is well accepted
7.1):
that language plays a pivotal
role in the communication of
+
H
O(l)
⇋
H
(aq)
+
OH
(aq)
2
knowledge. Language not only
+
[H
][OH
]
_
K
allows us to form descriptions
=
c
[H
O]
of our experiences, it can also
2
+
as
[H
O]
is
constant,
K
2
=
[H
14
][OH
]
=
1.0
×
10
at
298
inuence the way we view
K.
w
and subsequently understand
This
expression
is
the
ion
product
constant
for
water.
In
pure
water,
our interactions with the
_________
+
[H
14
]
=
[OH
]
√
=
1.0
×
7
10
=
1.0
×
universe. For what reasons is
10
it important to have only one
“scientic” language?
Worked examples: calculating pH
Example 1
+
A
solution
of
fresh
milk
has
a
pH
of
6.70.
Calculate
[H
]
and
[OH
].
Solution
+
[H
pH
]
=
10
6.70
=
7
10
=
2.0
×
10
+
K
=
[H
×
10
3
mol
dm
14
][OH
]
=
1.0
w
14
14
1.0 × 10
__
[OH
]
=
1.0 ×
10
__
=
+
7
[H
]
×
10
2.0
×
8
=
5.0
10
3
mol
dm
Example 2
2
Calculate
the
pH
of
a
1.0
×
3
10
mol
dm
solution
of
sodium
hydroxide.
Solution
Sodium
hydroxide
is
a
strong
base
that
completely
ionizes
in
water:
+
NaOH
(aq)
→
Na
(aq)
+
OH
2
[OH
]
=
1.0
×
10
3
mol
dm
+
K
=
[H
(aq)
14
][OH
]
=
1.0
×
10
w
14
1.0 × 10
__
+
[H
]
=
[OH
]
14
1.0 × 10
__
+
[H
]
=
12
=
1.0
×
10
3
mol
dm
2
1.0
×
10
+
pH
=
-log
[H
(aq)]
=
-log
(1.0
12
×
10
)
=
12.00
199
8
a c i d s
a n d
b a s e s
pH and acid–base titrations
The
analytical
laboratory
by
the
for
technique
the
addition
illustrates
the
of
past
an
features
data
be
(gure
can
titration
years.
indicator
progress
characteristic
that
of
200
of
of
an
the
collected
(sub-topic
Traditionally
(sub-topic
acid–base
titration.
using
a
pH
1.3)
a
18.3).
titration
These
probe
been
Plotting
and
curves
and
has
titration
its
is
a
pH
enables
are
used
in
the
monitored
curve
analysis
generated
associated
of
from
software
2).
Figure 2 A pH probe can be used to
collect data during an acid–base
titration
8.4 sto  wk   
Understandings
Applications and skills
➔
Strong and weak acids and bases dier in the
➔
Distinction between strong and weak acids and
extent of ionization.
bases in terms of the rates of their reactions
➔
Strong acids and bases of equal concentrations
have higher conductivities than weak acids and
bases.
➔
A strong acid is a good proton donor and has a
with metals, metal oxides, metal hydroxides,
metal hydrogencarbonates, and metal
carbonates, and their electrical conductivities
for solutions of equal concentrations.
weak conjugate base.
➔
A strong base is a good proton acceptor and has
a weak conjugate acid.
Nature of science
➔
Improved instrumentation – the use of advanced analytical techniques has allowed the relative
strengths of dierent acids and bases to be quantied.
➔
Looking for trends and discrepancies – patterns and anomalies in relative strengths of acids and bases
can be explained at the molecular level.
➔
The outcomes of experiments or models may be used as fur ther evidence for a claim – data for a
par ticular type of reaction suppor ts the idea that weak acids exist in equilibrium.
200
8 . 4
s T r O n g
a n d
W e a K
a c i d s
a n d
b a s e s
Predictions, patterns, and anomalies
Advances
in
techniques
computing
(including
between
research
this
looking
data
gain
greater
of
acids
to
be
strengths
institutes
for
bases
with
of
allows
high
different
chemical
behaviour
evidence
also
Science
trends
and
recognizes
modern
certainty.
Science
strengthened
be
by
developed
idea
there
an
that
are
in
allow
at
networking
Scientists
patterns
on
the
and
chemical
anomalies
molecular
acids
provides
human
exist
in
all
data
to
strengths
behaviour
in
the
of
relative
their
level.
in
analyse
anomalies
relative
explanations
the
weak
analytical
and
data.
their
and
limitations
inherently
advances
of
data
about
bases
of
meters),
wealth
Patterns
instrumentation
is
pH
Quantitative
and
the
that
a
predictions
acids
to
development
as
discrepancies,
reliability.
supports
however,
the
such
provide
understanding.
and
made
power,
sensors
Empirical
equilibrium.
measurements;
that
is
endeavour
close
which
to
is
technology.
Strengths of acids and bases
Sulfuric acid is a strong acid.
The
strength
of
an
acid
or
base
depends
on
the
degree
to
which
it
ionizes
However, the rst ionization
or
dissociates
in
water.
A
strong
acid
is
an
effective
proton
donor
is complete but the second
that
is
assumed
to
completely
dissociate
in
water.
Examples
include
dissociation is incomplete so
hydrochloric
acid,
HCl,
sulfuric
acid,
H
SO
2
,
and
nitric
acid,
HNO
4
:
3
an equilibrium sign is used:
+
HCl(aq)
+
H
O(l)
→
H
2
O
(aq)
+
Cl
(aq)
H
3
SO
2
(aq) + H
4
O(l) →
2
+
HSO
+
H
SO
2
(aq)
+
H
4
O(l)
→
H
2
O
(aq)
+
HSO
3
(aq)
(aq) + H
4
O
(aq)
3
4
+
HNO
(aq)
+
H
3
O(l)
→
H
2
O
(aq)
+
NO
3
HSO
(aq)
(aq) + H
4
3
O(l) ⇋
2
2
SO
These
to
go
base.
reactions
to
represented
completion.
For
which
are
The
hydrochloric
has
almost
no
by
chemical
conjugate
acid,
the
afnity
base
of
conjugate
for
a
proton
a
equations
strong
base
in
is
that
acid
the
aqueous
is
are
a
very
chloride
+
(aq) + H
4
assumed
O
(aq)
3
weak
ion,
Cl
solution.
stuy t
A
weak
donor.
acid
The
dissociation
equilibrium
acid
dissociates
(sub-topic
molecules
higher
have
afnity
for
a
only
of
a
partially
weak
7.1).
At
is
water;
a
than
The
only
conjugate
does
the
it
is
reversible
equilibrium
dissociated.
proton
acid
in
a
a
poor
reaction
small
base
conjugate
proton
of
that
reaches
proportion
a
weak
base
of
a
acid
of
the
has
strong
a
acid.
The terms “ionization”
and “dissociation” are
interchangeable and are
both equally acceptable in
examination answers.
+
CH
COOH(aq)
+
H
3
O(l)
⇋
H
2
O
(aq)
+
CH
3
COO
(aq)
3
+
H
CO
2
In
the
reactions
accepting
The
(aq)
terms
distinct
+
H
3
a
O(l)
⇋
H
2
of
both
(aq)
+
HCO
strong
and
(aq)
3
weak
acids,
water
is
acting
as
a
base,
proton.
“strong”
from
and
“weak”
“concentrated”
when
and
cott
sto
O
3
applied
“dilute”.
to
Table
acids
1
bases
some
are
quite
examples.
dut
3
6 mol dm
and
gives
3
HCl
0.5 mol dm
HNO
3
3
Wk
10 mol dm
3
CH
3
COOH
0.1 mol dm
H
2
CO
3
T
able 1 Some examples of strong, weak, dilute, and concentrated acid solutions
201
8
A C I D S
A N D
B A S E S
A
amot: Species that
strong
base
hydroxides
are
also
all
completely
soluble
in
dissociates
water
and
in
are
water.
good
The
group
examples
of
1
metal
strong
bases:
behaves both as an acid and a
+
NaOH(aq)
→
Na
(aq)
+
OH
(aq)
base eg aluminium hydroxide:
+
KOH(aq)
→
K
(aq)
+
OH
(aq)
Acting as a base:
A
Al(OH)
metal
hydroxide
does
not
act
as
a
Brønsted–Lowry
base
because
it
(s) + 3HCl(aq) →
3
does
AlCl
(aq) + 3H
3
not
have
the
capacity
to
accept
a
proton.
However,
in
solution
the
O(l)
2
hydroxide
ion
acts
as
a
Brønsted–Lowry
base,
accepting
a
proton:
Acting as an acid:
+
OH
(aq)
+
H
O
(aq)
→
2H
3
Al(OH)
O(l)
2
(s) + NaOH(aq) →
3
Ammonia
Na[Al(OH)
is
an
example
of
a
weak
base.
In
the
reaction
with
water,
](aq)
4
ammonia
accepts
a
proton
and
effectively
undergoes
ionization.
amot: One type of
+
NH
amphoteric species are
amphiprotic molecules. These
can act as Brønsted-Lowry acids
(aq)
+
H
3
In
this
O(l)
⇋
NH
2
reaction
water
Brønsted–Lowry
(aq)
+
OH
(aq)
4
displays
acid,
its
donating
amphiprotic
a
nature
by
acting
as
a
proton.
(proton donors) or Brønsted-
Lowry bases (proton acceptors)
eg water and amino acids.
Experimental determination of the strength of
acids and bases
A
number
of
techniques
concentration,
one
so
that
can
they
be
can
used
be
to
compare
assigned
an
acids
order
of
and
bases
strength
of
equal
relative
to
another.
Conductivity
All
acids
and
conductivity
present.
and
graphite
can
current
passing
they
the
acids
a
be
be
through
and
a
degree
in
a
connected
identical
bases
higher
to
solution
measured
for
of
are
of
water
to
an
on
and
the
experiment
ammeter
solution
so
solutions
using
(gure
that
of
create
ions.
The
concentration
any
different
a
1).
of
power
The
voltage
difference
acids
or
ions
pack
in
bases
ions.
strong
conductivity
comparison
in
depends
simple
each
aqueous
concentration
display
example,
dissociate
aqueous
electrodes
must
Strong
strong acid. The lamp gives a qualitative
This
an
applied
reects
Figure 1 Testing the conductivity of a
bases
of
the
electrolytes
than
weak
conductivity
(sub-topic
acids
of
and
19.1),
bases.
equimolar
so
For
solutions
reading of current; an ammeter would
of
hydrochloric
acid
and
ethanoic
acid
would
demonstrate
that
the
provide quantitative readings
hydrochloric
higher
weak
acid
degree
of
gives
a
higher
dissociation
ammeter
than
reading
ethanoic
acid,
and
so
which
has
is
a
a
acid.
Energy changes on neutralization
Neutralization
reaction
is
exothermic
neutralization
The
strong
enthalpy
water
202
for
a
neutralization
reaction,
A
occurs
so
or
(ΔH
strong
the
base
is
consideration
from
hydrogen
<
an
0,
acid
reaction
driving
acid
when
acid
sub-topic
is
almost
removes
reaction
to
completely
in
and
this
and
a
base
5.1).
react
The
identical
ionized
to
species
together.
enthalpy
that
for
from
change
a
the
The
weak
of
acid.
dissociation
completion.
dissociated
reaction
hydroxide
is
the
ions.
in
solution
exothermic
so
the
only
formation
of
8 . 4
For
a
neutralization
reaction
involving
a
weak
s T r O n g
acid
or
a n d
base
W e a K
there
a c i d s
a n d
b a s e s
are
TOK
other
enthalpy
undissociated
considerations.
forms
in
These
aqueous
species
solution.
exist
The
predominantly
ionization
of
a
in
weak
their
acid
We use our senses – sight,
or
base
is
mildly
endothermic.
Therefore
the
enthalpy
of
neutralization
touch, smell, taste, and
for
a
strong
base–weak
acid
reaction
will
be
slightly
less
exothermic
than
hearing – to perceive the
that
for
a
strong
base–strong
acid
reaction.
The
weaker
the
acid,
the
world and develop our
more
endothermic
the
dissociation
reaction
becomes
and
thus
the
lower
understanding. Technology
the
enthalpy
change
of
neutralization.
allows us to extend our
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
senses, unveiling new
O(l)
2
1
ΔH
=
−57.1
kJ
knowledge and experiences
mol
neutralization
that may challenge the
boundaries of our current
HCl(aq)
+
NH
(aq)
⇋
NH
3
Cl(aq)
4
understanding. How does
1
ΔH
=
−53.4
kJ
mol
neutralization
this increased sensory
perception aect our view
CH
COOH(aq)
+
NaOH(aq)
→
CH
3
COONa(aq)
+
H
3
O(l)
2
of the world? How might
1
ΔH
=
−56.1
kJ
mol
neutralization
reasoning based on evidence
and discussion help us to
CH
COOH(aq)
+
NH
3
(aq)
⇋
CH
3
COONH
3
(aq)
decide if this new knowledge
4
1
ΔH
=
−50.4
kJ
mol
changes or reinforces our
neutralization
view of the world?
Monitoring the rate of a reaction
3
The
reactions
metals,
metal
carbonates
all
of
strong
produce
(topic
through
observation
and
weak
acids
hydrogencarbonates,
reaction
analysis
and
6)
can
a
be
gas.
The
and
rate
determined
(gure
quantitatively
2)
by
The
with
metal
of
of
the
by
which
gas
is
evolved
monitoring
through
loss
of
hydrogencarbonate
of
the
mass.
reaction
shown
of
such
experiments
enables
a
strong
distinguished
from
a
weak
1
mol
ethanoic
sodium
acids
strong
solutions
with
carbonate,
demonstrate
by
dm
and
the
zinc
and
sodium
different
weak
rates
acids.
rate
acid
3)
the
reactions
enables
mass
on
data
an
to
electronic
be
balance
collected
over
to
time.
be
of
A
(gure
series
and
powdered
Performing
at
reactions
hydrochloric
granules,
qualitatively
followed
individual
Graphing
these
results
illustrates
the
differences
acid.
in
the
initial
rate
of
reaction
(sub-topic
16.1).
Figure 2 Obser vation provides qualitative
Figure 3 Monitoring the loss of mass provides
data for a reaction that evolves a gas
quantitative data for a reaction that evolves a gas
203
8
a c i d s
a n d
b a s e s
8.5 a oto
Understandings
Applications and skills
➔
Rain is naturally acidic because of dissolved
Balancing the equations that describe the
➔
CO
and has a pH of 5.6. Acid deposition has a
2
combustion of sulfur and nitrogen to their
pH below 5.6.
oxides and the subsequent formation of H
SO
2
➔
Acid deposition is formed when nitrogen or
H
SO
2
sulfur oxides dissolve in water to form HNO
, H
2
➔
SO
2
and H
4
4
,
3
and HNO
2
3
,
3
HNO
, HNO
Distinction between the pre-combustion and
➔
SO
2
3
post-combustion methods of reducing sulfur
Sources of the oxides of sulfur and nitrogen
oxides emissions.
and the eects of acid deposition should be
Deduction of acid deposition equations for acid
➔
covered.
deposition with reactive metals and carbonates.
Nature of science
➔
Risks and problems – oxides of metals and non-metals can be characterized by their acid–base proper ties.
➔
Acid deposition is a topic that can be discussed from dierent perspectives.
➔
Chemistry allows us to understand and reduce the environmental impact of human activities.
Acid deposition
Acid
are
deposition
deposited
on
is
the
the
economicdevelopment
increasing
emissions
acidrain,
Acid
the
deposition
deforestation,
acid
levels
plants,
of
most
toxic
shing
in
metals
industry
marble,
ofthe
the
and
in
by
many
the
and
of
of
and
of
rivers,
pH
the
of
metal
in
other
world
ledto
that
ways.
soils
of
and
rapidly
cause
health,
from
can
soil
by
uptake
affect
corrosive
and
include:
elevated
increased
which
and
bridges,
to
minerals
systems,
life
These
leading
toxic
marine
people’s
oxides
many
river
buildings,
have
deposition.
from
and
pollutants
industrialization
sulfur
uptake
lake
and
the
acid
minerals
ultimately
and
parts
form
acid-forming
Increased
environment
shellsh
limestone,
which
nitrogen
leaching
lakes
reduction
in
by
surface.
prevalent
affects
the
process
Earth’s
the
effects
on
vehicles.
Acid rain
Pure
water
presence
H
CO
2
.
A
has
of
a
pH
of
dissolved
typical
pH
7.0.
Rainwater
carbon
value
of
dioxide
is
naturally
which
rainwater
is
forms
5.6.
3
CO
(g)
+
H
2
O(l)
⇋
H
2
CO
2
(aq)
3
+
H
CO
2
(aq)
⇋
H
(aq)
+
HCO
3
HCO
(aq)
3
204
(aq)
3
+
⇋
H
(aq)
2
+
CO
(aq)
3
acidic
weak
due
to
the
carbonic
acid,
8 . 5
Acid
rain
has
deposition
a
are
pH
less
sulfur
than
5.6.
dioxide,
The
SO
major
and
pollutants
nitrogen
that
oxides,
cause
NO
and
and
are
the
products
of
natural
decomposition
of
d e p O s i T i O n
acid
NO
2
These
a c i d
.
a oto: a
2
occurrences
vegetation,
as
such
well
as
as
volcanic
eruptions
man-made
o om
primary
Acid deposition, a secondary
pollutants
from
the
combustion
of
fossil
fuels
containing
high
levels
pollutant, can take many
of
sulfur
impurities
(optionC.2).
Acid
rain
results
principally
from
the
dierent forms including rain,
formation
of
two
strong
acids,
nitric
acid,
HNO
,
and
sulfuric
acid,
H
3
SO
2
4
snow, fog and dry dust. The
and
can
be
considered
as
a
major
global
environmental
problem.
components of acid deposition
For
a
example,
car
or
oxide,
a
jet
at
high
engine,
nitrogen(II)
temperature
nitrogen
oxide
gas
in
the
reacts
(nitrogen
internal
with
combustion
oxygen
gas
to
engine
form
of
the
(the primary pollutants) may
be generated in one country
monoxide):
and depending on climate
patterns may be deposited
High
Temperature
in neighbouring countries or
N
(g)
+
O
2
(g)
→
2NO(g)
even dierent continents.
2
There are no boundaries
On
reaction
with
oxygen,
the
oxide,
nitrogen(IV)
oxide
(nitrogen
for acid deposition. For
dioxide)
can
form:
example in Europe industrial
2NO(g)
+
O
(g)
→
2NO
2
Nitrogen(IV)
air
pollution
USA
and
(g)
conurbations in countries
2
oxide
causes
which
Mexico
is
the
often
brown
observed
colour
in
of
cities
smog,
such
as
a
common
Los
type
Angeles
in
of
the
such as Germany and the
UK may act as the source of
acid rain but due to factors
City.
such as prevailing winds,
The
reaction
and
nitrous
between
water
and
nitrogen(IV)
oxide
produces
nitric
acid
acid deposition may occur in
acid:
Scandinavian countries fur ther
2NO
(g)
+
H
2
O(l)
→
HNO
2
(aq)
+
HNO
3
nor th such as Norway and
(aq)
2
Sweden. Hence the eects
Another
oxide
of
nitrogen,
NO,
is
easily
oxidized
to
nitrogen(IV)
oxide
of acid rain may occur away
by
atmospheric
oxygen:
from the actual source leading
2NO(g)
+
O
(g)
→
2NO
2
(g)
to widespread deforestation
2
and pollution of lakes and
Nitrous
acid
can
be
also
oxidized
by
atmospheric
oxygen:
river systems. National and
2HNO
(aq)
+
O
2
Therefore,
(g)
→
2HNO
2
all
oxides
(g)
regional environmental
3
of
nitrogen
eventually
produce
nitric
acid,
protection agencies throughout
HNO
3
the world collaborate in an
Sulfur
dioxide
combines
with
water
to
form
sulfurous
acid:
eor t to better understand
SO
(g)
+
H
2
O(l)
⇋
H
2
SO
2
and control acid deposition.
(aq)
3
The US Environmental
+
H
SO
2
(aq)
+
H
3
O(l)
⇋
HSO
2
(aq)
+
H
3
O
(aq)
3
Protection Agency and the Acid
Some
coal
can
contain
almost
3%
sulfur.
On
combustion,
sulfur
dioxide
forms:
Deposition Monitoring Network
in East Asia (EANET) websites
provide data that can be used
S(s)
+
O
(g)
→
SO
2
(g)
2
in the discussion of secondary
On
subsequent
sulfur
trioxide
2SO
(g)
+
reaction
is
O
2
Sulfur
SO
oxygen
in
the
atmosphere,
the
oxide,
generated:
(g)
⇋
2SO
2
trioxide
sulfuric
with
can
pollutants and their political
implications.
(g)
3
then
react
with
rain
in
the
atmosphere,
to
form
acid:
(g)
3
+
H
O(l)
2
→
H
SO
2
(aq)
4
205
8
A C I D S
A N D
B A S E S
Pre- and post-combustion technologies
Pre-combustion
techniques
used
methods
on
fuels
mineral
beneciation
reduces
the
different
of
amounts
technologies
dioxins
and
CaO(s)
to
their
sulfur
and
methods
emissions
combustion.
crushing
sulfur
methods
remove
from
atmosphere.
dioxide
of
reduce
coal,
other
followed
impurities.
result
in
the
refer
Physical
by
to
cleaning
otation
Combinations
removal
of
up
to
or
that
of
80–90 %
sulfur.
Post-combustion
and
involves
pre-combustion
inorganic
to
before
For
the
+
SO
(g)
it
→
ue
CaSO
2
several
complementary
nitrogen
gases
calcium
from
on
dioxide,
combustion
example,
remove
focus
sulfur
before
oxide
or
oxides,
they
lime
are
will
heavy
metals
released
react
with
into
the
sulfur
gases:
(s)
3
The eects of acid rain on buildings
Limestone
are
of
and
commonly
signicant
the
world.
marble
used
in
cultural
Both
are
building
monuments
importance
contain
calcium
materials
and
that
buildings
throughout
When
calcium
neutralization
gradually
eroded,
(s)
+
H
3
only
in
their
structural
is
exposed
occurs
causing
and
to
the
signicant
acid
rain,
building
a
is
damage.
carbonate,
CaCO
differing
carbonate
reaction
SO
2
(aq)
→
CaSO
4
(s)
+
4
CO
(g)
+
H
2
O(l)
2
composition.
The role of chemists in studying acid deposition
Science,
to
and
chemistry
understand
occurs
and
environment.
is
wide
interactions
206
Acid
–
which
its
of
cross
G:
particular,
in
of
study
and
option
Environmental
5.8:
ways
extent
The
ranging
Geography
Global
the
the
in
acid
impact
this
on
and
The
societies
interest
in
reects
the
deposition
about
the
evidence
HL–
change;
and
sub-topic
this
acid
decisions
subject
to
reduction
of
deposition.
on
economic
needed
concern
that
deposition
(IB
environments;
environmental
systems
us
phenomenon
disciplinary
Urban
deposition).
enables
informs
how
on
to
the
of
acid
wider
Scientic
community
research
discussions
reduce
the
between
success
in
deposition.
which
impact
environment.
cooperation
achieve
the
of
Political
nations
the
provides
lead
and
is
control
to
acid
also
and
Q U e s T i O n s
Questions
1
Consider
the
equilibrium
6
below:
An
aqueous
reacts
CH
CH
3
COOH(aq)
+
H
2
O(l)
⇋
CH
2
CH
3
COO
with
solution
of
magnesium
which
of
the
following
metal?
(aq)
2
+
+
H
O
(aq)
A.
Ammonia
B.
Hydrogen
C.
Potassium
D.
Sodium
3
Which
A.
species
CH
CH
3
B.
H
represent
COOH
and
a
conjugate
H
2
O
acid–base
chloride
pair?
hydroxide
O
2
and
CH
2
CH
3
hydrogencarbonate
[1]
COO
2
IB,
May
2003
+
C.
H
O
and
H
3
O
2
+
D.
CH
CH
3
IB,
COO
and
H
2
May
O
[1]
3
7
2011
Which
property
aqueous
A.
2
Which
is
not
a
conjugate
acid–base
HNO
and
CH
C.
H
of
acids
in
Acids
react
with
ammonia
hydrogen
gas
solution
and
a
to
salt.
NO
3
B.
characteristic
pair?
produce
A.
is
solution?
3
COOH
B.
and
CH
3
Acids
COO
react
oxygen
3
with
gas,
a
metal
salt,
oxides
and
to
produce
water.
+
O
and
OH
C.
3
Acids
2
D.
HSO
and
IB,
May
SO
4
react
hydrogen
[1]
D.
Acids
a
Which
species
behave
as
Brønsted–Lowry
the
a
metals
to
produce
salt.
react
following
reversible
salt,
with
metal
hydrogen
and
carbonates
to
gas,
water.
[1]
acids
IB,
in
reactive
and
4
2011
produce
3
with
gas
May
2010
reaction?
2
H
PO
2
(aq)
+
CN
(aq)
⇋
HCN(aq)
+
HPO
4
(aq)
4
8
A.
HCN
and
A
solution
of
acid
A
has
a
pH
of
1
and
a
CN
solution
of
acid
B
has
a
pH
of
2.
Which
2
B.
HCN
and
HPO
statement
must
be
correct?
4
2
C.
H
PO
2
D.
and
HPO
4
HCN
4
and
H
PO
2
IB,
May
A.
Acid
B.
[A]
C.
The
A
is
stronger
than
acid
B
[1]
4
>
[B]
2010
+
concentration
than
in
of
H
of
H
ions
in
A
is
higher
B
is
twice
B
+
D.
4
Explain,
using
the
Brønsted–Lowry
theory,
The
concentration
ions
in
how
+
the
water
case
IB,
can
act
identify
May
either
the
as
an
acid
conjugate
or
acid
a
or
base.
base
In
concentration
of
H
ions
in
A
[1]
each
IB,
formed.
November
2010
2011
3
9
100
cm
of
a
NaOH
solution
of
pH
12
is
mixed
3
with
5
Which
of
the
following
is/are
formed
when
900
oxide
I)
A
reacts
metal
Water
III)
Hydrogen
I
only
B.
I
and
C.
II
and
D.
I,
II
a
dilute
IB,
is
the
pH
of
the
solution?
A.
1
B.
3
C.
11
D.
13
gas
May
[1]
2009
only
III
and
What
acid?
IB,
II
water.
salt
II)
A.
with
of
a
resulting
metal
cm
only
III
November
[1]
2003
207
8
A C I D S
10
Black
a
pH
A N D
coffee
of
8.
B A S E S
has
a
pH
Identify
of
5
which
and
is
toothpaste
more
acidic
has
III)
Use
and
each
battery
solution
and
lamp
in
a
and
circuit
see
with
how
a
bright
the
+
deduce
11
how
the
more
IB,
May
many
acidic
times
the
[H
is
greater
product.
100
the
pH
of
the
solution
of
0.50
mol
dm
and
II
only
b)
I
and
III
c)
II
and
d)
I,
II,
with
IB,
Specimen
only
III
only
and
III
[1]
HCl(aq)
200
3
cm
of
0.10
mol
paper
dm
[5]
16
May
I
3
cm
NaOH(aq).
IB,
a)
resulting
3
mixed
glows.
[2]
3
is
lamp
in
2011
Determine
when
]
Describe
two
different
properties
that
could
be
used
2011
3
to
distinguish
of
a
between
a
1.00
mol
dm
solution
3
strong
monoprotic
acid
and
a
1.00
mol
dm
3
12
Which
0.10
mol
dm
solution
would
have
the
solution
highest
of
a
weak
monoprotic
acid.
[2]
conductivity?
IB,
A.
HCl
B.
NH
C.
CH
3
17
May
2011
Ethanoic
acid,
CH
COOH,
is
a
weak
acid.
3
COOH
3
a)
D.
H
CO
2
IB,
Dene
May
the
term
weak
acid
and
state
the
[1]
3
equation
for
the
reaction
of
ethanoic
acid
2011
with
b)
water.
Vinegar,
[2]
which
contains
ethanoic
acid,
3
13
A
student
sodium
has
equal
hydroxide
volumes
and
of
1.0
ammonia
mol
dm
can
solutions.
be
used
carbonate
kettles.
Which
statement
about
the
solutions
is
Sodium
hydroxide
has
a
lower
B.
Sodium
than
hydroxide
concentration
Sodium
ethanoic
May
has
than
a
of
calcium
of
electric
acid
equation
with
for
calcium
the
reaction
carbonate.
[2]
2009
higher
hydrogen
The
equations
hydroxide
has
of
two
acid–base
reactions
are
ammonia.
given
C.
the
deposits
elements
ammonia.
18
ion
State
the
electrical
IB,
conductivity
clean
correct?
of
A.
to
from
a
higher
pH
below.
than
Reaction
A
ammonia.
+
NH
D.
Sodium
ion
hydroxide
concentration
has
than
a
higher
hydroxide
ammonia.
(aq)
May
reaction
reactants
the
Which
list
contains
only
strong
CH
B.
HCl,
COOH,
H
3
CO
2
,
H
,
H
OH
(aq)
mixture
in
3
3
2
A
consists
mainly
HNO
,
H
lies
to
B
4
(aq)
+
H
O(l)
⇋
NH
2
,
H
(aq)
+
OH
(aq)
3
reaction
in
B
consists
mainly
of
4
SO
2
mixture
SO
2
products
3
equilibrium
3
3
HNO
the
PO
The
COOH,
HCl,
+
CO
3
D.
(aq)
4
because
2
CH
NH
left.
NH
HNO
3
C.
⇋
acids?
Reaction
A.
O(l)
2
2010
of
14
H
[1]
The
IB,
+
3
because
the
equilibrium
lies
to
the
[1]
4«
right.
IB,
May
2009
a)
For
each
whether
15
Which
methods
will
distinguish
solutions
of
a
strong
base
and
water
reactions
is
acting
A
as
and
an
B,
acid
deduce
explain
your
or
a
base
answer.
[2]
a
b)
strong
the
between
and
equimolar
of
In
reaction
B,
identify
the
stronger
base,
acid?
NH
or
OH
and
explain
your
answer.
[2]
2
I)
Add
magnesium
to
each
solution
and
c)
look
for
the
formation
of
gas
In
reactions
A
and
B,
identify
the
stronger
bubbles.
+
acid,
NH
or
4
II)
Add
aqueous
sodium
hydroxide
to
your
solution
and
measure
the
change.
(underlined)
and
explain
3
answer.
temperature
IB,
208
NH
each
November
2009
[2]
9
R E D O X
P R O C E S S E S
Introduction
Redox
lie
at
reactions
the
chemical
centre
and
applications.
different
be
on
many
reduction
everyday
biochemical)
In
ways
considered
based
of
this
both
and
and
chapter
we
oxidation
introduce
and
have
the
(both
numerous
explore
and
oxidation
processes
the
reduction
idea
of
state,
which
oxidation
be
problems
will
examine
also
chemical
can
can
titration
of
and
will
be
useful
tool
volumetric
the
energy
electrical
electrochemistry.
cells
a
in
solving
conversion
energy,
Both
in
which
voltaic
redox
chemistry.
and
is
We
between
the
basis
electrolytic
introduced.
9.1 O ao a o
Understandings
Applications and skills
➔
Oxidation and reduction can be considered in
➔
Deduction of the oxidation states of an atom in
terms of oxygen gain/hydrogen loss, electron
an ion or a compound.
transfer, or change in oxidation number.
➔
➔
Deduction of the name of a transition metal
An oxidizing agent is reduced and a reducing
compound from a given formula, applying
agent is oxidized.
oxidation numbers represented by Roman
➔
Variable oxidation numbers exist for transition
numerals.
metals and for most main-group non-metals.
➔
➔
The activity series ranks metals according to
reduced and the oxidizing and reducing agents
the ease with which they undergo oxidation.
➔
The Winkler method can be used to measure
Identication of the species oxidized and
in redox reactions.
➔
biochemical oxygen demand (BOD), used as a
Deduction of redox reactions using half-
equations in acidic or neutral solutions.
measure of the degree of pollution in a water
➔
Deduction of the feasibility of a redox reaction
sample.
from the activity series or reaction data.
➔
Solution of a range of redox titration problems.
➔
Application of the Winkler method to calculate
BOD.
Nature of science
➔
How evidence is used
changes in the denition
oxidation numbers is a good example of the
of oxidation and reduction from one involving
way that scientists broaden similarities to
specic elements (oxygen and hydrogen), to
general principles.
one involving electron transfer, to one invoking
209
9
R E D O X
P R O C E S S E S
Redox reactions
Three
of
the
main
●
acid–base
●
precipitation
●
redox
A
redox
Both
of
reaction
that
occur
in
chemistry
are:
reactions
reactions
reactions.
reaction
reduction
different
The
types
ways,
different
involves
and
and
ways
two
oxidation
all
of
three
processes,
can
in
terms
of
specic
●
in
terms
of
electron
●
in
terms
of
oxidation
these
elements
reduction
considered
descriptions
describing
●
be
have
and
a
merit
processes
oxygen
in
and
oxidation .
number
in
their
of
own
right.
are:
hydrogen
transfer
number.
Oxidation: Combining with oxygen
At
a
the
simplest
substance
2Mg(s)
level
oxidation
combines
+
O
(g)
with
→
can
be
oxygen.
considered
Examples
as
a
reaction
in
which
include:
2MgO(s)
2
2CH
OH(l)
+
3O
3
▲
(g)
→
2CO
2
(g)
+
4H
2
O(l)
2
Figure 1 The Statue of Liber ty, New York, USA.
It took many years after restoration in 1986
4Fe(s)
+
3O
(g)
→
2Fe
2
O
2
(s)
3
for the copper to be oxidized and for the statue
Fe
to reform the green patina
O
2
(s),
iron(III)
oxide,
is
rust.
Rusting
is
an
example
of
the
process
of
3
corrosion
Global examples of corrosion
The
deterioration
electrochemical
described
as
of
metals
processes
caused
(redox
used,
by
reactions)
is
the
addition
polymer
polytetrauoroethene
(PTFE):
corrosion.
F
F
C
C
F
F
Oxidation of iron and copper: The Statue
of Liber ty
n
The
Statue
restored
in
of
Liberty
1986
as
it
in
New
was
York,
found
USA
that
was
corrosion
PTFE
had
occurred
between
the
wrought
is
iron
structural
support
Shellac
resin
and
the
outer
copper
,
secreted
by
the
female
lac
on
trees
in
Thailand
and
India)
inserted
and
the
as
iron
an
but
insulator
over
between
time
the
an
failed
the
and
the
renovation
iron
work
supports
a
rusted.
different
As
its
brand
as
a
name
non-stick
cooking
pans.
in
outer
the
Statue
green
of
Liberty
coating
called
oxidized
the
to
patina.
the
restoration
work
on
the
Statue
of
was
completed
the
statue
was
brown
in
part
insulator
It
has
taken
many
years
for
it
to
was
fully
210
by
used
the
colour.
of
also
insulation
Liberty
had
for
copper
When
copper
known
also
was
form
originally
is
bug
The
found
which
skin.
coating
(a
commonly
®
Teon
and
reform
the
patina
(gure
1).
oxidize
9 . 1
O x i d A t i O n
substances
zarda),
of
a
in
added
betel
person.
mouth
cause
nut,
is
of
to
r e d u c t i O n
it
lime
tobacco
of
the
the
tobacco
found
and
(shada/
damage
with
have
corrosive
corrosion
as
can
Paan
Scientists
highly
the
such
and
Chewing
cancer.
Paan
A n d
this
that
is
health
can
cause
the
the
lime
primary
bridge.
Avy
Can you think of some initiatives that governments
and city councils can adopt to inform the general
▲
public of the health eects related to chewing Paan in
Figure 2 Howrah bridge in Calcutta, India
countries where this is par ticularly prevalent?
Steel and Paan: The Howrah bridge
Another
Howrah
2010
interesting
bridge
the
bridge
corrosion
is
a
India.
of
example
Calcutta,
was
caused
mixture
(calcium
in
in
found
by
an
betel
The
leaf
to
be
areca
chewed
itself
corrosion
unusual
leaf,
hydroxide)
of
India
is
not
(gure
the
nut,
in
Paan.
and
harmful,
Paan
slaked
millions
of
but
In
contrast,
digestive
In
undergoing
agent,
by
is
2).
lime
people
other
many
and
of
is
these
a
that
substance
down
useful
used
the
as
a
aphrodisiac,
research
properties
nature
slows
also
recent
antioxidant
A
Reduction: Removal of oxygen or
India
Paan
stimulant,
of
rate
the
at
mnemonic
may
Paan.
delays
has
for
due
to
tonic
that
the
antioxidant
onset
which
nerve
suggested
be
An
post-meal
and
of
oxidation
oxidation
remembering
is
or
occurs.
this
is
OILRIG:
addition of hydrogen
Oxidation
Reduction
oxygen,
may
for
NiO(s)
be
considered
as
the
removal
Reduction
example:
+
C(s)
→
Ni(s)
+
Let
CO(g)
us
with
In
this
carbon
reaction
to
give
nickel(II)
metallic
oxide
is
reduced
may
also
be
return
oxygen
Magnesium
considered
as
the
Loss
of
electrons
Is
to
Gain
the
gas
to
of
electrons
reaction
form
of
magnesium
magnesium
metal
oxide.
by
nickel.
earth
Reduction
Is
of
is
a
metals),
member
and
has
of
the
group
2
electron
(alkaline
conguration
addition
2
[Ne]3s
of
hydrogen.
An
example
of
such
a
reaction
the
WO
(s)
+
3H
3
(g)
→
W(s)
+
3H
2
.
It
mirrors
the
its
previous
two
valence
electrons
to
attain
noble
gas
core
conguration,
[Ne]:
O(g)
2+
2
Mg
This
loses
is:
interpretation
→
Mg
+
2e
of
2
[Ne]3s
reduction,
oxide
in
as
the
oxygen
is
removed
from
[Ne]
tungsten(VI)
Oxygen
process.
is
a
is
a
member
non-metal.
2
[He]2s
It
has
of
group
the
16
(chalcogens)
electron
and
conguration
4
2p
and
gains
two
electrons
to
attain
the
Oxidation and reduction in terms of
noble
gas
conguration
[Ne]:
electron transfer
2
O
In
terms
of
electron
transfer,
oxidation
+
and
2e
2
[He]2s
reduction
can
be
dened
as
Oxidation
involves
the
loss
of
electrons
involves
the
2
2p
[Ne]
or
[He]2s
6
2p
magnesium
gain
of
is
oxidized
(loses
electrons)
and
and
oxygen
reduction
O
follows:
Hence
●
→
4
is
reduced
(gains
electrons).
electrons.
211
9
R E D O X
The
P R O C E S S E S
overall
reaction
is:
example,
in
2Mg(s)
+
O
(g)
→
the
complete
form
of
combustion
coal)
in
oxygen
of
solid
yields
carbon
carbon
(eg
dioxide:
2MgO(s)
2
C(s)
We
can
consider
the
reaction
as
being
However,
covalent
2+
→
2Mg
+
O
(g)
→
CO
2
processes:
2Mg
+
two
(g)
2
carbon
bonds,
dioxide
so
no
is
molecular,
ionic
bonds
are
with
formed.
We
4e
cannot
describe
this
combustion
reaction
as
a
redox
2
O
+
4e
→
2O
process
2
in
electrons
Considering
redox
processes
in
terms
of
is
a
common
interpretation
must
approach;
be
applied
however
with
are
of
lost
electron
or
gained
transfer
and
as
in
carbon
theory
dioxide
no
is
a
electron
neutral
transfer
terms
species!
The
original
denition
of
oxidation
this
caution.
as
the
as
carbon
addition
of
oxygen
is
more
appropriate
here
For
is
clearly
oxidized
in
this
process.
An application of redox chemistry from optometry
Optometrists
often
prescribe
glasses
with
The
chlorine
atoms
formed
by
the
exposure
+
photochromic
lenses.
These
lenses
darken
in
the
light
are
reduced
by
the
Cu
The
2+
presence
of
ultraviolet
light
(from
sunlight);
this
are
oxidized
to
to
+
ions.
Cu
ions
2+
Cu
ions.
These
Cu
ions
then
+
change
is
based
on
a
redox
reaction.
oxidize
silver
atoms
to
2+
Ordinary
glass
photochromic
is
composed
lenses
of
contain
silicates
while
copper(I)
Cu
and
silver
chloride,
ions:
+
+
Ag
→
Cu
+
+
Ag
chloride,
The
CuCl,
Ag
lenses
then
become
transparent
again
and
the
AgCl.
+
silver
The
on
chloride
exposure
ions
to
are
oxidized
ultraviolet
to
light
chlorine
atoms
and
and
Cl
chlorine
atoms
return
to
the
initial
Ag
species.
( hν).
hν
Cl
→
Electron
cation
Cl
+
e
transfer
to
be
then
reduced
takes
to
place
metallic
causing
silver
the
silver
atoms.
+
Ag
The
+
silver
making
The
e
→
atoms
the
inhibit
lenses
darkening
chloride
Ag
process
allowing
transparent
turn
the
the
transmittance
of
light,
dark.
is
reversed
lenses
to
by
copper(I)
become
again.
▲
When
the
following
lenses
are
reaction
takes
+
Cu
removed
from
the
light,
Figure 3 Photochromic lenses
the
place:
2+
+
Cl
→
Cu
+
Cl
Electron book-keeping
Chemists
developed
which
reactants
and
the
denition
specic
that
can
to
of
one
scientists
be
products
an
used
(oxygen
invoking
broaden
electron
to
during
oxidation
elements
transfer,
212
have
reactions
a
and
and
track
book-keeping
the
number
chemical
reduction
process.
from
hydrogen),
oxidation
similarities
states
to
of
to
is
a
general
a
model
The
example
principles.
of
involving
involving
good
redox
in
development
denition
one
for
electrons
electron
of
the
way
9 . 1
O x i d A t i O n
A n d
r e d u c t i O n
Oxidation and reduction in terms of oxidation states
The
a
oxidation
molecule,
In
terms
●
of
or
state
an
oxidation
Oxidation
and
is
the
apparent
charge
of
an
atom
in
a
free
element,
ion.
state:
describes
reduction
a
process
describes
a
in
which
process
in
the
which
oxidation
the
state
oxidation
increases
state
decreases.
rs fo assgg oao sas
1
The oxidation state of an atom in a free element is 0;
, O
for example, S
8
, P
2
5
The oxidation state of uorine is
, and Na all have atoms with an
4
, and LiF. For the other
2
oxidation state of 0.
group 17 halogen elements the oxidation state is
usually
2
1 in all its
compounds, for example HF, OF
1 in binary compounds (HI, NaCl, KBr) but in
Group 1 metals always have a +1 oxidation state in
combination with oxygen in oxoanions and oxoacids
their ions and compounds.
the oxidation state is positive (for example, in HClO
4
Group 2 elements always have a +2 oxidation state in
chlorine has a +7 oxidation state).
their ions and compounds.
6
In
Aluminium, which is a member of group 3, has an
a
neutral
states of
oxidation state of +3 in the majority of its compounds.
the
sum
equals
3
molecule
all
of
the
the
the
the
atoms
is
oxidation
overall
sum
z ero.
of
In
the
a
states of
cha rge
of
the
oxidat ion
poly ato mic
all
ion.
the
For
ion
at om s
ex amp le,
The oxidation state of hydrogen is +1 when hydrogen
in
is bonded to a non-metal, such as in HCl and HNO
the
NH
.
oxidation
state of
nitrogen
is
3
and
3
3
hydrogen
is +1;
the
sum
of
the
oxida tion
states
is
However, when hydrogen is bonded to a metal,
3
+
(3
×
+1)
=
0,
which
equals
the
net
cha rge
for example in a metal hydride such as NaH, the
on
oxidation number of hydrogen is
the
ammonia
molecu le.
1.
+
In
4
The oxidation state of oxygen is usually
the
ammonium
O and H
2
SO
2
,
NH
the
individual
4
oxidation
in H
cation,
2, such as
states
of
nitrogen
and
hy drogen
are
the
. The main exception is in a peroxide
4
same
(a species with an
O
O
O
2
and
NH
the
sum
of
the
oxid a tion
states
3
is
3
+
(4
×
+1)
=
+1
which
equals
the
net
1. A typical example of such a
charge
compound is H
in
linkage); here the oxidation
now
state of oxygen is
as
on
the
ammoniu m
ca tion.
, hydrogen peroxide.
2
An application of redox chemistry at the hair salon
Proteins
have
a
number
of
functions,
one
International directives
of
which
involves
having
a
structural
role
in
In
the
body.
Proteins
are
polymers
composed
the
European
peroxide
monomeric
units
called
amino
acids
(see
B.2).
The
protein
molecules
in
hair
restricted
thiol
groups,
and
hydrogen
peroxide
these
to
sulfonic
acid
groups,
skin,
(EU)
and
the
oral
use
of
hydrogen
hygiene
to
maximum
products
concentrations
of
12 %,
and
0.1%,
respectively.
As
stipulated
by
the
can
EU
oxidize
hair,
contain
4%,
SH
in
sub-
is
topic
Union
of
SO
H.
Cosmetics
Directive
such
products
must
be
This
3
labelled:
oxidation
of
the
of
the
proteins
thiol
and
groups
hair
can
changes
become
the
“Contains
hydrogen
peroxide.
Avoid
contact
structure
more
with
eyes.
Rinse
with
them.”
immediately
if
product
comes
in
contact
brittle.
Avy
The
Directive
Do people with bleached hair use particular conditioners?
products
Find out what a suitable conditioner might be for heavily
should
states
containing
be
that
when
hydrogen
using
hair
peroxide
gloves
worn.
bleached hair. What might its chemical components be?
213
9
R E D O X
P R O C E S S E S
Variable oxidation states
d b w
As
mentioned
in
the
r ul es
for
a ss i gn i n g
ox ida ti on
s t a t es
ab o ve ,
oao mb a
although
oao sa
many
compounds,
group
“Oxidation number ” and
exist
“oxidation state” are used
the
interchangeably in many
2
such
alkaline
for
many
transition
variable
textbooks.
metals.
el e me nts
as
e a r th
me ta ls
elements
range
s ta te s
of
xe d
g ro up
ma i n- g r oup
oxidation
The
the
ha ve
1
ox id at i on
al ka li
(e g
+2
for
non-met a l s
(also
ar e
called
a
met a l s
Ca),
a nd
the
(e g
+1
for
f or
and
o x i d a ti on
of
the
ions
Na)
fo r
the
and
the
s ta te s
most
metals).
p ro pe rt y
s t at e s
t h e ir
pa r t ic ul a r
transition
ox ida t io n
in
variable
in
cha r ac t er is t i c
d i ff er e nt
sta t e s,
of
Indeed
tr a ns i t ion
d- b l oc k
Note however IUPAC
elements
is
shown
in
g ur e
4,
whic h
is
g ive n
in
s ec t ion
14
of
the
recommends the use of
Data
booklet.
IUPAC
describes
tr a ns i ti on
e le me nts
as
el e me n t s
wh o se
Roman numerals for oxidation
atoms
have
an
inco mp le te
d- subs he ll
or
wh i c h
ca n
g i ve
ri se
to
c at i ons
numbers. Interestingly,
with
an
incomple te
d - s ubs he ll .
In
t he
 r st - r ow
d- blo c k
el e m e n t s,
oxidation states in some
the
transition
eleme nts
ar e
Sc
to
Cu
in c l u si ve
(but
n ot
Z n,
wh i c h
is
compounds may not have
explained
in
topic
13).
integer values; they may be
fractional, for example the
oxidation state of oxygen in the
S
t
V
c
M
F
co
n
c
Z
+1
+1
+1
+1
+1
+1
+1
1
is –
superoxide anions, O
2
.
2
+2
+2
+2
+2
+2
+2
+2
+2
+2
+3
+3
+3
+3
+3
+3
+3
+3
+3
+4
+4
+4
+4
+4
+4
+4
+5
+5
+5
+6
+6
+2
O zg a g
ags
+6
An ozg ag causes
another species to be oxidized,
+7
and is itself reduced in the
process. A g ag
type
A:
type
B:
type
C:
causes another species to be
Sc,
Ti,
and
V
Cr
and
Mn
Fe,
Co,
Ni,
Cu,
and
Zn
reduced, and is itself oxidized
in the process.
▲
Figure 4 Oxidation states of the rst-row d-block metals. The most stable oxidation states
are marked in green
Worked example
Deduce
the
Solution
oxidation
states
of
each
x
a)
K
Cr
2
O
2
7
x
atom
(marked
)
in
each
of
the
following
2(+1)
+
2x
+
7(–2)
species:
x
a)
K
x
x
Cr
O
2
2
b)
Mn
7
=
+6
O
4
x
x
c)
Mg
x
N
3
d)
[NH
]
4
214
[Fe
2
Mn
O
4
8
x
e)
b)
S
2
(H
O)
2
][SO
6
x
+
4(
x
=
+7
]
4
2
2)
=
–1
=
0
9 . 1
O x i d A t i O n
A n d
x
c)
Mg
x
N
e)
[NH
2
3
3(+2)
]
4
+
2x
=
0
To
[Fe
=
(H
2
O)
2
answer
this
knowledge
x
r e d u c t i O n
of
][SO
6
question,
the
]
4
2
you
charges
of
should
use
your
ammonium
( +1),
–3
water
(0)
and
sulfate
(
2)
species
2(
2)
=
(see
sub-
x
d)
S
topic
8
x
=
0,
since
this
is
a
free
element.
4.1).
2(+1)
x
=
+
x
+
6(0)
+
0
+2
Qk qsos
x
1
Deduce the oxidation states of each atom (marked with an
) in each of the
following species:
x
a)
C
x
O
b)
HCl
O
2
4
x
)
Na
P
x
O
3
)
O
4
3
x
)
P
x
H
f)
I
Cl
3
x
g)
x
(SO
Fe
2
4
)
h)
H
3
C
2
O
2
4
x
)
N
O
3
2
In the following balanced equation:
Cl
(aq) + 2KI(aq) → 2KCl(aq) + I
2
(aq)
2
a)
Deduce the oxidation states of chlorine and iodine in the reactants
and products.
b)
State which element is oxidized and which element is reduced.
)
Identify the oxidizing agent and the reducing agent.
Oxidation states and the nomenclature
of transition metal compounds
As
is
stated
that
previously
they
can
Traditionally,
used
to
the
name
numbers.
The
one
have
of
Roman
such
the
variable
characteristics
oxidation
numeral
compounds
system
is
called
states
system
and
the
this
of
of
transition
in
their
nomenclature
system
Stock
is
elements
compounds.
based
on
nomenclature
has
been
oxidation
system .
In
Sy ps
the
1
Stock
system,
Roman
numerals
(I,
II,
III
etc.)
are
used
to
indicate
Remember when writing
the
oxidation states the
oxidation
number.
charge goes before the
number and not after
it. For example, the
The Stock nomenclature system
oxidation number of
hydrogen in HBr is +1
In
KMnO
,
often
called
by
its
old
name
potassium
permanganate
by
4
and not 1+.
many
chemists,
manganese
has
an
oxidation
state
of
+7.
However,
7+
from
is
a
purely
highly
Stock
electrostatic
improbable.
nomenclature
Its
perspective
name
system.
is
the
presence
potassium
of
an
Mn
manganate(VII)
cation
using
the
2
The oxidizing and
reducing agents are
always the reactants.
215
9
R E D O X
P R O C E S S E S
in -ate, and the one with the greatest number of
Sy ps
oxygens will be prexed by “per ” and end in -ate. The
1
When deducing the name of a transition metal
four oxoanions of chlorine, bromine, and iodine follow
compound using the Stock system, do not be
this system.
tempted to use the subscript representing the
number of atoms of the other element in the
Foma of ooao
no-sysma am
compound to write the oxidation number in Roman
2
carbonate
CO
3
is iron(II)
numerals. For example, the name of FeCl
2
2
C
chloride because iron is deduced to have the +2
oxidation state [x + 2(
ethanedioate (oxalate)
O
2
4
1) = 0,
nitrite
NO
2
so x = +2]. It is coincidental that 2 matches the
nitrate
NO
3
number of chlorine atoms in the formula. In the
2
sulte
SO
compound FeO this becomes clearer: the correct
3
name is iron(II) oxide [x + (
2) = 0, so x =
+2].
2
sulfate
SO
4
2
In working out the names of many transition metal
3
phosphite
PO
3
compounds, knowledge of the non-systematic
3
phosphate
PO
4
names and charges of the various oxoanions can be
useful (table 1).
hypochlorite
ClO
chlorite
ClO
In naming oxoanions, a good rule of thumb is as follows:
2
chlorate
ClO
If there is only one oxoanion, the ending will be -ate.
●
3
If there are two oxoanions, the one with the smaller
●
perchlorate
ClO
4
number of oxygens will end in -ite and the one with
hydroxide
OH
the greater number of oxygens will end in -ate.
4
or thosilicate
SiO
4
If there are four oxanions, the one with the lowest
●
T
able 1 Formulas and non-systematic names of some
▲
number of oxygens will end in -ite and be prexed
oxoanions
by “hypo”, the next will end in -ite, the third will end
tOK
Qk qso
Chemistry has developed
Using the Stock nomenclature system, deduce the name of each of the
a systematic language that
following transition metal compounds:
has resulted in older names
a)
CoF
)
Cu(OH)
3
)
Cu
2
O
2
becoming obsolete. What
b)
V
has been lost and gained in
O
2
)
MnO
3
2
this process?
Nomenclature
In
theory
the
as
names
stated
one
new
is
not
all
set
of
inorganic
state,
many
such
In
guidelines
and
the
Although
systematic
include
required.
oxoanions
oxoacids.
216
of
could
previously
oxidation
state
of
one
additive
the
compounds.
elements
as
the
IUPAC
the
new
names,
system,
has
huge
in
only
+1,
so
published
systematic
corresponding
state
However,
have
potassium,
2005
for
oxidation
the
a
naming
inorganic
based
merit,
on
IUPAC
does
completely
such
as
acid,
etc.
called
be
recognize
eliminate
carbonate,
In
this
that
old
system,
),
are
acid,
),
nitrate,
and
formed?
acid
nitric
Can
you
to
names
nitric
would
nitrate
carbonic
hydroxidodioxidonitrogen.
names
unrealistic
carbonate
trioxidocarbonate(2
trioxidonitrate(1
is
non-systematic
carbonic
dihydroxidooxidocarbon,
these
it
be
would
would
acid
be
would
suggest
be
how
9 . 1
O x i d A t i O n
A n d
r e d u c t i O n
Expressing redox reactions using half-equations in acidic
or neutral solutions
Half-equations
complex
redox
represents
processes.
the
can
be
separate
The
very
reactions.
useful
Each
in
balancing
oxidation
following
general
balance
redox
and
reduction
working
Step
the
be
used
to
a
reaction
State
and
states.
neutral
to
In
the
balance
an
IB
syllabus
equation
you
in
are
acidic
4:
the
reactant
oxidation
and
product
states
for
each
for
process.
Balance
of
Step
5:
write
the
these
electrons
half-equations
lost
equals
the
so
that
number
the
of
gained.
Add
Step
6:
atom
the
two
overall
Check
product
Assign
oxidation
only
media.
1:
the
or
Working method
Step
for
half-equation
method
electrons
required
half-equation
corresponding
involving
number
oxidation
the
the
reduction
Step
can
3:
process
half-equation
half-equations
redox
the
together
to
reaction.
total
charge
on
the
reactant
and
sides.
in
species.
+
Step
7:
Balance
the
charge
by
adding
H
and
H
O
2
to
Step
2:
which
Deduce
species
which
is
species
is
oxidized
the
appropriate
sides.
and
reduced.
Worked example
1
Iron
The
tablets
iron
in
are
often
the
prescribed
tablets
is
to
patients.
commonly
Step
The
as
anhydrous
iron(II)
sulfate,
FeSO
.
to
determine
the
oxidation
2+
in
of
iron
in
such
tablets
percentage
involves
of
Fe
changes
from
+2
3+
to
Fe
+3
a
in
Fe
,
so
the
oxidation
state
by
increases,
mass
state
An
4
experiment
2:
present
indicative
of
oxidation.
The
oxidation
redox
state
of
Mn
changes
from
+7
in
MnO
to
4
reaction,
shown
in
the
following
unbalanced
2+
+2
in
Mn
,
indicative
of
so
the
oxidation
state
decreases,
equation:
2+
Fe
3+
(aq) + MnO
(aq) → Fe
(aq)
Step
4
a)
Deduce
the
balanced
redox
3:
equation
Oxidation
in
acid
and
identify
the
reduction.
2+
(aq) + Mn
oxidizing
(loss
of
2+
Fe
reducing
3+
(aq)
→
Fe
(aq)
+
e
agents.
Reduction
b)
electrons):
and
Consider
the
oxidation
state
(gain
of
electrons):
2+
of
MnO
(aq)
+
5e
→
Mn
(aq)
4
manganese
in
the
permanganate
anion,
Step
MnO
.
Comment
on
the
4:
following
4
Oxidation
(loss
of
electrons):
statement:
2+
5Fe
“If
oxidation
apparent
element
state
charge
has
in
is
considered
that
an
an
ion,
atom
then
as
of
the
3+
(aq)
→
5Fe
(aq)
+
5e
the
Reduction
an
oxidation
(gain
of
electrons):
2+
MnO
(aq)
+
5e
→
Mn
(aq)
4
state
of
manganese
here
must
signify
the
Step
presence
of
the
corresponding
5:
ion!”
2+
Oxidation:
5Fe
3+
(aq)
→
5Fe
(aq)
+
Solution
5e
2+
Reduction:
MnO
(aq)
+
5e
→
Mn
(aq)
4
a)
Step
1:
2+
Overall:
5Fe
(aq)
+
MnO
(aq)
→
4
2+
Fe
:
Fe,
MnO
:
x
=
Mn,
3+
+2
x
+
5Fe
4(
4
O,
x
=
2)
=
–1,
so
x
=
+7;
Step
2+
(aq)
+
Mn
(aq)
6:
–2
Total
charge
on
reactant
Total
charge
on
product
side
=
9+
3+
Fe
:
Fe,
x
=
+3
side
=
17+
2+
Mn
:
Mn,
x
=
+2
217
9
R E D O X
Step
P R O C E S S E S
7:
b)
Possible
a)
+
To
balance
this
equation
8H
must
be
the
response
oxidation
the
reactant
However,
+
MnO
(aq)
+
8H
(aq)
3+
Water
need
can
whichever
part
in
anion
was
found
to
be
the
+7.
oxidation
be
to
balance
included
side
of
the
at
the
+
Mn
(aq)
state
very
equation
it
last
is
of
ionic
hydrogens.
the
different
states
meanings,
and
and
ionic
an
charges
oxidation
2+
(aq)
stage
+7
are
required:
does
charge
properties
on
of
of
Based
not
7+.
ions,
theoretical
real.
2+
5Fe
in
→
have
5Fe
we
question:
manganese
+
(aq)
4
Next
of
side:
2+
5Fe
NOS
inserted
permanganate
on
to
state
on
signify
Ionic
a
corresponding
charges
whereas
constructs
and
electrostatic
are
real
oxidation
are
states
not
considerations
+
(aq)
+
MnO
(aq)
+
8H
(aq)
→
4
the
3+
5Fe
oxidizing
agent
of
a
7+
ionic
charge
is
most
2+
(aq)
+
Mn
(aq)
+
4H
O(l)
unlikely.
2
The
presence
is
MnO
(aq)
and
the
bonds
Oxidation
are
ionic.
states
assume
However,
in
that
MnO
4
the
4
2+
reducing
agent
is
Fe
(aq).
manganese–oxygen
bonds
are
covalent
in
nature.
The activity series
The
activity
they
those
lower
series
is
it
is
a
most
series
undergo
down
primarily
non-metal.
reactive
(table
oxidation.
from
based
The
metals
2)
ranks
Metals
solutions
on
are
em
of
metals,
series
is
metals
higher
at
their
according
in
in
the
the
is
section
top
of
the
of
ease
series
salts.
often
25
the
to
activity
respective
hydrogen
given
found
up
Data
which
displace
Although
included
the
with
can
the
even
booklet.
though
The
series.
dasg avy
eas of oao ass
lithium
potassium
sodium
magnesium
aluminium
manganese
zinc
iron
lead
hydrogen
copper
silver
mercury
gold
▲
T
able 2 The activity series
Let’s
consider
some
examples:
2+
●
Zn(s)
Zinc
+
Cu
metal
is
2+
(aq)
→
above
Zn
(aq)
copper
+
metal
Cu(s)
in
the
series,
so
therefore
it
is
more
2+
reactive
218
and
can
displace
the
Cu
ions
in
solution
to
form
copper
metal.
9 . 1
●
Zn(s)
+
2HCl(aq)
→
ZnCl
(aq)
+
H
2
O x i d A t i O n
A n d
r e d u c t i O n
(g)
2
Qk qsos
Zinc
the
metal
is
above
hydrogen
ions
hydrogen
in
in
the
hydrochloric
series.
acid
to
Therefore,
form
it
can
hydrogen
displace
1
gas.
Deduce the oxidizing and
reducing agents in the reaction of
●
2Al(s)
+
Fe
O
2
Aluminium
(s)
→
2Fe(l)
+
Al
3
O
2
metal
is
above
iron
(s)
3
in
potassium bromide with chlorine.
the
series.
Hence
molten
iron
can
2
form
according
to
the
reaction
Table 4 shows reactions involving
above.
aqueous solutions of halogens
●
2Na(s)
+
2H
O(l)
→
2NaOH(aq)
+
H
2
with aqueous potassium iodide
(g)
2
solution. Copy and complete the
In
this
reaction,
hydrogen
is
displaced
from
water
by
the
very
table, and in each case:
reactive
alkali
metal,
sodium,
to
liberate
hydrogen
gas
in
the
process.
a)
A
reactivity
series
can
also
be
written
for
the
group
17
state whether a reaction will
elements,
occur or not
uorine,
chlorine,
bromine,
and
iodine
(table
3).
b)
identify the colour of the
halide solution after reaction
Gop 17 m
Aom as
eogavy
(pm)
iasg
(Pag sa)
χ
)
avy
deduce the balanced
p
equation for any reaction that
uorine
60
4.0
chlorine
100
3.2
occurs.
bromine
117
3.0
iodine
136
2.7
Haog
c
(aq)
2
B
(aq)
2
a) rao
wh
T
able 3 Reactivity series for group 1
7 elements
▲
Ki(aq)
Again
the
more
2KBr(aq)
+
reactive
Cl
(aq)
elements
→
are
2KCl(aq)
found
+
Br
2
The
atomic
chlorine
a
is
greater
is
by
oxidation
in
the
state
losing
states.
oxidation
of
of
bromine
chlorine
is
smaller
electronegative.
gaining
in
the
series.
(aq)
2
attraction
reduced,
oxidized
radius
more
higher
for
an
an
an
electron
electron
electron
Chlorine
state
to
to
from
than
form
of
1
than
a
does
the
0
so
bromine,
bromine,
chloride
to
so
is
so
the
an
b) coo
has
of ha
chlorine
Bromine
change
there
process.
there
so
therefore
ion.
Note
1,
reduction
0,
of
nucleus
bromine.
from
to
that
chlorine
form
changes
indicative
changes
The
is
a
The
soo
is
in
decrease
oxidation
increase
in
the
) Baa
oxidation
state
indicative
of
an
oxidation
process.
qao
In
the
laboratory,
potassium
colourless
the
solution
formation
reactivity
when
bromide
of
series
chlorine
there
of
a
can
gas
is
bubbled
corresponding
potassium
aqueous
so
is
bromide
bromine.
displace
to
a
solution
change
from
yellow/orange,
Chlorine
bromide
through
colour
ions
is
higher
from
up
in
of
the
indicating
the
potassium
bromide
to
▲
form
bromine
(see
topic
T
able 4
3).
219
9
R E D O X
P R O C E S S E S
Uses of chlorine in everyday life
Chlorine
widely
is
a
used
Calcium
powerful
as
a
oxidizing
disinfectant
hypochlorite,
agent
and
Ca(OCl)
,
is
and
is
antiseptic.
often
used
in
2
hospitals
their
hands
NaOCl,
homes
The
is
as
healthcare
(gure
another
is
the
which
a
can
syndrome
that
of
professionals
Sodium
and
T he
US
Preve nti o n
disinfection
of
used
syr i ng es
f acto r
a cq uir ed
(AIDS) .
and
often
in
the
imm uno de cie ncy
to
disinfect
in
our
bleach.
ne e d l e s
lead
to
hypochlorite,
disinfectant,
contributo r y
human
Control
5).
household
sharing
users
of
by
am on g
vi ru s
i mmune
C ente rs
(C DC)
s y r i ng e s
and
( H I V) ,
d e c i en c y
for
ha s
d rug
t r an s m i ss i on
D i se a se
r e por t e d
ne ed le s
▲
with
household
b l e a ch
ma y
go
s om e
way
Figure 5 A solution of calcium hypochlorite acts
to
as both a disinfectant and an antiseptic. Can you
alleviating
this
ris k .
explain the dierence between these two terms?
Use of chlorine and ozone as disinfectants
H
in drinking water
Access
to
United
over
C
a
supply
Nations
one
billion
fundamental
of
as
a
clean
drinking
fundamental
people
resource
worldwide
to
water
human
do
mankind.
has
been
right,
not
yet
have
Water
the
supplies
recognized
it
is
luxury
are
by
estimated
of
the
that
such
disinfected
a
using
Cl
Cl
strong
oxidizing
agents
such
as
chlorine,
Cl
or
ozone,
O
2
pathogens.
be
added
In
in
the
USA,
three
chlorine
forms:
is
chlorine
used
gas,
for
Cl
;
to
kill
microbial
3
this
purpose.
sodium
Chlorine
hypochlorite,
can
NaOCl;
2
Cl
and
calcium
hypochlorite,
Ca(OCl)
.
All
three
of
these
solutions
yield
2
hypochlorous
The
use
object
of
to
chlorine
acid,
chlorine
the
can
taste
also
HOCl,
can
and
react
trichloromethane,
which
cause
with
,
the
problems
general
CHCl
is
odour
other
of
antibacterial
for
the
general
residual
chemicals
commonly
to
known
agent.
form
as
public.
chlorine
in
toxic
Some
water.
products
chloroform
people
Residual
(gure
such
as
6).
3
In
Europe,
water
Nice
supplies.
in
1906
chlorine
▲
France
for
was
The
for
rst
this
water
one
of
the
rst
industrial
purpose.
countries
ozonation
Table
5
to
plant
compares
use
ozone
was
the
to
disinfect
established
use
of
ozone
in
and
treatment.
Figure 6 The structure of
trichloromethane, CHCl
3
Avy
Ozo
cho
can be used to treat viruses
cannot be used to treat viruses
leaves no unpleasant residual
leaves a residual taste and unpleasant odour
taste or odour
In your country, nd out
fewer toxic by-products
can form toxic by-products, often carcinogenic
more expensive
cheaper
whether chlorine or ozone is
used to disinfect municipal
water supplies.
▲
T
able
5
Advantages
water supplies
220
and
disadvantages
of
using
ozone
and
chlorine
in
the
treatment
of
9 . 1
O x i d A t i O n
A n d
r e d u c t i O n
Redox titration reactions
Moay
In
topic
eld
of
1
titrations
volumetric
titrations,
those
were
introduced,
analysis.
involving
In
and
addition
redox
these
to
play
reactions
reactions
are
also
a
pivotal
role
involving
extremely
in
acid
the
base
Note that for convenience
concentration is often
useful.
termed the “moay”
In
order
to
solve
to
recall
some
titration
questions
involving
redox
reactions,
you
need
(the unit is sometimes
of
the
formulae
used
in
volumetric
analysis:
abbreviated to M), but it is
1
Amount
of
substance
(in
mol)
=
best practice to use the unit
n
3
mol dm
m
_
n
in calculations.
=
M
1
where
m
=
mass
in
g;
M
=
molar
mass
in
g
mol
3
2
n
=
volume
(in
dm
3
)
×
concentration
(in
mol
dm
3
volume
(in
cm
)
3
)
×
concentration
(in
mol
dm
)
_____
=
1000
3
because
1
=
1000
cm
1
_
1
_
(n
3
3
dm
)
=
(n
A
ν
)
and
hence
B
ν
A
B
1
_
1
_
(V
×
c
A
ν
)
=
(V
A
×
c
B
ν
A
)
B
B
3
V
=
volume
of
reactant
A
(in
dm
)
A
3
c
=
concentration
of
reactant
A
(in
mol
dm
)
A
3
V
=
volume
of
reactant
B
(in
dm
)
B
3
c
=
concentration
of
reactant
B
(in
mol
dm
)
B
ν
and
ν
A
are
the
stoichiometry
coefcients
B
Working method
Step
1:
Deduce
Step
2:
From
given
from
the
V
,
c
A
to
be
the
balanced
redox
information
,
V
A
,
and
c
B
determined.
equation,
given,
and
state
identify
using
which
the
oxidation
three
fourth
pieces
variable
states.
of
data
that
B
Identify
the
stoichiometry
coefcients
ν
and
ν
A
from
the
Step
3:
balanced
Set
up
following
expression
and
ll
×
c
A
)
=
A
(V
4:
×
B
ν
A
the
known
data:
c
)
B
B
Solve
for
the
unknown
variable
(V
,
A
Step
in
1
_
(V
ν
Step
B
equation.
the
1
_
are
needs
5:
Answer
concentration
in
any
riders
particular
to
the
question
c
,
V
A
,
or
B
(such
c
).
B
as
expressing
a
units).
Worked example
Sy p
1
Consider
the
following
balanced
equation
for
the
reaction
of
This
potassium
manganate(VII)
with
ammonium
iron(II)
type
of
frequently
2+
5Fe
(aq)
+
+
MnO
(aq)
+
8H
3+
(aq)
→
5Fe
a
titration
to
+
Mn
(aq)
+
4H
the
in
O(l)
2
determine
appears
2+
(aq)
4
In
question
sulfate.
concentration
of
a
Question
1
of
Paper
2.
potassium
3
manganate(VII)
solution,
28.0
cm
of
the
potassium
221
9
R E D O X
P R O C E S S E S
manganate(VII)
Sy p
solution
solution
3
25.0
reacted
completely
with
3
cm
of
a
0.0100
mol
dm
solution
of
ammonium
iron(II)
Notice that all variables
3
sulfate.
Determine
the
concentration,
in
g
dm
,
of
the
potassium
are given correct to three
manganate(VII)
solution.
signicant gures, hence
the nal answer for the
Solution
concentration should also
Step
1:
Deduce
the
balanced
redox
equation,
using
oxidation
states.
be expressed correct to
three signicant gures.
This
step
Step
are
2:
is
not
From
given
required
the
from
V
to
be
as
information
,
c
A
needs
here
,
V
A
,
and
equation
given,
c
B
determined.
the
and
state
is
given
which
identify
the
in
the
three
question.
pieces
fourth
of
variable
data
that
B
Identify
the
stoichiometry
coefcients,
ν
and
A
ν
Sy p
,
from
the
balanced
equation.
B
2+
Although molar masses are
A
represents
Fe
and
B
represents
MnO
4
given correct to two decimal
2+
V
=
volume
of
Fe
3
=
0.0250
dm
A
places in the Data booklet,
2+
c
the number of signicant
=
concentration
of
Fe
3
=
0.0100
mol
dm
A
gures in the nal answer
3
V
=
volume
of
MnO
B
=
0.0280
dm
4
is determined from the
c
experimental data given
=
concentration
of
MnO
B
:
this
is
what
must
be
calculated
4
in the question and not
ν
=
5
=
1
A
from published data. The
ν
B
same would apply to any
constants used to answer
Step
3:
Set
up
the
following
expression
and
ll
in
the
known
data:
questions.
1
_
1
_
(0.0250
×
0.0100)
=
(0.0280
×
c
)
B
5
Step
1
4:
Solve
for
the
unknown
variable
(V
,
c
A
,
V
A
,
or
B
c
).
B
3
c
(concentration
of
MnO
B
Step
)
=
0.00179
mol
dm
4
5:
Answer
concentration
any
in
riders
to
particular
the
question
(eg
expressing
a
units).
3
To
calculate
1
mol
the
of
concentration
KMnO
≡
in
(39.10)
g
+
dm
,
we
(54.94)
+
use
dimensional
4(16.00)
≡
analysis.
158.04
g
4
So:
0.00179
158.04
mol
__
g
_
×
3
=
0.283
g
dm
3
1
1
dm
mol
An environmental application of redox chemistry:
The Winkler method
Aquatic
dioxide
life
and
depends
oxygen
on
gases
dissolved
such
in
as
the
carbon
water
The
in
solubility
dependent.
of
At
oxygen
273
K
(0°
in
water
C)
the
is
temperature
solubility
is
3
order
to
survive.
Oxygen,
O
,
is
a
non-polar
14.
6
mg
7.6
mg
dm
(or
14.6
ppm),
compared
with
just
2
3
molecule,
but
water,
H
O,
is
polar.
Therefore
the
dm
(7.6
ppm)
at
293
K
(20
°C).
Clearly
2
solubility
of
oxygen
in
water
will
be
very
low.
as
the
gas
222
temperature
decreases.
increases,
the
solubility
of
the
9 . 1
The
O x i d A t i O n
degree
water
can
oxygen
dipole moment
as
the
of
be
organic
demand
of
matter
in
over
units
environmental
or
a
measured
In
pollution
by
the
BOD.
oxygen
temperature
in
r e d u c t i O n
measured
amount
organic
A n d
period
of
a
This
is
of
to
water
of
sample
of
dened
required
sample
a
in
biochemical
5
oxidize
at
days.
a
denite
BOD
is
ppm.
science,
ppm
is
often
used
as
O
the
standard
maximum
toxic
H
or
unit
of
concentration
allowable
carcinogenic
upper
limit
to
of
a
indicate
the
potentially
(cancer-causing)
substance.
For
H
example,
according
to
recommendations
from
the
Figure 7 Water is a polar molecule. The vectorial sum
▲
World
Health
Organization
(WHO)
the
maximum
of the two individual OH polar bonds results in a net
2+
allowed
concentration
of
lead(II)
cations,
Pb
(aq),
in
dipole moment for the molecule
3
drinking
water
is
0.001
mg
dm
or
0.001
ppm.
coaos  pa s p mo
Avy
The concentration of very dilute solutions is often
measured in pa s p mo, ppm
Go to the WHO website (http://www.who.int/en/)
and try to nd data about the maximum allowed
concentration in ppm
concentrations of other metals in drinking water.
mass of component in solution
_______
=
6
Compare this data with the limits set by the
× 10
total mass of solution
government of the country where you live or by
mass of solute in mg
_____
directives set by a wider union of countries (eg the
=
3
volume of solution in dm
The
a
amount
of
barometer
water.
The
to
dissolved
indicate
Winkler
oxygen
the
is
quality
method,
European Union).
often
of
based
a
used
body
on
as
of
Typical values of BOD
redox
Pure
reactions,
is
one
technique
that
can
be
used
water
Water
measure
the
amount
of
dissolved
oxygen
in
from
general,
a
high
concentration
of
indicates
a
low
level
of
▲
be
considered
from
a
river
considered
a
BOD
has
a
less
BOD
than
of
1
1
ppm.
ppm
very
with
a
clean.
BOD
However,
of
20
ppm
water
would
poor
quality
(table
6).
BOd (ppm)
pure water
less than 1
untreated domestic sewage
350
euent from a brewery
500
water from an abbatoir
3000
T
able 6 Typical biological oxygen demands for water samples
When
organic
water,
it
material
Figure 8 The WHO is the directing and coordinating authority
of
eamp so
present.
for health within the United Nations
has
that
pollution.
be
▲
river
dissolved
taken
oxygen
a
water.
would
In
generally
to
matter
provides
The
water
The
carbon
hydrogen
in
is
is
is
discharged
source
bacteria
into
and
a
break
compounds
a
series
of
oxidized
oxidized
of
down
such
as
carbon
water,
into
for
the
a
any
body
of
bacteria
organic
carbon
oxidation
to
to
food
dioxide
reactions.
dioxide,
and
any
the
nitrogen
223
9
R E D O X
present
is
P R O C E S S E S
oxidized
to
nitrate,
NO
.
The
The
bacteria
procedure
developed
by
Winkler
is
an
indirect
3
multiply
more
and
their
dissolved
processes.
If
increased
oxygen
the
is
uptake
levels
used
of
for
mean
these
oxygen
by
one,
that
oxidation
the
react
faster
than
the
rate
at
which
dissolved
with
replaced
from
the
atmosphere
and
the
depleted
body
of
of
water
oxygen.
will
Under
the
hydrogen
bacteria
sulde,
H
will
S,
produce
NH
2
amines),
and
PH
.
the
gas
commonly
then
the
in
Hydrogen
left
dark.
to
with
rotten
eggs;
it
is
also
often
was
for
a
The
measure
the
water
water
is
taken
from
discharged
a
into
location
a
rst
saturated
with
marina
oxygen
period
of
Winkler
5
days
method
at
293
was
K
carried
the
dissolved
sample
before
oxygen
and
content
after
the
5-day
sulde
incubation
associated
of
efuent
UAE,
the
period.
The
following
is
the
series
of
odour
reactions
from
sample
treated
such
3
is
reagent.
(and
3
phosphine,
directly
anaerobic
products
ammonia,
not
eventually
such
out
as
cm
Dubai,
in
conditions
redox
does
from
and
become
the
oxygen
3
50.0
in
photosynthesis,
dissolved
oxygen
where
is
the
bacteria
A
is
as
liberated
related
to
the
method:
from
2+
volcanoes.
Due
odour
its
and
to
its
characteristic
potential
source
H
Mn
unpleasant
S
is
often
(aq)
+
2OH
(aq)
sewer
(s)
referred
2Mn(OH)
(s)
+
O
2
as
Mn(OH)
2
2
to
→
(g)
→
2MnO(OH)
2
(s)
2
gas.
+
MnO(OH)
(s)
+
4H
(aq)
+
2I
(aq)
→
2
em
Sbsa
Sbsa po
po 
 aaob
aob oos
oos
2+
Mn
(aq)
+
I
(aq)
+
3H
2
2
I
(aq)
+
2S
2
carbon
CO
CH
2
O
2
O(l)
2
2
(aq)
→
2I
(aq)
+
S
3
O
4
(methane,
(aq)
6
3
It
4
was
found
that
5.25
3
cm
of
a
0.00500
mol
dm
commonly known as
solution
of
sodium
thiosulfate,
Na
S
2
O
2
(aq)
was
3
marsh gas)
required
hydrogen
H
O
CH
2
, NH
4
, H
3
S, and H
2
NH
NO
2
sulfur
H
SO
, amines
the
iodine
produced.
Determine
the
oxygen,
ppm,
in
concentration
in
the
of
sample
dissolved
of
water.
S (hydrogen
b)
2
4
with
O
3
3
react
2
a)
nitrogen
to
Deduce
the
BOD,
in
ppm,
of
the
water
sample,
sulde)
assuming
3
phosphorus
PH
PO
oxygen
in
the
the
maximum
water
is
9.00
solubility
ppm
at
of
293
K.
3
4
▲
(phosphine)
that
T
able 7 Substances produced by bacteria under aerobic and
c)
Comment
on
the
BOD
value
obtained.
anaerobic conditions
Solution
The
the
reduction
depletion
in
of
d i s s o l ve d
s h
oxy ge n
s to cks .
If
the
c an
B OD
re s u lt
is
in
Step
1:
Deduce
The
than
the
dissolved
co nte nt
in
the
wa t er,
cannot
surviv e .
Ty p i ca ll y
s h
re qu i re
3
ppm
of
diss o l v e d
ox yg en
in
series
w a t e r,
in
sustain
a
healthy
a q ua tic
of
dissolve d
e nv ir onme n t
o x yg en
in
wat e r
below
6
equation.
redox
The
equations
important
is
point
the
correct
stoichiometric
is
ratio
a nd
oxygen
and
thiosulfate;
this
is
needed
the
s h ou ld
the
calculation.
Careful
examination
of
the
not
three
fall
balanced
question.
determine
for
content
of
the
between
to
redox
at
to
least
balanced
aq u at i c
given
life
the
g r e at e r
reactions
gives
the
following
ratio:
ppm.
1
mol
O
(g)
→
2
mol
MnO(OH)
2
(s)
→
2
2
4
mol
S
O
2
(aq)
3
Worked example: measuring BOD
Step
using the Winkler method
three
2:
From
pieces
the
of
information
data
are
given
given,
from
V
state
,
A
In
the
Winkler
method,
an
iodine/thiosulfate
c
and
identify
the
fourth
variable
that
c
,
A
which
V
and
B
needs
to
be
B
redox
titration
dissolved
is
oxygen
carried
out
present
in
to
a
measure
water
the
sample.
determined.
ν
A
224
and
ν
,
B
Identify
from
the
the
stoichiometry
balanced
equation.
coefcients,
9 . 1
O x i d A t i O n
A n d
r e d u c t i O n
2
A
represents
S
O
2
and
B
represents
=
volume
of
S
A
5:
Answer
any
riders
to
the
question.
2
O
2
3
3
2
V
Step
O
3
=
0.00525
In
dm
order
to
calculate
the
concentration
in
g
dm
,
3
we
use
dimensional
analysis.
2
c
=
concentration
of
S
A
O
2
3
3
=
0.00500
mol
1
dm
mol
of
O
≡
2(16.00)
≡
32.00
g
2
3
V
=
volume
of
O
B
=
0.0500
So:
dm
2
4
c
=
concentration
of
O
B
;
this
must
be
1.31
calculated
×
10
32.00
mol
__
2
g
_
×
3
1
ν
=
4
=
1
1
dm
mol
A
3
ν
=
4.19
×
10
=
4.19
mg
3
g
dm
B
3
Step
the
3:
Set
known
up
the
following
expression
and
ll
dm
=
Hence
data:
the
oxygen
(BOD) = 9.00
1
_
4.19
ppm
in
used
by
the
bacteria
4.19 = 4.81 ppm.
3
(0.00525
dm
×
0.00500)
This
BOD
value
shows
reasonable
water
4
quality
1
_
=
(0.0500
×
c
for
discharge
effective
Step
4:
Solve
for
the
unknown
variable
(V
,
c
A
,
taken
at
the
point
in
sewage
Dubai,
efuent
suggesting
treatment
plan
that
must
be
an
in
V
A
B
place.
c
sample
)
A
1
or
the
Typically
untreated
domestic
sewage
has
).
B
a
4
c
(concentration
of
O
B
)
=
1.31
×
10
BOD
in
the
range
100–400
ppm.
3
mol
dm
2
Activity
Chemistry
Discuss
is
this
full
of
commenting
on
ionic
and
charge,
domain
abstract
statement
may
be
with
aspects
a
concepts,
theories,
reference
to
such
negative
as
oxidation
charge
preferable
the
term
centres.
to
and
assumptions.
thiosulfate
numbers,
Suggest
negative
oxoanion,
formal
why
charge
charge,
electron
centre
in
this
context.
225
9
r e d O x
P r O c e S S e S
9.2 eohma  s
Understandings
Applications and skills
Voltaic (Galvanic) cells:
Construction and annotation of both types of
➔
➔
Voltaic cells conver t energy from spontaneous,
electrochemical cells.
exothermic chemical processes to electrical
Explanation of how a redox reaction is used
➔
energy.
to produce electricity in a voltaic cell and how
➔
Oxidation occurs at the anode (negative
current is conducted in an electrolytic cell.
electrode) and reduction occurs at the cathode
Distinction between electron and ion ow in
➔
(positive electrode) in a voltaic cell.
both electrochemical cells.
Electrolytic cells:
Performance of laboratory experiments
➔
➔
Electrolytic cells conver t electrical energy
involving a typical voltaic cell using two metal/
to chemical energy, by bringing about non-
metal-ion half-cells.
spontaneous processes.
Deduction of the products of the electrolysis of
➔
➔
Oxidation occurs at the anode (positive
a molten salt.
electrode) and reduction occurs at the cathode
(negative electrode) in an electrolytic cell.
Nature of science
➔
Ethical implications of research – the desire to produce energy can be driven by social needs or prot.
Energy
Energy
The
created
There
due
is
law
or
are
to
the
of
capacity
destroyed
many
motion),
energy,
to
do
conservation
heat
but
is
different
The
energy
converted
forms
potential
energy,
work.
of
of
energy
nuclear
SI
from
energy,
(stored
energy,
unit
states
of
that
one
form
such
or
sound
energy
energy
as
to
is
energy,
joule
(J).
be
another.
kinetic
positional
the
cannot
energy
energy),
chemical
(energy
light
energy,
and
Figure 1 Voltaic cells conver t energy
electrical
energy.
from spontaneous exothermic chemical
processes to electrical energy
Electrochemical cells
In
an
take
electrochemical
place,
which
electrochemical
1
Voltaic
(or
electrical
cell
go
chemical
in
either
energy–electrical
direction.
There
energy
are
two
galvanic)
chemical
Electrolytic
cells
Voltaic
cells
–
–
cells
these
processes
these
convert
convert
to
electrical
convert
chemical
energy
from
energy to chemical energy, bringing about
a non-spontaneous process
226
types
bringing
about
a
to
energy.
electrical
non-spontaneous
energy
spontaneous,
energy
Figure 2 Electrolytic cells convert electrical
energy,
conversions
main
cell:
energy.
exothermic
2
can
process.
to
chemical
of
9 . 2
e l e c t r O c H e M i c A l
c e l l S
Sy p
Early ideas about electricity
An easy way to remember
Electrochemistry
explores
energy
conversions
between
chemical
and
the energy conversion in an
electrical
energy.
The
Italian
physician,
physicist,
and
philosopher
Luigi
electrolytic cell
Galvani
origins,
(1737–
1798)
whereas
the
considered
Italian
electricity
physicist
essentially
Alessandro
Volta
biological
in
is “ee”:
its
(1745–1827)
electrolytic cell – electrical to
did
not.
The
initial
discovery
of
electrochemical
cells
resulted
from
chemical.
serendipitous
driven
by
demand
observations
potential
for
of
prots
energy
has
scientists,
from
driven
but
improved
innovation
recent
developments
technology.
in
devices
The
and
have
been
increasing
processes.
Sy p
An easy way to remember
Electrodes
Electrons
are
conductor
circuit,
cell
In
of
such
both
of
electricity
as
contains
redox processes at electrodes
carriers
the
two
voltaic
electric
used
solution
to
in
electrodes,
and
charge
make
a
cell
the
electrolytic
in
metals.
contact
(the
anode
An
with
a
electrode
non-metallic
electrolyte).
and
the
is
An
a
is the mnemomic “crOA”:
part
of
a
electrochemical
cathode – reduction and
Oxidation – Anode.
cathode.
cells:
●
oxidation
always
takes
place
at
the
anode
●
reduction
always
takes
place
at
the
cathode.
Sy p
An easy way to remember
The
In
a
polarity
of
voltaic
the
electrodes
differs
in
the
different
types
of
cell.
the polarities at electrodes
is “cnAP”: for the electrolytic
cell:
cell, cathode–negative;
●
the
cathode
●
the
anode
is
the
positive
electrode
Anode –Positive. If you can
is
the
negative
electrode.
recall this mnemonic for the
In
an
electrolytic
cell:
electrolytic cell, you know that
●
the
cathode
●
the
anode
is
is
the
the
negative
positive
electrode
the opposite is true for polarities
in a voltaic cell.
electrode.
The voltaic cell
A
voltaic
(the
cell
anode)
consists
and
are
different
ion
electrodes,
ion
electrode.
required
to
types
of
metal
For
be
of
two
reduction
half-cells.
occurs
electrode
ions
the
IB
familiar
in
at
used
two
with
in
other
voltaic
different
Chemistry
only
Oxidation
the
cells,
at
as
states,
programme,
metal/metal-ion
one
(the
such
oxidation
Diploma
the
occurs
half-cell
half-cell
cathode).
There
metal/metal-
and
SL
the
gas-
students
are
electrode.
The metal/metal-ion electrode
A
metal/metal-ion
solution
type
of
electrode
containing
electrode
cations
consists
of
the
of
same
a
bar
of
metal.
metal
Typical
dipped
into
examples
a
of
this
include:
2+
●
Fe(s)|Fe
(aq)
(gure
3)
2+
●
Zn(s)|Zn
(aq)
bar of metallic iron, Fe(s)
2+
●
Cu(s)|Cu
(aq).
2+
solution of Fe
In
this
notation,
the
vertical
line
represents
a
phase
boundary
(aq),
or
called the electrolyte
junction
In
a
voltaic
cell
the
two
half-cells
are
separated
–
if
the
solutions
were
2+
Figure 3 The Fe(s)|Fe
allowed
to
mix
in
a
single
container,
a
spontaneous
reaction
would
(aq) electrode
occur
227
9
r e d O x
P r O c e S S e S
but
there
circuit,
would
and
be
hence
no
no
movement
current.
of
The
electrons
two
through
electrodes
are
the
in
external
electrical
contact
Activity
via
Discuss
the
origins
of
of
from
a
It
electrochemistry
Nature
perspective,
of
some
of
the
the
eld
working
to
the
It
provides
of
with
produce
has
a
number
of
functions:
reduction
of
the
cathode
processes,
and
anode
preventing
and
mixing
hence
of
the
continuity
–
a
path
for
the
migration
of
the
positive
(the
cations)
and
the
negative
ions
(the
anions)
in
the
cell.
It
reduces
the
liquid-junction
p ot e nt ial .
This
is
the
voltag e
ethical
when
tw o
d i ffe r ent
so lut i on s
co m e
in t o
con t a c t
wi t h
current
our
global
more
separation
and
electrical
other,
which
o ccur s
due
to
un e qu a l
ca t io n
and
a n i on
ever-
migration
increasing
This
in
each
research,
bridge.
solutions.
generated
implications
salt
original
●
this
a
discoveries
ions
scientists
called
physical
oxidation
two
●
of
junction
allows
the
Science
from
serendipitous
liquid
the
●
eld
a
desire
across
the
j unction.
to
energy.
A
salt
high
bridge
contains
concentration
a
of
concentrated
ions
in
the
solution
salt
bridge
of
a
strong
allows
ions
electrolyte.
to
diffuse
The
out
of
2+
it.
For
example,
the
Daniell
voltaic
cell
consists
of
the
Cu(s)|Cu
(aq)
2+
and
for
Zn(s)|Zn
this
cell
(aq)
could
electrodes.
be
sodium
Typical
sulfate,
compounds
Na
tOK
SO
2
KCl(aq).
Is energy real, or just an
react
The
with
ions
the
used
other
in
ions
the
in
salt
(aq)
used
or
in
the
potassium
salt
bridge
chloride,
4
bridge
must
be
inert
–
they
should
not
thesolution.
abstract concept used to
justify why cer tain types
The Daniell voltaic cell
of changes are always
In
the
Daniell
cell
(gure
4),
the
following
half-equations
show
the
associated with each other?
redox
●
processes
Anode
occurring.
(negative
electrode):
oxidation.
2+
Zn(s)
●
→
Cathode
(aq)
Zn
+
(positive
2e
electrode):
reduction.
2+
Cu
●
(aq)
Overall
+
cell
2e
→
Cu(s)
reaction:
2+
Cu
Once
the
of
the
as
it
2+
(aq)
+
cell
side
Zn
sulfate
coated
drawing
right-hand
→
connected,
copper(II)
becomes
When
is
Zn(s)
in
voltaic
as
in
(aq)
as
+
the
solution
more
cells,
gure
Cu(s)
redox
fades.
copper,
by
processes
The
and
occur
copper
the
convention
zinc
the
bar
bar
the
blue
increases
gets
cathode
is
colour
in
size
thinner.
drawn
on
the
4.
e
e
V
2
SO
4
+
Na
Zn anode
Cu cathode
salt bridge
(
)
(+)
2
2
SO
4
SO
4
cotton wool
2+
Zn
2
SO
ZnSO
CuSO
4
4
4
2
SO
2+
Cu
4
2+
Zn(s) →
Zn
2+
(aq) +
2e
Cu
(aq) +
movement of cations
movement of anions
2+
Figure 4 The Daniell cell: a cell consisting of Zn(s)|Zn
228
2+
(aq) and Cu(s)|Cu
(aq) half-cells
2e
→
Cu(s)
9 . 2
How
will
can
be
For
the
more
you
determine
reduced
Daniell
easily
in
a
cell,
which
voltaic
zinc
oxidized
–
is
the
metal
cell?
The
higher
zinc
up
will
be
answer
in
half-cell
the
acts
oxidized
lies
in
series
as
and
e l e c t r O c H e M i c A l
which
metal
the
activity
series.
than
copper,
so
the
c e l l S
it
is
anode.
Cell diagrams
Cell
cell.
on
diagrams
By
the
For
are
convention
right.
the
The
Daniell
used
the
salt
cell
a
bridge
the
2+
Zn(s)|Zn
as
anode
cell
convenient
is
is
always
shorthand
written
represented
diagram
would
by
be
on
to
the
two
represent
left
and
parallel
written
a
the
voltaic
cathode
vertical
lines.
as:
2+
(aq)
||
Cu
(aq)|Cu(s)
Sy p
When answering questions about voltaic cells, make sure you know the direction of
ow of the electrons and ions (gure 5).
e
e
V
+
Zn anode
Cu cathode
salt bridge
(
)
(+)
+
+
anode
cathode
(oxidation)
(reduction)
Figure 5 The direction of ow of electrons, positive ions and negative ions in a voltaic cell
Qk qso
2+
For a voltaic cell consisting of a Zn(s)|Zn
2+
(aq) half-cell and an Fe(s)|Fe
(aq)
half-cell:
a)
State the cell diagram for the cell.
b)
Write half-equations for the reactions occurring at the cathode and the anode.
)
Identify a suitable compound that may be used in the salt bridge.
)
Identify the direction of the movement of electrons and ion ow, both in
Ky pma wok
solution and in the salt bridge.
In this topic it is impor tant to
)
Explain why the cation and anion of the salt used in the salt bridge should have
have carried out laboratory
approximately the same size and charge. Identify using section 9 of the Data
experiments or seen videos or
booklet the ionic radii of the cation and anion of the compound given in (c).
simulations involving a typical
f)
Cotton wool is often used at the tips of the salt bridge. Suggest the function of
voltaic cell using two metal/
this.
metal-ion half-cells.
229
9
R E D O X
P R O C E S S E S
The global energy perspective
One
type
of
fuel
cell
is
the
biological
fuel
cell,
Fuel cells
which
The
combustion
of
fuels
such
as
oil,
uses
natural
gas
releases
heat
energy
which
into
electrical
energy
in
an
energy
waste
plant.
The
energy
loss
in
this
process
can
or
greater.
Fuel
cells
however
can
an
70%
electrical
energy.
based
redox
on
A
of
the
fuel
energy
cell
processes.
is
a
The
in
a
fuel
cell,
the
so
common
hydrogen–oxygen
reaction
between
hydrogen
fuel
and
fuel
is
a
the
(g)
+
O
2
Under
(g)
→
2H
2
acidic
methane
or
fuel
alternative
cell
to
is
the
non-polluting
internal
engine,
the
cell
uses
problem.
storage
The
of
hydrogen
methanol
liquid
methanol
rather
than
fuel
hydrogen,
type
cell ,
is
much
easier
to
transport.
Methanol
uses
be
produced
from
biomass
as
a
carbon-
oxygen:
fuel
(it
does
not
contribute
to
the
O(l)
2
conditions,
hydrogen
major
neutral
2H
as
is
can
the
such
into
cell
which
of
chemicals
convert
fuel
voltaic
most
in
materials.
efcient
combustion
approximately
from
be
and
67%
electricity
electric
Although
power
generate
is
organic
converted
to
coal,
chemical
or
bacteria
greenhouse
the
following
effect).
reactions
Fuel cells and the International
take
place
at
the
anode
and
cathode:
Space Station
●
anode
(negative
electrode):
oxidation
The
hydrogen–oxygen
fuel
cell
can
be
used
as
an
+
2H
(g)
→
4H
+
4e
2
energy
●
cathode
(positive
electrode):
reduction
Space
ve
+
O
(g)
+
4H
+
4e
→
2H
2
cells
the
only
cell
so
it
are
they
of
C.6).
are
One
by
lifetime
expensive
of
cell
cells
is
space
(ISS)
agencies
is
a
The
International
collaborative
representing
15
product
nations,
of
as
are
the
added.
of
6
that
the
very
the
not
are
cells
and
fuel
produce.
is
in
requires
does
Fuel
of
November
inhabited
by
and
humans
since
2000.
fuel
need
continuous
(gure
to
cells
purication
it
a
is
continuously
Another
reactants
disadvantage
fuel
Water
been
conventional
fuels).
that
provide
impurities
or
(unlike
fossil
expensive
of
devices.
hydrogen–oxygen
applications
very
poisoning
the
because
disadvantage
their
fuel
reactants
number
topic
a
Fuel
electricity
more
of
reactions
of
recharging.
a
Station
spacecraft.
O(l)
efcient
non-polluting,
advantage
up,
highly
product
is
combustion
of
in
2
has
Fuel
source
have
sub-
cells
is
that
Another
they
fuel,
supply
used
are
prone
which
complex
to
reduces
and
fuel.
Figure 7 The Japanese pressurized experiment module
for the International Space Station, shown here at its
manufacturing facility in Nagoya, Japan. The module, called
Kibo or “hope” in Japanese, is Japan’s rst human space
facility. Experiments in Kibo focus on space medicine,
biology, Ear th obser vations, material science, biotechnology,
and communications research
Figure 6 This bus in Reyjavik, Iceland, is powered by a fuel cell
that runs on hydrogen
230
9 . 2
e l e c t r O c H e M i c A l
c e l l S
Avy
1
2
Find out the cathode and anode half-equations and
with wind turbines. Consider a number of
the overall cell reaction for the direct methanol fuel
countries worldwide where wind farms are located
cell. Compare and contrast the methanol fuel cell with
and compare and contrast any government
the hydrogen–oxygen fuel cell from an environmental
regulations that may be in place regarding their
perspective.
construction.
a)
Discuss some aspects of what is commonly
termed the “hydrogen economy”. Your answer
might address aspects such as the advantages
and problems of using hydrogen as a fuel in
motor cars, the various methods for generating
hydrogen, and the use of renewable energy
sources such as wind and solar energy.
b)
Suggest how wind farms may be assisting
developing countries in dealing with their energy
needs, and so driving the global “hydrogen
economy”. Explore what problems wind turbines
may pose for rural communities. Research and
Figure 8 Example of an on-shore wind farm in Kilmore, Co.
discuss any possible health eects associated
Wexford, Republic of Ireland
Electrolytic cells
Electrolysis
energy
is
is
used
the
to
process
drive
a
by
which
●
electrical
cathode
(negative
electrode):
reduction
non-spontaneous
2+
Pb
chemical
reaction.
An
electrolytic
cell
is
used
●
this
purpose,
two
which
electrodes
(the
consists
cathode
of
a
and
single
the
overall
container,
anode),
(l)
+
2e
→
Pb(l)
for
cell
reaction:
a
PbBr
(l)
→
Pb(l)
+
Br
2
solution
be
(the
electrolyte),
considered
There
cell
are
but
at
as
many
SL
an
electron
different
you
and
will
be
a
battery
which
(g)
2
can
pump.
types
only
of
electrolytic
assessed
on
the
e
e
electrolysis
of
a
molten
salt.
Electrolysis of a molten salt such as
+
graphite electrode
graphite electrode
lead(II) bromide
e
e
In
the
PbBr
electrolysis
(l),
inert
of
molten
graphite
lead(II)
electrodes
bromide,
are
e
dipped
2
2+
into
the
PbBr
(l)
electrolyte.
The
following
Br
half-
Pb
2
equations
show
the
processes
that
take
place
at
the
e
Br
e
electrodes:
2+
Pb
Br
●
anode
(positive
electrode):
oxidation
PbBr
(l)
2
2Br
→
Br
(g)
+
2e
2
Figure 9 Electrolysis of molten lead bromide, PbBr
(l)
2
231
9
R E D O X
P R O C E S S E S
Working method for the electrolysis of a molten salt
Step
1:
Identify
all
species
present.
Step
4:
show
Step
2:
Identify
which
species
are
attracted
to
(negative
attracted
to
the
electrode)
anode
and
(positive
which
and
annotate
direction
of
the
the
electrolytic
movement
of
cell
and
electrons
the
and
cathode
Draw
the
species
the
direction
of
ion
ow.
are
electrode).
Step
5:
State
what
would
be
observed
at
each
electrode.
Step
3:
Deduce
place
at
the
cell
the
cathode
two
and
half-equations
anode
and
taking
the
overall
reaction.
Worked example
Describe
the
electrolysis
of
molten
sodium
chloride.
Step
4:
Solution
e
Step
e
1:
battery
iner t electrode
iner t electrode
+
NaCl
→
Na
+
Cl
Cl
(g)
2
+
So
Na(l)
and
Cl(l)
ions
are
present.
porous
separator
Step
2:
NaCl(l)
+
Cathode
(negative
electrode):
Na
+
(l)
Cl
Anode
(positive
electrode):
Cl
cathode
Step
(+)
3:
Anode
(positive
electrode):
(
)
oxidation:
Figure 10 Electrolytic cell for molten sodium chloride, NaCl(l).
This experimental set-up is used commercially in the dows
2Cl(l)
→
Cl
(g)
+
2e
2
 for the electrolysis of sodium chloride. The liquid sodium
Cathode
(negative
electrode):
reduction:
metal is less dense than the molten sodium chloride, so it
oats on the surface and is collected
+
Na(l)
This
+
needs
number
of
e
to
→
be
Na(l)
multiplied
electrons
from
by
the
2
to
balance
anode
the
Step
5:
equation:
At
the
anode
(positive
electrode):
bubbles
of
+
2Na(l)
Overall
+
cell
2e
→
2Na(l)
chlorine
reaction:
At
+
2Cl(l)
+
2Na(l)
→
the
liquid
2Na(l)
+
Cl
gas
are
cathode
sodium
observed.
(negative
electrode):
a
pool
of
forms.
(g)
2
Qk qsos
1
Explain why solid lead(II) bromide does not conduct
)
State a suitable material for each electrode.
)
Identify the direction of movement of electrons
electricity.
2
a)
Construct and annotate a diagram of the
and ion ow.
electrolytic cell for the electrolysis of molten
)
State what would be observed at each electrode.
f)
Discuss, with reference to dierences in
aluminium oxide.
b)
Identify the half-equations occurring at the
proper ties, why aluminium is used to replace iron
cathode and at the anode.
in many applications.
232
Q u e S t i O n S
Questions
1
Which
species
could
be
reduced
to
form
NO
?
5
Which
statement
about
the
electrolysis
of
2
molten
A.
N
sodium
chloride
is
correct?
O
2
B.
A.
A
B.
A
yellow-green
gas
is
produced
at
the
NO
3
negative
C.
electrode.
HNO
2
D.
NO
silvery
metal
is
produced
at
the
positive
[1]
electrode.
IB,
May
2011
C.
Chloride
ions
electrode
2
Consider
voltaic
the
overall
reaction
taking
place
in
D.
a
cell.
O(s)
+
Zn(s)
+
H
2
O(l)
→
2Ag(s)
+
Zn(OH)
2
What
is
the
role
of
The
positive
May
are
are
and
attracted
undergo
attracted
undergo
to
the
positive
oxidation.
to
the
negative
oxidation.
[1]
2011
(s)
2
zinc
in
the
cell?
6
A.
ions
electrode
IB,
Ag
Sodium
and
electrode
and
the
oxidizing
What
is
the
reducing
agent
in
the
reaction
agent.
below?
B.
The
C.
The
positive
electrode
and
the
reducing
agent.
2MnO
negative
electrode
and
the
oxidizing
(aq)
+
Br
(aq)
+
H
4
O(l)
→
2MnO
2
agent.
D.
The
(s)
+
2
BrO
+
2OH
(aq)
3
negative
electrode
and
the
reducing
agent.
[1]
A.
Br
B.
BrO
C.
MnO
D.
MnO
IB,
May
3
IB,
May
2011
4
[1]
2
3
What
BrO
happens
,
are
to
bromine
converted
to
when
bromine
bromate
ions,
molecules,
Br
3
A.
?
2
It
undergoes
state
B.
2012
It
undergoes
state
reduction
changes
from
1
and
to
oxidation
changes
from
1
and
to
its
oxidation
7
0.
Which
changes
electrode
its
oxidation
could
(cathode)
take
in
a
place
voltaic
at
the
positive
cell?
2+
I.
Zn
II.
Cl
III.
Mg(s)
A.
I
and
II
B.
I
and
III
(aq)
to
Zn(s)
0.
(g)
to
Cl
to
Mg
(aq)
2
C.
It
undergoes
reduction
and
its
oxidation
2+
state
D.
It
undergoes
state
4
changes
the
unknown
from
+
+5
following
metals
(aq)
+5
to
oxidation
changes
Consider
2XNO
from
X,
Y(s)
Y,
→
and
to
its
2X(s)
of
Y(NO
3
Y(NO
2XNO
(aq)
+
Z(s)
→
no
the
II
and
D.
I,
II,
IB,
May
III
and
only
III
[1]
2010
(aq)
2
reaction
2
(aq)
+
Z(s)
→
2X(s)
+
Z(NO
3
What
)
3
)
3
C.
only
three
Z.
+
only
oxidation
0.
reactions
and
(aq)
0.
)
3
is
the
metals
order
(least
A.
X
<
Y
<
Z
B.
X
<
Z
<
Y
C.
Z
<
Y
<
X
D.
Y
<
Z
<
X
IB,
May
of
increasing
reactive
(aq)
2
reactivity
of
rst)?
[1]
2011
233
9
R E D O X
8
Metal
P R O C E S S E S
A
voltaic
is
more
cell
is
reactive
made
as
than
shown
metal
B.
(gure
A
standard
10
Describe
how
the
concentration
11).
a)
V
a
car
factory
river
voltmeter
b)
A
a
in
after
farmer
dissolved
a
river
releases
using
puts
it
decrease
warm
for
large
oxygen
would
water
if:
into
the
cooling
[1]
quantities
of
a
fertilizer
B
salt bridge
on
IB,
11
a
eld
May
how
phosphates
solution containing
solution containing
2+
A
IB,
of
to
the
river.
[1]
2009
Describe
value
next
a
the
to
addition
water
water
November
can
of
nitrates
increase
the
or
BOD
sample.
[2]
2009
2+
(aq)
B
(aq)
Figure 11
12
Which
statement
is
correct?
The
nd
Winkler
the
method
uses
concentration
of
redox
reactions
oxygen
in
to
water.
3
100
A.
Electrons
ow
in
the
external
circuit
cm
of
analysed
Ato
using
was
this
taken
from
method.
The
a
river
and
reactions
B.
taking
B.
water
from
Positive
ions
ow
through
the
salt
place
are
summarized
below.
bridge
2+
from
A
to
Step
B.
1:
2Mn
(aq)
+
4OH
(aq)
+
O
(aq)
→
2
2MnO
C.
Positive
from
B
ions
to
ow
in
the
external
circuit
(s)
+
2H
2
O(l)
2
+
A.
Step
2:
MnO
(s)
+
2I
(aq)
+
4H
(aq)
→
2
2+
D.
Electrons
ow
through
the
salt
bridge
Mn
from
(aq)
+
I
(aq)
+
2H
2
B
to
A.
[1]
2
Step
3:
2S
O
2
IB,
November
O(l)
2
2
(aq)
+
I
3
(aq)
→
S
2
O
4
(aq)
+
2I
(aq)
6
2010
a)
State
what
happened
to
the
O
in
step
1
in
2
terms
9
A
0.1337
XIO
,
g
was
sample
of
dissolved
an
in
alkali
water,
metal
b)
iodate,
acidied,
State
of
electrons.
the
change
manganese
and
in
[1]
in
step
oxidation
number
for
2.
[1]
3
an
excess
of
potassium
iodide,
KI
added.
The
c)
0.0002
moles
of
I
were
formed
in
step
3.
3
resulting
the
iodine
sodium
solution
thiosulfate
required
36.64
pentahydrate
cm
of
Calculate
solution,
O
,
the
dissolved
amount,
in
the
2
3
Na
S
2
O
2
.5H
3
titration
the
starch
relative
identify
half-equations
234
(25.49
g
dm
)
for
complete
IB,
using
Calculate
hence
O
2
the
solution
atomic
metal.
involved.
as
an
mass
Deduce
indicator.
of
all
X
and
relevant
November
2009
in
moles,
water.
of
oxygen,
[1]
10
O R G A N I C
C H E M I S T R Y
Introduction
Organic
chemistry
containing
array
From
from
of
is
compounds
biological
foods
pesticides
the
compounds
to
and
and
systems
fuels,
chemistry
and
their
to
from
fertilizers,
of
fundamental
of
our
of
of
a
carbon-
organic
the
to
dyes
will
is
expansion
material
we
world.
be
the
In
and
their
alkenes,
and
develop
system
of
of
understanding
organic
rules
focus,
in
reactions.
will
The
of
the
The
nomenclature
to
functional
chemistry
of
halogenoalkanes,
be
of
compounds.
addition
important
alcohols,
benzene
an
of
IUPAC
main
identication
to
chemistry
the
topic,
application
biotechnology,
and
this
classication
vast
reactions.
paints
importance
understanding
studies
the
groups
alkanes,
polymers
explored.
10.1 Fnnts of ognc chst
Understandings
Applications and skills
➔
A homologous series is a series of compounds
➔
Explanation of the trends in boiling points of
of the same family, with the same general
members of a homologous series.
formula, which dier from each other by a
➔
Distinction between empirical, molecular, and
common structural unit.
structural formulas.
➔
Structural formulas can be represented in full
➔
Identication of dierent classes: alkanes,
and condensed format.
alkynes, halogenoalkanes, alcohols, ethers,
➔
Structural isomers are compounds with
aldehydes, ketones, esters, carboxylic acids,
the same molecular formula but dierent
amines, amides, nitriles, and arenes.
arrangements of atoms.
➔
➔
Identication of typical functional groups in
Functional groups are the reactive par ts of
molecules eg phenyl, hydroxyl, carbonyl,
molecules.
carboxamide, aldehyde, ester, ether, amine,
➔
Saturated compounds contain single bonds
nitrile, alkyl, alkenyl and alkynyl.
only and unsaturated compounds contain
➔
Construction of 3-D models (real or vir tual) of
double or triple bonds.
organic molecules.
➔
Benzene is an aromatic, unsaturated
➔
Application of IUPAC rules in the nomenclature
hydrocarbon.
of straight-chain and branched-chain isomers.
➔
Identication of primary, secondary, and
ter tiary carbon atoms in halogenoalkanes, and
alcohols and primary, secondary, and ter tiary
nitrogen atoms in amines.
➔
Discussion of the structure of benzene using
physical and chemical evidence.
Nature of science
➔
Serendipity and scientic discoveries – PTFE and
superglue.
➔
Ethical implications – drugs, additives, and
pesticides can have harmful eects on both
people and the environment.
235
10
O R G A N I C
C H E M I S T R Y
Introduction to organic chemistry
TOK
Organic
The theory of “Vitalism” was
based on the belief that
a vital force was involved
in the chemistry of living
organisms. Indeed the word
chemistry
compounds.
form
four
process
bonds,
bonds
by
to
which
is
the
eld
atoms
other
many
producing
chemistry
is
Carbon
of
have
atoms.
identical
a
wide
atoms
can
are
varied
studies
electrons
undergo
joined
branched,
and
that
valence
Carbon
straight-chain,
therefore
chemistry
four
or
eld
carbon-based
so
they
together
cyclic
can
catenation,
by
the
covalent
structures.
Organic
ofstudy.
“organic” originated from
Understanding
natural
and
synthetic
organic
compounds
requires
the
scientists’ understanding
study
of
chemical
bonding
and
nomenclature,
chemical
structure,
at that time, that organic
stoichiometric
relationships,
functional
groups,
and
reaction
mechanisms.
compounds could only be
The
energetics
and
the
of
reactions
and
their
role
in
industry,
chemical
kinetics,
synthesized within living
impact
of
synthetic
medicines
and
drugs
on
the
health
of
society
organisms. This belief
are
some
of
the
points
of
focus
of
organic
chemistry.
remained until the German
chemist Friedrich Wöhler
ar ticially synthesized
Homologous series
urea from the inorganic
Classication
is
a
common
human
activity.
Just
as
biology
uses
compound ammonium
scientic
cyanate, NH
taxonomy
to
classify
organisms
on
the
basis
of
shared
OCN.
4
characteristics,
Are there other examples in
group
science where vocabulary
of
and
chemists
name
utilize
compounds
a
unique
that
share
system
of
important
nomenclature
features
and
to
patterns
reactions.
has developed from a
A
homologous
series
is
a
series
of
compounds
that
can
be
grouped
misunderstanding that
together
based
on
similarities
in
their
structure
and
reactions.
A
was the product of the
homologous
series
has
the
same
general
formula
which
varies
from
one
technology of the day?
member
to
another
by
one
CH
(methylene)
group.
2
Language plays a vital
role in the communication
The
alkane
of knowledge and its
are
subsequent understanding.
alkenes
Therefore should language
that
series
has
the
general
formula
C
H
n
hydrocarbons
and
contain
(they
alkynes
contain
are
two
carbon–carbon
carbon
more
double
and
(table
2n
+
hydrocarbon
and
triple
1).
The
alkanes
2
hydrogen
only).
The
homologous
bonds,
series
respectively.
be universal so that
Homologous
series
that
contain
functional
groups
can
also
be
misnomers arising from
described
by
a
general
formula
and
also
show
similar
physical
and
misconceptions may be
chemical
properties
within
the
series.
The
functional
groups
are
the
eliminated?
reactive
as
parts
oxygen
the
of
and
the
nitrogen.
carbon–carbon
functional
groups
homologous
molecules
double
of
series
In
the
that
and
the
and
commonly
alkene
triple
series.
and
contain
alkyne
bonds
Table2
contain
elements
homologous
respectively
shows
the
make
structures
such
series
up
of
the
three
oxygen.
Physical properties of a homologous series
The
physical
gradually
boiling
as
points
apparatus
point
molar
rises
236
length
of
with
mass),
as
in
an
is
the
the
gure
a
of
in
the
1.
the
table
members
carbon
series:
liquid.
an
number
1.
of
This
a
chain
alkane
Such
increasing
seen
within
pentane
of
of
members
shown
temperature
while
properties
the
series
can
experiment
of
can
butane
homologous
increases.
is
carbon
be
a
seen
gas
at
be
atoms
the
room
change
example,
measured
shows
by
series
For
that
(or
the
using
the
increasing
state
at
the
boiling
room
temperature,
10 . 1
N
Fo
F u N d a m e N T a l S
O F
Conns stct fo
O r G a N i C
C H e m i S T r y
Stct fo
Bong ont / °C
H
methane
C
CH
CH
4
161
4
H
ethane
C
H
2
propane
C
C
4
pentane
C
5
hexane
C
6
▲
CH
8
H
CH
10
H
CH
12
CH
3
H
CH
14
CH
3
CH
2
C
H
H
H
89
2
H
CH
2
CH
2
H
H
H
C
C
C
H
H
H
H
42
H
H
H
H
C
C
C
C
H
H
H
H
H
0.5
3
CH
2
H
CH
2
CH
2
C
3
CH
2
H
CH
2
CH
3
H
3
CH
3
H
CH
3
H
3
butane
CH
6
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
36
3
CH
2
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
H
CH
H
69
3
T
able 1 The homologous series of alkanes
Hooogos
cohos
hs
ktons
ss
Gn
C
H
n
fo
C
OH
H
H
H
n
2n+1
H
O
C
2n
H
H
H
n
O
2n
H
O
H
C
C
C
O
H
C
C
C
C
H
OH
C
C
H
C
H
3
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
O
H
H
C
C
C
C
H
H
O
H
C
C
C
C
C
H
OH
C
C
C
H
C
H
4
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
O
H
H
C
C
C
C
C
H
H
H
H
O
C
H
C
C
C
C
C
OH
H
C
C
C
C
H
C
H
5
H
H
▲
H
H
H
H
H
H
H
H
T
able 2 The general formula and structural formulae of the homologous series of alcohols, aldehydes, and ketones
237
10
O R G A N I C
C H E M I S T R Y
This
trend
forces
becomes
chain
oil
is
in
boiling
(London
longer.
length
a
chain.
are
mixture
Trends
well
of
boiling
in
point.
fractions
A
in
points
to
by
that
is
density
the
vary
in
the
as
and
viscosity
the
the
length
separation
mixture
apparatus
non-volatile
strong
4.4)
petrochemical
physical
separate
distillation
long-chain,
a
increasingly
sub-topic
increasing
understood
simple
from
forces,
distillation
boiling
from
results
hydrocarbons
Fractional
differences
point
(dispersion)
can
intermolecular
carbon
of
their
process
into
in
carbon
industry.
a
Crude
carbon
that
fractions
effectively
compounds
chain
with
uses
of
separate
school
similar
volatile
laboratory.
thermometer
condenser
Homologous series have
similar chemical proper ties
due to the presence of the
water
same functional group; this is
to sink
responsible for their overall
water
chemical reactivity and
distillate
from faucet
the types of characteristic
reactions they undergo.
▲
Figure 1 Distillation apparatus incorporating a temperature probe
Qck qstons
1
Alkenes are impor tant star ting materials for a variety
b)
Applying IUPAC rules, state the names of
of products.
)
isomers B and C.
State and explain the trend of the boiling
[2]
iB, Nov 20 09
points of the rst ve members of the alkene
homologous series.
b)
[3]
Describe two features of a homologous series.[2]
iB, m 2011
2
The boiling points of the isomers of pentane, C
5
H
,
12
A
shown in gure 2 are 10 °C, 28 °C, and 36 °C, but not
necessarily in that order.
)
Identify the boiling point for each of the isomers
a, B, and C in a copy of table 1 and state a reason
for your answer.
iso
[3]
a
B
C
Bong ont
B
▲
238
T
able 3
▲
C
Figure 2
10 . 1
F u N d a m e N T a l S
O F
O r G a N i C
C H e m i S T r y
Chemical formulae of organic compounds
The
structure
different
In
of
ways
sub-topic
organic
providing
4.3
structures.
an
we
These
compound
varying
examined
are
useful
levels
the
to
may
use
of
of
be
represented
the
H
(electron-dot)
valence
electrons
present
molecular
compounds
and
polyatomic
H
C
C
H
H
H
in
▲
simple
H
several
information.
Lewis
visualize
in
Figure 3 Lewis structure
ions.
of ethane, C
H
2
Empirical
formulae
atoms
present
actual
number
formula
offer
in
a
of
(sub-topic
molecule.
atoms
little
or
The
present
no
1.2)
represent
molecular
in
the
information
the
formula
molecule.
about
simplest
the
Both
ratio
describes
these
possible
the
types
structure
O
of
of
H
larger,
more
Structural
complex
6
of
C
O
H
molecules.
formulae
take
three
forms:
full,
condensed,
and
▲
skeletal.
Figure 4 Lewis structure
of methanoic acid, HCOOH
●
Full
structural
showing
another
all
in
the
a
formulae
atoms
and
are
two-dimensional
bonds,
and
their
representations
positions
relative
to
one
compound.
In structural formulae a
●
In
a
condensed
structural
formula
all
the
atoms
and
their
relative
covalent bond between two
positions
are
represented
but
the
bonds
are
omitted.
atoms is represented by a
●
A
skeletal
formula
the
end
atoms
table
formula
where
of
each
present
the
the
carbon
line
in
is
and
most
and
each
functional
basic
representation
hydrogen
vertex
groups
atoms
represents
are
also
are
a
of
not
the
shown
carbon
included
as
structural
single line that describes two
but
atom.
shown
bonding electrons. For a double
The
bond two lines are used and
in
for a triple bond, three lines
4.
N
propane
(sub-topic 4.2).
F stct fo
H
H
H
H
C
C
C
Conns stct fo
H
CH
CH
3
H
H
H
H
H
H
H
C
C
C
CH
2
3
H
OH
propan-2-ol
CH
CH(OH)CH
3
H
O
Skt fo
3
H
H
H
H
C
C
O
O
propanal
H
C
CH
CH
3
CHO
2
H
H
H
H
O
H
C
C
C
O
propanone
H
H
CH
C(O)CH
3
H
3
H
H
H
C
propene
H
CH
H
C
H
▲
C
3
CH=CH
2
or CH
3
CHCH
2
H
T
able 4 Full, condensed, and skeletal structural formulae can all be used to represent organic compounds
239
10
O R G A N I C
C H E M I S T R Y
Nomenclature of organic compounds
The
International
world
authority
substance
of
needs
compound
substituents
parts
The
that
Union
on
to
which
form
the
the
and
Applied
nomenclature.
enough
the
functional
describe
alkanes
Pure
provide
from
and
of
chemical
information
chemical
groups
compound
backbone
is
the
name
to
derived,
present.
(gure
of
Chemistry
The
The
(IUPAC)
of
a
signpost
the
including
name
has
is
the
chemical
a
class
any
number
of
5).
IUPAC
rules
for
naming
organic
compounds.
The longest carbon chain that
includes the principal group
or the most complex cyclic
Sux indicating
or heterocyclic system
the principal group
-ane
a
b
c
-ene
Principal (parent) chain
-yne
Prexes
Suxes indicating
in alphabetical order
saturation or unsaturation
of the principal chain
▲
lngth of
Figure 5 Outline of the nomenclature of organic compounds
Nomenclature of alkanes
N
cbon chn
1
Examine
the
continuous
1
structure
carbon
of
the
chain.
compound
This
provides
and
the
determine
root
name
the
for
longest
the
alkane
meth-
(table
2
eth-
3
prop-
4
but-
5
pent-
6
hex-
2
If
5).
alkyl
the
substituents
branch
sufx
will
will
be
change
are
present,
determined
from
creating
by
“-ane”
to
the
branched
number
of
chains,
carbons
the
name
(table
5).
for
The
“-yl”.
H
methyl
H
▲
T
able 5 The IUPAC root names for the
C
substituent
H
H
H
H
C
C
C
C
H
H
H
H
alkane series
3
Sbsttnt
Conns
n
fo
When
numbering
substituent
must
numbering
from
on
carbon
2.
substituent
the
be
longest
the
left
to
right
Numbering
on
carbon
5.
CH
methyl
3
H
CH
ethyl
CH
2
3
H
propyl
CH
CH
2
C
H
CH
2
3
H
butyl
CH
2
CH
CH
2
2
CH
3
H
H
C
C
1
H
▲
T
able 6 Naming alkyl substituents
240
C
2
H
H
C
3
H
H
H
H
C
4
H
H
C
5
6
H
carbon
lowest
results
from
chain,
numbered
right
in
the
to
left
the
position
carbon.
methyl
would
In
this
of
any
example,
substituent
incorrectly
being
have
the
10 . 1
4
When
there
are
alphabetical
several
order
different
prior
to
the
F u N d a m e N T a l S
substituents,
root
arrange
O F
them
O r G a N i C
C H e m i S T r y
in
name.
H
ethyl
H
H
H
C
H
C
H
H
H
C
H
C
2
C
H
C
4
5
H
H
H
C
3
H
H
C
C
1
H
substituent
3-ethyl-2-methylhexane
6
H
H
H
methyl
substituent
H
alphabetically, ethyl comes before methyl.
5
Use
a
comma
6
Use
a
hyphen
7
The
number
prexes
8
to
to
of
in
words
table
are
1
mono
2
di
numbers.
separate
multiple
shown
Successive
separate
numbers
and
substituents
of
letters.
the
same
type
is
indicated
3
tri
4
tetra
5
penta
by
7.
merged
into
one
word.
▲
T
able 7 Numerical multipliers in the
IUPAC nomenclature system
To
or
demonstrate
structural
compounds
formula.
1
the
isomers
that
have
Isomers
Begin
by
application
have
drawing
carbon–carbon
of
the
the
of
rules
hydrocarbon
same
unique
the
these
chemical
physical
molecule
we
shall
hexane.
formula
and
with
Structural
but
chemical
the
examine
a
the
isomers
isomers
different
are
structural
properties.
longest
straight
chain
of
isos
atoms.
Isomers may dier from one
another in their physical
H
C
CH
CH
3
2
CH
2
CH
2
CH
2
3
proper ties. The ability of
hexane
molecules of one isomer to
pack closer together will result
2
Reduce
to
act
so
this
of
the
as
the
a
longest
methyl
derivative
chain
lowest
chain
numbered
one
substituent.
of
must
by
hexane
result
in
is
carbon
The
a
the
and
longest
chain
substituted
methyl
use
the
is
now
pentane.
group
removed
ve
The
branching
carbon
carbons
numbering
off
at
the
in increased intermolecular
forces and therefore an
increase in the boiling point.
Hexane molecules (boiling
carbon.
point 69 °C) can approach each
other more closely than those
CH
3
H
C
of the branched-chain hexane
CH
3
CH
CH
2
1
2
2
3
derivative 2,3-dimethylbutane
CH
3
4
5
(boiling point 58 °C).
2-methylpentane
3
Examine
another
is
the
not
the
isomer
carbon.
last
exist
of
as
If
the
it
is
it
to
is
see
if
substituted
the
the
moved
same
as
to
methyl
C3
pentane
substituent
another
isomers
isomer
can
is
be
moved
formed.
to
This
(4-methylpentane
does
2-methylpentane).
CH
3
H
C
3
CH
CH
CH
2
1
2
CH
2
3
4
3
5
3-methylpentane
241
10
O r G a N i C
C H e m i S T r y
4
In
a
similar
way,
substituted
Qck qston
now
butane
remove
another
compounds
from
carbon
the
atom
original
and
create
hexane:
Applying IUPAC nomenclature
CH
CH
3
CH
3
3
rules, state the name of each of the
H
molecules shown in gure 6.
C
C
3
a)
CH
3
CH
CH
2
CH
CH
2
2
CH
2
CH
H
CH
2
1
C
CH
3
3
3
4
CH
CH
3
1
2
3
4
CH
3
3
2,2-dimethylbutane
CH
2,3-dimethylbutane
3
CH
3
b)
CH
3
CH
C
CH
2
CH
3
CH
CH
3
2
CH
3
CH
Saturated and unsaturated hydrocarbons
Hydrocarbons
3
CH
c)
CH
2
CH
3
2
organic
atoms
only.
bonds
single
bonds.
or
CH
are
hydrogen
are
triple
In
a
compounds
saturated
Unsaturated
carbon–carbon
bonds.
The
consisting
compound
of
all
compounds
simplest
carbon
the
contain
example
of
and
carbon–carbon
a
double
and/
saturated
3
hydrocarbon
is
methane,
CH
,
a
member
of
the
alkane
family.
Alkanes
4
CH
3
CH
CH
3
are
CH
2
CH
aliphatic
The
CH
3
This
3
▲
naturally
compounds
occurring
(gure
7).
mixture
is
extracted
from
hydrocarbons
beneath
the
come
Earth’s
from
crude
surface,
re ned,
CH
C
3
and
CH
of
straight-chain
3
oil.
d)
majority
or
CH
2
separated
petroleum,
by
fractional
butane,
and
distillation
into
useful
substances
such
as
kerosene.
Figure 6
The
of
mixture
mainly
of
Cycloalkanes
(gure
hydrocarbons
alkanes,
8),
are
ring
whereas
consisting
of
that
cycloalkanes
structures
aromatic
alternating
makes
and
that
up
contain
hydrocarbons
single
and
crude
aromatic
double
oil
single
or
is
a
combination
hydrocarbons .
carbon–carbon
arenes
are
ring
carbon–carbon
bonds
structures
bonds.
Functional groups
Tens
is
of
millions
constantly
companies
and
organic
as
new
chemical
products
of
Natural
compounds
organisms.
reactions
All
compounds
▲
of
rising
these
compounds
exist
compounds
are
industries.
involving
found
containing
Synthetic
both
in
substances
natural
plants
are
specic
in
the
world
synthesized
and
man-made
animals
functional
into
are
number
are
the
compounds.
synthesized
classes
groups
the
pharmaceutical
compounds
and
organized
and
by
(table
of
by
organic
7).
Figure 7 Computer-generated 3D models
When
naming
compounds
that
contain
a
functional
group,
the
position
of methane, ethane, and propane, the
of
the
group
is
identied
by
giving
the
number
of
the
carbon
atom
to
first three members of the alkane series
which
it
groups
is
attached.
take
priority
When
over
numbering
substituents
the
and
carbon
atoms,
carbon–carbon
functional
multiple
bonds.
Unsaturated hydrocarbons
The
or
primary
triple
left
to
right,
double
Figure 8 Computer-generated model of
the
bond
numbering
▲
chain
in
unsaturated
carbon–carbon
methyl
would
is
from
be
bond.
H
H
H
H
C
C
C
C
C
hydrocarbons
the
substituent
at
right
H
If
C3.
to
molecule
would
However,
left
as
must
below
branch
the
off
double
include
was
from
bond
the
double
numbered
C2
takes
and
from
the
priority
shown.
4-methylpent-2-ene. Note the use of hyphens and the fact
cyclohexane, a cycloalkane
H
5
H
4
3
CH
3
242
2
H
1
H
that the substituent is named before the functional group
so
10 . 1
Css
Fncton go
Sx
alkanes
―
-ane
F u N d a m e N T a l S
O F
O r G a N i C
Gn fo
C H e m i S T r y
ex
Constructing 3-D models
C
Propane C
H
n
H
3
2n + 2
8
of organic compounds is
an excellent interactive
but-2-ene
alkenes
C
alkenyl
C
-ene
H
C
n
technique, enhancing
2n
CH=CHCH
CH
3
3
visualization of the molecule
C
C
and mutual orientation of
but-2-yne
alkynes
-yne
C
H
n
2n
2
alkynyl
individual atoms. Models can
C≡CCH
CH
3
3
enhance understanding of a
benzene
arenes
phenyl
―
C
variety of concepts from the
H
n
2n
6
H
C
6
naming of organic molecules,
6
visualizing stereoisomers
2-chlorobutane
―X
halogenoalkanes
―
C
H
n
(including optical isomers)
X
2n + 1
(X = F, Cl, Br, I)
CH
CH(Cl)CH
3
CH
2
3
to complex reactions
OH
mechanisms (sub- topic 20.1).
butan-2-ol
alcohols
-ol
ROH
CH(OH)CH
CH
hydroxyl
3
CH
2
3
O
ethanal
aldehydes
aldehyde
-al
C
RCHO
CH
CHO
3
H
C
propanone
ketones
-one
RC(O)R′
O
C(O)CH
CH
3
3
carbonyl
O
-oic
C
carboxylic acids
ethanoic acid
RCOOH
acid
OH
COOH
CH
3
carboxyl
O
methyl ethanoate
C
-oate
esters
RCOOR′
COOCH
CH
O
3
3
ester
O
ethoxyethane
ethers
―
ROR′
CH
ether
CH
3
OCH
2
CH
2
3
H
N
propan-1-amine
RNH
2
CH
H
CH
3
amines
-amine
N
CH
2
NH
2
2
RNHR′
N-methylethanamine
N
RN(R′)R′′
H
CH
CH
3
NHCH
2
3
amino
O
C
ethanamide
amides
-amide
RCONH
2
CONH
CH
NH
2
3
2
amido
C
N
nitriles
propanenitrile
-nitrile
RCN
CH
cyano
▲
3
CH
CN
2
T
able 7 A summary of classes of organic compounds showing their functional groups
243
10
O R G A N I C
C H E M I S T R Y
Aliphatic compounds
H
H
H
H
H
H
H
H
H
H
Cl
H
H
H
C
C
C
C
C
H
Cl
H
H
H
O
H
C
C
C
H
C
H
C
C
C
C
H
C
C
C
H
O
H
H
H
H
H
H
H
OH
H
H
H
butan-2-ol
Number
chain
so
that
group
the
from
the
is
numbered
butanal
carbon
right
to
The
left
for
functional
on
the
H
carbon
atom.
there
this
group
aldehyde
denition
lowest
carbon
functional
an
propanone
on
the
atom
is
no
is
has
the
state
be
general
on
In
For
this
group
multi-substituted
halogenoalkanes,
compound
functional
only
a
RC(O)R ′.
three-carbon
so
to
ketone
formula
terminal
(C1),
need
The
by
2,2-dichloropentane
carbon
can
along
C2.
number
with
signify
number.
the
halogens
in
H
H
H
H
C
C
H
H
O
C
C
C
H
O
H
halogens
they
are
H
H
C
C
H
H
are
listed
order.
H
H
C
C
H
H
C
C
O
H
of
If
H
H
H
to
number
alphabetical
O
H
used
prex
present.
different
present,
a
the
is
H
C
O
H
H
H
butanoic acid
Note
that
carbon
the
atom
functional
By convention, the functional
counted
group of the carboxylic acids,
carbon
aldehydes, esters and amides
functional
are positioned at the right-hand
the
end of the structural formulae
there
of organic compounds.
include
Esters
in
the
group
in
the
chain.
group
is
The
hydrogen
the
functional
COOH.
The
atom
are
The
on
is
O
R ′
the
group
group
C1
in
is
no
need
the
the
position,
substituent
to
rst,
number
the
name.
is
OCH
alkyl
followed
name
of
group
alkoxy
so
CH
ethoxy
named
as
alkanes.
called
2
terminal
named
substituted
alkyl
replaces
the
in
Ethers
acid
substituent
the
is
are
derivatives.
longest
As
ethoxyethane
methyl propanoate
is
the
3
group.
by
the
acid
anion.
PTFE, a fortunate discovery
PTFE
or
polytetrauoroethene
polymer
that
commercial
polymer
is
more
name,
which
is
commonly
Teon.
means
it
It
is
can
a
be
a
synthetic
known
by
its
thermoplastic
moulded
This
in
uoropolymer
1938,
a
product
polymerization
retaining
its
new
shape
upon
a
properties
low
coefcient
electrical
surface
all
of
and
PTFE
is
of
a
chemical
friction,
and
solvents.
as
high
thermal
qualities,
known
implements.
244
include
high
very
The
non-stick
low
most
resistance,
melting
insulation,
point,
non-stick
solubility
common
surface
on
discovered
iron-catalysed
tetrauoroethene
gas.
describes
fortunate
accidental
cooling.
discoveries
Its
accidently
the
when
Serendipity
heated,
of
was
of
in
use
cooking
stems
and
from
within
ashes
serendipity.
“superglue”
discovery.
risk
is
of
good.”
the
(IB
of
Methyl
and
scientic
Scientic
inspiration,
another
“Ethical
assessment
parts
science.
cyanoacrylate
example
discussions,
the
Chemistry
of
of
an
or
accidental
risk-benet
precautionary
way
endeavour
imagination,
addressing
syllabus.)
analyses,
principle
the
are
common
all
H
10 . 1
F u N d a m e N T a l S
O F
O r G a N i C
C H e m i S T r y
Classifying molecules: primary, secondary,
C o x tcton on
and tertiary compounds
th wo
We
shall
look
at
three
classes
of
compound
to
understand
the
effect
A large majority of the world’s oil
on
their
chemical
reactions
of
changing
the
position
of
the
functional
reserves are controlled by a small
group
on
the
carbon
chain.
We
shall
look
at
the
carbon
attached
to
the
number of countries. Countries
functional
group:
a
primary
carbon
atom
is
bonded
to
one
other
carbon
with oil reserves in decreasing
atom,
a
secondary
carbon
atom
to
two
other
carbon
atoms,
and
a
order of their magnitude include
tertiary
carbon
atom
to
three
other
carbon
atoms
(gure
9).
Saudi Arabia, Venezuela, Canada,
Iran, Kuwait, United Arab Emirates,
Classifying halogenoalkanes
Alkanes
undergo
(sub-topic
known
10.2)
as
and
the
reaction.
is
This
formed
idea
Whether
be
reactions
with
mono-substituted
depends
will
Russia, Libya, Kazakhstan, Qatar,
substitution
resulting
halogenoalkanes.
halogenoalkane
the
free-radical
a
on
primary,
the
developed
alkanes
secondary,
conditions
in
and
sub-topic
USA, China, Brazil, Algeria, and
halogens
Mexico. These countries dier
are
or
widely in their economic, social,
tertiary
mechanism
of
and political demographics.
20.1.
It is instructive to compare
countries that are net expor ters
H
3°carbon
with those that are net impor ters
1°carbon
C
H
2°carbon
H
of crude oil. Nt xo ts produce
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
Cl
H
H
H
more barrels of oil than they
H
Cl
H
H
H
C
C
C
H
H
Cl
H
consume, while nt o ts
consume more barrels of oil than
they produce. For many years
1
-chloropropane
2-chloropropane
2-chloro-2-methylpropane
the USA has been the largest net
▲
Figure 9 Primary (1°), secondary (2°), and ter tiary (3°) halogenoalkanes
impor ter of crude oil, but in 2013
it was over taken by China. China’s
wish to reduce its heavy reliance
Classifying alcohols
on non-renewable energy sources
Alcohols
The
can
position
when
the
be
classied
of
the
alcohol
potassium
in
the
hydroxyl
group
undergoes
dichromate(VI)
same
way
halogenoalkanes
determines
oxidation
or
as
in
potassium
the
the
(gure
products
presence
of
manganate(VII)
10).
formed
is reected in its domestic and
international policies – China
acidied
is a world leader in investment
(sub-topic
10.2).
in renewable energy sources,
par ticularly wind energy.
H
A country’s reliance on oil
1°carbon
H
2°carbon
C
H
3°carbon
shapes its global policies.
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
OH
H
H
H
Relationships with countries that
H
OH
H
H
H
C
C
C
H
OH
H
H
are net expor ters are of paramount
propan1
-ol
propan-2-ol
impor tance and the need to secure
energy supplies for a nation can
2-methylpropan-2-ol
result in governments overlooking
▲
Figure 10 Primary (1°), secondary (2°), and ter tiary (3°) alcohols
perceived political dierences of
trading par tners.
Biofuels
are
in
Alcohols
plants.
alternative
large-scale
it
to
substances
fuels
to
and
other
petrol
production
traditional
whose
fossil
of
energy
biofuels
(gasoline)
ethanol
fuels.
Fossil
is
are
and
from
fuels
derived
from
increasingly
diesel.
sugar
Brazil
cane
remain
the
for
carbon
being
has
xation
used
as
undertaken
decades,
primary
St t
adding
source
of
When classifying compounds as
energy
on
a
global
scale,
but
the
complex
mixture
that
makes
up
crude
primary, secondary, or ter tiary,
oil
contains
components
that
can
be
used
for
the
synthesis
of
various
focus on the carbon that is bonded
products
–
from
dyes
and
cosmetics
to
pesticides
and
polymers.
The
ever-
to the functional group.
increasing
in
the
combustion
depletion
materials
for
a
not
vast
of
only
array
valuable
of
of
the
non-renewable
fuels
themselves,
substances
that
are
a
fossil
but
part
fuels
also
of
of
our
could
result
valuable
daily
raw
lives.
245
10
O r G a N i C
C H e m i S T r y
Classifying amines
Octn nbs of fs
An
amine
is
classied
as
a
primary,
secondary
or
tertiary
amine
The fractional distillation of crude
depending
on
the
number
of
alkyl
groups
bonded
to
the
nitrogen
atom
oil or petroleum separates this
of
the
functional
group
(unlike
halogenoalkanes
and
alcohols,
which
mixture of hydrocarbons into
consider
the
carbon
atom
next
to
the
functional
group).When
naming
various fuels including natural
amines
the
root
loses
the
-e
and
is
replaced
by
“amine”.
“ N”
signies
that
gas, gasoline or petrol, kerosene
the
substituent,
namely
the
methyl
group
in
gure
11,
is
bonded
to
the
used in the aviation industry,
nitrogen
atom
rather
than
the
carbon
atom.
diesel fuel used in transpor t,
construction, and agriculture,
1°amine
2°amine
3°amine
and fuel oil for heating.
H
H
H
H
The octn nb is a standard
H
H
H
H
H
H
C
C
C
H
H
H
CH
CH
3
H
C
C
C
H
H
H
N
H
C
C
C
H
H
H
3
N
H
N
method of describing the
H
H
CH
3
performance of fuels used in cars
and aircraft. The octane number
propan1
-amine
N-methylpropan-
N, N-dimethylpropan-
1
-amine
1
-amine
is not indicative of the energy
content of the fuel but is rather
▲
Figure 11 Primary (1°), secondary (2°), and ter tiary (3°) amines
a way of describing its ability to
combust in a controlled manner
without causing excessive
Aromatic hydrocarbons
ngn knockng. In simple
The
alkanes,
alkenes,
and
alkynes
are
differentiated
by
the
presence
of
terms, engine knocking is the
single,
double,
or
triple
bonds
respectively.
Aromatic
hydrocarbons
result of uncontrolled detonation
are
characterized
by
the
presence
of
the
benzene
ring.
The
German
of the air–fuel mixture in a
scientist
August
Kekulé
(1829–96)
proposed
a
ring
structure
for
benzene
combustion engine. This is more
C
H
6
composed
of
six
carbons
bonded
together
by
alternating
double
and
6
common in fuels that have a low
single
bonds.
This
structure
would
result
in
an
unsymmetrical
molecule
octane number.
with
carbon–carbon
bonds
of
different
bond
lengths.
Automobile manufacturers
Benzene
crystallizes
upon
cooling
and
analysis
of
X-ray
diffraction
are multinational companies
patterns
generated
from
the
crystalline
substance
revealed
that
all
six
that produce and sell their
carbon–carbon
bonds
in
its
molecule
have
identical
bond
lengths
of
products globally. The octane
140
pm.
It
is
now
understood
that
the
carbon–carbon
bonds
in
benzene
rating system is expressed
are
intermediate
in
length
between
single
(154
pm)
and
double
(134
pm)
dierently in dierent countries,
carbon–carbon
bonds
and
thus
have
a
bond
order
of
1.5.
Electrostatic
making comparisons difficult.
potential
mapping
of
benzene
(gure
12)
conrms
that
all
the
carbon
The sch octn nb
atoms
have
equal
electron
density,
so
the
molecule
(RON) is the most common
method of determining the
octane number, though the
oto octn nb (MON)
is also used either in isolation
or in combination with the RON.
Thus methods of repor ting fuel
octane numbers at the pump can
vary from country to country,
leading to inconsistency in
communication to consumers.
▲
246
Figure 12 An electrostatic potential map of benzene
is
symmetrical.
10 . 1
F u N d a m e N T a l S
O F
O r G a N i C
C H e m i S T r y
2
Benzene
bonded
contains
to
one
six
sp
another
hybridized
and
each
carbon
carbon
is
atoms
(sub-topic
bonded
to
a
14.2)
single
hydrogen
2
atom
by
atoms
and
sigma
overlap
below
electrons
one
the
over
structures
of
bonds.
p
another,
plane
the
The
of
the
orbitals
forming
six
a
carbon
six
carbon
nuclei
benzene
shown
in
H
of
the
sp
hybridized
continuous
atoms.
can
gure
six
be
π
The
bond
carbon
that
lies
delocalization
represented
by
the
above
of
the
π
resonance
13.
The Kekulé structure of
H
H
H
H
H
benzene was drawn as a
series of alternating double
H
H
H
H
and single bonds.
H
The circle
H
within the six-member ring
H
structure represents a system
H
H
of delocalized pi electrons that
H
H
are evenly distributed between
H
▲
Figure 13 Resonance structures of benzene
Hydrogenation
in
the
presence
carbon–carbon
is
of
benzene
of
hydrogenation
is
as
in
the
the
fact
would
much
enhanced
benzene
topic
be
less
readily
20.1)
aliphatic
decolorize
benzene
but
and
does
cyclic
than
π
the
of
this,
or
the
and
A
do;
for
one
the
difference
this
reactions
as
isomers
suggested
reactions
other
for
if
by
from
of
the
stability
benzene
exists
that
known
evidence
resulting
example,
two
as
is
change
show
energy,
energy,
of
three
hydrogenation
enthalpy
in
substitution
isomer
be
bonds
of
hydrocarbons
contained
experiments
molecule
addition
would
double
unsaturated
benzene
consequence
electrophilic
Only
there
this
If
times
However,
benzene
show
water.
three
to
change
delocalization
electrons.
not
10.2).
enthalpy
and
compounds
compounds;
single
hydrogen
approximately
undergoes
bromine
alternating
bonds,
energy
the
of
(sub-topic
cyclohexene.
stability
of
addition
catalyst
of
resonance
delocalization
the
a
double
of
it
each of the six carbons.
is
that
(sub-
unsaturated
does
not
1,2-disubstituted
the
benzene
ring
had
Kekulé.
TOK
August Kekulé is best known for his discovery of the structure of benzene.
While many scientic discoveries are the product of reasoning suppor ted
by evidence obtained through observation and experimentation, some
discoveries are born from moments of inspiration or ashes of intuition, as well
as a healthy imagination with a high degree of creativity. Kekulé is said to have
visualized the cyclic structure of benzene in a dream. While the impor tance
of evidence is universally accepted in the scientic community, there is
an understanding of the role of less analytical ways of knowledge in the
acquisition of scientic knowledge. To what degree do these ways of knowing
play a par t in the acquisition of new knowledge, and can you think of some
recent examples?
▲
Figure 14 August Kekulé
247
10
O r G a N i C
C H e m i S T r y
10.2 Fncton go chst
Understandings
Applications and skills
Alkanes:
Alkanes:
➔
Alkanes have low reactivity and undergo free➔
Writing equations for the complete and
radical substitution reactions.
incomplete combustion of hydrocarbons.
Alkenes:
➔
➔
Alkenes are more reactive than alkanes and
Explanation of the reaction of methane and
ethane with halogens in terms of a free-
undergo addition reactions. Bromine water can
radical substitution mechanism involving
be used to distinguish between alkenes and
photochemical homolytic ssion.
alkanes.
Alkenes:
Alcohols:
➔
➔
Alcohols undergo nucleophilic substitution
Writing equations for the reactions of
alkenes with hydrogen and halogens and of
reactions with acids (also called esterication
symmetrical alkenes with hydrogen halides
or condensation) and some undergo oxidation
and water.
reactions.
➔
Halogenoalkanes:
➔
Halogenoalkanes are more reactive than
Outline of the addition polymerization of
alkenes.
➔
alkanes. They can undergo (nucleophilic)
Relationship between the structure of the
monomer to the polymer and repeating unit.
substitution reactions. A nucleophile is an
Alcohols:
electron-rich species containing a lone pair that
➔
Writing equations for the complete combustion
it donates to an electron-decient carbon.
of alcohols.
Polymers:
➔
➔
Writing equations for the oxidation reactions
Addition polymers consist of a wide range of
of primary and secondary alcohols (using
monomers and form the basis of the plastics
acidied potassium dichromate(VI) or
industry.
potassium manganate(VII) as oxidizing
Benzene:
➔
agents). Explanation of distillation and reux
Benzene does not readily undergo addition
in the isolation of the aldehyde and carboxylic
reactions but does undergo electrophilic
acid products.
substitution reactions.
➔
Writing the equation for the condensation
reaction of an alcohol with a carboxylic acid,
in the presence of a catalyst (eg concentrated
Nature of science
➔
sulfuric acid) to form an ester.
Use of data – much of the progress that has
Halogenoalkanes:
been made to date in the developments and
➔
Writing the equation for the substitution
applications of scientic research can be
reactions of halogenoalkanes with aqueous
mapped back to key organic chemical reactions
sodium hydroxide.
involving functional group interconversions.
248
10 . 2
F u N C T i O N a l
G r O u p
C H e m i S T r y
Converting one functional group to another
Chemical
form
the
research
the
reactions
basis
into
of
natural
in
a
to
both
advance
the
improve
their
to
to
the
A
of
has
their
and
and
of
so
on
the
they
to
utilize
can
various
food
supplies
populations,
develop
be
of
S o c i e t y ’s
ensure
global
to
involves
chemistry
molecules.
communities,
research
that
the
chemists
organic
developing
drives
subsequently
data
Scientific
production
synthesis
of
interconversions
compounds.
their
enabled
new
health
life
and
wealth
create
group
organic
structure
achieve
developed
quality
of
compounds
groups
pathways
functional
synthesis
l a b o r a t o r y.
functional
reaction
desire
of
pathways
produced
organic
involving
the
determination
design
for
for
new
and
organic
compounds.
Alkanes
Alkanes
are
the
simplest
hydrocarbons.
Wi t h
low
bond
polarity
and
1
strong
covalent
carbon–carbon
bonds
(bond
energy
346
kJ
mol
)
1
and
are
carbon–hydrogen
relatively
inert.
bonds
(bond
H o w e v e r,
energy
alkanes
do
414
kJ
undergo
mol
),
they
some
important
reactions.
The combustion of alkanes
Alkanes
in
are
commonly
combustion
liquid
to
Alkanes
and
gas)
decreases
used
as
fuels
as
fuels,
Volatility
as
the
tend
to
releasing
(the
length
be
large
tendency
of
the
to
carbon
short-chain
amounts
change
chain
molecules
of
energy
state
from
increases.
such
as
butane
octane.
Alkanes
undergo
oxygen.
This
and
used
reactions.
water.
complete
highly
Carbon
contributing
to
hydrocarbons
dioxide
global
with
combustion
exothermic
has
a
warming.
octane
reaction
in
signicant
Petrol
present
in
or
the
the
presence
produces
carbon
of
environmental
gasoline
highest
is
a
excess
dioxide
impact,
mixture
of
proportion.
1
_
C
H
8
(l)
+
12
O
18
(g)
→
8CO
2
(g)
+
9H
2
O(g)
∆H
=
-5470
kJ
2
2
Qck qston
Incomplete combustion
When
oxygen
combustion.
gas,
is
is
in
In
produced.
reducing
its
limited
this
It
supply
reaction
irreversibly
oxygen-carrying
alkanes
carbon
binds
undergo
monoxide,
to
Deduce balanced equations for
incomplete
which
hemoglobin
in
is
a
the
poisonous
blood
the complete combustion of:
thus
)
propane
b)
pentane
c)
hexane.
capacity.
1
C
H
5
(l)
12
+
5
O
2
(g)
2
→
5CO(g)
+
6H
O(g)
∆H
=
-1830
kJ
2
249
10
O R G A N I C
C H E M I S T R Y
Wtng ognc chnss
Gob o ts to c gnhos gs ssons
“Curly arrows” are used to
The work of environmental organizations has long focused on the ght
illustrate the movement of
to reduce greenhouse gas emissions and slow the rate of pollution that
electrons in organic reaction
accompanies economic development throughout the world. The US
mechanisms as bonds are
Environmental Protection Agency estimates that globally 80 million tonnes of
broken and made. A sh-
methane annually (28% of global methane emissions) can be attributed to
hook arrow is used to show
ruminant livestock . Countries such as Brazil, Uruguay, Argentina, Australia, and
hootc sson, breaking a
New Zealand contribute dispropor tionally large amounts of greenhouse gases
bond to produce two par ticles
for their levels of population and economic development, due to the scale of
that both have a single
their livestock industries.
unpaired electron, a radical.
Landll in developed countries contains an increasing amount of organic “green”
The half arrow represents the
waste and domestic kitchen waste. In anaerobic conditions common in landll
movement of a single electron
sites, microbes produce methane in vast quantities. This form of anaerobic
as the bond breaks:
respiration is known as thnognss. Governments and environmental
agencies are developing technologies to reduce these emissions, using the
gas to generate electricity for domestic power grids through methane capture
systems. Governments and local councils in some countries are collecting green
A full arrowhead shows the
waste to compost, avoiding the waste going to landll and contributing to the
movement of a pair of electrons
production of methane.
during htotc sson,
when both electrons move
together to form a new bond:
The halogenation of alkanes
Alkanes
to
When drawing mechanisms
are
increase
the
relatively
their
inert,
reactivity.
and
One
chemists
way
of
often
work
achieving
this
to
is
activate
to
alkanes
halogenate
alkane.
using curly arrows:
Common
●
addition,
originate from the exact
individual
location of the electrons
In
being moved.
single
an
from
●
and
atoms
addition
with
in
other
reaction
molecule,
the
studied
elimination.
while
two
organic
chemistry
Substitution
single
atoms
molecules
elimination
is
is
or
are
the
include
the
substitution,
replacement
with
added
a
small
together
removal
of
two
of
group
to
of
atoms.
produce
a
substituents
molecule.
The arrowhead must
accurately nish at the
exact destination of the
electrons.
●
reactions
The base of the arrow must
Alkanes
can
unsaturated
three
types
undergo
alkenes
of
free-radical
and
reaction
alkynes.
listed
substitution
Alkenes
and
and
elimination
alkynes
can
to
form
undergo
all
above.
The arrow commences at
an electron-rich region and
Free-radical substitution
ends at an electron-poor
An
example
of
free-radical
substitution
is
the
reaction
between
methane
region of the molecule.
and
chlorine
The
term
in
the
presence
of
UV
light.
Students are not required to
free-radical
refers
to
a
species
that
is
formed
when
a
molecule
make the distinction between
undergoes
homolytic
ssion:
the
two
electrons
of
a
covalent
bond
are
sh-hook arrows and full
split
evenly
between
two
arrowheads when drawing
have
a
single
electron:
reaction mechanisms. In this
text, only full arrowheads will
be used in mechanisms.
250
A
B
A
+
B
atoms
resulting
in
two
free-radicals
that
each
10 . 2
Heterolytic
electrons
ssion
involved
of
in
a
bond
creates
the
bond
are
a
cation
unevenly
and
split
an
F u N C T i O N a l
anion,
between
as
the
G r O u p
C H e m i S T r y
the
two
atoms:
+
A
+
A
B
When
methane
B
reacts
halogenoalkane
with
chlorine
chloromethane
H
is
in
the
presence
of
UV
light
the
produced:
H
hv
H
C
H
+
H
C
H
Cl
+
H
Cl
H
methane
There
chlorine
are
three
reactions:
chloromethane
stages
initiation,
involved
in
such
propagation,
free-radical
and
substitution
termination,
described
below.
Initiation
The
homolytic
light
produces
ssion
two
of
the
chlorine
chlorine
radicals
molecule
that
have
in
a
the
short
presence
of
UV
lifespan.
hv
Cl
Cl
2
Cl
initiation
Propagation
The
rst
propagation
chlorine
free-radical.
Cl
+
H
CH
H
stage
involves
Cl
a
reaction
between
methane
and
a
CH
3
3
propagation 1
The
production
chain
reaction
producing
chlorine
CH
+
the
the
set
methyl
up.
desired
radical
Cl
of
is
that
The
allows
radical
halogenoalkane,
can
Cl
radical
methyl
Cl
now
CH
3
take
+
part
the
reaction
reacts
with
a
chloromethane,
in
the
rst
to
continue
chlorine
along
with
propagation
as
a
molecule
a
reaction.
Cl
3
propagation 2
Termination
A
termination
mixture.
concentration
radicals,
Cl
Cl
step
3
C
the
reactions
concentration
become
of
the
hydrocarbon
slowing
the
rate
of
Cl
Cl
Cl
CH
Cl
CH
3
H
reduces
Termination
CH
3
more
begins
reaction
and
to
of
radicals
prevalent
decrease.
eventually
in
the
when
They
reaction
the
“mop
stopping
it
up”
the
completely.
3
H
3
C
CH
3
termination reactions
251
10
O R G A N I C
C H E M I S T R Y
Alkenes
Alkenes
are
unsaturated
carbon–carbon
alkenes
more
undergo
double
reactive
addition
hydrocarbons
bond.
than
The
the
that
presence
contain
of
the
corresponding
at
least
double
saturated
one
bond
makes
alkanes.
Alkenes
reactions.
Test for unsaturation
The
presence
using
the
of
a
double
addition
of
bond
bromine
in
a
hydrocarbon
water,
Br
(aq).
A
can
be
mixture
demonstrated
of
the
alkene
2
and
bromine
water
will
undergo
a
colour
change
from
brown
to
colourless:
C
H
2
(g)
+
Br
4
(aq)
→
2
mechanism
change
with
absence
of
of
Br
4
(aq)
2
colourless
this
bromine
the
H
2
brown
The
C
reaction
water
is
this
carbon–carbon
is
shown
a
in
gure
negative
double
1.
result,
If
there
is
no
indicating
the
CH
CH
bond.
+
CH
CH
2
CH
2
CH
2
2
2
2
δ+
Br
Br
Br
Br
Br
induced dipole
Br
δ
▲
▲
Figure 1 Testing for the presence of a C=C bond by the addition of bromine water
Figure 2 Bromine water is decolorized by gaseous ethene
252
colour
10 . 2
F u N C T i O N a l
G r O u p
C H e m i S T r y
Addition of hydrogen: hydrogenation
ethn n nng ft
Large
quantities
product
of
ethene
thermal
hydrocarbons,
organic
of
are
used
decomposition
ethene
is
an
in
or
the
chemical
catalytic
important
raw
industry.
cracking
material
in
of
The
Ethene is used in the food
long-chain
the
production
industry to accelerate the
of
ripening of fruit. Fruits are
polymers.
generally picked while they
In
the
presence
of
150
to
produce
of
a
nely
divided
nickel
catalyst
at
a
temperature
are still unripe to enable
°C,
ethene
will
undergo
an
addition
reaction
with
hydrogen
gas
them to be transpor ted from
the
saturated
alkane
ethane:
farm to supermarket without
Ni
C
H
2
(g)
+
H
4
→
(g)
2
C
H
2
∆
becoming damaged and
(g)
6
looking unappealing to the
This
type
of
reaction
is
very
important
in
the
food
industry.
The
addition
consumer. Ethene is a natural
of
hydrogen
to
unsaturated
fats
and
oils
occurs
in
the
manufacture
of
par t of food ripening, as it is
margarine.
Removing
the
carbon–carbon
double
bonds
increases
the
released by ripening fruit.
melting
point,
making
a
substance
that
is
solid
rather
than
liquid
at
Exposure of the fruit to ethene
room
temperature.
increases the rate of ripening,
The
partial
results
solid
oils
be
in
at
as
hydrogenation
an
elevation
room
they
stored
saturated
a
of
Partial
past
the
in
the
health
double
work
decade
been
bonds
of
the
have
introduced
consumer
(see
diet
also
trans-fats
to
avoid
Cis-
and
t
increase
which
Raised
results
length
for
the
are
levels
uses
the
in
the
of
the
LDLs
negative
trans-
fats
many
of
of
these
content
producers
time
a
are
in
the
is
a
hydrogenated
product
can
However,
concentration
conversion
dangers
use
of
while preventing the build-
which
up of ethene around fruit by
introducing carbon dioxide to
the container holding the fruit
slows the rate of ripening.
of
transport
associated
with
B.3).
community,
the
partially
blood
therefore
margarine
involved
of
oils
consumption.
trans-carbon–carbon
aware
food
the
vegetable
creating
industry
life,
sub-topic
into
point,
remain
(LDLs)
reduce
more
from
share.
shelf
recognized
to
food
and
scientic
multinational
market
The
the
blood.
risks
large
and
in
hydrogenation
carbon
the
oils
lipoproteins
cholesterol
increased
of
and
melting
prolonged
supermarket
fats
low-density
a
polyunsaturated
the
temperature.
have
in
of
of
have
are
trans-fats
of
cis-carbon–
bonds.
governments
harmful
and
and
discussed
resulting
in
detail
a
result
the
legislation
foods.
distanced
As
over
products
processed
also
publicity
of
double
and
has
make
Many
themselves
loss
in
of
sales
sub-topics
and
B.3
B.10.
Halogenation of alkenes
The
electrophilic
addition
of
halogenation
elemental
of
halogens
symmetrical
such
as
alkenes
chlorine,
Cl
or
2
resulting
C
H
4
H
in
(g)
a
dihalogenated
+
Br
8
(g)
→
2
H
H
H
H
C
C
C
C
H +
H
but-2-ene
H
4
Br
2
H
C
involves
the
bromine,
Br
,
2
alkane:
Br
8
(l)
2
H
H
H
H
H
C
C
C
C
H
Br
Br
H
H
2,3-dibromobutane
253
10
O R G A N I C
C H E M I S T R Y
The
addition
in
single
a
two
are
of
a
alternative
discussed
products
in
sub-topic
20.1).
C
+
H
4
H
hydrogen
halide,
mono-halogenated
(g)
this
are
topic
HBr(g)
→
possible;
C
H
4
H
H
H
C
C
C
C
H
H +
and
steam
a
C
H
in
(g)
petrol,
+
H
of
H
a
6–7
symmetrical
are
covered
at
alkenes
HL
in
H
H
H
H
C
C
C
C
H
H
Br
H
H
a
ethanol
of
a
is
achieved
catalyst,
by
reacting
phosphoric(V)
acid,
ethene
at
300
°C
MPa:
O(g)
→
C
H
2
variety
creating
of
presence
2
has
only
alkenes
results
alkene
Br(l)
HBr
production
the
4
Ethanol
alkene
unsymmetrical
2-bromobutane
pressure
2
symmetrical
an
9
H
large-scale
with
a
however,
but-2-ene
The
to
With
(unsymmetrical
8
H
HX,
alkane.
of
OH(g)
5
uses
biofuel
including
(sub-topic
as
an
additive
to
gasoline
or
10.1).
Polymerization of alkenes
The
plastics
world,
variety
they
of
can
have
a
a
The
monomer
H
2
one
of
→
the
range
Polymers
negative
ethene
largest
of
impact
is
on
the
industry,
the
the
quality
of
bodies
used
our
in
widely
lives
the
for
a
although
environment.
of
bond,
supplied
undergoes
polymers
reaction
double
monomer,
manufacturing
addition
improve
carbon–carbon
petrochemical
nC
is
broad
polymerization
contain
polymer.
the
a
purposes.
Addition
that
industry
producing
to
addition
many
linking
the
small
monomers
together
plastics
to
form
industry
polymerization
to
a
by
the
produce
polyethene:
[ CH
4
CH
2
]
2
n
St t
C
C
When drawing diagrams to
C
C
C
represent polymerization, it is
C
C
impor tant to draw continuation
C
bonds through the brackets.
C
C
C
C
C
▲
254
C
C
C
C
Figure 3 The polymerization of ethene
C
C
C
C
C
C
C
C
C
C
C
10 . 2
Any
monomer
that
polymerization.
structure
by
a
single
adjacent
of
of
contains
The
the
monomer
bond,
a
carbon–carbon
repeating
and
monomers.
the
For
structural
that
formed
electrons
example,
double
unit
it,
with
released
gure
of
4
the
the
bond
can
polymer
double
forming
shows
F u N C T i O N a l
new
the
G r O u p
C H e m i S T r y
undergo
reects
bond
the
replaced
bonds
to
the
polymerization
propene.
n CH
CH
CH
2
CH
2
CH
CH
3
3
n
▲
Figure 4 The polymerization of propene
Alcohols
Alcohols
form
applications
Alcohols
a
diverse
and
can
play
a
undergo
group
of
compounds
signicant
complete
part
in
that
have
synthetic
combustion,
a
wide
range
of
reactions.
releasing
carbon
dioxide
▲
and
Figure 5 Colour change during the
water:
2
reduction of orange Cr
O
(aq)
7
2
3+
C
H
2
OH(l)
+
3O
5
(g)
→
2CO
2
(g)
+
3H
2
O(g)
∆H
=
-1367
kJ
ions to green Cr
(aq) ions
2
Soc ctons of coho
Oxidation of alcohols
conston
Acidied
potassium
manganate(VII)
equations
to
for
construct
dichromate(VI)
can
these
these
be
used
for
oxidizing
equations
the
(gure
or
oxidation
agents
was
5)
are
as
of
potassium
alcohols.
follows.
developed
in
(The
sub-topic
The
half-
working
method
Excessive alcohol consumption
is a growing problem in many
countries. Bng nkng is
9.2.)
sometimes dened as drinking
+
MnO
(aq)
+
8H
2+
(aq)
+
5e
→
Mn
(aq)
+
4H
4
2
Cr
O
O(l)
+
(aq)
+
14H
3+
(aq)
+
6e
→
2Cr
(aq)
+
7H
7
2
5 (for a man) or 4 (for a woman)
2
alcohol units over a 2-hour period
O(l)
2
and increasing the blood alcohol
The
oxidation
products
of
alcohols
depend
on
the
type
of
alcohols
content above 0.08% by volume.
involved.
An alcohol unit varies from
country to country. In general,
one alcohol unit is equivalent
Primary alcohols
3
to approximately 10 cm
The
oxidation
of
a
primary
alcohol
is
a
two-stage
process
that
of
rst
pure alcohol. Having reached
produces
an
aldehyde
followed
by
a
carboxylic
acid.
Potassium
alarming proportions amongst
dichromate(VI),
K
Cr
2
manganate(VII),
O
2
is
a
milder
oxidizing
agent
than
potassium
adolescents and young adults in
7
KMnO
.
When
the
primary
alcohol
ethanol,
4
is
heated
with
acidied
This
K
aldehyde
H
2
Cr
2
produced.
C
O
2
can
,
the
aldehyde
ethanal,
7
be
CH
CHO
OH
5
further
oxidized
to
the
many western societies, binge
is
drinking is having a signicant
3
carboxylic
acid
impact on economies, social
ethanoic
acid,
CH
COOH.
structure, law and order, and
3
ultimately health systems.
H
H
Associated health risks include
H
H
O
O
physical and psychological
H
C
C
H
C
C
C
O
H
H
H
H
C
H
dependence on alcohol, liver and
H
brain damage, elevated risk of
ethanol
ethanal
ethanoic acid
cancer of the throat, mouth, and
esophagus, depression, anxiety
and social problems at work and
within the family.
255
10
O R G A N I C
C H E M I S T R Y
The
aldehyde
Carboxylic acids are capable
preventing
of forming dimers, paired
point
molecules held together by
forces:
hydrogen bonds (gure 6). The
carboxylic
increased size of the molecule
have
can
its
than
the
recovered
acids
have
have
boiling
by
oxidation.
carboxylic
aldehydes
higher
be
further
acid
weak
the
The
due
to
process
distillation ,
has
differences
dipole–dipole
stronger
of
aldehyde
a
in
lower
the
intermolecular
intermolecular
boiling
intermolecular
hydrogen
forces
bonds
while
and
so
points.
leads to stronger van der Waals’
If
the
carboxylic
acid
is
the
desired
product,
the
aldehyde
must
remain
forces and a higher boiling
in
the
reaction
of
time.
mixture
with
the
oxidizing
agent
for
a
longer
period
point.
Instead
Reuxing
δ
O
3
a
of
the
distillation
technique
that
apparatus
involves
the
a
reux
cyclic
column
is
evaporation
used.
and
δ+
H
condensation
O
does
CH
is
C
C
not
of
a
volatile
reaction
mixture,
preserving
the
solvent
as
it
evaporate.
CH
3
H
δ+
δ
Secondary alcohols
▲
Figure 6 A dimer of ethanoic
acid
The
oxidation
formation
of
H
of
a
a
secondary
alcohol
such
as
propan-2-ol
results
in
the
ketone:
H
H
H
H
2H
H
C
C
H
H
C
O
H
removed
C
C
H
H
H
C
O
H
H
oxidation
Upon
as
formation
the
carbon
attachedto
of
the
atom
of
ketone,
the
no
further
functional
oxidation
group
has
no
is
possible
hydrogens
it.
Condensation reaction of an alcohol and a
carboxylic acid
Esters
are
derived
applications
and
from
ranging
carboxylic
from
acids
avouring
acid
and
an
is
a
reversible
alcohol
concentrated
are
sulfuric
reaction
heated
in
the
a
variety
of
medications
to
solvents
CH
3
COOH(l)
IUPAC
carboxylic
7).
occurs
of
CH
OH(l)
SO
acid
names
the
CH
The
acid
esters
name
second
(in
a
carboxylic
catalyst,
this
of
CH
3
methanol
of
when
a
normally
(conc)
2
4
________
→
3
from
“methyl”).
(gure
+
2
propanoic
derived
that
presence
acid:
H
CH
256
have
and
explosives.
Esterication
The
and
agents
consist
the
word
case,
of
alkyl
is
COOCH
2
two
words.
The
rst
in
the
alcohol
composed
of
the
root
followed
by
+
H
O(l)
2
propanoate
chain
“prop”),
(l)
3
methyl
word
(in
name
the
sufx
this
of
is
example,
the
“anoate”
10 . 2
F u N C T i O N a l
G r O u p
C H e m i S T r y
methyl is derived
-anoate is the
from the alcohol,
methyl
proponate
sux for esters
methanol
-prop- signies three
carbons present in
the carboxylic acid
parent molecule
▲
Figure 7 Naming esters
▲
Figure 8 3D computer-generated image of
methylpropanoate.
Grey = carbon, white = hydrogen,
red = oxygen
Nucleophilic substitution reactions:
An introduction
Once
form
an
a
types
of
bond,
unreactive
reaction.
C
X,
the
electron-rich
a
lone
pair
aqueous
polarized
X
There
being
are
of
These
An
of
the
distinct
of
aqueous
(ethanol)
and
CH
3
now
carry
atom
a
reaction
part
polar
open
attack
Nucleophiles
full
in
to
other
carbon–halogen
to
negative
NaOH(aq),
charge
by
a
take
by
contain
charge.
18.1).
hydroxide,
susceptible
halogen
reaction
will
The
to
on
the
attack
the
sodium
releases
+
mechanisms
mechanism
present,
be
namely
discussed
nucleophilic
Cl(g)
2
it
is
(sub-topic
positive
can
contain
carbon
sometimes
sodium
makes
substitution
nucleophiles.
bases
partial
reaction.
mechanisms
CH
of
The
halogenoalkane
example
with
.
and
as
a
molecule
by
contains
carbon
the
nucleophile
the
atom
in
hydroxide
will
result
the
ion.
in
an
formed.
two
substitution
class
known
Lewis
bond
substitution
alcohol
as
solution
:OH
The
undergone
resulting
halogenoalkanes
electrons
act
nucleophile
C
Since
has
the
electron-decient
species
of
Nucleophiles
An
alkane
halogenoalkane,
OH
chloride
(aq)
→
This
ion,
CH
primary,
is
the
,
the
nucleophilic
depends
secondary,
of
produces
leaving
OH(aq)
on
or
the
tertiary.
20.1.
reaction
reaction
Cl
CH
3
this
occurs
sub-topic
substitution
hydroxide.
a
in
for
that
+
Cl
chloroethane
an
alcohol
group.
(aq)
2
257
10
O R G A N I C
C H E M I S T R Y
Electrophilic substitution reactions:
An introduction
As
discussed
addition
reactions.
accepting
While
the
in
An
an
than
illustrating
pair.
In
unique
of
an
acts
as
requires
a
Lewis
are
a
20.1
of
of
the
reaction
the
is
species
undergo
of
substitution
capable
(sub-topic
to
the
π
will
be
of
18.1).
electrons
leads
mechanism
to
of
in
substitution
the
nitration
developed,
benzene.
the
takes
base
readily
ring
benzene
properties
Lewis
acid
attracted
stability
substitution
This
not
electrophilic
electron-poor
reaction
reaction
does
undergo
sub-topic
bromine.
and
is
the
electrophilic
elemental
environment
ring,
substitution)
the
It
benzene
will
electrophiles
benzene
addition.
example
with
it
electrophile
electron-poor
(electrophilic
10.1,
Instead
electron
aromatic
rather
An
sub-topic
reactions.
reaction
place
(FeBr
or
in
of
an
AlBr
3
)
benzene
anhydrous
as
the
catalyst.
3
FeBr
3
C
H
6
One
+
___
→
Br
6
C
2
bromine
H
6
atom
Br
+
HBr
5
from
replaces
Br
a
hydrogen
atom
in
benzene;
2
the
remaining
product,
an
258
hydrogen
organic
products
bromine
and
bromide.
(non-aqueous)
are
omitted.
hydrogen
Please
atoms
note
environment,
that
so
form
this
the
the
inorganic
reaction
states
of
takes
by-
place
reactants
and
in
Q u e S T i O N S
Questions
1
Which
be
a
A.
three
compounds
homologous
CH
OH,
CH
3
can
be
considered
5
to
Alkenes
CH
3
OH,
are
important
series?
CH
2
CH
3
CH
2
a)
OH
The
an
reaction
CH
CH
3
OH,
CH
2
CHO,
CH
3
CH
CH
3
CH(OH)CH
2
)
3
,
CH
3
CH
3
CH
2
CH
2
CH
CH
2
CH
2
OH,
CH
2
CH
3
OCH
2
CH
2
a
test
)
CH
2
May
Deduce
the
colour
in
the
change
water
is
added
to
when
chloroethene.
[1]
the
Lewis
CHO
identify
the
structure
formula
of
of
chloroethene
the
repeating
[1]
2
of
the
polymer
poly(chloroethene).
[2]
2009
Besides
uses
the
IUPAC
name
for
CH
CH
3
A.
1,1-dimethylpropane
B.
2-ethylpropane
C.
unsaturation
,
c)
is
for
Describe
3
unit
What
water
COH
CH
3
2
bromine
OH,
and
IB,
with
3
3
(CH
chemically
2
b)
D.
alkenes
and
compounds.
COOH
bromine
(CH
of
organic
3
laboratory.
C.
of
2
provides
B.
economically
family
CH(CH
2
)CH
3
IB,
?
May
6
State
IB,
May
reactions
of
state
two
commercial
alkenes.
[2]
2010
and
explain
molecules
tertiary
3-methylbutane
polymerization,
the
3
2-methylbutane
D.
of
(gure
whether
9)
are
the
following
primary,
secondary,
or
halogenoalkanes.
[4]
[1]
2009
H
C
3
Which
conditions
are
required
to
obtain
is
yield
of
oxidized
a
carboxylic
using
acid
potassium
when
H
H
C
C
C
H
H
H
H
H
H
C
C
C
Cl
H
H
C
H
Cl
H
Cr
2
O
2
H
C
C
H
H
H
H
C
H
dichromate(VI),
H
C
Add
sulfuric
II.
Heat
III.
Distil
the
reaction
mixture
under
product
as
the
oxidizing
I
and
II
B.
I
and
III
C.
II
and
D.
I,
II,
Figure 9
IB,
May
2011
only
only
III
7
Consider
only
the
reaction
following
1
RCH
3
and
May
III
sequence
reaction
_____
→
RCH
IB,
c)
agent
added
A.
b)
reux
▲
the
H
acid
a)
is
H
(aq)?
7
H
I.
H
C
ethanol
H
K
H
a
H
good
H
_____
→
Br
of
reactions:
2
RCH
2
OH
2
[1]
reaction
3
_____
→
RCOOH
2009
RCH
is
an
unknown
alkane
in
which
R
3
represents
4
Alkenes
are
important
starting
materials
for
of
State
and
points
alkene
b)
of
explain
the
rst
the
ve
homologous
Describe
two
series.
The
trend
of
the
alkane
May
contains
members
of
the
series.
features
of
a
Determine
showing
boiling
b)
[3]
Equal
[2]
your
81.7 %
by
mass
of
of
alkane
mass,
signicant
its
formula,
[3]
carbon
are
gures,
at
dioxide
found
measured
pressure.
empirical
working.
volumes
unknown
same
homologous
and
IB,
group.
products.
carbon.
a)
alkyl
a
a)
variety
an
to
the
Deduce
an
to
have
the
the
accuracy
same
the
and
of
two
temperature
molecular
2011
formula
of
the
alkane.
[1]
259
10
O R G A N I C
c)
(i)
C H E M I S T R Y
State
for
(ii)
the
reagent
reaction
State
the
and
conditions
needed
1.
10
[2]
reagent(s)
and
Alkenes
for
reaction
3.
Reaction
1
Describe
giving
involves
the
a
free-radical
stepwise
equations
propagation
to
and
economically
of
mechanism.
mechanism,
The
represent
the
termination
in
by
reaction
of
the
provides
initiation,
steps.
when
test
and
chemically
compounds.
with
for
bromine
unsaturation
Describe
bromine
the
water
colour
is
added
to
chloroethene.
[1]
[4]
Deduce
the
Lewis
chloroethene
November
alkenes
a
laboratory.
change
b)
IB,
organic
[2]
water
d)
an
family
conditions
a)
needed
are
important
structure
and
identify
of
the
formula
2010
of
the
repeating
unit
of
the
polymer
poly(chloroethene).
8
a)
Identify
A
E,
the
formulas
formed
in
the
of
the
organic
reactions,
I
products,
CH
(CH
3
II.
)
2
(CH
)
3
OH
+
K
8
Cr
2
+
CBr
+
NaOH
of
H
__
→
O
2
Besides
polymerization,
commercial
+
H
__
I.
c)
IV:
→
A
[2]
uses
of
the
state
two
reactions
alkenes.
[2]
B
7
IB,
__
→
May
2010
C
3
+
H
__
III. (CH
)
3
IV
.
H
CHOH
+
K
2
Cr
2
C=CH
2
+
__
→
Br
2
→
O
2
D
7
11
E
Chloroethene,
H
C=CH
2
can
react
to
form
a
polymer.
of
polymer
and
draw
the
of
a
section
of
this
consisting
of
three
repeating
units.
Draw
the
Lewis
predict
c)
compounds,
C
H
4
Draw
a
A
and
D,
each
have
the
with
a
is
reacted
hydroxide
formula
of
to
C
to
C
with
produce
H
4
O.
potassium
Compound
sodium
D
is
of
C
H
4
react
with
Deduce
B
B
is
then
C
with
resists
a
manganate(VII)
formula
further
of
oxidation
by
the
May
manganate(VII).
to
O.
2011
dilute
aqueous
compound
Compound
potassium
structural
A,
with
produce
E
does
E
with
not
10
acidied
compounds
260
compound
Compound
potassium
reacted
hydroxide
formula
IB,
aqueous
10
compound
C
dilute
8
acidied
a
O.
acidied
Compound
with
produce
H
4
oxidized
of
B,
C,
manganate(VII).
formulas
D,
and
E
for
[5]
six
why
structure
H
–
C
of
organic
polymer
–
for
Cl
chloroethene
bond
angle.
[2]
poly(chloroethene)
of
May
2010
carbon
the
economic
disposal
IB,
sodium
Outline
is
Cl.
A
the
section
9
Compound
the
paper
containing
formula
important
[2]
b)
Specimen
Two
an
polymer
and
9
is
manufacture
structural
a)
IB,
Cl,
3
to
poly(chloroethene).
type
formula
used
Name
2
this
H
2
compound
b)
C
[5]
2
atoms.
[1]
polymerization
importance
plastics
is
a
and
of
alkenes
why
problem.
the
[2]
M E A S U R E M E N T
A N D
D ATA
11
P R O C E S S I N G
Introduction
Analytical
techniques
chemistry.
As
appreciate
the
we
must
limit
of
taken
power
realize
that
precision
into
lie
chemists,
the
very
only
analysis,
any
and
account
of
at
not
core
we
but
in
measurement
accuracy,
when
do
and
to
addition
has
this
evaluating
that
of
need
we
can
graphical
be
experimental
the
means,
analytical
(IR),
uncertainty
effectively
identication
a
must
underpin
how
mass
of
and
in
measurement,
represent
examine
organic
of
infrared
(MS),
and
by
spectroscopic
compounds,
techniques
spectrometry
data
the
looking
at
spectroscopy
proton
nuclear
1
results.
In
this
topic
we
will
explore
the
principles
magnetic
resonance
spectroscopy
(
H
NMR).
11.1 Uc as a os  masum
a sus
Understandings
Applications and skills
➔
Qualitative data includes all non-numerical
➔
Distinction between random errors and
information obtained from obser vations not
systematic errors.
from measurement.
➔
➔
Record uncer tainties in all measurements as a
Quantitative data are obtained from
range (+) to an appropriate precision.
measurements, and are always associated with
➔
Discussion of ways to reduce uncer tainties in
random errors/uncer tainties, determined by the
an experiment.
apparatus, and by human limitations such as
➔
Propagation of uncer tainties in processed data,
reaction times.
including the use of percentage uncer tainties.
➔
Propagation of random errors in data processing
➔
Discussion of systematic errors in all
shows the impact of the uncer tainties on the
experimental work , their impact on the results,
nal result.
and how they can be reduced.
➔
Experimental design and procedure usually
➔
Estimation of whether a par ticular source of
lead to systematic errors in measurement,
error is likely to have a major or minor eect on
which cause a deviation in a par ticular
the nal result.
direction.
➔
➔
Calculation of percentage error when the
Repeat trials and measurements will reduce
experimental result can be compared with a
random errors but not systematic errors.
theoretical or accepted result.
➔
Distinction between accuracy and precision in
evaluating results.
Nature of science
➔
Making quantitative measurements with replicates to ensure reliability – precision, accuracy,
systematic, and random errors must be interpreted through replication.
261
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
Qualitative and quantitative analysis
According to IUPAC, qualitative
and quantitative analysis can be
The
analytical
Analysis
distinguished as:
Quaav aayss
Substances are identied
can
chemist
be
●
qualitative
●
quantitative
of
is
two
often
described
as
the
chemical
detective.
types:
analysis
analysis.
or classied on the basis of
their chemical or physical
proper ties, such as chemical
reactivity, solubility, molar
Uncertainty in measurement
In
science
numerical
data
can
be
divided
into
two
types:
mass, melting point, radiative
●
data
involving
exact
numbers
(that
is,
the
values
are
known
exactly
–
proper ties (emission,
there
is
no
uncertainty)
absorption), mass spectra,
●
nuclear half-life, etc.
data
involving
degree
of
inexact
numbers
(for
these
types
of
numbers
there
is
a
uncertainty).
Quaav aayss
As
scientists,
when
we
carry
out
a
particular
experiment
involving
The amount or concentration of
measurement,
there
will
always
be
some
uncertainty
associated
with
the
an analyte may be determined
measured
data,
that
is
the
data
will
involve
inexact
numbers
to
some
(estimated) and expressed
degree.
Such
uncertainty
may
be
associated
with
factors
such
as
the
as a numerical value in
instruments
used
in
the
laboratory.
For
example,
the
mass
of
a
sample
of
appropriate units.
potassium
bromide,
KBr(s),
will
depend
on
the
type
of
balance
used.
In
‘Nomenclature in evaluation of
a
typical
school
laboratory,
top-pan
balances
often
read
to
0.01
g,
but
an
analytical methods including
analytical
balance,
to
0.0001
used
in
more
precise
analytical
experiments,
can
read
detection and quantication
at
least
g
or
better)
(gure
1).
Uncertainty
may
also
depend
capabilities’ Pure and Applied
human
error.
Chemistry, 67(1699), (1995),
p1701
Data may also be classied as
qualitative or quantitative data:
Quaav aa
Qualitative data includes all
non-numerical information
obtained from observations not
from measurement.
Quaav aa
Quantitative data are obtained
from measurements, and are
always associated with random
errors/uncer tainties (dened
shor tly) determined by the
apparatus and by human
Figure 1 (a) A top-pan balance used to measure mass in a typical school laboratory
limitations, such as reaction
times.
can read to 0.01 g. (b) An analytical balance used to measure mass to a high degree of
precision can often read to 0.0001 g. The shutters on the balance should be closed to
reduce both air ow and dust collecting which can both aect the reading
262
on
11 . 1
U n C e r t A i n t i e S
A n d
e r r O r S
i n
M e A S U r e M e n t
A n d
r e S U lt S
An example of a awed experiment in science
The OPERA experiment – a case-study involving CERN and LNGS
Neutrinos
neutral
and
are
that
are
one
make
as
unlike
electrical
OPERA
of
up
considered
but,
Gran
a
greater
The
In
betwe e n
that
international
the
do
L a b ora tor i
in
tha n
Gra n
to
ve loci ty
all
ove r
of
of
This
in
wor ld ,
of
that
the
nd i ngs
the
in
wa s
and
the re
In
in
of
the
imp o r ta nce
of
the
it
by
that
in
fa c t ,
whi ch
 a ws
l ed
v el ocity
li g ht .
the
ex pe ri me n t al
the
in
ar e ,
of
te s ti n g
to
s et
r e por te d
ne utri nos
und e r s ta nd ing
there
or i g in a l
wa s
of
Albe rt
S ub s eq ue n t l y,
wer e ,
2 0 12
wi th
not hi n g
s tau n c h ly
s a id
e q uip ment,
pr o b l e ms
a bou t
tha t
pos tula t e d
e x p e ri me nt.
val i d i ty
and
conce rned
as
velocities
the
and
science
l i g ht,
tha t
consistent
shows
v e ry
consi de r ed
ho w e ve r,
th e
timing
uncertainty
but
their
in
set-up
reliability
li g ht .
than
reported
indeed,
at
was
measurements.
Nature
ma de
the
the
of
aw
possible
po ur
I t a ly)
tr a vel
no
was
in
Na zi on al i
d is co ve ry
hea d l i ne s
it
t he
Ge n eva ,
Sa ss o,
a ppe a r
the
in
was
were
it
OPERA,
defensive
an
as
faster
Einstein.
r e se a rch
Euro p é e n
( C ERN),
scie nti c
c ar r y
fr om
physicists
ndings
travels
be
e le ct r on
not
r e sults
Co ns e il
(L NGS )
m ig h t
inte r na ti ona l
ne utr i no s
velocity
proposed
the
to
the
r e ac t ion s
pa r t ic l es
T he y
the y
2 0 11,
(an
Nuc l é a i r e
Sasso
suggested
s i m il ar
many
e le ct r ic a l ly
nucl ea r
f und ame nta l
ele ctr o n,
and
in
uni v e r se .
bein g
charge.
Switzerland,
del
the
experime nt
Recherche
s ma ll ,
p r o d uce d
the
the
collaboration
la
extr e me l y
particles
id ea
r e su l t s
of
me a s ur e me nt.
A xamp of  mpac of os fom sac spac
Cas of  Mas Cma Ob Spaccaf
Dr. Edward Weiler, NASA’s Associate Administrator for
Space Science stated the following:
In 1998, NASA launched the Mars Climate Orbiter, a
space probe designed to examine the climate on the
People sometimes make errors. The problem here was not
planet Mars. However in 1999, the spacecraft crashed
the error, it was the failure of NASA’s systems engineering,
and completely disintegrated as it approached Mars at
and the checks and balances in our processes to detect
an incorrect altitude (http://mars.jpl.nasa.gov/msp98/
the error. That’s why we lost the spacecraft.
orbiter/). The reason for the crash related to an error in
It was subsequently repor ted that one team used imperial
the transfer of information between the spacecraft team
units while the other team used SI units for a pivotal
based in Colorado, and the mission navigation team based
operation for the space probe, which led to the incorrect
in California, USA.
trajectory required to place the probe on Mars.
Dierence between precision and accuracy
In
order
need
to
to
understand
consider
the
the
idea
difference
of
uncertainty
between
in
measured
precision
Precision
According
of
to
under
by
we
accuracy.
Accuracy
IUPAC,
agreement
obtained
and
values,
precision
between
applying
stipulated
random
part
of
shortly)
which
the
the
closeness
test
experimental
conditions.
the
is
independent
The
experimental
results
procedure
smaller
errors
the
According
of
the
to
IUPAC,
agreement
measurement
(which
is
the
accuracy
between
and
a
true
particular
the
results,
the
more
value
of
quantity
closeness
of
the
to
a
measurand
be
measured).
for
the
presentation
of
results
precise
of
the
the
result
(dened
‘Nomenclature
affect
is
the
chemical
analysis’
Pure
and
Applied
Chemistry ,
procedure.
66(595),
(1994),
p598
263
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
Aaogy
Case 3: high precision and high accuracy
The analogy of the dar tboard is useful in appreciating the
dierence between precision and accuracy (cases 1–3).
Case 1: high precision and low accuracy
As
scientists
we
always
our
experimental
not
only
do
need
to
work .
strive
That
to
is,
replicate
for
any
Case
3
in
experiment
Case 2: low precision and low accuracy
we
we
be
need
able
each
time
In
experiment
an
and
low
with
calibration
are
of
a
to
the
be
same
pH
level
we
errors
often
accurate,
reproduce
where
accuracy,
instrument
to
of
have
the
in
addition
experiment
consistency.
high
associated
common
but
(for
precision
with
the
example,
poor
meter).
Signicant gures
Signicant
a
given
greater
the
quantity.
a
gures
certainty
In
order
measurement,
notation
SF
to
it
is
the
to
the
The
number
greater
the
of
digits
number
about
the
numerical
know
the
number
useful
(sometimes
represent
For
refer
measurement.
to
called
number
of
express
of
the
signicant
the
signicant
measured
exponential
signicant
of
value
of
the
reecting
precision
gures,
measured
gures
For
calculated
associated
parameter
notation).
or
in
of
the
with
scientic
convenience,
let
gures:
example:
Masum
Scc oao
numb of SFs
2
135.680 g
1.35680 × 10
3
0.00620 dm
6.00 kg
3
6.20 × 10
3
six
3
dm
6.00 kg
2.0600 m
g
three
three
3
2.0600 m
ve
1
0.2 mg
mg
2 × 10
one
2
300 kg*
3 × 10
kg*
one
*If a number is expressed with no decimal point, for example 300 kg, then it is assumed that the zeros
are not signicant. Hence 300 kg has just one SF.
264
11 . 1
In
numerical
gures
●
For
always
two
an
result
●
calculations
should
employing
be
simple
be
number
For
operation
be
number
of
decimal
e r r O r S
measured
account.
i n
M e A S U r e M e n t
quantities,
These
multiplication
based
signicant
are
A n d
r e S U lt S
signicant
handled
based
on
the
or
division:
measurement
with
The
the
gures
involving
expressed
of
with
into
involving
expressed
smallest
should
dealing
taken
A n d
rules:
operation
should
an
U n C e r t A i n t i e S
on
addition
the
or
subtraction:
measurement
with
the
The
result
smallest
places
Worked example: using signicant gures
The
mass
of
35.4200g.
a
sample
The
mass
bottle
of
the
and
a
piece
empty
of
sample
aluminium
bottle
is
metal
28.9200
is
g.
If
3
the
aluminium
displaces
2.41
cm
of
water,
calculate
the
density
of
3
aluminium,
in
g
cm
Solution
d
=
m
V
density
=
=
mass
of
of
volume
aluminium
aluminium
of
aluminium.
m
_
d
=
roug
V
Let
m
=
mass
of
empty
sample
bottle
=
28.9200
g
=
mass
of
empty
sample
bottle
+
aluminium
In calculations in chemistry
1
we often have to round o
Let
m
=
35.4200
g
2
numbers. If the digit to be
m
=
m
m
2
=
(35.4200
g
28.9200
g)
=
6.5000
g
removed for rounding purposes
1
is less than the number ve,
In
nding
m,
the
operation
is
subtraction,
so
the
value
of
m
is
the digit immediately before it
expressed
to
the
smallest
number
of
decimal
places,
which
is
four
for
will not be changed. However,
both
m
and
m
1
2
if the digit is equal to ve or
3
d
=
(6.5000
g)/(2.41
cm
)
greater than ve, the number
immediately before it will be
increased by one.
ve
SF
three
SF
3
For example 3.42 cm
rounded
3
Hence,
the
result
should
be
correctly
reported
as
d
=
2.70
g
cm
,
3
to two SFs would be 3.4 cm
as
the
operation
involves
division
and
to
nd
the
result
from
the
3
while 23.46 cm
quotient
we
used
the
smallest
number
of
signicant
gures
which
,
3
or 23.45 cm
will
rounded to three SFs would be
be
three
from
the
volume,
V
3
23.5 cm
Signicant
carefully
as
the
gures
in
power
composed
associated
calculations.
of
to
which
two
The
the
with
log
base
logarithms
of
can
a
(logs)
number
be
raised
parts:
to
is
need
dened
get
that
to
be
handled
mathematically
number.
A
log
is
Suy p
You can ignore SFs associated
with physical constants given
in the Data booklet for the
●
The
characteristic
●
The
mantissa
–
the
integer
part
purposes of calculations – only
–
the
decimal
part
consider SFs for measured data.
265
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
The
number
signicant
base
that
the
10
and
digits
logs
involve
natural
equation.
of
gures
to
base
log,
For
in
for
the
base
logs
is
mantissa
logarithmic
the
10
ln,
a
e,
are
that
is
used,
in
pH,
for
the
(this
natural
common
frequently
example,
indicates
entity
number
covers
logs,
both
ln).
whereas
example
in
of
logs
to
the
Calculations
in
kinetics
the
Arrhenius
in:
1
log
(2.7)
=
0.43
=
4.3
×
10
10
two
the
number
with
the
SFs
2.7
mantissa
contains
mantissa
(the
two
SFs,
decimal
so
the
part)
(decimal
answer
having
part)
should
two
SFs.
two
be
SF
reported
Here
is
another
example:
ln(6.28)
three
the
number
with
the
6.28
SFs
1.837
mantissa
contains
mantissa
=
(the
three
decimal
SFs,
(decimal
so
part)
the
part)
answer
having
three
three
SFs
should
be
reported
SFs.
Experimental errors
As
stated
already,
associated
types
Sysmac o
of
with
every
it.
This
experimental
●
systematic
●
random
single
is
measurement
termed
has
a
experimental
degree
error.
of
uncertainty
There
are
two
error:
error
error.
According to IUPAC, systematic
error is the mean that would
Systematic errors
result from an innite number
Systematic
of measurements of the same
design
measurand carried out under
the
or
errors
with
measured
are
the
associated
with
instrumentation
quantity
will
always
a
aw
used.
be
in
the
actual
Systematic
greater
or
less
experimental
errors
than
imply
the
that
true
value.
repeatability conditions minus
Systematic
errors
can
be
further
classied
into
three
types:
a true value of the measurand.
●
instrumentation
●
experimental
●
personal
errors
raom o
methodology
errors
According to IUPAC, random
error is the result of a
errors.
measurement minus the
Examples
of
systematic
errors:
mean that would result
from an innite number of
●
Faulty
●
Errors
gas
syringes
that
have
associated
leakage
( instrumentation
error).
measurements of the same
in
the
readings
taken
from
a
pH
meter
due
to
faulty
measurand carried out under
calibration
of
the
instrument
( instrumentation
error).
repeatability conditions.
●
Poorly
insulated
calorimeter
in
a
thermochemistry
experiment
International Vocabulary
(experimental
methodology
error).
of Basic and General Terms
in Metrology,
Second Edition, ISO, 1993.
266
●
Measuring
or
burette
the
volume
from
the
top
of
of
a
colourless
the
liquid
meniscus
in
a
instead
graduated
of
from
the
cylinder
bottom.
11 . 1
Such
a
systematic
manipulation
●
Evaporation
methodology
●
of
error
data
would
lead
(experimental
volatile
liquids
on
A n d
to
e r r O r S
greater
a
M e A S U r e M e n t
volumes
methodology
heating
i n
in
A n d
r e S U lt S
the
error ).
sample
( experimental
error).
Occurrence
being
of
U n C e r t A i n t i e S
of
side-reactions
measured
which
(experimental
can
interfere
methodology
with
the
parameter
error ).
Usfu souc
●
The
●
parallax
exact
colour
of
a
solution
at
its
end
point
( personal
error).
The BPIM document Evaluation of
error
associated
with
reading
a
graduated
cylinder
Measured Data – Guide to the Expression
incorrectly
of Uncer tainty in Measurement can be
accessed at http://www.bipm.org/en/
publications/guides/gum.html.
BiPM
The mission of BIPM (Bureau
International des Poids et Mesures,
correct reading of volume at
whose headquar ters are based in
bottom of meniscus of water
Paris, France) is to ensure worldwide
uniformity of measurements and their
traceability to the iaoa Sysm
of Us (SI).
With the authority of the Convention
of the Metre, a diplomatic treaty
between 55 nations, BIPM functions
through a series of consultative
Systematic
errors
can
often
be
reduced
by
adopting
greater
care
committees, whose members are the
to
the
experimental
design .
Such
errors
are
consistent
and
can
be
national metrology laboratories of the
detected
and
ultimately
corrected.
signatory states, and through its own
Systematic
errors
will
affect
the
accuracy
of
the
results.
experimental programmes. BIPM carries
out measurement-related research. Par t
Random errors
of the work of BIPM involves looking at
international comparisons of national
Random
errors
occur
because
of
uncontrolled
variables
in
an
measurement standards as well as
experiment
and
hence
cannot
be
eliminated .
They
can,
however,
performing calibrations for its member
be
reduced
by
repeated
measurements.
Random
errors
affect
the
states. As a result of collaboration
precision
of
the
results.
between seven international
Examples
of
random
errors:
organizations, including IUPAC and the
●
estimating
a
quantity
particular
instrument
apparatus
(for
which
eg
example,
lies
with
a
a
between
marked
graduations
spectrophotometer)
or
of
a
measuring
International Organization for Standards
(ISO), the Evaluation of Measurement
Data – Guide to to the Expression of
burette).
Uncer tainty in Measurement was rst
●
not
being
able
to
read
an
instrument
due
to
uctuations
in
readings
published in 1995. This was revised in
that
occur
during
measurements
due
to
changes
in
changes
in
the
2008 and has been widely adopted in
surroundings
(for
example,
temperature
variations,
airow,
changes
most countries; it has been translated
in
pressure)
into several languages.
●
reaction
time.
267
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
Absolute and relative uncertainty
Uc ay of masum
According to IUPAC, uncertainty of
measurement is a parameter, associated
with the result of a measurement, that
Suppose
m,
in
of
a
the
in
an
experiment
sample
of
laboratory.
aspirin
Your
you
use
a
top-pan
(2-acetoxybenzoic
recorded
mass
was
balance
acid)
m
=
that
3.56
to
measure
you
have
the
mass,
synthesized
g.
characterizes the dispersion of the values
As
stated,
there
will
be
a
degree
of
uncertainty
associated
with
every
that could reasonably be attributed to the
single
measurement.
So
here
there
is
an
uncertainty
associated
with
the
measurand.
instrument
used,
International Vocabulary of Basic and
been
between
General Terms in Metrology,
mass
should
in
3.55
be
this
g
case
and
reported
the
3.57
as
g.
top-pan
Hence,
follows
in
balance.
the
your
The
mass
uncertainty
laboratory
is
could
0.01
g
have
and
the
notebook:
Second Edition, ISO, 1993.
m
Any
=
(3.56
±
0.01)
experimental
experimental
Where
the
A
An
uncertainty
●
Absolute
the
●
result
result
represents
uncertainty
result
Relative
absolute
result,
g
in
=
the
A
can
(A
strictly
a
given
is
or
the
∆A,
to
the
experimental
speaking
the
in
form:
unit
the
result
and
absolute
∆A
represents
uncertainty
relative.
margin
of
measurement.
uncertainty
uncertainty,
is
reported
∆A)
absolute
uncertainty
from
±
be
measured
or,
be
should
ratio
the
uncertainty
Its
symbol
comparing
size
of
the
is
the
associated
with
∆A.
size
measured
of
the
experimental
A
∆A
_
relative
uncertainty
=
A
Example
3
A
calibrated
burette
has
an
absolute
uncertainty
of
±0.02
cm
.
During
Pcag (%) av uc ay =
3
a
titration,
the
volume
of
a
0.15
mol
dm
solution
of
hydrochloric
acid
______
∆A
relative uncer tainty
(
)
× 100 %
3
at
the
end
point
was
recorded
as
12.25
cm
.
Calculate
the
associated
A
relative
uncertainty.
Solution
3
absolute
uncertainty
(∆A)
=
±0.02
cm
3
measured
experimental
result
( A)
=
(12.25
±
0.02)
cm
3
∆A
_
relative
uncertainty
_
=
3
=
ru 1
2
×
10
3
A
When adding or subtracting measurements,
the absolute uncertainty associated with the
Note
that
each
other
relative
uncertainty
is
dimensionless,
since
the
units
cancel
out!
net measured parameter is the square root
The
relative
uncertainty
is
often
expressed
as
the
percentage
relative
of the sum of the squares of the absolute
_____
uncertainties that is
√
2
uncertainty,
so
in
this
example
the
percentage
relative
uncertainty
Σ∆A
would
be
0.2%
ru 2
When multiplying or dividing measurements,
Propagation of uncer tainty
the relative uncertainty associated with the
net measured parameter is the square root
of the sum of the squares of the relative
uncertainties.
After
identifying
quantities,
combine
the
to
give
propagation
268
the
next
the
of
uncertainties
step
is
to
associated
gure
resultant
out
how
uncertainty
.
uncertainties,
and
in
with
experimentally
these
This
order
is
to
different
what
do
is
this
measured
uncertainties
termed
two
the
rules
are
applied.
11 . 1
Percentage
U n C e r t A i n t i e S
A n d
e r r O r S
i n
M e A S U r e M e n t
A n d
r e S U lt S
error
Pcag o =
Percentage
relative
uncertainty
is
different
to
percentage
error.
literature value
experimental value
________
× 100%
For
example,
change
CaCO
of
the
the
(s),
literature
value
decomposition
was
found
to
be
for
the
reaction
+178.1
of
standard
calcium
enthalpy
literature value
carbonate,
kJ:
3
CaCO
(s)
→
CaO(s)
+
CO
3
The
(g)
∆H
=
+178.1
kJ
2
experimental
percentage
error
value
is
was
given
found
to
be
+172.0
kJ.
Hence
the
by:
178.1
172.0
__
percentage
error
=
ǀ
×
100%
100%
=
3.4%
ǀ
178.1
6.1
_
=
ǀ
ǀ
×
178.1
tOK
The vertical lines, |...|, used in the expression for percentage error represent the idea
of a modulus mathematically, that is any negative value is considered positive. This
is also used in describing the modulus of a complex number in mathematics, |z|,
which is the distance the complex number is, in the form z = x + iy, from the origin.
______
2
|z| is expressed as √ (x
2
+ y
) . This entity has to be positive.
tOK
The same symbols are often used to represent alternative meanings in dierent
Science has been described
scientic disciplines. For example, in chemistry we use ver tical lines in cell
as a self-correcting and
diagram notations to represent dierent phase boundaries. In both physics and
communal public endeavour.
chemistry, ver tical lines of dierent lengths represent a cell, used in a battery,
To what extent do these
that is | |
where the shor ter ver tical line represents the negative pole and the
characteristics also apply
longer ver tical line represents the positive pole. Equally, even in chemistry we
to the other areas of
often use the square brackets, [ ], symbol for dierent purposes, for example
knowledge?
concentration, idea of a complex, etc.
Worked examples
Example 1
The
mass
of
a
sample
bottle
metal
is
33.2901
g.
The
bottle
is
26.3505
g.
If
4.506
g
and
mass
of
a
piece
the
of
empty
titanium
V
is
the
sample
density
of
titanium
at
298
K,
calculate
the
volume
of
titanium.
=
V
is
3
cm
of
m
_
d
the
volume
The
mass
(m)
of
titanium
sample
is:
3
water,
in
cm
,
displaced
at
this
temperature
by
m
=
33.2901
g
26.3505
g
=
6.9396
g
themetal.
(this
operation
express
the
involves
reported
subtraction
result
based
on
so
you
the
Solution
smallest
●
d
is
the
density
of
number
of
decimal
places,
which
is
titanium;
four).
m
is
the
mass
of
titanium;
269
11
M E A S U R E M E N T
D ATA
6.9396
m
_
V
A N D
P R O C E S S I N G
g
__
=
Solution
3
=
=
1.540
cm
3
d
4.506
g
cm
+
pH
=
-log
[H
]
=
3.75
10
(this
operation
express
the
involves
reported
division
result
so
based
you
on
The
mantissa
here
contains
two
SFs.
the
+
[H
3.75
]
=
anti-log
of
3.75
=
e
10
smallest
number
of
SF
,
which
is
four).
4
=
expressed
Example 2
State
the
number
of
signicant
gures
1.8
as
×
two
10
3
mol
dm
SFs.
associated
Example 5
with
the
following:
A
a)
0.00390
kg
of
calibrated
burette
has
an
absolute
uncertainty
Cu(s)
3
of
±0.02
cm
.
During
a
titration,
the
volume
of
a
3
b)
136.250
g
of
NaCl(s)
0.10
mol
dm
solution
of
hydrochloric
acid
at
the
3
end
was
percentage
Solution
First
point
express
both
masses
in
scientic
recorded
relative
as
22.18
cm
.
Calculate
its
uncertainty.
notation
Solution
3
3.90
×
10
kg
of
Cu(s)
3
absolute
uncertainty
(∆A)
=
±0.02
cm
2
1.36250
×
10
g
of
NaCl(s)
measured
a)
three
b)
six
experimental
result
( A)
=
(22.18
±
3
SFs
0.02)
cm
3
SFs
(note
the
last
zero
is
∆A
_
signicant).
relative
uncertainty
(
_
)
=
(
3
)
A
4
=
9
×
10
Example 3
percentage
(%)
relative
uncertainty
=
3
Calculate
the
perchloric
pH
acid,
of
a
0.020
HClO
mol
dm
solution
of
4
(9
×
10
)
×
100%
=
0.09%
(aq).
4
Example 6
Solution
During
Perchloric
acid
completely
is
a
strong
dissociated
in
acid,
so
is
assumed
to
be
solution:
a
titration
the
following
titres
were
3
recorded
for
a
hydrochloric
0.10
acid
mol
from
dm
a
solution
of
burette:
+
HClO
(aq)
+
H
4
O(l)
→
H
2
O
(aq)
+
ClO
3
3
(aq)
initial
4
titre
=
(5.00
±
0.02)
cm
(21.35
±
0.02)
cm
3
+
pH
=
-log
[H
10
O
]
=
-log
3
nal
(0.020)
titre
=
10
3
3
0.020
mol
dm
2
=
2.0
×
3
10
mol
Calculate
dm
the
uncertainty
which
involves
pH
=
two
SFs.
volume
of
this
delivered,
in
cm
,
and
the
volume.
Therefore:
Solution
1.70
3
volume
(since
the
mantissa
must
contain
two
In
Also
note
also
there
are
no
units
for
pH
since
based
on
a
logarithmic
=
21.35
5.00
=
16.35
cm
order
to
obtain
the
uncertainty
in
this
volume
it
we
is
delivered
SFs).
need
to
use
the
expression:
expression.
_____
2
√
ΣΔA
Example 4
since
The
pH
of
a
carton
of
orange
juice
was
found
to
a
subtractive
operation
is
involved.
_______________
be
2
uncertainty
3.75.
Calculate
the
+
[H
270
],
hydrogen
ion
=
√ [(0.02)
2
+
(0.02)
]
=
0.03
concentration,
3
in
mol
dm
3
The
volume
would
be
reported
as
(16.35
±
0.03)
cm
11 . 1
U n C e r t A i n t i e S
A n d
e r r O r S
i n
M e A S U r e M e n t
percentage
Example 7
A n d
relative
r e S U lt S
uncertainty
in
concentration
=
______________
(13.3
±
0.1)
g
of
sodium
chloride
salt,
2
NaCl(s),
is
dissolved
of
water,
of
sodium
in
H
a
ask
O(l).
containing
Calculate
the
(2.0
±
0.1)
2
10
_
√[ (
3
dm
10
_
)
+
(
)
13.3
=
]
5%
2.0
concentration
3
2
concentration
=
6.7
g
dm
(±5%)
3
and
the
chloride,
percentage
concentration.
dissolved
in
in
g
relative
Assume
the
dm
,
in
the
solution
uncertainty
that
the
salt
is
of
The
this
absolute
uncertainty
subsequently
fully
from
can
also
be
found
this:
solution.
5 × 6.7
_
∆A
=
3
=
±0.3
g
dm
100
Solution
3
concentration
13.3
=
(6.7
±
0.3)
g
dm
g
_
concentration
=
3
=
6.7
g
dm
3
2.0
Notice
that
because
is
this
division
required
on
dm
answer
is
is
expressed
involved
calculation.
and
Check
as
two
rounding
this
using
Example 8
SFs
up
The
your
literature
change
of
value
for
combustion
the
of
standard
methanol,
enthalpy
CH
OH(l),
3
1
calculator!
When
was
multiplying
or
dividing
found
to
be
726.0
kJ
mol
:
measurements,
3
_
CH
the
percentage
the
sum
of
the
uncertainty
squares
of
is
the
the
square
root
percentage
of
OH(l)
+
O
3
(g)
→
CO
2
(g)
+
2H
2
O(l)
2
2
relative
1
∆H
=
-726.0
kJ
mol
c
uncertainties:
The
percentage
relative
experimental
value
was
found
to
uncertainty
1
be
in
mass
680.0kJmol
=
Calculate
the
percentage
error,
correct
to
two
0.1
_
∆A
_
×
100%
=
A
×
100%
decimal
places.
13.3
percentage
relative
uncertainty
Solution
in
volume
=
(
726.0)
(
680.0)
__
percentage
error
=
ǀ
(
ǀ
726.0)
×
100%
0.1
_
∆A
_
×
100%
=
A
×
100%
46.0
_
2.0
=
ǀ
ǀ
×
100%
=
6.34%
726.0
Writing thermochemical equations
Thermochemical
equations
can
be
written
with
non-integer
3
stoichiometric
coefcients
(for
example,
O
ratios.
molecules,
only
of
when
considering
2CH
molecule
of
an
are
equation
used,
oxygen
molecules
it
as
being
is
we
is
considered
would
broken
better
to
in
not
a
write
in
of
terms
consider
reaction!
the
of
two-
Hence,
equation
for
this
as:
OH(l)
3
when
integers
thirds
reaction
a
However,
indicative
2
2
mole
(g))
+
3O
(g)
2
→
2CO
(g)
2
+
4H
O(l)
2
271
11
M e A S U r e M e n t
A n d
d AtA
P r O C e S S i n G
11.2 Gapca cqus
Understandings
Applications and skills
Graphical techniques are an eective means
➔
Drawing graphs of experimental results,
➔
of communicating the eect of an independent
including the correct choice of axes and scale.
variable on a dependent variable, and can lead
Interpretation of graphs in terms of the relationships
➔
to determination of physical quantities.
of dependent and independent variables.
Sketched graphs have labelled but unscaled
➔
Production and interpretation of best-t lines
➔
axes, and are used to show qualitative trends,
or cur ves through data points, including an
such as variables that are propor tional or
assessment of when these can and cannot be
inversely propor tional.
considered as a linear function.
Drawn graphs have labelled and scaled axes,
➔
Calculation of quantities from graphs by
➔
and are used in quantitative measurements.
measuring slope (gradient) and intercept,
including appropriate units.
Nature of science
The idea of correlation can be tested in experiments, the results of which can be displayed graphically.
➔
Graphs and correlation
In
science
a
graph
representing
interpreted.
statistical
or
of
is,
versus
the
two
X,
We
as
an
plotted
considered
two
variables.
on
the
(that
way
is,
sets
In
a
the
already
joy
of
discovery
described
detective.
chemist
that
(hence
the
Part
to
analytical
of
data
(that
one
is
chemist
role
of
statistics,
all
about
within
detective
a
and
the
set
of
can
be
described
as
a
and
technique
that
272
direction
of
the
This
is
extent
one
where
parallel
is
of
the
to
one
second
idea
experiments
can
represented
coefcient
a
is
be
deduced
measure
relationship
between
represented
by
various
a
the
the
with
two
one
in
which
variable
correlation
whose
results
can
be
is
possible
a
on
a
useful
linear
of
two
scatter
points
by
from
the
the
plots
to
correlation
strength
variables.
graph.
way
the
symbol,
that
The
Data
show
This
the
are
the
often
scatter
correlation
quantify
relationship
r.
of
the
between
extent
the
two
statistical
indicates
relationship
while
coefcient,
of
in
correlation
versa.
of
change
graphically.
coefcient
analogy).
in
measure
decrease
increases
vice
a
Correlations
the
between
in
thedata
set.
The
degree
from
and
or
is
correlation
or
negative
tested
variables
measure
A
can
it
variables
positive
variable
be
is,
two
increase
is
of
Correlation
A
decreases
plotted
that
the
the
statistical
and
embedded
chemical
the
explore
chemistry
patterns
the
is
involve
analytical
of
which
another.
graph
effect)
variables;
to
variables
data
and
of
another.
any
of
variable
x-axis
of
be
displayed
chemical
relationships
data
is
useful
subsequently
independent
variable
analytical
the
is
very
y-axis.
have
the
a
can
between
random
the
cause)
the
be
which
Dependence
dependent
on
can
relationship
between
Y
data,
two
sets
1
to+1
(gure
1):
value
of
r
can
range
11 . 2
y
G r A P h i C A l
y
t e C h n i Q U e S
y
x
x
r = +1
x
r = 0
r = -1
Figure 1 Sketches of various scatter plots showing dierent correlation coecients, r. The independent variable is the
caus, represented on the x-axis. The dependent variable is the c, represented by the y-axis
●
r
=
+1,
is
indicative
relationship
(all
of
a
points
perfect
lie
on
a
positive
straight
lead
linear
and
●
r
=
0,
no
linear
complete
relationship
scatter
of
exists
(there
to
predictions,
government
line)
r
=
-1,
is
and
of
a
perfect
negative
variable
increases,
the
graphs,
barriers,
–
all
the
data
points
will
lie
on
line
but
the
gradient
will
be
used
in
representations
diverse
analysis,
Interpretation
●
Graphical
effect
lead
●
of
to
the
●
a
have
only
graphs
appropriate
are
a
an
by
have
of
units
Units
●
the
slope
●
the
intercept,
●
the
idea
or
of
a
as
of
health
largely
facilitate
transcend
communication
worldwide.
Chile,
where
At
the
a
research
language
presenters
is
of
of
benet
of
the
but
the
Spanish,
variable,
would
pictorially
and
tongue
is
presentation
research
communicating
dependent
native
a
chemist
clearly
Arabic
from
given
in
may
the
understand
ndings.
This
represented
not
Spanish
graphical
shows
the
scientic
data!
the
can
quantities.
unscaled
variables
generally
axes,
that
would
and
are
are
used
proportional
not
need
to
be
to
or
shown
variables.
and
scaled
Drawn
axes,
graphs
and
are
always
based
display
on
the
variables.
features
gradient
but
as
often
means
a
physical
such
labelled
for
on
in
whose
language,
data
can
effective
which
can
scientists
understand
population
trends
variable
measurements.
number
of
the
as
widely
modelling.
labelled
trends,
proportional.
quantitative
There
are
are
such
statistical
determination
sketch,
Drawn
elds
independent
graphs
data
climate
these
qualitative
inversely
on
of
and
techniques
an
Sketched
show
subject
nance,
of
setting
negative).
Oman
Graphical
the
such
a
used
straight
areas,
other
conference
decreases
underpins
linear
between
(one
so
many
education.
Charts
points)
indicative
relationship
in
is
language
●
and
policies
of
a
that
line,
you
are
required
to
know
for
graphs:
m
c
“best-t”
line
The slope or gradient of a line, m
Mathematically,
angle,
the
θ,
that
x-axis.
In
the
the
slope
line
order
to
of
makes
nd
a
line,
with
the
m,
the
slope
is
the
tangent
positive
of
a
line,
of
direction
you
need
the
of
to
273
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
choose
two
points
on
the
line,
ideally
well
separated
from
each
y
other,
(x
,
y
1
)
and
1
of
the
2
1
=
x
∆x
the
line
negative
-x
2
there
x-axis,
).
_
=
Where
y
-y
2
_
c
,
2
y
∆y
m
(x
is
1
an
incline
gradient
in
the
will
of
the
have
positive
a
line
in
the
positive
direction
of
positive
value;
the
direction
where
x-axis,
the
there
of
is
a
gradient
the
decline
will
have
a
value.
x
Figure 2 The intercept, c, is the point
where the line cuts the y-axis at x = 0
The intercept, c
The
intercept,
c,
is
the
point
where
the
line
cuts
the
y-axis
at
x
=
0
y
(gure
The
2).
intercept
can
be
found
●
using
extrapolation
●
using
the
by
two
methods:
c
equation
Sometimes
when
graph
appropriate
with
you
of
plot
a
line,
a
y
graph,
scales
=
it
where
mx
is
+
c
more
the
convenient
x-axis
scale
to
begins
draw
at
a
the
point
x
greater
than
zero
as
the
data
points
may
not
be
located
at
zero
on
the
Figure 3 The method of
x-axis.
If
this
is
the
case,
by
extrapolation
you
can
simply
extend
the
ex trapolation involves ex tending a
line
back
to
the
y-axis
to
ndc
(gure
3).
line back to the y-axis to nd c
Alternatively,
you
could
choose
some
point
on
the
line,
(x
,
y
c
),
and
use
c
y
the
equation
c
=
y
of
the
line
y
=
mx
+
c
to
nd
c,
as
long
as
you
know
m:
mx
c
c
The idea of a “best-t ” line
When
that,
lie
you
plot
although
exactly
on
data
there
the
obtained
is
line.
a
from
linear
For
this
an
experiment
relationship,
purpose,
it
is
not
you
all
best
to
may
the
nd
data
draw
a
points
line
of
x
best
t
(gure
4).
Remember,
this
line
may
not
necessarily
contain
all
Figure 4 Line of “best t”
experimental
data
points.
Worked examples: using graphs
3
Example 1
Beer’s
A
law
=
where
is
The
based
A
the
relationship:
0.1002
0.130
0.2008
0.270
is
the
c
is
absorbance,
the
ε
is
the
concentration,
0.2819
0.380
0.4000
0.540
0.5082
0.685
0.6000
0.810
Unknown sample
0.460
extinction
and
l
is
the
length.
following
standard
sulfate
data
solutions
concentration
274
on
Absobac, A
εcl
coefcient,
path
Cocao, c/mo m
using
of
was
in
an
atomic
recorded
order
to
using
six
determine
unknown
sample
absorption
of
the
copper(II)
spectroscopy.
the
11 . 2
a)
Explain
what
you
understand
by
a
standard
G r A P h i C A l
t e C h n i Q U e S
Example 2
solution.
The
b)
Draw
a
suitable
relationship
Beer’s
at
plot
low
showing
the
linear
concentrations
that
graph
reactant
proves
law.
the
A
table
→
below
A
is
versus
for
a
a
plot
time
of
concentration
based
on
decomposition
the
data
reaction
of
given
of
in
reagent
products.
3
c)
Calculate
the
concentration,
in
mol
dm
,
of
3
tm t/s
the
unknown
copper(II)
sulfate
[A]/mo m
solution.
2
d)
Calculate
the
slope,
m,
of
the
line,
and
3
4.00 × 10
state
2.30 × 10
itsunits.
3
3
1.00 × 10
2.00 × 10
3
a)
A
3
2.00 × 10
Solution
standard
solution
concentration
is
is
one
known
1.50 × 10
3
whose
3
3.00 × 10
1.00 × 10
exactly.
3
4
4.00 × 10
b)
Based
on
on
the
graph
data
paper
a
graph
(never
is
now
drawn
on
just
white
The
paper!)
or
using
a
such
Microsoft
computer
graph
have
a
title
and
Excel.
all
the
The
axes
can
be
expressed
mathematically
as:
programme
[A]
as
5.0 × 10
drawn
graph
must
=
-kt
+
[A]
must
be
o
labelled
where
k
represents
the
rate
constant
and
[ A]
the
o
with
the
scaled.
appropriate
A
“best-t”
units
line
is
and
appropriately
then
plotted.
initial
In
a)
this
case
all
the
data
points
lie
exactly
concentration.
Calculate
its
the
line,
so
absorbance
proving
is
Beer’s
directly
law;
that
proportion
to
is,
slope,
m,
of
the
graph
and
state
units.
the
the
b)
Determine
concentration.
state
Plot of
the
on
its
the
intercept,
c,
of
the
graph
and
units.
c)
Calculate
the
rate
d)
Deduce
the
state
units.
constant,
k,
and
state
its
units.
A versus c
0.9
initial
concentration,
[A]
,
and
o
(x
, y
2
)
2
0.8
its
0.7
Plot of [A] versus t
2.5
0.6
(x
3
A
0.4
, y
1
md lom
0.5
)
1
2.0
(x
, y
c
)
c
1.5
3
01
0.3
× /]A[
0.2
(x
, y
)
1
1
0.1
1.0
(x
0.5
, y
2
)
2
0
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0
0.7
1
2
3
4
5
3
3
t/ ×
c/mol dm
10
s
3
c)
When
d)
(x
,
y
1
(x
,
)
A
is
C
=
0.34
=
(0.1002,
0.130)
=
(0.6000,
0.810)
mol
Solution
dm
a)
and
(x
,
y
1
)
=
(0.400,
=
(4.00,
2.30)
and
1
1
y
2
)
(x
2
∆y
=
0.810
0.130
__
=
∆x
,
y
2
___
m
0.460,
∆y
0.680
_
0.1002
=
0.4998
1.36
mol
=
1.80
_
=
4.00
∆x
1
=
0.50)
0.500
2.30
__
_
m
=
0.6000
)
2
0.400
=
-0.500.
3.60
3
dm
Next
we
need
3
10
to
take
3
mol
dm
/10
the
units
3
of
m
into
6
s
=
10
account:
3
mol
dm
1
s
275
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
6
Therefore
m
=
-0.500
1
=
5.00
×
×
10
×
10
10
3
mol
3
10
mol
7
5.00
×
6
dm
3
mol
1
dm
s
Suy p
1
s
=
1
dm
s
Mathematically, any propor tionality sign, “∝”, in
a mathematical expression can be replaced by an
b)
Let
(x
,
y
c
)
=
(2.00,
1.50)
c
equality expression plus a constant, for example “= k”.
y
=
mx
c
c
+
c
c
=
y
=
2.50
mx
c
=
1.50
(
0.500
×
2.00)
c
3
×
This
3
10
mol
is
in
the
form
y
=
mx
+
c,
so
a
suitable
linear
dm
1
__
sketch
would
be
to
plot
p
on
the
y-axis
and
(
)
on
V
the
x-axis.
m
would
equate
to
the
constant
k.
As
Suy p
1
__
there
is
no
term
to
the
right
of
the
k(
)
part
of
the
V
Note that the units of c will always correspond to the
expression,
mathematically
the
intercept,
c,
will
units of the y-axis variable, which in this case is [A].
bezero.
c)
The
expression
[A]
=
-kt
+
[A]
is
in
the
Plot of p versus 1/V (n, T constant)
form
o
of
y
So
=
m
mx
=
+
c
-k,
and
hence:
7
k
=
-m
=
-(
5.00
×
10
3
mol
dm
1
s
)
p
7
=
5.00
×
10
3
mol
1
dm
s
3
d)
c
=
[A]
,
so
[A]
o
=
2.50
×
10
3
mol
dm
o
For HL students only. This data is based on a zero-order
1/V
reaction, discussed fur ther in topic 16.
Suy p
Example 3
Sketched graphs have labelled but unscaled axes, and
are used to show qualitative trends, such as variables
Boyle’s
law
is
one
of
the
gas
laws
and
states
that
that are propor tional or inversely propor tional. Units
with
the
temperature,
T,
and
the
amount
of
gas,
generally, therefore, do not have to be included in
n,
constant,
to
the
the
pressure
is
inversely
proportional
sketches. In the IB programme you should know the
volume.
State
how
you
would
represent
dierence between the command terms draw and
this
law
mathematically
as
an
expression
and
sketch. You should always use graph paper when
suggest
the
type
of
linear
graph
you
would
sketch
drawing a graph (complete with a title, labelled
to
illustrate
this
relationship.
and scaled axes, and units). For a sketch, you can
represent the graph on white paper (including a title
Solution
and labelled, but unscaled, axes).
The
key
phrase
here
is
inverse
proportionality.
Hence:
1
_
p
∝
V
1
_
p
=
k
(
)
V
where
276
k
is
a
constant
of
proportionality.
11 . 3
S P e C t r O S C O P i C
i d e n t i F i C A t i O n
O r
O r G A n i C
C O M P O U n d S
11.3 Spcoscopc  cao
o ogac compous
Understandings
Applications and skills
The degree of unsaturation or index of hydrogen
➔
Determination of the IHD from a molecular
➔
deciency (IHD) can be used to determine from
formula.
a molecular formula the number of rings or
Deduction of information about the structural
➔
multiple bonds in a molecule.
features of a compound from percentage
Mass spectrometry (MS), proton nuclear
➔
1
composition data, MS,
H NMR, or IR.
1
magnetic resonance spectroscopy (
H NMR),
and infrared spectroscopy (IR) are techniques
that can be used to help identify compounds
and to determine their structure.
Nature of science
Improvements in instrumentation – mass
➔
➔
Models are developed to explain cer tain
spectrometry, proton nuclear magnetic
phenomena that may not be obser vable –
resonance and infrared spectroscopy have made
for example, spectra are based on the bond
identication and structural determination of
vibration model.
compounds routine.
Degree of unsaturation or index of hydrogen
deciency (IHD)
The
can
or
degree
be
of
used
multiple
The
degree
and
π
to
bonds
of
bonds
●
a
double
●
a
triple
●
a
ring
●
an
The
unsaturation
determine
bond
is
is
can
be
ring
in
a
index
of
hydrogen
molecular
is
as
used
structure,
counted
is
a
formula
deciency
the
number
(IHD)
of
rings
molecule.
counted
counted
aromatic
IHD
a
unsaturation
present
bond
is
in
or
from
one
as
as
degree
worked
●
from
the
structure
●
from
the
molecular
out
as
two
calculate
the
number
of
rings
where:
one
two
counted
to
degree
degrees
of
of
of
unsaturation
unsaturation
unsaturation
four
degrees
of
unsaturation.
ways:
formula.
277
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
From the structure
Compou
Sucu
numb of gs,
ihd
oub bos, a
p bos
benzene
•
one ring
4
•
three double
bonds (in Kekulé
structure)
cyclobutane
•
one ring
1
cyclohexane
•
one ring
1
•
one ring
3
•
two double bonds
•
one ring
•
ve double bonds
•
one triple bond
(chair
conformation)
cyclopentadiene
2-acetoxybenzoic
6
OH
O
acid
C
(aspirin)
CH
O
3
C
O
ethyne
2
From the molecular formula
In
order
C H
c
use
N
h
to
O
n
X
o
the
=
for
=
IHD
is
a
for
the
halogen
generic
atom
=
8
n
=
0
o
=
2
x
=
0
IHD
=
Therefore
molecular
Cl,
Br,
or
formula
I),
we
can
C
expression:
(0.5)(2c
H
O
8
+
2
-
h
-
x
+
n)
:
2
(0.5)(8
the
There
+
2
8
molecule
are
several
0
+
contains
isomers
0)
=
illustration:
1
either
of
C
H
4
278
(F
,
4
h
ring.
X
x
4
c
the
where
following
IHD
Hence
deduce
,
one
O
8
.
2
double
Here
are
bond
just
or
one
three
for
11 . 3
isom of C
h
4
S P e C t r O S C O P i C
O
8
i d e n t i F i C A t i O n
O r
Sucu
O r G A n i C
C O M P O U n d S
ihd
2
methyl propionate
1
O
H
O
C
CH
3
2
3
1
ethyl ethanoate
O
H
C
O
CH
3
2
3
tetrahydro-3-furanol
1
OH
O
Let
us
take
deduce
four
their
compounds
IHD
using
the
with
different
molecular
formulas
and
formula:
Mocua fomua
ihd
Acvy
Using the ChemSpider RSC
C
H
17
NO
21
(cocaine)
8
4
database (www.chemspider.
C
H
27
O (cholesterol)
com) look at the structures of
5
46
these molecules and check to
C
H
6
N (aniline)
4
see if the above calculations
7
agree with what you expect the
C
H
15
ClN
10
O
3
(clonazepam)
12
3
IHD to be from the respective
structures.
Electromagnetic spectrum (EMS)
The
electromagnetic
spectrum
(gure
1)
is
given
in
section
3
of
the
Databooklet
energy
16
14
10
12
10
10
10
10
8
6
10
γ
UV
V
I
4
10
2
10
10
0
2
10
10
IR
B
G
Y
O
400
4
10
6
10
8
10
wavelength/m
radio waves
R
700
wavelength/nm
Figure 1 The electromagnetic spectrum
The
of
energy
the
of
electromagnetic
radiation
by
Planck’s
radiation,
E,
is
related
to
the
frequency,
ν,
equation:
hc
_
E
=
hν
=
λ
where:
34
h
=
Planck’s
Data
constant
=
6.63
×
10
J
s
(given
in
section
2
of
the
booklet )
279
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
E
=
energy
of
radiation
(measured
in
J)
tOK
ν
=
frequency
of
radiation
(measured
in
Hz)
Electromagnetic waves
8
c
=
speed
of
light
=
3.00
×
1
10
m
s
(given
in
section
2
of
Data
booklet)
can transmit information
beyond that of our sense
λ
=
wavelength
(measured
in
m).
perceptions. What are
From
this
relationship
it
can
be
seen
that
the
energy
is
directly
the limitations of sense
proportional
to
the
frequency
and
inversely
proportional
to
the
perception as a way of
wavelength,
that
is:
knowing?
1
_
E
∝
ν
and
E
∝
λ
The
various
regions
spectroscopy
radiation)
of
the
(which
and
is
various
EMS
the
are
study
techniques
the
of
basis
the
are
of
way
used
different
matter
to
types
of
interacts
identify
the
with
structures
of
substances:
●
X-rays
from
to
–
the
as
information
structure
●
Visible
and
and
and
hence
electronic
UV-vis
●
the
●
of
high,
the
of
these
of
X-ray
(UV)
light
an
atom
electrons
Diffraction
distances
spectroscopy
in
cause
atoms.
bond
basis
levels
radiation
and
to
be
patterns
bond
removed
can
angles
in
lead
a
crystallography
give
rise
gives
or
to
electronic
information
molecule.
This
transitions
about
is
the
the
basis
of
causes
stretch
and
groups
cause
certain
bend)
present.
molecular
bonds
and
This
as
is
in
such
the
rotations
a
molecule
provides
basis
and
of
can
IR
to
vibrate
information
on
spectroscopy
give
information
on
lengths.
Radiowaves
because
can
spin
states
environments
the
now
cause
radiowaves
spectroscopy
Let’s
as
the
type
energy
Microwaves
on
is
levels
ultraviolet
this
functional
their
such
form
example,
bond
energy
energy
spectroscopy
Infrared
(for
●
their
inner
to
is
of
consider
●
infrared
●
proton
●
mass
of
the
(IR)
nuclear
be
change.
based
atoms
connectivity
cornerstone
can
of
three
on
Nuclear
this
can
the
transitions
absorbed
be
and
different
in
a
certain
strong
nuclei,
magnetic
which
present
types
of
in
a
on
leads
of
to
eld
causes
(NMR)
different
chemical
information
molecule.
spectroscopy
identication
magnetic
which
resonance
information
deduced,
atoms
spectroscopic
by
that
organic
form
the
compounds:
spectroscopy
1
nuclear
magnetic
spectrometry
resonance
(
H
NMR)
spectroscopy
(MS).
x
Infrared spectroscopy
Unlike
UV
energy
F
to
spring model used to understand the
vibration of molecules
280
various
in
which
about
electronic
their
is
bonds.
the
IR
the
Hence,
can
be
but
vibration
using
identied
correspond
spring
radiation
transitions,
from
groups
transitions
spectroscopy
radiation,
result
functional
vibrational
IR
visible
result
vibrations,
molecules
Figure 2 Hooke’s law is the basis of the
and
to
can
of
in
not
a
have
cause
certain
this
denite
model
does
type
molecular
groups
of
of
spectroscopy
molecule.
energy
sufcient
The
levels.
The
basis
of
11 . 3
In
a
the
spring
spring
or
can
twisted,
vibration
F
That
∝
is,
exerted
be
a
covalent
(both
a
bond
is
considered
symmetrically
distortion.
law
to
the
of
directly
the
with
by
=
where
on
to
expansion
will
applied
F
rise
based
the
expressed
The
giving
is
every
stretched
i d e n t i F i C A t i O n
from
The
physics
and
force
called
as
a
O r
O r G A n i C
spring.
Such
asymmetrically),
required
to
Hooke’s
cause
C O M P O U n d S
bent,
the
law:
x
position
load
model,
be
S P e C t r O S C O P i C
a
the
length
of
proportional
spring
(gure
negative
sign
2).
a
to
spring
the
By
where
F
( x)
force
from
( F)
convention
represents
its
equilibrium
caused
the
the
law
by
is
the
typically
restoring
force
spring:
-kx
k
is
a
constant
fundamental
obeying
Hooke’s
of
proportionality,
frequency
law
can
be
of
the
called
vibration,
related
to
the
the
ν,
spring
based
mass,
m,
on
by
constant
a
the
system
expression:
__
k
_
1
_
ν
=
√
2π
Hence,
for
vibrate
at
the
vibration
higher
frequencies,
and
m
triple
ν.
of
an
atom
frequencies,
The
bonds).
same
If
you
ν,
it
can
be
and
heavier
for
multiple
applies
imagine
two
seen
that
atoms
bonds
atoms
lighter
will
(for
connected
atoms
vibrate
at
example
by
a
will
lower
double
spring
H
Cl
the
Figure 3 Stretching in HCl molecule
stronger
and
For
of
the
bond
therefore
a
more
diatomic
molecular
connecting
energy
molecule,
vibration
is
is
the
two
atoms
required
such
as
to
stretch
hydrogen
possible,
that
is
the
tighter
the
string
will
be
it.
chloride,
stretching
HCl,
only
(gure
one
form
3).
O
If
we
HCl
compare
has
the
the
frequencies
smallest
mass
and
of
HCl,
HBr,
greatest
and
bond
HI
we
enthalpy
nd
(see
that
because
section
11
of
H
the
Data
booklet)
it
will
have
the
greatest
H
frequency.
symmetric stretch
1
Mocu
H
Cl
H
Br
1
Bo apy/kJ mo
Wavumb/cm
431
2886
366
2559
O
H
H
298
I
H
2230
asymmetric stretch
T
able 1 Bond enthalpies and wavenumbers for selected HX molecules
Different
required
As
can
molecules
to
be
execute
seen
absorb
a
from
at
different
vibration
table
1,
IR
will
frequencies
depend
absorptions
on
are
the
because
bond
typically
the
energy
enthalpy.
cited
as
the
O
1
reciprocal
of
the
wavelength
(
)
.
This
is
the
wavenumber
and
has
λ
1
units
of
cm
H
For
polyatomic
For
example,
species,
the
water
there
may
molecule
be
has
several
three
1
●
a
●
an
●
a
symmetric
stretch
(3652
cm
stretch
(3756
cm
of
modes
of
vibration
vibration.
(gure
4):
symmetric bend
)
1
asymmetric
different
modes
H
Figure 4 Modes of vibration of the
)
water molecule. All three modes of
1
symmetric
bend
(1595
cm
).
vibration are IR active
281
11
M E A S U R E M E N T
A N D
D ATA
O
P R O C E S S I N G
However,
for
a
in
change
vibration
a
covalent
the
bond
molecular
mode.
Let
us
to
absorb
dipole
take
some
IR
radiation
moment
there
associated
must
with
be
the
examples:
symmetric stretch
Mocu
Poay of mocu
A bsopo of ir aao
non-polar
no (IR inactive)
non-polar
no (IR inactive)
non-polar
symmetric stretch – no (IR
H
O
2
asymmetric stretch
O
2
CO
2
(gure 5)
inactive)
asymmetric stretch – yes (IR
O
1
)
active) (2349 cm
symmetric bend – yes (IR active)
1
(667 cm
)
symmetric bend
Figure 5 Modes of vibration of the
The
CO
absorbance,
A,
of
a
sample
can
be
related
to
the
transmittance
by
molecule. The asymmetric
2
the
expression:
stretch and the symmetric bend
are IR active because of a change in
A
=
-log
T
10
the molecular dipole moment that
results from the molecular vibration
An
IR
spectrum
is
a
plot
of
the
percentage
transmittance,
%T,
versus
the
1
wavenumber
spectrum,
for
the
and
of
IR
(in
),
functional
for
where
groups
absorptions
different
booklet;
cm
of
the
can
various
functional
example,
%T
be
from
identied.
bonds
groups
C=C
ranges
are
in
The
different
given
absorption
0%
in
in
to
100%.
In
an
characteristic
classes
section
alkenes
26
of
of
ranges
molecules
the
typically
IR
Data
occurs
in
1
the
range
1620–1680
cm
,
etc.
100
)%( ecnattimsnart
50
0
4000
3000
2000
1500
1000
500
1
wavenumber/cm
Figure 6 IR spectrum (in liquid lm) of butanoic acid
In
gure
6,
note
the
following
absorptions:
1
●
strong,
O
282
H
broad
bond
of
peak
a
in
the
range
carboxylic
acid
2500–3000
cm
characteristic
of
the
11 . 3
S P e C t r O S C O P i C
i d e n t i F i C A t i O n
O r
O r G A n i C
C O M P O U n d S
1
●
strong
peak
in
the
range
1700–1750
cm
characteristic
of
the
C=O
group
1
●
peak
the
in
C
the
H
range
2850–3090
cm
characteristic
of
bond.
1
Typically
can
be
in
the
identied.
IR
spectrum.
IR
is
often
is
a
it
of
on
chemistry.
300–1400
is
termed
supporting
provides
bonds
little
technique
the
IR
or
her
the
analytical
to
structural
some
journey
is
often
into
also
of
the
organic
only
physics
at
for
in
the
sometimes,
for
of
of
groups
However,
(and,
point
work
an
absence
decisions
starting
in
the
or
functional
detective
used
as
of
information.
key
an
the
vibrations
region
presence
different
elucidation
is
complex
technique
the
with
making
and
spectroscopy
more
ngerprint
relates
other
in
structural
compound
his
cm
associated
provides
inorganic)
chemist
a
really
powerful
beginning
an
it
chemical
molecules–
it
This
termed
information
certain
region
the
organic
analytical
heat
sensors
and
remotesensing.
1
Proton nuclear magnetic resonance (
H NMR)
spectroscopy
1
H
NMR
spectroscopy
environments
the
most
chemist.
states.
As
eld
orientation.
two
may
with
the
the
align
nuclear
of
they
can
with
energy
external
anti-parallel
the
two
the
applied
the
difference
hydrogen
as
are
The
have
is
of
corresponds
magnetic
depends
on
eld
a
the
and
available
exist
is
to
in
the
two
energy
same
a
magnetic
or
against
alignment
lower
the
it.
of
different
is
in
random
chemical
spin
results
nuclear
region
of
in
spin
to
energy,
will
spin
no
the
compared
radiowave
∆E
and
eld,
This
difference
increased,
there
the
energy
organic
possible
the
in
chemical
possibly
When
This
to
is
different
magnets.
eld
parallel
eld
the
can
placed
conguration.
levels
on
molecule
tiny
magnetic
magnetic
alignment
EMS.
energy
the
a
atoms
states
nuclei
levels.
between
As
behave
spin
the
in
technique
hydrogen
two
When
information
atoms
structural
nuclei
such
gives
hydrogen
important
The
magnetic
states
of
the
∆E,
the
increase.
The
environments
of
atoms.
1
In
a
H
NMR
standard
δ,
is
spectrum,
expressed
assigned
in
as
0
environments
Data
the
position
(tetramethylsilane,
parts
ppm.
have
per
TMS)
million
Hydrogen
different
of
is
the
NMR
termed
(ppm),
nuclei
chemical
of
in
the
the
the
shifts
signal
proton.
different
(see
relative
chemical
δ
for
to
a
shift,
TMS
chemical
section
27
of
the
booklet)
O
1
Therefore
the
the
number
number
of
of
different
signals
on
chemical
a
H
NMR
spectrum
environments
in
shows
which
the
1
hydrogen
of
atoms
methanoic
the
two
are
acid,
different
found.
HCCOH,
chemical
For
example,
two
signals
in
are
environments
of
the
H
found,
the
NMR
spectrum
which
two
C
shows
A
atoms,
A
and
H
hydrogen
B
B.
283
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
typ of poo
Cmca sf (ppm)
CH
0.9–1.0
3
CH
R
1.3–1.4
2
R
CH
1.5
2
O
2.0–2.5
C
RO
CH
2
O
2.2–2.7
C
R
CH
2
CH
3
H
CH
Hal
2.5–3.5
1.8–3.1
3.5–4.4
2
CH
3.3–3.7
2
O
3.7–4.8
C
R
O
CH
2
O
9.0–13.0
C
R
O
HC
H
H
1.0–6.0
CH
4.5–6.0
2
OH
4.0–12.0
H
6.9–9.0
O
9.4–10.0
C
R
H
T
able 2 Typical proton chemical shift values (δ) relative to tetramethylsilane (TMS).
R represents an alkyl group, and Hal represents F
, Cl, Br, or I. These values may vary
in dierent solvents and conditions
284
11 . 3
The
signals
(gure
7)
can
typ of poo
S P e C t r O S C O P i C
be
assigned
as
i d e n t i F i C A t i O n
O r
O r G A n i C
C O M P O U n d S
follows:
Pc δ/ppm
Acua δ/ppm
(fom sco 27 of
1
(fom
h nMr spcum)
Data booklet)
hCO
9.4–10.0
8.06
COOh
9.0–13.0
10.99
A:
B:
12
1
Another
useful
feature
of
a
H
NMR
spectrum
is
that
it
contains
10
8
6
4
2
0
2
an
ppm
integration
trace
that
shows
the
relative
number
of
hydrogen
atoms
1
Figure 7
1
present.
In
the
case
of
the
H
NMR
spectrum
for
methanoic
acid
H NMR spectrum of methanoic acid,
this
HCOOH.
will
be
1:1.
1
An
important
the
fact
that
detected
by
the
protons
magnetic
dimensional
topic
application
view
of
of
in
H
NMR
water
resonance
organs
in
spectroscopy
molecules
the
imaging
human
is
within
(MRI),
body
associated
human
which
(discussed
with
cells
can
gives
a
be
three-
further
in
21).
Mass spectrometry (MS)
In
topic
2
we
introduced
the
principles
of
mass
spectrometry
(MS).
When
a
+
gaseous
ion
molecule
peak
in
compound.
mass
of
which
spectrum
a
masses.
ionized
the
the
be
are
molecular
energetic
molecule
ions.
28
The
can,
in
of
the
to
M
,
the
fact,
on
is
formed.
molecular
ionization
break
fragmentation
information
some
ion,
corresponds
highly
further
Section
Here
its
spectrum
to
will
provides
molecule.
their
mass
Owing
spectrometer,
some
in
a
is
Data booklet
lists
into
pattern
certain
mass
process
up
of
molecular
of
the
involved
smaller
groups
these
in
a
fragments,
observed
functional
some
The
in
a
mass
present
fragments
and
examples:
+
●
(M
15)
results
from
the
loss
of
CH
results
from
the
loss
of
OH
results
from
the
loss
of
CHO
r
3
+
●
(M
17)
r
+
●
(M
29)
or
r
the
loss
of
CH
CH
2
3
+
●
(M
31)
results
from
the
loss
of
OCH
r
3
+
●
(M
45)
results
from
the
loss
of
COOH.
r
100
80
ytisnetni evitaler
60
Figure 8 A nuclear magnetic resonance (NMR)
spectrometer. NMR spectroscopy measures
40
the resonance between an applied magnetic
eld and the magnetic moment of a molecule’s
20
atoms. It allows identication of molecules in
a sample. This 400 MHz Agilent Unity Inova
0
12
15
20
25
30
35
40
45
50
55
60
NMR Spectrometer is located at the Magnetic
m/z
Resonance Facility of the National Renewable
Figure 9 MS of propan1
-ol
Energy Laboratory (NREL). The NREL is based
in Golden, Colorado, USA
285
11
M E A S U R E M E N T
A N D
D ATA
P R O C E S S I N G
Some
of
the
peaks
for
propan-1-ol
(gure
m/z
8)
can
be
assigned
as
follows:
Pc δ / ppm
(fom sco 27 of Data booklet)
+
60
molecular ion peak , CH
CH
3
CH
2
OH
2
+
31 (high relative intensity)
(M
29)
from the loss of
CH
r
CH
2
3
+
29
(M
31)
from the loss of
CH
r
OH
2
Worked examples
Example 1
Deduce
the
Solution
index
of
hydrogen
deciency
of:
For
(CH
)
3
a)
Zanamivir
caused
by
Section
(an
the
37
of
inhibitor
inuenza
the
used
A
to
and
treat
B
infections
viruses)
a)
2
b)
9:1
COH:
3
using
Data booklet
For
CH
COOCH
3
b)
Carbolic
C
H
6
acid
which
has
the
molecular
:
3
formula
O.
a)
2
b)
1:1
c)
m/z
6
Solution
+
=
74
molecular
ion
peak
(CH
)
3
a)
Zanamivir
has
four
ring,
IHD
double
bonds
and
COH
3
one
+
m/z
so
its
=
=
57
(M
5.
17)
from
loss
of
OH.
from
loss
of
CH
r
+
m/z
b)
For
C
H
6
IHD
=
59
(M
O:
15)
r
3
6
=
(0.5)(12
+
2
6
0
+
0)
=
4
Suy p
In MS, don’t forget the + sign for species that remain
Carbolic acid is, in fact, phenol which
OH
after a fragment has been lost.
has the Kekulé structure:
Example 3
An
unknown
compound,
X,
of
molecular
formula
1
C
H
2
Example 2
For
each
of
IR
the
following
two
O,
has
the
following
IR
and
H
NMR
spectra.
4
spectrum
(in
liquid
lm):
compounds:
100
(CH
)
CH
COH
3
COOCH
3
a)
State
3
how
many
signals
each
compound
shows
1
in
b)
its
State
H
NMR
what
spectrum.
you
expect
the
integration
trace
)% ( ecnattimsnart
3
50
to
1
be
for
each
H
NMR.
0
4000
c)
For
the
MS
of
(CH
)
3
values,
286
giving
a
COH,
state
possible
3
reason
3000
2000
1500
m/z
for
your
answer.
1000
1
wavenumber/cm
500
11 . 3
S P e C t r O S C O P i C
i d e n t i F i C A t i O n
O r
O r G A n i C
C O M P O U n d S
1
H
NMR
spectrum
(in
CDCl
):
absorption
for
C=O
in
the
wavenumber
range
3
1
1700–1750
Data
cm
booklet.
,
based
Indeed,
on
there
section
is
a
26
strong
of
the
peak
at
1
approximately
presence
●
If
C=O
an
of
is
a
1727cm
C=O
present,
aldehyde
or
a
,
which
suggests
the
bond.
then
X
might
ketone.
An
be
either
aldehydic
1
proton
is
quite
spectrum,
with
9.4–10.0ppm,
Data
booklet.
single
●
10
9
8
7
6
5
4
3
2
1
If
X
an
CHO.
atoms
a
CH
of
X
showed
peaks
at
m/z
values
15,
29,
Deduce
given
the
and
the
and
Data
44
structure
any
other
booklet.
spectroscopic
structure
(other
of
For
peaks
of
X
were
using
additional
each
the
as
found).
δ
assign
possible
shift,
from
there
=
9.8
that
must
,
portion
the
δ,
does
H
in
section
NMR
the
27
appear
of
to
range
the
be
a
ppm.
of
means
contain
this
which
the
we
now
molecule,
remaining
one
suggests
number
carbon
indicates
have
a
that
is
of
and
methyl
X
that
three
group,
ethanal,
CHO.
3
information
information
spectrum
information
also
fact,
the
3
CH
of
asseen
with
As
hydrogens,
spectrum
chemical
aldehyde
identied
is
MS
a
in
0
ppm
The
In
signal
is
characteristic
O
H
from
as
based
much
on
C
H
the
X.
H
A
B
Solution
●
●
As
the
molecular
formula
of
X
is
given,
Let’s
on
is
worth
rst
nding
out
the
IHD,
now
test
this
proposed
structure
based
it
the
spectroscopic
data
gained
from
the
which
1
H
indicates
the
index
of
hydrogen
NMR
atoms
or
degree
of
spectrum.
Two
types
of
hydrogen
deciency
are
present
in
different
chemical
unsaturation
environments,
For the generic molecular formula C H
c
IHD = (0.5)(2c + 2
h
N
h
O
n
X
o
A
and
B.
:
x
typ of
Pc
igao
Acua
poo
δ/ppm
ac
δ/ppm
x + n)
(fom
(fom
For
C
H
2
O
the
IHD
is:
Sco
1
4
h nMr
27 of Data
(0.5)(4
+
2
4
0
+
0)
=
spcum)
1
booklet)
Therefore
double
●
We
the
bond
note
molecule
or
that,
one
contains
either
one
ring.
based
on
A:
the
molecular
9.4–10.0
ChO
9.8
1:3
formula,
B:
2.2–2.7
X
contains
just
one
oxygen
atom.
The
2.2
classes
COCh
3
for
(C
X
could
CO
be
C),
an
an
ether
(C
aldehyde
O
(C
C),
a
CHO),
ketone
or
an
●
alcohol
(C
Having
is
●
Based
on
spectrum
established
the
structure
of
X,
it
OH).
the
and
above
see
we
now
whether
examine
there
is
a
the
IR
strong
IR
to
worth
returning
conrm
range
for
the
the
to
the
additional
infrared
IR
spectrum
characteristic
absorption
due
to
287
11
M E A S U R E M E N T
the
CH
bonds
in
A N D
the
D ATA
P R O C E S S I N G
wavenumber
as
range
follows
(using
section
28
of
the
Data
1
2850–3090cm
IR
spectrum,
.
As
there
can
are,
be
seen
indeed,
from
booklet):
the
absorptions
■
m/z
=
15,
signies
the
presence
of
CH
,
3
within
this
range.
which
indicates
loss
of
CHO
from
molecule
+
●
F i n a l l y,
consider
the
MS.
There
should
X,
that
is
(M
29)
r
be
a
molecular
ion
peak
at
m/z
=
44,
+
■
corresponding
to
the
relative
molar
m/z
=
29,
signies
the
presence
of
CHO
,
mass
which
indicates
loss
of
CH
from
molecule
3
of
C
H
2
O,
calculated
as
44.06.
This,
indeed,
+
4
X,
that
is
(M
15)
r
is
present.
In
addition,
the
other
dominant
●
m/z
values
in
the
MS
can
be
assigned
This
conrms
compound
X
to
be
ethanal.
Usfu soucs
●
Spectral Database for Organic Compounds, SDBS, hosted by National Institute
of Advanced Industrial Science and Technology (AIST), Japan.
http://sdbs.
db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi
●
EURACHEM – a network of organisations in Europe having the objective
of establishing a system for the international traceability of chemical
measurements and the promotion of good quality practices. There is an
excellent guide on uncer tainty in measurement which might be useful for
your IA and other laboratory work . http://www.eurachem.org/
●
NIST – National Institute of Standards and Technology, USA. The chemistry
por tal is wor th accessing for spectroscopic data etc. http://www.nist.gov/
chemistry-por tal.cfm
288
Q U e S t i O n S
Questions
1
How
many
0.0200
signicant
gures
are
D.
in
g?
with
IB
A.
1
B.
2
Measuring
5
a
May
Which
gas
2
3
D.
5
A
burette
reading
is
recorded
as
27.70
±
volume
of
a
gas
produced
[1]
2010
are
likely
experiment
C.
the
syringe
is
to
be
reduced
repeated
A.
Random
B.
Systematic
C.
Both
D.
Neither
a
when
number
of
an
times?
errors
errors
random
and
systematic
errors
3
0.05cm
.
Which
of
the
following
could
bethe
random
nor
systematic
errors
[1]
actualvalue?
IB
November
2009
3
I.
27.68
cm
II.
27.78
cm
III.
27.74
cm
3
6
Deduce
the
IHD
for
codeine
using
section
37
3
ofthe
A.
I
and
II
B.
I
and
III
C.
II
booklet
only
only
7
and
Data
III
Deduce
the
IHD
formula
C
H
5
D.
I,
IB
II,
and
May
for
a
molecule
of
molecular
only
III
N
10
2
[1]
2011
1
8
The
H
NMR
formula
C
H
3
3
A
piece
of
metallic
aluminium
with
a
mass
spectrum
O
is
of
shown
X
with
molecular
below.
6
of
3
10.044
A
g
was
student
to
found
carried
determine
the
to
out
have
the
a
volume
following
of
3.70cm
calculation
density:
3
2
10.044
_
3
density
(g
cm
)
1
=
3.70
What
is
the
reportfor
best
the
value
density
the
of
student
could
aluminium?
3
A.
2.715
g
B.
2.7
C.
2.71
D.
2.7146
cm
3
g
cm
3
g
cm
10
9
8
7
6
3
g
cm
5
4
3
2
1
0
chemical shift/ppm
[1]
Source: SDBSWeb, http://sdbs.riodb.aist.go.jp (National
IB
May
2011
Institute of Advanced Industrial Science and Technology)
a)
4
Which
experimental
procedure
is
Deduce
which
compoundsis
likelytolead
to
a
large
systematic
of
the
following
most
X
and
explain
your
error?
answer.
A.
Determining
the
concentration
of
an
alkali
CH
[2]
CO
CH
3
by
titration
with
a
CH
=CH
CH
2
B.
Measuring
the
volume
of
a
solution
using
CH
CH
3
CHO;
2
OH
2
a
b)
volumetric
;
3
burette
Deduce
which
one
of
the
signals
in
pipette
1
the
C.
Determining
the
neutralization
in
enthalpy
a
change
HNMR
spectrum
of
occurinthe
spectrum
of
X
would
also
of
one
of
the
beaker
otherisomers,
giving
your
reasoning.
[2]
289
11
M E A S U R E M E N T
c)
The
infrared
werealso
(i)
Apart
C
C
A N D
and
D ATA
mass
P R O C E S S I N G
spectra
for
X
recorded.
from
absorptions
andC
H
absorption,in
wouldbe
bonds,
due
to
suggest
wavenumbers,
present
in
the
one
that
infrared
spectrum.
(ii)
Apart
C
C
[1]
from
absorptions
andC
H
absorption,in
infrared
inone
of
shownin
d)
IB
290
Suggest
of
two
in
the
May
the
2011
to
but
one
absent
present
compounds
a).
formulas
species
mass
spectrum,
other
part
due
suggest
wavenumbers,
inthis
the
bonds,
that
[1]
and
m/z
would
spectrum.
be
values
detected
[2]
12
AT O M I C
ST R U CT U R E
(AHL)
Introduction
The
is
quantized
related
atoms
to
and
nature
the
of
energy
molecules.
energy
states
In
of
this
transitions
empirical
electrons
spectra
topic
we
in
see
how
for
the
evidence
has
been
gained
used
existence
of
to
from
line
provide
energy
emission
strong
evidence
levels.
12.1 Eeto  to
Understandings
Applications and skills
➔
In an emission spectrum, the limit of
➔
Solving
problems
using
E
=
hv
➔
Calculation of the value of the rst ionization
convergence at higher frequency corresponds
to the rst ionization energy.
energy from spectral data which gives the
➔
Trends in rst ionization energy across periods
wavelength or frequency of the convergence
account for the existence of main energy levels
limit.
and sublevels in atoms.
➔
➔
Deduction
of
the
grou p
of
an
element
f rom
i ts
Successive ionization energy data for an
successive
ionization
energy
da ta.
element give information that shows relations
➔
Explanation of the trends and discontinuities in
to electron congurations.
data on  rst ionization energy across a period.
Nature of science
➔
Experimental evidence to suppor t theories – emission spectra provide evidence for the existence of
energy levels.
291
12
ATO M I C
S T R U C T U R E
( A H L )
CERN
Scientic
theories
experimental
(Conseil
Européen
Nucléaire)
Nuclear
near
and
is
Geneva
France
the
on
has
20
several
of
its
are
universe,
set
most
and
up
in
in
the
CERN
it
the
Europe.
has
1911
and
CERN
The
origins
of.
Our
focus
of
When
second
at
the
CERN
world
war
on
understanding
signicantly
nucleus
(sub-topic
an
main
concentrated
the
is
Scientists
progressed
of
involves
international
the
the
time
atom.
then
project
made
most
instruments
physics.
after
Switzerland
and
Europe,
research.
is
for
located
fundamental
extensive
discovery
Rutherford
largest
discover
the
laboratory,
across
outside
to
at
since
the
study
of
Organization
between
the
CERN
Recherche
scientic
The
1954
inside
matter
to
what
research
beyond
of
particle
trying
exploring
of
border
scientic
is
la
CERN
states
example
research
CERN
was
used
matter.
collaborative
of
The
by
accepted.
European
the
countries
supported
be
pour
some
member
excellent
be
to
sophisticated
world,
particles
over
the
Research.
advanced
in
must
evidence
by
Figure 1 The Large Hadron Collider (LHC) at
Ernest
CERN is the world’s largest and most powerful
2.1).
par ticle accelerator. It consists of a 27 km ring of
CERN
was
also
the
birthplace
of
the
World
Wide
superconducting magnets
Web
(the
scientist
internet),
Tim
invented
Berners-Lee;
it
by
the
was
British
originally
In
developed
so
that
scientists
and
1964
a
number
theoretical
around
the
globe
could
instantly
and
2013
the
the
preliminary
Large
CERN
Hadron
suggested
Higgs
by
a
the
Standard
universe
in
the
is
Collider
in
Model ,
from
In
gure
existence
people
is
at
of
the
is
an
termed
how
the
everything
plants
to
stars
was
is
considered
to
be
composed
of
just
a
no
until
who
2013.
blocks.
These
are
the
particles
of
boson
particles
Higgs
themselves
boson
is
one
of
are
17
governed
has
(another
are
sub-topic
2.2).
is
the
constituents
others
of
(including
Some
are
292
in
photon,
Higgs
responsible
nature,
main
for
matter
particle
for
excluding
its
is
Peter
the
described
for
named
Higgs,
existence
for
was
problem
that
its
after
one
of
originally
existence
as
the
linking
there
existence
British
six
in
of
the
greatest
theory
scientists
1964.
Higgs
scientic
with
evidence.
Nobel
Prize
in
Physics
2013
was
awarded
few
to
François
Englert,
Université
Libre
de
matter.
by
Belgium,
and
Peter
W
.
Higgs,
University
forces.
Edinburgh,
of
a
Scotland,
for
all
the
gravity.
mechanism
the
that
theoretical
contributes
example,
our
understanding
of
the
origin
of
mass
of
particles
subatomic
particles,
conrmed
through
and
which
was
recently
whereas
and
the
discovery
of
the
predicted
the
forces
particle,
by
the
that
experiments
occur
the
years
evidence
100years,
fundamental
photon)
a
existence
fundamental
fundamental
everyday
the
50
particle
been
in
experimental
to
see
the
to
discovery
particles
The
evidence
discovery
of
The
However,
almost
Professor
Bruxelles,
Such
for
proposed
jointly
building
boson.
experimental
Empirical
The
trees
Higgs
scientists
physicist
predicted
physics
model
to
1)
boson
describes
this
from
existence
Higgs
particle
which
constructed.
universe,
the
The
whose
model
results
(LHC,
for
particle.
particle
theoretical
experimental
evidence
boson
elementary
suggested
outlining
data.
for
In
physicists
mechanism
share
of
information
of
universities
at
CERN’s
LHC.
ATLAS
and
CMS
12 . 1
E l E c T r O n s
i n
a T O m s
TOK
In topic 2 we discussed a number of key discoveries
Theoretical scientists often work in elds where
associated with the structure of the atom. In this topic we
the application of their research to real life may
continue to look at historical developments in this eld.
be dicult to predict, or dicult to comprehend
by non-scientists.
de Broglie: Wave proper ties of electrons
In 1924 the French scientist Louis de Broglie brought the
wave–par ticle dual theory of the electron to the fore in
Should governments and funding bodies fund basic
theoretical research, or should they concentrate
on applied or strategic research where the end
the de Boge equto:
application may be more economically tangible? Can
h
_
λ =
p
you think of examples of scientic discoveries resulting
where:
from basic research that have resulted in unforeseen
applications?
λ = wavelength
h = Planck’s constant
Heisenberg’s uncer tainty principle can be expressed
p = momentum = mv = mass × velocity
mathematically as follows:
In this equation, wavelength is inversely propor tional
h
_
to momentum.
∆p × ∆q ≥
4π
You might nd that interpretation of this expression may
where:
warrant an exploratory journey of how we consider the
∆p = uncer tainty of momentum measurement
movement of particles. Suppose you have a particle of
large mass, such as a tennis ball. If the mass is large then
so will be the momentum,
∆q = uncer tainty of position measurement
p. The wavelength, λ, associated
h = Planck’s constant
with the tennis ball when moving at high velocity becomes
Suppose you wish to measure the location of a moving
negligible. This agrees with what we observe: you don’t see
electron. If the position is measured with high accuracy,
a tennis ball moving in a wave pattern across a tennis court!
then ∆q will approach zero. What then happens
However,
if
the
mass
of
the
par ticle
is
tiny,
s uch
as
simultaneously to ∆p? To explore this we can rearrange
31
the
electron
with
mas s m
=
9.109
×
10
kg,
then
the equation:
e
the
wavelength
will
be
la rge,
s ugges ting
th at
a
w av e
h
_
motion
can
be
associa ted
w ith
the
electron
in
an
atom.
∆p ≥
(
1
_
)
4π
So
de
wave
Broglie
suggested
proper ties,
but
so
that
d oes
not
only
does
li g ht
Δq
hav e
______
1
m at ter!
as ∆q → 0, then
→ ∞
Δq
The de Broglie equation shows that macroscopic par ticles
so ∆p → ∞
have too shor t a wavelength for their wave proper ties to be
observed.
that is, as the uncer tainty of position measurement
approaches zero, the uncer tainty of momentum
Is it meaningful to talk of proper ties that cannot be
measurement approaches innity so the momentum
observed with our senses?
becomes eectively undened.
Heisenberg’s uncer tainty principle
In 1927 another theoretical physicist, Werner Heisenberg
from Germany, published the ground-breaking theory
termed Heebeg’ ue tty ppe. Professor
Heisenberg was one of the pioneers in the eld of qutu
eh and the basis of his principle is as follows:
The more precisely the position is determined,
the less precisely the momentum is known in this
instant, and vice versa.
Heisenberg, uncer tainty paper, 1927.
Heisenberg said: “ What we observe is not nature itself,
but nature exposed to our method of questioning.”
Can our senses give us objective knowledge about
the world?
The idea of uncer tainty lying at the hear t of Heisenberg’s
thinking is an example of a historical journey of discovery
that embraces not just physics but the persona of an
individual as well.
Find out more about Heisenberg and consider what is
meant by this statement.
293
12
ATO M I C
S T R U C T U R E
( A H L )
The Schrödinger wave equation
The dual wave–particle nature of the electron has been one
shödge wve equto. See sub-topic 2.2 for more
of the great discussions in the history of subatomic particles.
information on the Schrödinger wave equation.
The Austrian physicist Erwin Schrödinger (1887–1961)
Schrödinger ’s wave equation accurately predicted the
was an advocate of wve eh, expressed in the
energy levels of atoms.
Emission spectra and ionization
The values of some rst
1
ionization energies, in kJ mol
,
are given in section 8 of the
Data booklet
In
sub-topic
Emission
atomic
In
2.2
energy
sub-topic
required
in
its
the
spectra
to
line
emission
provide
spectrum
experimental
of
hydrogen
evidence
for
the
was
introduced.
existence
of
levels.
3.2
ionization
remove
ground-state.
an
energy
electron
The
rst
was
from
dened
a
as
neutral
ionization
the
minimum
gaseous
energy
( IE
)
atom
of
a
energy
or
molecule
gaseous
atom
1
Uefu eoue
is
related
to
the
process:
Chemsoc Timeline – This is
+
X(g)
a visual exploration of key
→
X
(g)
+
e
events in the history of science
Successive
ionizations
are
also
possible;
for
example,
the
second
with par ticular emphasis on
ionization
energy
(IE
)
is
associated
with
the
process:
2
chemistry.
It was developed
+
X
2+
(g)
→
X
(g)
+
e
by Murray Roberston in
collaboration with ChemSoc,
th
The
n
ionization
energy
(IE
)
relates
to
the
process:
n
the chemical network of the
(n
1)+
X
n+
(g)
→
X
(g)
+
e
Royal Society of Chemistry
(RSC). You can even make
For
a
given
element
the
IE
increases
for
successive
ionizations,
in
the
order:
predictions for inventions or
IE
<
IE
1
<
IE
2
<
IE
3
<
IE
4
.
.
.
5
discoveries that you think will
be made in years to come!
This
is
because
removed
from
with
an
each
successive
increasingly
ionization
positive
species,
an
electron
and
hence
is
being
more
energy
is
http://www.rsc.org/chemsoc/
required.
For
example,
for
magnesium,
Mg:
timeline/pages/timeline.html
1
IE
=
737.7
kJ
mol
=
1450.7
kJ
mol
=
7732.7
kJ
mol
=
10542.5
=
13630
1
1
IE
2
1
IE
3
1
IE
kJ
mol
4
1
IE
kJ
mol
5
In
sub-topic
hydrogen
of
we
inuence
of
occurred).
shown
The
the
in
the
increase
frequency
of
can
2
the
be
in
the
merge,
can
have
the
in
at
any
principal
radiation
to
the
in
higher
forming
electron
represents
used
emission
converge
lines
nucleus:
gure
convergence
that
lines
electron
the
The
saw
the
convergence
continuum
∞
2.2
atom,
a
spectrum
of
energies.
At
quantum
the
so
outside
process
IE
.
of
In
is
no
the
spectrum
the
of
at
Lyman
the
under
the
(ionization
from
ionization
limit
Beyond
longer
atom
number
emission
determine
the
continuum.
energy,
is
the
n
=
the
the
series
1
to
has
n
=
atom.
limit
for
of
the
1
hydrogen
relates
to
returns
294
atom
the
to
(UV
energy
the
region),
given
the
out
ground-state,
n
frequency
when
=
1,
an
at
the
electron
asshown
in
limit
falls
gure
of
convergence
from
6
of
n
=
∞
and
sub-topic
2.2.
12 . 1
E l E c T r O n s
i n
a T O m s
outside the atom
continuum
n = ∞
inside the atom
n = 5
n = 4
n = 3
n = 2
n = 1
Figure 2 The ionization process for the hydrogen atom
Determining the wavelengths of lines in spectra: The Rydberg
equation
hc
_
The
Rydberg
equation
can
be
used
to
nd
IE
=
E
E
∞
=
hν
=
1
λ
the
wavelengths
of
all
the
spectral
lines
in
the
(since
emission
spectrum
of
hydrogen,
and
is
given
c
=
νλ)
by
8
the
c
expression:
=
speed
of
light
=
2.998
×
10
1
m
s
1
1
_
=
1
_
R
1
_
From
the
Rydberg
equation,
can
be
inserted
into
λ
(
H
λ
2
2
n
n
)
the
expression
for
IE
and
rearranged
to:
f
i
where:
IE
λ
=
=
Rydberg
=
constant
=
1.097
×
10
2
=
nal
)
2
∞
1
34
(6.626
×
8
10
state
J
7
(1.097
f
(
)
m
=
initial
1
___
1
__
(
1
H
i
hcR
H
7
R
=
wavelength
×
s)(2.998
×
10
1
m
s
)
1
10
m
)(1
0)
state
18
=
n
=
principal
quantum
2.179
×
10
J
number
–1
The
Note
that
n
is
greater
than
f
energy
in
kJ
mol
is
found
by:
n
i
18
IE
The
IE
can
then
be
determined
as
=
(2.179
=
1.312
=
1312
×
10
23
J)(6.022
6
ΔE
=
×
10
1
mol
)
follows:
×
10
1
J
mol
hν
1
kJ
mol
where:
This
value
calculated
for
the
rst
ionization
34
h
=
Planck’s
constant
=
6.626
×
10
J
s
energy
(IE
)
for
hydrogen
is
given
in
section
8
of
1
ν
=
frequency
the
Data
booklet
295
12
ATO M I C
S T R U C T U R E
( A H L )
study tp
●
Worked examples: determining energy
In any calculation you
Example 1
should use the data given
in the question or otherwise
Determine
the
energy,
in
J,
of
a
photon
of
red
light,
correct
to
four
data from the Data booklet.
signicant
gures,
In example 1 you should
given
that
the
wavelength
34
h
=
6.626
×
8
10
J
s;
c
=
2.998
×
10
λ
=
650.0
nm.
1
m
s
use the values of h and
c provided. Note that the
Solution
question requires an
hc
_
ΔE
=
hν
=
answer to a given number
λ
of signicant gures.
34
8
1
6.626 × 10
J s × 2.998 × 10
m s
____
ΔE
19
=
=
3.056
×
10
J
9
●
Always
read
carefu lly.
the
Also,
650.0
question
make
×
10
m
s u re
Example 2
you
include
the
units
1
Calculate
throughout
the
that
stages
of
your
the
rst
ionization
energy,
in
its
shortest-wavelength
line
in
the
34
the
to
will
correct
the
final
answer.
you
help
In
must
conver t
you
units
for
hydrogen
series
is
91.16
given
nm.
this
=
6.626
×
8
10
J
s;
c
=
2.998
×
10
1
m
s
23
;
N
=
6.022
×
1
10
mol
A
related
Solution
questi on
remember
to
,
Lyman
obta in
numerical
nm
mol
answer.
h
This
kJ
variou s
The
to
shortest-wavelength
transition
of
m.
n
=
∞
to
n
line
=
in
the
Lyman
series
corresponds
to
a
1.
hc
_
IE
=
hν
=
1
λ
34
8
1
6.626 × 10
J s × 2.998 × 10
m s
____
IE
18
=
=
1
2.179
×
10
J
9
91.16
×
10
m
×
(6.022
1
expressed
in
kJ
mol
:
18
IE
=
(2.179
=
1312
×
23
10
J)
×
10
1
mol
6
)
=
1.312
×
1
10
J
mol
1
1
kJ
mol
Periodic trends in ionization energies
1
Figure
group
metal
from
3
2
shows
ten
alkaline
titanium,
IE
to
IE
2
;
successive
earth
Ti.
In
the
metal
the
third
ionization
calcium,
case
of
Ca
ionization
Ca
energies,
and
there
is
energy
the
a
in
kJ
group
mol
4
signicant
corresponds
,
for
the
transition
jump
to
the
going
removal
3+
an
electron
do
not
from
the
fully
occupied
3p
sublevel.
As
a
result
Ca
species
Ca
Ti
occur.
This
supports
observations
that
for
the
group
2
metals
2+
20 000
there
is
one
contrast,
Ti
stable
oxidation
exhibits
state,
oxidation
+2
states
(forming
of
+2,
+3,
2+
ions,
and
+4
eg
Ca
(see
).
topic
In
13
1
lom Jk/I
15 000
and
Ti
section
is
+4.
In
14
of
the
gure3
Data
there
booklet).
is
a
The
large
most
jump
in
stable
IE
for
oxidation
Ti
going
state
from
of
IE
to
4
IE
10 000
,
corresponding
to
the
removal
of
the
fth
electron,
supporting
the
5
observation
that
species
with
Ti
in
the
+5
oxidation
state
do
not
occur.
5000
The
electron
deduced
as
congurations
follows:
0
2
0
1
2
3
4
5
6
7
8
9
10
11
Ca:
[Ar]4s
I No.
2+
Ca
:
[Ar]
Figure 3 Ten successive ionization energies for
calcium and titanium
2
Ti:
[Ar]3d
4+
Ti
296
of
3
25 000
:
[Ar]
4s
2
for
the
most
stable
ions
of
Ca
and
Ti
are
12 . 1
It
is
interesting
to
note
that
in
the
case
of
Ti,
the
ionization
E l E c T r O n s
i n
a T O m s
energies
1
IE ube
increase
the
3p
3d
and
and
The
more
4s
4s
gradually
orbitals,
than
which
for
are
Ca
as
electrons
much
closer
in
are
being
energy
removed
compared
to
1
alkali
metal
potassium,
K
(Z
=
19)
has
the
1
418.8
2
3052
3
4420
4
5877
5
7975
6
9590
7
11343
8
14944
9
16963.7
10
48610
11
54490
12
60730
13
68950
14
75900
15
83080
16
93400
the
orbitals.
group
IE/kJ o
from
electron
conguration:
2
K:
1s
2
2s
From
6
2p
this
2
3s
6
3p
1
4s
conguration
we
might
expect
a
large
jump
going
from
IE
1
to
IE
,
from
IE
2
to
IE
9
associated
principal
with
,
and
from
IE
10
the
removal
quantum
to
IE
17
number
of
n.
electrons
Table
1
.
These
signicant
jumps
are
18
from
shows
energy
the
rst
levels
19
of
IEs
different
for
K.
TOK
In plotting ionization energies, a ogth e allows all data points to be
plotted on a single graph.
1
The dierence between IE
and IE
1
for K is 2633.2 kJ mol
, while the dierence
2
1
between IE
18
and IE
is 31191 kJ mol
. Therefore it is dicult to represent all
19
19 ionization energies for K on a linear scale. Look at the unreasonably long
y-axis when comparing the plot of IE versus IE No. to the plot of log
17
99710
18
444870
19
476061
IE versus
10
IE No. in gure 4 (a) and (b).
Can you think of examples in chemistry or other sciences that present data in a
par ticular way to suppor t the scientist’s postulates, theories, and hypotheses?
T
able 1 Values of ionization energy
Where else in chemistry do we use logarithmic plots? Do you know the
dierence between log
10
and log
(IE) for the rst 19 ionization
(ln) and can you give examples of where
energies for potassium. Signicant
e
jumps are evident between IEs 1
each type of log is used in chemistry?
and 2, 9 and 10, and 1
7 and 18
Plot of log
Plot of I versus I No. for K
I versus I No. for K
10
6
500000
450000
5.5
400000
5
350000
4.5
300000
1
I gol
lom Jk/I
250000
4
200000
3.5
150000
3
100000
2.5
50000
0
2
0
5
10
15
20
0
5
I No.
10
15
20
I No.
Figure 4 (a) Plot of IE versus IE No. for potassium; (b) Plot of log
IE versus IE No. for potassium
10
297
12
ATO M I C
S T R U C T U R E
( A H L )
Worked examples
Example 1
Values
for
element
group
X
of
expect
successive
are
the
to
study tp
the
given
in
periodic
nd
X.
IEs
table
table
State
for
the
2.
of
an
unknown
Deduce
in
elements
name
of
you
this
When writing electron congurations, electrons in
which
individual orbitals should be presented as superscript:
2
would
1s
group.
2
2s
6
2p
1
3s
rather than 1s2, 2s2, 2p6, 3s1. In
addition, always take note of the type of electron
conguration requested – this question asks for the
1
IE ube
IE/kJ o
full electron conguration so you should not write a
1
condensed electron conguration such as [Ne]3s
IE
899
1
IE
1757
2
Solution
14850
IE
2
3
Na:
1s
2
2s
6
2p
1
3s
21005
IE
4
The
the
plot
in
easiest
gure
to
5
shows
remove.
that
This
is
the
rst
because
electron
it
is
is
furthest
Table 2 Ionization energies (IEs) for X.
from
the
nucleus,
occupying
Solution
is
a
large
the
being
the
outermost
increase
going
n
valence
=
from
3
electron
energy
IE
to
IE
1
The
largest
jump
in
IE
occurs
between
IE
level.
There
because
2
and
2
the
next
electron
is
removed
from
the
n
=
2
level.
1
IE
,
corresponding
to
ΔIE
=
13093
kJ
mol
.
This
3
However,
mustcorrespond
to
a
change
in
energy
X
must
be
in
group
2,
the
the
next
seven
electrons
the
small,
the
level;
gradual
therefore
for
increase
in
IE
reects
the
fact
that
all
eight
alkaline
electrons
occupy
the
same
n
=
2
energy
level.
There
earthmetals.
is
another
large
jump
in
IE
going
from
IE
to
IE
9
associated
with
the
removal
of
an
,
10
electron
from
the
Example 2
n
Figure
5
represents
the
successive
=
1
energy
nucleus
energies
of
sodium.
The
vertical
axis
plots
and
energy)
instead
of
ionization
so
This
will
be
electron
very
is
closest
difcult
to
energy
electron
also
comes
from
the
1s
data
unreasonably
to
be
long
represented
vertical
without
The
sublevel,
IE
shows
only
a
small
increase
over
IE
11
the
the
to
so
allow
to
remove.
log
eleventh
(ionization
level.
ionization
using
.
10
an
axis.
Example 3
6.0
Figure
6
energies
shows
for
the
the
variation
second-row
in
rst
ionization
elements
in
the
5.5
periodic
table
from
Li
to
Ne.
5.0
He
I gol
4.5
2400
Ne
4.0
2000
3.5
F
1600
N
1
3.0
1
2
3
4
5
6
7
8
9
10
11
number of electrons removed
lom Jk/I
2.5
O
H
1200
C
Be
800
Figure 5 Successive ionization energies of sodium
B
Li
State
and
the
full
explain
electron
how
the
conguration
successive
of
400
sodium
ionization
0
energy
data
for
sodium
are
related
to
its
0
electron
conguration.
2
4
6
8
10
[4]
Z
IB,
298
May
2010
Figure 6 Ionization energies for the rst 10 elements
12 . 1
a)
Explain
why
as
you
go
across
a
period,
E l E c T r O n s
i n
a T O m s
IEs
increase.
b)
Although
there
a
general
increase
in
IE
1
across
the
second
period
as
expected,
there
2
is
1
2s
1
2p
evidence
of
some
discontinuity.
This
is
to
as
a
dog-teeth
plot.
2p
y
z
often
For
referred
1
2p
x
Explain
nitrogen
the
most
loosely
bound
electron
is
why:
any
one
of
the
three
3p
electrons
which
are
all
1
i)
IE
for
oxygen
(1314
kJ
mol
)
is
lower
are
1
degenerate
(have
the
same
energy).
1
than
IE
for
nitrogen
(1402
kJ
mol
)
1
ii)
Again
start
by
drawing
orbital
diagrams
for
Be
1
ii)
IE
for
boron
(801
kJ
mol
)
is
lower
than
1
and
B:
1
IE
for
beryllium
(900
kJ
mol
).
1
2
Be:
[He]2s
Solution
a)
IEs
increase
across
●
decreasing
a
period
atomic
radii
for
two
across
reasons:
a
period
2
from
left
to
0
2s
right
increasing
nuclear
charge,
B:
Z
[He]2s
1
2p
b)
i)
First
consider
elements
2
O:
the
oxygen
[He]2s
2
2p
orbital
and
1
2p
x
diagrams
of
2p
y
0
2p
x
0
2p
x
2
●
0
2p
z
0
2p
y
z
the
nitrogen:
1
2p
y
z
2
1
2s
0
2p
The
the
most
2p
loosely
orbital
2p
y
bound
while
0
2p
x
for
z
electron
Be
it
is
for
one
B
of
occupies
the
two
x
2
2
2s
1
2p
1
2p
x
electrons
2p
y
in
the
2s
level.
It
will
be
easier
to
z
remove
the
electron
from
the
2p
orbital
in
B
x
The
rst
electron
to
be
removed
from
a
neutral
since
gaseous
atom
will
come
from
the
highest
2p
is
of
the
highest
energy
level.
In
in
energy
than
2s.
This
criterion
occupied
overrides
sublevel
higher
this
any
consideration
of
paired
electrons
case
in
1
an
orbital.
Hence
IE
for
B
(801
kJ
mol
)
is
lower
1
this
is
the
2p
sublevel
which
is
higher
than
2s
1
than
IE
for
Be
(900
kJ
mol
).
1
in
energy.
obvious
This
from
represented
of
the
(sub-topic
an
orbital
with
horizontally.
order
difference
the
energy
diagram
outermost
Remember
energies
in
of
that
the
may
which
not
is
energy
be
often
levels
just
shown
within an energy level
sublevels
is
s
<
p
<
d
the
<
f
2.2).
aogy
Suppose you have a two-story building and you
need to remove one oor in order to meet new height
regulations. Which oor would you remove? Obviously
the top oor (oor 2) – the building would collapse if
So
an
electron
will
be
removed
from
the
2p
you removed the ground oor (oor 1)! It is the same
sublevel.
Which
is
the
most
loosely
bound
electron?
when removing electrons from energy levels and
There
will
be
maximum
repulsion
in
an
orbital
that
sublevels – electrons are removed from the energy
contains
paired
electrons,
so
the
most
loosely
bound
level of highest principal quantum number n rst, and
electron
will
be
a
2p
electron
as
circled
in
the
orbital
x
from the sublevel with the greatest energy, within that
diagram
for
oxygen
above.
This
is
the
reason
why
energy level.
IE
is
lower
for
oxygen
than
for
nitrogen,
whose
1
orbital
diagram
2
N:
[He]2s
is
1
2p
shown
1
2p
x
on
the
here:
1
2p
y
z
299
12
ATO M I C
S T R U C T U R E
( A H L )
Questions
1
Figure
7
represents
remove
nine
atom
an
have
of
been
the
energy
electrons,
element.
one
Not
all
at
of
needed
a
time,
the
4
to
from
The
an
graph
against
electrons
of
the
atomic
elements
rst
ionization
number
shows
for
the
periodicity
energy
rst
(gure
plotted
twenty
8).
removed.
2500
1
lom Jk/ygrene noitazinoi tsr
I gol
number of electrons removed
▲
2000
1500
1000
500
0
Figure 7
0
1
2
3
4
5
6
7
8
9
10 11
12
13
14
15
16
1
7
18 19 20
atomic number
Which
element
could
this
be?
▲
A.
C
B.
Si
Figure 8
i)
Dene
state
C.
P
D.
S
IB,
May
the
what
term
is
rst
meant
ionization
by
the
energy
and
term
periodicity.
ii)
Explain
of
which
ionization
energies
of
main
there
be
the
greatest
energy
State
A.
Between
1st
B.
Between
2nd
C.
Between
3rd
and
4th
ionization
energies
D.
Between
4th
and
5th
ionization
energies
IB,
November
Which
of
A.
>
IE
the
2nd
ionization
and
3rd
ionization
energies
[1]
2009
following
is
correct?
4
B.
Molar
C.
The
ionization
third
energies
ionization
are
energy
measured
represents
in
kJ.
the
process:
2+
D.
→
X
Ionization
going
300
3+
(g)
(g)
+
e
energies
from
left
to
levels
is
meant
decrease
right.
energy.
energies
IE
3
X
this
and
the
existence
sublevels
[4]
by
the
term
second
difference?
IB,
3
from
for
atoms.
what
ionization
and
evidence
boron
iii)
will
information
provides
within
Between
how
2010
graph
2
[2]
[1]
across
a
period
[1]
May
2009
[1]
T H E
P E R I O D I C
TA B L E
–
T H E
13
T R A N S I T I O N
M E TA LS
( A H L )
Introduction
Transition
elements
incomplete
consider
d
one
complexes
of
have
sublevels
simple
of
characteristic
such
theory,
transition
metals.
crystal
metal
are
properties.
In
eld
often
this
These
topic
theory
we
which
properties
explore
can
help
can
these
us
in
be
associated
properties
and
understanding
with
in
the
particular
why
the
coloured.
13.1 Fi- -c 
Understandings
Applications and skills
➔
Transition elements have variable oxidation
➔
Explanation of the ability of transition metals to
states, form complex ions with ligands, have
form variable oxidation states from successive
coloured compounds, and display catalytic and
ionization energies.
magnetic proper ties.
➔
➔
Explanation of the nature of the coordinate
Zn is not considered to be a transition element
bond within a complex ion.
as it does not form ions with incomplete d
➔
Deduction of the total charge given the formula
orbitals.
of the ion and ligands present.
➔
Transition elements show an oxidation state of
➔
Explanation of the magnetic proper ties in
+2 when the s- electrons are removed.
transition metals in terms of unpaired electrons.
Nature of science
➔
Looking for trends and discrepancies –
transition elements follow cer tain patterns
of behaviour. The elements Zn, Cr, and Cu do
not follow these patterns and are therefore
considered anomalous in the rst-row d-block .
301
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
At
IUPAC ii f a
the
family
aii 
centre
of
chemical
of
M E TA L S
the
elements,
( A H L )
periodic
called
properties
often
table
the
of
elements
transition
play
a
key
role
lies
a
very
elements,
in
many
important
whose
facets
of
physical
and
everyday
life.
According to IUPAC, a aii
In
the
periodic
table
(gure
1)
the
rst-row
transition
elements
are
the
 is an element that has
elements
in
period
elements
below
4
from
scandium
(Sc)
to
copper
(Cu)
inclusive.
The
an atom with an incomplete
these
elements
in
periods
5,
6,
and
7
are
also
described
d-sublevel or that gives rise
as
transition
elements.
to cations with an incomplete
The
lanthanoids
are
the
elements
from
Z
=
57
to
Z
=
71
and
the
d-sublevel. These refer to
actinoids
are
the
elements
from
Z
=
89
to
Z
=
103.
La
(Z
=
57)
and
elements in groups 3–11.
1
Ac
(Z
=
89)
respectively
all
the
have
(so
other
do
not
The
transition
congurations
contain
lanthanoids
congurations.
inner
electron
and
f-electrons
actinoids
f-block
of
in
[Xe]5d
their
contain
elements
are
2
1
6s
outer
and
[Rn]6d
energy
f-electrons
sometimes
in
levels),
their
described
2
7s
electron
as
the
elements
1
18
2
1
He
H
hydrogen
[1.007; 1.009]
3
,
but
helium
13
2
14
5
4
15
6
16
1
7
8
9
7
4.003
10
Li
Be
B
C
N
O
F
Ne
lithium
beryllium
boron
carbon
nitroge n
oxygen
uorine
neon
[6.938; 6.997]
9.012
[10.80; 10.83]
[12.00; 12.02]
[14.00; 14.01]
[15.99; 16.00]
19.00
20.18
11
12
13
14
15
16
1
7
Na
Mg
Al
Si
P
S
Cl
Ar
sodium
magnesium
aluminium
silicon
phosphorus
sulfur
chlorine
argon
22.99
24.31
26.98
[28.08; 28.09]
30.97
[32.05; 32.08]
[35.44; 35.46]
39.95
3
4
5
6
7
8
9
10
19
20
2
1
22
23
24
25
26
27
28
29
30
31
32
33
34
35
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
potassium
calcium
scandium
titanium
vanadium
chromium
manganese
iron
cobalt
nickel
copper
zinc
gallium
germanium
arsenic
selenium
bromine
krypton
39.10
40.08
44.96
47
.87
50.94
52.00
54.94
55.85
58.93
58.69
63.55
65.38(2)
69.72
72.63
74.92
78.96(3)
79.90
83.80
37
38
39
40
41
42
43
44
45
11
18
46
12
47
48
49
50
51
52
36
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
rubidium
strontium
yttrium
zirconium
niobium
molybdenum
technetium
ruthenium
rhodium
palladium
silver
cadmium
indium
tin
antimony
tellurium
iodine
xenon
85.47
87
.62
88.91
91.22
92.91
95.96(2)
101.1
102.9
106.4
107
.9
112.4
114.8
118.7
12
1.8
127
.6
126.9
131.3
57-7
1
55
56
Cs
Ba
caesium
barium
Cs
132.9
87
Fr
francium
lanthanoids
74
75
76
77
78
79
80
81
82
83
84
85
86
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
hafnium
polonium
astatine
radon
tantalum
tungsten
rhenium
osmium
iridium
platinum
gold
mercury
thallium
lead
bismuth
1
78.5
180.9
183.8
186.2
190.2
192.2
195.1
197
.0
200.6
[204.3; 204.4]
207
.2
209.0
104
105
106
107
108
109
110
111
112
Rf
Db
Sg
Bh
Hs
Mt
actinoids
rutherfordium
dubnium
seaborgium
bohrium
hassium
meitnerium
Ra
radium
73
T
a
89103
137
.3
88
72
Hf
57
59
58
60
6
1
62
Ds
darmstadtium
63
114
116
Rg
Cn
Fl
Lv
roentgenium
copernicium
erovium
livermorium
64
65
66
67
68
69
70
7
1
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
lanthanum
cerium
praseodymium
neodymium
promethium
samarium
europium
gadolinium
terbium
dysprosium
holmium
erbium
thulium
ytterbium
lutetium
138.9
140.1
140.9
144.2
150.4
152.0
157
.3
158.9
162.5
164.9
167
.3
168.9
1
73.1
1
75.0
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
actinium
thorium
protactinium
uranium
neptunium
plutonium
americium
curium
berkelium
californium
einsteinium
fermium
mendelevium
nobelium
lawrencium
232.0
231.0
238.0
▲
Figure 1 IUPAC periodic table of the elements
The
as
elements
transition
of
group
elements
12,
that
is
according
Zn,
to
Cd,
Hg,
IUPAC
as
and
all
Cn,
four
are
not
classied
elements
have
10
d-sublevels
scandium
Ufu uc
containing
and
yttrium
ten
are
d-electrons
classied
as
(for
transition
1
an
incomplete
d-sublevel
3+
The electron congurations of
only
all the elements in the periodic
to
table can be found by accessing
(Sc
http://www.webelements.com/,
of
compiled by Professor Mark
metal
Winter at the University of
and
Sheeld, UK.
a
Sc
be
[Ar]3d
1
;
Y
,
[Ar]3d
elements
2
4s
Zn:
[Kr]4d
as
they
full
2
4s
).
Both
have
2
5s
).
In
1920,
when
3+
and
Y
compounds
non-transition
3+
,
(Sc,
example,
elements
were
known,
because
they
their
ions
were
widely
contained
considered
no
d-electrons
3+
[Ar],
these
Y
,
[Kr]).
elements
bonding.
Since
have
For
then
been
example,
many
lower
synthesized,
scandium
oxidation
most
can
of
exist
which
in
the
state
compounds
involve
+2
metal–
oxidation
state,
1
because
transition
compound
its
electron
element
in
which
conguration
according
scandium
to
is
the
in
is
[Ar]3d
IUPAC
the
+2
,
scandium
denition.
oxidation
An
state
is
considered
example
is
CsScCl
3
302
of
a
13 . 1
Collectively,
Ac)
are
comprise
These
the
referred
the
in
as
f-block
elements
f-block
elements
to
the
are
the
in
are
those
formal
periodic
groups
d-block
in
3–12
which
members
table
inclusive
elements.
(gure
of
the
The
4f
group
3
F I r s t- r o w
(including
elements
and
but
5f
La
e l e m e n t s
and
which
orbitals
they
d - b l o C k
form
are
a
lled.
separate
2).
1
18
1
1
2
H
He
2
13
14
15
16
1
7
2
3
4
3
4
5
6
7
8
9
2
1
22
23
24
25
26
27
10
28
11
29
12
30
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
39
40
41
42
43
44
45
46
47
48
s-block
p-block
elements
5
elements
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
57
72
73
74
75
76
77
78
79
Cd
80
6
La*
Hf
T
a
W
Re
Os
Ir
Pt
Au
Hg
89
104
105
106
107
108
109
110
111
112
7
Ac*
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
transition elements
d-block elements
58
59
60
6
1
62
63
64
65
66
67
68
69
70
7
1
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
lanthanoids
(*also includes La)
actinoids
(*also includes Ac)
f-block elements (sometimes called the inner transition elements)
main-group elements
group 1 (excluding H), group 2 and groups 13–18
transition elements
groups 3–11 (the f-block elements are sometimes described as the
inner transition elements)
s-block elements
groups 1 and 2 and He
p-block elements
groups 13–18 (excluding He)
d-block elements
groups 3–12 [including Z = 57 (La) and Z = 89 (Ac), but excluding
Z = 58 (Ce) to Z = 71 (Lu) and Z = 90 (Th) to Z = 103 (Lr), which are
classied as f-block elements]
f-block elements
elements from Z = 58 (Ce) to Z = 71 (Lu) and from Z = 90 (Th) to Z =
103 (Lr)
lanthanoids
elements from Z = 57 (La) to Z = 71 (Lu)
actinoids
▲
elements from Z = 89 (Ac) to Z = 103 (Lr)
Figure 2 Periodic table of the elements showing the main-group elements, the transition elements, the s-, p-, d-,
and f-block elements, the lanthanoids and the actinoids
The
metallic
described
as
nature
the
of
the
transition
transition
elements
means
they
are
often
metals.
303
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
Electron congurations of rst-row d-block elements
suy ip
and their ions
You need to be able to write the
The
electron congurations (both
full and condensed) and the
orbital diagrams for atoms and
following
are
some
congurations
of
the
corresponding
orbital
examples
rst-row
of
full
d-block
and
condensed
elements,
their
electron
ions,
and
their
diagrams:
ions up to Z = 36, that is to Kr.
●
For
vanadium,
2
1s
2
2s
6
2p
3s
3
[Ar]3d
The
2
V
(Z
6
3p
=
3
3d
23):
2
4s
(full
electron
conguration)
2
4s
orbital
(condensed
diagram
electron
conguration)
is:
[Ar]
2
3
4s
In
the
rst
3d
orbital
before
diagram,
lling
multiplicity,
●
For
2
1s
Hu’ u f
aiu uipiciy
see
nickel,
2
2s
Ni
6
2p
8
[Ar]3d
in
three
pairs,
sub-topic
(Z
2
3s
the
them
=
6
3p
ll
the
Hund’s
3d
orbitals
rule
of
singly
maximum
2.2.
28):
8
3d
d-electrons
following
2
4s
(full
electron
conguration)
2
4s
(condensed
electron
conguration)
Hu’ u f aiu
The
orbital
diagram
is:
uipiciy states that when
lling degenerate orbitals
(that is, of the same energy)
[Ar]
electrons ll the orbitals singly
before lling them in pairs.
2
8
4s
3d
2+
●
For
Ni
2
1s
(Z
2
2s
6
2p
=
28):
2
3s
6
3p
8
3d
(full
electron
conguration)
suy ip
8
[Ar]3d
(condensed
electron
conguration)
When removing electrons to
The
orbital
diagram
is:
form cations (positive ions)
electrons are always removed
from the level of highest
[Ar]
principal quantum number, n.
In the case of the rst-row
0
8
4s
3d
d-block elements this will be
the 4s level.
2+
In
the
the
3d
case
of
Ni
the
electrons
are
removed
from
the
4s
level
before
level.
Transition metals and the Auf bau principle
In
an
and
article
D.
Chemical
relative
304
by
L.G.
Devoghel
Education
energies
Vanquickenborne,
published
(71,
of
the
in
the
(1994),
3d
and
K.
Journal
Pierloot,
of
p469-471),
4s
orbitals
the
are
discussed
respect
to
and
the
ground-state
is,
why
an
explanation
lling
and
electrons
in
of
these
is
transition
are
given
orbitals
removed
metal
from
with
in
both
ions
the
the
(that
4s
level
13 . 1
before
this
is
the
3d
for
beyond
article
is
regards
to
rst-row).
current
interesting
misconceptions
with
the
the
to
that
IB
read
in
appear
what
scope
some
to
An
of
but
relation
in
orbital
The
syllabus,
more
some
need
an
to
earlier
based
the
sources
energies
F I r s t- r o w
on
e l e m e n t s
explanation
proposed
electron–electron
schematic
introductory
electron
be
d - b l o C k
in
nature
level
in
by
R.L.
Rich,
interactions,
and
can
be
was
useful
understanding
at
d-block
congurations.
considered.
An
orbital
However,
is
assumed
when
two
to
have
one
electrons
energy
occupy
an
level.
orbital,
Electron congurations involving
because
exceptions
be
In
the
rst-row
in
transition
terms
of
metals,
electron
there
are
have
to
be
careful
with:
Cr
(Z
=
24)
=
29).
In
the
case
of
Cr,
you
may
be
and
write
nuclear
is
the
electron
incorrect.
conguration
Chromium
has
a
as
conguration
of
[Ar]3d
charge
of
the
occurs
for
Cu.
One
(Z)
(both
additional
pairing
increases,
electrons:
from
[Ar]3d
orbitals.
2
4s
As
a
the
result
are
factor
energy,
P.
doing
1
4s
.
A
would
there
is
greater
d
orbitals
nucleus
to
the
are
not
same
shielded
extent
electrons
will
occupy
the
as
orbitals,
previously
which
in
is
earlier
what
we
topics
in
have
been
writing
For
example,
vanadium
(Z
electron
=
23)
to
be
[Ar]3d
10
electron
conguration
is
[Ar]3d
a
2
4s
.
The
attempts
are
often
made
to
already
.
At
a
terms
of
the
extra
stability
of
the
fully-lled
(d
)
d-sublevel.
half-lled
(d
is
far
too
simplistic
explanation
(in
the
However,
and
a
much
previously
stated,
cited
point
after
(note
that
Chem
Educ.,
paper
and
references
the
effect
of
of
increasing
the
4s
and
3d
nuclear
levels
lines
electrons
the
explanation
that
occupy
energies
of
involves
the
all
nding
electrons
with
Rich’s
work,
vanadium,
crossover
that
on
in
line
A
strong
correlation
experimental
data
and
on
advanced
same
the
as
energy
is
a
crossover
to
onto
chromium
copper,
point).
This
process
there
produces
the
subshell.
energies
As
Rich
of
the
points
individual
out,
is
followed
until
the
subshell
is
the
half-lled;
the
upper
the
line
outer
is
also
used.
has
electrons
are
Hence,
simply
for
given
orbital.
sum
their
lowest
energies
available.
As
can
be
interpreted
of
from
the
lled
(d
diagram,
this
does
not
lead
exactly
to
half-
respective
been
theoretical
computational
on
leading
represent
each
element
10
)
or
fully-lled
(d
0
)
(or
empty
d
)
subshells.
found
these
often
occur
because
some
additional
data
energy
based
the
the
However,
between
from
there
leading
nickel,
5
interactions.
seen
interactions
the
This
be
relates
every
between
can
more
therein)
charge
and
from
a
thereafter,
energies
as
mentioned
lower
to
but
after
another
electrons
J
2
4s
this
two
detailed
[Ar]3d
)
is
approach
of
this
10
and
conguration
simplistic
rationalize
5
in
electron
correct
1
4s
condensed
levels
level
has
the
3
conguration
s
lowest
similar
expect
9
electron
to
As
.
congurations.
anomaly
an
condensed
5
electron
the
repulsion
tempted
available
This
termed
is
Cu
4
to
there
that
(screened)
(Z
electrostatic
two
congurations
attraction
you
their
charged),
considered,
the
exceptions
of
negatively
is
required
to
go
beyond
them.
calculations.
1
+
2
3d
number of electrons
1
2
in half-subshell
electron
1
1
+
1
spin
2
+
+
Sc
2
2
+
Ti
+
V
+
Cr
E
1
+
Mn
+
Fe
3
1
+
Co
+
Ni
2
4s
1
1
+
Cu
+
Zn
1
2
1
2
1
5
1
1
1
1
1
1
5
1
1
1
1
5
1
Sc
2
3
5
1
4s
1
1
4
Ti
3
1
V
Cr
5
5
Mn
1
4
1
Fe
1
Co
5
5
5
5
Ni
1
+
5
5
Cu
Zn
5
5
5
(a)
1
1
1
(b)
2
3d
5
1
5
▲
2
Figure 3 Schematic representation of Rich’s interpretation of electron congurations for transition elements in terms of intra-orbital
repulsion and trends in subshell energies. In the rst diagram, the order in which the levels are occupied is presented. In both
diagrams key crossover points feature and in the second diagram one sees how an electron is removed from the 4s level before
one from the 3d level
305
13
T H E
In
on
this
approach,
the
lled
P E R I O D I C
model
and
of
TA B L E
the
a
invalid.
T H E
commonly
perceived
fully-lled
somewhat
–
held
extra
d-sublevels
In
the
case
T R A N S I T I O N
rationale
stability
could
of
the
be
of
based
the
half-
considered
formation
M E TA L S
of
A
( A H L )
more
found
detailed
in
Inorganic
G.L.
account
Miessler,
Chemistry
(5th
of
P
.J.
this
discussion
Fischer
Edition),
and
2013,
can
D.A.
be
Tarr.,
Prentice
Hall.
n+
the
transition
removed,
the
metal
cation,
overall
M
electron
,
when
electrons
repulsion
is
are
Reproduced with permission from R.L. Rich
decreased
(Periodic Correlations, W.A. Benjamin, Inc., 1965).
and
the
greater
energy
extent
of
the
d
orbitals
compared
to
that
is
lowered
of
the
s
to
a
orbitals.
Aciviy
suy ip
On closer examination of the
For the IB Chemistry Diploma programme, you are only required to know
electron congurations of the entire
anomalous electron congurations of the elements chromium and copper from
periodic table, other elements also
the rst-row transition metals. These congurations are: Cr, [Ar]3d
5
10
1
4s
, and Cu,
1
4s
. All other rst-row transition metals will have electron congurations
convey deviations from expected
[Ar]3d
patterns. Look at the webelements
as predicted based on their position in the periodic table.
website and try to nd four other
Note the following point, however. Once deduced, do not be tempted to modify electron
d-block elements with electron
5
congurations of cations further to follow this 3d
congurations that dier from what
6
has an electron conguration of [Ar]3d 4s
is expected.
10
and 3d
pattern. For example, Fe
2
2+
. However, the electron conguration of Fe
6
is [Ar]3d
5
[Ar]3d
. Do not then be inclined to rearrange this electron conguration further to
1
4s
2+
. This is an incorrect electron conguration for Fe
. In summary, just note the
two exceptions of Cr and Cu for the IB Diploma Chemistry programme.
Quic qui
1
Deduce the full electron
tok
congurations of:
3+
a)
Co;
)
Zn;
c)
Ti
The medical symbols for the female and male genders originate from the
symbols used for copper and iron by the alchemists. These symbols have
and explain why Zn is not
been used since Renaissance times (see the Royal Society of Chemistry (RSC)
described as a transition
Visual Elements Periodic Table, www.rsc.org/periodic-table/alchemy).
element, according to IUPAC
recommendations.
2
Deduce the condensed electron
congurations of:
2+
a)
3
V;
)
Mn;
c)
Mn
Deduce the orbital diagrams of:
3+
a)
Co
3+
;
)
Cr
+
;
c)
Cu
Mars symbol – symbolizes a male
Alchemist’s symbol for iron
organism
Venus symbol – symbolizes a female
organism
306
Alchemist’s symbol for copper
13 . 1
F I r s t- r o w
d - b l o C k
e l e m e n t s
r  by
Iron and copper are two of the seven metals of alchemy (gold, silver, mercury,
copper, lead, iron and tin). Alchemists are often considered as the rst chemists.
Rober t Boyle, who was born in Lismore
Alchemists developed a unique language to describe not only chemical
in Ireland in 1627, is often described
reactions, but also philosophical doctrines. Some commentators claim that the
as “ The Father of Chemistry”
pseudoscience of alchemy has played a key role in the development of modern
(see www.rober tboyle.ie/).
medicine and chemistry. Alchemists made a signicant contribution to the
chemical industries of that period, in areas such as the metallurgical industry, the
dye industry and the glass-manufacturing industry. Alchemists extracted metals
from their ores and tried to arrange the information known at that time of the
various substances. The original idea of a periodic table might therefore in par t be
attributed to the alchemists. During the early days of alchemy the astronomical
signs of the planets were used as alchemical symbols. Alchemical symbols were
used to represent some elements and their compounds until the 18th century.
Boyle was not only a devotee of
“natural philosophy”, an advocate of the
Characteristics of transition elements
As
mentioned
the
of
a
nuclear
these
increase
lower
In
in
electrons
the
the
the
can
case
rst
shown
to
be
enter
electrons
greater
going
the
of
IE
rst
and
across
of
the
the
the
valence
to
right
atomic
energy
elements,
period,
the
fact
the
orbital,
shell
that
effect
than
radii
( IE)
rate
of
experimental sciences, but also a key
the
will
there
valence
is
increase
a
is
a
table,
result
across
gradual
elements.
electrons.
This
the
elements,
electrons
This
have
founder of the Royal Society in England.
Boyle proved the inverse relationship
between the volume of a gas and its
pressure, known as by’ a (sub-
topic 1.3). Although Boyle made the
much
elements,
main-group
Inner-shell
As
increase
transition
for
periodic
decrease.
main-group
for
whereas
orbital.
across
although
corresponding
to
(screening)
left
ionization
transition
that
a
from
increases
inner-shell
enter
gure
3,
attributed
an
shielding
in
Z,
factors,
compared
difference
topic
charge,
two
period.
in
a
trend
is
4.
transition to modern science, much of his
thinking centred around what is termed
“chaici”, which has its basis as
the extension of knowledge by reasoning
and inference. All these attributes are still
important to the modern day chemical
practitioner. Have any of the principles of
the earlier alchemists also been carried
2500
He
through the ages to modern day scientic
methodology and chemical practice? You
Ne
might wish to reect on the importance of
2000
1
hypothesis and observation in chemical
lom Jk/ygrene noitazinoi tsr
experiments you carry out in the laboratory.
Ar
1500
Kr
tok
Xe
Rn
Hg
Zn
1000
Rober t Boyle was a scientist and a
Cd
philosopher. In many countries such
as France the study of philosophy is
Al
Ga
500
Li
mandatory at high school. In France
Ti
In
Ni
the philosophy curriculum aims
K
Rh
Cs
Fr
at producing enlightened citizens
capable of intelligent criticism.
10
20
30
40
50
60
70
80
90
100
Find out what other countries prescribe
the teaching of philosophy as mandatory
atomic number
at school and discuss the role and
▲
Figure 4 Trends in the rst IE for main-group and transition elements. Notice that the rate of
importance of taking a philosophical
increase in the rst IE across the period is much more gradual for the transition elements
compared to that for the main-group elements
view in scientic discourse.
307
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
Transition
M E TA L S
metals
have
a
( A H L )
number
of
key
characteristics:
laay ip
•
●
they
●
compounds
●
transition
metals
form
●
transition
metals
are
●
magnetic
have
variable
oxidation
states
When reading the meniscus for
potassium permanganate, the top of
of
transition
elements
and
their
ions
are
often
coloured
the meniscus should always be read
complexes
with
ligands
(B in gure 6), because it is convex
often
used
as
catalysts
upwards. This is because the deep
colour makes it very dicult to read the
meniscus. This is in contrast to normal
states
properties
and
of
transition
coordination
metals
depend
sodium,
where
on
their
oxidation
number.
practice for most clear solutions (A)
where the meniscus is read from the
Variable oxidation states
bottom, that is concave upwards.
In
▲
contrast
to
an
alkali
metal
such
as
the
oxidation
state
Figure 6 How to read a meniscus: A, for
is
always
+1
in
its
ion
and
compounds,
transition
metals
are
often
found
clear solutions; B, for KMnO
4
with
the
different
rst-row
oxidation
d-block
states.
elements
The
(see
range
of
sections
different
9
and
oxidation
14
of
the
states
Data
for
booklet)
A
can
be
B
be
seen
split
from
the
according
sc
to
ti
diagram
their
shown
oxidation
in
gure
states
into
5.
The
three
d-block
types
C
–
elements
A,
B,
V
C
m
F
ni
+1
+1
+1
+1
+1
+1
+1
and
Cu
+2
+2
+2
+2
+2
+2
+2
+2
+2
+3
+3
+3
+3
+3
+3
+3
+3
+3
+4
+4
+4
+4
+4
+4
+4
+5
+5
+5
+6
+6
can
C.
Z
+2
+6
+7
•
In carrying out a redox titration
type
involving potassium permanganate
Sc,
the colour change is typically
from purple to colourless (with a
▲
faint pink tinge, signifying the +7
A:
Ti,
type
and
V
Cr
B:
and
type
Mn
Fe,
C:
Co,
Ni,
Cu,
and
Figure 5 Range of oxidation states of the rst-row d-block metals. The most common
oxidation states are marked in green
to +2 oxidation state change for
manganese.). If, however, the colour
The
characteristics
of
Type
A
are
dominated
by:
changes from purple to brown, this
●
stable
high
oxidation
states
(for
example,
V
is
+5
in
VO
)
3
would signify the formation of the
4+
intermediate ion of manganese, Mn
●
,
unstable
low
oxidation
states.
with an associated oxidation state of
The
characteristics
of
Type
B
are
dominated
by:
+4, which is also a stable oxidation
●
stable
high
oxidation
state. This may occur if there is
insucient acid in the conical ask .
308
states
(for
example,
Mn
is
+7
in
MnO
,
4
2
is
+6
in
Cr
O
2
)
7
Cr
Zn
13 . 1
F I r s t- r o w
d - b l o C k
e l e m e n t s
2+
●
stable
low
oxidation
states
(for
example,
Mn
is
+2
in
[Mn(H
O)
2
]
,
6
3+
Cr
is
+3
in
[Cr(H
O)
2
The
characteristics
●
unstable
●
stable
high
of
]
).
6
Type
C
oxidation
are
dominated
by:
states
2+
low
oxidation
states
(for
example,
Fe
is
+2
in
[Fe(H
O)
2
Manganese
+7.
In
the
is
characterized
chemical
permanganate,
in
manganate(VII)
practice
deep
in
oxidation
+2,
the
state
which
is
redox
though
+7
almost
is
you
In
reduced
often
(more
compound
colour.
states
may
workplace!)
(purple)
of
oxidation
titrations
this
chemical
burgundy
by
laboratory
This
named
named
reagent
is
reactions,
manganese
from
reagent
+1
potassium
this
way
in
characterized
an
to
potassium
manganese
with
).
by
with
oxidation
a
an
state
of
colourless:
7+
2+
Mn
oxidation
rarely
range
the
correctly
is
redox
to
that
use
]
6
state:
+
5e
→
+7
Mn
+2
2+
species:
[MnO
]
[Mn(H
4
Another
states.
transition
In
yellow
its
metal,
highest
compounds,
chromium
in
a
+3
chromium,
oxidation
which
state,
can
oxidation
be
can
+6,
oxidation
The
state:
oxidation
alcohol
is
of
rst
oxidized
also
exist
to
green
in
various
forms
oxidation
orange
complexes
and
with
state.
3+
+
3e
→
+6
primary
]
6
chromium
reduced
6+
Cr
O)
2
Cr
+3
alcohols
into
an
is
a
two-step
aldehyde,
process.
which
in
turn
A
is
primary
oxidized
laay ip
further
into
the
corresponding
carboxylic
acid.
•
Primary
alcohols
can
be
oxidized
by
strong
oxidizing
agents
In the oxidation of a primary
such
alcohol, the aldehyde can
as
potassium
dichromate(VI),
K
Cr
2
O
2
,
in
sulfuric
acid,
H
7
SO
2
,
to
4
be isolated by iiig it o
form
the
corresponding
carboxylic
acid,
under
reux
(as
discussed
in
as it forms. Distillation is a
sub-topic
10.2):
technique used to separate
K
Cr
2
CH
CH
3
O
2
K
7
Cr
2
OH
liquids that have dierent
O
2
7
CH
2
COOH
3
3
+
+
H
boiling points (boiling point
H
of ethanal is 20.2°C; that of
ethanol
ethanal
ethanoic acid
(primary alcohol)
(aldehyde)
(carboxylic acid)
ethanoic acid is 118°C).
•
Alternatively, if a milder
Oxidation of a primary alcohol
oxidizing agent is used, such
as pyridinium chlorochromate
Secondary
alcohols
can
also
be
oxidized
by
potassium
dichromate(VI)
in
(PCC), with an organic solvent
sulfuric
acid
to
form
the
corresponding
ketone:
such as tetrahydrofuran (THF),
K
Cr
2
CH
CH
3
CH(CH
2
O
2
the aldehyde forms as the nal
7
)OH
CH
3
CH
3
C(O)CH
2
3
+
product of the reaction.
H
PCC
butan-2-ol
butan-2-one
CH
3
(secondary alcohol)
As
outlined
oxidation
in
and
+6
to
OH
topic
9,
each
reduction.
the
In
redox
the
chromium
process
case
is
of
CH
CHO
3
THF
involves
this
reduced
→
2
(ketone)
dichromate(VI),
of
CH
two
reaction
from
an
ethanol
ethanal
(primary alcohol)
(aldehyde)
half-reactions,
with
potassium
oxidation
state
+3.
309
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
Oxidation
M E TA L S
( A H L )
half-reaction:
+
CH
CH
3
OH(aq)
+
H
2
Reduction
O
2
→
CH
COOH(aq)
4H
+
(aq)
+
14H
(aq)
+
4e
3+
(aq)
+
6e
→
2Cr
(aq)
+
7H
7
Overall
O(l)
2
equation:
2
3CH
+
3
half-reaction:
2
Cr
O(l)
2
CH
3
OH(aq)
+
2Cr
2
O
2
+
(aq)
+
16H
(aq)
→
3CH
7
COOH(aq)
+
3
3+
4Cr
(aq)
+
11H
O(l)
2
bahay 
The redox reaction involving potassium dichromate(VI)
is the basis of the ahay  used by police
forces worldwide to determine if a driver of a vehicle has
consumed alcohol. In this test, crystals of potassium
dichromate(VI), which are orange/yellow in colour, change
3+
to green, which signies the formation of the Cr
species.
Since 2012, it has been required by French law that all
vehicles need to be equipped with a breathalyser. As seen
in gure 7, the simple version of this on-board vehicle
breathalyser test kit involves the colour change from
orange/yellow to green. This type of breathalyser does not
record the  ach ccai (BAC), which is
the concentration of ethanol in a person’s blood. BAC is the
3
mass, in milligrams, of ethanol per 100 cm
of blood.
•
In order to measure the BAC three devices can be used:
•
semiconductor oxide-based sensor
•
fuel-cell sensor
•
intoximeter, which is an IR spectrometer; this type
If the crystals are all yellow/orange, the result is
zero – you are clear to go!
•
If the crystals are green below the line, according
to the tube, you are under the maximum limit.
But you do have alcohol in your blood and your
judgement and reaction times will almost cer tainly
of technology is often used in large, table-top
be aected – you should consider waiting a while
breathalysers found at police stations.
and retesting
sicuc i 
•
These are relatively new to the market and have a
If the crystals are green beyond the line – you are
denitely over the limit. do not drIVe!
number of advantages, such as their low cost, low power
▲
Figure 7 Example of a simple breathalyser test kit used in
consumption, and por tability. The disadvantage of this
France. Notice the orange/yellow to green colour change,
type of breathalyser is that their sensors need to be
6+
which signies the Cr
3+
+ 3e
→ Cr
reduction caused
calibrated more frequently than fuel-cell based testers.
by ethanol
Incorrect calibration can result in systematic errors.
suppor t a legal case in a cour t of justice. For this reason,
Fu-c 
positive tests obtained by preliminary screening need to be
Another type of breathalyser is based on the fu c
conrmed by more advanced analytical techniques, such
Ethanol is oxidized initially into ethanoic acid and then
as gas liquid chromatography(GLC), in which a sample
into carbon dioxide and water. The fuel cell conver ts
is sent to a forensic science laboratory and the exact
chemical energy generated from the oxidation process
concentration of ethanol in the blood is determined. GLC is
into electrical energy. The electric potential is used to
used to analyse volatile substances.
determine the concentration of ethanol.
Ii
This type of fuel cell can also be quite basic and the results
typically determined may not be suciently accurate to
310
A third type of breathalyser is the intoximeter based on IR
spectroscopy. This is discussed in detail in option D.
13 . 1
F I r s t- r o w
d - b l o C k
e l e m e n t s
Explanation of the ability of transition metals to form variable
oxidation states from successive ionization energies
As
stated
they
metals,
an
already
exhibit
which
alkaline
have
earth
compounds.
In
states
+3,
of
types
+2,
of
one
variable
metal
of
the
only
metal
and
one
and
contrast,
is
key
oxidation
xed
occurs
the
+4.
related
This
the
is
reason
a
for
this
in
transition
stark
state.
+2
metal
patterns
of
in
oxidation
with
transition
The
to
characteristics
states.
For
titanium
to
state
occurs
difference
in
that
s-block
calcium
its
ion
with
between
successive
is
the
example,
oxidation
the
metals
contrast
is
and
oxidation
the
ionization
two
energies.
Coloured compounds of transition metals
and their ions
Transition
metal
compounds
and
KMnO
ions
are
often
burgundy
coloured,
for
example:
(purple)
4
2+
[Mn(H
O)
2
K
Cr
2
]
almost
O
2
colourless,
with
a
faint
pink
tinge
6
orange
7
3+
[Cr(H
O)
2
]
CuSO
5H
4
[NH
]
4
green
6
O
blue
2
[Fe(H
2
O)
2
Crystalline
][SO
6
]
4
hydrated
pale
green
2
copper(II)
sulfate,
CuSO
·5H
4
blue
in
colour
(gure
ofcrystallization,
8).
and
the
Upon
heating
solid,
the
anhydrous
O,
is
Mediterranean
2
compound
CuSO
loses
forms,
its
water
which
is
a
4
white
powder.
O
O
O
O
O
O
Cia ig
S
H
S
H
In coordinate bonding the
O
pair of electrons comes from
O
O
OH
H
H
2
O
O
the same atom, unlike typical
OH
2
2
2
covalent bonding where
Cu
Cu
the shared pair consists of
electrons that originate from
OH
O
H
2
H
2
O
OH
O
2
O
2
both atoms, A and B, which
O
form the covalent bond, as
H
H
S
S
O
O
discussed in topic 4. The older
O
O
name for coordinate bonding
O
O
was dative covalent bonding.
2+
▲
Figure 8 Structure of CuSO
·5H
4
O. Note the presence of hydrogen bonding and that Cu
The use of this older name is
2
has an octahedral stereochemistry, which may not be obvious from the formula
no longer recommended by
IUPAC.
As
stated
previously,
zinc
is
not
classied
as
a
transition
element,
as
it
IUPAC recommends the term
10
has
a
complete
d-sublevel,
[Ar]3d
2
4s
2+
.
Its
ion,
Zn
,
has
the
electron
coordination bonding but in
10
conguration,
[Ar]3d
this text we will use the more
Compounds
(explained
atoms
which
of
zinc(II)
below)
responsible
can
absorb
in
for
in
are
the
the
the
usually
complex
colourless,
have
absorption
visible
of
region
a
unless
the
chromophore
electromagnetic
of
the
widely used term coordinate
ligands
(group
of
radiation),
electromagnetic
bonding as applied in the IB
Chemistry guide.
spectrum.
311
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
Complexes of transition metals
Compounds
that
contain
transition
elements
and
in
which
the
central
n+
metal
ion,
M
molecules
or
complexes.
,
is
bonded,
via
ions
(termed
the
Such
compounds,
compounds
to
signify
the
coordinate
ligands)
are
often
bonding,
are
termed
described
coordinate
bonding
as
to
a
group
transition
of
metal
coordination
present
between
the
dcipi f a iga
ligand(s)
and
the
central
metal
ion.
A ligand is an atom, molecule,
Examples
of
species
with
coordinate
bonding:
or ion that contains a lone pair
of electrons (non-bonding
+
●
hydronium
cation,
[H
O]
●
carbon
●
a
monoxide,
CO
3
pair) that coordinates, through
cia ig, to a
+
central transition metal ion to
O
H
H
form a cp.
H
The term piga is often
+
used to describe a species
●
ammonium
cation,
[NH
]
transition
metal
complex,
for
4
2+
that has the ability to act as a
example
[Ni(NH
)
3
]
6
ligand in a complex, but is not
+
+
2
O,
yet coordinated. Hence, H
2
H
NH
3
because of its two lone pairs, is
H
NH
N
3
3
a neutral proligand, but in the
N
3+
complex [Cr(H
O)
2
]
Ni
H
, water
H
6
acts as a ligand.
H
H
N
NH
3
3
NH
3
)
a)
Bonding models of transition
+
1
+
3
OH
2
+
1
+
1
OH
2
H
metal complexes
H
O
electroneutrality
OH
principle
is
2
Fe
OH
2
Pauling’s
O
2
3
2
an
Fe
H
OH
O
2
2
approximate
method
of
estimating
how
H
charge
O
+
1
OH
2
2
+
1
OH
is
distributed
of
in
The
basis
any
individual
a
this
molecule
principle
atom
in
or
is
the
complex
that
the
2
ion.
charge
molecule
or
+
1
OH
2
on
ion
is
c)
)
1
OH
+
2
restricted
to
a
range
between
1
to
1+
and
2
ideally
H
the
charge
should
be
close
to
2
zero.
2
2
2
H
3+
9
shows
various
representations
of
O
OH
2
Fe
Figure
2
Fe
the
H
+
2
+
OH
1
OH
1
O
0
O
2
OH
3+
cationic
complex,
[Fe(H
O)
2
]
2
.
6
O
H
1
OH
2
2
1
+
OH
+
OH
2
2
2
2
1
+
●
In
gure
cationic
9(a)
a
typical
complex
is
representation
given.
As
the
of
lone
the
pair
2
of
▲
electrons
on
each
water
ligand
contributes
to
Figure 9 Various representation and bonding models for
3+
the cationic complex, [Fe(H
O)
2
]
. (a) Conventional
6
3+
the
coordinate
bond,
an
arrow
is
used
instead
representation of the cationic complex [Fe(H
O)
2
of
a
straight
]
. The lone
6
pair on each water ligand forms the coordinate bond with the
line.
3+
central Fe
●
If
we
were
to
adopt
the
model
proposed
ion. Square brackets here represent the complex,
in
which has an octahedral stereochemistry. The overall charge
gure
9(b),
it
would
mean
a
net
transfer
of
on the complex is 3+. (b) Charge distribution in the cationic
charge
from
each
water
ligand
to
the
metal
3+
complex [Fe(H
O)
2
centre.
The
charge
distribution
that
results
]
based on a 100% covalent bonding
6
3+
model. (c) Charge distribution in [Fe(H
O)
2
from
this
100%
covalent
bonding
model
]
based on a
6
100% ionic bonding model. (d) Approximate charge distribution
3+
would
confer
3
on
Fe
and
1+
on
each
water.
in [Fe(H
O)
2
312
]
6
based on Pauling’s electroneutrality principle
13 . 1
(b)
is
not
charges
a
valid
residing
model,
on
however,
metals
is
as
F I r s t- r o w
a
negative
target
In
gure
9(c),
a
100%
ionic
bonding
protein
shown.
ion
and
The
the
neutral.
3+
charge
water
This
resides
molecules
theoretical
on
stay
model
the
results
have
the
body,
quantum
performed
which
interact
on
with
the
the
mechanical
atoms
target
in
drug
classical
mechanics
was
used
to
simulate
iron
the
remainder
the
press
of
the
protein.
As
outlined
in
effectively
is
also
invalid,
shown
the
release
given
by
the
The
Royal
Swedish
as
Academy
experimental
in
were
e l e m e n t s
model
and
is
protein
calculations
atypical.
the
●
d - b l o C k
of
Sciences
on
the
2013
prize
they
remark
existence
that
“Todaythe
computer
is
just
as
important
a
3+
of
the
[Fe(H
O)
2
]
species
in
aqueous
solution;
6
tool
that
is,
it
remains
a
single
unit
in
for
chemists
realistic
In
●
gure
9(d),
principle
is
distribution
iron,
the
are
total
Pauling’s
applied
and
means
central
electroneutrality
the
that
metal,
approximate
now
the
should
as
the
test
tube .
Simulations
are
so
solution.
net
be
predict
the
outcome
of
traditional
experiments”
charge
charge
zero.
thatthey
As
Activity
on
there
Potassium
permanganate,
KMnO
,
is
frequently
4
a
of
six
water
ligands
in
the
compound,
used
as
an
oxidizing
agent.
The
manganate(VII)
3+
the
Fe
cation
needs,
effectively,
three
electrons
ion
has
the
formula
[MnO
]
.
4
to
confer
on
it
a
net
zero
charge.
The
charge
(i)
distribution,
then,
on
each
water
ligand
will
Comment,
VSEPR
1
three
electrons/six
ligands
=
+.
Hence,
giving
a
reason,
whether
or
not
deduce
the
be
theory
could
be
used
to
in
2
geometry
3+
this
model,
coordinate
bonds
in
[Fe(H
O)
2
be
50%
covalent
and
50%
is
a
good
to
try
example
to
of
evaluating
understand
the
structure
of
If
the
bonding
terms
of
a
Prize
in
Chemistry
2013
jointly
to
Martin
Karplus
in
this
100%
what
France
and
the
and
Harvard
Michael
of
Levitt
Medicine,
this
(Stanford
USA)
and
of
Southern
charge
for
the
The
California,
Los
development
chemical
used
model
American
of
multi-scale
systems.
Chemists
models
ranging
sophisticated
further
the
for
may
from
known
spheres
and
computational
structures
proteins.
was
of
remarkable
the
their
of
understand
Their
2013
about
Noble
models
both
research
the
Prize
combined
traditional
quantum
chemical
of
invalid
properties
two
simulating
For
how
of
the
the
Linus
Pauling,
is
well
development
but
of
Pauling
the
is
scale
of
perhaps
for
his
electroneutrality
electroneutrality
principle.
principle
to
the
manganate(VII)
ion,
was
suggest
the
be
approximate
on
each
charge
oxygen
if
distribution
manganese
help
in
a
net
charge
of
1 +.
Determine
on
What
this
basis
and
the
the
percentage
covalent
character
recipients
was
percentage
ionic
character.
that
approaches
mechanics
example
a
for
programmes
Chemistry
the
be
molecules,
processes.
work
classical
mechanics.
of
in
the
explain
and
resulted
chemists
on
ion.
chemist,
his
Pauling’s
might
complexes
be
and
models
what
explore
would
have
applied
to
model,
Angeles,
If
to
considered
Warshel
less
complex
sticks
was
bonding
atoms
electronegativities,
always
anion
ionic
University
Arieh
known
(University
for
the
angles).
University,
(iii)
USA)
identify
bond
oxygen
manganate(VII)
School
and
the
(Université
why
Strasbourg,
USA),
ion,
was
manganese
de
the
idea.
Nobel
awarded
of
(including
a
deduce
The
Draw
various
nature
in
scientic
ion.
]
(ii)
models
manganate(VII)
ionic.
geometry
This
the
6
the
would
of
drug
in
and
their
interacted
with
Classication of ligands
The
ion
number
depends
pairs)
in
the
coordinate
known
as
of
on
coordinate
the
ligand.
bond
of
formed
donor
Monodentate
with
chelate
bonds
number
a
metal
ligands)
ion
can
by
ligands
while
form
one
centres
ligand
(atoms
are
able
to
polydentate
two
or
more
with
with
a
metal
lone
form
ligands
such
electron
only
one
(also
bonds.
313
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
Monodentate ligands
O
N
Monodentate
ligands
contain
a
single
donor
atom
and
have
one
lone
H
H
H
H
pair
contributing
to
the
coordinate
bond
in
a
complex.
Typical
examples
H
include
▲
water,
ammonia,
and
the
halides
such
as
Cl
etc.
Figure 10 Notice that the water proligand
contains two lone pairs of electrons, but
Cl
only one contributes to the coordinate
bond in a transition metal complex
Polydentate (chelate) ligands
These
are
ligands
coordinate
A  f caui!
examples
which
bonds
are
with
given
have
a
two
or
transition
more
metal
donor
centre.
atoms
Some
that
form
common
below.
Monodentate does not refer to
the number of lone pairs present
1,2-ethanediamine
(en),
H
NCH
2
CH
2
NH
2
2
in the proligand, but refers to
the number of lone pairs from
the ligand that actually are
H
NH
N
2
2
involved in the formation of the
coordinate bond. Monodentate
M
ligands are often described as
“one-toothed ligands”.
The
structure
bidentate
transition
its
older
N
centre
in
in
it
a
section
has
two
16
of
donor
complex.
en
is
the
Data
atoms
still
booklet.
that
en
is
a
coordinate
sometimes
referred
to
to
the
by
ethylenediamine.
ligands
language,
are
often
meaning
described
crab-claw)
as
as
chelate
the
ligands
ligands
look
like
(coming
from
they
grabbing
are
the
N
the
N
N
metal
between
two
or
more
donor
atoms,
just
like
a
crab
can
grab
N
your
Fe
toes
on
a
beach!
The
complexes
formed
from
chelate
ligands
are
very
Fe
stable.
N
N
given
3
Greek
N
is
because
+
3
N
en
metal
name,
Polydentate
+
of
ligand
N
An
example
of
a
chelate
complex
is
[Fe(en)
N
number
N
]Cl
3
optical
of
the
isomers
iron
is
(two
six
as
each
en
ligand
non-superimposable
is
.
bidentate.
mirror
The
coordination
3
The
images,
complex
gure
has
11).
2
Ethanedioate
▲
Figure 11 Optical isomers of [Fe(en)
(ox),
(C
O
2
]Cl
3
)
4
3
-
-
M
Ethanedioate,
dianionic
often
referred
to
by
its
older
name,
oxalate,
ligand.
4
Ethylenediaminetetraacetate,
(EDTA)
O
O
-
-
O
C
CH
2
O
-
314
C
CH
2
CH
C
O
CH
C
O
2
N
N
2
-
is
a
bidentate,
13 . 1
F I r s t- r o w
d - b l o C k
e l e m e n t s
4
(EDTA)
bonds.
in
an
is
It
a
polydentate
has
the
octahedral
[Co(EDTA)]
,
ability
ligand
to
complex.
the
EDTA
that
wrap
For
acts
can
itself
example,
as
a
form
around
in
up
a
the
hexadentate
to
six
coordinate
transition
anionic
metal
centre
O
complex
O
N
Co
ligand.
O
EDTA
●
is
used
Removal
as
its
with
of
use
heavy
in
other
N
O
in:
metals.
the
The
treatment
metal
ions
ligand
of
lead
present
in
has
a
number
poisoning.
blood.
of
applications,
EDTA
When
Na
can
such
▲
Figure 12 Structure of [Co(EDTA)]
▲
Figure 13 Use of EDTA as a preser vative
coordinate
[Ca(EDTA)]
is
2
administered
to
a
patient,
lead
can
displace
calcium
to
form
the
2
anionic
complex
[Pb(EDTA)]
2
:
2+
[Ca(EDTA)]
+
2
Pb
→
2+
[Pb(EDTA)]
+
Ca
2
Once
●
formed,
Chelation
therapy.
potential
use
considered
of
the
the
lled
from
plaque,
Water
free
The
Food
the
is
loss
perceived
in
by
rancidity
and
chains
double
acids
of
are
of
a
lipids
oxidized
bond
in
to
are
be
Restorative
sculpture.
a
oxygen
have
catalysed
for
sculptured
metal
of
can
pieces.
the
risk
has
of
been
(“hardening
reduces
removes
reduce
cholesterol-
cardiovascular
therapy
softening
(which
added
has
shown
a
to
food
down
or
lipid.
food
that
no
with
to
fats
and
“gone
off”
because
In
across
the
ions.
It
is
of
hydrolytic
the
fatty
fatty
acid
carbon-to-carbon
aldehydes
The
oils.
components,
rancidity,
Volatile
(for
leading
in
their
metal
ensure
products
appearence.
odours.
or
to
precipitate
reactions
occurs
has
into
added
can
reason.
catalyse
oxidative
light
can
urine.
its
bloodstream
chelation
same
can
noxious
by
the
is
therapy
effectively
This
water
taste,
is
EDTA
the
process
EDTA
and
carboxylic
involves
acts
as
a
ions.
EDTA
artwork
coating
In
the
Rancidity
odour,
of
into
disease.
the
when
unsaturated
can
such
ions
broken
of
in
often
colour.
bad
and
the
which
reactions
develop
or
in
and
remain
for
kidneys
atherosclerosis
tissue.
used
is
ions
use
ions
Metal
senses
scavenger
of
also
for
EDTA
heart
shampoos
taste
the
the
for
the
Chelation
reduces
the
propan-1,2,3-triol.
form,
radical
is
Ca-EDTA
development
acids
benet
mayonnaise).
rancidity,
date
magnesium
used
of
calcium
potentially
EDTA
or
preservation.
example
presence
by
application
treatment
free
to
passed
surgery.
atherosclerotic
which
softening.
It
of
be
medical
by-pass
potential
limited
calcium
soaps).
●
a
However,
somewhat
the
as
can
Another
heart
arteries”).
problems.
●
in
concentration
calcium
●
[Pb(EDTA)]
also
Old
be
used
brass
insoluble
solid
or
in
the
copper
complex,
restoration
sculptures
brochantite,
2
CuSO
3Cu(OH)
4
form,
●
which
Cosmetics.
.
Upon
the
addition
of
EDTA,
[Cu(EDTA)]
can
2
is
soluble
EDTA
is
and
easily
sometimes
removed.
used
as
a
preservative
in
cosmetics.
in cosmetics
315
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
Coordination numbers
The
six
majority
of
(octahedral
transition
metal
geometry)
schiy
or
complexes
four
or
coordination
square
planar
numbers
90 (and 180)
eap
6
[Fe(H
O)
2
tetrahedral
109.5
4
square planar
90 (and 180)
4
K
K
]Cl
6
theory
ions.
The
of
X-ray
available.
distances
Many
the
and
transplatin,
of
deduce
such
complexes
the
features,
testicular
no
can
complexes
geometries.
Cisplatin,
cancer,
is
can
if
such
structure,
geometry
be
of
be
all
of
crystals
the
bond
elucidated
4
in
transition
the
planar.
by
the
angles
using
Its
the
metal
metal
structural
complex
and
this
of
bond
technique.
number
treatment
]
4
transition
of
coordination
used
of
the
determined
single
as
square
anticancer
the
d-sublevels
crystallography
in
shows
to
incomplete
platinum(II)
planar
bladder
used
the
Structural
present
of
square
be
of
structures
technique
are
cannot
because
]
[Ni(CN)
2
complexes
2
[CoCl
2
VSEPR
of
geometries).
Ciai u
b ag/°
octahedral
have
(tetrahedral
four
have
ovarian,
geometrical
isomer,
activity.
Transition metals as catalysts
Transition
are
some
sections
●
metals
examples
of
Haber
N
(g)
the
often
of
reactions
+
3H
catalyst:
(g)
⇋
2NH
catalysts
that
you
in
may
chemical
be
reactions.
familiar
with
Here
from
other
(g)
3
Fe(s)
decomposition
of
2H
O(l)
O
as
process:
2
2
used
programme.
2
●
are
(aq)
→
2H
2
hydrogen
+
O
2
catalyst:
MnO
peroxide:
(g)
2
(s)
2
●
hydrogenation
H
C=CH
2
(g)
+
of
H
2
alkenes:
(g)
→
Ni(s),
Pd(s),
hydrogenation
RCH=CHR’
+
H
of
or
Pt(s)
oils
(g)
→
RCH
2
catalyst:
product
a
is
●
oils
has
The
controlled.
Mono-
be
This
R’
2
hydrogenated,
is
greater
texture
The
and
saturated
316
can
liquid.
also
oxidation.
CH
2
Ni(s)
Unsaturated
of
(g)
3
ethane
catalyst:
instead
CH
3
ethene
●
CH
2
main
chemical
(that
is,
its
form
for
stability
fats
of
are
a
semi-solid
cooking
due
hardness
disadvantages
polyunsaturated
fats.
to
advantageous
to
and
a
reduced
plasticity)
hydrogenation
healthier
(or
purposes.
for
solid)
The
rate
of
the
of
product
are:
the
heart
than
13 . 1
●
Trans
fatty
acids
metabolize
tissues
of
the
lipoprotein
which
of
the
can
with
body.
(LDL)
may
formed
Trans
and
in
This
in
partial
therefore
fatty
cholesterol
result
arteries.
be
difculty
acids
be
hydrogenation.
may
increase
(colloquially
cardiovascular
will
the
levels
known
further
as
in
of
of
e l e m e n t s
the
fatty
low-density
“bad
because
in
d - b l o C k
These
accumulate
problems
discussed
F I r s t- r o w
cholesterol”),
the
sub-topic
narrowing
B.3.
Catalytic conver ters in cars
In
a
running
temperature
N
(g)
+
car
O
2
When
engine,
conditions
(g)
→
gaseous
(1500
nitrogen
°C)
to
form
and
oxygen
nitrogen
react
under
high-
monoxide:
2NO(g)
2
NO(g)
is
released
into
the
atmosphere,
it
combines
with
O
(g)
to
2
form
nitrogen
dioxide
NO
(g):
2
2NO(g)
+
O
(g)
→
2NO
2
Nitrogen
the
dioxide
brown
(g)
2
is
colour
a
of
secondary
pollutant
photochemical
that
smog.
is
primarily
Nitrogen
responsible
dioxide
is
toxic
for
and
▲
can
result
in
respiratory
Figure 14 Catalytic conver ter on the underside
problems.
of a car. Three- way catalysts conver t oxides of
Carbon
also
monoxide,
emitted
Most
from
modern
reduce
NO(g)
CO(g),
the
a
highly
exhaust
cars
now
and
NO
are
(g)
of
a
toxic,
car,
equipped
to
N
and
H
2
(g)
odourless,
as
well
with
while
as
and
colourless
unburned
catalytic
oxidizing
is
hydrocarbons.
converters
CO(g)
gas,
and
nitrogen, carbon monoxide and hydrocarbons
into nitrogen, carbon dioxide and water.
However, unleaded fuel has to be used in
that
unburned
vehicles tted with catalytic conver ters. If
2
leaded fuel is used (that is, fuel containing
hydrocarbons
to
CO
(g)
2
O(g),
which
are
less
harmful
substances:
2
added lead compounds used as anti-knocking
2NO(g)
+
2CO(g)
→
N
(g)
+
2CO
2
CH
CH
3
Ethane
In
one
Rh
and
(g)
propane
of
CO(g)
often
so
CuO
Cr
NO(g)
the
of
there
or
in
and
(g)
→
3CO
O
a
,
(g)
+
4H
2
exhausts
catalytic
exhaust
second
which
O(g)
2
can
result
converter
unburned
the
is
2
reducing
5O
agents) the catalyst can be poisoned
2
2
temperature
NO(g)
+
3
chamber
oxidize
the
of
CH
2
(g)
in
ozone
(gure
hydrocarbons.
gases
and
chamber
operates
at
a
14)
of
it
Pt,
a
lower
Pd,
and
increases
additional
contains
much
beads
However,
produces
that
formation.
amounts
different
catalyst,
temperature,
3
to
N
(g).
2
Hgu a
hgu caay
Catalysts in green chemistry
Hgu caay
Catalysts
American
and
play
an
important
Chemical
Society ,
implementation
eliminate
health
the
and
use
the
of
and
role
green
in
green
chemistry
chemical
chemistry.
is
products
generation
of
the
and
According
design,
the
development,
processes
substances
to
to
hazardous
reduce
to
or
A homogeneous catalyst is one that is in
the same phase or physical state as the
substances involved in the reaction that
human
it is catalysing.
environment.
Hgu caay
Biological catalysts
A heterogeneous catalyst is one that is
An
enzyme
is
a
biological
enzyme-catalysed
metals.
One
hemoglobin
oxygen
subunit
in
of
reactions
example
(Hb).
the
catalyst.
is
that
heme
The
hemoglobin
occur
(gure
Hemoglobin
blood.
In
contains
red
an
in
human
cells
15),
(gure
vibrant
the
is
colour
atom
and
which
16)
of
body
the
of
is
involve
the
to
in a dierent phase to the substances
many
centre
that
stems
which
are
involved in the reaction that it is
transition
iron
protein
blood
iron,
there
catalysing. Industrial catalysts that
of
involve transition metals are usually
transports
from
heme.
oxygen
Each
heterogeneous catalysts.
binds.
317
13
T H E
H
P E R I O D I C
TA B L E
C
2
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
CH
3
CH
2
O
H
C
3
N
N
O
rest of porphyrin not shown
N
Fe
N
H
N
Fe
N
N
C
3
N
CH
3
heme
N
protein (globin)
N
O
HO
O
▲
HO
Figure 15 Structure of heme. In heme, iron has
▲
a coordination number of four and a square
Figure 16 Structure of human oxyhemoglobin. Oxygen is carried
through the blood stream by the formation of a weak bond with
2+
planar geometry. The Fe
2+
ion is at the centre
heme. The O
Fe
bond is then broken relatively easily. When
2
2+
of a large nitrogenous heterocyclic ring called
O
bonds to Fe
2+
, O
2
functions as a monodentate ligand and Fe
2
a pphyi. Each Hb molecule contains four
then adopts an octahedral stereochemistry, with a coordination
heme groups. The iron can bind to one O
number of six, as heme is linked to the protein (the globin) via an
2
molecule and therefore a single Hb molecule
can transpor t up to four O
additional Fe
N bond.
molecules
2
Magnetic properties of transition metals
Magnetic
many
number,
contain
by
an
not
properties
factors,
and
external
magnetic
transition
the
geometry
electrons
magnetic
unpaired
elds.
compounds
318
the
unpaired
contain
of
including
are
Para-
of
the
that
eld.
further
complex.
as
contrast,
and
in
their
of
the
complexes
metal,
its
Paramagnetic
tiny
magnets
are
properties
sub-topic
A.2.
are
by
metals,
on
materials
attracted
materials
repelled
of
depend
coordination
and
diamagnetic
therefore
diamagnetic
discussed
and
state
behave
In
electrons
and
metals
oxidation
do
external
ions,
and
13 . 2
C o l o U r e d
C o m P l e x e s
13.2 C u cp 
Understandings
Applications and skills
The d-sublevel splits into two sets of orbitals of
➔
Explanation of the eect of the identity of the
➔
dierent energy in a complex ion.
metal ion, the oxidation state of the metal
Complexes of d-block elements are coloured,
➔
and the identity of the ligand on the colour of
as light is absorbed when an electron is excited
transition metal ion complexes.
between the d orbitals.
Explanation of the eect of dierent ligands on
➔
The colour absorbed is complementary to the
➔
the splitting of the d-orbitals in transition metal
colour obser ved.
complexes and colour obser ved using the
spectrochemical series.
Nature of science
Models and theories – the colour of transition
➔
Transdisciplinary – colour linked to symmetry
➔
metal complexes can be explained through
can be explored in the sciences, architecture,
the use of models and theories based on how
and the ar ts.
electrons are distributed in d-orbitals.
In
an
ion,
isolated
they
atom,
split
into
these
sublevels
light,
which
d
leads
are
orbitals
two
to
have
sublevels.
absorption
responsible
for
the
The
and
the
same
energy
electronic
emission
colour
of
but
in
a
transitions
of
the
complex
between
photons
of
visible
complex.
Theories on complexes
A
number
proposed
in
of
to
different
explain
complexes.
theories
the
These
have
bonding
theories
are
of
been
transition
d-block
listed
metals
below
Valence
bond
developed
which
had
theory
is
Ligand
by
theory
Linus
(VBT).
Pauling
hybridization
rarely
used
as
in
its
VBT
the
is
was
often
1930s,
basis.
not
This
and
today.
eld
Crystal
an
eld
theory
electrostatic
limitations,
order
(this
●
of
be
Molecular
theory,
for
(CFT).
model.
example,
ligands
will
theory
CFT
in
the
is
orbital
it
does
the
ligands
are
(LFT).
cannot
theory
explain
(MOT).
an
a
models.
with
discussed
in
differs
LFT
is
from
an
electrostatic
model.
combination
The
LFT
is
terms
bonding
more
of
CFT
of
the
as
it
LFT
is
CFT
description
detailed
electronic
and
can
energy
the
series
●
Angular
relative
in
In
between
involving
frontier
orbitals*.
its
later).
interactions
on
but
based
have
spectrochemical
considered
covalent
CFT
CFT,
considered
MOT
levels
on
of
based
associated
be
●
and
order:
extension
●
centre
in
●
chronological
metal
considered.
a
overlap
sizes
of
molecular
model.
orbital
orbital
In
this
energies
(MO)
model,
are
the
estimated
calculation.
this
the
319
13
T H E
●
P E R I O D I C
These
theories
many
of
the
complexes,
and
TA B L E
and
–
models
characteristics
such
magnetic
as
colour,
properties.
T H E
help
of
T R A N S I T I O N
us
explain
transition
electronic
The
M E TA L S
( A H L )
details
metal
of
spectra,
the
this
comprehensive
of
IB
these
book,
explain
models
Chemistry
we
the
shall
colour
are
beyond
Diploma
use
of
only
the
scope
programme.
the
transition
CFT
In
model
metal
to
complexes.
*As outlined in the IUPAC Gold Book (http://goldbook.iupac.org/), fi ia refer to the highest-energy occupied molecular
orbital (HOMO) (lled or par tly lled) and the lowest-energy unoccupied molecular orbital (LUMO) (completely or par tly vacant) of
a molecular entity. The IUPAC Gold Book is an invaluable source for chemists.
Crystal eld theory (CFT)
The
d-sublevel
and
d
consists
of
ve
d-orbitals
(gure
1)
d
,
d
xy
,
d
yz
,
d
xz
,
2
x
2
y
2
z
As
can
be
electron
seen
from
density
gure
pointing
1
at
three
45 °
to
of
these
the
orbitals
Cartesian
have
axes
their
(d
,
d
xy
contrast,
electron
the
remaining
density
two
pointing
orbitals
along
the
(
d
,
2
and
2
x
d
y
2
z
Cartesian
have
)
axes.
,
lobes
d
yz
).
of
In
xz
their
However,
lobes
in
the
of
free
n+
metal
ion,
d-orbitals
CFT
is
M
,
are
based
with
ligands
(L)
at
an
innite
distance
away,
these
ve
degenerate.
on
an
electrostatic
model,
where
the
ligands
are
considered
n+
as
point
eld
charges
created
by
symmetrical),
Ufu uc
will
increase
Look at the Orbitron website to
created
see the shapes of the d orbitals:
will
by
split
the
the
in
the
into
that
surround
ligand
point
energies
energy
ligand
two
of
of
metal
charges
the
d
uniformly.
point
sets
the
is
will
however,
is
degenerate
M
.
isotropic
orbitals
If,
charges
cation,
the
the
is,
electrostatic
spherically
degenerate
electrostatic
then
t
the
(that
remain
octahedral,
energy,
If
set
the
and
d
orbitals
the
e
2g
http://winter.group.shef.ac.uk/
Three
orbitron/
are
of
the
orbitals
(the
t
set)
will
decrease
in
but
eld
set.
g
energy
(that
is,
they
2g
stabilized)
and
two
of
the
orbitals
(the
e
set)
will
increase
in
energy
g
(that
is,
they
are
destabilized).
y
The
stabilized
orbitals
that
z
z
x
y
x
d
d
xy
d
xz
yz
third car tesian axis in each case is or thogonal (90°) to the 2D plane
y
z
x
z
2
x
320
x
y
d
▲
comprise
Figure 1 Five d-orbitals
2
y
d
2
z
the
13 . 2
t
set
are
the
d
2g
is
,
d
xy
associated
electron
,
and
d
yz
with
density
orbitals.
The
reason
for
this
C o l o U r e d
stabilization
xz
the
fact
lying
at
that
45 °
these
to
the
three
orbitals
Cartesian
have
axes.
In
their
lobes
contrast,
the
of
d
2
2
x
and
d
orbitals
2
(e
z
C o m P l e x e s
)
are
destabilized
because
their
lobes
of
y
electron
g
density
are
between
directed
the
two
along
split
the
Cartesian
degenerate
sets
axes.
of
The
orbitals
energy
is
separation
dened
as
Δ
,
the
o
crystal
eld
splitting
energy
n
For
the
rst
three
1
d
electron
will
occupy
the
t
set
of
and
will
the
orbitals
congurations,
degenerate
orbitals
d
in
2
,
d
an
3
,
and
d
,
the
octahedral
electrons
crystal
eld,
2g
ll
singly
before
lling
them
in
pairs,
following
3
Hund’s
rule
electron
of
has
a
maximum
choice
–
it
multiplicity.
can
either
However,
occupy
the
after
d
,
the
destabilized
fourth
e
level
or
g
else
occupy
the
stabilized
t
level.
Although
the
electron
would
enter
2g
a
stabilized
pair
the
electron
additional
So
energy
what
with
energy
are
parameter,
the
Δ
?
level,
is
to
do
termed
factors
First
so
another
the
that
of
all,
would
electron
pairing
affect
it
is
the
require
in
an
additional
already
energy,
crystal
important
to
lled
to
This
P
eld
stress
splitting
that
Δ
o
experimental
energy
orbital.
energy
is
an
o
quantity.
The
following
are
the
factors
that
affect
the
size
of
Δ
:
o
●
identity
●
oxidation
●
nature
●
geometry
of
the
metal
state
of
the
of
of
ion
the
metal
ion
ligands
the
complex
ion.
Symmetry
Science
which
of
is
peppered
form
science.
The
historically
never
just
should
part
of
with
the
origin
of
interesting.
accept
always
symbolic
universal
such
As
symbols
try
to
representations
symbols
chemists
at
grasp
their
the
●
language
can
we
face
origin
be
should
and
Part
of
the
IB
learner
to
to
a
a
triply
The
symbols
one
orbital
term
for
a
as
IB
learners
we
strive
to
be
prole
t
and
e
2g
notations
used
in
b
are
used
of
set
if
orbitals.
of
there
Ungerade
is
e
orbitals.
is
the
only
German
odd.
The
number
2
is
used
if
the
sign
of
the
is
changes
upon
rotation
about
inquirers.
the
The
or
set
degenerate
involved.
wavefunction
that
degenerate
doubly
such
●
representations.
refers
refers
value
of
t
an
let
octahedral
Cartesian
us
look
at
axes
(gure
what
2).
happens
For
to
example,
the
sign
of
the
g
crystal
eld
energy
splitting
diagram
have
their
wavefunction
with
respect
to
the
d
orbital
on
xy
origin
●
g
in
symmetry:
comes
the
from
rotation
the
wavefunction
upon
inversion;
parity
of
the
term
does
that
orbital.
is,
u
gerade,
not
meaning
change
there
comes
is
no
from
that
about
the
x-axis:
y
sign
change
the
in
change of sign from + to -
term
x
on rotation around x-axis
ungerade,
changes
is
u
a
meaning
sign
change
are
only
that
upon
in
the
used
if
the
wavefunction
inversion;
parity
a
of
that
the
is,
there
orbital.
geometrical
entity
g
d
and
has
xy
a
▲
centre
of
of
inversion.
inversion
(but
the
not
in
in
a
German
the
Hence,
octahedral
tetrahedron),
term
as
for
g
there
is
a
stereochemistry
is
used.
Figure 2 Explanation of the 2 symmetry label
centre
Gerade
is
As
is
the
sign
changes
from
+
to
,
the
number
2
used.
even.
321
13
T H E
P E R I O D I C
TA B L E
1
Gup 9 cp
Δ
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
Identity of the metal ion
/ c

The
identity
of
the
metal
ion
can
inuence
the
extent
of
the
crystal
eld
3+
[Co(NH
)
3
22900
]
splitting.
6
In
general,
Δ
increases
descending
a
group
with
the
metal
in
o
the
same
oxidation
state.
3+
[Rh(NH
)
3
34100
]
6
3+
[Ir(NH
)
3
41100
]
Oxidation state of the metal ion
6
For
a
given
metal,
Δ
increases
as
the
oxidation
state
increases.
Since
the
o
metal–ligand
1
Cp
Δ
/ c
the
metal
interaction
increases,
the
is
partly
electrostatic
distances
between
in
the
nature,
metal
as
and
the
charge
ligands
on
decrease

resulting
in
a
better
overlap
between
the
metal
orbitals
and
the
ligand
orbitals.
2+
[Co(NH
)
3
10200
]
6
Nature of the ligands
3+
[Co(NH
)
3
22900
]
6
Ligands
ligand,
may
NH
have
,
has
a
different
greater
charge
charge
densities.
density
For
example,
compared
to
the
water,
3
hence
the
ammonia
H
O,
and
2
crystal
eld
splitting
caused
by
ammonia
will
be
greater.
1
Cp
Δ
/ c

3+
[Co(H
O)
2
18200
]
6
spcchica i
3+
[Co(NH
)
3
22900
]
-
6
I
-
<
Br
-
<
Cl
2
<
F
< [C
O
2
]
4
≈ H
O < NH
2
< en < bpy < phen < NO
3
< CN
≈ CO
2
weak-eld ligands
strong-eld ligands
increasing ∆
0
In the case of weak-eld ligands, the conguration adopted involves a pi-f
conguration (gure 3), whereas in the case of strong-eld ligands, such as CN ,
the conguration adopted involves a pi-pai arrangement (gure 4).
The following is a guideline when
considering whether a conguration
involves a spin-paired or a spin-free
e
ygrene latibro
2+
•
m
: d
g
arrangement:
: In the spectrochemical
series, for complexes that
2
2
x
and d
y
2
z
orbitals destabilized
3
∆
0
5
barycentre
2
∆
0
5
: d
t
2g
, d
xy
, d
xz
yz
2+
, ligands to the
involve M
orbitals stabilized
right of NO
are designated
2
as g- iga (and
hence adopt a spin-paired
2+
conguration), whereas
▲
Figure 3 Crystal eld splitting for [Fe(H
O)
2
4
ligand. The t
complexes with ligands to the
]
, which involves the H
6
O weak-eld
2
2
e
2g
conguration adopted is a pi-f conguration
g
are designated as
left of NO
2
a- iga (and hence
adopt a spin-free conguration).
e
3+
m
series, for complexes that
3+
involve M
of H
: d
g
: In the spectrochemical
ygrene latibro
•
, ligands to the right
O are designated as strong-
2
2
2
x
and d
y
2
z
orbitals destabilized
3
∆
5
0
barycentre
2
∆
0
5
t
: d
2g
, d
xy
, d
xz
yz
eld ligands whereas complexes
orbitals stabilized
with ligands to the left of H
O
2
are designated as weak-eld
ligands.
3
▲
Figure 4 Crystal eld splitting for [Fe(CN)
]
, which involves the CN
strong-eld
6
5
ligand. The t
2g
322
0
e
conguration adopted is a pi-pai conguration
g
13 . 2
A
Japanese
arranged
Δ
.
The
chemist,
into
a
R.
Tsuchida,
suggested
spectrochemical
spectrochemical
series,
series ,
which
is
that
based
given
ligands
on
in
could
order
section
of
15
C o l o U r e d
C o m P l e x e s
be
increasing
of
the
Data
o
booklet
is
based
on
empirical
evidence.
The geometry of the complex ion
The
geometry
of
the
complex
ion
can
also
inuence
the
crystal
eld
4
splitting
parameter.
For
example,
Δ
for
a
tetrahedral
complex
is
∼
t
It
must
is
empirical
values
be
of
emphasized,
in
the
nature,
crystal
however,
as
a
eld
model
energy
Δ
o
9
that
because
CFT
cannot
splitting
the
spectrochemical
account
for
the
series
relative
parameters.
Explanation of the colour of transition metal
complexes
The
colour
of
transition
metal
ions
is
associated
2+
d
orbitals.
For
example,
conguration,
so
its
d
has
sublevel
an
is
partially
condensed
lled
and
thus
electron
one
would
2+
Cu
example,
[Ar]3d
incomplete,
2+
expect
with
9
Cu
ions
to
crystals
be
of
coloured.
CuSO
· 5H
4
Cu
O
ions
have
a
are
often
dominant
blue
in
colour.
Mediterranean
For
blue
2
2+
colour.
In
contrast,
conguration,
and
ions
so
of
the
zinc’s
d-block
d-sublevel
metal
is
zinc,
fully
Zn
lled.
10
,
As
have
a
an
result,
[Ar]3d
ions
of
2+
Zn
are
White
metal
be
typically
light
consists
complexes
transmitted.
the
colour
colourless.
of
The
the
of
all
absorb
the
colour
light
colours
some
of
of
these
wheel
transmitted,
the
visible
colours,
(gure
that
is
5)
can
the
spectrum.
allowing
be
Transition
other
used
to
colours
to
determine
complementary
colour
3+
of
the
absorbed
light.
For
example,
[Ti(H
O)
2
light.
The
complementary
colour
to
colour
wheel.
]
absorbs
yellow–green
6
yellow-green
is
red-violet,
which
lies
3+
at
the
opposite
side
of
the
Therefore,
[Ti(H
O)
2
a
appear
in
red-violet.
transition
metal
will
6
Let
us
now
complexes.
i
v
t
le
io
v
–
e
lu
b
b
lu
e
e
g
n
ra
o
–
w
llo
e
y
b
l
u
n
e
e
rg
–
w
o
lle
y
rg
e
e
n
w
o
l
l
e
▲
l
o
n
absorption
o
ra
n
g
e
y
and
ions
t
e
r
e
d
g
light
e
r
of
colour
te
lo
iv
–
d
e
r
e
nature
d
the
complementary
–
examine
the
o
r
transmit
]
e
–
g
r
e
e
n
Figure 5 The colour wheel
323
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
As
outlined
are
split
M E TA L S
previously,
into
two
sets
the
of
( A H L )
ve
d-orbitals
degenerate
in
orbitals
an
–
octahedral
the
crystal
stabilized
t
set
eld
and
the
2g
destabilized
e
set.
If
the
d
orbitals
are
partially
lled,
d-to-d
electronic
g
transitions
energy
t
can
set
of
occur.
In
orbitals
other
to
the
words,
an
electron
higher-energy
e
2g
set
can
of
jump
from
orbitals.
This
the
lower-
d-to-d
g
electronic
transition
is
the
origin
of
the
colour
of
transition
metal
complexes.
3+
In
the
case
of
[Ti(H
O)
2
yellow-green
absorbed
is
a
light
]
,
such
an
electronic
transition
requires
a
photon
of
6
to
be
measure
absorbed.
of
Δ
.
Since
The
frequency
of
the
ΔE
represents
the
yellow–green
energy
change,
light
the
o
frequency
of
the
light,
v
(or
f ),
is
related
to
ΔE
via
the
expression:
hc
_
ΔE
=
hv
=
λ
where:
34
h
=
Planck’s
constant
=
6.63
×
10
J
s
8
c
=
λ
speed
=
of
light
wavelength,
Since
ΔE
is
in
in
related
a
vacuum
=
3.00
×
10
1
m
s
m
to
Δ
,
the
actual
colour
of
any
complex
will
depend
o
on
all
the
factors
oxidation
state
geometry
of
described
of
the
the
previously,
metal
complex
ion,
the
including
nature
of
the
the
identity
ligands,
and
and
the
ion.
a)
e
: d
g
2
2
x
and d
y
2
z
ygrene latibro
orbitals destabilized
3
∆
5
0
barycentre
2
∆
5
0
: d
t
2g
, d
xy
, d
xz
yz
orbitals stabilized
b)
d-to-d electronic
transition hν
e
: d
g
2
2
x
and d
y
2
z
ygrene latibro
orbitals destabilized
3
∆
5
0
barycentre
2
∆
5
0
t
: d
2g
, d
xy
, d
xz
yz
orbitals stabilized
3+
▲
Figure 6 (a) Crystal eld splitting for ground-state [Ti(H
O)
2
]
1
, which involves a t
6
conguration. (b)
Crystal eld splitting for excited-state [Ti(H
O)
2
conguration
324
0
e
2g
3+
]
6
g
0
involving a t
2g
1
e
g
13 . 2
C o l o U r e d
C o m P l e x e s
Worked examples
base.
Each
water
molecule
acts
as
a
monodentate
Example 1
2+
ligand,
For
the
complex
K
[Fe(ONO)
3
],
forming
a
coordinate
bond
with
Ni
.
As
deduce:
6
there
are
six
water
ligands
involved,
the
geometry
2+
a)
The
oxidation
state
of
the
transition
metal
in
of
the
cationic
complex,
[Ni(H
O)
2
]
,
is
octahedral,
6
2+
the
with
complex.
Ni
having
perchlorate
b)
The
condensed
electron
conguration
of
metal
in
this
oxidation
coordination
are
in
the
number
lattice
and
do
of
6.
not
The
form
the
part
transition
ions
a
of
the
cationic
complex.
state.
OH
2
c)
The
coordination
number
of
the
metal
in
the
H
OH
O
2
2
complex.
Ni
d)
The
stereochemistry
e)
The
charge
(geometry)
of
the
[ClO
90°
]
4
2
complex.
H
OH
O
2
2
OH
2
on
the
complex.
Example 3
Solution
Consider
a)
Let
x
=
the
oxidation
state
of
iron
in
the
complex
[Ni(NH
the
)
3
]Cl
6
2
complex.
a)
Deduce
of
3(+1)
+
x
+
6(
1)
=
0,
so
x
=
the
the
transition
Fe
is
c)
Six,
[Ar]3d
2
4s
3+
,
so
Fe
a
in
its
conguration
associated
state
in
this
complex.
5
is
[Ar]3d
b)
assuming
metal
electron
+3.
oxidation
6
b)
condensed
monodentate
nitrito
State
the
geometry
of
the
transition
metal
ligand,
complex
and
draw
a
diagram
of
the
complex.
(ONO)
c)
d)
Octahedral,
assuming
a
monodentate
Identify
the
ligand,
the
nature
of
the
bonding
between
nitrito
ligand
and
the
transition
metal
ion
in
the
(ONO)
complex.
e)
Each
potassium
has
a
charge
of
1+,
so
be
3+,
meaning
the
net
d)
State
the
e)
Draw
denticity
of
the
ammonia
ligand.
+
charge
that
for
the
three
charge
K
ions
on
the
will
anionic
complex
a
diagram
d-sublevel.
(that
is
nested
in
the
square
brackets)
showing
the
splitting
of
the
part
will
be
Label
the
orbitals
involved
and
3
populate
f)
Explain
each
of
the
orbitals
with
whether
the
complex
is
electrons.
paramagnetic
suy ip
or
diamagnetic.
Oxidation states are written with the sign rst and then
the number (for example, here iron has an oxidation
Solution
state of +3); ions are written with the number rst and
3+
a)
ion).
then the charge (for example, Fe
Let
x
=
the
oxidation
state
of
nickel
in
the
complex.
x
+
6(0)
+
2(
1)
=
0,
so
x
=
+2;
Example 2
8
Ni
Ni(ClO
)
4
reacts
with
water
to
form
the
O)
2
of
][ClO
6
an
]
4
is
2+
,
Ni
8
is
[Ar]3d
.
Explain
this
reaction
Octahedral,
acid–base
formed
CN
=
6.
in
2
theory,
and
outline
how
the
NH
2+
bond
2
4s
0
[Ni(H
terms
[Ar]3d
2
b)
ion
is
complex
between
Ni
and
H
3
O.
2
H
NH
N
3
3
Ni
Cl
90°
2
Solution
A
Lewis
acid
is
an
electron-pair
acceptor
and
a
H
NH
N
3
3
NH
3
2+
Lewis
acts
base
as
the
is
an
electron-pair
Lewis
acid
and
H
donor.
O
acts
Hence
as
the
Ni
Lewis
2
325
13
T H E
P E R I O D I C
TA B L E
–
T H E
T R A N S I T I O N
M E TA L S
( A H L )
6
c)
Coordinate
bonding.
e)
t
2
e
2g
d)
NH
has
one
lone
pair
involved
in
the
f)
g
Paramagnetic
since
there
are
two
unpaired
3
coordinate
bond
to
Ni
so
it
is
monodentate.
electrons.
e
: d
g
2
and d
2
x
y
2
z
orbitals destabilized
3
ygrene latibro
∆
0
5
barycentre
2
∆
0
5
t
: d
2g
, d
xy
, d
xz
yz
orbitals stabilized
326
Q U e s t I o n s
Questions
1
Which
of
the
transition
following
elements
is
not
a
7
What
of
the
following
can
act
as
a
ligand?
element?
I.
PH
II.
H
III.
NO
A.
I
and
II
B.
I
and
III
C.
II
and
D.
I,
II,
3
A.
Fe
O
2
B.
Cu
2
C.
D.
2
Sc
only
Zn
What
is
the
condensed
electron
only
III
only
conguration
and
III
2+
of
Co
?
2
A.
[Ar]4s
B.
[Ar]4s
7
3d
8
2
Which
electron
transitions
are
responsible
for
5
3d
the
colours
of
transition
metal
compounds?
7
C.
[Ar]3d
D.
[Ar]4d
A.
Between
B.
Among
d
orbitals
and
s
orbitals
7
3
What
is
the
condensed
electron
the
C.
From
D.
Between
the
IB,
May
attached
metal
ion
ligands
to
the
attached
ligands
conguration
d
orbitals
[1]
2+
of
Fe
?
2
A.
[Ar]4s
B.
[Ar]4s
1
5
3d
9
6
C.
2009
6
3d
Which
salts
dissolved
2
D.
form
coloured
solutions
when
[Ar]3d
[Ar]4s
in
water?
4
3d
I.
ZnCl
II.
FeBr
III.
Co(NO
2
4
What
[NH
is
]
4
the
ligand
[Fe(H
2
O)
2
in
the
][SO
6
]
4
2
complex
?
2+
A.
Fe
B.
[SO
)
3
3
2
2
A.
I
and
II
only
B.
I
and
III
only
]
4
C.
H
C.
II
and
D.
I,
II
III
only
O
2
+
D.
[NH
and
III
]
4
10
5
What
is
the
oxidation
state
of
iron
in
Which
the
complex
Na[Fe(EDTA)]
3H
of
the
following
statements
is
correct
for
the
complex
[Cr(H
O)
2
O?
]Cl
6
?
3
2
A.
A.
It
is
paramagnetic.
B.
It
is
diamagnetic.
C.
The
+1
B.
+2
C.
+3
ion
D.
coordination
is
number
of
the
chromium
3.
+6
D.
H
O
acts
as
a
bidentate
ligand
in
the
2
complex.
6
What
is
the
total
charge,
n,
in
the
following
n
complex
of
Ni(II),
[Ni(NH
)
3
]
?
6
11
A.
0
B.
1+
Explain,
energies,
by
why
states,but
C.
2+
D.
3+
referring
Ca
Ti
to
forms
only
successive
variable
occurs
in
ionization
oxidation
the
+2
oxidationstate.
327
13
T H E
12
P E R I O D I C
Explain
why
TA B L E
[Ni(H
O)
2
–
T H E
][BF
6
]
4
T R A N S I T I O N
is
coloured.
M E TA L S
14
( A H L )
In
an
article
Journal
13
Consider
the
complex,
K
written
by
W
.B.
Jensen
in
the
2
[Fe(C
4
O
2
)
4
].
3
of
Chemical
p1182-3),
it
quantities
of
was
Education
reported
HgF
have
(85,
that
been
9,
(2008),
minute
detected,
4
using
a)
Deduce
the
condensed
matrix
of
the
transition
metal
extreme
State
why,
the
metal
geometry
of
the
complex
Identify
in
and
draw
a
diagram
the
ligand
the
nature
and
the
of
the
bonding
transition
metal
State
Draw
the
and
the
denticity
a
diagram
d-sublevel.
populate
Explain
of
the
the
be
rst
considered
metal.
In
the
publication,
Label
each
of
the
the
or
the
a
however,
challenges
this
claim.
counterargument
Explore
may
have
why
the
splitting
orbitals
orbitals
complex
diamagnetic.
is
merit,
in
view
of
conventional
ion
ethanedioato
showing
whether
paramagnetic
328
in
now
thinking
on
what
complex.
electrons.
f)
basis,
might
between
ligand.
e)
this
of
considers
d)
K
complex.
Jensen’s
the
4
transition
Jensen
c)
on
mercury
transition
the
at
conditions.
complex.
instance
b)
non-equilibrium
in
Suggest
this
techniques,
electron
under
conguration
isolation
of
involved
with
as
a
transition
element.
IUPAC
C H E M I C A L
B O N D I N G
14
A N D
S T R U C T U R E
( A H L )
Introduction
More
and
in-depth
closer
often
require
theories
topic
we
of
and
of
more
bonding
expand
introduced
ve
explanations
analysis
six
in
bonding
to
4
electron
be
to
concepts
considered.
principles
of
explore
domains
systems
arrangements
sophisticated
the
topic
of
structural
In
and
this
VSEPR
species
and
Theory
involving
discuss
the
roles
play
that
in
bonding.
as
a
formal
such
a
charge
Hybridization
mathematical
hybridization
examination
around
a
and
treatment
the
central
is
and
can
be
number
interior
delocalization
structure
also
model
schemes
of
of
and
introduced
we
see
how
deduced
of
atom
from
electron
in
a
an
domains
species.
14.1 Ft sts of ot o  stt
Understandings
Applications and skills
➔
Covalent bonds result from the overlap of atomic
➔
Prediction whether sigma (σ) or pi (π) bonds
orbitals. A sigma bond (σ) is formed by the direct
are formed from the linear combination of
head-on/end-to-end overlap of atomic orbitals,
atomic orbitals.
resulting in electron density concentrated between
➔
Deduction of the Lewis (electron dot)
the nuclei of the bonding atoms. A pi bond (π) is
structures of molecules and ions showing all
formed by the sideways overlap of atomic orbitals,
valence electrons for up to six electron pairs on
resulting in electron density above and below the
each atom.
plane of the nuclei of the bonding atoms.
➔
➔
Application of FC to ascer tain which Lewis
Formal charge (FC) can be used to decide which
(electron dot) structure is preferred from
Lewis (electron dot) structure is preferred from
dierent Lewis (electron dot) structures.
several. The FC is the charge an atom would
➔
Deduction using VSEPR theory of the electron
have if all atoms in the molecule had the same
domain geometry and molecular geometry with
electronegativity. FC = (Number of valence
___
1
electrons) -
ve and six electron domains and associated
(Number of bonding electrons) –
2
bond angles.
(Number of non-bonding electrons). The Lewis
(electron dot) structure with the atoms having
➔
to dissociate oxygen and ozone.
FC values closest to zero is preferred.
➔
Exceptions to the octet rule include some species
Explanation of the wavelength of light required
➔
Description of the mechanism of the catalysis
of ozone depletion when catalysed by CFCs
having incomplete octets and expanded octets.
and NO
➔
Delocalization involves electrons that are shared
x
by more than two atoms in a molecule or ion as
opposed to being localized between a pair of atoms.
➔
Nature of science
Resonance involves using two or more
Lewis (electron dot) structures to represent
➔
Principle of Occam’s razor – bonding theories
a par ticular molecule or ion. A resonance
have been modied over time. Newer theories
structure is one of two or more alternative
need to remain as simple as possible while
Lewis (electron dot) structures for a molecule
maximizing explanatory power, for example the
or ion that cannot be described fully with one
idea of formal charge.
Lewis (electron dot) structure alone.
329
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
Theories of bonding and structure
To
study
large
structures
explanations
of
sophisticated
concepts,
bonding
we
look
than
at
and
bonding
we
have
species
consider
systems
models,
met
based
so
on
and
far.
ve
this
and
six
should
more
theories
In
underpinning
in-depth
requires
maintaining
of
and
consider
(and
challenge)
models
and
theories
used
in
for
capacity
and
chemical
theories
theories
as
for
is
is
that
a
theory
possible
while
maximum
discovery
of
application.
have
need
to
been
modied
remain
as
over
simple
as
time.
possible
bonding
maximizing
their
explanatory
power.
In
this
structure.
chapter
The
principle
simple
of
while
and
as
electron
some
New
the
a
understanding
topic
Bonding
domains,
this
remain
principle
the
of
Occam’s
development
different
elds
of
of
razor
theories
knowledge.
is
a
in
The
and
blueprint
a
number
several
bonding
associated
of
models
are
and
theories
presented,
each
of
with
structure
its
pros
and
limitations.
philosophy
Formal charge
TOK
In
As we shall see throughout
this topic, covalent bonding
can be described using
valence bond theory
are
as
obey
having alternative ways
of describing the same
(FC)
can
of
be
a
the
structure
theory. To what extent is
4.3
a
is
the
we
in
a
the
considered
of
A
as
a
idea
molecular
useful
in
the
process
worked
of
showing
is
to
(electron
the
or
a
structures
in
or
valence
the
ion.
be
ion.
drawn
which
formal
The
electronic
dot)
electrons
polyatomic
can
deciding
determine
molecule
as
Lewis
species
Lewis
involving
out
a
how
approach
appropriate
present
charge
the
of
different
rule.
most
atoms
way
covalent
number
octet
hypothetical
introduced
convenient
distributed
Sometimes
all
or molecular orbital
sub-topic
structure
that
Lewis
charge
calculation
of
book-keeping;
it
FC
is
a
follows:
phenomena a strength or a
1
_
FC
=
(number
of
valence
electrons)
-
weakness?
(number
of
bonding
2
electrons)
For
example,
in
the
-
(number
molecule
of
non-bonding
tetrachloromethane,
electrons)
(gure
CCl
1)
the
4
Cl
formal
charge
on
each
atom
is
calculated
as
follows:
1
_
FC(C)
C
Cl
=
(4)
(8)
Cl
0
=
0
2
1
_
FC(Cl)
=
(7)
(2)
6
=
0
2
Cl
2
In
the
case
of
the
carbonate
anion,
CO
(gure
2),
the
FCs
on
the
3
▲
Figure 1 Lewis structure of
carbon
and
oxygen
atoms
are:
tetrachloromethane, CCl
4
1
_
FC(C)
=
(4)
(8)
O
0
=
0
2
A
1
_
FC(O
-
2
)
=
(6)
(2)
6
=
-1
(4)
4
=
0
A
2
O
1
_
FC(O
)
=
(6)
B
2
C
If
O
O
there
rule,
●
are
the
the
a
number
most
one
of
possible
reasonable
with
FC
one
Lewis
will
difference
structures
( ΔFC
=
FC
FC
max
O
B
▲
to
O
0,
all
obey
)
closest
min
and
A
Figure 2 Lewis structure of the
●
the
one
that
has
the
negative
2
carbonate anion, CO
3
330
that
be:
electronegative
atoms.
charges
located
on
the
most
the
octet
14 . 1
For
BF
example,
.
For
gure
structure
3
F u r T h e r
shows
two
a S p e c T S
Lewis
O F
c O v a l e n T
structures
for
boron
b O n d i n g
a n d
S T r u c T u r e
triuoride,
St t
(a):
3
To check if you have calculated formal
1
FC(B)
=
(3)
(6)
0
=
0
2
charges correctly, nd the sum of the
1
FC(F)
=
(7)
(2)
6
=
0
FCs for the molecule or ion. For a neutral
2
molecule the sum of the FCs = 0 (for
ΔFC
=
0
example, for CCl
: 0 + [4 × 0]
= 0). For
4
For
structure
(b):
a polyatomic ion the sum of the
1
FC(B)
=
(3)
(8)
0
=
FCs = charge on the ion (for example, for
-1
2
2
CO
1
FC(F
)
=
(7)
A
: 0 + 2[
1] + 0 = 2
).
3
(2)
6
=
0
(4)
4
=
+1
2
1
FC(F
)
=
(7)
B
ΔFC
2
=
FC
FC
max
Since
ΔFC
is
=
(+1)
(
1)
=
+2
min
closest
to
zero
for
Lewis
structure
(a),
this
is
the
F
most
A
reasonable
representation
of
BF
.
3
Although
preferred
Lewis
based
incomplete
Species
than
see
8
can
in
FC
octet
also
valence
later
structure
on
be
of
obeys
electrons
found
electrons
this
(b)
the
considerations
with
octet
even
(fewer
than
expanded
surrounding
the
rule,
though
8
structure
boron
valence
octets
central
of
(a)
has
is
F
F
B
B
an
electrons).
electrons
atom),
as
(more
we
shall
topic.
F
F
F
F
Dierent interpretations of “charge”
F
(a)
The
idea
state,
and
of
charge
formal
we
has
charge,
need
to
many
ionic
interpret
connotations
charge,
the
partial
intended
in
chemistry
charge,
meaning
total
(b)
F
A
B
(oxidation
charge),
depending
▲
Figure 3 Two possible Lewis structures for
boron triuoride, BF
on
3
the
To
context.
distinguish
charge,
and
the
ionic
terms
oxidation
charge,
consider
state,
the
formal
charge,
hydrogen
partial
uoride
molecule,
H
HF
(gure
F
4).
▲
Figure 4 Lewis structure for hydrogen
Oxidation states
uoride, HF
hydrogen:
+1
uorine:
1
St t
Formal charges
1
FC(H)
=
(1)
(2)
0
=
In writing oxidation states, the sign goes
0
2
before the number (eg +2). For charges,
1
FC(F)
=
(7)
(2)
(6)
=
0
2
ΔFC
=
FC
the sign goes after the number (3
FC
max
=
0
).
Formal charges do not represent the actual
min
charge on the ion so hence signs are put
Par tial charges
From
H
=
on
section
2.2
and
uorine
hydrogen.
8
of
for
to
F
be
One
before the number (eg +1). Oxidation
the
=
Data
4.0,
more
rather
booklet ,
so
you
negative
the
electronegativity
would
than
approximate
expect
the
but
the
partial
simple
for
partial
charge
way
of
charge
on
calculating
numbers are represented by Roman
numerals (eg I, II) whereas oxidation
states are represented by Arabic numerals
(eg +1, +2).
331
14
c h e M i c a l
b O n d i n g
H
a n d
S T r u c T u r e
the
F
as
+
partial
charge
based
-
electronegativity
values
is
(2.2)
__
H:
=
(2.2
+
on
follows:
0.30
0.30
( a h l )
+
0.35
of
the
charge
of
the
bonding
pair
4.0)
-
δ
δ
(4.0)
__
▲
F:
Figure 5 The par tial charge in a
=
(2.2
+
0.65
of
the
charge
of
the
bonding
pair
4.0)
molecule of hydrogen uoride, HF,
based on simple approximations
Hence,
since
electrons,
uorine
neutral
on
HF
atom
uorine
The
and
6.
the
F
The
is
1.86
The
length
1.30e.
to
charge
a
on
of
0.30
is
than
partial
hydrogen.
which
two
Therefore
charge
experimentally
shows
the
nding
of
=
consists
negative
moment,
shows
of
2e
HF
HF
a
charge
is
represented
calculated
a
polar
by
the
dipole
D.
region
region
in
equates
partial
5.
×
more
electrostatic
method
bond
This
dipole
bond
0.65
0.30e
0.30+
gure
red
experimentally
have
have
net
in
HF
blue
Another
H
a
a
molecular
gure
covalent
uorine.
and
with
for
will
will
of
shown
moment
single
uorine
in
molecule
vector
the
92
the
area
area
partial
pm
calculated
potential
of
lowest
10
of
HF
involves
Data
1.86
is
shown
electron
electron
the
moment
for
greatest
charges
(section
dipole
(MEP)
of
density.
taking
booklet)
the
and
the
D:
F
μ
_
Q
=
H
r
where:
Q
=
apparent
μ
=
dipole
r
▲
Figure 6 The molecular electrostatic potential
We
=
H
can
F
use
charge
on
each
end
of
the
molecule
moment
bond
the
length
following
conversion:
(MEP) on the van der Waals surface of the
30
1
D
=
3.34
×
10
C
m
hydrogen uoride molecule
hence:
30
(1.86
×
3.34
×
10
C
m)
___
Q
=
12
(92
×
10
m)
20
=
6.75
×
C
10
19
The
the
charge
Data
of
an
booklet),
isolated
electron
therefore
the
=
1.60
partial
×
10
charge
δ
C
is
(section
given
4
of
by:
20
(6.75
×
10
C)
__
δ
=
=
0.42
19
(1.60
Computer
accurately,
has
δ +
=
×
10
programs
and
from
0.54+
and
C)
can
the
F
calculate
MEP
has
δ-
partial
model
=
0.54
Net charge on the HF molecule
The
332
net
charge
on
the
HF
molecule
is
charges
shown
0.
in
much
 gure
in
density
6,
more
H
14 . 1
F u r T h e r
a S p e c T S
O F
c O v a l e n T
b O n d i n g
a n d
S T r u c T u r e
Molecular geometries based on ve and
St t
six electron domains
In
topic
and
six
4
four
we
electron
electron
using
examined
the
domains.
domains
procedure
eqto and x positions
molecular
are
geometries
Molecular
summarized
outlined
in
topic
based
geometries
in
table
1,
on
two,
based
and
on
can
ve
be
occur only for geometries
three,
based on ve electron
and
domains, not those based
deduced
on six electron domains. We
4.
usually do not refer to axial
For
molecular
geometries
based
on
ve
electron
domains,
remember
or equatorial positions for an
that
lone
pairs
rst
instance.
(non-bonding
pairs)
occupy
equatorial
positions
in
the
octahedral geometry.
order
of
We
=
shall
is
based
interactions
LP|LP
(LP
This
>
LP|BP
lone
>
pair;
examine
is
as
the
follows
fact
that
(topic
in
terms
of
repulsion,
the
4):
BP|BP
BP
the
on
=
bonding
reason
for
pair)
this
in
nm of
eto om
to oms
omt
a
worked
example
later
in
this
topic.
Mo omt
nots
trigonal bipyramidal
equatorial
F
position
B
a
F
AB
B
5
e
120°
B
A
P
e
5
trigonal bipyramidal
5 BPs
B
e
90°
F
example: PF
B
a
5
F
axial position
F
see-saw
F
B
a
AB
F
E
4
B
e
5
trigonal bipyramidal
less than 120°
A
4 BPs and 1 LP
S
B
e
example: SF
4
F
B
a
less than 90°
F
T-shaped
F
B
a
AB
3
B
E
2
A
e
5
trigonal bipyramidal
3 BPs and 2 LPs
Cl
F
B
a
example: ClF
less than
3
90°
F
333
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
nm of
eto om
to oms
omt
( A H L )
Mo omt
nots
Linear
I
B
a
AB
E
2
5
3
180°
A
trigonal bipyramidal
2 BPs and 3 LPs
I
example: I
B
a
3
I
octahedral
F
F
F
B
AB
S
B
6
6
B
octahedral
A
6 BPs
90°
B
B
example: SF
6
B
F
F
F
square-based pyramidal
F
AB
6
B
octahedral
E
5
B
F
B
5 BPs and 1 LP
F
Br
A
B
example: BrF
B
5
F
F
square planar
F
AB
E
4
B
2
B
F
6
octahedral
A
4 BPs and 2 LPs
90°
B
B
F
example: XeF
4
Xe
F
▲
T
able 1 Electron domain geometries and molecular geometries based on ve and six electron domains. B
= axial substituent; B
a
=
e
equatorial substituent; E (in formula) = lone pair of electrons; BP = bonding pair of electrons; LP = lone pair of electrons
Overlap of atomic orbitals: Sigma (σ) and
pi (π) bonding
In
topic
and
a
bond
The
A
4
discussed
consists
single
double
between
334
we
multiple
bond
of
bond,
two
difference
as
a
atoms
double
electrons
represented
covalent
two
the
such
bond
A
and
shared
by
a
of
triple
a
is
a
four
single
bond.
between
stick,
consists
B.
between
or
two
sigma
A
covalent
single
atoms
bond
electrons,
two
A
bond
covalent
and
B:
(σ).
pairs,
shared
14 . 1
F u r T h e r
a S p e c T S
O F
c O v a l e n T
b O n d i n g
a n d
S T r u c T u r e
σ
The
A
double
triple
bond,
represented
covalent
between
two
bond
atoms
A
by
two
consists
and
sticks,
of
six
is
a
sigma
electrons,
plus
or
pi
three
bond
pairs,
(σ
+
π).
A
B
shared
B.
σ
+
π
A
The
triple
bonds
A
(σ
Lewis
bond,
+
represented
by
three
sticks,
is
a
sigma
plus
2
two
2π).
structure
is
a
simple
model
showing
how
the
valence
(outer-
σ
shell)
B
pi
electrons
are
distributed
in
a
molecule
or
polyatomic
ion.
+
2π
In
A
sub-topic
14.2
quantum
mechanics
helpful
in
we
shall
look
called
visualizing
the
at
a
more
sophisticated
molecular
difference
orbital
between
a
theory
theory
sigma
based
(MOT),
bond
and
which
a
pi
is
bond.
▲
For
atomic
relatively
be
orbitals
close
identical.
X
in
to
overlap
energy,
atomic
and
orbitals
and
the
form
molecular
symmetry
combine
to
of
form
orbitals
the
x
atomic
new
they
must
orbitals
molecular
B
on
Figure 7 Covalent bonds
be
must
orbitals.
i o of o t:
There
are
three
possible
outcomes:
A
1
bonding
orbital:
2
anti-bonding
sigma
(σ)
or
pi
(π)
B > A=B > A≡B
orbital
i o of o stt:
orbital:
sigma
star
( σ*)
or
pi
star
(π*)
orbital
A≡B > A=B > A
3
non-bonding
Table
2
shows
a
B
situation.
number
of
combinations
of
atomic
comto of tom ots
orbitals.
Mo ots fom
T
σ
+
bonding
s + s
σ*
+
anti-bonding
σ
+
bonding
s + p
x
σ*
+
anti-bonding
+
NB
+
NB
s + p
non-bonding
y
s + p
non-bonding
z
σ
+
bonding
p
x
+ p
x
σ*
+
anti-bonding
335
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
comto of tom ots
( A H L )
Mo ots fom
T
π
+
bonding
p
+ p
y
y
π*
+
anti-bonding
π
+
bonding
p
+ p
z
z
π*
+
anti-bonding
p
+ p
x
p
+
NB
+
NB
+ p
x
NB
+ p
y
▲
non-bonding
z
+
p
non-bonding
y
non-bonding
z
T
able 2 Combination of atomic orbitals (LCAOs). Black represents the positive wavefunction, Ψ
, and white represents the negative
+
wavefunction, Ψ
Delocalization and resonance
dsto of  sm 
Figure
8
shows
two
Lewis
structures
for
the
nitrite
oxoanion,
NO
that
2
 o
have
In the formation of a sm o
identical
arrangements
of
atoms
but
different
arrangements
electrons.
there is a direct head-on overlap
-
of the atomic orbitals along the
-
internuclear axis and the electron
N
N
density is located along this axis.
In the formation of a  o
there is a sideways overlap
of the atomic orbitals and the
▲
Figure 8 Lewis structures of the nitrite oxoanion, NO
2
electron density is located above
and below the internuclear axis.
Each
Lewis
double
two
336
structure
bond,
but
structures.
the
An
shows
one
position
ion
with
of
one
N
O
the
of
single
double
these
bond
bond
and
is
structures
one
N=O
different
would
in
have
the
of
14 . 1
one
In
shorter
fact
the
N=O
bond
(114
experimentally
intermediate
between
F u r T h e r
a
pm)
and
measured
single
a S p e c T S
and
one
longer
bond
double
O F
N
lengths
NO
c O v a l e n T
O
are
bond.
bond
both
Both
b O n d i n g
(136
125
a n d
pm).
rso
pm,
Lewis
S T r u c T u r e
structures,
In so, two or more
called
resonance
called
a
forms,
contribute
to
the
electronic
structure
which
is
Lewis structures can represent
resonance
hybrid.
This
idea
of
resonance
is
shown
by
linking
a par ticular molecule or ion.
the
contributing
resonance
forms
by
a
double-headed
arrow
(gure
9).
A so stt is one
of two or more alternative
N
Lewis structures for a molecule
N
O
or ion that cannot be described
O
fully with one Lewis structure
alone.
Figure 9 Resonance in the nitrite oxoanion, NO
▲
2
Note that resonance structures
The
resonance
in
the
NO
ion
can
also
be
represented
as
shown
in
are purely hypothetical
2
gure
10.
The
dashed
curve
conveys
delocalization.
As
specied
by
species that do not actually
IUPAC,
delocalization
is
a
quantum
mechanical
concept
used
to
describe
exist.
the
A
pi
bonding
conjugated
represented
as
in
a
conjugated
system
a
is
system
a
of
system.
molecular
alternating
entity
single
whose
and
structure
multiple
can
bonds.
be
Conjugation
is
the
interaction
bond.
bond
A
a
NO
one
conjugated
and
a
conjugated
has
of
p
“fractional
oxoanion
is
may
containing
not
double
the
orbital
system
orbital
system
p
with
also
a
bond
bond
form
lone
localized
another
of
between
character”
order
the
pair
of
each
intervening
interaction
atoms
bond
O
an
electrons.
two
or
N
across
but
order.
bond
is
of
The
1.5,
a
sigma
double
bonding
instead
In
the
since
in
each
case
both
of
a
N
link
the
bonds
are
2
equivalent
and
intermediate
between
a
single
and
a
double
bond:
▲
total
number
of
NO
bonding
pairs
3
_
____
bond
order
in
NO
=
=
=
1.5
Figure 10 An alternative representation
of resonance in the NO
oxoanion
2
2
total
Another
oxoanion
that
has
number
resonance
of
NO
structures
is
the
carbonate
2
CO
2
positions
oxoanion,
2
.
Three
resonance
structures
can
be
written
for
CO
3
(gure
11).
3
2
2
O
O
doto
doto involves
C
C
electrons that are shared by
O
O
O
O
more than two atoms in a
molecule or ion as opposed to
being localized between a pair
of atoms.
2
O
C
O
O
2
▲
Figure 11 Resonance in the carbonate oxoanion, CO
3
St t
2
In
CO
the
three
C
O
bond
lengths
are
equivalent
(129
Covalent bond lengths (both
pm),
3
intermediate
bond
length
between
(122
a
pm).
single
The
C
bond
O
bond
order
is
(143
pm)
worked
and
out
as
a
double
C =O
single and multiple) are
provided in section 10 of the
follows:
Data booklet
total
order
in
CO
number
of
CO
bonding
pairs
4
_
____
2
bond
=
=
=
1.33
3
total
number
of
CO
positions
3
337
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
Worked examples
(iv)
To
draw
the
Lewis
structure,
complete
the
Example 1
octets
Consider
a)
the
following
BrF
b)
six
d)
c)
e)
the
(ii)
the
F
]
[ICl
]
4
each
(i)
atoms:
2
4
For
[ICl
5
SOF
uorine
species:
IF
3
on
species,
deduce:
electron
domain
molecular
Br
F
geometry
geometry
F
(iii)
the
(iv)
a
approximate
valid
Lewis
bond
angle(s)
(electron
dot)
b)
structure.
Ball-and-stick
diagram
of
the
IF
molecule
5
(ignoring
bond
angles):
Solution
F
a)
Ball-and-stick
diagram
of
the
BrF
molecule
3
(ignoring
the
bond
F
angles):
F
I
F
F
F
Br
For
F
For
F
I:
number
of
valence
number
of
sigma
electrons
bonds
=
7
=
5
=
12
=
6
Br:
total
number
of
valence
number
of
sigma
electrons
=
7
=
3
=
10
number
number
bonds
(i)
total
number
of
valence
electrons
of
of
valence
electron
From
table
electrons
domains
1,
electron
domain
geometry:
octahedral.
number
of
electron
domains
=
5
(ii)
This
is
an
example
of
an
AB
E
system,
so
5
(i)
From
table
trigonal
(ii)
This
is
1,
electron
domain
geometry:
the
bipyramidal.
an
example
six
of
an
AB
E
3
one
lone
pair
equivalent
is
located
in
any
of
the
positions.
system,
2
F
so
the
two
equatorial
lone
pairs
are
located
in
the
positions:
F
F
I
F
F
Br
F
Molecular
(iii)
less than 90°
F
geometry:
(iii)
angles:
Bond
experimental
cannot
338
be
than
value
predicted
is
angles:
angles
T-shaped.
less
geometry:
Bond
of
Molecular
F
90 °.
angles
(The
86.2 °,
from
but
F
Br
F
this
VSEPR
theory.)
are
these
is
square-based
six
less
F
than
I
90°.
experimental
80.9°.)
F
F
pyramidal.
bond
(Each
I
F
one
bond
14 . 1
(iv)
To
draw
octets
the
on
F u r T h e r
Lewis
uorine
a S p e c T S
structure,
O F
complete
c O v a l e n T
b O n d i n g
(iv)
the
To
draw
octets
atoms:
a n d
the
on
S T r u c T u r e
Lewis
chlorine
structure,
complete
the
atoms:
-
F
Cl
F
F
I
I
F
F
Cl
c)
Ball-and-stick
diagram
of
the
[ICl
]
anion
2
(ignoring
bond
angles):
St t
-
When drawing the Lewis structure of an anion or cation,
Cl
include square brackets and the respective charge.
Don’t forget to complete the octets on the terminal
atoms in a Lewis structure, unless the terminal atom is
I
hydrogen which has only the two electrons in a bonded
Cl
pair, so attaining the noble gas conguration of helium.
For
I:
d)
Ball-and-stick
diagram
of
the
SOF
molecule
4
number
of
valence
number
of
sigma
electrons
=
7
=
2
(ignoring
bond
maximum
bonds
cannot
ve
charge
=
1
member
S
total
number
number
of
of
valence
electron
electrons
=
domains
be
must
valency
the
of
be
angles);
period
the
of
central
3,
note
oxygen
atom.
can
central
that
is
in
so
its
From
table
trigonal
(ii)
This
is
1,
electron
domain
5
F
the
an
example
three
equatorial
lone
O
of
an
pairs
AB
positions
bipyramidal
of
electron
E
so
F
F
O
S
system,
located
the
trigonal
S
F
F
3
are
domain
octet,
geometry:
bipyramidal.
2
so
a
SOF
F
(i)
oxygen
as
4
10
=
2
Sulfur,
expand
atom
the
in
F
F
geometry:
For
S:
number
of
valence
number
of
sigma
electrons
=
6
=
5
=
-1
=
10
=
5
Cl
1
pi
bonds
bond
I
total
number
number
of
of
valence
electron
electrons
domains
Cl
p os  vSepr to
Molecular
geometry:
linear.
Pi bonding is o-axis bonding. Because the shape of a
(iii)
The
bond
pairs
are
angle
remains
arranged
180 °,
as
the
symmetrically.
lone
molecule is controlled by the sigma-bonding framework
along the internuclear axis, when counting the valence
electrons we subtract 1 for each pi bond present.
339
14
C H E M I C A L
(i)
(ii)
From
B O N D I N G
table
1,
A N D
electron
geometry:
trigonal
Molecular
geometry:
S T R U C T U R E
( A H L )
(i)
domain
From
bipyramidal.
trigonal
table
1,
electron-domain
geometry:
octahedral.
bipyramidal:
(ii)
This
is
an
example
of
an
AB
E
4
the
F
two
lone
as
possible
to
minimize
pairs
from
are
each
other
system,
so
2
located
as
(i.e.
far
at
apart
180 °)
repulsion:
F
less than 120°
-
S
O
F
Cl
greater than 90°
Cl
90°
I
F
Cl
(iii)
Bond
angles:
less
than
120 °
for
F
S
F
eq
greater
than
90°
double
bond
for
O=S
F
.
;
eq
The
axial
F
S
F
120°.
for
bond
The
F
To
is
F
;
90.7°
the
on
Molecular
space
decreased
for
values
O=S
so
(iii)
from
are
geometry:
square
planar.
the
Bond
pairs
110 °
so
F
eq
draw
octets
angle
more
experimental
S
eq
(iv)
occupies
Cl
angle(s):
present
all
bond
90°.
are
The
two
arranged
angles
remain
sets
of
lone
symmetrically
90 °
axial
Lewis
uorine
structure,
and
complete
oxygen
(iv)
the
To
draw
octets
atoms:
the
on
Lewis
chlorine
structure,
complete
the
atoms:
-
F
Cl
Cl
F
I
S
O
Cl
F
Cl
F
Example 2
e)
Ball-and-stick
diagram
of
the
ICl
anion
4
Deduce
(ignoring
bond
the
molecular
polarities
of
the
following:
angles):
a)
SF
6
-
b)
cisplatin,
Pt(NH
)
3
Cl
Cl
2
2
Cl
c)
transplatin,
Pt(NH
)
3
Cl
2
2
I
Solution
Cl
Cl
a)
Since
SF
has
a
symmetrical
octahedral
6
structure
For
I:
of
individual
and
number
of
valence
number
of
sigma
electrons
=
7
=
4
=
1
(see
the
table
S
F
molecule
1),
the
bonds
as
a
dipole
cancel
whole
moments
one
another,
remains
non-polar.
bonds
b)
Cisplatin,
[Pt(NH
)
3
ve
charge
From
section
8
of
electronegativities
total
number
of
valence
electrons
=
of
electron
domains
=
=
3.0.
Hence
are
the
is
square
Data
planar.
Pt
booklet,
=
2.2,
Cl
the
Pt
Cl
bond
=
3.2,
(difference
in
6
electronegativity
Pt
340
]
2
12
N
number
Cl
2
the
N
bond
=
1.0)
(difference
is
=
more
0.8).
polar
than
the
14 . 1
To
deduce
the
polarity
of
F u r T h e r
the
a S p e c T S
complex
O F
c O v a l e n T
b O n d i n g
a n d
S T r u c T u r e
we
cst s   ttmt
consider
the
moments.
is
direction
In
cisplatin
approximately
using
This
the
90°,
in
a
cisplatin
the
the
so
parallelogram
results
below;
of
Cl
we
dipole
is
polar
Pt
Pt
add
law
net
a
two
Cl
Cl
bond
the
from
dipole
two
angle
vectors
mathematics.
moment
as
shown
complex:
Cisplatin is used to treat ovarian and testicular cancer.
In the body the two Cl ligands are replaced by the DNA
base guanine, and the geometrical arrangement of
the resulting complex is a perfect t to the two double
helix strands in DNA . Transplatin is ineective in the
treatment of cancer due to the trans conguration
of the Cl ligands. The two complexes can be
H
N
Cl
3
distinguished by their dierent molecular polarities.
Pt
Example 3
H
N
Cl
3
a)
Below
the
are
two
dinitrogen
alternative
oxide
Lewis
molecule,
structures
N
for
O:
2
St t
In deducing molecular polarities based on molecular
N
O
A
B
geometries you should nd the vectorial sum of
Based
the individual dipole moments using the
Lewis
parallelogram law.
›
For two vectors
on
formal
structure
charge
is
considerations,
which
preferred?
›
v
and
v
1
both star ting from the same
b)
Below
are
three
alternative
Lewis
structures
2
›
point, the sum of the two vectors
v
for
can be found
the
cyanate
anion,
OCN
by drawing and completing the parallelogram. The
›
diagonal gives
›
v
›
= v
-
v
(gure 12):
›
C
+ v
1
2
A
→
v
1
→
v
-
→
C
v
2
B
▲
Figure 12 The parallelogram law for adding vectors
-
c)
Transplatin,
Pt(NH
)
3
Cl
2
is
also
square
planar
2
C
but
the
Cl
vectorial
is
0
since
sum
of
and
bond
the
are
along
complex.
moment
Cl
they
directions
the
Pt
two
equal
the
Cl
Therefore
hence
H
angle
Pt
Cl
but
Pt
is
in
Cl
there
dipole
The
moments
C
opposite
linear
is
transplatin
N
180°.
no
is
axis
net
in
dipole
non-polar:
Based
which
on
Lewis
charge
structure
considerations,
is
deduce
preferred.
Cl
3
Solution
Pt
a)
Table
Lewis
Cl
formal
3
shows
the
structures
formal
A
and
charges
(FC)
for
B.
NH
3
341
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
on
atom
Fc =
ΔFc = Fc
the
more
electronegative
element,
in
this
case
- Fc
mx
m
oxygen.
The
preferred
Lewis
structure
is
therefore
C.
Lewis structure A
Example 4
1
terminal N
(4)
(5)
(4) = -1
Explain
2
why
bromine
triuoride,
BrF
has
its
lone
3
pairs
of
electrons
located
in
equatorial
positions.
1
central N
(5)
(8)
(0) = +1
(+1)
(
1) = +2
2
Solution
1
O
(6)
(4)
(4) = 0
2
In
example
1(a)
the
molecular
geometry
of
BrF
3
Lewis structure B
was
arrangement
1
terminal N
(4)
(5)
deduced
to
is
be
T-shaped.
based
on
an
The
AB
geometrical
E
3
(4) = -1
system.
Below
2
2
are
(+2)
(
(6)
structures
for
BrF
with
the
lone
3
pairs
(8)
Lewis
1) = +3
1
O
three
surrounding
Br
located
in
different
positions:
(0) = +2
2
F
▲
T
able 3 Formal charges (FCs) for the dinitrogen oxide
F
molecule, N
O: Lewis structures A and B
Br
F
2
F
F
Br
Since
ΔFC
preferred
b)
Table
B,
for
A
Lewis
4
and
is
shows
C
less
than
structure
for
the
the
ΔFC
will
FCs
for
cyanate
be
for
B,
the
Lewis
anion,
structures
Fc =
F
A
B
C
both lone pairs
one lone pair equatorial
both lone pairs
equatorial
and one axial
axial
A,
OCN
Table
atom
F
F
A.
Br
F
ΔFc = Fc
- Fc
mx
5
shows
possible
Lewis
the
interactions
for
these
three
structures.
m
Lewis structure A
ittos
Lewis
Lewis
Lewis
structure
structure
structure
A
B
C
posto of o
both
equatorial
both axial
s
equatorial
and axial
120°
90°
180°
6 × 90°
1
O
(6)
(6)
(2) = +1
2
1
C
(4)
(8)
(0) = 0
= (+1)
(
2) = +3
2
1
N
(5)
(2)
(6) = -2
2
a t w
Lewis structure B
lp|lp
1
O
(4)
(6)
C
(4) = 0
2
a t w
1
lp|bp
(4)
(8)
(0) = 0
(0)
(
1) = +1
4 × 90°
3 × 90°
2 × 120°
2 × 120°
2 × 90°
1 × 120°
2
a t w
1
N
(5)
(4)
3 × 120°
(4) = -1
2
bp|bp
2 × 90°
Lewis structure C
▲
T
able 5 Interactions for three Lewis structures A, B, and C for
1
O
(2)
(6)
(6) = -1
bromine triuoride, BrF
2
. (LP = lone pair; BP = bonding pair)
3
1
C
(4)
(8)
(0) = 0
(6)
(2) = 0
= (0)
(
1) = +1
Remember
that:
2
1
N
(5)
2
▲
LP|LP
T
able 4 Formal charges (FCs) for the cyanate anion, OCN
>
LP|BP
>
BP|BP
:
The
greatest
bond
angle
for
LP |LP
interactions
Lewis structures A, B, and C
occurs
Based
are
B
atom
2.6).
on
and
C.
the
two
However,
preferred
O
is
(electronegativities:
Since
second
342
ΔFC
both
criterion
B
is
and
that
C
the
O
=
have
the
Lewis
most
3.4;
the
N
structures
electronegative
=
3.0;
same
negative
FC
C
ΔFC,
=
the
should
reside
Lewis
in
structures
structure
interaction:
while
will
both
the
A
C
gives
preferred
occupying
and
Next
structure
structure
be
B.
A
C,
so
we
considering
gives
only
6
4
structure,
equatorial
×
×
can
discard
LP |BP
90°
90°
with
positions.
bond
angles
angles,
the
so
lone
A
pairs
14 . 1
F u r T h e r
a S p e c T S
O F
c O v a l e n T
b O n d i n g
a n d
S T r u c T u r e
An environmental perspective: Catalysis of ozone depletion
Ozone,
O
is
a
V-shaped
(bent)
molecule
with
where:
3
a
bond
angle
of
116.8°
and
its
two
O
O
bond
34
lengths
are
resonance
shown
in
equal
forms
gure
(128
can
pm).
be
Two
written
h
=
Planck’s
for
ozone,
as
ν
=
frequency
=
6.63
×
of
the
10
J
s
radiation
13.
8
c
=
λ
+
constant
contributing
O
O
speed
=
of
light
wavelength
=
of
3.00
the
×
1
10
m
s
radiation
+
Worked example
▲
The
average
362
kJ
bond
enthalpy
in
ozone
is
Figure 13 Resonance forms of ozone, O
3
1
mol
section
The
bond
order
for
the
O
O
bond
in
ozone
as
and
bond
order
in
section
required
number
of
O
O
bonding
oxygen
pairs
enthalpy
given
data
in
given
11
of
the
Data
=
number
wavelength,
booklet,
in
nm,
calculate
of
the
UV
the
radiation
of
O
O
to
break
and
the
the
O
O =O
O
double
bond
in
bond
in
ozone.
3
_
_
_
_
_
=
total
bond
relationships
O
3
total
the
the
follows:
maximum
each
Using
is
in
calculated
1
.
=
1.5
2
positions
Solution
The wavelength of light required to
Oxygen:
The
bond
enthalpy
for
O =O
is
1
498
dissociate oxygen and ozone
kJ
mol
photons
In
topic
6
we
mentioned
that
in
the
ozone
radiation
layer
from
absorbs
the
over
95 %
J.
Next
First
calculate
convert
the
kJ
to
energy
J
and
of
then
stratosphere,
use
the
in
.
of
harmful
Avogadro’s
constant,
L:
UV
1
sun:
498 × 1000 J mol
__
E
=
19
=
23
6.02
×
8.27
×
10
J
1
10
mol
hν
O
(g)
→
O
3
(g)
+
O•(g)
hc
__
2
E
=
,
so
by
cross-multiplication
Eλ
=
hc.
We
λ
O
(g)
+
O•(g)
→
O
2
(g)
+
then
heat
make
λ
the
subject
of
the
equation:
3
34
There
is
a
net
energy
conversion
from
UV
radiation
to
heat
(6.63
hc
_
λ
=
×
8
10
J
s)(3.00
×
10
1
m
s
)
____
=
19
energy.
E
8.27
×
10
J
7
=
The
progressive
allows
more
surface,
cancers
depletion
of
the
radiation
to
reach
UV
resulting
in
an
(melanomas)
ozone
increased
and
the
risk
bonds
radiation
in
ozone
( hv).
The
can
10
m
9
of
1
nm
=
7
10
m,
so
λ
=
(2.41
×
10
m)
×
9
skin
(1
nm/10
m)
=
241
nm
cataracts.
be
bond
×
layer
Earth’s
Ozone:
The
2.41
broken
order
in
by
ozone
The
bond
enthalpy
for
the
O
O
bond
in
1
UV
ozone
is
362
kJ
mol
(1.5)
1
is
lower
than
the
bond
order
in
oxygen
(2),
362 × 1000 J mol
__
so
E
=
the
O=O
double
Radiation
of
bond
shorter
in
oxygen
wavelength
is
is
6.02
stronger.
break
the
stronger
bond
in
×
10
oxygen.
=
is
that
the
energy,
E,
of
a
photon
of
10
J
(6.63
×
10
8
J
s)(3.00
×
10
1
m
s
)
____
=
19
E
reason
×
mol
34
λ
The
6.01
1
required
hc
_
to
19
=
23
6.01
×
10
J
light
7
=
is
inversely
proportional
to
the
3.31
×
10
m
wavelength
7
λ,
so
the
greater
wavelength
and
the
vice
energy,
the
shorter
the
λ
=
The
hc
_
E
=
hν
(3.31
×
10
9
m)
×
(1
nm/10
m)
=
331
nm
versa.
depletion
greater
detail
of
in
the
ozone
topic
layer
was
discussed
in
6.
=
λ
343
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
Models and theories of structure and bonding
You
come
she
to
knocks
nature
them
with
all
all
her
theories,
and
at.
Another
of
Pierre-Auguste
approach
structure
Renoir
(1841–1919),
with
is
the
S
to
simply
having
a
draw
a
Lewis
complete
octet
electrons:
French
impressionist
artist.
2
O
Chemical
models
bonding
and
and
theories
structure
which
can
are
riddled
often
be
with
interpreted
O
in
specic
should
ways
be
used
to
suit
with
our
requirements.
caution
in
science.
Models
Let
us
take
O
an
example
and
we
its
of
a
Lewis
associated
mean
by
structure
molecular
of
an
geometry
oxoanion
to
see
what
Here
this.
of
It
is
the
theory
that
decides
what
we
can
O
although
shown
physicist
services
Einstein
who
to
the
(1879–1955),
Nobel
theoretical
Prize
physics,
in
German
Physics
especially
in
for
theoretical
1921
his
for
his
at
the
the
law
of
the
obeys
the
octet
rule,
However,
as
the
S
valency
is
same
by
four
electron
geometry
for
domains
the
we
oxoanion,
arrive
namely
tetrahedral.
discovery
Both
of
S
incorrect.
observe.
surrounded
Albert
is
photoelectric
models
result
in
each
S
O
bond
being
effect.
equivalent,
a
single
S
intermediate
O
bond.
The
between
question
a
double
then
is
and
which
A question of approach: Expanded octets,
Lewis
structure
is
the
most
valid.
octets, or formal charge considerations
If
In
topic
4
we
looked
at
one
method
we
we
determining
the
shape
of
an
oxoanion
the
sulfate
oxoanion,
get
the
now
to
formal
results
shown
charge
in
considerations
table
6.
such
2
as
turn
of
Table
SO
6
shows
that
in
both
Lewis
structures
4
the
more
negative
FC
resides
on
the
more
O
electronegative
oxygen
atom,
so
this
is
not
a
109.5°
factor
here.
ΔFC
is
closer
to
zero
for
the
Lewis
S
O
O
structure
in
octet,
this
so
which
sulfur
supports
has
the
an
expanded
suggestion
that
this
is
O
the
Using
this
method
correctly
the
sulfate
is
can
represented
molecular
worked
out
as
geometry
preferred
structure.
of
tetrahedral
and
atom
Fc =
ΔFc = Fc
- Fc
mx
be
structures.
Try
equivalent
in
to
by
write
six
contributing
these!
All
S
O
m
resonance
bonds
are
Lewis structure: expanded octet
length.
1
The
nal
step
involves
completing
the
octets
S
on
(6)
(12)
(0) = 0
2
the
four
terminal
oxygen
atoms,
in
order
to
write
1
six
valid
Lewis
structures
for
sulfate.
Here
are
O
two
(6)
(2)
(6) = -1
(4)
(4) = 0
(0)
(
1) = +1
2
of
these
as
an
example:
1
O=
(6)
2
O
O
S
S
Lewis structure: octet
O
O
etc.
O
O
1
S
O
(6)
(8)
(0) = +2
(2)
(6) = -1
2
O
(+2)
(
1) = +3
1
O
(6)
2
This
of
model
shows
electrons:
in
that
each
of
S
has
the
an
six
expanded
Lewis
octet
structures,
▲
T
able 6 Formal charges (FC) for the Lewis structures of the
2
the
central
S
atom
is
surrounded
by
12
electrons.
sulfate oxoanion, SO
4
344
14 . 2
A
similar
analysis
with
the
phosphate
oxoanion,
evidence
points
h y b r i d i z a T i O n
to
two
shorter
S =O
bonds,
3
[PO
]
reveals
that
the
Lewis
structure
in
which
supporting
a
Lewis
structure
for
the
sulfate
4
phosphorus
based
on
has
FC
an
expanded
calculations.
octet
is
However,
favourable
oxoanion
with
an
expanded
octet:
theoretical
H
quantum
mechanical
calculations
suggest
that
the
O
most
with
favoured
the
single
octet
of
Lewis
electrons
structure
and
is
many
the
one
H
chemists
S
argue
that
Lewis
structures
for
the
obeyed
when
Lewis
choosing
for
structure
rather
than
between
oxoanions,
the
in
it
which
alternative
is
better
the
expanded
octet
octet
to
rule
O
is
alternative.
As
we
said
theories
If
we
go
one
molecular
step
further
geometry
of
and
examine
sulfuric
acid,
O
O
opt
at
with
the
an
outset,
open
approach
scientic
models
and
mind!
the
experimental
14.2 h to
Understandings
Applications and skills
➔
A hybrid orbital results from the mixing of
3
Explanation of the formation of sp
➔
2
, sp
, and sp
dierent types of atomic orbitals on the same
hybrid orbitals in methane, ethene, and ethyne.
atom.
Identication and explanation of the
➔
relationships between Lewis (electron dot)
structures, electron domains, molecular
geometries, and types of hybridization.
Nature of science
➔
The need to regard theories as uncer tain –
Quantum mechanics involves several theories
hybridization in valence bond theory can help
explaining the same phenomena, depending on
explain molecular geometries, but is limited.
specic requirements.
Models, theories, assumptions, and deductions
What
look
we
see
depends
mainly
on
what
and
we
theories
critical
for.
Theories
Sir
John
Lubbock
(1834–1913),
biologist,
banker
and
model
each
theory
the
our
merits
in
chemistry
has
of
a
its
own
We
its
own
limitations.
particular
perspective.
has
theory
perhaps
should
merits
and
depend
treat
as
not
with
Samuel
be
seen
the
same
Beckett
in
black
degree
(gure
and
of
1).
we
should
also
look
for
white
shades
of
–
grey
politician.
Sometimes
may
should
scientists
and
Each
chemistry
FRS,
as
English
in
perspective
on
models
question
with
a
every
experimental
combination
to
see
the
possible
of
bigger
with
assumption,
evidence,
models
picture.
this
and
challenge
and
consider
theories
“Real
to
science”
theories
using
allow
us
becomes
perspective!
345
14
C H E M I C A L
In
topic
dot)
4
we
B O N D I N G
saw
structure
in
the
A N D
value
providing
of
a
a
S T R U C T U R E
Lewis
simple
( A H L )
(electron
model
Valence bond theory (VBT)
which
The
shows
how
valence
or
outer-shell
electrons
VBT
each
distributed
in
a
molecule.
Sub-topics
4.3
and
other
how
the
number
of
electron
domains
from
deduced
from
a
Lewis
structure,
considers
enabling
form
a
that
atoms
molecule.
The
approach
bond
the
overlap
of
atomic
formed
orbitals
can
resulting
be
to
14.1
results
showed
model
are
in
a
bonding
orbital
with
the
electrons
the
localized
between
assumed
that
their
own
Figure
atoms
2
and
A
form
the
H
,
two
the
respective
shows
H
the
when
atoms.
atoms
atomic
VBT
it
is
they
retain
orbitals.
interaction
each
In
interact
having
of
two
one
hydrogen
electron,
to
B
the
diatomic
molecule
H
.
The
two
atoms
2
approach
each
interaction
there
to
is
the
a
other
between
decrease
interaction,
and
subsequently
their
in
a
the
electrons
energy
chemical
of
bond
there
and
the
is
is
an
nuclei.
system
If
due
formed.
+
H
H
A
▲
H
H
B
Figure 2 Formation of the H
diatomic molecule based on VBT.
2
The two electrons are localized between the two atoms and
each atom involved in the bonding retains its own respective
▲
Figure 1 The Irish playwright Samuel Beckett. John Calder,
atomic orbital
author of The Theology of Samuel Beckett, said to Beckett “It
We
is a ne day,” to which Beckett replied “So far!”
can
represent
potential
prediction
of
the
ultimately
the
electron-domain
molecular
geometry
geometry
using
and
the
two
energy
atomic
this
bonding
versus
nuclei
the
model
distance,
(gure
by
d,
a
sketch
of
separating
3).
VSEPR
Z
theory.
VSEPR
of
the
bonds
Neither
theory)
detailed
or
even
of
these
models
however
offers
electronic
why
(Lewis
an
+
Y
explanation
structure
chemical
and
of
bonds
X
covalent
exist.
To
0
we
explain
need
to
look
phenomena
scale,
with
sub-atomic
forms
the
the
can
formation
to
quantum
be
models
of
explored
that
particles.
chemical
explain
the
the
Schrödinger’s
launch-pad
for
the
bonds
mechanics.
on
eld
ygrene laitnetop
fully
Physical
microscopic
behaviour
wave
of
of
equation
bond energy
quantum
bond length
mechanics.
bonding
shapes
Two
It
and
of
helps
us
provides
understand
an
chemical
understanding
of
the
internuclear distance, d
molecules.
theories
in
quantum
mechanics
that
aid
an
▲
understanding
of
molecular
geometries
Figure 3 Interaction energy diagram showing potential energy
are:
versus internuclear distance, d, for the H
molecule
2
●
valence
●
molecular
We
346
shall
bond
theory
orbital
discuss
each
(VBT)
theory
one
On
gure
●
At
3:
(MOT).
separately.
X
the
the
potential
hydrogen
energy
atoms
are
is
too
essentially
far
apart
zero,
to
as
interact.
14 . 2
●
At
Y
The
the
hydrogen
electron
on
atoms
atom
A,
approach
e
,
is
each
attracted
other.
to
the
orbitals
the
h y b r i d i z a T i O n
overlapping,
formation
of
but
new
the
overlap
orbitals
called
results
in
molecular
A
nucleus
on
atom
B,
H
.
Simultaneously,
e
B
e
repel
each
other
nuclei
H
B
and
H
A
each
other.
energy
There
going
is
from
a
X
and
orbitals.
decrease
to
Y
as
molecular
than
the
electrons
orbitals,
are
and
assigned
associated
to
these
with
the
whole
B
in
the
repel
molecule
potential
attraction
rather
than
individual
atoms.
is
Figure
greater
The
A
and
4
shows
the
formation
of
the
molecular
repulsion.
orbitals
for
orbitals
on
the
H
molecule.
The
two
1s
atomic
2
●
At
Z
the
minimum
achieved.
of
H
.
This
The
H
potential
represents
H
bond
energy
the
length
most
is
74
is
stable
pm
state
(see
two
new
which
is
the
hydrogen
molecular
additive,
atoms
orbitals.
results
in
combine
One
a
to
form
combination,
bonding
2
section
10
of
the
Data
booklet).
molecular
energy,
orbital
1
Explain
and
difference,
Question
why
signicantly
the
to
potential
the
left
of
energy
point
Z
orbital
the
other
results
sigma
sigma
in
star
(σ),
which
combination,
an
is
which
is
of
lower
which
anti-bonding
(σ*),
of
is
a
molecular
higher
energy.
rises
on
gure
3.
+
St t
bonding orbital σ
Remember from chapter 5 that bond breaking is
an endothermic process (ΔH positive) and bond
node
formation is an exothermic process (ΔH negative).
In
VBT,
the
greater
the
degree
of
orbital
overlap
anti-bonding orbital σ*
(interaction),
the
stronger
the
bond
will
be.
▲
Figure 4 Molecular orbital theory (MOT) model for the H
2
VBT
can
be
applied
to
other
homonuclear
molecule. There is a build-up of electron density in the bonding
diatomic
molecules
such
as
F
and
to
molecular orbital, σ, between the two nuclei. If we think of a
2
heteronuclear
or
HF
.
One
model
the
is
inherent
the
a
of
of
the
be
the
used
the
chemical
applied
the
in
VBT
not
that
together, the electron density is the negative “glue” that holds
the
to
the two positively charged nuclei together. In the anti-bonding
molecular orbital, σ*, there is a nodal plane (node) between the
explain
bonds.
occur
two nuclei, which means that there is zero electron density here
VBT
on
We
can
represent
this
orbital
diagram
as
We
shall
For
hybridization,
bonds
that
an
atom
explain
can
form
the
and
molecular
conditions
the
of
these
discussing
The
hybridization
note
that
VBT
its
The
limitations
and
has
been
superseded
as
by
molecular
5.
orbitals
must
be
to
be
formed
two
met.
atomic
for
orbitals
effective
must
be
relatively
close
in
overlap.
orbital
symmetry
the
sign
of
of
the
the
atomic
orbitals,
wavefunction
Ψ,
that
must
a
be
model
gure
also
is,
has
molecular
bonds.
2
Before
in
a
spatial
energy
orientation
shown
by
number
1
of
model
electronic
molecules.
to
covalent bond as the “glue” that holds atoms in a molecule
bond.
to
model
HCl
considered
covalent
energy
as
structure
are
does
between
in
such
Lewis
bonds
model
polyatomic
expand
concept
of
changes
also
structure
shortly
so
molecules
covalent
differences
formation
can
all
and
considers
VBT
limitation
that
same
diatomic
identical.
For
example,
the
signs
of
both
theory.
Ψ
could
be
positive
(both
Ψ
),
and
sum
to
+
form
the
σ
bonding
molecular
orbital.
In
Molecular orbital theory (MOT)
contrast
VBT
assumes
that
when
bonds
are
formed
the
subtractive
combination
would
from
have
one
wavefunction
positive,
Ψ
,
and
one
+
the
overlap
of
atomic
orbitals,
the
original
nature
wavefunction
of
the
atoms
is
retained.
MOT
involves
negative,
Ψ
,
resulting
in
the
σ*
atomic
anti-bonding
molecular
orbital.
347
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
In
MOT
x
atomic
orbitals
combine
to
form
x
new
1
molecular
orbitals;
for
example,
two
1s
atomic
σ*
orbitals
combine
orbitals,
Many
ygrene
each
σ
and
theories
theory
mechanics
to
form
two
new
molecular
σ*.
have
has
its
provides
same
their
own
advantages
constraints.
several
explain
the
specic
requirements.
theories
phenomena,
but
equally
Quantum
that
can
depending
on
σ
1
H : 1s
▲
1
H
H : 1s
2
Figure 5 Molecular orbital diagram for the H
molecule
2
Hybridization
hto 
VBT
uses
the
concept
of
hybridization
to
provide
an
electronic
 ots
description
of
polyatomic
molecules
such
as
CH
and
4
NH
,
but
it
can
also
3
hto is a term used to
account
for
the
geometries
of
molecules.
describe the mixing of atomic
We
shall
now
consider
three
different
types
of
hybridization:
orbitals to generate a set of
new hybrid orbitals that are
3
1
sp
2
sp
3
sp
as
seen
in
methane
as
seen
in
ethene
equivalent. Hybridization is a
2
mathematical procedure.
as
seen
in
ethyne.
A  ot results from
the mixing of dierent types of
atomic orbital on the same atom.
The
hybridization
domains
around
type
the
can
be
central
determined
from
the
number
of
electron
atom.
TOK
hto is a mathematical tool that allows us to relate the bonding in
a molecule to its symmetry. What is the relationship between the natural
sciences, mathematics, and the natural world?
Smmt is key in several dierent areas of knowledge. In the eld of
mathematics, for example, there is great emphasis on the concept of
symmetry. Early Greek mathematicians such as Pythagoras and Euclid used
the idea of the symmetry of specic shapes and objects to develop their
theories. The structure of DNA is an excellent example of the symmetry of the
helix, and it was this symmetry that ultimately led Crick and Watson to unravel
the structure of DNA.
Ssmmt (SUSY) is an impor tant theory met in par ticle physics. What
is meant by supersymmetry, why is it discussed so much, and to which
elementary par ticles does it relate? How are architects, musicians, ar tists, and
dancers inuenced by symmetry in their creativity?
348
14 . 2
h y b r i d i z a T i O n
3
The formation of sp
Methane,
CH
is
a
hydrocarbon
hybrid orbitals in methane
and
is
one
of
the
shaped)
combined
with
25 %
of
the
characteristics
4
3
major
the
components
electron
of
natural
conguration
of
gas.
Let’s
consider
carbon:
electron
conguration:
1s
2
electron
spherical
2s
conguration:
orbital
s
is
orbital.
shown
orbital
in
shape
gure
of
an
sp
6.
2p
[He]2s
2
2p
+
ground-state
The
2
2
condensed
the
hybrid
2
full
of
=
diagram:
3
▲
[He]
Figure 6 One of four sp
orbitals which each have 75% p
character and 25% s character. This hybrid orbital is aligned in
the z
2
1
1
2p
2s
direction
0
2p
x
2p
y
z
3
The
1
Hydrogen
has
the
electron
conguration
1s
.
a
might
expect
electrons
in
that
its
since
2p
carbon
sublevel,
it
has
two
would
four
sp
hybrid
orbitals
point
to
the
corners
of
You
tetrahedron
(gure
7).
unpaired
form
two
3
sp
bonds
with
hydrogen
in
its
ground-state,
hybrid orbitals
forming
109.5°
CH
rather
than
CH
2
extremely
The
.
Although
CH
4
does
exist
it
is
central atom
2
unstable.
tetrahedral
hybridization
nature
of
its
of
one
methane
2s
and
involves
three
2p
the
orbitals
on
central carbon
the
C
atom
in
an
excited-state
as
follows.
3
▲
Figure 7 (a) The tetrahedral orientation of the four sp
Step 1
orbitals on carbon. Each of these sp
One
of
the
electrons
ground-state
to
the
hybrid
3
the
2s
orbital
of
con guration
of
carbon
is
vacant
2p
in
orbital,
to
form
an
hybrid orbitals contains
one electron. (b) For simplicity, the smaler of the two lobes is
the
often omitted on diagrams
promoted
excited-state:
z
Step 3
[He]
The
nal
step
involves
overlap
of
each
carbon
3
sp
orbital
with
a
hydrogen
1s
atomic
orbital.
1
Hydrogen
This
arrangement
gives
four
potential
C
H
has
a
1s
electron
conguration
and
the
bonds.
1
s
However,
we
know
from
topic
4
that
methane
orbital
is
spherically
symmetrical.
Each
1s
atomic
has
3
orbital
a
tetrahedral
molecular
geometry
with
H
C
the
bond
angles
of
109.5 °.
the
excited-state
the
geometry
orbitals
are
is
at
So
accounts
although
for
incompatible
90°
to
each
the
this
four
because
other
(topic
model
C
H
the
on
hydrogen
and
each
sp
hybrid
orbital
on
H
central
carbon
contains
one
electron
(gure
8).
for
bonds,
three
H
2p
2).
central
109.5°
carbon atom
3
sp
Step 2
The
four
atomic
orbitals
2s,
2p
,
2p
x
,
and
hybrid
2p
y
z
H
combine
to
form
The
new
a
set
of
four
new
hybrid
orbitals.
3
four
equivalent
sp
and
hybrid
each
one
orbitals
are
consists
of
H
entirely
25 %
s
1
H
character
and
75%
p
character,
since
they
1s
atomic
were
orbitals
formed
from
one
s
orbital
and
three
p
orbitals.
3
▲
Figure 8 In methane, each carbon sp
orbital overlaps with a
3
The
shape
75%
of
of
the
each
sp
hybrid
characteristics
of
orbital
a
p
will
orbital
have
hydrogen 1s orbital to form a tetrahedral geometry
(dumbbell
349
14
C H E M I C A L
Any
molecule
geometry
on
B O N D I N G
with
its
a
A N D
S T R U C T U R E
tetrahedral
interior
central
electron
atom
( A H L )
domain
(based
Step 2
on
To
four
electron
domains)
would
be
predicted
account
for
the
120°
approximate
bond
angles
to
(based
on
three
electron
domains),
the
next
step
3
have
sp
hybridization.
involves
2s,
2p
hybridization
and
2p
x
.
These
of
three
of
combine
the
to
atomic
form
a
set
orbitals
of
three
y
2
St t
new
sp
hybrid
orbitals.
The
2p
orbital
remains
z
unhybridized.
Hybridization is deduced from the electron domain
geometry rather than the molecular geometry. For
2
The
has a tetrahedral electron
example, ammonia, NH
three
atoms
3
domain geometry, corresponding to four electron
are
33.3%
domains (three bonding pairs and one non-bonding
point
or lone pair). Its molecular geometry is trigonal
s
sp
entirely
character
to
(gure
new
the
hybrid
orbitals
equivalent
and
corners
of
66.7%
a
on
and
p
the
each
carbon
one
character.
trigonal
planar
has
They
system
10).
pyramidal (with a bond angle of 107° due to the
p
repulsion between the lone pair and bonding pairs,
z
which reduces the bond angle from 109.5°). The
2
3
sp
since it is based on
hybridization of ammonia is sp
the four electron domains.
2
sp
C
2
The formation of sp
hybrid orbitals
2
sp
in ethene
Ethene,
C
H
2
an
alkene
,
is
another
hydrocarbon.
Ethene
is
4
2
Figure 10 Each of the three sp
▲
with
one
C=C
double
bond.
hybrid orbitals contains one
Around
electron. The 2p
unhybridized atomic orbital also contains
z
each
carbon
there
are
three
electron
domains
so
one electron. As before, for simplicity of representation, the
the
electron
domain
geometry
(and
molecular
smaller of the two lobes is omitted
geometry)
angle
is
(gure
is
trigonal
121.3°
and
planar.
the
H
The
C
H
H
C=C
bond
bond
angle
is
117°
Step 3
9).
The
12
1.3°
H
next
bond
H
step
along
involves
the
the
formation
internuclear
axis
by
of
the
a
sigma
overlap
of
2
two
C
sp
orbitals,
one
on
each
carbon
atom.
11
7°
C
A
pi
the
H
bond
two
H
p
is
formed
from
unhybridized
the
sideways
atomic
overlap
orbitals,
with
of
the
z
overlap
▲
hybrid
regions
above
and
below
the
internuclear
Figure 9 The molecular geometry of ethene. The double
axis
bond occupies more space so the H
C
(gure
11).
H bond angle
is reduced from the predicted 120° for a trigonal planar
p
p
z
z
pi
geometry
2
sp
2
sp
To
deduce
the
hybridization
scheme
of
the
sigma
carbon
orbital
for
atom
in
diagram
ethene,
of
we
carbon
start
as
with
shown
the
previously
C
C
methane.
2
2
sp
sp
2
sp
2
sp
Step 1
One
of
the
electrons
in
the
2s
orbital
of
conguration
of
carbon
is
the
▲
ground-state
Figure 11 The formation of the C=C double bond: a sigma bond
promoted
2
forms from the overlap of two sp
to
the
vacant
2p
orbital
z
to
form
an
excited-state,
the overlap of two p
z
as
350
for
methane
above.
orbitals
orbitals, and a pi bond from
14 . 2
and
Step 4
2p
.
These
h y b r i d i z a T i O n
combine
to
form
a
set
of
two
hybrid
orbitals.
The
remaining
2p
and
2p
y
The
new
x
sp
nal
step
involves
overlap
of
each
z
remaining
orbitals
remain
unhybridized.
2
sp
orbital
orbital
on
carbon
(gure
with
a
hydrogen
1s
atomic
The
12).
two
atoms
new
are
sp
hybrid
entirely
orbitals
equivalent
on
and
the
carbon
each
one
pi
consists
H
H
2
of
They
point
Each
of
50%
in
s
character
opposite
and
50%
directions
p
character.
(gure
14).
2
sp
sp
2
these
sp
hybrid
orbitals
contains
one
2
sp
sp
electron.
C
The
remaining
orbitals
are
the
2p
C
and
y
2
2
sp
sp
2p
unhybridized
atomic
orbitals,
also
containing
z
sigma
one
H
electron
each.
H
p
p
z
z
2
▲
Figure 12 In ethene, four carbon sp
orbitals overlap with four
hydrogen 1s orbitals to form a trigonal planar geometry on
sp
C
sp
each carbon atom
Any
molecule
with
a
trigonal-planar
electron
domain
p
y
2
geometry
will
be
predicted
to
have
sp
hybridization.
p
z
The formation of sp hybrid orbitals in
▲
Figure 14 Each of the two sp hybrid orbitals contains one
electron. The 2p
and 2p
y
unhybridized atomic orbitals also
z
ethyne
contain one electron each
Ethyne,
C
H
2
bond.
As
,
is
an
alkyne
and
has
one
C≡C
triple
2
seen
in
sub-topic
14.1,
around
each
carbon
Step 3
atom
there
are
two
electron
domains
so
the
electron
The
domain
geometry
(and
molecular
geometry)
next
sigma
linear,
with
a
180°
bond
angle
(gure
step
involves
the
formation
of
a
is
bond
along
the
internuclear
axis
by
the
13).
overlap
of
two
sp
hybrid
orbitals,
one
on
each
180°
carbon
H
the
atom.
sideways
Two
pi
bonds
overlap
of
are
the
formed
two
p
and
from
two
y
unhybridized
▲
atomic
orbitals,
p
z
with
the
overlap
Figure 13 The molecule of ethyne is linear
regions
To
deduce
atom
in
the
hybridization
ethyne,
again
we
scheme
start
with
of
the
the
carbon
(gure
above
and
below
the
internuclear
axis
15).
orbital
pi
diagram
of
carbon
as
shown
for
methane.
sp
sp
Step 1
One
of
the
electrons
ground-state
to
the
as
for
vacant
the
2s
orbital
of
conguration
of
carbon
is
2p
in
orbital
to
form
an
the
sigma
promoted
excited-state,
z
pi
p
p
y
y
methane.
p
p
z
Step 2
▲
z
Figure 15 The formation of the C≡C triple bond in ethyne: a
sigma bond forms from the overlap of two sp orbitals, a pi
To
account
for
the
180°
bond
angle
(based
on
bond from the overlap of two p
two
electron
domains),
the
next
step
from the overlap of two p
hybridization
of
two
of
the
atomic
orbitals
2s
orbitals, and another pi bond
y
involves
orbitals
z
351
14
C H E M I C A L
B O N D I N G
A N D
S T R U C T U R E
( A H L )
H
Step 4
The
sp
nal
step
orbital
orbital
on
involves
carbon
(gure
overlap
with
a
of
each
hydrogen
H
O
H
remaining
1s
atomic
H
C
O
H
16).
A
H
H
pi
H
B
▲
sp
Figure 1
7 Methyl propanoate
sp
C
H
C
C
H
Solution
sigma
pi
p
ito tom
A
B
C
nm of
4
3
4
tetrahedral
trigonal
tetrahedral
p
y
y
p
p
z
z
to oms
▲
Figure 16 In ethyne, two carbon sp orbitals overlap with two
eto om
hydrogen 1s orbitals to create a linear geometry on each
omt
carbon atom
Any
molecule
geometry
on
predicted
to
with
its
a
linear
interior
have
sp
electron
central
domain
atom
would
planar
Mo
be
tetrahedral
trigonal
V-shaped
planar
(bent)
omt
hybridization.
bo (s)
H
C
H:
C
C
109.5°
Summary
O:
C
120°
O
C:
less than
109.5°
nm of
eto om
to oms
omt
4
tetrahedral
hto
3
hto
2
3
sp
sp
sp
3
sp
▲
T
able 2
2
3
trigonal planar
2
linear
sp
Example 3
sp
The
▲
T
able 1 Electron-domain geometry and hybridization
condensed
(traditional
structural
name
formula
aniline)
is
C
H
6
a)
Using
Worked examples
the
Example 1
Deduce
the
interior
atom
VSEPR
domain
and
carbon
theory,
phenylamine
.
2
deduce
the
molecular
and
of
NH
5
electron
geometries
nitrogen
atoms
phenylamine.
hybridization
of
the
central
nitrogen
b)
in
A
model
of
the
molecule
is
shown
in
ammonia.
Solution
As
seen
in
electron
topic
4,
domains
because
in
there
ammonia,
are
the
four
hybridization
3
must
be
sp
Example 2
For
each
molecule
deduce
interior
of
the
molecular
atom
methyl
electron
A,
B,
and
propanoate
domain
geometry,
the
C
in
a
(gure
geometry,
bond
angles,
17),
the
and
the
▲
hybridization
352
of
in
state.
Figure 18 Molecular model of phenylamine
gure
18.
14 . 2
In
this
model
nitrogen
of
the
the
appear
two
to
be
hydrogen
atoms
above
horizontal
the
attached
to
For
h y b r i d i z a T i O n
C:
plane
number
of
valence
number
of
sigma
electrons
=
4
=
3
=
-1
=
6
=
3
molecule.
Figure
19
shows
phenylamine,
electrostatic
a
space-lling
which
is
potential
the
map
model
basis
of
one
an
showing
pi
bond
electron
total
charge
bonds
of
number
of
valence
electrons
density.
number
electron
of
domain
molecular
bond
electron
domains
geometry:
geometry:
angle:
trigonal
trigonal
planar
planar
120°
predicted
hybridization
domains:
sp
based
on
3
electron
2
Ball-and-stick
bond
diagram
for
nitrogen
(ignoring
angles):
C
H
For
N:
number
of
valence
number
of
bonds
total
▲
number
of
electrons
valence
electrons
=
5
=
3
=
8
=
4
Figure 19 Space-lling model of phenylamine
number
Deduce
gure
two
c)
the
18
hybridization
and
from
of
gure
the
19
nitrogen
and
electron
from
comment
on
of
electron
domain
domains
geometry:
tetrahedral
the
predicted
molecular
predicted
H
geometry:
trigonal
pyramidal
models.
A
of
theoretical
study
phenylamine
of
the
found
electronic
the
H
N
H
N
H
bond
angle:
less
than
109.5 °
structure
bond
predicted
hybridization
domains:
sp
based
on
4
electron
3
angle
is
in
very
phenylamine
close
to
the
to
be
112.79 °,
experimental
which
value
from
See
gas-phase
microwave
studies.
Discuss
gure
20.
what
H
you
may
conclude
about
the
H
molecular
3
nitrogen sp
N
geometry
around
the
nitrogen
in
the
NH
2
group
in
the
structure
of
phenylamine,
H
and
H
2
deduce
its
hybridization
state
on
this
carbon sp
basis.
H
H
Solution
H
a)
Using
VSEPR
theory
to
consider
the
geometry
3
▲
at
carbon,
start
by
drawing
the
Figure 20 VSEPR theory suggests sp
hybridization at nitrogen
ball-and-stick
in phenylamine
diagram
(ignoring
bond
angles):
b)
C
C
The
model
in
gure
18
shows
a
trigonal
C
3
pyramidal
H
at
geometry,
nitrogen
as
suggesting
outlined
above,
sp
hybridization
while
that
in
353
14
C H E M I C A L
gure
19
around
B O N D I N G
shows
a
nitrogen
A N D
trigonal
with
a
S T R U C T U R E
planar
120°
( A H L )
on
geometry
bond
pi
angle,
N
and
system
increases
on
the
the
electron
aromatic
ring.
density
A
more
of
the
planar
2
suggesting
sp
geometry
hybridization.
is
therefore
adopted
around
the
NH
2
group.
Although
VSEPR
theory
is
a
useful
model
We
structure,
in
some
cases
it
does
such
the
In
with
cases
the
it
geometry
cannot
be
found
used
to
experimentally;
accurately
the
electron
domain
around
N
in
one
of
these
three
dipolar
structures:
in
deduce
number
of
valence
number
of
sigma
electrons
=
5
=
3
=
-1
=
-1
=
6
=
3
hybridization.
phenylamine
the
NH
functional
group
type
attened
geometry
interactions
resulting
about
with
the
N,
in
due
a
trigonal
to
aromatic
resonance
ring.
1
pi
bond
1
+ve
planar-
The
p
interacts
with
the
pi
system
on
the
charge
orbital
total
N
number
resulting
electron
pair
structure
of
in
delocalization
(lone
pair).
The
phenylamine
is
of
the
most
of
in
Lewis
gure
with
three
dipolar
resonance
electron
+
NH
2
domains
domain
geometry:
trigonal
planar
geometry:
trigonal
planar
is
consistent
with
the
planar
geometry
+
NH
2
electron
structures.
This
+
NH
electrons
21
molecular
along
valence
non-bonding
stable
shown
of
aromatic
number
ring
bonds
is
2
slightly
on
out
not
resonance
agree
work
for
geometry
predicting
can
NH
2
2
2
from
the
space-lling
model,
suggesting
sp
hybridization.
c)
The
experimental
suggests
▲
a
H
geometry
trigonal
pyramidal
trigonal
planar
N
H
bond
somewhere
(109.5 °
bond
angle
113°
between
angle)
and
Figure 2
1 Lewis structures for phenylamine
(120°),
and
a
hybridization
2
state
Delocalization
reduces
the
of
the
electron
non-bonding
density
in
the
pair
p
on
somewhere
N
orbital
usf so
The “ChemEd DL” website contains numerous digital resources. It is a
collaboration with the Journal of Chemical Education, the education division of
the American Chemical Society, and the ChemCollective Project, initiated through
the National Science Foundation in the USA. It has an excellent library of three-
dimensional models.
354
of
between
sp
3
and
sp
.
Q u e S T i O n S
Questions
1
What
are
the
formal
charges
on
P
and
O
in
the
5
What
is
the
molecular
geometry
of
[PF
]
?
6
Lewis
(electron
oxoanion
dot)
structure
represented
in
of
gure
the
phosphate
A.
Trigonal
planar
B.
Trigonal
bipyramidal
C.
Square
D.
Octahedral
22?
3
pyramidal
O
O
P
O
6
Which
of
the
following
molecules
is
O
non-polar?
▲
Figure 22
A.
SF
B.
ClF
C.
BrCl
D.
SeF
4
A.
B.
P
P
is
1
is
and
+5
C.
P
is
0
D.
Both
O
and
and
O
O
are
is
is
is
3
0
5
2
6
0
3
7
Which
of
orbitals
2
What
are
oxygen
ΔFC,
formal
atoms
in
carbon
the
the
and
Lewis
dioxide
charges
the
formal
(electron
shown
on
in
the
carbon
charge
dot)
gure
following
in
combinations
gure
24
results
in
of
a
atomic
sigma
bond?
and
difference,
structure
the
shown
I.
+
of
23?
II.
+
O
▲
C
O
III.
Figure 23
+
FC(C)
FC(O)
ΔFC
▲
3
A.
+4
2
B.
+4
+4
Figure 24
+6
0
A.
I
and
II
I
and
III
C.
+4
+2
+2
B.
D.
0
0
0
C.
II
and
D.
I,
II,
What
is
the
electron
domain
geometry
only
only
III
and
only
III
of
2
the
sulte
oxoanion,
[SO
]
?
8
3
What
in
A.
Trigonal
planar
B.
Trigonal
pyramidal
C.
D.
is
the
hybridization
of
the
oxygen
hybridization
of
the
carbon
atom
ethanol?
A.
sp
Tetrahedral
B.
sp
V-shaped
C.
sp
D.
dsp
2
3
(bent)
3
4
What
is
the
molecular
geometry
of
BrF
?
5
9
A.
Octahedral
B.
Square
C.
T-shaped
D.
Square
What
atom
is
in
the
methanal?
planar
A.
sp
B.
sp
C.
sp
D.
dsp
2
pyramidal
3
3
355
14
C H E M I C A L
10
How
many
molecule
B O N D I N G
sigma
of
and
propyne,
pi
H
A N D
S T R U C T U R E
bonds
are
present
( A H L )
in
a
12
Consider
the
following
+
3
[NO
]
[SiF
3
sigma
pi
A.
5
3
B.
6
2
C.
7
1
D.
8
0
IB,
May
For
a)
each
i)
its
electron
the
following
NH
4
ii)
its
molecular
iii)
the
Draw
geometry
geometry
bond
angle(s).
an
appropriate
each
and
Lewis
calculate
the
(electron
formal
dot)
charge
atom.
2
The
following
reactions
take
place
in
the
ozone
species,
by
the
absorption
of
ultraviolet
light.
Deduce:
I
O
→
O
3
i)
its
electron-domain
ii)
its
molecular
iii)
bond
O
→
and
hybridization
state
of
of
central
Draw
molecular
an
structure
on
each
O•
explain,
the
by
reference
reactions,
wavelength.
atom
IB,
its
+
polarity.
appropriate
and
Lewis
calculate
atom.
the
I
or
II,
to
the
bonding,
requires
a
the
shorter
v)
O•
angle
which
the
O•
2
geometry
State
iv)
+
2
geometry
II
356
domain
CS
3
layer
b)
4
species:
13
a)
SCl
[1]
on
each
]
4
2011
Consider
For
[IF
species:
structure
CCl
]
6
Deduce:
b)
11
species:
CCCH?
(electron
formal
dot)
charge
May
2011
[2]
E N E R G E T I C S
A N D
15
T H E R M O C H E M I S T R Y
( A H L )
Introduction
In
sub-topic
compounds
lattice
cycles,
the
4.1
structures.
derived
calculation
determined
While
we
form
some
examined
and
This
from
of
their
topic
Hess’s
energy
directly
from
reactions
go
how
begins
Law
values
with
(topic
that
empirical
to
reach
ionic
three-dimensional
energy
5)
be
evidence.
completion,
others
as
associated
disorder,
for
cannot
equilibrium
products,
behind
Gibbs
with
can
and
of
7.
and
Entropy,
randomness
as
chemical
enables
the
reactants
topic
considered
energy
spontaneity
in
molecular
be
physical
free
between
discussed
the
or
driving
changes
chemists
to
force
and
assess
the
reaction.
15.1 Ee  les
Understandings
Applications and skills
+
➔
Representative equations (eg M
(g) →
➔
Construction
of
Born– Ha ber
cy cles
for
+
M
(aq)) can be used for enthalpy/energy of
group
1
and
2
oxides
a nd
chlorid es .
hydration, ionization, atomization, electron
➔
Construction
of
energy
cy cl es
from
hy dration ,
anity, lattice, covalent bond, and solution.
lat tice,
➔
and
solution
e ntha lpy.
For
example,
Enthalpy of solution hydration enthalpy and
dissolution
of
solid
Na OH
or
NH
Cl
in
water.
4
lattice enthalpy are related in an energy cycle.
➔
Calcu lation
Haber
Nature of science
➔
Relate
or
enthalpy
dissolution
size
hydration
➔
of
and
cha nges
energy
charge
of
from
Born–
cycles.
ions
to
lat tice
and
enthalpies.
Making quantitative measurements with
replicates to ensure reliability – energy cycles
allow for the calculation of values that cannot
be determined directly.
➔
Perform
include
laboratory
single
aqueous
ex periments
replacem ent
which
reactions
cou ld
in
solutions.
357
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
TOK
Models for nding enthalpy changes
Subject-specic terminology
of reaction
with precise denitions
We
summarize
many
chemical
and
biological
processes
in
a
single
plays a central role within
chemical
equation
representing
one
pathway
from
reactants
to
the areas of knowledge:
products.
These
processes
in
reality
often
take
place
involving
several
the natural sciences,
reactions
with
alternative
pathways.
We
can
use
these
alternative
mathematics, history,
pathways
to
determine
changes
in
enthalpy
values
that
cannot
be
human sciences, the ar ts,
measured
directly.
ethics, and religious and
indigenous knowledge
In
systems. Communication
taking
using this specic
models
terminology may shape
reactants
understanding of concepts
Using
within these areas of
methodology
doing
this
place
are
we
develop
within
in
a
models
system.
agreement
that
are
empirical
broken
data
in
to
that
We
with
and
can
represent
assess
empirical
bonds
conrm
or
data
that
are
modify
the
the
energy
extent
by
to
focusing
formed
proposed
to
changes
which
on
bonds
make
models
these
in
products.
is
a
central
science.
knowledge. How impor tant
is this terminology and what
role does it play?
The Born–Haber cycle and enthalpy of formation
The
standard
represented
enthalpy
by
a
single
of
formation
of
an
ionic
compound
can
be
equation:
Stu ti
1
Na(s)
+
You need to know the individual
steps of the Born–Haber cycle
in order to apply the method to
a variety of situations. However,
1
Cl
An
a
application
series
of
(g)
→
of
ΔH
of
an
=
-411
kJ
mol
f
Hess’s
reactions
formation
NaCl(s)
2
2
law
that
ionic
(sub-topic
can
be
5.2),
combined
to
the
Born–Haber
determine
the
cycle,
enthalpy
is
of
compound.
rather than writing these steps
for specic elements and
compounds it is permissible to
Constructing a Born–Haber cycle
write representative equations
The
when answering questions, for
several
Born–Haber
steps
in
cycle
the
combines
formation
of
the
an
enthalpy
ionic
changes
compound,
associated
dened
with
below.
+
example, M(g) → M
(g) + e
.
This approach will be used in
Lattice enthalpy
this chapter.
The
lattice
occurs
on
enthalpy
the
is
formation
dened
of
1
as
mol
the
of
standard
gaseous
enthalpy
ions
from
the
change
solid
that
lattice:
+
MX(s)
→
M
(g)
+
X
(g)
∆H
>
0
lat
Lattie ethalies are often
quoted as negative values
The
process
298K
can
is
be
endothermic.
found
in
Experimental
section
18
of
the
values
Data
of
lattice
enthalpy
at
booklet.
that represent the exothermic
formation of the lattice from
Enthalpy of atomization
gaseous ions. However, in this
The
enthalpy
of
atomization
ΔH
is
the
standard
enthalpy
change
at
textbook we shall consider the
that
occurs
on
the
formation
of
1
opposite process (endothermic
element
in
its
standard
state:
formation of gaseous ions from
M(s)
→
M(g)
ΔH
lattice), which is consistent
with the denition given in the
X
2
Data booklet
358
>
0
>
0
at
1
(g)
2
→
X(g)
ΔH
at
mol
of
separate
gaseous
atoms
of
an
15 . 1
E n E r g y
c y c L E S
Ionization energy
As
introduced
in
topic
3,
the
ionization
energy,
ΔH
,
is
the
standard
IE
enthalpy
1mol
ions
of
change
atoms
with
third
that
or
positively
multiple
ionization
occurs
valence
energies
on
the
removal
charged
ions
electrons
are
the
in
of
1
the
rst,
mol
of
gaseous
second,
electrons
phase.
and
from
For
metal
sometimes
dened.
+
IE
:
M(g)
→
M
(g)
+
e
∆H
+
IE
>
0
>
0
IE
1
:
M
2+
(g)
→
M
(g)
+
e
∆H
IE
2
2
Electron anity
The
electron
afnity,
∆H
,
is
the
standard
enthalpy
change
EA
on
the
addition
gaseous
of
1
mol
of
electrons
to
1
mol
of
atoms
in
the
phase:
X(g)
+
e
→
X
(g)
∆H
<
0
EA
As
discussed
are
in
exceptions,
topic
such
3,
as
electron
the
afnity
electron
is
typically
afnity
for
negative,
but
there
helium.
Constructing the Born–Haber cycle
The
lattice
and
the
and
nd
enthalpy,
electron
the
Born–Haber
energy
afnity
enthalpy
cycle
relationships
each
the
of
enthalpy
are
1)
the
atomization,
combined
formation
(gure
between
of
of
focuses
on
individual
to
an
the
construct
ionic
the
steps
ionization
the
Born–Haber
compound.
processes
rather
The
involved
than
energy,
the
cycle
standard
and
the
magnitude
of
change.
+
M
∆H
(g)
+
X(g)
+e
(X)
at
∆H
(X)
EA
Stu ti
1
+
M
X
(g)
2
(g)
e
2
Values for lattice enthalpies
can be found in the Data
+
∆H
M
(M)
+
(g)
X
(g)
booklet (section 18), along
IE
with enthalpies of aqueous
solutions (section 19), and
1
+
M(g)
X
2
(g)
2
enthalpies of hydration
∆H
lat
(section 20) which will be used
∆H
later in this topic. The Data
(M)
at
booklet will be available during
∆H
f
1
+
X
2
the examination, except in
(g)
2
Paper 1.
Figure 1 A generalized Born–Haber cycle
359
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
Worked examples
Example 1
Write
an
Example 2
equation
formation
of
Born–Haber
for
the
potassium
cycle
to
enthalpy
bromide.
calculate
change
Construct
the
lattice
of
Use
a
the
the
Born–Haber
enthalpy
of
cycle
in
formation
gure
for
3
to
calculate
magnesium
oxide.
enthalpy
2+
Mg
for
this
+
(g)
O(g)
+2e
compound.
Solution
1
∆H
(O)
+249 kJ mol
at
∆H
K(s)
+
1
(O)
141 kJ mol
EA(1)
1
Br
(l)
→
KBr(s)
2
2
1
2+
Mg
O
(g)
2e
(g)
2
2
+
K
+
(g)
Br(g)
+e
2+
Mg
+
(g)
O
(g)
1
∆H
(Mg)
+1451 kJ mol
IE(2)
1
∆H
(Br)
∆H
+97 kJ mol
1
∆H
(Br)
325 kJ mol
+798 kJ mol
1
+
Mg
+
(g)
O
EA
+e
(g)
2
2
1
+
K
1
(O)
EA(2)
at
Br
(g)
e
(l)
2
2
2+
1
∆H
(Mg)
Mg
+738 kJ mol
2
+
(g)
O
(g)
IE(1)
+
1
∆H
(K)
K
+419 kJ mol
+
(g)
Br
(g)
IE
1
+
Mg(g)
O
(g)
2
2
1
∆H
1
(MgO)
+3795 kJ mol
lat
+
K(g)
Br
(l)
2
2
1
∆H
x kJ mol
lat
1
∆H
(Mg)
+148 kJ mol
at
1
∆H
(K)
+89 kJ mol
∆H
at
(MgO)
f
1
+
O
(g)
2
2
1
x kJ mol
∆H
f
1
+
Br
(l)
2
2
1
Figure 3 Born–Haber cycle to calculate the enthalpy of
392 kJ mol
formation for magnesium oxide
Figure 2 Born–Haber cycle to calculate the lattice enthalpy for
potassium bromide
Solution
To
for
determine
the
potassium
(endothermic)
bromide,
follow
lattice
the
enthalpy
pathway
∆H
(MgO)
=
∆H
f
(Mg)
+
∆H
at
+
on
(Mg)
IE(1)
∆H
(Mg)
+
∆H
IE(2)
gure
2
and
add
up
the
values
for
the
enthalpy
+
∆H
-
∆H
(O)
+
∆H
EA(1)
changes
shown,
taking
note
of
their
sign.
(O)
at
(O)
EA(2)
(MgO)
lat
∆H
(KBr)
=
∆H
f
(K)
+
∆H
at
+
(K)
=
IE
∆H
(Br)
+
∆H
at
(Br)
+
∆H
EA
(148
+
(KBr)
(
+
738
141)
+
+
1451
798
+
249
3795)
lat
1
-
392
=
89
=
-
+
419
+
97
+
[
325]
+
=
x
-
552
kJ
mol
1
672
kJ
mol
Figure 4 The lattice structure of crystalline magnesium oxide
2+
Grey: Mg
360
2
Red: O
15 . 1
E n E r g y
c y c L E S
collaboatio i the sieti ommuit
“ Why is it impor tant for countries to collaborate to combat global problems like
global warming?” Chemistry Syllabus sub-topic 15.1
The scientic community brings together various scientic disciplines and also other
elds such as engineering, technology, and mathematics. Some notable examples
of international collaboration include the Human Genome Project, CERN, and the
Manhattan project.
Scientists look to the past and the future to understand the patterns in the Ear th’s
climate. The Intergovernmental Panel on Climate Change (IPCC) was established
in 1988 by the United Nations Environment Programme (UNEP) in conjunction
with the World Meteorological Organization (WMO), to coordinate data collection,
independently analyse it, and publish repor ts. Complex models make projections
about the Ear th’s future climate using indirect indicators of global warming. These
include data from ice cores, cores from ancient coral formations, ocean and lake
sediments, borehole temperatures, evaporation and precipitation cycles, glacial
recession patterns, and receding polar ice caps.
More accurate data results from linking repeated measurements on a global scale.
Global collaboration is essential to investigate the causes and eects of global
warming.
Variations in lattice enthalpy values
The
by
magnitude
both
An
the
increase
oppositely
the
For
ionic
the
17
lattice
on
the
ionic
is,
the
from
this
is
and
the
lattice
radius
of
the
by
in
anion
iodine.
a
compound
a
greater
of
the
directly
attraction
required
(table
affected
the
to
as
in
you
distance
electrostatic
decrease
break
between
apart
1).
increases
As
is
radii.
energy
enthalpy
strength
reected
a
ionic
result
the
the
to
for
the
will
increasing
uorine
increases,
and
enthalpy
ions
charge
ions,
that
halides,
ions
decreases
(table
in
group
bonded
the
charged
lattice,
metal
down
of
charge
the
move
between
the
attraction
lattice
enthalpy
2).
∆H
(theoetial value)/
Vaiatio i hae a ioi
lat
1
aius omae with nacl
kJ mol
MgO
3795
greater ionic charge
NaCl
769
—
KBr
671
increased ionic radius
T
able 1 Variations in the lattice enthalpy with ionic radius and ionic charge
1
∆H
(theoetial value)/ kJ mol
lat
NaF
910
NaCl
769
NaBr
732
NaI
682
T
able 2 Lattice enthalpy of metal halides
361
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
Enthalpy changes in solution
Many
reactions
therefore
between
the
The
to
the
lattice
studied
consider
in
the
enthalpy
chemistry
enthalpy
change
of
take
of
place
in
solution
solution,
the
solution.
and
the
hydration
It
is
useful
relationship
enthalpy,
and
enthalpy.
standard
enthalpy
change
of
solution ,
∆H
,
is
the
change
in
sol
Enthalpy changes of solution
enthalpy
when
1
mol
of
a
substance
is
dissolved
in
a
large
excess
of
a
can have either positive or
pure
solvent:
negative values.
1
+
NH
Cl(s)
→
NH
4
(aq)
+
Cl
(aq)
∆H
=
+14.78
kJ
mol
=
-48.83
kJ
mol
sol
4
+
LiBr(s)
→
1
(aq)
Li
+
Br
(aq)
∆H
sol
It
is
or
possible
by
ionic
using
solid
to
an
and
calculate
energy
the
the
cycle
enthalpy
that
subsequent
change
involves
hydration
the
of
solution
lattice
enthalpy
empirically,
enthalpy
of
the
of
the
gaseous
ions
produced.
The
enthalpy
change
of
hydration ,
∆H
,
for
an
ion
is
the
enthalpy
hyd
The enthalpy change of
change
when
1
mol
of
the
gaseous
ion
is
added
to
water
to
form
a
dilute
hydration always has a negative
solution.
The
term
solvation
is
used
in
place
of
hydration
for
solvents
value.
other
than
water.
+
M
+
(g)
→
M
1
(aq)
∆H
=
-
kJ
mol
hyd
1
X
(g)
→
X
(aq)
∆H
=
-
kJ
mol
hyd
Worked example
Find
the
enthalpy
change
of
solution
∆H
for
sodium
hydroxide
sol
using
the
enthalpy
cycle
in
gure
5.
+
Na
(g) +
OH
(g)
1
∆H
=
(
424)
+
(
519) kJ mol
hyd
1
∆H
=
+900 kJ mol
lat
+
Na
(aq) +
OH
(aq)
1
∆H
=
x kJ mol
sol
NaOH(s)
Figure 5 Enthalpy cycle to calculate the enthalpy change of solution for sodium
hydroxide
Solution
+
∆H
=
sol
∆H
(NaOH)
+
∆H
lat
(Na
)
+
∆H
hyd
1
=
900
=
-43
+
(
424)
-1
362
kJ
mol
+
(
519)
(OH
hyd
kJ
mol
)
15 . 1
E n E r g y
c y c L E S
Solvation, dissolution, and hydration
+
δ
+
δ
Three
terms
commonly
used
when
describing
the
interactions
between
+
solvents
briey
and
solutes
and
the
subsequent
solutions
formed,
are
described
δ
δ
here:
+
δ
δ
Solvation
is
interaction
described
of
a
solute
by
the
and
IUPAC
the
Gold
solvent
or
Book
a
as
“any
similar
stabilizing
interaction
of
+
solvent
δ
δ
with
groups
of
an
insoluble
material.
Such
interactions
generally
involve
+
δ
+
electrostatic
specic
forces
effects
and
such
as
van
der
Waals’
hydrogen
forces,
bond
as
well
as
chemically
δ
more
+
formation.”
δ
+
δ
Water
is
oxygen
a
polar
and
solvent.
hydrogen,
molecule
(bent)
electrons
on
due
the
negative
on
hydrogen
the
their
partial
shell,
also
(gure6).
new
The
repulsive
atom,
on
Water
as
a
phase
is
the
in
oxygen
cations
polar
shell
and
is
known
lone
the
solvent
with
a
charges
so
that
solvation
is
water
liquid
solution.
positively charged sodium ion
of
having
themselves
the
(solvent) molecules surround a
pairs
positive
forming
mixed
as
the
Figure 6 Solvation shell: the water
water
molecule
anions,
when
between
the
partial
orientate
and
of
between
this
hydroxide
formed
geometry
atom
molecules
hydration
sodium
electronegativity
forces
result
the
surround
solid
with
in
water,
This
is
a
the
dissolution
enthalpy
released
of
known
When
of
the
atoms.
homogeneous
process
to
charges
charges
difference
combined
oxygen
partial
The
of
during
hydration
is
hydration
the
is
process
inuenced
a
of
by
way
of
quantifying
solvation.
the
charge
1
catio
∆H
the
amount
The
magnitude
and
size
of
the
of
ion
energy
enthalpy
(table
3).
1
Aio
/kJ mol
of
the
∆H
h
/kJ mol
h
+
Li
538
F
504
424
Cl
359
1963
Br
328
+
Na
2+
Mg
3+
Al
4741
I
287
T
able 3 Enthalpies of hydration (more data is available in section 20 of the Data booklet)
As
you
move
down
hydration
decreases
hydration
enthalpy
group
17.
decrease
For
in
a
group
as
in
cations,
size
results
the
in
group
an
in
the
ionic
1
while
increase
a
periodic
radius
in
table
the
increases.
uorine
charge
signicantly
has
on
larger
enthalpy
Lithium
the
the
has
highest
ion
of
the
combined
enthalpy
of
greatest
value
in
with
a
hydration.
Quik questio
Calculate the enthalpy change of solution of barium chloride given the
following data:
2+
BaCl
(s) → Ba
1
(g) + 2Cl
2
2+
Ba
(g)
∆H
= +2069 kJ mol
lat
2+
(g) →
Ba
1
(aq)
∆H
= -1346 kJ mol
hyd
1
Cl
(g) → Cl
(aq)
∆H
= -359 kJ mol
hyd
363
15
E n E r g E T I c S
A n d
T H E r M O c H E M I S T r y
( A H L )
15.2 Eto a sotaeit
Understandings
Applications and skills
➔
Entropy (S) refers to the distribution of
Prediction of whether a change will result in an
➔
available energy among the par ticles. The more
increase or decrease in entropy, by considering
ways the energy can be distributed the higher
the states of the reactants and products.
the entropy.
Calculation of entropy changes (ΔS) from
➔
➔
Gibbs free energy (G) relates the energy that
standard entropy values (S
).
can be obtained from a chemical reaction to the
Application of ΔG
➔
= ∆H
-
T∆S
in
change in enthalpy (ΔH), change in entropy
predicting spontaneity and calculation of
(ΔS), and absolute temperature (T).
various conditions of enthalpy and temperature
➔
Entropy of gas > liquid > solid under the same
that will aect this.
conditions.
Relation of ΔG to position of equilibrium.
➔
Nature of science
➔
Theories can be superseded – the idea of entropy has evolved through the years as a result of
developments in statistics and probability.
Sotaeous haes
Chemists
work
reactions
will
systems
A
to
reaction
is
said
or
intervention.
may
place
be
a
enthalpy
just
one
The
and
in
conservation
of
energy
to
the
system.
the
will
of
the
(S)
and
movement
us
of
to
it
set
chemical
control
moves
of
chemical
can
exothermic.
are
occur
be
external
different
that
do
rates
not
take
non-spontaneous
positive
examining
usually
either
without
at
Reactions
to
whether
when
are
said
towards
conditions
or
the
negative,
is
spontaneity
spontaneous
thermodynamics
focuses
Spontaneous
which
and
but
there
of
a
are
rule.
concerns
to
when
spontaneous
or
applications
is
under
modify
given
reaction,
a
of
on
The
of
original
reactions
predict
the
greater
to
the
shift
and
the
of
the
gain
total
the
law,
the
of
the
total
higher
in
an
chemical
being
the
the
the
of
chance
entropy
total
reaction.
change
of
of
entropy
understanding
entropy
of
second
available
energy
lower
law
The
spontaneity
increase
can
fundamental
world.
from
particles,
an
of
rst
distribution
we
direction
are
The
physical
and
state
If
quantify
the
the
the
the
lead
surroundings.
so
in
entropy
amongst
their
and
chemistry.
energy
measure
widespread
system
allow
a
particles.
returning
a
conditions
laws
energy,
being
particles
this
of
are
reactions
this
practical
Entropy
the
freedom
364
of
second
between
localized
within
to
can
outcomes.
under
that
of
thermodynamics
reactions.
desired
considered
Exothermic
exceptions
importance
law
be
conditions
they
spontaneous
set
change
to
the
that
endothermic
given
aspect
rst
be
Reactions
The
many
to
the
so
equilibrium
either
under
reaction.
understand
achieve
completion
and
to
proceed,
for
a
of
this
system,
15 . 2
E n T r O p y
A n d
S p O n T A n E I T y
Changes in entropy
Figure
1
a
containing
glass
shows
difference
and
the
that
glass)
condensation
iced
exists
and
water.
on
between
the
the
The
outside
that
of
gas
temperature
the
system
surroundings
(iced
water
under
is
greater
than
the
system)
results
in
thermal
energy
from
to
and
the
surrounding
glass
equilibrium.
entropy
of
its
With
the
contents,
this
until
thermal
water/ice
they
energy
mixture
of
a
of
a
the
liquid
entropy
which
in
of
a
turn
is
solid.
S
is
is
a
state
function,
determined
by
the
so
a
change
difference
in
between
atmosphere
its
the
that
conditions,
that
being
entropy
transferred
same
than
(everything
Entropy,
outside
the
greater
will
reach
nal
and
initial
values:
an
transfer,
increase
∆
the
(reaction)
=
∑∆
(products)
298
298
while
∑∆
(reactants)
298
the
entropy
of
the
surroundings
will
decrease
as
The
energy
is
transferred
from
it.
The
condensed
conditions
entropy
on
the
surface
of
the
glass
is
lower
in
entropy
water
vapour
in
the
change,
be
and
specied
the
for
subscript
a
particular
“298”
refers
to
than
a
the
must
water
temperature
of
298
K.
atmosphere.
The
second
chemical
in
the
law
reactions
entropy
When
the
of
the
of
rather
in
which
it
∆S
is
=
in
the
than
is
of
in
∆S
an
are
the
+
overall
that
increase
spontaneous.
universe
is
found
a
describes
(table
says
equilibrium.
describing
opposite
written
in
universe
this
the
total
result
universe
entropy
non-spontaneous,
spontaneous
that
system
entropy
negative,
thermodynamics
the
overall
unchanged,
overall
of
remains
If
to
reaction
a
reaction
direction
to
the
be
that
is
that
the
is
way
1).
∆S
system
surroundings
Figure 1 Changes in entropy are associated with every
chemical and physical process
∆S
spontaneous
> 0
total
∆S
Predicting changes in entropy
Simple
representations
of
particles
in
the
equilibrium
= 0
total
∆S
different
non-spontaneous
< 0
total
states
of
matter
show
an
increasing
entropy
as
T
able 1 The second law of thermodynamics allows us to
the
particles
gain
more
freedom
of
movement
predict the direction of a reaction
and
more
particles
ways
move
of
distributing
from
solids
the
energy
through
as
liquids
the
to
gases.
An
increase
in
the
system
the
particles,
heat
will
energy
result
leading
in
to
(enthalpy)
increased
greater
within
movement
disorder
and
of
an
increase in entropy (S)
increase
liquid
the
the
spontaneity
gas
in
entropy
changes
●
solid
in
the
both
Exothermic
of
of
a
reactions
as
this
enthalpy
greater
are
system.
and
chemical
spontaneous,
and
the
enthalpy
affect
reaction.
more
leads
Therefore,
entropy
to
a
stability
likely
to
be
reduction
of
the
in
reaction
products.
Figure 2 Entropy increases from the solid through to the liquid
●
An
increase
in
entropy
makes
reactions
more
to the gaseous phase
likely
leads
Achieving
a
change
of
state
from
solid
to
liquid
to
to
is
sometimes
absorbed
particles
which
described
results
increasing.
In
in
in
terms
the
terms
of
kinetic
of
energy
energy
entropy
we
spontaneous,
more
uniform
as
greater
distribution
disorder
of
energy
to
within
gas
be
the
system.
being
of
can
the
say
This
will
be
revisited
in
greater
depth
later
in
thistopic.
365
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
reteati laies: Iteatioal ollaboatio
The impor tance of being able to obtain data from around the globe is highly
signicant. Glacier recession is an indirect indicator of global warming. Why is
it impor tant for countries to collaborate to combat global problems like global
warming? Chemistry Syllabus sub-topic 15.1
Glaciers around the world are retreating to higher altitudes as global temperatures
steadily rise. Ice cores from glaciers provide vital information to climate scientists,
enabling them to build up a picture of the Ear th’s climate and signicant
events, such as volcanic eruptions, over thousands of years. Studies into the
consequences of the melting of glaciers have stimulated widespread discussions
within the media, government, and scientic organizations.
Figure 3 Jostedalsbreen glacier, Nor way.
calulati eto haes
The
entropy
change
thermodynamic
The
standard
temperature
To
calculate
nd
the
total
and
of
of
a
which
entropy
standard
to
●
be
can
provided
values,
entropy
between
the
S
,
be
in
calculated
section
relate
to
12
from
of
the
standard
Data
booklet.
conditions
(reaction)
the
change
total
associated
entropy
of
=
∑∆S
(products)
-
performing
the
∑∆S
298
with
a
reaction
products
and
entropy
(reactants)
298
change
calculations
the
following
that
values
for
entropy
are
specic
for
points
different
1
matter,
for
example,
70.0
The
J
K
S
(H
O(g))
=
188.8
J
K
mol
states
of
while
S
(H
O(l))
2
1
mol
coefcients
molar
366
need
1
2
1
●
we
the
considered:
Remember
=
of
reactants:
298
When
system
is
pressure.
difference
entropy
∆S
molar
the
ΔS
data
entropy
used
values
to
balance
when
the
equation
calculating
the
must
overall
be
applied
entropy
to
change.
15 . 2
Examine
●
reaction
degree
be
the
to
of
used
chemical
have
disorder
to
reaction
positive
check
in
or
the
your
and
predict
negative
products
nal
whether
entropy
and
you
change
reactants.
E n T r O p y
expect
based
This
on
A n d
S p O n T A n E I T y
the
the
prediction
can
calculation.
Stu ti
Worked example
Standard molar entropy has the unit
Calculate
the
standard
entropy
change
for
the
following
reactions:
1
J K
1
a)
H
(g)
1
+
O
2
(g)
→
H
2
2
S
O(l)
(H
298
)
130.7
J
for standard enthalpy of formation. These
1
b)
NH
Cl(s)
→
NH
4
(g)
+
S
HCl(g)
(O
298
1
; compare this with kJ mol
mol
2
2
1
mol
1
K
)
205.1
J
K
1
values are combined in Gibbs free energy
mol
2
3
calculations (see later in this topic). When
combining these quantities be sure to
Solution
convert units appropriately.
1
a)
H
(g)
+
O
2
(g)
→
H
2
2
O(l)
2
1
∆S
=
[∆S
298
(H
298
O)]
[∆S
(H
298
2
)
+
2
∆S
(O
298
2
)]
2
1
=
[70.0]
[130.7
+
×
205.1]
2
1
=
The
-163.3
negative
J
entropy
K
change
associated
with
this
chemical
reaction
1
indicates
a
decrease
in
disorder
(greater
order),
with
1
mol
of
gas
in
large
2
changing
b)
NH
into
Cl(s)
1
mol
→
NH
4
of
(g)
a
liquid.
+
HCl(g)
3
∆S
=
[∆S
298
(NH
298
)
+
∆S
=
[192.5
+
=
+284.55
(HCl)]
-
[∆S
298
3
186.9]
(NH
298
Cl)]
4
[94.85]
1
Transforming
increase
data
in
can
1
mol
of
disorder,
be
found
J
K
a
solid
hence
in
into
the
section
2
mol
large
12
of
of
a
positive
the
Data
gas
results
entropy.
a
Thermodynamic
Booklet.
Quik questios
1
Predict whether the following reactions will have a
positive or negative entropy change, ∆S
a)
NH
NO
4
(s) → N
3
O(g) + 2H
2
.
Table 2 shows the standard entropy values of the
substances in the reaction above.
O(g)
Substae
2
CO
(g)
2
b)
N
(g) + 3H
2
)
N
O
2
)
(g) → 2NH
2
(g) → 2NO
4
CaCO
2C
H
2
2
1
S
214
131
(g)
H
4
186
O(g)
2
189
(g)
Calculate the standard entropy change for the
2
6
1
mol
CH
Table 2
(s) → CaO(s) + CO
(g) + 7O
/J K
(g)
2
(g)
2
3
e)
(g)
3
H
(g) → 4CO
2
(g) + 6H
2
O(l)
reaction.
[3]
2
The equation for the reaction between carbon dioxide
Explain how the sign can be predicted from the
equation for the reaction.
and hydrogen is shown below.
[2]
IB, nov 20 07
CO
2
(g) + 4H
2
(g) → CH
4
(g) + 2H
O(g)
2
367
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
Sustaiable ee
“Sustainable energy is a UN initiative with a goal of doubling of global sustainable
energy resources by 2030.” Chemistry Syllabus sub-topic 15.1
“Sustainable energy for all” is a United Nations (UN) initiative that aims to reduce the
inequalities that exist in the provision of clean and ecient energy services, improve
energy eciency thereby reducing energy demand, and increase the proportion of
energy that comes from renewable resources (http://www.sustainableenergyforall.
org/objectives/universal-access). The project has three main objectives:
1
Ensure universal access to modern energy services which focuses on
improving the lives and economic conditions of people throughout the world.
Approximately one-fth of the world’s population do not have electricity in
their home and almost 40 per cent utilize fuel sources such as animal waste,
charcoal, and wood to provide heat for cooking. Toxic products from this form
of combustion result in the deaths of over 2 million people annually, mainly
women and children. “Electricity enables children to study after dark. It enables
water to be pumped for crops, and foods and medicines to be refrigerated.”
2
Energy eciency is the par t of the project that looks at countries, including
the way in which we use power. From industry to households, oce and
accommodation buildings to transpor tation, lighting to electrical appliances,
a variety of people, agencies, and governments are being encouraged to
both educate and legislate, with the aim of decreasing the global electricity
demand. Energy-saving light bulbs, energy-ecient televisions, buildings
that require less energy to heat and cool, and the use of information
technology in industry to better manage power usage are all examples of
how the global community is reducing the demand for power. This ultimately
saves governments, individuals, and businesses money and lessens the
impact of coal-red power stations on the environment.
3
Renewable
renewable
The
cost
of
energy
energy
the
development
appreciably
over
governments,
renewable
–
the
has
of
can
dams
in
play
set
to
a
target
global
renewable
decades
businesses,
energy
hydroelectric
UN
contributed
and
and
a
Brazil
energy
now
role
generate
doubling
in
sources
Where
power
83% of
the
production
represents
individuals.
major
of
energy
a
has
share
by
decreased
viable
option
resources
generation.
the
of
2030.
country’s
are
For
for
available,
example,
electricity.
Figure 4 Itaipu dam, built between Brazil and Paraguay, is the second largest
hydroelectric power plant in the world
368
15 . 2
E n T r O p y
A n d
S p O n T A n E I T y
gibbs fee ee
The
H,
Gibbs
free
entropy
S,
importance
now
look
of
at
a
G
of
is
a
absolute
entropy
the
temperature
For
energy
and
in
state
dening
relationship
the
spontaneous
function,
temperature
the
T.
along
with
Having
spontaneity
between
total
enthalpy
established
of
entropy,
a
the
reaction,
enthalpy,
we
and
shall
the
system.
Reactions that are
reaction:
spontaneous and are
∆S
=
∆S
+
total
∆S
>
sys
0
surroundings
therefore thermodynamically
favourable can sometimes be
A
chemical
reaction
may
be
either
exothermic
or
endothermic:
the
kinetically improbable, due
transfer
of
heat
across
the
system/surroundings
boundary
is
directionally
to the existence of very high
dependent
on
the
change
in
enthalpy.
For
an
exothermic
reaction
in
an
activation energies (see
open
system,
heat
is
transferred
from
the
system
to
the
surroundings.
sub-topic 16.2).
This
results
The
the
impact
an
that
low
and
energy
to
melt
in
a
as
each
the
has
will
of
to
G
=
H
=
Gibbs
reaction
-
a
a
change
at
high
60
less
state
of
a
the
reaction
two
The
a
the
level
transfer
water
marked
entropy,
function
as
the
a
on
the
the
the
in
The
entropy
the
will
increases,
ice
the
so
at
of
ice
amount
ice
begin
water
additional
level
of
entropy.
temperature
of
system
Gibbs
free
at
of
resulting
hot
the
of
system.
one
block
same
However,
with
and
called
of
molecules
effect
on
systems,
such
effect.
entropy.
compared
has
existing
separate
different
of
surroundings.
conditions
temperature,
°C.
of
the
of
the
into
have
enthalpy,
new
on
disorder
much
of
at
will
in
entropy
energy
energy
change
energy,
can
be
G:
TS
∆H
free
-
T∆S
energy
system
spontaneity.
have
one
signicant
dene
∆G
heat
water
kinetic
a
the
enthalpy
system
have
in
dependent
and
combination
used
The
bowl
signicant
energy
The
a
into
already
is
transferring
temperature
°C
increase
the
surroundings
Imagine
0
in
at
For
negative
a
provides
constant
reaction
value
( ∆G
an
effective
temperature
to
be
<
0).
way
and
spontaneous
∆H
∆S
positive (> 0):
positive (> 0):
endothermic
more disorder
of
focusing
pressure
the
to
Gibbs
on
a
determine
free
energy
its
must
∆G
Sotaeit
Exlaatio
negative at high T
dependent on
spontaneous only at high
positive at low T
temperature
temperatures when
T∆S > H
positive (> 0):
negative (< 0):
endothermic
more order
always positive
never
reverse reaction
> 0
spontaneous
spontaneous at all
temperatures
negative (< 0):
positive (> 0):
exothermic
more disorder
always negative
always
forward reaction
< 0
spontaneous
spontaneous at all
negative at low T
dependent on
spontaneous only at low
positive at high T
temperature
temperatures when
temperatures
negative (< 0):
negative (< 0):
exothermic
more order
T∆S < H
T
able 3 Factors aecting ∆G and the spontaneity of a reaction
369
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
It
is
be
not
always
possible
spontaneous
increasing
not
with
to
will
of
3).
always
whether
be
>
0.
The
chemical
reactions
spontaneous,
increasing
the
a
Exothermic
of
∆G
temperature
predict
(table
reactions
spontaneous,
the
or
disorder
endothermic
on
( A H L )
order
will
spontaneity
with
∆G
always
of
reaction
that
other
be
will
involve
<
0.
Similarly,
non-
reactions
depends
system.
gibbs fee ee hae of fomatio
The
Gibbs
free
energy
change
of
formation ,
ΔG
,
represents
the
free
f
energy
under
change
when
standard
ΔG
=
1
mol
conditions
ΣΔG
r
of
of
(products)
a
compound
298
-
ΣΔG
Gibbs
free
energy
change
of
a
formed
pressure
from
of
100
its
elements
kPa:
(reactants)
f
from ΔG
r
the
and
f
Worked example: nding ΔG
Calculate
K
is
values
f
reaction,
Solution
ΔG
for
the
combustion
of
ethanol,
C
r
H
2
give
CO
(g)
and
H
2
OH
to
5
C
O(g).
H
2
OH(l)
+
3O
5
(g)
ΔG
=
ΣΔG
r
Substae
C
H
2
OH(l)
H
5
O(g)
CO
2CO
(g)
+
3H
2
O(g)
2
2
(products)
-
ΣΔG
f
(reactants)
f
(g)
2
=
[2ΔG
(CO
f
∆g
→
2
2
)
+
3ΔG
(H
f
2
O)]
2
values/
[ΔG
f
(C
f
175
228.6
H
2
394.4
OH)]
5
1
kJ mol
=
[2
×
-394.4
=
-1299.6
+
3
×
-228.6]
[
175]
1
Substae
kJ
mol
1
∆g
/kJ mol
f
Quik questio
SO
(g)
371.1
3
Calculate the Gibbs free energy change for the following reactions. Values for
H
SO
2
(l)
690.0
4
can be found in section 12 of the Data booklet; additional data is listed in
∆G
f
NH
Cl(s)
202.9
table 4.
4
CaCO
(s)
1129.1
a)
SO
CaO(s)
604.0
(g) + H
3
3
b)
O(l) → H
2
2NH
SO
2
(l)
4
Cl(s) + CaO(s) → CaCl
4
CaCl
(s)
748.1
)
2
NH
C
H
2
(g)
(s) + H
2
(g) + H
4
O(l) → C
2
H
2
O(l) + 2NH
2
(g)
3
OH(l)
5
16.5
3
T
able 4 ∆G
values not found in
f
section 12 of the Data booklet
calulati the gibbs fee ee hae of
a eatio fom ethal a eto ata
To
determine
need
to
standard
calculate
spontaneity
the
conditions
formation
370
the
calculate
Gibbs
(298
of
free
K
and
of
reactants
and/or
the
enthalpy
and
a
reaction
energy
100
kPa).
products
entropy
from
change
If
are
∆G
for
the
=
the
Gibbs
unknown,
changes
for
the
∆H
reaction
free
we
T∆S
energies
need
reaction.
,
we
under
rst
of
to
15 . 2
Worked example: calculating ∆G
Standard
2C
enthalpy
H
2
(g)
H
+
O
(g)
combustion
4CO
(g)
+
→
2H
on
3O
the
(g)
→
6H
O(l)
are
from ∆H
given
H
2
(g)
=
-3120
∆H
=
-572
2CO
(g)
+
2H
ΔH
=
-1411
kJ
kJ
information,
change
in
O(l)
C
6
H
2
(g)
kJ
2
b)
calculate
enthalpy,
∆H
,
for
The
an
the
sign
+
H
4
for
increase
number
reaction:
→
the
of
the
in
stating
disorder
moles
or
c)
for
the
is
gas
entropy
evident
is
as
increases
positive:
the
from
1
to
2
(g)
a
reason,
whether
the
sign
∆S
=
[∆S
(C
298
of
[∆S
above
reaction
would
be
H
2
(C
298
∆S
of
in
reaction.
298
Predict,
change
2
c)
b)
- T∆S
below:
∆H
in
C
S p O n T A n E I T y
2
2
above
standard
following
+
O(l)
2
Based
(g)
2
reactions
A n d
2
4
the
→
2
2
a)
(g)
of
2
2
C
7O
6
(g)
2H
+
change
E n T r O p y
)(g)
+
∆S
2
(H
298
4
H
)(g)]
2
)(g)]
6
positive
=
[220
=
120
+
131]
[230]
negative.
Calculate
the
standard
entropy
change
for
J
K
1
the
reaction.
d)
∆G
=
∆H
T∆S
=
+137
120
_
d)
Determine
the
value
of
∆G
for
the
reaction
(
298
×
)
1000
at
298
K.
=
e)
Determine
the
temperature
at
which
In
reaction
will
occur
+101
this
calculation,
Nov
2009
by
1000.
Gibbs
is
Solution
Rearrange
the
three
combustion
reactions
the
standard
The
rst
change
in
The
free
joules
entropy
positive
energy
examining
the
to
value
kilojoules
is
value
indicates
This
for
positive
the
that
can
by
the
be
value
dividing
change
in
reaction
predicted
for
change
to
in
nd
from
non-spontaneous.
by
a)
the
spontaneously.
converted
IB,
kJ
this
enthalpy
(endothermic)
and
the
low
enthalpy.
temperature.
equation
will
occur
in
the
same
e)
direction
but
only
half
of
the
stoichiometry
To
determine
reaction
needed
so
halve
the
enthalpy
the
temperature
at
which
this
is
will
occur
spontaneously,
we
make
value:
the
assumption
that
the
value
for
Gibbs
free
1
C
H
2
(g)
+
3
6
O
(g)
→
2CO
2
2
(g)
+
3H
2
O(l)
energy
2
=
∆H
-1560
second
equation
needs
to
be
reversed
zero
and
solve
for
T.
kJ
∆G
The
is
=
∆H
T∆S
0
=
∆H
T∆S
T
=
and
halved:
137
__
∆H
_
1
H
O(l)
→
H
2
The
(g)
+
2
third
O
equation
(g)
∆H
=
+286
needs
to
be
(g)
+
2H
2
O(l)
→
C
2
H
2
standard
∆H
C
H
2
=
these
-1560
(g)
6
of
enthalpy
(g)
+
→
C
1142
K
10
reaction
becomes
temperatures
(g)
greater
spontaneous
than
1142
at
K.
2
=
+1411
determines
kJ
the
change:
+
H
2
3O
4
equations
×
reversed:
∆H
Summation
=
3
120
∆S
The
2CO
=
kJ
2
2
286
(g)
4
+
+
1411
H
(g)
=
∆H
+
=
137
kJ
+137
kJ
2
371
15
E N E R G E T I C S
A N D
T H E R M O C H E M I S T R Y
( A H L )
gibbs fee ee a hemial equilibium
We
and
have
established
pressure
reversible
the
free
As
towards
the
From
is
Gibbs
which
5
we
time
reverse
reaches
a
can
the
As
is
in
between
constant
∆G
the
as
Gibbs
that
a
the
as
the
its
reaction
the
will
same
Gibbs
be
In
way
as
reactants
at
to
the
(A)
the
the
Gibbs
the
change
detail
in
products
moves
point
energy
point
of
of
a
topic
(B).
reaction.
G(reactants)
∆G
r
ygrene eerf sbbiG
A
G(products)
B
equilibrium
0 mol reactant
1 mol product
Figure 5 How the Gibbs free energy changes as the reaction proceeds
372
The
again
17.
0 mol product
free
equilibrium
reaction
1 mol reactant
5).
during
energy
forward
of
forward
increases,
free
(for
(gure
Gibbs
the
non-spontaneous
the
during
in
At
a
equilibrium,
equilibrium
free
then
when
reaction
or
region
temperature
time
reaches
proceeds
and
energy
examined
it
the
Gibbs
continues,
and
the
decreases.
this
becomes
free
constant
reactions)
minimum
reaction
of
and
reaction
spontaneous
the
ratio
energy
the
at
From
where
minimum
minimum.
a
0.
increases
free
reached
place
<
point
non-reversible
reaches
then
to
products
see
forward
minimum
equilibrium
has
energy
reaction
relationship
the
taking
when
changing
of
towards
favoured.
free
is
(for
system
decreases
reaction
the
the
gure
energy
energy
amount
reactions)
equilibrium
reactions
commences
completion
reversible
that
spontaneous
reaction
Gibbs
alters.
are
and
The
the
Q U E S T I O n S
Questios
1
The
be
in
lattice
enthalpy
calculated
gure
from
of
magnesium
the
chloride
Born–Haber
cycle
II)
can
The
shown
of
ionic
charge
of
lithium
is
less
than
that
calcium.
6.
A.
I
only
B.
II
C.
I
D.
Neither
IB,
May
2+
Mg
III
=
+738
+
1451
(g) +
2e
kJ
+
2Cl(g)
IV
2+
Mg(g) +
2Cl(g)
Mg(g) +
Cl
Mg
(g) +
2Cl
(g)
only
and
II
I
nor
II
[1]
2004
II
H
3
Which
reaction
occurs
with
the
largest
increase
(g)
2
∆H
I
=
+148
(MgCl
lat
kJ
in
entropy?
A.
Pb(NO
)
2
)
3
Mg(s) +
(s)
+
2KI(s)
→
PbI
2
(s)
+
2KNO
2
(s)
3
(g)
Cl
2
B.
CaCO
C.
3H
D.
H
(s)
→
CaO(s)
+
CO
3
V
=
-642
(g)
2
kJ
(g)
+
N
2
MgCl
(g)
→
2NH
2
(g)
3
(s)
2
(g)
+
I
2
IB,
Figure 6
a)
Identify
by
I
the
and
V
enthalpy
in
the
changes
Use
the
[2]
energies
given
in
above
and
further
data
4
The
booklet
lattice
c)
The
calculate
of
theoretically
lattice
is
to
enthalpy
enthalpy
+2326
kJ.
a
from
value
magnesium
calculated
of
and
Explain
the
for
value
theoretically
for
[4]
value
the
The
the
is
reaction
correct
spontaneity
of
this
reaction
at
temperatures?
be
spontaneous
at
all
B.
It
will
be
spontaneous
at
high
but
not
at
low
temperatures.
temperatures
temperatures.
between
and
It
will
be
spontaneous
at
low
temperatures
the
value.
experimental
certain
will
not
at
high
temperatures.
[2]
D.
d)
a
statement
It
but
experimental
for
A.
chloride
difference
calculated
values
Which
the
chloride.
magnesium
∆S
positive.
the
C.
the
2004
∆H
both
different
Data
[1]
the
about
cycle
2HI(g)
labelled
cycle.
ionization
→
2
May
are
b)
(g)
lattice
enthalpy
It
will
not
be
spontaneous
at
any
of
temperature.
magnesium
oxide
is
given
in
section
18
IB,
the
Data
oxide
booklet.
has
a
magnesium
IB,
November
Explain
higher
why
lattice
[1]
of
May
2004
magnesium
enthalpy
than
chloride.
[2]
5
The
following
temperatures
2010
CaCO
(s)
→
reaction
above
CaO(s)
850°
+
3
Which
2
The
lattice
enthalpy
values
for
lithium
calcium
uoride
are
shown
∆H
=
+1022
kJ
only
at
C.
CO
(g)
2
combination
1000
is
correct
for
this
reaction
°C?
below.
1
LiF(s)
spontaneous
uoride
at
and
is
ΔG
ΔH
ΔS
A.
-
-
-
B.
+
+
+
+
+
-
-
mol
1
CaF
(s)
∆H
=
+2602
kJ
mol
2
Which
of
explain
less
the
why
than
following
the
that
for
value
statements
for
calcium
lithium
help(s)
uoride
to
is
C.
uoride?
D.
+
[1]
I)
The
ionic
radius
of
lithium
is
less
than
that
IB,
of
May
2007
calcium.
373
15
E N E R G E T I C S
6
Explain
which
can
in
terms
both
7
May
of
ΔH
sometimes
sometimes
IB,
A N D
be
Dene
the
formation
spontaneous
question,
Data
use
reaction
are
for
positive
and
of
undergoes
enthalpy
your
state
nitric
relevant
booklet
standard
illustrate
including
formation
Propyne
the
term
and
equation,
C
a
values
2004
this
as
why
[4]
from
b)
,
ΔS
( A H L )
not.
information
the
ΔG
and
Throughout
a)
T H E R M O C H E M I S T R Y
change
answer
symbols,
of
with
an
for
acid.
[4]
complete
combustion
follows:
H
3
(g)
+
4O
4
(g)
→
3CO
2
Calculate
the
given
following
the
(g)
+
2H
2
enthalpy
O(l)
2
change
additional
of
this
reaction,
values:
[4]
1
ΔH
of
CO
f
(g)
=
-394
kJ
mol
2
1
ΔH
of
H
f
c)
Predict
ΔS
for
and
the
negative,
IB,
374
May
O(l)
=
-286
kJ
mol
2
explain
close
2005
whether
reaction
to
in
zero,
part
or
the
(b)
value
would
positive.
of
be
[3]
16
C H E M I CA L
K I N ET I C S
(AHL)
Introduction
In
this
topic
we
mathematical
rate
can
of
a
only
many
explore
equations
chemical
be
cases
limited
various
that
reaction.
determined
are
the
relate
Rate
the
the
and
slowest
a
in
step
reaction.
reaction
equations
empirically
by
the
to
reaction
in
of
terms
and
the
A
detailed
mechanism
and
of
understanding
allows
optimize
yield,
the
reaction
environmental
chemists
reaction
time,
of
to
the
control
conditions
product
cost
impact.
16.1 Ra rn an racn can
Understandings
Applications and skills
➔
Reactions may occur by more than one step
➔
Deduction
of
the
rate
eq u a tion
from
and the slowest step determines the rate of
experimental
data
and
s olvi ng
problems
reaction (rate determining step/RDS).
involving
➔
the
rate
equ a tion.
The molecularity of an elementary step is the
➔
Sketching,
identifying,
a nd
a na lys ing
number of reactant par ticles taking par t in
graphical
representations
for
zero,
first ,
and
that step.
second
➔
order
reactions .
The order of a reaction can be either integer or
➔
Evalu ation
of
proposed
rea ction
mecha nis ms
fractional in nature. The order of a reaction can
to
be
consistent
with
kinetic
and
describe, with respect to a reactant, the number
stoichiometric
data.
of par ticles taking par t in the rate-determining
step.
➔
experimentally.
➔
➔
Nature of science
Rate equations can only be determined
➔
Principle of Occam’s razor – newer theories
The value of the rate constant (k) is aected by
need to remain as simple as possible while
temperature and its units are determined from
maximizing explanatory power. The low
the overall order of the reaction.
probability of three-molecule collisions means
Catalysts alter a reaction mechanism,
stepwise reaction mechanisms are more likely.
introducing a step with lower activation energy.
375
16
C H E M I C A L
K I N E T I C S
( A H L )
Rate equation
In
topic
6
we
differential
example,
consider
xA
+
yB
where
x,
y,
The
introduced
expression
rate
→
q,
=
Ra quan
m
rate = k[A]
rate
p
is
the
d[A]
[A]
rate
∝
[B]
rate
∝
[A][B]
rate
=
k[A][B]
equation
in
terms
as
of
the
mathematical
concentration.
For
1
_
=
d[C]
=
q
on
1
_
_
+
dt
depends
coefcients
follows:
_
y
reaction
as
-
dt
∝
rate
rate
d[B]
1
_
_
rate
a
stoichiometry
expressed
=
a
of
pD
-
of
idea
expresses
reaction:
are
equation
x
The
+
and
1
_
rate
the
qC
the
that
the
d[D]
_
+
p
dt
dt
concentrations
of
the
reactants:
n
[B]
where:
k = rate constant
This
[A] = concentration of reactant A
is
the
rate
general,
can
account
the
equation
be
(sometimes
expressed
exponents
as
m
shown
and
n,
in
the
called
the
the
box
orders
rate
to
the
with
law)
left,
respect
and,
in
taking
to
into
each
[B] = concentration of reactant B
reactant,
which
convey
how
sensitive
the
rate
of
reaction
is
to
changes
m = exponent in rate equation
in
the
concentrations
of
A
and
B.
described as the rr
The
overall
and
n
order
of
the
reaction
is
then
dened
as
the
sum
of
the
m
w rc  racan A
exponents:
n = exponent in rate equation
overall
reaction
order
=
m
+
n
described as the rr
w rc  racan B
Rate
equations
orders
Let’s
can
take
the
(g)
NO
N:
can
only
+
be
only
be
determined
deduced
following
CO(g)
→
The
and n) cann be worked
carbon
because
the
reaction:
NO(g)
+
CO
2
The orders (for example, m
experimentally
empirically.
(g)
2
rate
equation
for
the
reaction
of
nitrogen
dioxide,
NO
(g),
with
2
monoxide,
out from the stoichiometry
CO(g),
has
been
found
experimentally
to
be:
2
rate
=
k[NO
]
2
coecients (for example, x
m
Hence
the
rate
equation
rate
=
k[A]
n
2
[B]
corresponds
to
rate
=
k[NO
and y) of a par ticular reaction.
]
.
2
This
means
that
m
=
2
and
n
=
0;
that
is,
the
order
with
respect
to
NO
(g)
2
is
m
two
and
of
and
n,
are
which
One
a
order
not
are
However,
implies
the
with
deduced
respect
from
to
the
CO(g)
is
zero.
stoichiometry
Notice
how
coefcients
the
x
orders,
and
y,
both
1.
the
overall
order
second-order
method
of
of
the
reaction,
given
equation
is
by
m
+
n
=
2
+
0
=
2,
reaction.
deducing
the
rate
to
use
the
method
of
Typically orders with respect
initial
rates,
the
principle
of
which
we
introduced
in
topic
6.
The
to reactants are either two, one
value
of
the
rate
constant,
k,
is
affected
by
temperature
and
its
units
are
or zero order, but orders can
determined
from
the
overall
order
of
the
reaction.
in fact be fractional or even
negative! In this book , only the
Catalysts
reactions with whole-number
As
discussed
in
topic
6,
a
catalyst
is
a
substance
that
increases
the
rate
orders will be discussed.
of
a
chemical
catalyst
reaction,
provides
activation
an
energy,
is
not
alternative
E
(gure
a
376
but
10
consumed
pathway
in
for
sub-topic
in
the
the
reaction
reaction
6.1).
itself.
and
A
lowers
the
16 . 1
R A t e
e x p R e s s i o N
A N d
R e A C t i o N
m e C h A N i s m
The contact process
Uful rurc
The
rst
catalyst
used
in
industry
was
for
the
production
of
sulfuric
acid.
The American Chemical Council (ACC)
●
In
this
process,
called
the
contact
process,
elemental
sulfur,
S(s),
is
has developed C AB, a Ccal Ac 
rst
reacted
with
oxygen
gas,
O
(g),
to
form
sulfur
dioxide
gas,
SO
2
S(s)
+
O
(g)
→
SO
2
(g):
Barr, which is an economic
2
indicator that predicts peaks and
(g)
2
troughs in the overall economy in the
●
Sulfur
dioxide
then
reacts
with
oxygen
gas
to
produce
sulfur
trioxide,
USA and highlights potential trends in
SO
(g):
3
other industries. The barometer ser ves
V
O
2
2SO
(g)
+
O
2
(s)
5
(g)
2SO
2
as a pivotal tool in predicting broader
(g)
3
economic health in the USA . CAB is
The
catalyst
used
is
vanadium(V)
oxide,
O
V
2
heterogeneous
(s),
which
is
a
5
a leading index of overall industrial
catalyst.
production and has a number of
dierent indicators including chemical
Can you see why this is a heterogeneous catalyst?
company stock data, etc. (For example,
have a look at the video on their website
●
Sulfur
trioxide
is
next
absorbed
into
concentrated
sulfuric
acid,
(http://www.americanchemistry.com/
H
SO
2
(l).
This
produces
oleum,
H
4
S
2
O
2
(l).
Oleum
reacts
with
water
to
7
Jobs/CAB) for a greater insight into the
produce
aqueous
sulfuric
acid,
H
SO
2
(aq).
4
impor tance of chemistry to the global
SO
(g)
+
H
3
H
The
2
S
2
O
2
for
a
war,
O(l)
S
→
the
2
2H
oleum
trioxide
For
the
(l)
economy). CAB was the rst of its kind
7
SO
formed
is
rst
too
mirrored
during
the
decreased,
reversed
developed globally.
(aq)
4
water
closely
acid
sulfuric
is
and
example
trend
O
2
sulfuric
of
H
2
production
of
production
H
why
time.
production
+
sulfur
acid
→
2
reason
long
world
(l)
(l)
4
7
between
Sulfuric
SO
and
a
there
that
the
country’s
rst
but
is
direct
reaction
vigourous.
and
economic
second
immediately
was
a
world
after
dramatic
health
wars
the
the
second
increase
in
the
acid.
What are some current indicators of a country’s economic health?
Molecularity and rate-determining step
(slow step) of a reaction
The
sequence
reactants
to
of
the
reaction
kinetics.
This
reaction
mechanism
or
sequence
elementary
step
is
classied
molecules
or
steps
formation
of
any
its
atoms
events
unimolecular:
●
bimolecular:
is
or
single
two
is
the
a
step
is
elementary
as
molecule
molecules
or
the
reaction
process.
in
as
In
the
an
of
turn,
the
in
In
elementary
an
a
step
elementary
number
of
reaction:
elementary
involved
chemical
mechanism.
elementary
in
from
aspect
an
represents
involved
atoms
pathway
important
described
which
reactants
reaction
very
termed
molecularity,
involved
●
outlining
products
individual
reaction
by
of
collision
step
in
an
elementarystep
●
termolecular:
elementary
Each
molecules
or
atoms
involved
in
collision
in
an
step.
elementary
energy,
three
step
has
its
own
rate
constant,
k,
and
its
own
activation
E
a
Let’s
return
NO
(g)
2
to
+
the
reaction
CO(g)
→
of
NO(g)
nitrogen
+
CO
dioxide
with
carbon
monoxide:
(g)
2
377
16
C H E M I C A L
K I N E T I C S
( A H L )
The
Rar f trlcular
Racn
reaction
events
elementary
Termolecular reactions are
step
mechanism
leading
from
representing
reactants
to
the
products
sequence
is
of
actually
molecular
composed
of
two
steps:
1:
NO
(g)
+
(g)
2
very rare as it is very unlikely
→
NO(g)
+
(g)
2
step
1
is
3
bimolecular
that three par ticles would
(g)
+
CO(g)
→
(g)
3
collide simultaneously with
+
CO
2
(g)
2
each other in the correct
orientation. For example, have
overall
reaction:
NO
(g)
+
CO(g)
→
NO(g)
+
CO
2
(g)
2
you ever seen three snooker
In
this
mechanism,
NO
(g)
is
described
as
a
reaction
intermediate ,
as
3
balls colliding at the same
it
is
formed
in
step
1
and
then
is
consumed
subsequently
in
step
2.
time when watching a World
Championship snooker match
Therefore,
on TV?
step
reactions
determines
may
the
rate-determining
occur
rate
of
step
the
by
more
than
reaction.
one
The
step
slow
and
step
is
the
slow
termed
the
(RDS).
Deduction of a rate equation from a proposed
reaction mechanism
In
order
1
to
deduce
Decide
2
on
equal
to
From
(1)
the
which
the
rate
deduce
rate
step
of
equation
is
this
the
the
RDS.
slow
rate
from
a
The
proposed
rate
of
reaction
the
overall
mechanism:
reaction
is
step.
equation
for
the
RDS.
Anal
For
temperatures
the
reaction
less
than
498
K,
the
experimental
rate
equation
for
Passengers having arrived
just
discussed
has
been
found
to
be:
through passpor t control at
2
rate
an airpor t have to follow a
=
k[NO
]
2
sequence involving two stages
●
In
effect,
the
reaction
mechanism
is
essentially
a
hypothesis
of
the
in order to exit the airpor t. The
sequence
of
events
that
has
led
to
the
overall
reaction
converting
the
rst step involves collecting
reactants
into
products.
There
might,
therefore,
be
a
number
of
possible
luggage at the baggage carousel
reaction
mechanisms
that
equate
with
the
experimental
rate
equation.
and the second step involves
For
the
example
just
discussed,
here
is
a
proposed
reaction
mechanism.
exiting the arrivals area. The
step that determines the rate at
Consider
step
1
as
the
slow
step
(so
is
which the passengers can be
the
RDS)
and
step
2
as
the
fast
step:
k
__
→
1
step
1:
NO
(g)
+
(g)
2
on their way out of the airpor t
NO(g)
+
(g)
2
(slow)
3
k
is determined by the rate at
(g)
+
CO(g)
2
__
→
(g)
3
+
CO
2
(g)
(fast)
2
which their luggage arrives on
overall
reaction:
NO
(g)
+
CO(g)
→
NO(g)
+
CO
2
(g)
2
the carousel. This is the ra-
rnn  of the two-
Hence:
step sequence of events and
rate
of
overall
is analogous to the idea of the
reaction
=
rate
of
the
slow
step
(in
this
case
step
1)
2
=
k[NO
]
2
lw  in chemical kinetics.
where
k
represents
mechanism
If step 1 is the slow step, the
●
activation energy for this step,
In
is
the
rate
consistent
contrast,
at
rate
equation
rate
=
constant
with
the
temperatures
for
the
for
the
overall
experimentally
greater
reaction
just
than
498
discussed
reaction.
This
determined
K,
has
the
rate
proposed
equation.
experimental
been
found
to
be:
(1) will be large. If step 2 is
E
a
k[NO
][CO]
2
(2) will be
the fast step, then E
a
small .
A
proposed
reaction
mechanism
here
might
be
a
single-step
bimolecular
process:
single
step:
NO
(g)
2
378
+
CO(g)
k
_
_
→
NO(g)
+
CO
(g)
2
(slow)
16 . 1
R A t e
e x p R e s s i o N
A N d
R e A C t i o N
m e C h A N i s m
Hence:
rate
of
step)
overall
=
k[NO
reaction
=
rate
of
the
slow
step
(in
this
case
the
single
][CO]
2
This
proposed
rate
equation.
mechanism
is
consistent
with
the
experimentally
determined
toK
A reaction mechanism can be suppor ted by indirect
Cancer research, for example, is all about identifying
evidence. What is the role of empirical (experimental)
mechanisms for carcinogens as well as for cancer-
evidence in the formulation of scientic theories? Can
killing agents and inhibitors.
we ever be cer tain in science?
Worked example: deduction of the rate equation from experimental
data and solving problems involving the rate equation
1
Consider
the
balanced
stoichiometry
equation,
coefcients
of
and
the
note
reactants
the
5
Deduce
the
overall
products.
For
+
yB
reaction
→
qC
+
y,
q,
and
p
are
the
stoichiometry
Determine
these
coefcients
2
Write
the
order
the
rate
rate
equation,
where
m
and
(1,
values
deduce
down
of
the
reaction:
=
m
+
constant,
k,
n
for
each
pD
experiment
x,
order
example,
6
xA
overall
and
the
to
2,
3,
etc.).
give
the
appropriate
Find
mean
units
the
mean
value
for
of
k
of
and
k
n
Example 1
represent
the
orders
with
respect
to
each
Consider
reactant:
m
rate
3
From
=
k[A]
the
ratios
(as
A(g)
data
for
deduce
each
each
of
of
Based
the
the
1)
(rate
2)
(rate
_
,
3)
for
the
,
2)
(rate
3)
,
(rate
pairs
of
concentration
rate
does
data
not
Deduce
→
C(g)
orders
and
the
from
D(g)
with
overall
i niti al
one
where
change
experiment
respect
the
each
orders
ratio
to
m
obtained
and
n.
Use
logs
may
be
tools
with
helpful
Deduce
the
rate
equation.
in
one
of
them
[B()]
inal ra
another.
in
of
step
some
respect
here,
each
order.
the
3,
3
/ l 
for
3
/ l 
1

deduce
fundamental
2
mathematical
to
reaction
3
From
d at a
etc.
/ l 
4
rate
4)
[A()]
going
+
expe r i me nta l
the
reactant
_
●
Look
on
B(g)
appropriate):
_
(rate
+
below:
following
●
(rate
reac ti o n:
n
[B]
given
experiments,
the
to
indices
Experiment 1
1.00 × 10
Experiment 2
2.00 × 10
Experiment 3
2.00 × 10
2
1.00 × 10
3
4.20 × 10
and
example:
0
x
log
=
2
1
(XY)
=
log
X
+
log
2
1.00 × 10
3
8.40 × 10
Y
X
_
log
(
Y
)
=
log
X
log
Y
2
2
2.00 × 10
2
3.36 × 10
p
log
X
=
plog
X
379
16
C H E M I C A L
●
Calculate
for
the
K I N E T I C S
the
value
reaction
of
from
( A H L )
the
rate
constant,
experiment
3
and
k,
Then
substituting
the
data
state
from
experiment
2
(3.36
×
3
10
mol
3:
1
dm
s
)
_____
its
k
units.
=
2
(2.00
●
Determine
the
rate
of
the
reaction
=
3.00
×
×
10
mol
dm
and
3
[B(g)]
4.20
×
dm
)(2.00
2
10
6
mol
×
10
3
mol
dm
1
dm
The
dm
units
were
worked
s
out
as
follows:
3
1
mol dm
s
____
units
of
k
=
3
Solution
In
order
to
working
from
mol
solve
thi s
method
the
to
method
q ue s ti on
d e duce
of
i ni tia l
we
the
ca n
ra te
u se
dm
×
2
t he
units
of
k
=
3
mol
mol
6
3
dm
×
mol
dm
1
dm
s
eq u at i on
ra tes :
The orders may also be deduced by inspection. By
●
There
are
equation
two
so
reactants
the
rate
m
rate
=
k[A]
in
the
equation
chemical
is
given
keeping [B] constant in experiments 1 and 2 and
by:
doubling [A], the initial rate is seen to double. Hence
the order with respect to A will be one. Likewise, by
n
[B]
keeping [A] constant in experiments 2 and 3, [B]
●
You
to
next
use.
data
not
have
In
in
order
which
change
down
to
to
–
just
choose
to
decide
one
this
one
of
the
helps
the
this,
reduce
order.For
k(0.010)
for
the
ratios
pairs
doubles. However, this time the initial rate is seen
of
to increase by a factor of four, meaning that the
does
order with respect to B is two. This is a quick way of
problem
deducing the orders, but with more dicult numbers
example:
nding the orders by this method might be quite
n
tricky – following the working method using ratios
(0.010)
0.00420
_
__
=
=
m
rate
look
concentrations
m
rate 1
_
appropriate
2
will always allow you to nd t