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Contents
Structure 1. Models of the particulate nature of matter
2
s
s
Structure 1.1
Structure 1.2
Structure 1.3
Structure 1.5
r
Structure 2. Models of bonding and structure
94
Structure 2.2
y
Structure 2.4
t
i
s
228
Structure 3.1
Structure 3.2
Tool 1:
Reactivity 1. What drives chemic al reactions?
Reactivity 1.2
Reactivity 1.4
n
U
Reactivity 1.3
Reactivity 2. How much, how fast and how far?
i
t
a
r
o
Reactivity 3. What are the mechanisms of chemical change?
Reactivity 3.2
Reactivity 3.3
l
Reactivity 3.4
460
536
u
f
x
Reactivity 3.1
386
o
Reactivity 2.2
d
Reactivity 2.1
C
Reactivity 1.1
o
i
n
v
Tool 3:
p
Tool 2:
308
a
O
Cross-topic exam-style questions
652
The inquiry process
655
v
(authored by Maria Muñiz Valcárcel)
The internal assessment (IA)
E
Index
(authored by Maria Muñiz Valcárcel)
Periodic Table
Answers:
y
r
e
Tools for chemistry
O
Structure 3. Classic ation of matter
n
l
Structure 2.3
y
P
Structure 2.1
Reactivity 2.3
e
Structure 1.4
668
686
708
www.oxfordsecondary.com/ib-science-support
iii
Introduction
The diploma programme (DP) chemistry course is aimed at students in the 16
The curriculum seeks to develop
of the nature of science,
chemistry and
a conceptual understanding
working knowledge of fundamental principles of
practic al skills that
c an be applied in familiar and unfamiliar
As with all the components of the DP,
this course fosters the IB learner
prole attributes (see page viii) in the members of the school community.
r
denitely true.
throughout
the book and
NOS features
but
they c annot
This has led
science throughout
are encouraged to come up
p
Observations and experiments
contain assumptions or unrealistic simplic ations,
but
lead
to serendipitous results.
U
Measurements
•
all data are prone to error. It is important to
takes observed
These predictions may conrm a
theory (within observable limitations) or may falsify
it.
•
t
a
r
o
i
o
d
know the limitations of your data.
Evidence
explanation that
patterns and hypotheses and uses them to generate
predictions.
•
to reduce its limitations.
A theory is a broad
Measurements c an be qualitative or quantitative,
but
and
Theories
n
•
and
of science is to increase the complexity
of the model,
Sometimes the observations in experiments are
unexpected
the aim
C
i
n
•
models as simplied
explanations of their observations. Models oen
o
aspects:
v
into the following eleven
to paradigm shis in
history.
Models
Scientists construct
NOS c an be organized
be proved to be
y
•
the programme.
r
e
with further examples of your own as you work through
O
evidence,
entire course.
suggested
n
F alsic ation
Hypotheses c an be proved false using other
are specic to science.
NOS is a central theme that is present across the
You will nd
l
y
•
is concerned with methods,
outcomes that
t
i
s
purposes and
Science as a shared activity
Scientists learn to be sceptic al about their
observations and
be fully supported
by evidence.
Patterns and trends
u
f
x
•
Recognition of a pattern or trend
part
Patterns lead
on a convention for clear
forms an
•
Global impact of science
Scientists are responsible to society for the
consequences of their work,
environmental,
to a possible explanation. The
E
hypothesis is this provisional view and
further veric ation.
iv
such as peer review of work before
public ation or agreement
communic ation.
a
v
Hypotheses
collaboration,
of the scientist’s work whatever the
science.
•
l
O
important
Scientic activities are oen c arried out in
they require their knowledge to
it requires
knowledge must
and
fairly.
whether ethic al,
economic or social. Scientic
be shared
with the public clearly
y
P
Nature of science
Nature of science (NOS)
e
contexts.
s
s
to 19 age group.
Syllabus structure
Topics are organized
into two main concepts:
structure and
reactivity.
This is shown in the syllabus roadmap
The skills in the study of chemistry are overarching experimental,
skills that
are integrated
experimental work,
into the course.
inquiries and
Chemistry is a practic al subject,
technologic al,
mathematic al and inquiry
so these skills will be developed
through
investigations.
Reactivity
why
simple to more complex forms
chemic al reactions occur
l
Structure determines reactivity, which in turn transforms structure
What
enthalpy changes
chemic al
Structure 1.2 — The nuclear atom
of matter
O
particulate nature
Reactivity 1.1 — Measuring
drives
Reactivity 1.2 — Energy cycles in
reactions?
reactions
Structure 1.3 — Electron
Reactivity 1.3 — Energy from fuels
particles by mass: The mole
Reactivity 2.
How much,
how fast and
U
Structure 2.2 — The covalent
structure
Structure 2.3 — The metallic
d
Structure 2.4 — From models to
i
matter
extent
of chemic al change
Reactivity 3.
Reactivity 3.1 — Proton transfer
What
reactions
Classic ation of elements
are the
mechanisms
Reactivity 3.2 — Electron transfer
u
of chemic al
reactions
change?
Structure 3.2 — Functional
groups:
Reactivity 3.3 — Electron sharing
Classic ation of organic
reactions
a
compounds
v
E
roadmap
above,
Reactivity 3.4 — Electron-pair
sharing reactions
For example,
“Structure determines reactivity,
You are therefore encouraged
help
Reactivity 2.3 — How far? The
table:
Chemistry concepts are thoroughly interlinked.
new and
of chemic al change
Reactivity 2.2 — How fast? The
Structure 3.1 — The periodic
l
O
f
x
t
a
Classic ation of
r
o
materials
amount
rate of chemic al change
o
model
Structure 3.
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Course book denition
The IB Learner Prole
The IB Diploma Programme course books are resource
The aim
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materials designed
minded
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IB mission statement
The International Bacc alaureate aims to develop
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as the representation of the ideas
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ix
Experience
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and
proportions
theorized
equalled
c arbon
mass
of
of
idea
Hydrogen
the
mass
formed.
1.33 g
2.66 g
the
that
combined
Other
oxygen
oxygen
of
to
to
and
was
oxygen
hydrogen
c an
and
experiments
through
form
matter
composed
form other substances
react to
oxygen
showed that
combustion
c arbon
to
form
dioxide.
r
It was proposed that elements, such as hydrogen, oxygen or carbon, are the
internal
structure and
primary constituents of matter, and they cannot be chemically broken down into
characteristics of atoms will be
simpler substances. The idea of denite proportions suggested that particles of one
discussed in
Structure 1.2
y
simple ratio, and that atoms of one element have a dierent mass than atoms of a
dierent element. This, and other experimental evidence, led to the atomic theory.
be
reactions.
states
or
Physic al
that
all
matter
destroyed,
and
but
chemic al
is
composed of atoms. These atoms
they
are
properties
rearranged
of
during
arrangement of these atoms.
that
8th
of
that
one
smaller
the
is
parts
E
What
on
to
into
their
must
What
is
is
snap
as
to
“kana”.
said
a
to
have
seashell
further.
atomic
D alton.
to
dierent
powder
“atomos”, “not
any
of
John
experiments
evidence
theories?
broken
development
classied
based
be
particles
producing
known
a
be
until
units,
credited
knowledge
evidence.
atomic
in
the
Democritus
successively
not
conservation
“elements”,
Scientic
could
later,
could
c alled
l
mass
atoms
could
v
O
from
stage
years
He
BCE,
indivisible
that
due
proposed that “particles too
u
next
2000
of
f
x
splittable”,
The
century
increasingly
composed
are
i
into
5th
wor ld
parti cles.
t
a
observed
Āruni
natu ra l
matte r
They
mass together into the substances and
experience”.
in
the
r
o
Similarly,
BCE,
seen
in
these
that
pa rticl es.
o
objects
be
chan ges
betwe en
century
to
re a sone d
indi visible
d
small
tiny,
s age
n
interactions
Leuci ppus,
of
In dian
U
postu late d
In
and
up
the
philos ophers
theory,
D alton
over
drew
propose that
types
known as
masses.
be
supported
was
used
evidence?
to
Is
by
veriable
develop these
evidence
shaped

by
our
y
made
them
Gre ek
o
w as
the
C
Democ ritus
among
and
i
n
Ā runi
v
atomists,
Uddāl ak a
p
Evidence
Ancie nt
chemic al
matter depend on the bonding
r
e
and
theory
created
O
atomic
c annot
t
i
s
The
n
l
element, called atoms, would combine with atoms of another element in a xed,
y
P
The
e
1.0 g
was
and
of
denite
s
s
but
It
in
Figure 3
Top:
Āruni lived
in what
is now modern day
perspective?
Northern India,
depicted
by the Ganges river.
Bottom:
Democritus is
in a Renaissance-era painting
5
Structure
1
Models
of
the
particulate
nature
of
matter
Chemic al symbols
In
modern
which
example,
the
chemic al
of
one
atoms
or
chemic al
symbol
and
two
symbol
for
elements
letters
iron
is
for
Fe
and
are
are
hydrogen
(the
rst
represented
derived
is
two
H
from
(the
letters
rst
of
by
the
the
letter
the
same
element
of
L atin
symbols,
names.
ferrum
“iron”).
Name
Common
chemic al
elements
and
their
symbols
are
listed in table1; the full list is
given in the data booklet and in the periodic table at the end of this book.
H
hydrogen
oxygen
Na
sodium
Atoms
are
the
properties.
and
form
smallest
While
iron
is
another
magnesium
Figure 4
bound
atomic
formula
magnesium
is
a
species,
one
type
composed
Mg
sulde.
of
of
atom,
sulfur
compound,
and
as
S
Mg.
atoms
it
Similarly, sulfur
only.
consists
of
(gure4). MgS is the
p
C
sulfur (middle) and magnesium sulde (right)
single
more elements
In
contrast,
two
dierent,
chemic al
o
v
i
n
(le),
n
U
M agnesium
or
example, magnesium metal is an
only
chemic al
contain atoms of a
two
y

(MgS)
chemic ally
of
substance
For
of
O
Common chemic al elements
sulde
forces.
contains
atoms
t
i
s
Table 1
elementary
it
contain
r
e

as
chemic al
Pure substances and mixtures
o
Figure 5
i
E

6
t
a
v
sulfur (S)
classied
as
a
pure
substance
or
a
mixture, depending on the type
(gure 5).
– any substance that
definite and
mixture
u
e.g.,
a
atoms,
be
arrangement
matter
composition
l
O
of
magnesium (Mg),
c an
particle
compound
–
composed of one
kind
of
occupies space and has mass
– has a
uniform chemical
element
d
r
o
f
x
pure substance
M atter
of
two
or
atoms
–
composed
more kinds of
in
a
fixed
ratio,
e.g., magnesium sulfide
(MgS),
How matter is classied
– a
combination of two or more pure
substances that retain their indiidual properties
water (H
)

homogeneous
– has
uniform composition
and
properties
throughout,
water,
according to the arrangement of particles
e.g.,
metal
sea
alloy
heterogeneous
– has
nonuniform composition
and
e.g.,
arying
paint,
properties,
salad
dressing
y
Fe
(S)
chemic al
certain
n
chlorine
by
substance,
possess
Elementary substances
chemic al compounds
together
elementary
still
l
Cl
bound
that
individually, they tend to combine together
P
sulfur
matter
exist
y
magnesium
S
of
c an
chemic al substances.
element, while
Mg
units
atoms
e
c arbon
O
r
C
For
hydrogen), and
s
s
the
Symbol
chemistry,
consist
Structure
1.1
Introduction
to
the
particulate
nature
of
matter
Pure substances cannot be separated into individual constituents without a chemical
reaction, which alters their physical properties. In contrast, mixtures can be separated
into individual components that retain their respective physical properties.
A
student
had
two
pure
substances,
observations
were
A
made
and
and
B.
They
were
heated
in
separate
crucibles
and
some
qualitative and
recorded in table 2.
r
e
quantitative
s
s
Data-based questions
B
were
n
l
and
a
to
O
is
y
A
ice
p
4.
change
qualitative
changes
y
P
y
Melting
the
v
3.
the
a
E
State
of
M ass
of
crucible
crucible
Change
and
substance
contents
in
Observations
mass / g
heating
aer
aer
/ g
heating
/ g
26.12
±
0.02
26.62
±
0.02
Black colour
27.05
±
0.02
25.76
±
0.02
Black colour
heating substances A and B
a
C alculate
2.
u
Results from
l
1.
Table 2
colour
o
C
M ass
and
heating
colour
Green
O

f
x
Red
n
i
t
a
r
o
before
o
Appearance aer heating each of the two substances
Observations
Substance
B
Substance B
v
i
n
U
d

A
t
i
s

Substance A
r
e

in
mass
for substances A and B.
observation
physic al
change
substances
both
pure
from
A
the
while
and
B
experiment
rusting
iron
represented
substances,
not
a
mixtures.
is
performed on A and B.
a
chemic al
physic al
Discuss
change.
change
or
whether
a
the
Explain,
chemic al
using
the
observations, whether
change.
experiment
shows
that
A
and
B
are
elements.
5.
Both
A
same
and
B
turned
black
on
heating.
C an
it
be
concluded
that
the
heating
of
these
two
substances
produced the
substance?
7
Structure
1
Models
of
the
particulate
nature
of
matter
Melting point determination
Melting
purity
point
of
points,
t e m p e ra t u r e
v a l u e.
over
a
The
its
c an
which
that
presence
melting
used
of
to
they
matches
impurities
point
a ss e ss
substances
me ans
closely
t e m p e ra t u r e
and
melt
the
in
c auses
a
Method
the
h av e
at
a
sharp
(Your teacher will provide specic instructions, depending
on
specific
the oretic al
substance
melting
to
occur
2.
Prepare
3.
being
two
samples
of
each
record
sucient
amounts
relevant
relevant
Following
4.
qualitative
Prepare,
your
of
in
a
two
third
solids
B
and
the
the
melting
organic
solids,
for
Record
t
i
s
1.
known
relevant
example,
appropriate
aspirin
and
salol
(phenyl
2-hydroxybenzoate)
2.
Comment
S afety
points
•
Note
that
•
You
protection.
the
teacher
melting
will
give
3.
point
you
apparatus
further
gets
safety
very hot.
this
environmentally
salol
and
aspirin
point
of
a
substance
not
are
contain
more
chemic ally
Mixtures
discussed in the
c an be
than
If
the
the
particles
the
top,
mixture,
is
structural
are
which
extent
the
to
and
evenly
reveals
of
its
could
a
,
which
of
where
the
melting
of
an
samples
quantitative data in an
impure substances.
dierence in their
point
organic
compound
the
air
distributed,
that
is
and
with
separated
in
by
particles
of
is
data
be
used to
synthesis?
no
xed
ratio, which
physic al methods.
are
evenly
distributed.
milk
is
a
as
in
a
its
and
80%
nitrogen and 20%
mixture of two solids, then
Natural
milk
heterogeneous
maintains
explosive,
roughly
sampled.
such
heterogeneous.
mixture
hydrogen, H
or
be
composition
regardless
not
c an
in
three
formulas of A and B and use
explain
success
element
so
your
sample of the
oxygen, and small amounts of other gases. Air is a
referred to as
component
example,
nitrogen,
consistent
mixture
u
f
x
to
is
i
oxygen
one
and
homogeneous,
of
o
mixture
homogeneous
E ach
The most common homogeneous
a
t
a
r
o
d
chapter.
is
what
analyse
bonded
Tools for chemistry
Air
the
C
for determining the
are
melting
To
of
small
results, comparing the melting
substances
information
n
U
Mixtures
Methods
4.
are irritants
hazardous).
the
o
v
example,
qualitative
melting points.
i
n
and
(for
on
pure
Research
prec autions,
depending on the identity of the solids being
analysed
a
y
eye
for
separate
format.
p
Wear
of
r
e
•
tube,
point
mixture).
apparatus
two
two
together.
c apillary
Questions
of
B)
n
S amples
in
O
•
point
and
y
C apillary tubes
(A
l
Determine
(A,
Melting
solid
mixture of the two solids.
quantitative data
5.
•
solids
teacher ’s instructions, mix small
the
Materials
•
analysed.)
organic
physic al
and
oxygen, O
l
2
will
have
the
cream rise
mixture.
chemic al
properties.
For
, supports combustion.
2
a
O
mixtures, aqueous solutions, will
When
these
substances
are
present
in
a
mixture,
their
properties
stay
the
same.
be discussed in Reactivity 3.1, and
In
contrast,
water, H
O,
is
not
a
mixture
of
hydrogen
and
oxygen
but
a
chemic al
2
the properties of metal alloys in
E
8
v
Structure 2.4.
compound
The
new
gas,
is
with
its
water
formed
substance
not
by
explosive,
own
bonding
has
and
properties
without
a
none
it
and
chemic al
of
two
the
does
the
hydrogen
properties
not
support
hydrogen
reaction,
which
atoms
of
with
combustion.
and
oxygen
creates
one
hydrogen
or
It
is
c annot
oxygen atom.
oxygen. It is not a
a
pure substance
be
separated
new substances.
from
y
Inquiry
and
solids
of
P
•
observations
the
analysis.
skills
and
of
samples
c apillary tubes.
Tool 1: Melting point determination
Identify
identity
Obtain
ra n g e.
•
2:
the
1.
r
Relevant
be
Pure
e
l ow e rs
data
s u b s t a n c e.
s
s
melting
a
Structure
1.1
Introduction
to
the
particulate
nature
of
matter
Separating mixtures
Mixtures
mixture
c an
has
separated
of
using
property
magnetic
the
to
does
components
c an
c an
be
it
usually
is
be
a
a
of
magnetic
them.
sugar
sand
if
we
and
s u ga r
It
sulfur
is
powders
not.
iron(II)
maintains
pure
This
sulde,
none
of
c an be
dierence
FeS, is not
the
properties
substance.
understand
sugar
and
placed
each component of the
sulfur
compound
sugar
is
and
while
bec ause
between
bec ause
iron
smell.
individual
separated
from
The
sulfurous
new,
attractions
m i x tu re
is
of
their
will
intermolecular
dissolve
in
forces.
water, due to
water.
in
water
and
t he
s u ga r
Intermolecular
di ss o l ve s .
The
solution
c an
th e n
be
po u re d
t hro u g h
fi l t e r
paper
placed
i n s i de
p a ss
th e
by
will
and
p a ss
e v a p o ra te s
e v a p o ra t i n g
th ro u gh
l e avi n g
th e
f i l te r
re m a i n
on
t hro u gh
w a te r
pa p e r.
th e
t he
behind
f ro m
S u ga r
(fi gu re
fi l t e r
f i l te r
th e
t he
6 ).
p a p e r,
p a p e r.
pu re
fil t ra t e
c r ys ta l s
The
—
the
fo r m
sand
w he re a s
Th e
sand.
will
l a rg e
wet
Th e
t he
sand
s u ga r
solution
in
t hi s
p a r ti c l e s
sugar
is
be
which
pa ss e d
pro c e ss
a
residue
p
(We define
discussed
o
C
i
n
residue
that
remains
aer
n
U
as a substance
evaporation,
similar
or
any
process)
i
filtrate
u
Filtration apparatus
a
O
l
f
x
t
a
r
o
d
filtration
o
distillation,
evaporating
th e
v
filter funnel
are
y
filter paper
Figure 6
and
forces
Structure 2.2
o bt a i n e d
c r yst al li za t i on
7 ).

d i ss o l ve d
dried,
c an
will
r
e
(fi gu re
the
fil t ra t i on
O
w a te r
th ro u gh
w a te r
c alled
t
i
s
in
p ro c e ss
n
not
a
y
funnel,
l
in
a
y
solid
have
means
mixture
P
Th e
as
A
Iron
separate
not
separated
intermolecular
physic al
r
the
magnet.
used
and
solids
S and
is
a
by
properties.
e
Two
separated
s
s
in
be
unique
sugar solution
basin
E
v
solution from
evaporating basin
cold tile
leave for a
few days
t
Figure 7
The crystallization
heat
for sugar to crystallize
process
9
Structure
1
Models
of
the
particulate
nature
of
matter
Distillation can be used to separate miscible liquids with dierent boiling points,
such as ethanol and water. Ethanol has a lower boiling point and will evaporate rst.
Once the vapours rise up a cooling column, they can be condensed to a liquid. As
shown in gure 8, cold water surrounds the condenser and allows the vapours to
condense to liquid ethanol. The water remains mostly in the distillation ask.
s
s
r
e
thermometer
water out
Distillation apparatus
flask
water
water in
y
distillate
p
heat
r
e
and
O
t
i
s
ethanol
n
y
condenser
l
Figure 8
(mostly ethanol)
o
v
i
n
chromatography will
C
Paper chromatography can be used to separate substances such as components
Paper
in inks. A piece of chromatography paper is spotted with the mixture. The bottom
be
discussed
in
more detail in
of the paper, below the spot, is placed in a suitable solvent as in gure 9(a).
substances
in
the
n
The
U
Structure 2.2
mixture
phase) and the paper (the
dierent
anities
forces of attraction
between
and
or
9(c)
ve
the
solvent
o
intermolecular
for
the
solvent (the
mobile
paper.
Figure
the
shows
a
pure
substances
mixture
that
was
in
the
mixture
composed of
pure substances.
i
(a)
the
(b)
(c)
paper
some
some
u
l
a
E
v
O
f
x
t
a
The stages in 2D paper
chromatography
10
d
Figure 9
r
o
u
have
stationary phase). The anity depends on the
hours
hours
later
later
solvent
drop of
mixture
turn
and
paper
use
a
90°
clockwise
different
solvent
Data-based questions
Look
at
gure
9.
1.
Which colour dot had the strongest anity for both solvent 1 and solvent 2?
2.
Which
colour
3.
Which
had
a
dots
had
stronger
a
stronger
anity
for
anity
solvent
for
2
solvent
than
1
than
solvent 1?
solvent 2?
y
P
distillation
u
Structure
Table
3
shows
a
summary
of
the
separation
techniques
1.1
Introduction
to
the
particulate
nature
of
matter
discussed.
Components
Technique
Description
removed
ltration
is
le
s
s
mixture
poured
through a paper lter or
solid(s)
soluble
insoluble
substance(s)
substance(s)
solvent, the solution
more soluble
less soluble
cools
substance(s)
substance(s)
porous material
or
an
is
added
organic
mixture
is
to
water
solvent
dissolved in
e
mixture
(solvation)
r
dissolution
P
Activity
hot
water
or
an
organic
Suggest
a
suitable
method
down, and the
heated up
solid(s) and/
volatile
one
or
more of its
or
distillation
non-volatile
liquid(s)
components
vaporize(s)
liquid(s)
placed on
a.
salt and pepper
b.
several
c.
sugar
d.
For
less soluble
a
piece of paper; one
more soluble
side of the paper is
chromatography
submerged
component(s)
water or
move(s) faster
solvent; components
3
Summary of separation techniques

Figure 10 An advanced
each
describe
technique
and
component
is
the
outline
isolated.
place
n
o
i
t
a
a
by fossil fuels.
mixture,
u
l
v
E
provided
d
r
o
f
x
O
ltration technique c alled
water for millions of people.
how
slower
stay(s) in
C
Ta b l e
U

i
n
move along the paper
or
dyes
water
o
a
move(s)
v
in
water-soluble
and
separation
component(s)
p
paper
following
iron and copper lings
each
y
is
r
e
mixture
the
mixtures:
evaporation or
until
of
O
is
are
ltration
t
i
s
mixture
by
each
n
formed
isolated
y
crystals
separating
for
l
crystallization
y
liquid(s)
other
However,
Why might it
reverse osmosis extracts salt
from seawater,
this process requires vast amounts of energy,
be important
most
providing fresh
of which is currently
to consider alternative energy sources?
11
Structure
1
Models
of
the
particulate
nature
of
matter
Planning experiments and risk assessments
Relevant skills
Tool
1:
Separation
•
Tool
1:
Addressing
of
mixtures
safety
of
self,
others
and
the
environment
s
s
•
Instructions
1.
Using
the
c alcium
you
Determine
•
Identify
suitable
time,
try
methodology
the
separation.
the
masses
are
before
mix
all
and
the
them
dry,
iron
you
physic al and
hazards and complete
to
mass
of
c alculate
component.
your
teacher
method
and
school’
s
by
c alcium
health
should
validate
comparing
c arbonate)
the
mass
before and
each component prior to mixing
c arry
measure
your
with
beforehand.
your
together,
and
aer
of
lings,
C
i
n
components
salt,
Measure
Then
the
eectiveness
(sand,
that
assessment
o
together.
Remember
risk
aligned
p
the
them
out!
and
v
aer
evaluate
component
the
allow
powdered
y
could
each
identify
methods
r
e
of
the
you:
control measures
disposal
it
and
risk
relevant
Extension
You
consider
hazards
level of
would
four substances.
method,
which
safety policies.
have
must
that
lings
O
you
your
a
in
you
these
iron
n
•
on
protocol
so,
of
method
salt,
l
Assess
If
each
a
sand,
y
Identify the
•
the
doing
of
decided
•
devise
t
i
s
3.
have
assessment
and
In
properties
chapter,
containing
the
the
out
your
mass
of
separation,
each
percentage
again.
recovery
make
sure
Compare
of
each
n
U
Linking questions
factors
are
i
components
t
a
How
c an
How
do
l
u
are
contain
of
considered
a
mixture?
products
of
intermolecular
between
Why
the
o
d
r
o
f
x
a
E
v
O
12
What
two
a
alloys
generally
choosing
reaction
forces
substances?
metallic
in
be
method
to
separate the
purified?
influence
the
type
(Tool 1)
of
mixture
that
forms
(Structure 2.2)
considered
bonding?
a
(Tool 1)
(Structure
to
be
2.3
mixtures,
and
even though they often
Structure 2.4)
y
risk
this
P
a
in
mixture
r
Once
a
c arbonate.
chemic al
2.
ideas
separate
e
to
Structure
1.1
Introduction
to
the
particulate
nature
of
matter
States of matter (Structure 1.1.2)
Solids, liquids and gases
Matter is composed of particles. The types of interactions between these particles
determine the state of matter of a substance: solid, liquid or gas. All substances
s
s
can exist in these three states, depending on the temperature and pressure.
The
states
of
matter
for
of
solid,
substances
(l)
is
a
solid
Water
is
a
liquid
below
•
Water
is
a
gas
are
and
0 °C: H
shown
(g)
for
by
gas.
letters
For
in
brackets aer the
example:
O(s)
2
0
and
O(l)
100 °C: H
P
between
2
above
100 °C: H
O(g).
2
For example, the expression “NaCl(aq)” tells us that sodium chloride is dissolved
t
i
s
in water while “NaCl(s)” refers to the pure compound (solid sodium chloride). The
solid
liquid
gas
•
forces
•
around
Steam,
liquid
but
reached.
A
further
c arbon
water
This
states
it
the
of
until
vaporizes
of
conditions,
change
CO
of
(s),
move
around
liquid
ice are the three states of water
matter
heated.
ice
vibrate,
and
in a
a
as
The
they
(changes
accelerates
and
absorb
particles
temperature
melts
temperature
changes
dioxide,
is
violently,
point,
in
E
the
these
certain
melting.
this
increase
eventually
reverses
At
more
as
v
is
their
energy
rotate,
move
water and
a
positions,
Under
change
absorb
negligible
particles
around faster than
l
will
are
•
u
f
x
O
ice
forces
rotate, and
Changes of state
Substances
vibrate,
i
Figure 11
weaker than
particles
attractive
between particles
t
a
move
•
c an be
•
those in solids
r
o

not
forces
o
are
vibrate in
fixed positions but
do
attractive
volume
compressed
between particles
strong
particles
fixed shape
•
d
are
fixed
no
compressed
between particles
•
no
•
n
attractive
U
•
c annot be
•
y
c annot be
compressed
volume
fixed shape
o
no
C
fixed
•
•
volume
p
r
e
•
fixed shape
v
fixed
•
i
n
•
O
properties of the three states of matter are summarized in gure 11.
n
l
y
A special symbol, (aq), is used for molecules or other species in aqueous solutions.
y
Water
•
liquid
r
•
for
e
formula: (s)
becomes
a
release
to
energy. Solid
vibrate
in
xed
known as the melting point
its
the
or
continue
state
from solid to liquid).
movement of particles, and
gas.
The
decrease
in
temperature
state.
solid
state,
substances
known as
gure12),
c an
turn
into
sublimation,
which
is
is
commonly
gases
directly, without
typic al
used
for
for
dry
ice
(solid
refrigerating ice
2
cream
and
biologic al
samples.

Figure 12
Sublimation of dry ice
13
Structure
1
Models
of
the
particulate
nature
of
matter
The
process
water
opposite
vapour
in
the
to
air
sublimation
solidies
and
is
c alled
forms
deposition.
snowakes
of
At
low
temperatures,
various shapes and sizes
(gure13).
When
when
are
a
substance
energy
a
solid
particles
or
a
gas,
from
and
the
when
from
energy
or
to
a
solid,
the
a
to
less
the
forces
and
condensed
become
when
a
You
becomes
of
state
occurring
in
these
will
maize
known as
non-Newtonian uids,
typic al liquids. The
varies
make
starch
depending
a
viscosity
on
the
2.
force
a
non-Newtonian uid commonly
slime
or
“oobleck”,
and
3.
and
record
relevant
eye
•
250
•
Powdered
•
Water
large spatula
3
cm
t h re e
to
the
fo u r
h e a pe d
b e a ke r.
N o te
i ts
of
a
condensed
molecular
solid.
the
each
of
The
process of
water
•
the
maize
Suppose
you
of
in
gure 14.
starch and mix.
the
by
mixture
adding
achieves
more maize
needed.
exploring
the
harden
if
properties of
tapped,
and
ow
slowly.
the
following:
maize
starch
starch–water
were
relating
Consider
shown
properties and identify the state of
powdered
•
as
are
maize
until
Adjust
should
stirred
matter
question
asked
to
a
mixture.
to
maize
develop
a
research
starch–water
mixture.
possible independent and dependent
variables.
3.
Research
non-Newtonian uids and identify other
examples of these substances.
4.
How
has
about
this
states
experience
of
matter
changed
and
their
the
way
you think
properties?
Reect on
ma i z e
this,
a pp e a ra n c e
completing
the
following
sentence
starters:
and
•
I
•
Now, I think...
used to think...
E
v
Linking questions
Why
are
some
conditions?
Why
are
substances
solid
while
others
are
fluid
under
standard
(Structure 2.4)
some
(Structure
14
2.
a
O
c o n s i s t e n c y.
spoons
l
s ta rc h
or
starch
u
Ad d
maize
f
x
Method
beaker
t
a
r
o
or
o
Spoon
i
d
•
if
the
water,
It
Describe
•
protection.
Materials
1.
1.
qualitative
to
water
time
mixture.
smoothly
n
S afety
more
some
Questions
U
observations
Wear
more
a
o
Identify
or
C
2:
your
water
adding
consistency.
Spend
v
Inquiry
i
n
variables
thick
starch
explore its
Inquiry 1: Identify dependent and independent
add
Continue
applied
Relevant skills
•
Slowly
of non-
properties.
•
a
process.
transformations
p
as
for
n
them.
known
condensed
O
to
uids
to
and,
y
Newtonian
like
less
becomes a gas. These
state
y
changes
r
e
behave
liquid
exothermic
t
i
s
substances,
a
stronger. This happens when a gas
liquid
surroundings is an
Non-Newtonian uids
not
to
the product of
The
do
a
surroundings
deposition of water
Some
state
surroundings. This happens
2,
changes
of
state
Reactivity 1.2)
endothermic
and
some
exothermic?
y
energy
condensed
l
A snowake,
changes
lose
intermolecular
liquid
more
P
releasing
the
a
liquid
a
particles
processes.
substance
becomes
a
the
e
a
the
substance,
Figure 13
becomes
from
by
r
state,
changes
absorbed
endothermic
When

is
s
s
state,
Structure
1.1
Introduction
to
the
particulate
nature
of
matter
gas (g)
s
s

Endothermic and
Figure 15
Orange growers spray their fruit
exothermic
with water on cold nights.
t
i
s
Freezing of water is an exothermic process that
heat) to the fruit,
protecting it against cold
Kelvin temperature sc ale (Structure 1.1.3)
the
particles
of
a
faster,
a
gas
heated,
changes
added
energy
to
is
between
a
they
there
liquid
used
the
to
is
and
lattice
move
no
increase.
particles.
more,
As
particles
Temperature
substances
in
a
faster.
liquid
is a
absorb
vibrate
energy,
more and
temperature change during the periods when
when
disrupt
of
the
a
liquid
solid
changes
lattice
molecules in the liquid.
and
to
a
gas
(gure 16). The
overcome
the
intermolecular
n
U
forces
is
in
in
particles
o
water
solid
while
vibrate
of
energy
C
a
solid
energies
kinetic
i
n
When
the
v
move
rises,
average
p
of
y
temperature
measure
r
e
As
releases energy (in the form of
O
changes of state
n
liquid (l)
Figure 14
l
y
solid (s)
y
P
melting

e
r
freezing
vaporization
m
c
+
d
water
steam
s
om
t
a
u
K
A
a

Figure 17
are kilogram
length,
energy input
The seven base SI units
(kg) for mass,
second
(s) for time,
electric current,
meter (m) for
ampere (A) for
kelvin (K) for temperature,
Graph of the heating curve for water
E
were
accepted
l
freezing
ice
Figure 16
There
water
v
O

f
x
erutarepmet
0
+
water
melting
kg
condensation
i
/
r
o
C°
ice
steam
o
d
100
many
mole (mol) for amount
of substance, and
c andela (cd) for luminous intensity. All units
attempts
to
measure
relative
temperature,
but
the
rst widely
of measurement
temperature
sc ale
was
introduced
by
the
Polish-born
Dutch
c an be derived
from these
physicist
seven base units
D aniel
Gabriel
F ahrenheit.
You
The
kelvin
is
the
base
unit
of
temperature
of
Units
measurements
(SI).
c an
There
be
are
will
learn
more about the mole
measurement in the International
in
System
Structure 1.4.
seven base units, and all other units of
derived
from
these
(gure 17).
15
Structure
1
Models
of
the
particulate
nature
of
matter
Measurement
3
M aking,
recording,
and
communic ating
volume (m
measurements
2
greatly
benets
from
the
Bureau
French
mesures),
agreed
of
Weights
in
the
and
Measures
=
(BIPM,
rene
late
19th
which
–3
),
density
(kg m
),
energy
( joule,
J,
where
The
century, is an
base
)
and
units
including
so
on,
are
several
are
dened
that
Boltzmann constant,
standards.
derived
from
the
seven base
constant,
N
;
and
you
k;
the
according
will
to
seven constants,
recognize, such as the
speed of light,
Plank constant,
c;
the
Avogadro
h
A
International
System
of
Units
(SI,
from
the
r
The
French
The use of universal and precisely dened units is very
Système international d’unités) is the most commonly
seven
units:
(second,
amount
of
current
substance
m),
mass
blocks
important, as it allows scientists from dierent countries
are the
to understand one another and share the results of their
(kilogram, kg),
(ampere,
A),
studies. What other advantages are there to internationally
temperature
shared and continuously updated measurement systems in
(mole, mol) and luminous
the natural sciences? You might want to look up the Mars
(c andela, cd). All other units, such as those of
Climate Orbiter.
each
several
with
universal
dierent
temperature
reference
points.
v
summarized in table 4.
n
O boiling point = 212°
2
H
O
freezing point = 150°
2
o
H
O boiling point = 0°
2
1700s
O
H
freezing point = 0°
2
i
H
O boiling point = 100°
2
t
a
l
u
CGPM
1950s
Triple
BIPM
2018
Kelvin
Table 4
Examples
a
Temperature
unit
kg,
and
for
m
historic al
is
energy,
and
and
practic al
of
related
s.
It
v arious
to
has
practic al
reasons
( J),
been
water
=
273.16 K
k.
energy
which
are
decided
reasons”.
could
of
dened in terms of the Boltzmann
t e m p e ra t u r e
thermal
joules
zero = 0
point
constant,
the
temperature = 12°
to
What
sc ales
and
in
as
turn
keep
do
you
such
it
could
be
expressed
dened in terms of the base
kelvin
think
as
an
some
SI
of
base
these
unit
“for
historic al
be?
bec ame obsolete in
Look
c arefully
at
table
4
above.
Identify
one
thing
you
see, one thing it makes
when the kilogram and all other SI
units were redened
think
about,
as exact quantities
your
16
1700s
H
Absolute
in
points
body
1800s

are
freezing point = 32°
Kelvin
you
based
O
been
these
freezing point = 0°
2
H
have
of
2
Celsius
units
cylinder
to dene a kilogram of
This standard
2019,
v
mass.
E
A platinum–iridium
in the US was used
d
r
o
f
x
O
Figure 18
O
Human
1700s
Delisle

Reference
H
sc ales
Some
o
U
F ahrenheit
D ate
1700s
C
Newton
i
n
Sc ale
y
history,
developed,
p
Throughout
r
e
Thinking skills
ATL
O
t
i
s
intensity
electric
(metre,
building
n
K),
s),
length
Its
y
(kelvin,
measurement.
l
time
base
of
on physic al constants
class.
and
one
thing
it
makes
you
wonder.
Share
your
ideas with
y
system
P
used
1 J
–2
s
units.
seeks to set up and
measurement
1 kg m
e
organisation
continuously
sc ales. The
Bureau international des poids et
established
international
upon
s
s
International
from
Structure
Kelvin
is
temperature
considered
Absolute
any
an
Under
c annot
normal
get
in
proportional
any
that
at
collisions.
colder.
temperature
pressure,
373.15 K.
zero
this
average
kinetic
temperature
M atter
An
of
water
Absolute
the
the
particulate
nature
of
matter
energy of particles and
sc ale.
implies
on
to
to
at
increase
1
degree
boils
on
at
the
absolute
in
the
particles
zero
c annot
temperature
Celsius.
0 °C
is
of
1
c annot
lose
kelvin
equal
to
transfer
heat and
is
equivalent
273.15 K.
100 °C, so that makes the boiling point of
Celsius
sc ale
is
–273.15 °C.
Figure 19
The Celsius and
Kelvin
rounded
to whole numbers)
400 K
water
l
373 K
boils
273 K
freezes
40 °C
v
solid CO
2
more about the
C
o
i
n
100 K
U
liquid
191 °C
learn
energy of particles in
Reactivity 2.2.
150 K
150 °C
will
kinetic
195 K
100 °C
82 K
air
50 K
250 °C
o
0 K
zero
i
Kelvin
Celsius
u
l
a
v
O
f
x
t
a
r
o
d
absolute
n
200 °C
273 °C
You
p
200 K
dry ice
78 °C
y
r
e
250 K
50 °C
O
t
i
s
300 K
water
0 °C
n
y
350 K
50 °C
y
P
sc ales for temperature (all values are
100 °C
r
t
e
water
(0 K)
energy
increase
is
absolute
Introduction
s
s
to
zero
kinetic
hence
an
1.1
Linking questions
is
sample
E
What
the
graphic al
at
fixed
a
distribution
temperature?
of
kinetic
energy
values of particles in a
(Reactivity2.2)
What must happen to particles for a chemical reaction to occur? (Reactivity 2.2)
17
Structure
1
Models
of
the
particulate
nature
of
matter
End-of-topic questions
5.
Which
changes
of
state
are
opposite
to
each other?
Topic review
1.
Using
your
answer
the
knowledge
guiding
from the
question
as
Structure 1.1
fully
as
melting
and
s
s
A.
condensation
topic,
B.
vaporization and deposition
C.
deposition and sublimation
D.
sublimation
possible:
e
How can we model the particulate nature of matter?
Which
of
the
and
freezing
following
statements
is
r
6.
incorrect?
Exam-style questions
A.
solids
and
liquids
are
almost
incompressible
2.
Which
of
the
following
are
examples
of
particles
liquids
D.
particles
in
both
and
solids
gases
and
have
no
liquids
are mobile
l
B.
C.
homogeneous
xed shape
mixtures?
elements
Steel
III.
Aqueous
KMnO
potassium manganate(VII),
A.
oxygen
B.
hydrogen
(aq).
I and II only
D.
I, II and III
B
exothermic
CO
C
endothermic
CO
(s)
→
CO
(s)
→
C(g) + O
2
(s)
2
following
sodium
could
I and II only
C.
I and III only
E
B.
a
I only
v
O
distillation
be
u
ltration
I, II and III
A.
decrease
B.
increase
C.
decrease
D.
increase
(g)
Explain
why
by
by
temperature
increase
in
on
the
Celsius
temperature
by
sc ale is
20 K?
20 °C
20 °C
by
by
293.15 °C
293.15 °C
the
proportional
Celsius
to
Kelvin
temperature
average
temperature
temperature
is
kinetic
not,
increment
is
even
the
is
directly
energy but the
though
same
in
a
1-degree
each
sc ale?
[2]
used to
10.
Ionic
salts
c an
unbalanced
lead(II)
be
ionic
broken
down
equation
for
in
electrolysis. The
the
electrolysis of molten
bromide is:
2+
Pb
a.
+
Br
One
→
of
Pb + X
the
formula
b.
Balance
c.
The
of
products
product
the
point
and
of
is
lead,
Pb. State the
X.
[1]
equation.
electrolysis
c arriedout
matter
18
in
the
from a solution of sodium
l
evaporation
II.
C(g) + O
2
water?
I.
III.
D.
chloride
→
(g)
2
2
methods
f
x
in
(s)
9.
2
CO
i
solid
chloride
the
CO
→
(g)
t
a
obtain
A.
r
o
endothermic
of
to
oxygen in dry ice
Extended-response questions
(g)
2
d
2
Which
change
equivalent
n
CO
o
process
exothermic
U
endothermic?
D
magnesium and sulfur in magnesium sulde
Equation describing the
A
water
by
o
Exothermic or
in
other
D.
describes the sublimation of dry ice
dioxide)?
oxygen
vibrate
c arbon
Which
and
c an
each
C.
C
(c arbon
i
n
correctly
8.
from
y
III only
C.
gases
nitrogen in air
and
p
B.
and
r
e
II only
v
A.
and
separated
methods?
4
4.
liquids
be
O
physic al
c an
n
Which
t
i
s
II.
What
solids,
Air
7
.
3.
in
y
I.
at
of
molten
380 °C.
boiling
each
of
Write
equation
gave
data,
(b).
bromide is
reference to melting
species
state
in
lead(II)
With
point
the
temperature.
you
[1]
deduce the state of
in
the
symbols
equation at this
in
the
balanced
[2]
y
P
Multiple-choice questions
Structure
11.
The
kinetic
mass
×
energy
the
square
of
of
particles
the
is
c.
equal to half of their
Once
to
2
mv
=
.
Determine
how
much
the
the
obtain
Introduction
excess
removed,
velocity of the particles:
1
E
1.1
the
the
copper(II)
student
pure
to
particulate
oxide
needed
crystals
of
had
to
nature
matter
been
gure
copper(II)
of
out
sulfate
how
from
speed of
k
2
the
molecules
in
a
pure
gaseous
substance
will
solution.
couldfollow
the
Kelvin
temperature
is
doubled.
to
obtain
pure, dry copper(II)
s
s
when
Describe a method the student
increase
[2]
sulfatecrystals.
12.
Pure
c aeine
is
a
white
[3]
powder with melting point
14.
Study
the
gure
below.
235 °C.
e
vaporization
the
melting
chemist
is
point
of
c aeine
investigating
the
in
kelvin.
ec acy
of
[1]
three
r
State
A
C°
a.
b.
100
condensation
all
once
and
melting
0.960 g.
collects
point
of
the
the
She
uses
following
each
data
for the
product:
water
ice
+
Method 3
0.229
0.094
0.380
t
i
s
Method 2
O
freezing
ice
Mass of c aeine
obtained / g
Melting point of
188–201
a.
Explain
why,
C alculate
ii.
C alculate
the
mean
and
range
of
the
mass
constant
c aeine
student
word
a
product.
minimize
the
random
oxide
acid
is
insoluble
equation
for
this
sulfate
with
in
solution
excess
water.
reaction
is
by
copper(II)
as
f
x
acid
powder
could
was
was
be
chemic al
for
this
a
The
balanced
heated,
added
v
O
b.
a
symbols,
l
state
observed
E
use
to
a
the
iodine
sample
of
energy
remains
time.
chloride
atmospheric
in
[2]
100.0 g of
temperature and
properties:
–3 °C
similar
to
the
curve
one
for
a
in
gure 16
sample of this
solution.
exists
temperature
solid
cold
that
When
surfaces
pressure,
as
[2]
diatomic
molecules, I
.
At
and
readily
cooled,
without
solid
iodine
pressure,
forms
it
is
violet
a
lustrous purple-
fumes
when
heated
gaseous iodine deposits on
condensing.
melts
at
Under
114 °C
to
increased
form
a
deep-
equation, including
violet liquid.
[2]
copper(II)
Formulate
was
in
equations
that
represent all changes of
oxide
mentioned
above.
[3]
excess and
b.
State
the
melting
c.
Suggest
how
gaseous
iodine.
point
of
iodine
in
kelvin.
[1]
suspended in the solution,
reason, a method the student could
remove
period
sodium
heating
black
gently.
quickly sinking to the bottom of the beaker. Suggest,
giving
increasing
the
101 °C
graph
room
state
it
point:
point:
the
a
following
sodiumchloride
Elemental
a.
then
of
standard
the
a
show
for
the
of
2
follows:
reaction.
until
boiling
to
15.
u
Write
melting
of
oxide.
sulfuric acid + copper(II) oxide → copper(II) sulfate + water
a.
has
Sketch
0 °C
5.00 g

b.
at
(at

[1]
[1]
copper(II)
sulfuric
pressure)
t
a
The
c aeine
r
o
Copper(II)
[2]
experiment.
prepares
dilute
to
of
water
reason, which method
i
reacting
way
solution
pure
o
A
one
this
purest
a
d
13.
in
gures.
giving
A
appropriate number
n
error
an
U
Suggest
the
to
spite
o
signic ant
Determine,
gave
c.
answer
C
of
iii.
your
[2]
percentage yield of Method 1.
i
n
Give
obtained.
the
v
of
in
temperature
p
input,the
i.
energy input
y
229–233
r
e
190–220
caeine product / °C
steam
water
melting
0
Method 1
+
n
and
is
water
y
yield
c ases
theoretic al
l
method
three
The
y
in
methods.
erutarepmet
yield
extraction
P
/
c aeine
excess
copper(II)
oxide.
liquid
iodine
c an
be
obtained
from
[1]
[2]
19
The nuclear atom
Structure 1.2
Structure 1.2.1
The
answer
to
this
question
was
obtained
by
nucleus
years
of
brilliant
how we know
the
late
in
the
atoms
idea
that
electrons
that
Structure 1.2.2
matter
dierent
was
chemic al
Isotopes
of
are
atoms
gaining
relative
—
M ass
atomic
spectra
masses
of
are
electricity
study
the
and
popularity. The
radioactivity
structure
used to determine
elements
of
the
allowed
atom
from their isotopic
itself.
O
n
of
to
same element with
t
i
s
scientists
the
(known as
composition.
discovery
of
neutrons.
l
was
—
numbers
y
theory)
Negatively
occupy the space outside the nucleus
were indivisible and
reactions
the
atomic
charged, dense
(nucleons).
more fascinating than
Structure 1.2.3
the
neutrons
LHA
of
rearranged
positively
The structure of the atom (Structure 1.2.1)
atom
contains
neutrons
occupy
electrons
are
very
2.
It
is
a
It
an
has
highly
a
the
which
Atoms
nucleus.
nucleus
structure
to
the
The
contains
contain
protons,
protons
electrons,
neutrons and
atom
containing
itself.
virtually
all
the
mass of the atom.
charge.
designed
particles
made
are:
itself
also
are
by
Ernest
were
given
in
red
Rutherford
toward
a
in
1911,
sheet
of
positively
gold
charged
foil. The main
gure 1.
i
Rutherford’
s explanation
detector
source
+
+
u
l
a
E
v
O
large
of
t
a
f
x
alpha
outside
comparison
dense
alpha
observations
movable
in
o
d
r
o
beam of
the
positive
experiment
radioactive
alpha particles
of
small
n
In
factors
U
3.
nucleus,
nucleons).
o
is
region
as
C
It
vast
v
key
1.
the
charged
known
known as subatomic particles.
i
n
The
positively
p
which
a
(collectively
y
and
r
e
An
Most alpha
particles
are
undeflected
atom
+
gold
foil
Some alpha
vacuum
+
particles
are
deflected
slightly
A
few alpha
undeflected
+
particles
slight
bounce
+
deflection
deflection
off nucleus

20
Figure 1
Rutherford’s gold
foil experiment
y
composed
is
what is known
1800s,
a
and
P
In
contain
protons
r
the question of
of
research. Sometimes, the
charged
question of
Atoms
composed
e
100
—
over
s
s
Understandings
How do nuclei of atoms dier?
Structure
1.2
The
nuclear atom
F alsic ation
experiment
vulnerable
The
plum-pudding
claim
present
throughout.
red
at
the
undeected.
model,
this
gold
were
foil
the
would
Rutherford’
s
paving
the
way
that
blob
atom
was
that
never
electrons
be
stands
proven
c ase, all alpha
knowledge
have
degree
results
for
the
with
the
gone
through its
of
new model of the atom.
is
falsiable.
up
to
true
means
C an
a
Determine
in
the
which
gold
foil
experiment and the
observation
is
explained
Property
nucleus
has
a
positive
Occ asionally, some of the alpha
The
nucleus
is
particles
comparison to the size of the atom.
straight through the gold foil.
repelled
The
nucleus
virtually
all
nucleus.
is
very
the
dense, containing
mass of the atom.
model,
of
the
the
in
the
solar
entire
atom.
by
way
electrostatic
electrons
planets
the
orbit
atomic
instead
of
the
by
l
a
E
Figure 2
experiments
orbit
Sun.
by
claim?
proposing the
Rutherford model
the
Just
positively
as
contains
gravity,
steer it in
the
the
over
Sun
contains
99.9%
electrons
(gure2).
charged atomic
are
of
held
the
99.8%
mass of
around the
u
–
his
nucleus
attraction.
v
O

as
mass,
However,
f
x
nucleus
same
system’s
charged
of
t
a
nucleus
negatively
results
known as the
i
this
the
also
o
summarized
r
o
In
c an
n
U
Rutherford
planetary model of the atom,
d
1911,
evidence
a
by a
nature of scientic
o
are
approaching the
C
particles
closely
very small in
i
n
alpha
straight back.
v
bounced
charge.
p
The
c an
y
r
e
Observation
In
by
property.
when
accompanied
falsify
but
y
observations
are
n
the
nucleus.
they
l
show
the
Nearly all the alpha particles went
The
further
strong
O
which
is
y
below
of
that
certainty. Scientic
provisional
t
i
s
lists
properties
always
counterexample
Activity
The
testing
absolute
The
that
means
P
directions.
This
contradicts them. A scientic
severe
therefore
new
single
that
with
uncertainty.
knowledge
contradicted the
development of a
are
evidence
r
existing
If
suggested
charged
to
e
particles
atoms
model
positively
the atomic model
s
s
preceded it, namely the “plum-pudding model”.
amorphous
falsied
claims
gold
that
an
foil
Scientic
The
+
–
electron
–
proton
–
neutron
+
+
– nucleus
The Rutherford model of the atom
21
Structure
1
Models
of
the
particulate
nature
of
matter
Models
Scientists
use
models
to
represent
Atoms
natural phenomena. All
themselves
are
extremely small. The diameter of
–10
models
have
2.
The
useful
size
Consider
of
model
the
of
which
the
nucleus
the
should
depiction
is
be
of
the
exaggerated
most
identied and
atom
but
it
in
gure
unit
serves as a
atoms
used
to
is
in
the
range
1 × 10
–10
to
5 × 10
m. The
describe the dimensions of atoms is the
picometre, pm:
–12
nuclear atom.
1 pm = 10
In
the nucleus is hard to fully appreciate. Rutherford’
s native
X-ray
m
crystallography
dimensions
the
a
commonly
angstrom,
used
unit
–10
1 Å
= 10
m
the
atomic
radius of the uorine atom is
the centre of the pitch from the top row of seats. If a golf
–12
60 × 10
m
(60 pm).
To
convert
this
toÅ
ball were placed at the centre of the eld, the distance
using
the
we
conversion
c an use
factors
between you and the golf ball would represent the
above:
–12
10
relative
volume
of
open
space
in
the
atom
is
vast, and
m
60 pm ×
1 Å
t
i
s
The
×
=
–1
0.60 Å
=
6.0 × 10
Å
–10
our
simple
representation
of
Rutherford’s atomic model
In
in
gure
2
is
obviously
spite
a
tiny
of
its
of
much
volume of the atom and the diameter of an atom is
Rutherford
100 000 times the diameter of the nucleus.
is
of
our
Rutherford’
s
thinking
rumoured
to
on
have
the
work
has
formed the
structure of the atom.
said to his students:
y
r
e
approximately
m
limitations,
unrealistic. The nucleus occupies
basis
10
O
1 pm
n
y
given
distance between the electron and the nucleus.
l
dimensional analysis,
All science is either physics or stamp collecting!
p
C
o
v
i
n
n
U
o
i
t
a
u
l
a
22
Eden Park,
v
E
Figure 3
of the eld
d
r
o
f
x
O

Auckland,
New Zealand.
If the atom were the size of the stadium,
the nucleus would
look like a golf ball in the centre
y
example,
P
being at Eden Park stadium (gure 3) and looking down at
For
for atomic
symbol Å:
r
New Zealand is a great rugby-playing nation. Imagine
is
e
The vast space in the atom compared to the tiny size of
s
s
understood.
limitations,
Structure
1.2
The
nuclear atom
TOK
All
the
could
we
argued
physicists,
Switzerland
Binnig
that
gave
Prize
in
the
scientists
in
and
assume
only
that
“real”
Heinrich
sc anning
generates
Physics
are
atoms
when
Rohrer,
tunnelling
are
they
real.
c an
working
microscope
However, it
be
at
seen. In 1981
IBM in Zurich,
(STM),
an
electron
three-dimensional images of surfaces at the atomic
the
ability
1986
was
to
observe
awarded
individual
to
Binnig
atoms
and
directly. The
Rohrer
for their
r
groundbreaking work.
c an
nd
an
atomic
sc ale
lm
created
by
IBM
c alled
A Boy and his Atom
on
the internet.
are
material
tools
of
Relative
nucleus
same
mass
particles
also
as
are
l
O
charge
and
it
relative
charge
a
negligible
–1
value
particles
are
a
of
single
electron
approximately
is
known as the
1.602 × 10
of
a
the
charge
proton as +e.
electrons
and
of
The
commonly
protons
an
expressed
electron
symbol
have
charges for
electron
elementary
The
C.
The
in
elementary
actual
e
is
c an
oen
charges
of
be
represented as –e,
omitted,
–1
and
so
+1,
it
is
and
customary
masses
and
charges of
charges of
charge
units.
the
to
particles
c an
be
found in the
For
data
example,
neutron and
outside nucleus
these
subatomic
Relative masses and
nucleus
–19
has
Table 1
Loc ation
+1
0
by
neutral subatomic
The
the proton,
1
c arried
a
proton.
shown in table 1.
Relative
a
(e)
E
electric
v
electron
contains
the
t
mass
1
neutron
charge
the
subatomic
proton
The
the
u
Particle
that
nearly
the
production or acquisition of
o
charges
shown
with
observations of the
i
and
have
neutron,
f
x
masses
the
the
t
a
experiments
particle,
r
o
Other
in
d
knowledge?
O
important
make
n
How
to
y
c apacity
p
human’
s
o
extended
C
technology
natural world?
n
l
y
t
i
s
r
e
A still from A Boy and his Atom
U
Has
v
Figure 4
i
n

y
P
You
e
This
Nobel
Gerd
discussed
objects
invented
microscope
level.
have
that
s
s
two
models
be
booklet.
charge
say that
respectively.
23
Structure
1
Models
of
the
particulate
nature
of
matter
How small is small?
Relevant
skills
Tool
3:
Apply
•
Tool
3:
Use
and
and
use
SI
prexes and units
interpret scientic notation
s
s
•
Instructions
1.
A
variety
objects
of
small
but
in
lengths
rather
order
of
are
based
size,
shown in table 2. Without looking at their
on
what
from
you
smallest
stop,
atom,
267 pm
diameter
0.30 mm
diameter
fullerene,
C
250 µm
length
150 pm
diameter
0.71 nm
60
y
Lengths of various small items
p
2.
Table 2
r
e

O
full
bond,
t
i
s
c arbon
0.10 mm
n
iodine-iodine
printed
thickness
diameter
Convert the length values into metres and state them in standard form to
v
two signicant gures. Refer to the following conversion factors:
micro, µ: 10
•
nano, n: 10
•
pico, p: 10
•
femto, f: 10
the
4.
Provide
ATL
a
web
than
the
i
5.
question
Conduct
smaller
–15
values
following
the
full
your
1
in
table
values
to
school’s
nd
given
reference
t
a
u
Atomic number
2
in
order
of
increasing
size.
Was
the
list
you
correct?
search
o
d
r
o
f
x
for
three
in
your
citing
more
table
and
2,
values to add to the list: one
one
larger,
information
referencing
and the nuclear
and
one
intermediate.
sources in question 4,
system.
symbol
As of 2023, there are 118 known elements, given atomic numbers 1 to 118. The atomic
l
a
E
v
O
24
for
–12
n
gave
length
–6
–9
C
List
i
n
•
U
3.
milli, m: 10
o
–3
•
number of an element is also the number of protons in the nucleus of that atom. Gold,
atomic number 79, has 79 protons, while carbon, atomic number 6, has 6 protons. As
all the relative mass is in the nucleus, the dierence between the atomic number and
mass number is the number of neutrons in the element. Gold has atomic number 79 and
mass number 197
. Therefore, it has 197 – 79 = 118 neutrons. Each element is neutral,
with no charge, so the number of electrons in a neutral atom must equal the number
ofprotons.
y
paper,
cell,
0.84 fm
l
of
Length
radius
y
onion
each item, list these
P
sheet
charge
about
largest.
r
Item
proton,
know
to
e
lengths,
Structure
1.2
The
nuclear atom
Activity
Determine
Atomic
the
missing
symbol
values
Atomic
from
the
table.
number
M ass
number
Protons
Neutrons
Electrons
s
s
O
8
13
27
85
37
35
27
32
r
120
100
use
nuclear
symbol notation,
X, to denote the number of
isotope,
for
protons
Z
is the
example,
and
electrons in an atom.
atomic
with
mass
number,
number
and
197
X
is
and
A
the
represents
chemic al
atomic
the
mass number of the
symbol
number
79,
(gure 5). Gold,
would
number
of
chemic al
where
neutrons
compounds
are
protons.
For
the
compound
ionic
example,
the
ion
nucleus
is
sharing
magnesium
a
2+
two
greater
protons
+
also
number:
12
displayed
E
(12
oxygen
react
with
the
number
number
the
or
loses
of
of
nuclear
As
fewer
oxygen
M agnesium
the
in
electrons.
more
to
produce
electrons
positively
negatively
2
to
charged
charged
notation
24
24
result, these
atoms
two
symbol
a
electrons than
electrons
below:
+
charge: 2+
(12
with
a
atom
2–
protons
–
10
electrons)
Mg
12
chemic al element: Mg
protons)
gains
the
(magnesium)
two
electrons
lost
by
magnesium
to
produce
an
16
ion
form
protons
electrons)
atomic number: 12
The
as
than
a
mass
(12
is
having
atoms
charge,
l
charge
v
O
resulting
transferring
oxide.
remaining (10).
The
or
neutral,
magnesium
with
(12)
by
longer
u
f
x
a magnesium
in
no
i
form
sometimes
t
a
atoms
r
o
Atoms
The nuclear symbol notation
o
Figure 5
=
protons
d

of
n
U
atomic number
C
i
n
X
Z
number
symbol
for the element
o
=
N
v
N
+
Z
nuclear
p
mass number
=
a
y
Au.
79
A
have
r
e
197
symbol notation of
O
t
i
s
Z
neutrons,
n
frequently
l
y
A
Chemists
y
207
69
80
P
Pb
e
80
negative
charge.
The
nuclear
symbol
for
the
oxide ion is
oxide
2–
O
.
8
25
Structure
1
Models
of
the
particulate
nature
of
matter
The overall chemical equation for the reaction between magnesium and oxygen is
1
2+
O
Mg +
Ionic
in
bonding
is
→
Mg
2–
+ O
2
discussed further
2
Structure 2.1
2+
2–
Mg
+ O
ions
Ionic
is
more
result
bonds
in
a
hold
commonly
force
the
of
ions
written
attraction
together
as
MgO,
between
to
form
as
the
them
solid
opposite
charges on the
known as an
magnesium
s
s
two
ionic bond.
oxide.
e
Activity
Linking questions
the
21
an
ion
with
electrons,
What determines the different chemical properties of atoms? (Structure 1.3)
24
and
28
How
does
the
atomic
number
relate to the position of an element in the
neutrons.
periodic
table?
(Structure
3.1)
l
n
y
t
i
s
Isotopes (Structure 1.2.2)
O
Isotopes are dierent atoms of the same element with a dierent number of neutrons.
As a result, they have dierent mass numbers, A, but the same atomic number
, Z.
35
37
Cl, and one
17
y
r
e
Chlorine, for example, has two isotopes: one with mass number 35,
Cl. They have similar chemical properties, as they are both
with mass number 37
,
17
p
chlorine atoms with the same number of electrons, but dierent physical properties,
such as density, because atoms of one isotope are heavier than atoms of the other
.
is
hydrogen-2
radioactive,
consists
of
(deuterium).
so
it
does
two
The
U
C
(gure6),
hydrogen
i
n
and
not
o
occurring
(protium)
v
Naturally
stable
third
occur
in
isotopes,
isotope
nature
in
of
hydrogen-1
hydrogen, tritium
signic ant quantities.
Activity
the
table
below
n
Copy
and
complete
it
by
deducing
the
nuclear
symbols and/
Isotope
i
hydrogen-1
Z
symbol
N
A
(protium)
t
a
H
1
(deuterium)
hydrogen-3
(tritium)
u
l
a
A portable tritium light
Nuclear
1
hydrogen-2
Atomic
v
E
Figure 6
d
r
o
f
x
O

o
or composition of these isotopes.
numbers
of
1
3
isotopes
are
oen
omitted
in
nuclear
symbol
notation.
For
37
example,
‘Cl’
17
,
tells
so
you
isotope
the
including
written
A ,
the
with
listed
a
for
of
chlorine
isotope
the
chlorine
atomic
hyphen,
each
is
number
such
element
with
as
on
mass
and
is
therefore
not
37
must
necessary.
chlorine-37,
the
number
periodic
or
c an
is
not
written as
Cl.
have an atomic number of
These
Cl-37
. The
table
be
a
isotopes
c an also be
relative atomic mass,
whole
number
bec ause it
r
is
the
weighted
average of all isotopes of that element.
Natural abundance (NA)
of
an
isotope
is
the
percentage of its atoms among
source.
The radioactive dec ay of tritium produces
high-energy electrons (beta particles).
all
atoms
of
the
abundances
given
for
all
element
isotopes
of
found
an
on
our
element,
planet.
we
c an
If
we
know
c alculate
the
the
natural
average
A
of that
r
These electrons hit
a uorescent material
element.
The
opposite
task
(c alculation
of
natural
abundances
from
A )
r
and
26
make it
glow in the dark
only
if
the
element
is
composed
of
two
known isotopes.
is
possible
y
for
protons,
symbol
P
notation
nuclear
r
Deduce
Structure
1.2
The
nuclear atom
Worked example 1
C alculate
the
A
r
for
iron
Isotope
using
N atural
the
values
in
the
following
table.
abundance (NA)/ %
54
5.845
Fe
91.754
s
s
Fe
56
57
Fe
2.119
Fe
0.282
We
know
A
r
=
average
abundance
of
the
natural
values
add
abundance
to
57
×
100%
of
so
each
we
isotope
divide
by
multiplied
100
to
by
obtain
their
the
mass
numbers.
average.
r
×
5.845
+
56
×
91.754
+
2.119
+
58
×
y
54
A
0.282
=
= 55.91
100
two
stable
of
isotopes
each
of
chlorine:
isotope
given
Cl-35
that
A
and
for
Cl-37.
chlorine
C alculate
is
×
NA
of isotope 1)
+
(A
of isotope 2
(35 × NA
of
Cl-35) + (37 × NA
of
Cl-37)
=
x
=
NA
of
Substituting
Cl-35, then 100
in
the
above
equation
35.45
the
brackets
35.45
100
rearrange in terms of
3700
x
=
x:
22.5.
x
terms
gives:
Therefore,
the
natural
abundance
of
Cl-35
is
77.5%
and
Cl-37 22.5%.
a
actual
x =
l
77
.5 and 100
v
O
The
resolving the
3545
2
x =
gives:
u
f
x
Then
and
Cl-37.
t
a
2x
=
r
o
3700
of
i
100
Expanding
NA
o
x)
=
=
d
35x + 37(100
x
of isotope 2)
n
Let
35.45
U
100
NA
C
Therefore:
×
i
n
100
o
of isotope 1
v
(A
=
r
natural
p
Solution
A
the
35.45.
r
y
(NA)
r
e
are
abundance
O
t
i
s
Worked example 2
There
l
Therefore:
n
up
y
natural
P
The
r
Solution
e
58
35
natural abundances of
37
Cl
and
Cl
are
75.8
and
24.2%,
respectively.
Average
A
values
for all elements
r
The results of our calculations are slightly dierent because we used mass numbers,
are
E
which
rounded
are
35
values
for
the
actual
masses of the
given in the data booklet and in
37
Cl
and
Cl
atoms.
the periodic table at the end of this
book.
27
Structure
1
Models
of
the
particulate
nature
of
matter
s
s
at
e
Density
Melting
Boiling
point / °C
point / °C
Compound
–3
r
4 °C / g cm
1
H
0.00
1.106
3.82
H
2
Figure 7
A pellet
of enriched
properties
of
uranium
235
used
(increase
as fuel in nuclear reactors
nuclear
in
only
0.72%
one
type
of
of
and
in
a
U)
at
particular
progress
of
t
a
(gure8).
Her
bomb
stating
“I
will
to
work
with
is
(Mt),
a
technology
as
such
well
Frisch
In
later
being
up
is
U. The
enrichment
(gure7), as most
while
c an
This
the
natural
also
make
oen
have
nuclei
as
the
aer
uranium
it
possible
referred to as
to
years,
the
in
large
It
from
atoms
releasing
led tothe
bomb.
the
discovery
the
has
Meitner,
Meitner
developed
ethic al,
consequences.
of
atomic
Lise
doctorate
led
may
economic
development.
named
physics
Otto
1939.
is
U,
substance
and
splitting
one
fuel
235
applic ations
cultural
energy,
receive
Nature in
atomic
you
nuclear
history
published in
C an
of
their
involves
energy,
o
in
of
meitnerium
i
woman
109,
which
C
amounts
and
social,
n
U
ssion,
colossal
Element
l
u
Austrian-Swedish physicist
Lise Meitner in 1906
a
E
v
O
28
d
r
o
Figure 8
f
x

Nuclear
development
science
politic al,
for the
o
i
n
in
environmental,
nuclear
reactions.
Global impact of science
Developments
of
least 3% of
v
isotope labelling.
over
with
isotope.
isotope
mechanisms
238
U and
used
238
U
uranium
this
are
p
the
require
isotopes
y
Enriching
track
proportion of
these
r
e
contains
to
the
reactors
235
uranium consists of two main isotopes,
physic al
O

in
n
occurring
dierences
heavy water
t
i
s
Naturally
Physic al properties of normal and
101.4
l
Table 3
y

100.0
was
US.
the
second
University of Vienna
of
nuclear
ssion,
invited to work on
She
declined, famously
have nothing to do with a bomb!”
think
of
other
ethic alimplic ations?
scientic
developments
that
have had important
y
1.000
O
P
O
2
2
Structure
1.2
The
nuclear atom
LHA
Practice questions
Linking question
1.
State
the
nuclear
symbols
numbers
of
2.
Naturally
occurring
protons
and
for
potassium-39
neutrons
in
the
and
copper-65.
nucleus
of
each
Deduce the
How
isotope.
c an
provide
sulfur
has
abundances:
C alculate
33
S(95.02%),
the
isotopes
average
A
with
the
following
34
S(0.75%),
value
for
mechanism?
36
S(4.21%)
and
isotope
tracers
evidence
for
a
reaction
natural
s
s
32
four
(Reactivity 3.4)
S(0.02%).
sulfur.
r
3.
The actual
A
value
of
sulfur
is
32.07.
Suggest
why
your
answer to the
question
diers
from
this
value.
r
(gure
in
a
9)
is
an
instrument
used
to
detect
the
relative
sample.
positive
ions
are
(stage 5)
(deflected most)
accelerated
to
(stage 3)
vaporize
magnet
sample
are
atoms
c ations.
into
then
lose
For
some
Cu
of
(g) + 2e
their
and
with
vaporized
high-energy
electrons
copper
l
atoms
to
form
c an
be
(stage 1). The atoms
electrons
positively
ionized
as
(stage
2).
As
charged ions,
follows:
a
O
→
instrument
example,
+
Cu(g) + e
the
bombarded
least)
o
the
known as
injected
u
result,
is
sample
f
x
a
the
t
a
sample
within
of a mass spectrometer
r
o
The
Schematic diagram
S
i
Figure 9
(stage 2)
d

beam to
sample
(deflected
n
ionize
U
electron
heaviest particles
C
N
o
inject
sample
v
to
i
n
inlet
(stage 4)
(stage 1)
y
filament
electric
p
heating
the
r
e
field
in
O
t
i
s
detector
lightest particles
n
isotopes
y
of
l
mass spectrometer
abundance
y
P
M ass spectrometry (Structure 1.2.3)
The
e
r
previous
The resulting ions are then accelerated by an electric eld (stage 3) and deected by
v
a magnetic eld (stage 4). The degree of deection depends on the mass to charge
ratio (m/z ratio). Particles with no charge are not aected by the magnetic eld and
E
therefore never reach the detector
. The species with the lowest m and highest z will
be deected the most. When ions hit the detector (stage 5), their m/z values are
determined and passed to a computer
. The computer generates the mass spectrum
of the sample, in which relative abundances of all detected ions are plotted against
their m/z ratios(gure 10).
29
Structure
LHA
u
1
Models
Figure 10
of
the
particulate
nature
of
matter
M ass spectrum of a
sample of copper
100
80
s
s
ytisnetni
60
r
e
evitaler
40
62
64
66
68
t
i
s
m/z
operational
examination
details
of
of
11
shows
boron
a
from
mass
this
spectrum
mass
from
a
sample
of
boron.
C alculate
spectrum.
relative
80.1
then
derive
of
11,
a
which
c alculate
A
10
12
of boron
the
has
o
number
c an
to
which
i
We
10,
E
mass
need
of
u
we
number
8
a
Solution
First,
6
m/z
M ass spectrum
mass,
19.9
l
Figure 11
4
v
O

2
assessed in
t
a
r
o
f
x
0
d
evitaler
0
be
n
U
ytisnetni
50
atomic
not
o
C
i
n
100
the
v
r
will
y
,
spectrometer
p
A
mass
r
e
Worked example 3
Figure
the
papers.
O
The
information
relative
has
by
a
from
graph.
abundance
relative
finding
the
of
abundance
sum
of
The
19.9%.
the
of
peak at
The
m/z
peak at
=
10
m/z
represents
=
11
an
isotope
with
a
mass
represents an isotope with a
80.1%.
relative
abundance
of
each
isotope
multiplied
by
its
mass
r
number. The relative abundance values add up to 100%, so we divide the result by 100 to obtain the average.
11
×
80.1
+
10
×
19.9
=
100
30
10.8
n
60
l
0
y
0
y
P
20
Structure
1.2
The
nuclear atom
LHA
Data-based questions
1.
Estimate
atomic
the
relative
mass,
A
,
for
abundance
this
of
element
each
and
isotope
identify
from
the
gure
12.
Use
your
estimates
to
c alculate
the
relative
element.
r
Figure 12
s
s
t
M ass spectrum
of unknown element
6
e
5
r
204
203
206
205
207
208
209
p
m/z
spectrometry
of
is
used
cosmic
for
discovering
origin.
For
(gure14).
and
nickel
of
have
these
similar
two
properties
metals
are
(gure13).
they
elements
are
relative
c an
in
common
atomic
easily
be
masses.
of
iron
by
meteorites
the
mass
isotopic
spectrometry
n
100
ytisnetni evitaler
The
samples,
However,
distinguished
cobalt
60
40
20
0
58
60
0
62
58
60
62
m/z
m/z
the
relative
atomic
abundance
mass,
A
and
of
hence
each
isotope
deduce
for
whether
nickel.
cobalt
Use
or
your
nickel
estimates
has
the
actual
A
value
for
nickel
to
c alculate
larger
r
3.
geologic al
components
M ass spectra of cobalt (le) and nickel (right)
Estimate
relative
identic al
so
o
Figure 13
E

i
0
specic
nickel
a
v
0
of
and
80
u
20
nearly
dierent,
nickel
l
O
f
x
ytisnetni evitaler
60
cobalt
t
a
r
o
80
40
d
100
and
very
U
compositions
presence
C
Cobalt
the
example,
o
those
i
n
including
v
M ass
y
r
e
0
O
1
n
y
t
i
s
2
2.
l
evitaler
3
y
P
ytisnetni
4
its
A
r
is
58.69.
Suggest
why
your
result
in
question
2
is
dierent.
r
31
1
Models
of
the
particulate
nature
of
matter
LHA
Structure
s
s
e
r
to
c an
be
practice
in
various
databases
average
i
n
Relevant skills
Tool
2:
Identify
•
Tool
3:
Percentages
and
extract
U
C
•
data
atomic
on
the
mass
internet,
values
giving
you
from authentic
o
data.
found
c alculating
y
spectra
chance
v
a
in 1864 in the S ahara Desert
p
r
e
Mass spectra
M ass
found
n
l
iron meteorite,
O
Tamentit
from databases
Instructions
Using
a
database
n
1.
of
your
choice,
search
for
the
mass
spectra
of
three
dierent elements.
From
the
mass
element.
i
3.
Compare
t
a
booklet.
o
d
r
o
u
l
f
x
a
E
v
O
32
2.
your
spectra,
c alculate
c alculated
Comment
on
the
relative
any
relative
atomic
dierences
atomic
mass
you
to
mass
that
of
each
stated in the data
observe.
Linking question
How does the fragmentation pattern of a compound in the mass spectrometer
help in the determination of its structure? (Structure 3.2)
y
P
y
Figure 14
t
i
s

Structure
1.2
The
nuclear atom
End-of-topic questions
5.
Which
of
the
following
statements
are
correct?
Topic review
1.
Using
your
knowledge
from the
Structure 1.2
Nearly
all
mass
of
the
atom
is
s
s
I.
contained within
topic,
its nucleus.
answer
the
guiding
question
as
fully
as
possible:
II.
The
mass
number
shows the number of
protons in an atomic nucleus
Isotopes
of
the
same
element
numbers
correct
for
2+
Cu
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
?
Neutrons
Electrons
34
27
B
29
34
31
6.
Which
of
34
63
31
34
29
27
A.
values
are
the
same
for both
2
H
and
H
2
of combustion
IV.
density
C.
II and III only
D.
I, II and III
Li.
Which
shows
abundances
for
the
correct
lithium
approximate
lithium?
u
a
10
l
35
D
Li
75
v
O
C
The
at
gold
gold
page
foil
foil.
experiment
This
involved ring alpha particles
experiment
is
depicted
in
gure 1 on
20.
An alpha particle is a helium nucleus. State the
nuclear
b.
symbol
Suggest
that
the
would
for
an
results
have
of
been
alpha
the
particle.
gold
foil
observed
in
[1]
experiment
are
following
Li and
each of the
alternative scenarios:
percentage
i.
Atoms
of
ii.
are
positive
Atomic
instead
hard,
dense, solid balls
charge.
nuclei
are
[1]
instead
negatively
7
abundance of
Li
charged.
[1]
25
39
8.
There
are
two
stable
isotopes
of
potassium:
K and
50
41
K. The
A
of
potassium
is
39.10.
Use
this
information
LHA
f
x
6
50
7
.
Percentage
abundance of
B
cobalt-58 and cobalt-59
6
of
Percentage
A
cobalt-59 and nickel-58
D.
t
a
7
isotopes
C.
a.
i
occurring
numbers
Extended-response questions
o
I and IV only
cobalt-58 and nickel-59
n
U
I and III only
B.
naturally
protons
d
A.
The
of
equal
o
number
contain
cobalt-58 and nickel-58
B.
C
∆H
III.
i
n
II.
r
o
4.
boiling point
v
I.
?
2
species
p
Which
following
r
e
1
3.
the
neutrons in their nuclei?
y
C
D
of
O
29
t
i
s
A
n
Protons
y
29
y
is
equal
l
63
What
protons.
P
Multiple-choice questions
of
have
r
III.
Exam-style questions
2.
e
How do nuclei of atoms dier?
r
65
to
determine
isotopes
90
and
the
relative abundances of the two
sketch
the
mass
spectrum
of
potassium
metal.
E
9.
“Dutch
14%
[3]
metal”
zinc.
oen
used
Dutch
mass
is
This
for
metal
an
alloy
alloy
making
c an
be
spectrometry.
composed of 86% copper and
closely
resembles gold, so it is
costume
jewellery.
distinguished
Explain
how
from gold using
[2]
33
Electron congurations
Structure 1.3
s
s
How c an we model the energy states of electrons in atoms?
This question is complex with many layers. What are electrons? How do we know they exist in energy states? What various
e
models about these energy states are there?
of
modern
these
no
views,
analogues
clouds
electrons
in
depend
our
on
are
quantum
everyday
the
life,
energies
of
objects
we
c an
that
behave
visualize
electrons,
as
both
electrons
which
c an
in
have
particles
atoms
only
as
certain,
energy
are
produced
electrons
in
by
excited states
Structure 1.3.5
—
state
electron
for
The
line
emission
spectrum of
provides
evidence
for
the
space
existence of
E ach
c an
there
is
has
two
a
xed
a
high
in
discrete
energy
levels,
which
converge at
Structure 1.3.6
—
The
main
energy
level
is
given an
of
2
integer
number,
n,
and
c an hold a maximum of 2n
convergence
higher
C
i
n
Structure 1.3.4 — A more detailed model of the atom
Structure 1.3.7
for
an
element
describes the division of the main energy level into s, p, d
an
orbitals,
regions
emission
spectrum, the limit
frequency
corresponds to
—
Successive
give
ionization
information
about
energy data
its
electron
conguration.
n
U
o
i
Much
of
our
understanding
studies
that
involving
sunlight
prism.
This
c an
u
in
which
be
l
example
of
a
a
A
gaseous
pure
glow —
prism,
E
as an
it
400 nm
in
to
the
within
spectrum
a
prism
and
1c)
the
is
subjected
series
a
continuous
(gure
into
and
spectrum
words,
a
into
congurations
In
the
1600s,
dierent
next,
the
as
and
atoms
Isaac
has
come
Newton
from
showed
coloured components using a
(gure
appears
in
Sir
a
no
rainbow.
1a).
This
type
continuous
gaps
The
are
of
series
visible.
spectrum
of
The
colours,
classic
wavelength of visible light
700 nm.
emission spectrum
between
down
merges
element
other
produces
light.
continuous spectrum
continuous
from
electron
with
wavelengths,
colour
ranges
will
broken
generates a
each
of
interaction
contains light of all
v
O
f
x
t
a
r
o
d
Emission spectra (Structure 1.3.1)
lines
34
In
at
ionization.
electrons.
and f sublevels of successively higher energies.
—
of
probability of nding an
o
v
Structure 1.3.3
energy
chemic al
it
of
will
lines
(gure
source
to
emit
a
high
light.
against
1b)
of
spectrum
In
dark
contrast,
visible
will
a
voltage
When
light
appear.
under
this
light
reduced
passes
background.
when
of
all
This
a
cold
This
gas
pressure
through a
is
is
known
placed
wavelengths, a series of dark
is
known as an
absorption
LHA
energies.
and
electrons of opposite
p
higher
dened
number
electron.
electrons
a
conguration
hold
contain
where
orbital
y
of
hydrogen
and
Sublevels
r
e
—
given
environment,
levels.
spin.
Structure 1.3.2
a
n
lower
spectra
when
and
v alues.
l
to
Emission
photons
such
shapes
O
return
—
Although
The
y
emitting
t
i
s
atoms
waves.
clouds.
predened
Understandings
Structure 1.3.1
and
fuzzy
y
sizes
to
has
P
behaviour
r
According
Structure
1.3
Electron
congurations
a
continuous
spectrum
s
s
spectrum
r
hot gas
e
emission
b
gas (c) visible light
of all wavelengths
C
n
U
o
i
t
a
u
l
a
v
E
Figure 2
d
r
o
f
x
O

y
of all wavelengths (b) a heated
o
from (a) visible light
p
r
e
The spectra generated
v
Figure 1
passing through a cold gas
i
n

spectrum
n
l
absorption
O
cold gas
y
P
y
t
i
s
c
The aurora borealis (Northern Lights) in Lapland,
drawn by the E arth’s magnetic eld
to the polar regions,
Sweden.
Charged
high-energy particles from
where they excite atoms and
the Sun are
molecules of atmospheric gases,
c ausing them to emit light
35
Structure
1
Models
of
the
particulate
nature
of
matter
Emission spectra
Emission
spectra
handheld
be
lamps
when
a
observed
by
contain
voltage
through a simple
holding
it
up
low-pressure
is
to
a
gases
light
Method
source.
which
1.
Observe natural light through the spectroscope. Note
2.
O bs e r ve
are
down the details of the spectrum you observe.
applied.
L E D.
Relevant
a r ti fi c i a l
No te
d ow n
l i g ht
t he
f ro m
a
details
c o m pu te r
of
th e
s c re e n
s pe c tr u m
Inquiry
3:
Construct
graphs
and
draw lines of best t
3.
2:
Identify
and
record
relevant
Observe
sucient
relevant
the
Identify
and
describe
patterns,
2:
colours,
Q uestions
Assess
accuracy
Sketch
2.
Describe
the
spectra
each
as
you
a
observed.
continuous,
protection.
3.
lamps
will
get
very
hot.
Look
the
with
up
the
Handle them
c are.
theoretic al
•
Further
safety
prec autions
will
be
given
by
your
on
teacher,
depending
on
the
exact
emission
discharge
the
lamps
and
number,
nature of the
n
eye
discharge
emission or
spectra of the elements in
you
O
The
spectrum.
t
i
s
Wear
•
observe,
y
absorption
you
l
1.
S afety
•
lines
wavelengths and number of lines.
observed.
observed
Compare the
emission lines, commenting
colours and positions of the
emissionlines.
Next,
you
will
wavelengths
of
Discharge lamps
•
Handheld
spectroscope
orange
light
has
its
with
Like
i
emission
own
on
the
characteristic
For
example,
wavelengths
colour
c an
be
in
of
observed
through
accuracy
a
used
shop
to
line
a
graph
wavelength.
your data.
of
the
in
and
a
that
sodium
observed
test
be
of
used
chemic al
which
atoms
589.6 nm
ame
c an
identify
t
a
the line emission spectrum of sodium (right)
spectrum,
excited
589.0
appe ars
a
streetlights (le) and
t
observed
wavelength data.
barcodes
spectra
best
vs
and
Construct
Comment
o
substance.
of
wavelength
lines.
6.
element.
yellow-orange
line
theoretic al
Comment on the relationship shown in your graph.
u
l
Sodium
v
E
36
Figure 3
d
r
o
f
x
O

same
the
a
the
emission
o
element
identify
Draw
the
5.
n
to
theoretic al
C
i
n
U
E ach
v
•
compare
of
y
4.
M aterials
p
r
e
discharge lamps.
c an
emit
(gure3,
any
to
be
used
yellow-
right).
The
sodium-containing
identify
elements.
products,
line
y
Inquiry
discharge lamps. Note
emission
trends and
relationships
•
the
P
2:
various
of
quantitative data.
including
Inquiry
from
details
r
and
light
qualitative
down
observations
e
o bs e r ve.
Tool
•
or
yo u
skills
•
•
s
s
Discharge
ionized
c an
spectroscope
Structure
1.3
Electron
congurations
Observations
Chemists
oen
properties
through
instruments.
boundaries
of
our
through
in
the
Sodium
seen
lamps
is
spectrum
region
also
of
of
is
in
expand the
to
vapour lamps
a
strong
spectrum. The light
the
more
naked
complex

eye but the
Figure 4
to a higher energy level by the heat of the ame. When
•
they fall back to a lower energy level, photons of certain
•
p
of
salts
(e.g.
LiCl,
NaCl,
o
A
metal
)
2
Clean the end of the ame test wire by dipping it into
the HCl solution and placing it in a non-luminous Bunsen
burner ame. Repeat until no ame colour is observed.
2.
n
•
, CuCl
2
Method
1.
heatproof mat
various
Dip
the
end
Bunsen
metal
of
and
burner
in
the
the
ame
place
it
in
ame,
salt
and
test
the
noting
the
wire
into
one
of
the
salt
edge of the non-luminous
down the identity of the
colour(s)
you
observe.
o
Dilute
, SrCl
nichrome)
qualitative
protection.
prec autions
3.
Clean the wire again and repeat with other salt samples.
4.
Clear
around open ames.
up
as
instructed
by
your
teacher.
hydrochloric acid is an irritant.
E
•
relevant
a
eye
suitable
v
Take
i
record
l
O
Wear
and
or
hydrochloric acid
t
a
and
S afety
•
dilute
of
2
C
i
n
U
u
Identify
observations
•
burner
C aCl
(platinum
of
samples
samples,
d
r
o
f
x
2:
Small
wire
colours for dierent elements
Relevant skills
Inquiry
Bunsen
KCl,
v
visible region of the spectrum.
r
e
wavelengths are emitted. Some of these photons are in the
test
portion
n
Small
O
Flame
•
y
•
ame tests is atomic emission. Electrons are promoted
•
l
y
Materials
identify the presence of some metals. The principle behind
Flame test
emission spectrum
t
i
s
Flame testing is an analytical technique that can be used to
Figure 5
Helium
(gure 4).
Flame tests

natural
gure 3, observing
reveals
the
orange
helium
a
revealing otherwise
spectroscope
yellow
observing
(oen sight), or with
technology
detail.
As
between
directly and with the aid of an instrument?
y
helium
emission
a
or
light.
dierence
P
from
features
the
e
light
in
is
phenomenon
c an be made
senses
observations,
orange-yellow
emission
human
What
from observing the
r
the
the
data
Observations
Advancements
imperceptible
emit
generate
matter.
s
s
directly
of
variety
which
•
Dispose
•
Further
of
dierent
are
of
chloride
salts
will
be
used, some
Q uestions:
irritants —
all
avoid contact with the skin.
1.
substances
Look up the emission spectra of the metals you tested.
appropriately.
Compare these to the colours you observed. Comment
safety
prec autions
will
be
given
by
your
on any similarities and dierences.
teacher,
depending
on
the
identity
of
the
salts being
2.
Explain
why
the
dierent
metals
show
dierent
analysed.
amecolours.
37
Structure
1
Models
of
the
particulate
nature
of
matter
TOK
One
of
the
ways
knowledge
up”:
they
take
involves
specic
developed
drawing
observations
is
through
conclusions
and
build
reasoning.
from
general
inductive
Reasoning
experimental
principles
reasoning
c an
be
deductive
observations.
or
Inductive
(“bottom-up”
a
red ame test.
gives
a
red ame test.
to
apply
are
your
“top
make
down”:
scientic
the
conclusion
they
infer
knowledge
in
what
grounds
might
On
what
grounds
might
we
we
classied
of
into
gives
each
these
a
type
two
the
following
of
claim
reached
through
inductive
claim
reached
through
deductive
E
Visible light is one type of
microwaves,
The
energy
1
E
of
∝
λ
the
of
premises:
reasoning?
a
part
existing
observation
types?
a
all
approach):
red ame test.
doubt
are
You do this all the time
pattern
doubt
light,
38
includes
bromide
disadvantages
neatly
v
On
and
lithium
premises.
o
be
a
always
o
i
advantages
that
general
4.
t
a
propose
red ame tests.
n
U
red ame tests.
could
reasoning
knowledge
salt.
give
from
(“top-down”
3.
u
the
give
scientic
lithium
conclusions
hypothesis
d
are
salts
you
your
a
l
C an
O
What
this,
is
f
x
Lithium
From
suppose
bromide
r
o
example,
Lithium
salts
context.
reasoning
theory
2.
For
new
lithium
C
1.
specic
a
all
i
n
deductive
that
y
asked
c an
v
when
you
p
Deductive arguments
salts:
r
e
observations,
lithium
reasoning?
reasoning?
electromagnetic (EM) radiation. In addition to visible
infrared
radiation
electromagnetic
the
radiation
is
(IR),
ultraviolet
(UV),
X-rays
and
gamma
spectrum.
inversely
proportional
to
the
wavelength,
λ:
rays
y
gives
iodide
about
O
sulfate
observations
red ame test.
n
following
Lithium
these
a
the
Lithium
From
gives
l
make
chloride
e
might
P
you
pattern
y
example,
theory
hypothesis
observation
Lithium
“bottom
approach):
t
i
s
For
are
r
3.
1.
arguments
from them.
4.
2.
inductive.
s
s
Inductive reasoning
is
Structure
Electromagnetic
waves
all
travel at the
speed of light,
8
of
light
is
approximately
frequency
of
the
equal
radiation,
f,
to
by
3.00 × 10
the
c,
in
a
vacuum.
The
1.3
Electron
congurations
speed
–1
m s
following
.
Wavelength
is
related to the
equation:
c = f × λ
energy
EM
waves,
such
as
gamma
rays,
have
short
s
s
High
wavelengths and high
frequencies while low energy waves, such as microwaves, have long wavelengths
and
low
frequencies.
λm
r
1
10
4
10
14
10
(γ
rays)
10
ultraviolet

10
00
1

10
14
10
infrared
00
(IR)
4
10
1
10
v

microwaves
10
10
p
10
Activity
00
waves
o
radio
U
C

10

10
i
n
0
10

10
y
ygrene
(UV)
r
e
elbisiv
10
O
t
i
s
10
n
y
400
10
X-rays
1
l
λnm
1
10
0
10
y
rays
P
gamma

10
e
f
Compare
green
in
the
colours
red and
gure 6. Determine which
colour has:
a.
the
highest
wavelength
b.
the
highest
frequency
c.
the
highest
energy
4
10
4

Figure 6
The wavelength (λ) of electromagnetic radiation is inversely
the
spectra
below.
Explain
how
o
at
radiation
i
Data-based questions
energy of that
we
know
that
stars
are
partly
composed
of
hydrogen.
u
l
a
E
v
O
f
x
t
a
r
o
d
proportional to both frequency and
Look
n
10
3900
4000

Figure 7
7600
4500
5000
5500
The hydrogen emission spectrum (top) and
6000
6500
the absorption spectrum generated
7000
7500
from the Sun (bottom)
39
Structure
1
Models
of
the
particulate
nature
of
matter
The line emission spectrum of hydrogen
(Structure 1.3.2 and 1.3.3)
E ach
line
which
A
idea
that
photon
radiation
=
emission
is
as
to
a
spectrum
specic
electromagnetic
a
quantum
of
of
an
element
amount
radiation
energy,
of
comes
which
is
has
energy.
in
a
specic
This
is
proportional
to
the
the
specic
energy
possessed
by
the
r
E =
photon,
expressed in joules, J
=
constant,
frequency
of
the
6.63 × 10
radiation,
J s
expressed
The
electron
When
that
for
a
theory
only
the
of
the
orbit
returns
rst
to
the
a
stationary
orbits
the
lowest
moves
lower
energy
attempt
to
U
making
any
i
spectra.
of
the
to
of
of
model
transitions
the
a
energy
based on its
around the nucleus.
level absorbs a photon
energy
level
and
remains
level, it emits a photon of light.
dierence
between
the
main
Classic al
the
two
levels.
problem of the
electrodynamics
predicted
energy and quickly fall into the nucleus,
atoms
of
energy
higher
overcome
when
wavelengths,
measuring
energies
energy
Bohr
their
specic
By
impossible.
Bohr
postulated that
staying in stationary orbits.
the
atom
could
have only certain, well-
between stationary orbits could absorb or emit
producing
wavelengths
of
characteristic lines in the atomic
these
lines,
it
was
possible
to
c alculate
electrons in stationary orbits.
u
l
a
E
v
O
f
x
hydrogen
atom,
the
electron
energy (E
)
in
joules
could
be
related to the
n
energy
level number (n)
by
a
simple
equation:
1
E
= –R
n
H
2
n
–18
where
R
≈
2.18 × 10
J
is
the
Rydberg
constant.
This
equation
clearly
represents
H
the
quantum
only
nature
discrete,
of
the
quantized
half-integer
parameters,
number (n)
c an
numbers
40
in
radiate
existence
radiate
energies,
photons
a
not
electrons
dened
t
a
r
o
d
Since
would
prolonged
did
o
electrons
electrons
n
orbiting
atom
were:
Rutherford model of the atom (Structure 1.2).
that
inverse
discrete energy levels
with
it
hydrogen
theory
time.
represents
was
in certain
energy,
short
electron
photon
the
his
C
Bohr ’s
the
in
amount
level
of
of
o
When
This
electron
right
a
p
at
an
exist
model
postulates
associated with
v
the
i
n
3.
of
c an
are
or
y
2.
orbits
main
r
e
These
proposed
The
Hz,
O
1.
Bohr
spectra.
hertz,
n
Niels
t
i
s
1913,
emission
in
y
–1
seconds, s
l
f
Planck’
s
mean
take
values.
where
These
known as
only
higher
atom,
positive
energy.
the
energy
values
are
of
an
electron
characterized
quantum numbers. The
integer
values
(1,
2,
3,
c an
by
have
integer or
principal quantum
…),
where
greater
y
P
h =
For
frequency of the
follows:
–34
the
quantization:
discrete packets, or quanta.
h × f
Where
In
wavelength,
c alled
e
E
the
s
s
the
in
corresponds
Structure
The
most
electron
stable
has
ground state
c alled
return
state
the
of
lowest
of
the
atom.
excited states.
to
the
the
ground
hydrogen atom is the state at
possible
In
contrast,
Atoms
state
energy.
by
in
the
excited
emitting
This
energy
energy
states
of
=
is
unstable
specic
1,
Electron
congurations
where the
known as the
levels with
are
photons
n
level
1.3
n
=
and
2,
3,
…
are
spontaneously
wavelengths
(gure8).
s
s
+energy
e
+
+
p
e
+
p
p
dec ay
r
between
amount
electron
c an
level.
of
rungs
amount
of
be
of
of
ladders
between
a
ladder.
energy,
with
varying
energy
Jumping
and
levels,
up
distances
between the
much
how
each
jumping
rung
down
a
like
or
rung
level
or
you
c annot
requires a
level
releases
energy.
excited
Electrons
hydrogen
to
any
energy
returning to
n
=
2
level,
will
n,
and
return
to
any
lower
produce distinct lines in the visible
(gure 9).
v
line. The energy of the photon released is lower when an electron falls from n = 3 to
i
n
two of the allowable energy states of the electron in the hydrogen atom.
486
n
= 6
n
= 5
n
C
cyan
434
n
blue
410
= 4
i
u
l
n = 3
= 6
n
= 5
n
= 4
n
= 3
n
= 2
n
red
656
=
1
a
E
v
O
f
x
t
a
r
o
n
o
from
violet
d
transition
U
colour
wavelength / nm
o
n = 2, than from n = 6 to n = 2. In both cases, it represents the dierence between
p
Note that the red line has a longer wavelength and lower frequency than the violet
y
spectrum
exist
r
e
energy
the
discrete
same
resemble
c annot
O
An
atoms
t
i
s
specic,
the
in
n
stand
levels
Electrons
y
rungs.
l
Energy
Electrons returning to lower energy levels emit a photon of light, hf
y
Figure 8
hf
P

e
excitation
◂
Figure 9
The visible lines in the
emission spectrum of hydrogen
show electrons returning from higher
energy levels to energy level n = 2
41
Structure
1
Models
of
the
particulate
nature
of
matter
n = 7
Electron
transitions
to
the
ground
state,
n =
1,
release
higher
energy, shorter
n = 6
wavelength
ultraviolet
light,
while
electrons
returning to
n =
3
produce lines in
n = 5
the
infrared
region
It
important
of
the
electromagnetic
spectrum
(gure 10).
n = 4
is
required
IR
to
note
that
electrons
between
will
allowable
absorb
energy
or
rele ase
states.
Any
only
the
excess
exact
will
not
not
= 2
and
if
an
insucient
amount
of
energy
is
supplied
the
electrons
move.
visible light
levels
closer
to
the
nucleus
hold
fewer
e
Energy
holds
3
has
a
level,
electrons, at
maximum
of
18
n
n, is 2n
=
2
. For
there
example,
could
electrons, and
n
=
be
4
a
has
the
energy
maximum
a
radiation
32
us
convey
concepts,
our
ideas
we
more
sometimes
Prepare
a
written
explanation
of
atomic
n = 2 is much smaller than that
any
diagrams.
Exchange
it
with
a
electrons.
Order
•
Whether
have
to
diagram
whether
or
are
important
shared
make
to
voc abulary
ideas
each
it
other ’
s
improvements
accompany
not
given
concepts
adds
to
your
the
to
are
missing
other
feedback,
from
the
explanation.
feedback, spend some time using the
your
work.
explanation.
Finally,
Discuss
choose
why
a
graph, image
you chose it and
explanation.
n
U
C
or
any
i
n
feedback
which
o
you
scientic
in
v
When
of
each
p
Use
•
Give
y
r
e
•
graphs or images to
emission that does not include
partner.
n = 1
concentrating on:
diagrams,
clearly.
The energy dierence between
between n = 2 and
use
O
help
Linking questions
qualitative
i
such
as
gas
t
a
from
o
d
r
o
f
x
gaseous
How
l
u
How
does
period
and
quantitative
discharge
do
an
tubes
elements
different
a
E
v
O
42
What
emission
and
element’s
in
the
data
c an
prisms
in
be
the
collected
study
of
from instruments
emission
spectra
from light? (Inquiry 2)
spectra
elements?
number
and
provide
evidence
for
the
existence of
(Structure 1.2)
highest
periodic
=
n
explaining
t
i
s
When
the electron moves further away from the
y
Communic ation skills
ATL
Notice how the allowable
energy levels get closer together when
n = 3 and
of
n
occupied
table?
main
(Structure
energy
3.1)
level
1
electrons,
Electron transitions for the
hydrogen atom.
nucleus.
level with
eight
l
Figure 10
of
maximum
1
UV

=
two
energy
relate to its
y
=
to
any
P
n
n
up
in
r
electrons
will
electrons. The maximum number
2
of
energy
be
radiation
absorbed,
n
to
move
s
s
n = 3
Structure
1.3
Electron
congurations
The quantum mechanic al model
of the atom (Structure 1.3.4)
The
Bohr model
atoms.
of
was
levels.
narrow
The
model
It
assumed
the
the
Bohr,
incorrect
levels.
electron
It
of
states
electrons
spectra
these
However,
lines
this
of
of
electrons in
existed
in
hydrogen
discrete
consisted
corresponded to the
model
was
limited
by
assumptions:
predict
electron.
energy
that
emission
wavelengths
energy
not
the
the
idea
the
was
was
a
emission
only
spectra of elements containing
successful
subatomic
with
particle
in
the
a
hydrogen atom.
xed orbit about the
not
account
for
the
eect
of
y
could
electric and magnetic elds on the
spectral lines of atoms and ions.
Heisenberg’s
could
not
explain
molecular
bonding
and
geometry.
The
uncertainty
principle
states
that
it
is
impossible
to
principles
the
loc ation
and
momentum
of
an
electron
simultaneously.
and
stated
that
electrons
exhibited
xed
momentum
in
specic
these
limitations,
quantum
the
Bohr
theory
has
been
quantum
the
mechanics
combines
the
less
more
we
it
the
about
know
about
is
possible
not
trajectory
c alculate the
its
of
an
probability
f
x
region
know
of
simultaneously.
the
position
momentum,
to
pinpoint
electron
of
nding
in
the
an
an
of
and
This
an
electron,
vice
versa.
we
in
u
l
boundaries
are
E
What
of
the
limits
of
Wave–particle duality
subatomic
species
to
is
the
The
of
released
even whole atoms and
interference
(bending
through
or
nature.
around
obstacles),
all
(combination
obstacles) and
of
which
are
waves.
Einstein
(1879–1955)
described
by the
in
equation
duality
of
the
Schrödinger
1926
by
the
electron
is
equation,
Austrian
quantitatively
which
was
physicist Erwin
(1887–1961). Solutions to the Schrödinger
give
a
series
known as
states
and
of
three-dimensional
wave functions,
energies
of
mathematic al
which
describe the
electrons in atoms.
knowledge?
ability
behave
characteristics
to
wave–particle
as
of
both
The
concept
that
objects
these
of
wave–particle
duality
illustrates the fact
electrons and other
particles
and
species,
such
as
of
study
do
not
always
fall
neatly into the
waves.
discrete
Certain
diraction
and
of
absorbed
phenomena of light, but together they do.
possible
human
c apable
be
particulate
separately neither of them fully explains the
functions,
knowledge?
electrons,
are
to
their
We have two contradictory pictures of reality;
Schrödinger
(1908–1974)
implic ations of this uncertainty principle on
tendency
suggest
(passing
characteristic
formulated
a
the
v
O
the
Bronowski
are
the
photons,
waveforms),
Albert
has been to prove that this aim is unattainable.
and
entities
molecules,
tunnelling
each
exact picture of the material world. One achievement …
Jacob
of
c an
One aim of the physical sciences has been to give an
What
small
loc ation or
atom,
electron
space.
means
t
a
predict
particle
r
o
Although
we
a
However,
states that it is
accurately both the momentum
i
the
of
discrete
n
that
position
momentum
as
o
the
determine
d
and
to
idea of
following key principles.
Heisenberg’s uncertainty principle
impossible
the
U
with
by
o
modern
quantization
superseded
C
i
n
TOK
The
eventually
mechanic al model of the atom.
are
Structure 2.2
y
of
modern
v
the
p
Bec ause
r
e
circularorbits.
molecular
geometry
Bohr ’
s
explained in
model
behind
precisely
bonding
know
O
It
5.
t
i
s
4.
l
nucleus.
y
3.
could
one
to
the
n
It
than
and
explain
P
more
2.
bec ause
allowable
problems
to
r
1.
attempt
quantization:
e
several
in
an
on
According
lines
dierences
was
based
s
s
energy
It
c ategories
we
have
developed.
What
is
the
role
mass,
of
c ategorisation
in
the
construction
of
knowledge?
43
Structure
1
Models
of
the
particulate
nature
of
matter
Schrödinger ’s
probability
wave
density,
electrons
are
path,
theory
this
of
region
in
There
four
space
space
are
two
uncertain. Instead
gives
at
several
electrons.
atomic
a
types
order
of
a
electrons
electron
probability
orbitals,
has
an
follow
will
from the nucleus. An
high
atomic
orbital
are
a
that
that
and
of
increasing
theoretic al,
and
energy
these
orbital
shape
are
are
a
dened
found
an
energy.
labelled
labelled
number,
These
n,
energy
introduced
levels
are
by
split
Bohr
alphabetic ally
model
sublevels
represents the
comprised of
and
3s.
As
n
For
example,
increases,
the
s
for
n
=
orbitals
1,
2
are
and
3,
further
the
to
the
12
shows
away
nucleus
that,
from
for
1s,
there
is
The
is
a
There
is
zero
the
is
true
for
nucleus
with
the
two
the
high
of
that
of
zero
probability
an
nding
highest
regions
probability
reaches
highest
electron
the
of
nding
when
is
at
an
electrons close
move further
be
further
away,
found closer to the
between
even
the
two
peaks.
greater distance
o
C
n
50
i
o
pm
t
a
u
l
50
100
pm
average
radius
a
E
v
O
f
x
0
3s
50
100
pm
Figure 12
are 1s, 2s
probability.
0
0

we
somewhat
could
electron
probability
zero
v
U
d
r
o
2s
2s,
probability
probability
3s,
and
i
n
1s
For
small
a
never
p
from
nucleus.
there
same
probability
orbitals
from the nucleus.
y
nucleus.
the
this
r
e
although
and
atomic
O
Figure
t
i
s
The s orbital c an hold two
electrons
s
distanced
n
orbitals.
y
atomic
where there is a 99% probability of nding
44
rst
An s orbital is spheric al. The
sphere represents the boundary space
an electron.
the
into
The
s, p, d, and f.
The plots of the wavefunctions for the rst three s orbitals
150
y
Figure 11
quantum
levels.
is a
l

energy
travel
specic
c an hold a maximum
and
P
principal
main
a
electron.
(g, h, i, k and so on).
The
in
atomic orbital
nding
each
characteristic
be
r
orbitals
in
of
is
electrons in atoms in terms of their
idea that the momentum and position of
saying
distance
there
E ach
of
probability
certain
orbitals,
Subsequent
the
where
the
e
of
describe
Heisenberg’s
s
s
region
functions
using
Structure
Imagine
that
8.00am.
they
At
could
you
are
8.15am,
be.
a
Some
possibly
students
no
from
might perhaps be at the airport
•
might
even
the
have
exact
be
school
region
where
perimeter,
the
the
town
region
of
or
teacher
the
is
is
where
could
a
town
99%
with
be
a
the
it
drawn
your
library
possible
is
a
nding
is
of
draw
this
cluster
them.
an
a
probability
This
loc ated,
Similarly,
probability
to
high
around
of
school
airport.
high
is
there
chance
where
includes
space
unknown,
areas
surface
there
that
the
c ar park
Pole!
showing
boundary
space
around
represents
or
a
atomic
nding
an
(gure13).
t
y
y
all
describe
There
and
z
are
axes
boundaries
the
orbitals,
These
each
are
described with
labelled p
, p
x
highest
and
y
probability of nding
t
Figure 14
The three p atomic orbitals
are dumbbell shaped,
aligned along the
u
z
z
x, y and
z axes.
There is zero probability
of nding the electron at
the intersection
of the axes between the two lobes of the
x
x
a
x
orbital
l
p
v
O
f
x
z
y
with
p
14).
t
a
electrons in these orbitals.
x
three
(gure
y
shapes
y
o
These
z
x,
i
.
shaped
r
o
p
dumbbell
parallel to the
o
is
x
a sphere that encloses 99%
n
U
p orbital
orientations
d
A
of the dots (right)
C
i
n
z
z
Representation of a 1s atomic orbital as
a cluster of dots (le) and
p
r
e
v
x
Figure 13
O
electron
A
the
or
school
centre
North
dots
the
t
i
s
orbital
the
of
of
in
n
certain
teacher.
region
the
laboratory,
where
teacher:
y
might
the
a
of
town
or
begin at
y
dene
to
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cluster
the
oce
the
to
wonder
l
nding
to
in
gone
chemistry
that
lesson
you
P
three-dimensional
of
house
suggest
so
r
Although
their
class
chemistry
teacher,
congurations
e
may
•
at
the
principal’
s
DP
your
Electron
s
s
•
be
room,
your
of
your
could
school
for
sign
is
the
sta
waiting
still
•
in
the
is
•
be
in
student
there
1.3
dumbbell.
E ach of the p
orbitals c an hold
two electrons
y
y
p
y
orbital
p
z
orbital
E
45
Structure
1
Models
of
the
particulate
nature
of
matter
Theories and models
Current
before.
natural
theories
are
theory
are
world.
are
evolved
Contrary
substantiated
amassed,
from
previous
comprehensive
to
by
documented
the
vast
and
systems
use
of
the
amounts
models,
of
ideas
word
of
each
that
“theory”
observations
communic ated
by
a
superseding
model
in
and
everyday
and
the
one
explain
tested
an
that
c ame
aspect of
s
s
the
atomic
Theories
language, scientic
hypotheses, which
large number of scientists.
r
+
+
+
+
+
+
kana
1913
ohr model
1930
uantum
pudding” model
y
o
C
n
theories
c an
you
are found
of
probabilit
model
in regions of high probability
i
of
think
u
l
f
x
a
E
v
O
46
regions
t
a
examples
r
o
other
and
The atomic theory has seen the idea of atoms evolve from indestructible spheres to the quantum mechanic al
model where electrons have specic energies and
What
1926
Heisenberg’s uncertaint
o
Figure 15
d

1912
Rutherford’
s model
model
p
r
e
“billiard ball”
v
i
n
U
1803
D alton’s
mechanic al
model
O
Democritus’ atomos
1897
Thomson’
s “plum
t
i
s
Āruņi’
s
y
800–400 BCE
n
l
+
+
of ?
Linking question
What
is
the
periodic
relationship
table?
between
(Structure
3.1)
energy
sublevels
and
the
block
nature of the
y
+
+
P
+
e
+
+
Structure
1.3
Electron
congurations
Electron congurations (Structure 1.3.5)
E ach
atomic
spheric al
oriented
it
has
type
the
dierently.
in
energy
has
There
than
s
a
characteristic
lowest
or
are
possible
ve
d
shape
energy.
orbitals
and
There
and
energy. The s orbital is
are
seven
f
three
p
orbitals,
orbitals,
and
each
these
are
s
s
higher
orbital
and
p.
z
r
x
s
z
y
For
the
s,
p,
Type
d,
principal
only
n
=
the
3,
and
there
of
u
f
x
per
2
v
E
3
p
a
s
l
O
s
three
Total
orbitals
sublevel
1
are
For
types:
n
s,
number
orbitals
per
energy
x
=
n,
2,
p,
p
x
and
d.
n
x
f
2
orbitals need
c an hold
there
y
f
1
number,
exists.
f.
Number
of
(n)
quantum
orbital
of
quantum
number
s
t
a
Principal
p.
1,
y
f
Only the shapes of s and
O
and
by
=
p
s
types:
n
f orbitals.
i
four
For
and
z
o
orbital:
are
dened
1).
d
C
level
(table
p,
x
0
2
z
y
f
o
of
The shapes of the s,
r
o
there
–1
x
d
z
n
energy
orbitals
types
x
f
d
of
i
n
x
x
1
y
U
E ach
z
y
–2
y
d
v
z
f
Figure 16
x
0
y
x
y
d
–1
z
y
x
d
z
–3

y
r
e
x
f
z
y
–2
y
1
z
y
z
p
0
z
d
t
i
s
z
x
p
–1
n
x
p
y
x
y
l
y
y
z
P
z
e
y
3
to be known
types
are two
For
n = 4,
M aximum
number
electrons
of
within
type
1
2
level
(n
2
)
energy
level
1
2
4
8
9
18
(2n
)
1
3
s
1
p
3
d
5
t
Table 1
c an hold 2n
sublevels,
s
1
p
3
d
5
to n.
(s,
4
E ach energy level,
dened
by
2
n,
16
p,
electrons. The number of
or atomic orbital types,
is equal
For n = 4 there are four types of orbitals
d,
and
f ) with 16 atomic orbitals in total
32
2
occupied
by a maximum of 2(4)
= 32 total
electrons
f
7
47
Structure
1
Models
of
the
particulate
nature
of
matter
Activity
State
the
following
for
the
energy
level with
sublevel types
b.
the
number
c.
the total number of atomic orbitals
d.
the
of
maximum
atomic
orbitals
number
of
in
each
electrons
at
n = 5:
sublevel
that
energy
level.
r
convention,
represent
s
sublevel
(one
box
in
notation
arranged
orbitals
is
Arrows are drawn in the boxes to represent
so each box has a
of two “arrows”
representing
the
three p orbitals p
, p
x
, and p
y
)
z
five
d
orbitals)
n
o
sublevel
boxes
i
d
(seven
boxes
representing
the
seven
f
orbitals)
u
Atomic
l
a
E
v
O
f
x
t
a
r
o
f
the
o
U
(five
representing
C
sublevel
boxes
used
c alled electron conguration
v
d
(three
i
n
sublevel
is
(gure 17). The
y
of two electrons
p
orbital diagram
orbitals
p
maximum
r
e
A maximum
c an occupy each orbital,
c alled an
atomic
representing an s orbital)
shows
the number of orbitals for each sublevel.
electrons.
in
n
each box
This diagram
electrons
box”
are
O
In orbital diagrams,
in
t
i
s
Figure 17
represents an orbital.
of
“arrow
electrons
y
arrangement
u
an
how
l
to
orbitals
electrons.
two
electrons
mechanics
pair
of
regions
are
should
solves
electrons
directions.
one
are
Electrons
Hence
downwards
not
this
with
of
space
charged
be
able
problem
opposite
each
orbital
half-arrow,
where
there
negatively,
to
by
occupy
using
spins
box
is
(gure
and
a
±
the
is
a
like
high
same
spin
probability of nding
charges
region
notation
for
repel
of
each
space.
each
other, so
Quantum
electron. A
behave like magnets facing in opposite
shown
18).
This
with
is
an
upwards
known as the
half-arrow,
, and
Pauli exclusion
principle:
Only two electrons c an occupy the same atomic orbital and those electrons
must have opposite spins.
48
y
P
Orbital diagrams
For
e
the
s
s
a.
Structure
N
t
S
Figure 18
1.3
Electron
congurations
Electron spin is represented
by an arrow pointing up
(positive spin) or
down (negative spin)
s
s
N
in an orbital
what
of
our
shows
the
mathematics
extent
does
limits
as
the
atomic
mathematics
the
orbitals
same
of
the
energy
life
degenerate
3s
electrons in
2p
Unfortunately, neither
c an
and
basis:
around its
be
c an
visualized
in
any
way,
be
2s
expressed only in
degenerate
of
human
support
same
are
perception
type
of
and,
at
the
same
1s
time,
science.
1
knowledge
2
3
development in the
n

Figure 19
The three 2p
orbitals are
degenerate as they have the same energy.
in
one
referred to as
sublevel
are
degenerate
of
equal
orbitals
energy.
(gure 19).
These three degenerate atomic orbitals
have lower energy than the three 3p orbitals
a
E
v
O
with
physic al
rotate.
electrons
language
sciences?
the
no
everyday
u
of
Orbitals
in
l
E ach
analogues
f
x
natural
behaviour
has
c annot
electron
form. This lack of visualization does not undermine the quantum
rather
of
wave
the
i
To
but
power
a
of
t
a
the
wave-like
and
r
o
theory
interpretation
waves,
rotation
3p
ygrene
no
mathematic al
the
o
have
the
this
as
3d
degenerate
o
they
nor
like
interpreted
C
However,
behave
spin
oen
d
as
is
n
axis.
atoms
the
spin
U
Electron
own
v
i
n
TOK
n
of opposite spin
y
representing
p
electrons
r
e
half-arrows
O
t
i
s
magnet analogy
y
S
l
S
e
r
N
P
N
y
S
49
Structure
1
Models
of
the
particulate
nature
of
matter
An atom of boron (B) has ve electrons, and its orbital diagram is drawn as follows:
1
2p
s
s
2
2s
boron
(B)
e
2
1s
single
equal
show
lemost
that
every
occupy
any
orbitals
are
equivalence.
it
is
a
matter
of
personal
occupied
same
with
spin
an
in
each
electron
of
of
that
them
opposite
gure
which
of
A.
the
the
reason
1s
2s
1s
2s
i
1s
f
x
u
l
doubly occupied
for
the
2s
2s
four
orbitals
orbitals
below
Hund’s
drawn
are
are
fully
and
incorrect
the
a
any
correct
by
the
c an
electrons.
2p
orbitals?
electron
exclusion
being
orbital
(gure 20).
before
Pauli
diagrams
spin
occupied
lled
represents
rule
three p orbitals must
principle.
wrong.
2p
2p
2s
2p
1s
2s
2p
1s
2s
2p
t
a
2p
across the three degenerate
and
and
diagrams
on
is
D.
The electrons are evenly
orbitals in nitrogen before an orbital is
o
C.
1s
1s
n
U
2s
Figure 20
distributed
2p
r
o
1s

2p
d
2s
the
based
B.
1s
The
C
i
n
conguration
State
20.
think
o
State
you
v
2.
at
do
p
Why
half-arrow
y
r
e
Look
have
all electrons in singly
the
before
Practice questions
1.
the
they
joined
n
the
as
boxes
O
doubly
with
by
preference.
and that
means
t
i
s
become
electron
orbitals,
degenerate orbital in a sublevel is singly
occupied orbitals have the same spin. This
one
three
Traditionally,
of
occupied before any orbital is doubly occupied
have
the
represented
l
states
energy
although
2p
y
Hund’
s rule
c an
degenerate
their
box,
boron
E.
a
E
v
O
The Aufbau principle states that as electrons are added to atoms, the lowest
available energy orbitals ll before higher energy orbitals do. The third and
fourth energy levels contain d and f orbitals (gure 21). These orbitals are typically
lled aer the s orbitals of the following levels because they are higher in energy.
As shown in gure 21, the 3d sublevel has a higher energy than 4s but lower than
4p, so 4s is lled with electrons rst, followed by 3d and nally 4p. For the same
reason, 4d orbitals are lled aer 5s, and 4f orbitals are lled only aer 6s.
50
y
to
The
in
P
the
electron
energies.
together
in
2p
r
The
Structure
t
Figure 21
energy and
1.3
Electron
congurations
The 4s sublevel has a lower
will ll before the 3d
sublevel
4f
6s
5p
s
s
4d
5s
ygrene
4p
3d
e
4s
3p
r
3s
Ca
with
have
experimental
electrons
in
the
data
4s
that
show
that
potassium, K, and
sublevel, not in 3d.
t
Figure 22
Potassium
orbital lling diagram
electron in the 4s orbital bec ause 3d
3d
order
is
o
following
i
the
observed:
Activity
u
f
x
t
a
Generally,
d
r
o
1s
o
U
2s
orbitals are higher in energy
n
ygrene
2p
C
i
n
3s
v
3p
p
4s
showing the outermost
y
r
e
4p
n
consistent
O
is
c alcium,
t
i
s
This
l
y
1s
y
P
2p
2s
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p …
Copy
the
orbital
diagram
from
O
l
In the IB Diploma Programme, only the electron congurations of atoms and ions
gure21
and
complete
it
for the
up to Z=36 will be assessed. Electrons in these species can ll sublevels up to 4p.
sharing
There
1.
Full
so
it
are
is
important
three
E
ion.
electron
transfer
v
reactions,
and
a
Electron
ways
are
to
to
following
know
show
the
the
to
electron
electron
understanding
Condensed
Orbital
The
orbital
ground
a.
aluminium, Al
b.
chlorine, Cl
c.
iron,
conguration of an atom or an
Fe
conguration
electron
lling
lling
their
conguration:
Refer to the periodic table at the
of
this
diagram
diagram
book
to
deduce the
conguration
number
3.
in
chemic al
back
2.
elements
states:
fundamental
(“arrows
for
in
of
electrons
in
each atom.
boxes” notation)
potassium
is
given
in
gure 22.
51
Structure
1
Models
of
the
particulate
nature
of
matter
Full electron congurations
To
write
the
a
full
electron
electrons
Rule
and
the
in
conguration,
successive
Pauli
orbitals
exclusion
we
use
the
according
periodic
to
the
table,
Aufbau
and
“build
principle,
principle.
s
s
Worked example 1
Determine
the
full
up”
Hund’s
electron
conguration
for
the
c alcium
atom.
e
Solution
r
The Aufbau principle states that as electrons are added to atoms, the lowest
exclusion principle, we know that each orbital will have a maximum of two
l
electrons.
n
y
The atomic number of calcium is 20. Let’
s split the 20 electrons evenly across
each orbital, starting with the lowest energy rst. When writing electron
O
t
i
s
congurations, the number of electrons within each sublevel is given in
superscript, next to the sublevel:
2
2s
orbital
•
The
three
•
The
3s
•
The
three
has
orbital
2p
has
conguration
full
brings
electron
next
So,
for
for the phosphorus
us
up
lowest
have
three
18
electrons,
electrons
with
two
2
the
full
electron
6
each, six in total: 2p
2
electrons: 3s
energy, 4s: 4s
c alcium,
atom.
to
electrons
U
C
the
two
o
the
i
n
Determine
2
electrons: 2s
v
This
3.
two
have
two
orbitals
Practice question
electrons: 1s
has
orbitals
orbital
3p
two
also
y
1s
The
le
p
The
•
r
e
•
6
each, six in total: 3p
over to go into the orbital with
2
conguration is 1s
2
2s
6
2p
2
3s
6
3p
2
4s
Condensed electron congurations
the
gets
mostly
electron
c an
congurations
core
group
18
an
electrons
as
is
element
by their
inner
to
the
the
full
electron
conguration
write. The chemistry of atoms and
valence electrons, that is, the outermost
the
same
A
valence
electron
more
convenient
electrons
and
conguration
way of writing
represent the
as
the
previous
noble gases) element in the periodic table:
u
Condensed electron conguration = [previous noble gas] + valence electrons
Table 2 shows some more examples of full and condensed electron congurations.
l
a
Examples of full and
condensed
Condensed
Atomic
Element
Full
electron
conguration
electron
E
v
number
conguration
2
O
8
1s
Ne
10
1s
Mn
25
1s
2
2s
2
35
1s
2
4
2p
[Ne]
6
2p
2
2s
2
[He] 2s
6
2p
2s
2
Br
4
2p
2
2s
2
52
to
core electrons.
highlight
having
(known as the
increases,
time-consuming
determined
t
a
inner
of
be
rather than the
electron congurations for selected
elements
it
o
is
electrons,
number
and
i
d
r
o
Table 2
f
x
O
u
ions
atomic
longer
n
As
2
3s
6
2p
6
3p
2
3s
2
4s
6
3p
5
4s
2
3d
2
[Ar] 4s
10
3d
5
4p
5
3d
2
[Ar] 4s
10
3d
5
4p
y
P
available energy orbitals ll before higher energy orbitals do. From the Pauli
Structure
the
condensed
electron
conguration
for
the
c alcium
Determine
electron
Solution
the
condensed
conguration
example
1,
2
c alcium to be 1s
we
2
2s
determined
6
2p
2
6
3s
the
full
electron
conguration of
2
3p
4s
.
The previous noble gas in the periodic table is argon, which has an atomic
2
2s
6
2p
2
3s
e
2
number of 18. Argon has an electron conguration of 1s
6
3p
, and we
2
also
conguration.
sometimes
The
be
condensed
shortened
orbital
by
using
diagrams
for
a
condensed
periodic
oxygen and
manganese
are
shown
t
i
s
in
2p
3d
the
Write
key
ve
answer
of
the
should
that
test
key,
concepts
how
to
than
a
and
skills:
electron
sheet
of
A4
from the quantum
congurations.
paper.
is
discussed further
chapter.
know
your
then
write
try
from
this
chapter.
understanding
them
out
on
of
the
ideas in this
o
an
to
longer
from
you
questions
M ake
no
takeaways
voc abulary
brief
range
one
i
t
a
r
o
d
chapter.
summary,
key
a
atom,
This
n
List
chapter
span
the
C
three
of
i
n
a
Write
topic
U
Write
this
model
sublevel
o
v
Self-management skills
in.
of
electrons of elements
y
4s
in
structured
Structure 3.1.
p
r
e
manganese: [Ar]
ideas
is
type
O
2s
mechanic al
the
n
y
valence
appear
oxygen: [He]
The
table
below.
that
ATL
to
l
according
y
c an
The
electron
P
diagrams
.
r
can therefore write the condensed electron conguration of calcium as [Ar] 4s
Orbital
for the
phosphorus atom.
s
s
worked
congurations
atom.
4.
In
Electron
Practice question
Worked example 2
Determine
1.3
of
your
peers.
Exceptions to the Aufbau principle
The
Aufbau
elements.
the
for
Mn,
with
These
3
3d
at
.
with
are
conguration
the
(Sc)
valence
periodic
to
oxidation states
are
3d
order
of
lling
electrons
to
of
atomic
form
ions,
some
table
copper
2
[Ar] 4s
are
orbitals
the
lost
for
electrons
rst.
5
3d
,
the
4s
electrons will be lost
2+
electrons
known as the 3d
E
look
sc andium
There
the
lose
5
,
with
electron
conguration
[Ar] 3d
,
a
2
elements
you
electron
gives the manganese ion, Mn
v
O
This
ions.
atoms
sublevel with the highest principal quantum number ( n)
not [Ar] 4s
All
predicts
when
l
So,
rst.
correctly
However,
u
in
principle
f
x
most
(Cu).
at
tend
to
lose
two
4s
transition elements, or
the
back
These
of
this
transition
book,
metals
electrons
to
form 2+
transition metals. If
these
c an
elements
also
have
are
from
variable
in compounds.
exceptions.
With
only
one
electron
in
its
3d
orbital,
sc andium
3+
readily
forms only Sc
ions,
by
losing
this
3d
electron
and
the
two
4s
electrons.
53
Structure
1
Models
of
the
particulate
nature
of
matter
The
Ionization
and
oxidation
ground
those
covered in
Structure 2.1
state
congurations
of
copper
and
chromium
are
also
dierent
from
are
predicted
by
Aufbau
principle.
and
2
Structure 3.1
The predicted electron conguration of copper is [Ar]4s
9
3d
, as the Aufbau principle
suggests that the lower-energy 4s orbital should be lled rst. However
, the observed
2
chromium, the predicted conguration is [Ar]4s
10
3d
s
s
1
ground-state electron conguration for copper is [Ar]4s
(gure 23). For
4
1
3d
and the observed is [Ar]4s
Activity
5
3d
(gure 23). In each case, promoting a 4s electron to a 3d level leads to a more stable
e
electron conguration. In the case of copper
, this gives a full d sublevel, and in the
Deduce
the
electron
conguration
case of chromium, there are no paired electrons but rather six half-occupied orbitals,
2+
of the Cu
c ation.
r
each containing an electron with the same spin.
The expected
Cu (Z = 29)
Cr ( Z = 24)
electron
l
observed
expected
[Ar]
configuration
chromium
9
2
3d
4s
2
4s
O
[Ar]
[Ar]
10
1
are
that
lled
you
of
need
chromium
to
with
(Cr)
and
copper
(Cu)
electrons
according
to
the
general
1
4s
are the only
know. In all other elements up to
Z=36, the
order.
i
n
o
v
sublevels
congurations
y
exceptions
3d
p
electron
two
r
e
The
5
4s
3d
LHA
4
3d
t
i
s
observed
configuration
n
y
[Ar]
congurations of copper and
The
quantum
mechanic al
the
minimum
molecule
X(g)
The
periods.
18.
The
electrons
for
p,
the
the
atom
helps
to
to
eject
an
electron
explain
the
elements.
out
of
a
trends and
Ionization
energy
neutral atom or
→
X
(g) + e
f
table
the
c an
be
(gure
element
u
energy (IE
)
are
known
periodic
shown
24).
are
as
The
also
table,
four
as
groups,
the
groups
blocks
sublevels
and
are
the
rows
corresponding
holding
are
numbered
the
to
known
from 1 to
the
outermost
four
valence
shown.
generally
decreases
down
the
groups of the periodic
1
l
and
a
Going
increases
down
electrons
energy
are
a
shielding
group,
and
the
number
from
(so-c alled
therefore
the
“inner
less
of
pull
sublevels
of
the
electrons”).
energy
is
increases. The outermost
nucleus
by
The
required
the
more
to
electrons
sublevels,
remove
in
the
the
electrons
lower
greater
from the
E
sublevel.
ac ross
a
outermost
nucle ar
across the periods.
shielded
sublevels
outermost
Going
period,
the
ele ctrons
charge.
At
constantbe c ause
54
of
state.
periodic
and
each
required
across
table
d,
ionization
table
the
in
model
ionization energies (IE) of
+
energy
periodic
s,
the
energy
Going
sublevels
First
v
O
f
x
t
a
Structure 3.1
as
+
columns
discussed further in
rst
itsground
i
are
in
the
o
energy and periodic
trends
r
o
table
d
Ionization
in
n
is
U
discontinuities
C
Ionization energy (Structure 1.3.6 and 1.3.7)
the
the
number
are
held
s ame
time,
number
of
of
protons
closer
the
inner
to
the
in
the
nucleus
nucleus
shielding
ele ctrons
by
ee ct
does
the
inc re ases,
inc re ase d
remains
not
ne arly
change.
so
y
Figure 23
and
P
u
Structure
1.3
Electron
congurations
LHA
s-block
1
18
transition elements
1s
1s
2s
14
15
4
5
6
7
8
4f
5f
trend
period
is
re quire d
of
ac ross
decreasing
shown
remove
ionization
gure 25:
2
Period
outermost
period.
energy
down
a
ele ctrons,
group
3
Period
4
2500

Xe
Al
Na
Rb
K
10
2
Kr
a
E
0
u
l
Li
v
sriF
O
 o i a  i  o i
f
x
gree
B
500
i

O
Be
5
Ar
N
H
1000
Period
t
a
r
o
lo
1500
increasing
o
1
Ne
2000
d
He
and
so
n
Period
in
to
the
U
a
is
inc re ases
p
general
across
energy
o
more
energy
p, d and f
C
ionization
to the sublevels s,
i
n
Therefore,
The blocks of the periodic table correspond
v
Figure 24
y
r
e
f-block
7p
y
6d
6p
n
7s
l
5d
O
6s
4p
r
4d
12
P
5s
11
y
3d
10
t
i
s
4s
9
e
3p
3
The
17
2p
d-block
3s

16
s
s
13
2
18
36
54
Aoi ber

Figure 25
Plot
of rst
ionization energy against
atomic number for the elements from
hydrogen to xenon
55
1
Models
of
the
particulate
nature
of
matter
LHA
Structure
There
are
two
clear
discontinuities
1.
Between the group 2
The
valence
The
paired 2s
across the period:
and group 3 elements
2
electron
conguration
of
2
beryllium is 2s
while
for
boron it is 2s
1
2p
.
2
the
electron
shield
slightly
the
easier
single
to
2p
electron
in
boron
from the nucleus,
s
s
making
electrons
remove.
e
+
Be
Be
0
0
2p
2p
1
2s
r
2
2s
B
0
2p
presence
results
that
of
to
and
The
—
not
t
in
be
any
trend
period.
aluminium,
the
allows
s ame
c an
be
observe d
in
comparing
group
2
to
group
2
collect.
For
so
example,
the
rst
the
3s
ionization
ele ctrons
energy
magnesium.
of
further
3
elements
1
shield
the
aluminium
is
lone
3p
lower
ele ctron
than
that
in
of
y
conclusions
patterns
discrepancies
do
pattern
for
they
r
e
overall
out
data
n
—
look
the
O
The
in
y
trends
2s
t
i
s
Scientists
2
2s
Patterns and trends
l
1
2p
2
Suppose you have a two-story building and you need to remove one oor to meet
drawn.
new height regulations. Which oor would you remove? Obviously, it will be the
p
What can be inferred from the
top oor, as the building would collapse otherwise! The same reasoning can be
v
patterns in successive ionization
applied to the ionization of atoms — electrons are removed rst from the highest
o
energies?
occupied energy level, and from the highest energy sublevel within that level.
Between the group 15 and 16 elements
group
15
conguration
Nitrogen
has
a
same
p
three
and
(gure
region
t
a
the
is
stable
o
nitrogen
there
more
sublevel,
i
the
p
16
also
2
of
electrons
space
in
2s
the
drop
This
and
2p
more
is
in
ionization
energy.
The
3
2p
electron
therefore
26).
a
2
nitrogen is 1s
2
while
energy
is
the
increased
orbitals
oxygen it is 1s
conguration
bec ause
have
for
do
not
than
required
paired
to
remove
electrons
into
in
close
N
3
2
2p
2p
2
2
2s
2s
+
O
O
4
3
2p
2p
2
2s
2
2s
electron
in
occupy
nitrogen,
proximity.
+
N
an
oxygen
However,
sublevel is
sublevels with 2 or 4
4
2p
oxygen as it has a half-
repulsion.
come
electron
2
2s
a
E
v
O
56
from
u
A half-lled
l
electrons
d
r
o
f
x
Figure 26
more stable than p
lled
to
of
n
U
From
u
C
i
n
2.
y
P
+
B
Structure
1.3
Electron
congurations
6
most
stable
p
orbital
conguration is p
,
a
completely
lled
p
LHA
The
sublevel,
3
followed
by p
,
a
half-lled
10
example, d
why
sublevel.
This
is
generally
true
for
other
sublevels.
For
5
and d
chromium
and
electron
copper
congurations
do
not
obey
the
are
also
Aufbau
stable,
which
principle
partly
explains
(gure 23).
s
s
C alculating ionization energy from spectral data
As
the
principal
lines
the
quantum
levels
converging
in
number
converges
the
to
hydrogen
of
a
energy
levels
continuum.
emission
increases, the distance
This
c an
spectrum,
be
observed
shown
in
by
gure
spectral
27.
r
5
level
4
Figure 27
Ultraviolet and visible light
transitions in hydrogen and
O
high energy
low energy
p
visible light
C
i
n
o
light
v
ultraviolet
y
r
e
level 1
n
2
y
level
t
i
s
3
the resulting
l
emission spectrum
level
y
t
∞
level
P
level
e
between
–8
The spectral lines in the hydrogen emission spectrum converge at 9.12 × 10
m, or
912 Å (gure 28). This represents the wavelength of light at which the hydrogen
wavelength
c an
be
used
to
energy
of
hydrogen.
o
i
a
1000
ionization
t
Figure 28
Hydrogen is ionized at
the wavelength where the spectral lines
converge in the emission spectrum
u
l
950
rst
t
a
r
o
f
x
E
v
O
900
the
d
912 Å
c alculate
n
This
U
atom is ionized.
1050
1100
1150
1200
1250
Wavelength / Å
57
1
Models
of
the
particulate
nature
of
matter
LHA
Structure
Worked example 3
8
Spectral
lines
converge
at
9.12 × 10
m
in
the
emission
spectrum
of
the
1
hydrogen
atom.
C alculate
the
rst
ionization
energy
of
hydrogen
in
kJ
mol
s
s
Solution
First,
c alculate
the
frequency
of
radiation using
to
3.00
8
of
light,
approximately
1
× 10
m s
=
c alculate
f
the
=
2.18 × 10
×
(s
9.12 × 10
energy
×
using
3.29
Planck’s
constant
and
the
s
values
of
the
J
two
steps
c an
be
merged
into
The
constant
are
one
h
Avogadro’
s
λ
This
given in the data
mole
and
represents
absorbed
Avogadro’s
level,
discussed further in
or
in
the
energy of a single photon of light which would be
exciting
removing
the
one
electron
electron
in
a
hydrogen
atom
to
the
energies
are
usually
given
1
energy
and
kJ mol
the
using
in
following
kJ mol
.
You
ionization
in
equation:
needed
electron
23
–1
mol
i
n
1000
3
1
U
C
kJ mol
1.31 × 10
one
1000
J × 6.02 × 10
=
=
remove
from an atom)
× N
A
o
–18
2.18 × 10
to
v
=
kJ mol
p
(energy
energy
convert the ionization
, the number of atoms
A
1
The
c an
Avogadro’s constant (N
y
1mol)
to
r
e
in
value
f
convergence
from the atom.
1
Ionization
Structure 1.4.
×
by using the
O
booklet.
and
=
t
i
s
are
=
n
constant
E
y
h × c
constant
E
speed of light,
equation
Planck’
s
equation
1
× 10
speed
l
The
the
m
15
J s
these
is
)
18
Alternatively,
c
Worked example 4
rst
ionization
booklet.
the
emission
helium
i
the
atom,
the
spectral
–8
lines
converge
at
f
x
C alculate
the
5.05 × 10
m.
The
rst
in
ionization
590 kJ mol
wavelength
Å,
for
energy of
–1
is
the
of
.
nd
from
kJ
Na
is
496
of
kJ
mol
as
given
convergence
for
by
the
the
IB
data
sodium
atom
the
J
energy
and
Then
of
ionization
dividing
it
by
1
23
/
6.02 × 10
for
one
atom
by
converting
the
given
value
Avogadro’
s constant.
1
mol
19
=
8.24 × 10
J
h × c
c alculate
the
wavelength of light using
E
=
λ
–34
6.63
× 10
8
J s × 3.00
× 10
–1
m s
19
8.24 × 10
J
=
λ
C alculate
7
λ = 2.41 × 10
m
convergence,
=
2410 Å.
c alcium atomic
spectrum.
This
corresponds
to
the
UV
region
in
the
electromagnetic
spectrum.
E
Successive ionization energies
It
requires
atom
while
58
to
a
the
, of helium.
v
O
c alcium
kJ mol
l
6.
in
u
energy,
First,
496 000 J mol
rst ionization
–1
of
wavelength
Solution
spectrum of
t
a
In
r
o
5.
Å.
the
o
d
Practice questions
in
1
energy
C alculate
spectrum
n
The
more
bec ause
the
energy
the
to
remove
number
of
electron–electron
the
protons
repulsion
second and
exceeds
the
decreases.
successive
number
of
electrons
remaining
from an
electrons
y
6.63 × 10
λ, where
×
P
=
f
1
Hz
34
E
=
r
Then,
3.29 × 10
c
1
m s
8
=
15
f
× 10
e
8
3.00
equal
Structure
a
the
so
result,
electron
increased
that
only
remove
the
clouds
electrostatic
the
stable
next
are
noble
electron
pulled
attraction.
gas
closer
Once
to
all
conguration
increases
sharply,
as
the
the
nucleus
valence
remains,
shown
the
in
and
held
electrons
energy
gure
tighter
are
Electron
congurations
LHA
As
1.3
by
removed
required to
29.
Figure 29
ygrene
from
s
s
t
Removing 10 electrons
magnesium
gives the noble-gas
2
conguration 1s
or [He].
There is a
1–
lom Jk(
remove the 11th electron
r
2
3
4
5
6
7
o
8
9
10
eletron
11
12
remoe
The
rst
ve
successive
ionization
energies
for
an
unknown
element
X
have
1
the
group
of
values:
the
403,
periodic
2633,
table
3860,
in
which
5080
and
element
6850 kJ mol
X
is
likely
to
be
increase
in
energy
occurs
from
the
rst
ionization
second
removed
a
stable
energy
to
group
level
1
of
).
This
noble
of
the
the
means
gas
that
the
second
conguration
element
periodic
contains
table.
of
the
one
the
periodic
5.5
v
3.5
E
3.0
2427,
3660,
values:
25 026 and
–1
32 827 kJ mol
.
Deduce the
group of the periodic table in
which this element is likely to
befound.
the
two
large
jumps
in
the
a
gol
l
EI
O
4.0
unknown element
following
u
f
x
5.0
4.5
explain
sodium.
i
6.0
table,
for
Therefore, the
801,
the
successive ionization
an
t
a
r
o
energies
o
and
) to
of
n
U
30
ionization
d
gure
successive
ve
electron, so the element
Data-based question
Using
rst
have
1
(403 kJ mol
electron is likely to be
atom.
The
energies
C
outermost
belongs
(2633 kJ mol
from
i
n
the
7
.
o
1
Practice question
found.
v
largest
Deduce
p
Solution
The
.
y
following
r
e
the
O
t
i
s
Worked example 5
n
Numer
y
1
l
0
y
P
n o it a z i n o I
e
)
considerable increase in energy required to
2.5
1
2
3
4
number

Figure 30
5
of
6
7
electrons
8
9
10
11
removed
Successive ionization energies for sodium
59
LHA
Structure
1
Models
of
the
particulate
nature
of
matter
Ionization energy data
Relevant
Part 3: Graphing the logarithm of the ionization
skills
Tool
2:
Extract
•
Tool
2:
Use
represent
•
data
energies
from databases
6.
spreadsheets to manipulate data and
data
in
graphic al
Title
the
third
(ionization
form
column
energy)”
in
as
your
spreadsheet “log
shown
below:
T
ool 3: Construct and interpret graphs
Part 1: Data collection
C
element name:
1
r
1.
B
Identify a database that contains successive
2
ionization energy data for the elements (for example,
Choose
one
of
the
following
elements:
(ionization
energy/
sulfur,
l
–1
chlorine,
successive
ionization
labelling
the
energy data in a
columns
as
4
1
follows:
5
2
B
7
.
Compute
using
a
ionization
energies
5
2
6
etc
sure
sc ales
your
and
graph
a
10.
title.
that
outermost
electrons.
energy
increases
l
the
v
E
does
properties
trend in
of
metals
large
a
increases
change
in
in
ionization
main
12.
each
graph
that
electrons
and
“zoom
will
in
in”
allow
energy
IE
values
and
across
non-metals?
C an
you
see
any
to
of
period
is
later
the
role
of
how
and
(Structure
down
LHA
60
do
patterns
variable
of
successive
oxidation
states
of
ionization
these
data
is
oen
into
of
2.
Enlarge
increases
a
group
explain
transition
(Structure
3.1)
graphic al
of
the
elements
relate to the
organized into tables
graphic al
scientic
knowledge?
employed
in
other
trends
help
to
forms.
What
representations in the
] and ionization energies? (Tool 3, Reactivity 3.1)
energies
=
increases in
3.1)
elements?
n
the
they
+
the
energy
level.
you to closely
inspect
large
Explain
transformed
advancement
Why are log scales useful when discussing [H
How
energy)
sublevels.
Experimental
and
unusually
energy?
representations
a
energy
level
a
O
in
graph
existence
with
a
the
ionization
existence of main energy levels in the atom.
How
successive
closely.
11.
Explain how the graph provides evidence for the
Linking questions
Construct
the
u
f
x
electron.
of
(ionization
explain
Are
subject
graphic al
areas?
KOT
ionization
correspond to the
i
values
logs
log
following questions:
the
indic ate
examine
suitably, with
descriptive
the
plotting
energy
energies?
vs ionization.
t
a
why
r
o
IE
and
successive
c.
present
energy
following questions:
Identify the
Explain
you
suitable
the
innermost
b.
ionization
number.
ionization
LOG10) function.
Part 4: Evidence for the existence of sublevels
o
labels,
Answer
a.
that
of
d
axis
graph
by
(or
Why is it useful to plot the logs of the ionization
n
M ake
line
the
Identify
b.
U
a
a.
that
Part 2: Graphing successive ionization energies
Plot
Answer
each
showing
o
1
ionization
C
i
n
4
v
9.
–1
kJ mol
graph
of
LOG
p
energy/
ionization
logarithm
Construct
vs
5.
the
spreadsheet
y
8.
ionization
3
the
r
e
2
4.
etc
element name:
1
O
6
A
energy)
kJ mol
ionization
3
c alcium.
t
i
s
spreadsheet,
or
n
Collect
potassium
y
3.
argon,
y
P
log
ionization
WebElements).
2.
e
A
Instructions
s
s
•
Structure
1.3
Electron
congurations
End-of-topic questions
5.
What
is
the
maximum
possible
number
of
electrons in
Topic review
1.
Using
your
answer
knowledge
the
guiding
from the
question
as
Structure 1.3
fully
as
third
energy
s
s
the
level?
topic,
A.
3
B.
6
C.
9
D.
18
possible:
e
How can we model the energy states of electrons
inatoms?
the
is
the
electron
ground
conguration
state?
Multiple-choice questions
7
A.
Which
row
is
correct
for
the
following
[Ar] 3d
regions of the
2
[Ar] 4s
D.
[Ar] 4s
4
3d
1
(UV)
Infrared
low
energy
wavelength
energy
1
(IR)
low
7
.
A.
Which
of
frequency
A.
IE
5
4p
the
>
3
frequency
low
energy
long
short
high
long
C.
energy
wavelength
long
low
low
frequency
wavelength
frequency
energy
The
the
following
an
III.
gas
an
alkali
metal
salt
to
Which
produce
electron
a
line
in
n = 4 to
n = 1
B.
from
n = 4 to
n = 3
C.
from
n = 3 to
n = 2
D.
from
n = 5 to
v
from
E
A.
n = 3
Bunsen burner
energy
levels
n = 4 and
produces a line in the visible
transition
the
a
O
likely
between
atom
a
8.
UV
Ionization
Which
are
energy
3+
(g)
→
X
energies
statement
nitrogen
o
spectrum.
transition
isolated
u
an
l
f
x
in
in
i
II and III only
I, II and III
placed
2+
produce a
A.
IE
and
(N)
<
IE
1
B.
IE
of
measured in kJ.
the
atom
X
represents
–
(g) + e
decrease
across a period going
(N)
IE
>
in
IE
(N)
its
a
<
(O)
IE
(O)
(N)
higher
>
IE
1
The
bec ause
is
ionization
energies of
correct?
partly
oxygen
lled
has
two
paired
bec ause
two
paired
sublevel
lled
oxygen
has
sublevel
bec ause
oxygen
loses
an
electron
an
electron
1
rst
sublevel
(O)
than
bec ause
nitrogen
oxygen
loses
1
from
9.
its
partly
1
IE
rst
1
from
D.
(O)
the
atoms
1
in
1
C.
about
oxygen
electrons
electrons
t
a
C.
D.
will
prism?
d
I and II only
I and III only
r
o
A.
electron
a
discharge tube
B.
n = 2
light
inc andescent lamp
ame
An
of
behind
process:
X
D.
energies
ionization
from le to right.
n
a
II.
sources
placed
U
I.
when
third
C
of
spectrum
i
n
Which
line
4.
C.
ionization
o
wavelength
high
v
frequency
D.
3.
Molar
the
correct?
4
B.
p
high
wavelength
is
y
low
energy
r
e
high
B.
following
IE
LHA
short
t
i
s
Ultraviolet
high
5
3d
n
[Ar] 4s
C.
spectrum?
O
B.
y
electromagnetic
chromium (Z = 24)
l
2.
of
y
in
P
Exam-style questions
What
r
6.
a
higher
ve
unknown
sublevel
successive
element
are
than
nitrogen
ionization
578,
1817,
energies
2745,
for an
11 577 and
–1
in
the
same atom is
spectrum?
14 842 kJ mol
this
element
A.
1
B.
2
C.
13
D.
14
is
.
In
which
likely
to
group of the periodic table is
be
found?
61
Structure
1
Models
of
the
particulate
nature
of
matter
16.
Extended-response questions
Sketch
an
number
10.
Explain,
in
your
own
words,
why
orbital
of
lling
unpaired
diagram
for
Al
and
deduce the
electrons.
[2]
gaseous atoms
3+
produce
line
spectra
instead
of
continuous
spectra.
[3]
17
.
A
transition element ion, X
,
has
the
electron
5
conguration [Ar] 3d
11.
State
the
full
and
condensed
electron
congurations
for
following
species
in
their
atom
silicon
d.
Ti
the
condensed
germanium
and
orbital
lling
containing
atom
one
or
more
electrons.
[1]
19.
3+
c ation
Describe,
in
your
own
words,
how
of
an
atom
c an
be
determined
anion
[1]
spectrum.
12.
Determine
which
of
the
congurations
below is
why
it
c annot
exist.
Using
the
2
2
2s
7
2p
2
3s
Deduce
and
2
1s
The
the
3p
shape
diagram
of
below
energy
4s
an
s
(not
levels
10
3d
2
in
state.
5
4p
10
4p
to
sc ale)
the
to
[1]
represents some of the
= 6
X
the
the
n
1
3
= 5

o
represent
of
successive
element
are
ionization
given
found.
2
= 4
n
= 3
n
= 2
n
=
1
lowest
emission
[1]
n
n
i
to
region
be
4
t
a
hydrogen.
visible
four
Table 3
C
U
diagram
the
= 7
n
u
of
the
in
n
n
d
spectrum
62
on
trend and discontinuities in
in
energies
table3.
[3]
for an
Deduce the
group of the periodic table in which element X is likely
5s
a
arrow
transition
rst
unknown
1
hydrogen atom.
l
v
E
an
energy
The
1
orbital.
r
o
f
x
O
Draw
21.
[2]
5s
6
3d
electron
i
n
electron
3s
ground
2
4s
6
following
p
15.
2p
6
3p
2
the
general
rst ionization
from sodium to
energies.
o
Sketch
2
3s
6
a
of
v
14.
2s
represents
6
2p
2
which
the
the
elements,
emission
[2]
y
2s
2
these
r
e
2
1s
Explain
6
3p
explain,
explore
3
5
2
3s
congurations
period
its
O
13.
6
2p
booklet,
the
t
i
s
2
2s
of
3p
argon.
2
1s
rst ionization
from
n
energies
1s
data
[2]
y
Explain
[2]
l
20.
impossible.
for
[1]
–1
IE
/ kJ mol
n
738
1451
7733
10543
Successive ionization energies for element X
y
P
energy
S
the
[1]
2–
e.
diagram
deduce the total number of p orbitals
[1]
LHA
c.
Sketch
[1]
atom
e
selenium
r
titanium
b.
[1]
ground states:
18.
a.
X.
s
s
element
the
. Determine the atomic number of
Counting particles
Structure 1.4
by mass: the mole
s
s
How do we quantify matter on the atomic scale?
atoms
of
the
in
a
glass
oce ans
substance,
the
of
of
so
any
these
water
than
combined.
mole,
physic al
particles.
The
enables
glasses
unit
of
chemists
object
comfortably
There
the
of
the
to
are
water
same
relative
amount
for
with
time,
large
the
molecular
expressing
numbers
concepts
masses
masses
of
of
of
very
molar,
allow
the
atomic
small
particles.
relative
use
of
atomic
small
substance.
mole
entities
contains
given
by
the
Structure 1.4.4
exactly the number of
gives
Avogadro constant.
in
the
that
M asses
mass (A )
and
C
and
of
relative
are
atoms
are
expressed
compared on a
formula
as
relative atomic
mass (M ).
by
r
–1
Structure 1.4.3 — Molar mass (M) has the unitsg mol
i
n
so
small
lead,
Pb,
–16
3.4 × 10
balance.
This
is
too
At
the
small
same
to
time,
be
the
about
chemists
and
a
unit
very
century
,
that
large
and
which
is
allows
hard
to
them
imagine,
to
of
work
even
Pb
let
bec ame
one
of
the
on
of
all
and
the
present
present
gives the actual
in
a
is
molecule.
determined
volume of solution.
law
measured
pressure
element
concentration
Avogadro’s
and
each
formula
element
molar
solute
—
gases
temperature
states
under
contain
the
that
equal
same conditions
equal
numbers
with
mole,
useful
in
1 g
directly.
have
most
count.
comfortably
most
measured
would
the
atoms
alone
numbers of atoms. This unit, the
quickly
be
element,
weighed
u
19th
need
f
x
masses
2.9 × 10
c annot
stable
number
21
huge,
masses
t
a
analytic al
g.
their
heaviest
i
only
r
o
of
that
the
volumes
C
are
of
of
molecular
each
The
formula of a compound
of
n
atoms
o
molecules
amount
of
—
atoms
ofmolecules.
U
million
a
d
and
Even
the
Structure 1.4.6
of
The mole (Structure 1.4.1)
Atoms
atoms
The
o
v
r
of
Structure 1.4.5
of
p
relative to
empiric al
y
—
12
sc ale
The
ratio
compound.
number
Structure 1.4.2
—
simplest
r
e
elementary
— The mole (mol) is the SI unit of amount
One
O
of
n
y
t
i
s
Structure 1.4.1
numbers
species.
de al
Understandings
At
and
l
of
all
small,
number
y
in
huge
P
more
extremely
a
e
are
contains
r
Atoms
of
a
mass
sensitive
lead is
Therefore,
both
was
concepts
very small
devised in the
in
chemistry.
l
The mole (with the unit “mol”) is the SI unit of amount of substance that contains
an
will
use
atom,
the
a
a
be
elementary
molecule,
v
O
23
6.02214076 × 10
c an
rounded
value
of
entities
an
of
electron
the
that
or
substance. An elementary entity
any
other
species. In this book, we
23
mole:
1mol
=
6.02 × 10

Figure 1
One mole quantities
of dierent substances (le to right):
E
aluminium,
water,
copper,
sucrose and
sodium chloride
63
Structure
1
Models
of
the
particulate
nature
of
matter
Avogadro’s constant
Prex
Symbol
(N
)
is
the
conversion factor linking the number of particles
A
F actor
–1
and
amount of substance
in moles. It has the unit of mol
:
–12
pico
p
10
23
N
–9
nano
n
10
micro
µ
10
milli
m
10
centi
c
10
=
–1
6.02 × 10
mol
A
–6
In
chemic al
other
we
–2
c alculations,
conversion
need
to
factor
multiply
substance (n)
into
the
the
Avogadro’s
(table1).
mass
in
number
For
kg
of
constant
is
example,
by
1,000.
atoms
or
used
to
in
Similarly,
any
the
convert
other
same
way
kilograms
to
10
to
multiply
that
amount
by
N
:
A
3
k
10
mega
M
10
N =
n×N
6
A
In
10
chemistry
texts,
the
term
“amount
of
substance”
is
oen
abbreviated to just
“amount”.
Table 1
Decimal prexes
×
10
amount
atoms
of
of
this
lead
(Pb),
element.
Solution
we
c an
rearrange
n
=
N
A
n(Pb)
=
i
n
According
is
signic ant
discussed in the
3
≈
Tools for
containing
as
follows:
mol
3
mol, so 4.8
10
sample
×
10
mol
=
4.8 mmol.
3
mol and 4.8 mmol) have been
n
o
i
mole
is
t
a
are
).
so
a
small.
convey
just
Choose
huge
how
one
approximate
number,
Measuring
of
and
it
amounts
is
of
useful
for
everyday
counting
objects
particles
in
moles
bec ause they
c an help use to
large this number is.
the
following
and
conduct
the
necessary
research
to
reach an
answer.
•
How
many
moles
of
grains
•
How
many
moles
of
water
of
sand
are
molecules
in
are
a
desert
in
a
of
large
your
sea
or
choice?
ocean
of
choice?
molecules in
What
the number of atoms in
•
How
tall
0.25 mol
•
How
many
water
v
water
E
64
The
One
of
N
a
Research skills
ATL
•
of
=
21
•
0.25 mol
c.
of
a
number
l
O
the
1 mmol
×
in
rounded to two signicant gures, the same as in the least precise value used in
u
the number of atoms in
2.5 mol of copper metal
b.
d
r
o
f
x
a.
table1,
the division (2.9 × 10
Activity
4.8 × 10
23
10
In this example, both answers (4.8 × 10
chapter.
C alculate:
to
n
10
×
3
U
chemistry
correct
=
A
C
gures
of
N
21
×
6.02
use
equation
o
v
2.9
Therefore,
the
mmol,
p
N
and
y
n,
mol
r
e
To nd
in
O
the
21
2.9
t
i
s
C alculate
n
y
Worked example 1
The
l

mole
is
of
the
is
human
age
a
of
cells
the
represents
universe,
stack
of
one
moles
of
air
mole
are
in
in
of
roughly
moles
of
how
many
seconds?
sheets of paper?
your school building?
people?
your
y
G
need
P
9
giga
r
kilo
we
e
d
any
grams,
convert the amount of
structural units (N),
–1
deci
as
into
s
s
–3
Structure
Relative molecular
mass and molar
1.4
Counting
particles
by
mass: the mole
mass
(Structure 1.4.2 and 1.4.3)
In
Structure 1.2,
we
introduced the concept of
relative atomic mass,
A , which
r
is
the
ratio
the
Similarly,
mass
of
a
certain
atom
relative molecular
to
mass,
one-twelh
M ,
is
the
of
ratio
the
of
mass
the
of
mass
a
c arbon-12
of
a
s
s
atom.
of
molecule
r
or
other
A
and
multiatomic
M
are
ratios,
so
they
to
one-twelh
of
the
mass
of
a
c arbon-12 atom. Both
have no units.
r
To nd the
M
of
a
molecule,
we
need to add together the
A
r
that
values
for all atoms
r
r
in
molecule.
e
r
species
l
M
for
a
molecule
of
water.
r
Water, H
O,
is
composed
of
two
hydrogen atoms (A
2
atom (A
16.00).
Therefore
M (H
r
should
always
use
the
O)
=
2
×
1.01
+
1.01)
16.00
=
and
actual
(not
rounded)
values of
A ,
which
the
keep
data
all
booklet
signic ant
and
the
gures
periodic
in
table
at
c alculated M
the
end
values
of
and
this
book.
never
of
ions
(C aCl
)
is
an
ionic
compound
instead
of
formula unit.
that
consists
of
as
many
bec ause
ions.
the
masses
r
hydrates
is
compounds
sulfate
large,
ve
as
of
follows:
means
water.
a
O) =
2
A (Cu) +
r
E
v
O
•5H
4
O”
=
63.55
+
=
249.72
+
contains
masses
with
the
in
ions.
that
one
Ca
which
neutral atoms
The
in
=
composition
ionic
and
compounds
will
be
discussed
Structure 2.1.
110.98.
water
molecules
O.
Copper(II) sulfate
2
Activity
(gure2). The
stoichiometric
C alculate the
M
values
for the
r
M
value
for
this
hydrate
c an
following
r
species:
a.
ammonia, NH
b.
sulfuric acid, H
3
4×A (O) + 5×M (H
r
r
r
O)
2
SO
2
32.07
structure of
formula unit of copper(II) sulfate is
Therefore, the
A (S) +
) and
one
One of the most common
•5H
crystals
as
(2×35.45)
4
l
c alculated
40.08
compounds
2.2)
deep-blue
2
molecules
same
pentahydrate, CuSO
clear,
before “H
=
(C a
2+
the
r
hydrates:
(Structure
c ations
unit
negligible.
u
“5”
with
M (CuSO
r
forms
f
x
coecient
be
form
bonds
copper(II)
pentahydrate
bound
r
2+
c alcium
formula
i
ionic
coordination
are
A (C a) + 2×A (Cl)
2
many
smallest
t
a
form
) =
Its
approximately
electrons
r
o
M any
of
have
).
o
M (C aCl
ions
(Cl
d
Therefore,
The
anions
for that substance
c alcium chloride
n
and two Cl
chloride
M
r
example,
U
2
twice
molecules, the
For
p
composed
o
is
C
substance
c alculated using the smallest
given
v
a
number.
i
n
If
is
integer
are
Similarly,
round them to the
r
nearest
oxygen
y
r
in
one
18.02.
2
r
e
You
=
r
=
r
O
t
i
s
Solution
n
the
y
C alculate
y
P
Worked example 2
+
(4
×
16.00)
+
(5×18.02)
c.
sodium
sulfate
Na
•10H
SO
2
4
4
dec ahydrate,
O
2
65
Structure
1
Models
of
the
particulate
nature
of
matter
s
s
e
r
l
M,
is
of
a
chemic al
numeric ally
substance
equal
to
is
relative
the
mass
of
molecular
O
2
1 mol
mass
of
n
mass,
mass
•5H
4
y
Molar
p
r
e
Molar
Crystals of copper(II) sulfate pentahydrate, CuSO
O
Figure 2
that
substance.
(for substances with
molecular and ionic structures) or relative atomic mass (for substances with atomic
For
example,
M(Na)
=
–1
22.99 g mol
and
–1
M(H
O)
=
18.02 g mol
2
o
i
n
v
structure).
Science as a shared endeavour
shared
understanding of common terminology helps scientists to
Hi s to r i c a l l y,
c o n ta i n e d
16
to
This
was
as
t he re
as
entities
w e re
is
th e
constantly
amount
(a to ms ,
a to ms
in
u
l
a
E
v
O
f
x
here
m e a s u re
physic al
that
v alue
as
m a ss
2018,
all
constants
being
updated.
substance
mo l e c u l e s ,
0.0 12 k g
of
th e
th e
mo l e
(o r
t ha t
ions,
12 g )
( a p prox i m a t e l y
i m prove me n ts
with
gre a te r
scientists
from
of
e l e c tro n s
c a r b o n -1 2 .
SI
base
instead
units,
of
6 .0 2  × 10
i n s t r u me n t a t i o n
had
the
objects.
in
Versailles,
mole,
were
France.
It
was
dened in terms of
Following these changes, one
23
mole
of
entities
The
no
a
substance
of
2018
that
through
match
is
equals
two
the
now
dened
exactly
as
6.02214076 × 10
12 g
exact
SI
numeric al
of
the
mole
exactly.
As
quantities,
values
of
me ans
a
that
result,
the
their
the
kilogram
the
mass
experimentally
dierences
determined
and
respective
between
these
mass
of
numeric al
a
A
the
or
are
1 mol
values
mole)
M
no
of
of
c arbon-12
M
(dened
longer
(dened
through
r
c arbon-12
v alues
of
numeric al
r
the
elementary
substance.
redenition
longer
so
atom).
small
However,
the
(approximately
–8
4 × 10
%)
Why
are
How
do
existing
66
)
a l l ow e d
pre c i s i o n .
including
physic al
in
more than 60 countries met at the
General Conference on Weights and Measures
agreed
of
23
n u me r i c a l
f re qu e n tl y,
November
terminology
defined
e l e me n ta r y
o
the
re v i s e d
s c i e n ti s t s
On
ma ny
i
be
mo l e
pa r t i c l e s)
How e ve r,
to
as
t
a
r
o
d
other
eectively.
th e
n
U
communic ate
or
C
A
that
they
constants
scientists
c an
and
be
achieve
denitions?
ignored
values
a
for
all
continuously
shared
practic al
being
purposes.
revised
and
updated?
understanding of changes made to
y
P
y
t
i
s

Structure
The amount (n),
mass (m)
and
molar
mass (M)
of
any
substance
are
1.4
Counting
particles
by
mass: the mole
related as
follows:
m
n
=
M
all
is
probably
the
stoichiometric
the
masses
of
most
common
c alculations.
chemic al
expression
Although
substances
are
the
in
chemistry,
base
SI
traditionally
unit
of
as
it
is
mass
expressed
in
s
s
This
used in almost
is
the
kilogram,
grams, and
e
–1
molar
masses
in
g mol
r
sugar
sucrose.
H
O
22
.
sold
is
an
in
the
form
organic
of
cubes
compound
that
with
are
made
the
almost
molecular
C alculate:
11
b.
the amount
c.
the number of oxygen atoms in one cube of sugar
of sucrose in one cube (2.80 g) of sugar
r
H
12
O
22
)
=
12
×
12.01
+
22
×
1.01
+
11
×
16.00
H
12
O
22
)
=
342.34 g mol
11
=
M
2.80 g
H
n(C
12
O
22
)
=
≈
11
0.00818 mol
1
342.34 g mol
One
mole
=
11
of
×
sucrose
n(C
H
12
11
×
n(O)
11
0.00818 mol
×
N
=
≈
23
×
C alculate:
a.
the
H
molar
mass
of
sulfuric
acid,
SO
4
oxygen atoms, so
0.0900 mol
0.0900mol
A
of
6.02 × 10
o
=
22
11 mol
)
d
=
N(O)
contains
O
Activity
2
n
n(O)
U
c.
o
n
C
i
n
m
b.
342.34
v
1
M(C
=
11
y
M (C
p
r
e
Solution
O
the molar mass of sucrose
t
i
s
a.
a.
entirely
formula
n
12
oen
y
C
is
Sucrose
l
of
y
Table
P
Worked example 3
–1
mol
b.
the
amount
1.00 g
c.
the
of
of
substance
sulfuric
number
of
in
acid
hydrogen
22
≈ 5.42 × 10
atoms
in
1.00 g
of
sulfuric
acid
i
t
a
r
o
u
l
f
x
a
E
v
O
19

Figure 3
There are more oxygen atoms in one sugar cube than the estimated
total insect population on E arth (10
) and
total grains of
21
sand
on E arth’s beaches (10
)
67
Structure
1
Models
of
the
particulate
nature
of
matter
Empiric al formula, molecular
formula and
chemic al analysis (Structure 1.4.4)
The
composition
represented
each
element
formula
present
shows
in
the
empiric al
of
in
the
molecule
simplest
substance.
c an
be
formula
ions
the
in
the
the
of
ratio
The
identic al
is
substance
formula,
dierent
as
compound
the
a
molecular
of
the
and
O
2
O
H
2
2
r
e
H
6
12
H
2
5
CH
6
O
22
11
C
O
2
H
12
O
22
11
o
v
empiric al formulas of selected substances
n
U
C
i
n
Molecular and
C
p
H
C
Table 2
HO
O
12
O
y
C
O
2
10
formula
O
H
H
C
4

Empiric al
t
i
s
3
sucrose
represents the simplest
formula
2
glucose
which
n
H
are
same
l
water
that
the
O
O
of
For ionic compounds, the
unit,
y
ozone
peroxide
formulas
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O

Figure 4
is used
The
Sodium
uoride is an ionic compound
in some countries as a food
number
of
atoms
element in mol
(N
=
of
a
certain
n×N
).
with the empiric al formula NaF. It
supplement to prevent tooth dec ay
element
Therefore,
is
the
proportional to the amount of that
empiric al
formula
also
shows the
A
mole ratio
water, H
of
O,
elements
contains
in
two
a
chemic al
atoms
of
compound.
hydrogen
and
For
example,
one
atom
of
one
molecule of
oxygen, so the
2
atomic
68
ratio
of
hydrogen
to
oxygen
in
water
is
2:1.
Similarly,
one
mole
of
water
y
Molecular
O
c an be
empiric al
elements
P
Substance
hydrogen
contrast, the
(gure4).
oxygen
butane
In
dierent
empiric al
(table2).
formula
structure
shows the actual number of atoms
substance.
atoms
molecular
or
same
that
of
with
which
r
ratio
chemic al
e
substance
a
molecular
s
s
of
of
by a
Structure
contains
mole
The
two
ratio
of
moles
elemental
mass,
which
ratio
hydrogen
to
composition
is
c an
atoms
oxygen
of
a
in
and
compound
referred to as the
be
c alculate
to
mole
of
particles
the
is
oen
expressed
in
percent
percentage composition,
by
ω. The
percentage composition of a compound.
percentage
composition
of
water.
r
Solution
O)
=
1 mol, then
n(H)
=
2 mol and
n(O)
=
1 mol. Using
m
=
n
×
M:
2
1
m(H)
2 mol
=
1 mol
×
1.01 g mol
=
2.02 g
l
=
1
m(O)
×
16.00 g mol
=
16.00 g
O)
=
1 mol
×
18.02 g mol
×
100%
=
18.02 g
2
2.02 g
≈
11.2%
18.02 g
ω (O) = 100%
chemists
empiric al
experimental
masses
In
a
techniques,
typic al
products
mass
or
the
are
as
percentages
ω (Fe) = 72.36%, then
Let
m(Fe
O
x
compounds
contains
72.36%
ω (O) = 100%
100 g, then
m(Fe)
=
y
to
determine
72.36 g
≈
=
the
be
or
Practice question
in
by
elemental
and
weights
elements
C alculate
measured.
oxygen,
These
by
decomposed, and
determined
excess
(iron
determined
the
the
percentage
composition of sulfuric acid, H
SO
2
4
various
analysis.
the
volatile
are then
original
sample.
of
oxides).
Deduce
the
empiric al
iron.
72.36%
72.36 g and
amount
of
=
27.64%.
m(O)
=
27.64 g
each element:
a
M
l
m
n =
n(Fe)
=
v
O
Use
)
c an
in
chemic al
be
are
combustion
weighed.
u
f
x
If
of
deducing
o
Solution
several
that
burned
products
i
form
oxide
is
and
of
t
a
an
sample
c an
combusted
d
oxygen
of
a
automated
sample
trapped
r
o
and
formula
is
decomposition
in
problem
percentage composition or
composition
compound
fully
Worked example 5
Iron
its
n
into
the
elements
such
opposite
from
U
converted
of
the
percentage
which
experiment,
combustion
The
combustion
percentages
analytic al
face
C
mass
the
in
oen
compound
i
n
The
data.
analysis,
of
a
o
the
for
v
destruction
more
formula
p
other
88.8%
y
practice,
the
=
r
e
In
11.2%
O
t
i
s
ω (H) =
n
y
1
m(H
y
n(H
P
Let
e
the
mass: the mole
oxygen atoms, so the
Worked example 4
C alculate
by
water is also 2:1.
commonly
used
one
Counting
s
s
mole
of
hydrogen
1.4
1.296 mol
–1
55.85 g mol
E
27.64 g
n(O)
=
≈
1.728 mol
–1
16.00 g mol
The
mole
ratio
Therefore,
the
x : y
=
1.296 : 1.728
empiric al
formula
of
≈
1 : 1.333
the
oxide
≈
is
3 : 4
Fe
O
3
.
4

Figure 5
Fe
O
3
magnetite,
is the main component
of the mineral
4
a common iron ore
69
Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 6
Hydroc arbons
unknown
to
are
organic
hydroc arbon
produce
26.41 g
of
compounds
has
undergone
c arbon
dioxide,
of
c arbon
and
combustion
CO
,
and
in
hydrogen.
excess
13.52 g
of
An
oxygen
water,
H
2
the
empiric al
formula
of
the
O.
2
s
s
Deduce
hydroc arbon.
Solution
e
1
M(CO
)
=
12.01
+
2×16.00
=
44.01 g mol
2
n(CO
)
=
≈
r
26.41 g
0.6001 mol
2
–1
44.01 g mol
n(CO
)
=
0.6001 mol
2
1
M(H
O)
=
2×1.01
+
16.00
=
18.02 g mol
2
l
13.52 g
O)
=
≈
0.7503 mol
2
–1
18.02 g mol
n(H)
=
2
O)
=
2×0.7503 mol≈1.501 mol
the
hydrogen
hydroc arbon, C
H
x
The
mole
ratio
Therefore,
Multiplying
this
U
whole
number
whole
the
ratios.
ratio
we
divide
term
the
you
ratio
ratio
each
of
1
:
1.333.
should
by
of
3,
in
Then,
multiply
and
then
the
ratio
you
Whole
To
numbers
are also
c alculated
was
convert it to a whole
by the smallest number in the
c an
ratio
5
initially
1.296 and 1.728.
from
H
2
number
5,
originate
2 : 5
hydroc arbon is C
values:
ratio
which
as
example
=
products
y
by
a
worked
1 : 2.5
o
factor
you
gives
≈
the
C
a
ratio,
This
of
combustion
non-integer
i
n
ratio.
two
formula
formulas
In
v
number
empiric al
of
0.6001 : 1.501
empiric al
integers.
comprised
=
the
p
express
known as
x:y
in
r
e
We
the
atoms
, so:
y
O
n(H
and
t
i
s
×
c arbon
2
All
n
y
n(H
to
use
trial
obtain
subsequently
and
the
rounding
error to determine
whole
the
number
result,
ratio.
gives a
3 : 4.
n
The molecular formula of a compound can be deduced from the empirical formula
if we know the molar mass of the compound. For example, you might determine
The
masses of
discussed in
+
of
twice
–1
(1.01
5)
29.07 is
the
=
29.07 g mol
roughly
number
of
half
atoms
of
as
58.12,
the
therefore
empiric al
the
molecular
formula: C
suggests
be
sure
that
about
this
l
a
E
v
O
be
with
distinguished
comparing
their
it
hydroc arbon
without
the
by
same
further
could
be
analysis,
butane, C
molecular
as
formula.
there
is
Butane
formula must
H
4
methylpropane,
c an
×
10
H
4
c annot
Structure 1.5.
2)
value
have
Table2
is
o
×
. The molar mass of the empirical formula can be calculated:
i
molar
(12.01
t
a
the
substances
–1
58.12 g mol
u
gaseous
d
r
o
f
x
Determining
experimentally that the molar mass of the hydrocarbon in worked example 6 is
.
However, we
10
another
and
hydroc arbon,
methylpropane
measuring their boiling points ( Structure 1.1) or
infrared
spectra (Structure 3.2).
Practice questions
1.
Deduce
a.
an
b.
a
the
empiric al
oxide
of
formulas
manganese
hydroc arbon
that
of
that
the
following compounds:
contains
produces
36.81%
5.501 g
of
of
c arbon
oxygen
dioxide
and
2.253 g of
water upon complete combustion
2.
Deduce
the
molecular
–1
is
70
42.09 g mol
formula
of
the
hydroc arbon
from
1b
if
its
molar
mass
y
=
P
n(C)
Structure
1.4
Counting
particles
by
mass: the mole
Experimental determination of empirical formula
Relevant
skills
Tool
1:
•
Tool
3:
Instructions
Measure
C arry
mass
out
c alculations
involving
decimals
1.
Weigh
2.
Obtain
andratios
and
•
Tool
3:
Use
•
Tool
3:
Construct
and
•
Inquiry
3:
realistic
piece
from
dry
of
crucible.
magnesium
your
teacher.
Explain
and
relevant
4.
Heat
the
crucible.
improvements
the
crucible,
with
its
lid
enter
Periodic ally
on,
the
li
over
crucible.
equipment
prec autions
will
get
around open ames.
very
hot.
around it and do not touch it while it
burns
with
a
heating until the magnesium no longer
Then,
up.
6.
When
7
.
Heat
to
the
the
directly at it.
remove
for
Repeat
a
the
heat
source
and
minute.
this
its
contents
Allow
to
cool
strongly
and
Q uestions
balance
1.
•
pipeclay triangle
•
tripod
•
heat-proof mat
•
tongs
•
magnesium ribbon
n
i
o
d
mass
graph
the
empiric al
formula
oxide.
experimental
data
of
determine
cycle until the
from
mass
of
other
empiric al
members
magnesium
formula to the
of
oxide
your
vs
class.
mass of
6.
Identify
t
line
Explain
of
any
on
what
Explain
why
the
Identify
and
applic able)
and
draw a best
a
graph
shows about the composition
oxide.
you
until
(if
graph.
magnesium
crucible
7
.
anomalies
the
repeatedly
constant
explain
two
heated
mass
was
major
and
weighed the
achieved.
sources
of
error in this
procedure.
8.
Suggest
that
realistic
could
improvements to the methodology
minimize
the
sources
of
error
you
have
l
a
Figure 6
u
Bunsen burner
The experimental set-up
E
v
O
f
x
t
a
r
o
ribbon
to
for an
magnesium.
5.
coiled magnesium
a
your
one.
Obtain
Plot
4.
crucible
Compare
actual
3.
data
C
U
lid
2.
the
magnesium
o
v
Bunsen burner

Process
of
i
n
•
(±0.01 g)
y
crucible and lid
•
p
r
e
•
allow the
re-weigh.
heating-cooling-weighing
mass is constant.
M aterials
and
few minutes.
crucible is cool, weigh it.
crucible
additional
very bright light. Do not look
cool
O
M agnesium
allow
lights
t
i
s
ishot.
to
Continue
crucible
Take suitable
lid
y
prec autions
roaring
n
The
5.
protection.
a
crucible
l
•
eye
suitable
the
P
to
ame.
y
Take
0.3 g
mass.
Twist the magnesium into a loose coil and place it
air
Wear
exact
3.
graphs
S afety
•
its
e
interpret
investigation
•
(between
approximation and estimation
Bunsen
•
ribbon
Measure
r
an
clean,
a
1.0 g)
inside
to
a
s
s
•
identied.
9.
Reect
on
empiric al
round
C an
to
the
role
formula
the
of
approximation
c alculations.
nearest
whole
and
rounding in
When is it suitable to
number?
When is it not?
you come up with a rule of thumb of when to
round
and
when
not
to
round?
71
Structure
1
Models
of
the
particulate
nature
of
matter
Measurement
Atoms,
molecules
impossible.
As
with
to
all
Consider
and
ions
concept
mass,
which
mass
of
a
so
the
c an
measurements,
the
are
of
be
is
easily
mass
sample
small
mole
has
of
that
counting
powerful
them
bec ause
directly is virtually
it
relates number of
s
s
particles
The
measured.
an
uncertainty
c alcium
associated with it.
c arbonate,
C aCO
,
is
found to be
3
up
to
g
±
0.001
0.001
does
c alculation
a
in
it
represent?
nd
particles,
measurement
you
out.
is
How
You
quite
that
is
clearly
particles
see
that
in
a
c an
minuscule
does
moles
it
ever
negligible?
involve
the
DP
making
If
so,
inaccurate
mass.
uncertainty
O
properties
and
solutes.
heterogeneous
colourless
mixtures
are
discussed in
sugar
(from
the
For
the
so
sugar
a
solutions.
substances
more
the
solvent.
example,
than
the
of
or
two
properties
The
other
solution
(white
of
Solutions
solvent
more
The
of
the
crystalline
in
meaning
“water ”),
in
in
easier to
bec ause
chemic al
components.
whole
water
powder),
which
are
used
it
c an
reactions.
E ach solution
solvent is usually the major
components
sugar
is
participate
or
solutes.
a
of
is
solution
the
so
the
are similar
solution
more
like
water
solute. In this topic, we will consider only
aqua
is
are
water
the
(clear
solvent
aqueous solutions
solvent
is
water.
o
i
d
solute
t
a
r
o
u
l
f
x
a
E
v
O
72
is
L atin
solution,
of
liquid)
while
Structure 1.1.
the
or
in
Sometimes
mixtures
one
n
U
c alled
Homogeneous
of
properties
and
out
C
the
solvent
gases.
dissolved
homogeneous
i
n
component
of
c arried
or
o
are
consists of a
are
solids
v
Solutions
to
than
tiny, but in
course, particularly when
y
the
reactions
mix
p
aect
and
is
measurements.
t
i
s
chemic al
handle
r
e
M any
by
many
when? Think about these
chemistry
Solutions and concentration
(Structure 1.4.5)
How
represent? Do a quick
the
large.
through
be
n
experiments
many
will
uncertainty
proceed
This
measurement
l
doing
as
direction.
mass
y
questions
it
the

In
some
c ases,
the
ethanol
and
water
present
it
is
is
not
water ”
the
identity
water,
in
major
rather
Figure 7
each
the
of
of
the
these
mixture,
component.
than
How a solution is formed
“4%
it
solvent
liquids
is
For
solution
of
is
unclear:
c an
be
traditionally
example,
for
c alled
example, if we mix
a
solvent.
regarded
we
as
the
However, if
solvent,
even if
say “96% solution of ethanol in
water in ethanol”.
y
Is
of
g
and
either
means
P
terms
This
r
moles
g.
e
3.500
Structure
Solutions
solute
solute,
small
the
are
and
oen
and
so
has
proportion
term
much
a
of
high
ratio
solute,
“concentrated”
per
100 g
less
than
of
the
10 g
according
to
the
concentrated solution
of
of
and
refers
solvent,
the
solute
so
has
to
the
per
or
mole
a
ratio
with
term
100 g
of
much
of
the
to
more
particles
solvent.
than
has a
Generally,
10 g of the
refers to solutions with
solvent.
precisely
would
dened
c all
a
precise denitions and their choice and
dependent.
and
The
should
solution
of
be
terms
used
“concentrated”
with
c are.
5 g of sulfuric acid (H
For
SO
in
much
higher
laboratories.
At
proportions
the
in
same
100 g
time,
of
a
acid
solution
water
would
of
in
water
5 g
be
)
100 g
of
water
4
to
of
are commonly
potassium
considered
very
4
concentrated
by
permanganate
The
in
antiseptic
concentrations
extent
does
above
than
could
quantity
composition
to
of
solutions
concentration, c,
the
is
also
volume of the solution:
common
units
for
molar
be
very
dilute
smaller
1 × 10
of
1 × 10
molar
i
–6
=
–3
mol dm
concentrations
are
sometimes
–3
(for
of
Note
the
that
whole
“the
solution
For
of
is
of
is
to
to
say
is
the
same
–3
µmol dm
)
c an
–3
a
as
M
(for
“2.5 MNaOH”
sodium
refers
incorrect
ratio of the
mol dm
)
means that
hydroxide.
specic
that
“the
substance, not the
concentration of a
–3
1.0 mol dm
”,
as
chloride
it
is
or
not
clear
water.
The
whether
we
are talking
correct statement
–3
oen
of
sodium
chloride
represented
by
in
a
square
solution
brackets
is
1.0 mol dm
”.
around the solute
–3
example,
ammonia.
is
sodium
concentration
concentration
formula.
it
of
the
(which
or
abbreviated
expression
2.5 mol
a
be
solution
the
concentration”
example,
concentration
E
Molar
chloride
the
“molar
For
example,
contains
v
O
would
solution
term
solution.
sodium
about
the
For
u
each dm
).
l
f
x
3
mmol dm
t
a
units
r
o
–3
–3
mol dm
is
o
–3
=
1 µmoldm
mM
mol dm
(mmol dm
d
–3
or
are
–3
units
used:
1 mmoldm
The
solutions,
molarity,
n
also
For
water.
–3
concentration
–1
).
U
most
mol L
of
numeric ally.
C
solute
V
solution
potassium
expressed in terms of
=
as
100 g
expressed
known as
solute
The
be
per
numeric ally help or hinder the
n
c
0.1 g
knowledge?
Molar
solute
a
less
of
p
a
expressing
are
concentrations
i
n
of
examples
typic al
o
the
concentration.
amount
of
the
solutions
as
v
Quantitatively,
in
worker,
y
what
communic ation
medic al
r
e
To
any
O
)
sulfuric
t
i
s
permanganate (KMnO
of
n
used
as
“dilute”
y
2
“dilute”,
and
example, most
l
chemists
have
y
not
not
context
P
are
is
e
do
r
words
interpretation
mass: the mole
between the
TOK
Some
by
proportion of
dilute solution
solute
“dilute”
ratio
large
solvent, while a
low
solutions
and
solute
to
a
mass
contains
Counting
s
s
solute
classied
solvent. A
1.4
the
expression [NH
]
=
0.5 M
refers
to
a
0.5 mol dm
3
–
Similarly,
the
expression
[Cl
]
refers to the molar
concentration of chloride ions in a solution.
73
Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 7
–3
C alculate
a
the
solution
molar
concentration
prepared
by
dissolving
of
sodium
3.60 g
of
chloride,
N aCl(s)
in
in
mol dm
water
to
,
in
make
3
25.0 cm
of
the
nal
solution.
s
s
Solution
First,
c alculate
the
molar
mass of sodium chloride:
e
1
M(NaCl)
=
22.99
+
35.45
=
58.44g mol
m
n
=
to
r
Then use
c alculate the amount of solute:
M
3.60 g
0.0616 mol
–1
58.44 g mol
3
Convert
the
volume to dm
by
dividing
by
1,000:
25.0 cm
3
=0.0250 dm
n
C alculate
the
mass of sulfuric acid,
Use
c
=
to
c alculate
the
concentration:
SO
2
,
in
50.0 cm
of a
solution
0.0616 mol
4
c(NaCl)
–3
where [H
SO
2
]
=
≈
=
1.50 mol dm
of
a
ρ
solution
,
of
the
solute
sometimes
solute
V
solution
C
i
n
Worked example 8
C alculate
the
mass
example
concentration
acid,
in
mol dm
ρ(H
SO
2
f
x
mass
mass of the solute to
sodium
chloride
in
the
solution
of
the
solute
concentration
and
as
the
volume
of
the
solution,
we
3
=
=
144 g dm
3
the
mass
and
concentration
molar
mass,
using
of
NaCl
the
c an
be
found
relationship
ρ
from its molar
=
solute
c
×
most
common
a
mass,
as
follows:
ρ
solute
c
and
=
solute
M
solute
units
molar
for
mass
concentration
concentration
of
the
are
same
M
solute
3
ρ(NaCl) = c(NaCl)× M(NaCl) = 2.46 mol dm
concentration
c an
follows:
:
solute
1
×58.44 g mol
–3
The
from
3.60 g
concentration
u
l
E
v
O
74
of
the
0.0250 dm
Alternatively,
) =
4
–3
0.150 g cm
mass
mass
i
t
a
sulfuric
the
the
ρ(NaCl)
concentration
–3
of
in a solution with
know
o
d
molar
r
o
the
we
of
7.
Solution
c alculate
ratio
n
U
worked
C alculate
the
o
v
m
=
solute
If
expressed as the
is
volume of the solution:
ρ
Activity
which
p
the
is
solute,
y
composition
concentration,
r
e
The
3
2.46 mol dm
3
0.0250 dm
4
O
t
i
s
V
3
H
n
=
l
3
V(solution)
y
Activity
g dm
solute
3
≈ 144 g dm
–3
and
are
g cm
related
.
M ass
by molar
y
≈
=
P
n(NaCl)
Structure
1.4
Counting
particles
by
mass: the mole
Worked example 9
A
standard
solution
was
prepared
by
dissolving
6.624g
of
sodium
c arbonate,
Na
CO
2
a
,
in
deionized
water
3
3
using
3
250 cm
volumetric
ask.
An
analytic al
pipette
was
used
to
transfer
10.0 cm
sample
of
this
solution
to
a
3
100cm
volumetric
ask,
and
the
ask
was
topped
up
to
the
graduation
mark
with
deionized
water.
C alculate
in
moldm
,
of
sodium
c arbonate
in
the
new
solution.
Solution
we
need
to
nd
the
concentration
of
sodium
c arbonate
in
the
e
First,
standard solution:
1
M(Na
CO
2
)
=
2×22.99
+
12.01
+
3×16.00
=
105.99 g mol
3
r
6.624 g
n(Na
CO
)
=
≈
3
0.06250 mol
–1
105.99 g mol
3
V
=
3
250 cm
=
0.250 dm
standard
that
the
accuracy
of
a
typic al
volumetric
ask
is
three
signic ant
gures.
3
(Na
standard
CO
2
)
=
=
3
0.250 mol dm
3
we
First,
need
c alculate
to
c alculate
the
amount
the
of
concentration
Na
CO
2
in
=
10.0 cm
of
sodium
sample.
c arbonate
Remember
to
in
the
new solution.
convert
3
3
V
the
3
=
0.0100 dm
sample
3
(Na
CO
)
2
3
CO
)
=
c
(Na
sample
CO
2
)
=
0.250 mol dm
3
3
n
(Na
sample
the
2
sample
0.250 mol dm
3
×0.0100 dm
=
3
is
diluted
with
deionized
water
to
produce
(Na
sample
CO
2
)
=
n
3
work
out
(Na
new
the
CO
2
)
=
0.00250 mol
3
concentration
of
Na
CO
2
of
the
new solution:
3
V
=
3
100 cm
=
0.100 dm
new
0.00250 mol
(Na
new
CO
2
)
=
=
3
common
solutions
following
=
sample
c
×
sample
two
of
sample
n
=
new
new
know that
gives
n
=
sample
the
following
c
E
sample
=
new
V
n
,
so
the
of
solute
amount
Na
not
CO
2
form
the
of
change.
by the
3
standard
concentrated solutions
required
of
Practice question
concentration when
the
solute
are
c alled
3.
A
standard
by
copper(II)
solution
in
worked
example
solution
dissolving
was
prepared
2.497 g of
sulfate
pentahydrate,
9,
CuSO
• 5H
O,
4
2
using
a
in
deionized
3
water
100 cm
volumetric
3
ask.
A
5.00 cm
sample of this
3
solution
was
C alculate
diluted
the
to
250.0 cm
.
concentration, in
–3
mol dm
nal
you
of
does
c an
substitute
equation
1
into
, of copper(II) sulfate in
the
solution.
equation 2. This
new
expression:
× V
sample
c
a
You
V
v
O
c
l
new
2.
the
the
to
concentration
c alculations:
V
in
them
known
dividing
u
n
a
concentration
f
x
1.
the
with
chemic als
dilute
t
a
did
the
and
r
o
determine
we
store
i
Stock
standard solutions.
To
to
d
needed.
practice
stock solutions)
by
amount
n
a
solution
3
o
is
(so-c alled
new
the
0.0250 mol dm
3
0.100 dm
It
the
3
U
c
in
solution,
C
volume
i
n
c an
new
o
n
you
the
v
Therefore
Now
0.00250 mol
p
When
=
3
volumes to dm
y
standard
all
r
e
c
O
Then
t
i
s
0.250 dm
n
y
0.06250 mol
c
l
Note
y
P
2
the
s
s
–3
concentration,
The
process
solutions
is
for
preparing
standard
discussed in the
Tools
new
Therefore,
need
to
to
c alculate
know
the
solution,
and
the
c
c
V
×
1
V
=
1
×
2
for chemistry
the
original
volume
concentration
concentration
of
the
of
of
a
solute
the
in
solute,
a
new
the
solution,
chapter.
you just
volume of the original
new solution. In summary:
2
75
Structure
1
Models
of
the
particulate
nature
of
matter
C ase study: spectrophotometry and c alibration curves
Spectrophotometry
the
is
intensity
of
commonly
is
visible,
used
for
an
analytic al
ultraviolet
technique
and
determining
based
near-infrared
concentrations
on
the
radiation.
of
measurement of
This
technique
coloured substances in
s
s
solutions.
A
spectrophotometer
through
a
of
value
standard
solutions
substance
are
producing a
determining
or
a
c alibration
electric al
unknown
the
absorbed
are
result
of
measured.
the
unknown
curve
relates
conductivity)
of
c an
measurement
be
on
0.50
i
r
o
3
A typic al c alibration curve
O
a
was obtained using a series of
l
The calibration curve in gure8
u
f
x
Data-based question
coloured
(such as
concentration of
measuring
that
property
curve.
t
a
Figure 8
the
plotted
c alibration
y
0.40
by
the
property
c alibration
p
0.30
found
to
of
are
The
o
0.20
concentration/mmol dm

C
0.10
(gure8).
solution
several
o
0
d
0
absorbances
measurable
the
Initially,
by serial dilution
n
0.10
a
sample.
concentration
the
r
e
i
n
U
ecnabrosba
0.20
v
0.40
0.30
These
passes
measures the
absorbance. Absorbance
the
c alibration curve
concentration
the
by
prepared
studied solution.
c ase,
pH
The
plotting
for
converts it into the
light
which
photodetector
O
solute.
and
used
the
general
absorbance,
the
absorbances
wavelength,
The
n
in
studied
of
certain

Figure 9
A series of standard
solutions of potassium permanganate
Ideally, the calibration curve should be linear, pass through the origin and have a
tilt of approximately 45°. If the curve does not meet any of these requirements, it
standard solutions of potassium
should be constructed again using a slightly dierent wavelength of light and/or
. A solution
v
permanganate, KMnO
4
dierent set of standard solutions. Sometimes linearity can only be achieved within
with unknown concentration of
E
KMnO
a narrow range of concentrations. In this case, the studied solution can be diluted,
has an absorbance of
4
so the concentration of the studied substance falls within the range of calibration
0.285. Determine the concentration
of KMnO
curve. In the last case, some additional calculations will be required to relate the
in that solution.
4
concentrations of the studied substance in the diluted and original solutions.
76
y
then
and
a
solution.
y
the
their
light
amount
the
of
t
i
s
In
of
light
studied
l
is
the
concentrations,
substance
the
P
curve
and
of
transmitted
describing
(gure9),
against
the
produces
sample
r
is
small
e
intensity
a
Structure
Another
technique,
spectrophotometry
colorimetry,
but
“spectrophotometry”
correct
but
is
based
on
the
1.4
Counting
particles
by
mass: the mole
same principles as
limited to visible light. The terms “colorimetry” and
are
oen
used
interchangeably,
which
is
not
entirely
very common.
s
s
Concentration uncertainty of a standard solution
standard
solution
this
activity,
of
copper(II)
you
is
a
will
sulfate,
solution
prepare
each
by
of
known
two
using
concentration.
M aterials
standard solutions
dierent
equipment.
Wash
•
Weighing boats (2)
•
100 cm
•
Stirring
•
Funnels (2)
•
Pipettes
•
Spatula
•
Reagent
•
Blank labels
•
Colorimeter
•
Cuvettes
•
C alibration
bottle
containing
distilled
water
3
By
propagating
will
the
determine
a
measurement
the
of
allow
You
your solutions
you
to
assess the
1:
Measuring
Tool
1:
Standard
3:
volume
solution
C alculate
and
and
mass
preparation
interpret
percentage
error and
curve
relating
sulfate and absorbance
•
Tool
3:
Express quantities and uncertainties to an
•
appropriate
number
of
signic ant
3:
Record
•
Tool
3:
Propagate uncertainties
•
Inquiry
Assess
accuracy
and
precision
Wear
•
Solid
copper(II)
sulfate
is
an
irritant
and
environment
Dispose
of
all
solutions
pentahydrate, CuSO
toxic to
3
•
100 cm
•
Milligram
•5H
4
O
2
volumetric ask
balance
(three
decimal
places)
Additional equipment for solution 2:
3
•
100
•
Centigram
n
•
protection.
U
the
eye
sulfate
C
i
n
S afety
•
Copper(II)
Additional equipment for solution 1:
v
2:
gures
measurement uncertainties
p
Tool
o
•
concentration of copper(II)
y
r
e
percentage uncertainty
bottles (2)
O
Tool
beakers (2)
rods (2)
n
skills
•
Tool
then
you will
concentrations.
•
•
concentration
will
values.
t
i
s
Relevant
the
This
uncertainties,
concentration
y
of
the
actual
colorimeter.
accuracy
of
l
using
the
precision
P
assess
y
•
r
In
e
A
cm
measuring
balance
cylinder
(two
decimal
places)
Instructions
appropriately.
1.
Use
the
equipment
sulfate
provided
to
prepare two
standard solutions, both with
–3
concentration
i
solution
1,
milligram
0.020 mol dm
you
should
balance.
use
For
.
the
When
preparing
volumetric ask and
solution
2,
use
the
measuring
meniscus of the solution
etched
line
cylinder
2.
indic ating
u
volume,
l
a
E
v
O
f
x
t
a
r
o
d
o
copper(II)
balance.
measurements
you
make
along
the
way,
3
e.g.
250 cm
3.
Following
the
your
teacher ’s
colorimeter,
instructions
measure
the
on
how to use
absorbance
of
your
solutions.
4.
fixed
volume of solution
when the meniscus is on
the
the
centigram
including their uncertainties.
etched
Refer
to
the
c alibration
concentration
volumetric flask contains
a
Record
and
of
curve to determine the actual
your solutions.
Q uestions
1.
line,
Determine
the
uncertainty
of
the
concentrations of
solutions 1 and 2.
3
e.g.
250 cm
2.
C alculate
the
percentage
error
of
the
concentrations
of solutions 1 and 2.
3.
Assess
the
precision
and
accuracy of the
concentrations of solutions 1 and 2.
77
Structure
1
Models
of
the
particulate
nature
of
matter
5.
4.
Consider
the
way
you
have
presented
The
construction
of
c alibration
curves
involves
your
ATL
preparing
c alculations
for
the
questions
above.
Do
samples
concentrations.
think
they
convey
your
thinking?
Do
reader
would
be
able
to
easily
solutions
that
cover
Instead
of
measuring
a
and
range of
dissolving
you think
a
a
of
you
follow
certain
mass
of
solute
the
way
you
have done
your
s
s
here, chemists oen start with a stock solution and
thought
process?
How
could
you
improve
perform a
the
presentation
of
your
c alculations?
want
to
look
serial dilution.
the
advantages
disadvantages of using a serial dilution in the
through some of the
examples
in
this
textbook
for
of
samples
for
a
c alibration
ideas.
r
Avogadro’s law (Structure 1.4.6)
Amedeo
same
Avogadro
temperature
has
been
pressure
conrmed
that
equal
contain
in
many
of
two
a
substance
amount
reacting
proportional
to
the
of
gaseous
V
turn,
2
the
amounts
other
volume
of
any
gaseous
gas
a
and
consumed
substances
c an
be
S(g)
+
O
C alculate
the
combusted
this
→
of
2H
O(l)
2
volume of
+
under
the
proportional
Therefore, the
same conditions
are
chemic al
proportional to their
equation.
produced
found
without
H
S,
in
the
As
a
result, if we
reaction,
the
volumes
c alculating their amounts.
O(l)
+
2SO
proceeds
as
follows:
2
(g)
2
oxygen,
O
(g),
consumed
and
sulfur
dioxide,
2
the
volume
volumes
are
of
hydrogen
measured
under
sulde
the
combusted
same
was
conditions.
The
ratio of the stoichiometric coecients of H
you
c an
S and O
2
is
2 : 3.
Therefore,
2
3
multiply
the
volume
of
combusted H
S
by
to
nd
the
volume of
2
2
combusted O
:
2
3
V(O
)
=
3
V(H
2
S)
=
The
3
×
0.908 dm
3
≈
1.36 dm
2
2
2
ratio of the stoichiometric coecients of H
S and SO
2
is
1 : 1.
Therefore,
2
2S(s)
the
volume
of
combusted H
S
is
the
same
as
the
volume
of
produced SO
2
=
V(H
2
S)
=
0.908 dm
2
reaction
was
Note
that
the
volume
of
liquid
water
c annot
be
found
in
the
same
manner, as
consumed in
Avogadro’s
law
applies
to
gases
only.
3
1.25 dm
Linking question
Avogadro’s
behaviour
78
:
2
3
)
V(SO
hydrogen sulde if the
oxygen
E
volume
(g)
2
a
S(g)
2
v
O
2H
produces
instead of sulfur
l
dioxide:
o
i
sulfur
if
are
volume.
2
2H
of
its
measured
or
sulde,
particles
to
known as
Solution
u
f
x
sulde
elemental
All
t
a
r
o
d
3
.
→
volumes
produced
0.908 dm
Incomplete combustion of
hydrogen
(g),
hydrogen
(g)
2
the
2
Practice question
4.
of
3O
2
C alculate
SO
+
n
U
combustion
2H
of
molecules.
now
species:
products
balanced
Worked example 10
The
these
C
i
n
of
the
species
of
reactants
coecients in
number
o
know
of
v
stoichiometric
the
proportional
p
In
is
is
y
1
=
2
and
gas
amounts
V
1
n
a
r
e
n
of
the
of
and
n
volumes
are
amount
other,
any two gases
O
the
each
t
i
s
to
of
numbers
experiments
Avogadro’s law
Since
volumes
equal
y
hypothesis
suggested
and
law
of
a
applies
real
gas
to
ideal gases. Under what conditions might the
deviate
most
from
an
ideal
gas?
(Structure 1.5)
y
the
l
1811,
at
P
In
This
curve.
e
preparation
worked
Discuss
You
and
may
Structure
1.4
Counting
particles
by
mass: the mole
End-of-topic questions
C alculate:
Topic review
Using
your
knowledge
from the
Structure 1.4
The
molar
)
4
the
guiding
question
as
fully
as
of
potassium alum,
topic,
KAl(SO
answer
mass
•12H
2
O.
[1]
2
possible:
c.
The
d.
The
amount
potassium
total
of
substance,
in
mol,
of
atoms
in
alum.
the
number
of
oxygen
atoms
pentahydrate,
in
CuSO
0.400 mol of
by
•5H
potassiumalum.
4
O?
complete
water,
B.
9
D.
5.40
×
e.
10
The
×
10
of
c alcium
To
visualize
nitrate,
,
is
dissolved
in
water
to
a
volume
of
200 cm
composition,
alum.
the
mole,
is
the
concentration of NO
6.02 × 10
needed
–
What
up
.
2
ions
in
this
solution?
3
a
chemistry
by
B.
7.0 g dm
–3
C.
0.021 mol dm
D.
0.043 mol dm
50 kg
–3
–3
9.
4.
which
the
molecular
CH
3
CH
empiric al
formula
the
CH
2
OH
C.
CH
2
CH
3
COOCH
3
CH
2
D.
CH
3
CH
2
student
Which volume of a 5.0 mol dm
sulfuric acid (H
?
3
C.
3
B.
student
obtained
an
M ass
of
tin

M ass
of
oxide
mass=
B.
tin
aer
of
these
data,
tin?
SnO
SnO
2
=
empiric al
E
Alums
are
XAl(SO
)
4
salt
the
contains
following
an oxygen-containing organic compound, 5.00 g of
A
standard
59.95%
of
oxygen
[1]
[2]
solution
of
potassium
sulfate, K
SO
from
volumetric
8.714 g
ask.
of
the
C alculate
solid
the
salt
mass
using
a
was
250 cm
concentration, in
–3
–3
,
and
potassium
11.
,
4
3
prepared
g dm
The
ve
formula of
the
molar
sulfate
c alibration
standard
potassium
in
concentration,
the
curve
in
solutions,
nal
mol dm
, of
solution.
gure7
in
in
was
which
the
permanganate, KMnO
,
[2]
constructed using
concentration of
varied
from
0.100
4
–3
to
0.500 mmol dm
.
Describe
how
you
would
prepare
1.78 g
these
12.
solutions
C arbon
using
monoxide,
produces
serial
c arbon
CO,
is
dilution.
dioxide,
a
[3]
toxic gas. Its combustion
CO
(g).
2
what
is
the
correct
formula of
a.
Deduce
of
C.
the
c arbon
balanced
equation
[1]
3
3
b.
D.
for the combustion
monoxide.
SnO
C alculate
the
volumes, in dm
,
of
consumed
c arbon
SnO
5
monoxide
and
oxygen
if
the
combustion
produced
3
2.00 dm
Extended-response questions
7
.
that
for
heating to a constant
2.26 g
to
the
a
A.
oxide
heating
l
the
of
v
O
According
before
of
u
f
x

10 cm
following data during an
determination
oxide of tin:
cm
t
a
experimental
the
sulfur
[3]
formulas
2
3
D.
grain
shovel
of water upon complete combustion.
10.
i
A
cm
r
o
6.
0.0050
5.0
c an
of a
3
cm
average
which produces 9.55 g of carbon dioxide and 5.87 g
o
0.010
d
A.
an
an
)
n
U
0.10 mol dm
of
if
student
b.
4
solution whose concentration of hydrogen ions is
–3
oxide
the
minute.
compounds:
3
stock solution is required to prepare 0.50 dm
per
decided to
sand. Estimate the time
project
and
empiric al
3
SO
2
this
5 mg,
[2]
a.
COOH
3
sand
the
CH
2
–3
5.
same as
of
Deduce
C
B.
the
complete
o
CH
is
formula?
i
n
A.
molecule
v
For
to
weighs
of
[1]
mass, of
y
3.5 g dm
grains
p
r
e
ofsand
–3
A.
produced
1.00 g of
23
pile
3
)
3
be
O
8.
0.70 g
percentage
potassium
24
C a(NO
[1]
c an
n
2.16
containing
of
y
C.
t
i
s
3.60
sample
that
2
A.
A
mol,
decomposition
24
3.
in
l
sulfate
of
P
is
amount
[1]
1.00 g of
r
potassium
Multiple-choice questions
copper(II)
1.00 g of
alum.
number
Exam-style questions
What
in
y
The
e
b.
How do we quantify matter on the atomic scale?
2.
s
s
a.
1.
hydrates
•12H
2
O,
of
the
where
X
measured
general
is
an
of
c arbon
under
dioxide.
the
same
All
volumes
conditions.
are
[2]
formula
alkali metal or other
2
singly-charged
c ation.
decompose
follows:
XAl(SO
as
)
4
•12H
2
When
O(s)
2
→
heated, most alums
XAl(SO
)
4
(s)
2
+
12H
O(l)
2
79
Structure 1.5
Ideal gases
s
s
How does the model of ideal gas behaviour help us to predict the behaviour of real gases?
As
with
a
any
theoretic al
simplic ation
many
c ases,
it
predicts
sucient
model,
has
for
its
the
the
concept
properties
most
of
an
ideal gas
at
advantages and limitations. In
practic al
of
real gases with a
purposes.
low
real
temperatures
gases
ideal
gas
deviates
model
and
high
pressures
signic antly
c annot
be
from
used under these conditions.
gas
consists
are
moving
particles
with
—
Real
Structure 1.5.3
—
The
molar
Structure 1.5.4
—
The
relationship
deviate
volume
from
of
an
the
ideal
ideal
between
gas
gas
the
is
a
model,
pressure,
=
no
V
nRT
and
the
combined
gas
a
law
V
2
=
T
low
temperature
temperature
temperature
p
1
at
specic
intermolecular
and
and
amount
and
pressure.
pressure.
of
an
ideal
gas
is
shown in
2
T
1
high
forces. All
y
pV
and
r
e
equation
at
volume,
1
gas
particularly
constant
p
ideal
volume
O
Structure 1.5.2
the
negligible
t
i
s
gases
of
considered elastic.
n
ideal
particles
l
An
2
p
o
v
Assumptions of the ideal gas
i
n
The
ideal
gas
model
states
that
C
model (Structure 1.5.1)
an
ideal
gas
collide
inelastic
sound.
perfectly
molecules
with
collisions
However,
elastic
to
the
following
ve
another
are
gas
not
stationary.
molecule
or
They
the
move
side
of
a
in
straight lines
container.
and
of
larger
the
no
objects,
collisions
energy
is
energy
between
lost
from
c an
be
transferred
molecules
the
in
an
ideal
as
heat
gas
are
system.
The volume occupied by gas molecules is negligible compared to the
volume of the container
l
Vaporized
E
v
a
(0
°C)
and
water
100 kPa
gas
conditions.
gaseous
space
in
occupies
phase
but
is
the
which
the
about
In
both
same
volume
the
1600
times
the
volume
of
liquid
water
at
273.15K
pressure (standard temperature and pressure, STP).
the
same
they occupy
occupies
Nitrogen
changed,
are
that
o
they
i
or
u
f
x
O
forces
conforms
Collisions between molecules are perfectly elastic
In
4.
Intermolecular
means
t
a
r
o
d
until
3.
gas
Molecules of a gas are in constant random motion
This
2.
n
1.
U
assumptions:
gas
650
and
of
times
c ases,
the
the
There are no intermolecular
size
gas
molecules
the
is
are
the
volume
number
of
to
of
liquid
nitrogen under
molecules in liquid and
individual
>99.9%
free
of
molecules has not
empty
space. This is the
move.
forces between gas particles
studied in
For an ideal gas, the intermolecular forces are negligible compared to the kinetic
Structure 2.2.
energy of the molecules. As such, an ideal gas will not condense into a liquid.
80
y
P
—
between
y
collisions
behaviour of
prediction, so the
However,
Understandings
Structure 1.5.1
the
the
r
precision
that
e
is
Structure
5.
1.5
Ideal gases
The kinetic energy of the molecules is directly proportional to Kelvin
temperature
This
relationship
is
studied in
Reactivity 2.2
Pressure–volume relationships
Boyle
pressure
of
(1627–1691)
a
given
now
of
known as
a
gas
is
that,
at
constant
inversely
Boyle’s law,
c an
temperature, the
proportional
be
expressed
to
as
its
volume. This
e
relationship,
established
amount
s
s
Robert
follows:

Figure 1
An ideal gas consists of
1
p
or
∝
pV = k (a
constant)
or
p
V
1
particles that
V
2
collide elastic ally,
r
1
=p
2
V
intermolecular forces and
gure
1,
walls
pressure.
so
the
the
volume
every
is
is
gas
are constantly striking and bouncing o
force
halved,
second
pressure
a
The
there
these
there
are
doubled
of
are
twice
impacts
twice
as
as
produces
many
a
volume when compared
molecules
in
each unit
many impacts with the container
(gure 2).
Standard
databases
temperature
and
of
temperature
for
to the behaviour of ideal gases.
in
pasc al (Pa),
comparative
100.0 kPa
where
used
mercury
and
y
built on certain assumptions related
between pressure and
What
commonly
millimetres
o
C
(atm),
are
o
in
phenomena. The ideal gas model is
volume, 1/V
a
found
E
(psi).
pressure
representations of natural
–2
pressure is the
of
atmosphere
inch
i
t
a
l
of
v
unit
units
u
f
x
O
SI
other
TOK
Models are simplied
reciproc al
Graphs showing the inverse relationship
volume of an ideal gas
The
,erusserp
d
r
o
,erusserp
Figure 3
doubled
n
U
p
p
volume, V

pressure
Halving the volume of a container doubles the pressure
halved
p
r
e
v
i
n
Figure 2
O
t
i
s
volume

to the volume of
the gas (the container)
measurable
n
walls,
If
so
of
container.
l
space,
molecules
the
y
of
the
of
y
the
P
In
have no
occupy negligible
in
1 Pa
=
1 N m
–3
=
1 J m
.
M any
dierent countries, including the
(mm Hg),
pressure
purposes.
bar,
and
conditions
STP
for
pounds
(STP)
gases
is
are
0 °C
per
square
frequently
or
the
is
the
role
of
assumptions
development of scientic
models?
What
not
are
the
implic ations of
acknowledging
a
model’s
limitations?
273.15 K
pressure.
81
Structure
1
Models
of
the
particulate
nature
of
matter
Worked example 1
3
A
weather
is
released
balloon
at
sea
lled
level.
with
The
32.0 dm
balloon
of
helium
reaches
an
at
a
pressure
altitude
of
of
100.0 kPa
4500 m,
where
3
the
atmospheric
helium
at
in
that
the
altitude.
balloon
is
57.7 kPa.
Assume
remain
that
C alculate
the
the
volume,
temperature
and
in
the
dm
e
× p
1
it
follows that
V
1
=
, so:
2
r
law,
p
2
3
32.0 dm
× 100 kPa
≈
55.5 dm
2
57.7 kPa
certain
volume
contains
of
altitude,
a
0.250 m
weather
16.0 g
of
.
balloon
C alculate
helium,
the
has
a
temperature
pressure,
in
of
n
a
3
a
O
At
t
i
s
1.
l
y
Practice question
–35.0 °C and
kPa, inside the balloon if
He(g).
y
r
e
Real gases vs ideal gases (Structure 1.5.2)
of
between
the
the
against
no
of
become
pressure.
pressure
pressure
decreases
and
volume
longer
signic antly,
the
container.
signic ant.
This
means
volume
for
a
real
halves
n
U
doubling
gas
proportion
forces
reducing
pressure
real
C
relationship
no
and
molecules
so
a
real
applies.
an
ideal
gas,
gas.
For
volume.
erusserp
t
a
l
u
0.5p
a
E
v
V
2V
0
volume
0

Figure
an
ideal
4
gas
Doubling
but
not
for
the
a
pressure
real
gas
halves
the
the
Figure
ideal gas
p
may
little space to
decreases the number
for
longer
gas
the
This
that,
the
With
real gas
o
i
d
r
o
f
x
O
82
a
large
intermolecular
collisions,
graph
of
a
i
n
of
volume
occupy
o
move,
the
to
v
begin
p
When
volume
for
4
inverse
shows a
the
real gas,
y
P
3
=
V
the
of
constant.
V
Boyle’s
of
amount
Solution
From
,
s
s
balloon
pressure
Structure
For
a
gas
to
deviate
intermolecular
the
molecules
from
forces
ideal
and/or
themselves.
gas
a
behaviour,
signic ant
This
there
volume
commonly
must
of
occurs
the
at
a
be
Ideal gases
detectable
gas
low
1.5
must
be
occupied
by
temperature and high
pressure.
is
reduced.
form
and
As
molecules
At
the
may
high
not
pressure,
considered
to
be
an
c annot
low
negligible.
At
so
high
intermolecular
molecules
an
ideal
pressure,
the
space
of
gas
there
far
are
the
attraction
to
apart
only
no
and
behaviour
occupied
temperature,
forces
part
of
in
the
the
a
reduced
space
longer
very
by
few
the
prevent
are
low
molecules
molecules
molecules
are
interaction
per
unit
moving
too
fast
assumptions
from
how
ideal
each
of
behind
pressure
the
and
behaviour.
of
the
following
might
behaviour
low
high
gas
at
low
pressure
temperature
or
at
high
E
or
which
is
more
likely
to
exhibit
HBr(g)
(g)
4
dec ane, C
e.
ideal
HF(g)
bromide,
methane, CH
the
a
or
hydrogen
d.
temperature
fluoride,
v
O
hydrogen
l
gas
c.
lead to
u
f
x
b.
at
predict
reason:
pressure
or
gas
pairs,
a
i
at
give
of
to
t
a
gas
and
likely
validity
volume
following
r
o
a.
the
the
o
ideal
molecular
of
forces
affect
are
n
L arge
each
intermolecular
d
For
Strong
ideal gasmodel.
temperature
o
conditions
U
b.
4.
for
y
main
gasmodel:
a.
allow
C
Consider
to
form.
i
n
3.
volume
p
the
what
deviations
of
themselves is
r
e
Outline
Discuss
between
pressure and high
v
2.
between them,
inverse, so the gas is
Activity
1.
space.
volume of the gas.
n
At
container,
is
attraction
O
the
molecules
signic ant
volume
molecules
of
elastic ally.
compressed,
and
gas
t
i
s
in
more
the
forces
l
for
temperature.
are
of
y
keep
them.
conditions
be
rebound
a
energy
intermolecular
ideal gas.
Ideal gas conditions
The
there
becomes
pressure
kinetic
y
not
between
the
another,
P
relationship
one
necessarily
molecules
themselves
the
with
e
of
temperature,
r
As
volume
low
collide
molecules
High pressure:
The
At
they
s
s
Low temperature:
H
10
(g)
22
propanone, CH
COCH
3
(g)
3
or
butane, C
H
4
(g)
10
83
Structure
1
Models
of
the
particulate
nature
of
matter
Real gases
Gases
that
deviate
from
the
ideal
gas
model
are
known as
n
real gases.
(
2
) (
p + a
V
V – nb
)
nRT
=
s
s
Relevant skills
measured
Tool
2:
•
Inquiry
Use
spreadsheets to manipulate data.
pressure
1:
Select
sufficient
and
relevant
correction
sources of
correction
information.
forces
for
for
of
between
volume
molecules
r
molecules
•
Inquiry
1:
Demonstrate
creativity in the designing,
measured
presentation
of
the
investigation.
volume
Instructions
real
gases
is
volume, amount and
modelled
by
the
parameter
van der
and
b
for
b
corrects
various
equation:
a
/
6
× 10
ammonia, NH
Pa m
OH
20.94
3
2
6
H
ethanol, C
OH
5
hydrogen
uoride,
4
3
Ne
H
5
3
E
H
3
O
11
8
propan-1-ol, C
2
OH
a
12
H
propane, C
xenon,
l
5
v
O
H
pentane, C
pentan-1-ol, C
u
f
x
OH
methanol, CH
t
a
methane, CH
HF
r
o
Kr
HBr
HCl
o
chloride,
i
bromide,
hydrogen
d
hydrogen
OH
12.56
Van der Waals parameters, a and
b,
for a selection of gases
strength and
Values of
–3
b
/
× 10
3
m
0.0371
0.0320
0.1164
0.1326
0.0648
0.0651
0.0871
0.0346
0.0238
4.500
0.0442
3.700
0.0406
9.565
0.0739
5.193
0.0106
2.303
0.0431
9.476
0.0659
0.208
0.0167
19.09
0.1449
25.88
0.1568
9.39
0.0905
16.26
0.1079
5.537
0.0305
4.192
0.0516
7
Xe

Table 1
5.580
n
U
2
He
7.566
C
i
n
Cl
o
9
H
ethane, C
v
H
4
force
volume.
shown in table 1.
p
10
butan-1-ol, C
water, H
are
13.89
chloromethane, CH
neon,
molecular
1.355
H
4
krypton,
intermolecular
y
Ar
butane, C
helium,
gases
r
e
argon,
for
–2
mol
4.225
3
for
O
–1
Substance
84
a corrects
t
i
s
Waals
of
pressure,
a
n
temperature
between
y
relationship
l
Parameter
The
–1
mol
y
or
P
implementation
e
•
Structure
1.
Use
of
a
the
selection
factors
instance,
of
the
data
affecting
the
in
table
1
values of
to
a
explore some
and
You
b. For
will
how
you could look at:
to
need
to
analyse
choose
to
decide
it.
how
explore,
you
may
and/or look up additional
intermolecular force strength and the value of a
•
molar
•
the
mass
and
the
value of
b
2.
Consider
on
data
which
need
to
to
select, and
option
perform
you
c alculations
data.
how
you
could
present
your data
ATL
graphic ally.
effect
of
volume
on
the
deviation
from
Prepare a one-page summary of
ideal
exploration
to
share
with
your
behaviour.
r
conditions,
do
some
gases
deviate
more
from
ideal
volume of an ideal gas
Avogadro’s
temperature and pressure contain equal numbers of particles. The molar volume
p
of an ideal gas is a constant at specied temperature and pressure. For example,
3
at STP
, the molar volume of an ideal gas, V
=
0.0881 mol
has
a
Molar volume of an ideal gas
1
70.90 g mol
compared with a soccer ball
pressure
mass
of
2.47 g.
gas.
u
a
2.47 g
=
STP
the
= 22.7 dm
0.0881 mol
l
n
1
mol
v
O
=
=
3
22.7dm
m
M
3
2.00 dm
=
m
at
of
o
f
x
V
V
,
C
g mol
Solution
=
gas
1
in
n
unknown
mass,
3
V
o
an
i
of
molar
t
a
sample
the
covered in
28.3 cm

Figure 5
1
a given temperature and
d
r
o
3
2.00 dm
.
2
32.00 g mol
Molar volume of any gas is identic al at
Determine
n
U
1
16.05 g mol
–1
mol
Cl
2
1
4.00 g mol
Worked example 2
A
O
4
1
2.02 g mol

Figure 6
i
n
CH
He
v
, is equal to 22.7dm
m
2
is
y
r
e
Avogadro’
s law states that equal volumes of any two gases at the same
H
law
Structure 1.4.
(Structure 1.5.3)
O
t
i
s
The molar
why
(Structure 2.2)
n
others?
y
than
l
comparable
behaviour
y
P
Linking question
Under
class.
e
your
gas
Ideal gases
s
s
•
much
Depending
1.5
1
28.0 g mol
Practice question
E
2.
Determine
the
–1
molar
mass,
in
g mol
that has a density of 3.12 g dm
,
of
an
elemental
gaseous substance
–3
at STP. Identify the substance if its molecules
are diatomic.
85
Structure
1
Models
of
the
particulate
nature
of
matter
Hypotheses
Amedeo
contain
pressure.
postulated
numbers
This
of
that
particles
bec ame
equal
under
known
as
volumes
the
of
same
Avogadro’
s
dierent gases would
conditions
of
temperature
hypothesis.
A hypothesis is a tentative and falsiable explanation or description of a
phenomenon,
used
What
to
test
the
predictions
which
predictions
c an
be
deduced.
Predictions
hypothesis.
might
be
derived
from
Avogadro’s
hypothesis?
r
gas
equation
a
gas
by
c an
be
used to determine the
collecting
a
known
Instructions
volume of it
1.
under
known
conditions
of
temperature
and
Measure
ambient
Alternatively,
practic al
you
will
experimentally determine the molar
of
butane
found
in
disposable
plastic
lighters.
the
day
you
Relevant skills
2.
3:
range
Record
to
an
uncertainties
appropriate
in
Half-fill
measurements as a
precision
and
propagate
Fill
processed data.
the
Inquiry
C alculate
2:
3:
Assess
Identify
Inquiry
3:
and
and
the
and
discuss
random
Evaluate
percentage
sources and impacts of
error.
implic ations
limitations
and
of
3:
Explain
realistic
assumptions on
and
clamp
4.
an
relevant
and
5.
improvements
protection.
o
i
d
r
o
eye
Wear
•
Butane gas is flammable. Keep away from open flames
•
Disposable plastic lighter
for
100 cm
•
Balance (±0.01 g)
•
Clamp and stand
•
Thermometer
example
3
measuring
(if
large
plastic
cylinder
water.
to
the
Measure the
brim
with
water
trough so that its mouth is under
correctly,
of
the
measuring
cylinder
water. Hold it in this position with a
(figure 7).
dry
Hold
it
the
the
the
lighter
thoroughly
in
water, then take it out again
with
a
paper
towel.
Weigh the
lighter
lighter
under
to
water
and
press
the
button
release the gas so that it bubbles up
inside the measuring cylinder (figure 7). Continue until
3
you
have
exact
collected
around
100 cm
of
gas.
Record the
volume.
6.
Release
7
.
Dry
8.
If
the
the
gas
lighter
in
as
a
well-ventilated
thoroughly
as
area.
possible and
reweighit.
you
have
time,
repeat
to
get
three
sets
of
results.
trough
3
100 cm
cylinder
available)
water

Figure 7
86
with
for
cylinder
v
E
Barometer
a
experiment.
water.
data
loc ation on
a
•
•
container,
l
L arge
O
•
u
f
x
Materials
t
a
•
and sparks.
full
weather
geographic
lighter.
investigation.
S afety
done
be
Submerge
on
to
If
should
n
Inquiry
water.
methodologic al
conclusions.
•
error.
precision.
U
weaknesses,
interpret
the
C
systematic
•
and
accuracy
i
n
•
3:
in
barometer.
o
Inquiry
v
Tool
•
it
a
loc al
your
trough
the
measuring
andinvert
•
of
in
p
in
the
plastic
temperature
3.
uncertainties
the
pressure
do
with
search
y
Tool
r
e
•
c an
O
atmospheric
mass
you
t
i
s
this
pressure
pressure. In
n
of
y
mass
l
ideal
molar
Experiment
apparatus
measuring
y
P
Experimental determination of the molar mass of a gas
The
natural
c an then
e
be
from
s
s
and
Avogadro
equal
Structure
1.5
Ideal gases
Questions
1.
Design
a
suitable
results
table
for
8.
your data.
Suggest
realistic
determining
2.
Process
your
molar
data
mass
of
to
obtain
an
experimental
value
9.
The
the
Propagate the uncertainties.
4.
Compare
pressure
sum
partial
by
the
6.
Discuss
the
accuracy
to
percentage
and
precision
the
theoretic al
the
relative
impacts
of
systematic
and
additional
need
to
on
butane.
processing
data
and
Consider
alternative
least
two
major
cylinder is in fact
water and the
How
to
could
account
information
methods
random
for
do
If
you
have
time,
show
your
ideas
sources of
presented
the
(Tools
2
as
sketches
advantages
and
3,
or
as
and
accurately
limitations
Reactivity 2.2)
plotted
of
each
o
C
i
n
v
representation?
are
p
What
your
y
be
O
t
i
s
c an
points.
r
e
data
to
n
y
error.
Linking question
Graphs
this?
you
for determining the
teacher and try them out.
experimental
you
mass of a gas that could be done in a school
results.
at
of
l
Comment
the
measuring
pressure
research?
laboratory.
7
.
for
y
your
of
data
What
molar
on
your
the
vapour
your data.
10.
errors
inside
the
pressure
adjust
error.
of
method
r
Assess
c alculating
value
of
P
5.
experimental
this
e
value
your
to
mass of a gas.
for
butane.
3.
improvements
molar
s
s
the
the
Pressure, volume, temperature and amount
U
There
are
four
variables
of
ideal
volume the gas occupies,
3.
The
absolute
4.
The amount of the gas,
He
law:
a
he
gas
these
This
performed
observed
kept
that
E
v
proportional.
variables
constant.
were
u
of
of
two
a
changed.
two
other
Boyle’s
temperature
V
T
n
l
with
any
p
temperature of the gas,
f
x
O
up
of
the
the
each other:
an
is
on
each
what
other
Robert
experiment
constant,
pressure
o
The
aect
i
2.
eect
by the gas,
that
t
a
pressure
keeping
gas
d
The
r
o
1.
The
exerted
an
n
of an ideal gas (Structure 1.5.4)
but
and
the
c an
Boyle
be
investigated
did
when
he
by
c ame
where the amount and the
volume
volume
of
of
the
the
gas
container
were
was
inversely
87
Structure
1
Models
of
the
particulate
nature
of
matter
Graphing the gas laws
Online
simulations
relationships
ideal
of
gas
gas.
In
pressure
this
simulation,
direct
data
and
to
easily
task,
and
you
which
analysis
inverse
and
temperature
will
will
collect
allow
skills,
as
3.
explore the
volume,
you
well
as
for
data
to
Using
a
for
fixed
a
five
from an
simulation,
pressure
certain
amount
different
vary
and
the
record
of
gas.
temperatures
temperature at a
the
resulting
Collect
in
a
data
suitable
ideas
Compute the temperature values in both °C and K.
Construct
two
graphs of
V
vs
T; one with
Use
data
from simulations.
6.
Using
the
simulation,
vary
Understand
direct
and
inverse
Inquiry 1: Identify dependent, independent and
certain
five
amount
different
the
of
Collect
gas.
volumes
in
variables.
Construct
Simulation
and
allows
you
temperature
to
for
change
8.
Use
an
ideal
gas.
It
must
have
9.
option
to
hold
one
variable
constant
and
p
spreadsheet
Construct
vary the
graph of
a
graph of
p
a
vs
to
suitable
Spreadsheet
1.
software
the
for
a
certain
and
amount
different
vary
the
record
of
gas.
temperatures
temperature at a
the
resulting
Collect
in
a
data
suitable
table
Construct
a
graph of
p
vs
2.
Describe
as
direct
least
in
orother.
3.
When
1
for
.
1
the
c ase?
Which
relationship
proportionality,
studying
temperature
T.
this
is
the
variables
shown
inverse
in
were
each
graph
proportionality,
i
o
units
for
c an
it
into
is
SI
important
units
temperature,
vary
to
convert all
(kelvin).
Discuss
whereas
depending
on
the
why
pressure and
source.
The combined gas law
We
have
seen
u
proportional
l
a
E
v
O
f
x
t
a
r
o
d
volume
gases,
values
c ase
that
to
pressure
absolute
is
inversely
proportional
to
volume
temperature.
1
p
∝
;
p
∝
T
V
Combining
the
two
relationships
gives:
pV
pV ∝ T
= k (a
or
constant) or
T
p
V
p
1
V
2
1
2
=
T
T
1
This
88
values
your
n
spreadsheet.
2.
at
each
controlled?
pressure
for
U
five
simulation,
volume
your
your dependent and independent
in
C
constant
i
n
Using
in
for
least
vs
V
were
variables
at
V
o
v
Instructions
1.
What
table
compute
p
•
two.
pressure
for
V
Questions
other
data
y
an
your
pressure,
r
e
volume
that
a
resulting
O
7
.
Materials
t
i
s
spreadsheet.
•
volume at a constant
record
n
•
a
proportionality.
in °C and
l
Tool
and
y
•
controlled
the
spreadsheets to manipulate data.
temperature
3:
in K.
equation
is
known as the
2
combined gas law
and
directly
y
Generate
2:
T
T
r
2:
Tool
your
5.
P
Tool
least
in
4.
the other with
•
at
table
proportionality.
Relevant skills
•
volume
for
spreadsheet.
practice
reinforce
the
constant
volume and
e
spreadsheet
about
and
you
pressure
s
s
temperature,
amount
allow
between
Structure
1.5
Ideal gases
Experiments
The
gas
laws
variables
arose
were
experiments in which certain
Inspect
the
while
others
apparatus
were
shown
c arefully
in
gure8.
pressure gauge
What
might
is
independent
the
be
explored
with
variable?
this
set-up?
What
thermometer
What
variables must
250
controlled?
What
is
the
purpose
of
mL
round-bottomed flask
e
be
s
s
manipulated.
from
controlled,
each of the
containing air
itemsdepicted?
r
water bath
lled
released
35,000 m,
–50 °C.
C alculate
where
level.
the
of
The
helium
balloon
pressure
is
at
25 °C
475 Pa
and
the
3
the
volume,
in
m
,
of
the
gas
conditions
of
the
gas
in
the
100.0 kPa
3
=
32.0 dm
1
to
convert
temperature to kelvin:
=
25
1
the
conditions
to
make
the balloon.
sure
f
x
the
units
= –50
into
the
=
298.15 K
weather
balloon
at
35,000 m.
0.475kPa
=
+
2
unknown
273.15
combined
p
a
v
O
E
the
=
2
numbers
the
2
V
T
273.15
in
are consistent with the initial conditions of
p
p
V
1
=
223.15 K
gas
law:
V
2
1
2
=
T
T
3
×
Practice question
2
1
100.0 kPa
Rearranging
+
gas
u
the
the
l
Substitute
of
t
a
Remember
r
o
list
i
d
T
Then,
is
these
o
Remember
under
weather balloon.
=
1
V
an
n
p
pressure
temperature
balloon
U
the
the
a
reaches
C
Solution
List
in
i
n
conditions.
and
eventually
o
of
32.0 dm
sea
v
altitude
with
at
p
is
n
balloon
100.0 kPa
O
weather
y
3
of
r
e
Worked example 3
A
l
into the behaviour of a gas
y
P
Apparatus for conducting an
t
i
s
experiment
y
u
Figure 8
0.475kPa
32.0 dm
×
V
2
3.
=
A
sample
of
an
ideal gas has
223.15 K
298.15 K
3
a
expression in terms of
V
volume
of
1.00 dm
at
STP.
3
gives:
C alculate
the
volume, in dm
,
2
3
V
≈
5.04
×
10
3
dm
3
=
of
that
sample
at
50.0 °C and
5.04 m
2
50.0 kPa.
89
Structure
1
Models
of
the
particulate
nature
of
matter
TOK
Throughout
were
chapter
alpha
particles
you
through
with
have
explored
observations
gold
atoms,
to
of
the
models
the
related
natural
world
manipulation
of
to
or
the
particulate
obtained
gases
in
the
nature
through
gas
laws,
of
matter.
M any of these concepts
experimentation:
to
from
the
interaction
s
s
of
this
developed
explorations of subatomic particles
atCERN.
e
r
l
n
O
y
used
by Joseph Louis Gay-Lussac to investigate the thermal expansion of gases (le)
CERN used
to investigate elementary particles (right)
p
v
How do scientists investigate the behaviour of particles that are too small to be observed directly? How have advances in
o
technology inuenced scientic research into what matter is made up of ?
C
i
n
Ideal gas equation
combined
three
gas
law
parameters,
suggests
p,
pV
expression
remains
T
n
the
U
The
V
or
T,
that
for
aects
constant.
The
o
i
d
or
in
such
a
way that the
value of that constant must be
n:
R is the
nR
universal gas constant, or simply
known as the
u
=
=
T
ideal gas equation,
which
is
gas constant.
traditionally
The
written
as
last
expression
follows:
nRT
The value and units of
l
a
E
v
O
f
x
t
a
r
o
where
pV
two
pV
∝ n
T
is
given gas, the change in one of
other
exact
proportional to the amount of the gas,
pV
any
the
R depend on the units of
p,
V, T and
n. If all
four parameters
3
are
expressed
in
–1
R
≈
The
8.31 J K
same
standard SI units (p
in
Pa,
V
in m
,
T
in K
and
n in mol), then
–1
mol
.
value and units of
R
c an
be
used
if
pressure
3
volume in dm
is
expressed
in
kPa and
3
,
as
the
two
conversion
factors (10
–3
for
kPa
to
Pa and 10
3
for dm
3
m
)
c ancel
each other out.
Linking question
How
c an
the
experimental
90
ideal
gas
data?
law
be
used
to
(Tool 1, Inquiry 2)
c alculate
the
molar
mass
of
a
gas
from
to
y
P
y
t
i
s
The set-up
The ATLAS detector at
r
e

Figure 9
Structure
1.5
Ideal gases
Worked example 4
A 3.30 g sample of an unknown organic compound
was vaporized at
T
=150 °C and
p =101.3 kPa to produce
3
1.91 dm
3
of a gas.
The gas was combusted
in excess oxygen to produce 3.96 g of water,
2.49 dm
of c arbon dioxide
3
and
1.25 dm
of nitrogen at
STP.
s
s
Determine the following for the compound:
molar mass
b.
empiric al formula
c.
molecular formula
r
Solution
determine
the
molar
mass,
we
need
to
nd
out
the
amount
of
the
compound
using
the
ideal
pV
=
T
=
n
=
gas
equation:
l
n
RT
273.15
=
423.15 K
3
101.3 kPa
×
1.91 dm
0.0550 mol
–1
mol
× 423.15 K
3.30 g
–1
Therefore,
M
=
=
60.0 g mol
0.0550 mol
c arbon,
amounts
hydrogen
of
these
in
atoms
c arbon
in
the
dioxide,
combustion
water
and
=
=
≈
0.220 mol
–1
M
=
2
×
18.02 g mol
O)
n(H
=
2
×
0.220 mol
=
0.440 mol
3
)
=
=
≈
2
3
V
22.7 dm
0.110 mol
–1
mol
M
=
)
n(CO
=
0.110 mol
2
3
V
n(N
)
1.25 dm
=
=
≈
0.0551 mol
2
3
V
×
n(N
)
=
2
×
0.0551
2
The
m(C)
=
=
0.440 mol
0.110 mol
m(total)
+
1.32 g
organic
+
=
original
sample:
oxygen.
and
To
check
nitrogen)
with
this,
the
we
mass
need
of
to
the
compare
original
the
total
mass
sample:
1.32 g
≈
1.54 g
1.54 g
compound
a
the
the
0.444 g
≈
14.01 g mol
l
N)
4
so
the
E
M(CH
≈
–1
0.110 : 0.440 : 0.110
empiric al
contain
c arbon
–1
1.01 g mol
12.01 g mol
v
O
The
=
×
0.444 g
Therefore,
x : y : z
(hydrogen,
organic compound, so the
in
–1
0.110 mol
=
0.110 mol
u
=
×
×
f
x
m(N)
c.
elements
=
also
the
those
t
a
m(H)
three
could
i
the
compound
r
o
of
original
o
= 2
mol
d
n(N)
–1
22.7 dm
M
from
as
n
U
n(C)
same
C
2.49 dm
V
n(CO
i
n
2
originate
the
o
n(H)
are
v
2
products
nitrogen
p
O)
nitrogen
3.96 g
m
n(H
and
elements
y
All
r
e
b.
O
t
i
s
≈
–1
8.31 J K
n
+
y
150
y
To
P
a.
e
a.
≈
did
3.30 g
not
contain
12.01
+
so
its
formula
c an
be
represented as C
H
N
y
.
z
1 : 4 : 1
formula of the compound is CH
molecular
oxygen,
x
=
N.
4
–1
4
+
1.01
formula
of
+
the
14.01
=
30.06 g mol
compound
will
have
–1
.
This
twice
value
the
is
half
number
the
of
experimental
atoms
of
value
(60.0 g mol
each element: C
H
2
N
8
),
.
2
91
Structure
1
Models
of
the
particulate
nature
of
matter
End-of-topic questions
0.58
Topic review
×
8.31
×
373
B.
1.
Using
your
knowledge
from the
Structure 1.5
6
×
10
250
×
10
topic,
3
answer
the
guiding
question
as
fully
as
s
s
3
100 ×
0.58 ×
100
0.58 ×
100
×
10
6
×
250
×
10
×
10
C.
possible:
8.31
×
100
×
10
e
3
How does the model of ideal gas behaviour help us to
6
×
250
D.
predict the behaviour of real gases?
6
Which
Multiple-choice questions
of
the
following
are
shows
volume
assumptions
of
the
ideal gas
B.
P
by the gas particles is
negligible
II.
There
are
no
intermolecular
forces
V
between
r
e
gas particles
Zero
particle
movement
II and III only
I, II and III
The
temperature
is
A
gas
syringe
will
at
27 °C.
contains
volume
constant
40
of
of
gas
an
be
pressure?
3
cm
3
cm
3
44
cm
84
cm
3
D.
A
0.58
has
a
g
E
5.
sample
volume
expressions
0.58
of
is
×
of
an
equal
8.31
ideal
×
cm
to
.
the
100
3
92
10
6
×
gas
gas
it
is
100
250
×
10
molar
of
at
mass
and
balloons
atoms,
at
contains
constant
the
largest
temperature
pressure?
H
NH
(g)
2
(g)
CH
3
(g)
HF(g)
4
the
3
3
What
3
2 dm
2 dm
are
the
3
1 dm
conditions
for
the
ideal
4 dm
gas
behaviour of
gases?
A.
Low
temperature
and
low
B.
Low
temperature
and
high
pressure.
C.
High
temperature
and
low
D.
High
temperature
and
high
pressure.
pressure.
100 °C
following
of
following
hydrogen
27 °C.
8.
kPa
the
the
of
warmed to
real
at
Which
of
number
and
3
250
A.
100 ×
ideal
aer
Which
doubled and
a
3.6
v
O
C.
1.9
l
A.
B.
is
3
cm
the
7
.
What is the
u
57 °C
the
pressure
tripled?
f
x
What
is
the
i
1527 °C
1800 °C
gas
t
a
C.
D.
aer
1
V
o
162 °C
450 °C
ideal
d
A.
B.
an
gas
r
o
4.
volume
of
the
n
the
of
U
temperature
1
V
C
D.
P
o
I and III only
C.
v
B.
i
n
3.
I and II only
p
P
A.
V
D.
between
y
C.
III.
O
occupied
t
i
s
volume
relationship
ideal gas, at constant
n
A.
P
The
the
an
temperature?
model?
I.
of
l
Which
correctly
and
y
2.
graph
pressure
gas?
pressure.
y
the
373
P
Exam-style questions
×
r
8.31
Structure
9.
Which
are
of
the
following
statements
about
an
14.
ideal gas
At
forms
constant
temperature,
the
several
gaseous
empiric al
compounds
p
I.
C arbon
Deduce
correct?
using
and
the
compounds
molecular
data
from
the
1.5
with
formulas
table
Ideal gases
uorine.
for these
below.
[3]
= constant
V
3
C arbon /
M ass
of
1.00 dm
p
At
constant
volume,
s
s
Compound
II.
= constant
mass
%
at
STP / g
T
X
13.65
Y
24.02
Z
17
.40
3.88
V
III.
At
constant
pressure,
= constant
A.
I and II only
D.
I, II and III
15.
An
organic compound
9.1%
of
hydrogen
vaporized
and
sample of
A
contains
36.4%
A
with
of
54.5%
a
oxygen
mass
of
of
by
c arbon,
mass. A
0.230g occupies
l
3
a
volume
of
own
words,
behaviour
at
low
why
real gases deviate
a.
Determine
b.
Determine
c.
Using your answers to parts a and b, determine the
temperatures and high
pressures.
the
empiric al
[2]
the
relative
formula of
molecular
molecular formula of A.
11.
A
c ar
tyre
inated
to
2.50 bar
(250 kPa)
at
10 °C contains
of
compressed
air.
Aer
a
long
journey, the
A
closed
gas
tyre
temperature
increases
to
25 °C
and
the
pressure
and
Determine
the
tyre
volume under these
mixture
Assume
that
there
was
no
air
readily
when
)
4
(s),
heated:
CO
2
(s)
→
2NH
3
(g)
+
CO
3
(g) + H
2
gases
the
produced
vessel
heated
vessel
that
with
question12
to
a
each
were
of
1.50 dm
C alculate
the
,
initial
The
[1]
[1]
of
hydrogen
volume of the
temperature of the gas
pressure,
in
kPa, of the gas
cylinder.
gas
mixture
rises
to
[1]
is
completely
cylinder
inside
ignited,
and
800
the
°C.
both
reactants
are
temperature inside
C alculate
the
cylinder
at
that
the
pressure
moment.
[2]
–3
An
unknown gas
4.00 g
sample of
X
has
X
was
a
density
of
2.82 g dm
combusted
in
excess
at
STP. A
oxygen to
3
produce
and
the
c arbon
2.50 g
dioxide
of
at
hydrogen
uoride
and
2.84 dm
of
STP.
vessel
Determine
the
following
for
X:
pressure in the
Assume that the gases do not
other.
the
the
initial
the
0.32 mol
gas.
[2]
A.
transferred to a
3
volume
200 °C.
temperature.
u
l
a.
molar
b.
empiric al
mass
c.
molecular
[1]
[2]
formula
formula
[2]
[1]
a
E
v
O
f
x
react
at
with
up
o
was
in
17
.
[2]
t
a
sealed
produced
the
2.25 g of
i
gases
of
When
the
in
consumed
O(l)
contains
oxygen
and
A.
mass of
2
STP, of the individual
decomposition
c arbonate.
r
o
The
on
at
d
ammonium
13.
volumes, in dm
b.
n
3
Determine
C alculate
mixture
decomposes
3
U
(NH
CO
2
a.
C
)
4
i
n
c arbonate, (NH
25 dm
25 °C.
o
[2]
Ammonium
is
is
loss during the
journey.
12.
v
conditions.
of
p
261 kPa.
cylinder
3
cylinder
to
steel
0.16 mol
y
r
e
16.
3
12.0 dm
n
ideal
your
p=102 kPa.
O
from
in
t
i
s
Explain,
T=95 °C and
y
Extended-response questions
10.
at
0.0785 dm
y
II and III only
r
I and III only
6.08
P
B.
C.
4.41
e
T
93
s
s
e
r
O
n
l
y
P
y
t
i
s
y
p
r
e
n
U
C
o
v
i
n
structure
and
bonding
of
Models
2
Structure
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
Structure 2.1
The ionic model
s
s
What determines the ionic nature and properties of a compound?
the strong electrostatic attractions between oppositely
of positive and negative ions, which attract each other
charged ions. Once liquid, however, ionic compounds are
electrostatically. In solid ionic compounds, these ions
electrical conductors due to the presence of mobile ions.
are arranged in rigid crystalline lattices. Melting these
Due to their charge, ions interact strongly with polar water
solids requires a large amount of thermal energy due to
molecules, so ionic compounds are oen water-soluble.
r
e
Ionic compounds are characterized by the presence
l
c ations.
electrons,
they
Structure 2.1.2 —
form
The
atoms
negative
ionic
lose
electrons, they
Structure 2.1.3 —
When non-metal atoms
bond
ions
is
dimensional
c alled anions.
formed
formulas.
by
electrostatic attractions between oppositely charged ions.
by
compounds
the
anion.
are
The
named
anion
with
the
c ation
adopts
the
sux
rst,
“ide”.
exist
c an
The
them
rise
is
bonded
varying
to
.
are
atoms
arrangements
certain
nitrogen, N
They
to
dierent
However,
in
connected
of
of
help
properties.
agriculture,
crops
grow.
to
that
This
of
is
the
bec ause
the
nitrogenous
and
For
together
type,
several
to
atoms
features
of
the
example,
nitrogen
structure
in
or
fertilizers
and
compounds
78%
dierent
bonds
of
are
bonding
dierent
of
the
empiric al
between
air
around
of
added to soils
nitrogen in air
o
dierent
d
are
same
atoms
2
to
the
three-
by
n
us
give
isolation.
U
elements.
in
be
p
rarely
Atoms
C
ways.
i
n
Atoms
as
o
v
Introduction to bonds and structure
exist
represented
y
ionic
followed
compounds
structures,
r
e
Binary
Ionic
lattice
O
gain
ions
n
metal
c alled
positive
y
When
form
t
i
s
Structure 2.1.1 —
y
P
Understandings
found
in
fertilizers.
Atoms are held together by chemical bonds. This chapter discusses three dierent
i
r
o
bonding models: ionic, covalent and metallic. These lead to four types of structure:
t
a
ionic, molecular covalent, covalent network and metallic. Y
ou may be wondering
why there are four types of structure, given that there are only three types of bonds.
u
f
x
This is because covalent substances can be found in two arrangements: a continuous
3D network, or discrete groups of atoms known as molecules.
ionic
bond
ionic
E
v
types of
structure
covalent
metallic
a
O
l
types of
molecular
covalent
covalent
network
metallic
metal ion
deloc alised
electron
t
Figure 1
bonds and
There are three types of
four types of structure
95
Structure
2
Models
of
bonding
and
structure
Chemic al bonds
Models
Chemic al
Structure
models
2 .1 ,
of
bonding
models
represent
so
are
All
charged
on
strong
species
which
help
we
This
bonding
does
are
attraction that hold atoms or
occur
due
negatively
to
electrostatic
charged
us
to
The electrostatic attraction between…
c annot
is
one
of
models
Type of bonding
Positively charged
are
inadequate, but it
Negatively charged
species
ionic
c ations
covalent
atomic
metallic
c ations
species
As
All bonding types involve a positively charged
you
limitations
of
these
of
the
the
are electrostatic ally attracted
sections,
strengths and
various bonding
a negatively charged
to each other
compounds
compounds
instance,
However,
also
sodium
elemental
chlorine is a
poor
when
are
are
electric al
molten
very
chloride,
sodium
is
are
or
ionic compounds. They
characteristic of ionic compounds.
conductors when solid, but good
dissolved.
The
reactions
and
properties of
dierent to those of their constituent elements.
the
a
examples of
properties
main
so
poisonous gas.
ingredient
metal
that
in
reacts
table
salt,
violently
is
water-soluble.
with
water, and
n
U
C
For
ionic
i
n
these
are
conductors
sulfate
which
o
electric al
copper(II)
brittle,
v
Ionic
and
and
y
chloride
crystalline
p
are
r
e
Sodium
o
i
d
t
a
r
o
u
l
f
x
a
v
O

Figure 3
Sodium
Sodium
chloride and
chloride crystals on a tree branch and
copper(II) sulfate crystals.
copper(II) sulfate are ionic compounds
Oshore oil platform in
USA.
E
C alifornia,
species and
electrons
electrons
Ions (Structure 2.1.1)
models.

Figure 2
of
n
model.
pair
O
some
shared
deloc alized
y
through
identify
nuclei
t
i
s
a
species that
work
anions
understand the

Table 1
of
between
involved (table 1).
necessarily make
to
together
l
models
important
ions
attractions
species. The type of bonding
have limitations.
not
weaknesses
96
and
species
of
bonds
What
examples of structure
bonding are present
in the photo?
Before
rst
discussing
look
into
ionic
what
ions
bonds
are.
and
the
characteristics
of
ionic
structures, we will
y
models
the
and
forces
chemic al
useful.
This
is
positively
depends
and
bonds
substance.
P
All
re asons
a
r
the
s t r u c t u r e.
in
phenomena.
that
d i r e c t l y.
d i s c u ss
e
observe
and
models
things
2.3
simplify
c o m p l ex
Sometimes
visualize
and
s
s
Scientific
2.2
Structure
2.1 The ionic model
a
C ations and anions
b
+
Sodium chloride contains sodium ions, not sodium atoms. Sodium atoms and
sodium ions have dierent numbers of electrons, and therefore behave dierently.
Na
Na
+
notice
1.
number
2.
electron
3.
charge.
of
three
dierences
between
Na
and
Na
:
electrons

Figure 4
are
neutral.
Sodium
ions
have a 1+
charge,
indic ated
r
atoms
by a
+
sign
next
to
the
symbol:
Na
.
ions
have
a
1+
charge.
In
a
protons
charge is 11
a
sodium
protons
10
ion
=
there
(charge
electrons
=
(charge
11
0
are:
11+)
=
charge is 11
10
10
2
a
the
slightly
atom.
Figure
denoted
dierent
This
is
protons
electrons
You
c an
sodium
atom
of
ignore
now,
we
and
a
sodium
a
atoms
(c alled
these
are
b
of
means
that
sodium
protons
ions
is
c ations
S
are
have
(a) sulfur atom (b) sulde ion
greater than
11
protons and
Activity
ions.
They
contain
a
greater
number
of
electrons
Show
that
the
sulfur
atom
is
neutral
2
the
name:
superscript
the
rst
part
by the sux
in
the
symbol S
corresponds
to
. Note that anions adopt
the
name
of
their
parent
ide.
and
of
the
2–
by
sulde
only
consider
monatomic
ions.
You
will
look
at
charged
polyatomic ions)
in
a
later
ion
counting
particles.
has
their
Determine
of
the
a
charge
subatomic
the
electron
sulfur
atom
groups
and
of
as
ion.
shows a sulfur atom and a sulde ion. The sulde ion has a
by
followed
will
charge.
neutrons in ionic
c alculations
conguration
For
charge
1–
2–
This
charge
As
a
uncharged.

Figure 5
electrons.
have
learned
1+
S
electrons.
positive
a
charge is 1+
charged
5
than
combined
charge
overall
negatively
protons.
charge,
negative
E
than
are
the
a
electrons,
more
as
l
combined
Anions
with
v
O
10
ions
charged,
u
are
positively
the
1
3s
6
2p
f
x
C ations
6
2p
2
2s
and
charge
you
have
n
2
2s
2
: 1s
a
i
2
+
Na
of
protons
t
a
r
o
Solution
Na: 1s
conguration
o
electron
1+
d
the
)
=
Worked example 2
Deduce
)
U
O verall
11
=
Structure 1.2,
that
C
11
(charge
11+)
In
o
In
=
are:
i
n
b.
(charge
electrons
O verall
there
v
11
atom
p
11
sodium
y
a.
r
e
Solution
O
atoms are neutral
sodium
t
i
s
sodium
b.
n
Determine the number of subatomic particles to show that
a.
l
y
Worked example 1
y
+
superscript
atom (b) sodium ion
P
Sodium
(a) sodium
arrangement
e
will
s
s
You
sulde
ion.
section.
97
Structure
2
Models
of
bonding
and
structure
Predicting the charge of an ion
The
18.
main
The
group
elements
electron
corresponding
are
in
congurations
c ations
are
periodic
for
shown
table
some
main
groups 1, 2, 13, 14, 15, 16, 17 and
group element atoms and their
below:
s
s
1e
+
Na
2
Na
2
1s
6
2s
1
2p
2
3s
2
1s
6
2s
2p
e
+
Na
has
the
same
electron
conguration
as
neon,
Ne.
Two
dierent
species with
electron
conguration
are
c alled
r
+
same
isoelectronic.
Therefore,
isoelectronic.
1
2
2
1s
1s
2
2s
6
2
2p
3s
6
3p
2
is
isoelectronic
with
helium,
Ca
He.
this
all
formation
formed
parent
the
resulting
atoms
gain
gain
they
i
n
1s
2
2s
6
2p
2
3s
U
2
O
Cl
5
2
3p
1s
in
is
isoelectronic
with
argon,
r
o
the seven outermost electrons.
l
O
requires a large amount of energy
a
while the addition of a single
2Na(s)
the
+ Cl
u
f
x
of the positively charged nucleus
1s
3p
electrons
of
formation
Consider
noble gas conguration if it lost
electrons from the attractive pull
gain
t
a
The
Chlorine would also have a
3s
o
that
formation
a chlorine atom gains an electron.
However, the removal of so many
2p
2
6
Ar.
Atoms
the
2s
2
i
d
To obtain a noble gas conguration,
6
2
anions
of
an
is
ionic
formation
(g)
Noble
ions,
2s
6
3p
full
(or
achieve
the
the
achieve
and
electrons
charged.
loss
of
examples
a
noble
C ation
electrons.
below
gas
electron
+2e
2
O
2
1s
2
2s
6
2p
2–
O
sodium
is
anions.
reduction
compound
of
have
oen
2p
become
a
involves
at
gases
2
3s
have all done so by losing
positively
to
Ar.
they
electrons,
Look
order
argon,
6
2p
4
n
–
Cl
2
it
electrons.
electrons
lost
are
bec ause
C
+1e
have
c ations
of oxidation
when
atoms
As
conguration:
Cl
form
p
are
the
example
electrons.
o
where
charged,
an
congurations.
elements
conguration. The atoms above
valence
v
Anions
is
gas
group
with
y
negatively
noble
main
r
e
outermost
have
When
noble gas electron
their
are
c ations
sublevels.
isoelectronic
2
2s
O
resulting
“closed”)
is
t
i
s
The
1s
n
2+
+
Li
2+
Ca
2
4s
y
2s
are
l
2
1s
Ca
Ne
y
2e
+
Li
and
P
1e
Li
Na
isoelectronic
As
reduction
is
with
the
neon,
gain
of
Ne.
electrons,
process.
from its elements is a
chloride
redox
reaction.
from its elements:
2NaCl(s)
2
+
Sodium
chloride,
anions, Cl
other
is
a
.
The
NaCl,
half
reduction
is
made
equations
and
up
are
therefore
of
sodium
shown
the
c ations,
below.
formation
The
of
Na
rst
NaCl
is
, and chloride
an
oxidation and the
from its elements is a
redoxreaction.
electron releases energy. This is
v
+
2Na
+
2Na
2e
Electron
loss
=
oxidation
why chlorine instead will gain an
E
electron to become a chloride ion.
+
Cl
2e
2Cl
Electron gain
=
reduction
2
The energetics of these processes,
Once
you
have
learned about
oxidation states
(Structure 3.1),
you should also be
called ionization energy and
able
to
see
that
the
sodium
is
undergoing
oxidation
bec ause
its
oxidation state
bec ause
its
oxidation state
electron anity, are discussed in
increases
(from 0 to
+1)
and
Structure 3.1 and relevant in the
decreases
construction of Born–Haber cycles
(Reactivity 1.2).
98
(from 0 to
1).
the
chlorine
is
reduced
Structure
Atoms
tend
to
achieve
a
noble
gas
electron
conguration
2.1 The ionic model
through gaining,
TOK
losing,
or,
to as the
as
we
will
octet rule.
see in
It
is
Structure 2.2,
c alled
the
octet
sharing
rule
electrons.
bec ause
This
most
is
noble
oen
gases
referred
have
General
eight
rules
in
chemistry
(such
electrons in their outer shell.
as
exists
element
a
and
relationship
its
periodic
between
table
the
group.
charge
In
of
the
ion
formed
by
a
main
group
general:
octet
rule)
exceptions.
have
to
oen
How
exist
for
a
have
many
rule
s
s
There
the
exceptions
to
cease to
be useful?
in
groups
1,
2
•
Elements
in
groups
15,
•
Elements
in
group
and
16
13
and
form 1+, 2+
17
form
3–,
and 3+
2–
and
ions,
1–
respectively
ions,
respectively
The
18
(noble
gases)
do
not
e
Elements
electron
conguration of
2
form ions
c arbon, 1s
r
•
2
2s
2
2p
, suggests that
The
relationship
between
periodic
table
group
and
ionic
charge
is
illustrated in
four
electrons
in
order
to
achieve
a
noble
18
result
4+
of C
13
14
15
16
more
17
3
6
Cs
2+
2
Te
2+
Ba
is
simply
a
electron
gaining
hydrogen
in
one.
nucleus:
a
The
charge density of a H
combine
with
other
species.
ion
is
One
O
,
formed
when
hydrogen
f
x
3
thus
atoms
forming
c an
also
gain
an
–
hydride
anions, H
.
l
a
E
v
O
+
u
Hydrogen
very
example
+
hydronium ions, H
sublevel.
proton
therefore
such
1s
loss
with
t
a
+
it.
the
Electron
i
which
one
or
r
o
,
only
electron
to
process
discussed in
c arbon
forms compounds
c alled
does
covalent
not
involve
Covalent bonding is
Structure 2.2
n
have
electron
o
that
d
atoms
I
U
The charges of some common ions
losing
Br
i
n
2+
+
H
Se
Sr
+
either
2
Ca
+
Hydrogen
Cl
possible,
which
y
Rb
2
S
is
formation.
p
5
+
3
P
ion
formation
respectively.
o
K
3+
Al
C
4
2+
Mg
a
bonding,
F
v
Na
O
r
e
+
3

Figure 6
2
N
this
commonly
through
+
Li
2
the
ions,
O
t
i
s
2
in
4–
or C
Although
1
conguration. This
n
y
would
1
gas
l
gure 6.
y
P
c arbon atoms could lose or gain
no
high,
is
the
to
form
the
electrons
so
these
ions
by
formation of
surrounding
c ations
readily
formation of acidic
c ations
achieve
They
leads
a
bond
noble
with
gas
water.
conguration,
Hydride
bases.
anions
You
bases in
will
are
very
learn
strong
more about
Reactivity 3.1
hydrogen
–
+
+
e
–1
ion, H
+
1
e
hydride
hydrogen
+
ion, H
atom
+

Figure 7
The formation of H
and H
99
Structure
2
Models
of
bonding
and
structure
Practice questions
1.
Determine
the
charge
magnesium, Mg
c.
aluminium, Al
d.
uorine, F
e.
nitrogen, N
f.
selenium, Se
g.
barium,
ion
formed
each
3.
Complete the table:
name
of
ions
d,
e
and
f
above.
of
15
2p
6
conguration
2
3s
5.
noble
group
dierent
charges.
i
(gure 8).
t
a
u
a
E
v
O
l
3d
is
an
elements,
o
main
gases
do
not
n
transition element
to
d
r
o
f
x
Fe: [Ar]
100
why
U
A
Explain
given,
6
3p
2
a
element
o
1s
2
2s
C
i
n
b.
v
electron
2
18
p
+
H
three
2
isoelectronic
a
partially
element
c an
lled
d
sublevel.
example,
iron
commonly
In
contrast
form multiple ions with
2+
For
species:
form ions.
with
transition
identify
0
y
r
e
8
1s
Charge
electrons
O
+
K
each
Electron
conguration
t
i
s
beryllium
n
Number
of
protons
a.
following elements.
l
Symbol
y
Number
For
the
forms
Fe
3+
and
Fe
ions
+
2
–
–
Fe
: [Ar]
e
2
3d
4s
iron(II) ion
4s
–
3
e –
+
3
iron atom
Fe
: [Ar]
3d
4s
iron(III) ion

Figure 8
Iron atoms c an form ions with a 2+ charge and
ions with a 3+ charge
y
State
4.
of
Ba
2.
Name
by
P
the
the
e
b.
of
Li
s
s
lithium,
r
a.
Structure
Consider
the
electron
congurations
of
the
rst-row
transition elements.
As
2+
Most
of
them
areformed.
contain
This
two
helps
to
4s
electrons,
explain
why
which
most
of
are lost when the M
these
2.1 The ionic model
elements
seen in chapter
the
ions
commonly
3d
form
4s
sublevel
lls
Structure 1.3,
up
before the
sublevel.
2+c ations.
s
s
t
Table 2
Electron congurations of the
Electron conguration
Symbol
rst-row transition elements
Element
Atom
[Ar] 4s
Ti
titanium
[Ar] 4s
2
2
iron
[Ar] 4s
Co
cobalt
[Ar] 4s
Ni
nickel
[Ar] 4s
5
[Ar] 3d
5
6
7
8
the
bec ause
successive
ionization
to
to
a
oset
certain
the
previous
subsequent
in
sublevels
data
by
other
then
are
lost
before
It
it
have
energy data.
variable
in
energy,
important
to
Deduce
the
abbreviated
of
electron
each of the
following:
2+
a.
Mn
b.
V
c.
Cu
d.
Cu
oxidation
as
3+
shown
+
realize that
energy,
release
small
could
energy,
amount
be
but
of
this
such
energy
as
additional
lattice
7
.
energy
Zinc
a.
only
forms 2+ ions.
Deduce
the
full
electron
2+
energetic ally
favourable if it
conguration of Zn
b.
Explain
.
why zinc is not a
i
transition element.
t
a
3
30
6.
2+
absorbs
that
a
ionization
together
is
Practice questions
conguration
8.
The
has
ion
of
mass
a
transition metal
number
55,
electron
5
conguration
l
[Ar] 3d
and a
charge of 2+.
a.
Write
its
symbol using
nuclear notation.
v
E
n o it a z i n o i
a
01 / ygrene
u
lom Jk
r
o
3
f
x
O
3d
9).
They
process.
40
10
close
requires
ionization,
4.
Ionization
b.
Identify a 1+ ion that has the
same
as
electron
the
conguration
above.
0
0
2
4
number

Figure 9
are
processes
only
exothermic
20
sublevel.
successive
period
(gure
isolation.
ionization
4s
o
compared
3d
in
examining
elements
energy
happens
usually
a
and
the
electrons
many of these elements
oxidation states (Structure 3.1). This
d
If
4s
to
4s
in
n
is
formation.
leads
the
ionization
rarely
investment
transition
by
energy
the
occur
C
the
variable
explored
in
U
by
on
have
be
similar
ionized,
i
n
focus
c an
is
are
ionizations
o
elements
characteristic
states
sublevel
elements
successive
[Ar] 3d
v
Transition
3d
transition
Further
9
3d
p
bec ause
Let’
s
row
[Ar] 4s
10
y
rst
electrons.
8
[Ar] 3d
r
e
the
3d
7
[Ar] 3d
3d
1
the
6
[Ar] 3d
3d
2
When
[Ar] 3d
3d
2
copper
5
3d
2
Cu
4
3d
2
y
Fe
1
n
[Ar] 4s
l
[Ar] 4s
manganese
O
chromium
Mn
[Ar] 3d
P
Cr
3
3d
y
[Ar] 4s
2
[Ar] 3d
3
t
i
s
vanadium
1
[Ar] 3d
3d
2
V
2+ ion
1
3d
r
sc andium
e
2
Sc
D ata for the rst
of
6
8
electrons
12 ionization energies of iron.
electrons are very close together in energy.
10
As you c an see, the 4s and
The large jump
electrons occurs bec ause the 9th electron is removed
12
removed
between the 8th and 9th
from the 3p
energy level, which is
closer to the nucleus
101
Structure
2
Models
of
bonding
and
structure
Communic ation skills
ATL
You may have noticed that we c an refer to charge using dierent formats
depending on context. When using chemical symbols, charges appear as
3+
writing, we say “the ion has a
. In speech or
s
s
a superscript number followed by + or –, for example, Fe
3+ charge”. Charge is related to oxidation state
(Structure 3.1), where the + or – sign is given rst followed by the magnitude.
Roman
numerals
are
also
used
to
and
indic ate
oxidation states in the names of
Reactivity 3.2).
For
example,
r
compounds (Structure 3.1
the
2+
copper(II) ion is Cu
own
to
charge
example
ionic
to
is
help
2+,
you
and
its
symbol
remember
these
distinct
charge.
ways of
l
the
its
the
an
element
trend
in
variable
3.1)
successive
oxidation
in
the
periodic
ionization
states?
table
relate to the
energies
of
transition elements
(Structure 1.3)
o
v
p
their
of
(Structure
y
does
explain
position
ion(s)?
r
e
How
of
O
does
charge
n
y
How
t
i
s
Linking questions
for a
oxidation state is +2.
their
and
bond
bonding
to
way
look
at
to
a
attraction
element
estimate
measure
electrons.
periods
u
of
has
an
so
the
in
of
Within
and
element,
Values
forms
results
c ations,
in
to
the
and
each
other
bec ause of
formation of ionic bonds.
another
forms
anions,
they
c an
whether
up
it
the
the
the
has
in
a
ability
groups.
high
Pauling
of
periodic
sc ale
electronegativity
bond
an
between
atom
table,
This
to
sc ales
are
value
to
attract
that
attract
used
4.0.
a
given elements is ionic is to
the
by
The
of
is
of
Electronegativity
covalently
increases
the
most
covalently
chemists
and
two.
pair
uorine
pairs
dimensionless
of
two
between
electronegativity
means
tendency
electronegativity
the
a
electronegativity
is
electronegative
bonded
c alled the
range
from
electronegativity
0.8
of
bonded
across the
electrons.
Pauling sc ale.
to
4.0. Fluorine
c aesium, one of the
If two elements have an
a
electronegativity dierence greater than 1.8,
the bonding between them will have a high
E
v
ionic character
Electronegativity and other
periodic
is
One
l
O

Figure 10
This
form an ionic compound.
dierence
i
1.0
( χ)
the
t
a
r
o
2.0
f
x
ytivitagenortcele
One
character
0
trends
greater detail in
are
discussed in
Structure 3.1
least
electronegative
electronegativity
The
larger
the
compound,
bonding
than
the
of
102
given
ionic ally
o
ionic
a
electrostatic ally attracted
Electronegativity (χ)
greater
d
ecnereffid
ionic
if
are
charges.
n
U
Therefore,
3.0
anions
opposite
C
C ations
i
n
Ionic bonds (Structure 2.1.2)
1.8
is
dierence
the
main
greater
assumed
(gure
type
bonding
elements,
10).
of
is
0.8.
Noble
gases
are
generally
not
assigned
values.
to
In
in
ionic
occur
reality,
bonding
present.
electronegativity
the
in
character
when
the
bonding
the
between two elements in a
of
the
bond
dierence
occurs
compound
is
in
across
ionic,
between them. Ionic
electronegativity
a
continuum,
but
there
so
is
greater
above 1.8
may be other types
y
your
referring
its
P
Write
,
e
For example, “oxygen has an oxidation state of –2”.
Structure
2.1 The ionic model
Data-based question
Predict
which
of
the
compounds
sodium
Dierence
χ
uoride,
(Na)
∆
NaF
χ
chloride,
χ
ionic
structure.
electronegativity
0.9 and
∆
χ
(F)
=
(∆
χ
)
4.0
3.1
=
χ
0.9 and
(Cl) = 3.2
= 2.3
(Al) =
1.6 and
χ
(Cl) = 3.2
= 1.6
3

Table 3
Electronegativity dierences for selected metal chlorides
l
at
c an
the
of
dierences
from
in
approximate
constituent
how
electronegativity
each
ionic
elements
are
in
a
compound
the
periodic
generally
found
will
be
by looking
table. Elements with
at
a
greater horizontal
other.
p
o
i
n
v
Worked example 3
y
distance
its
r
e
large
qualitatively
positions
O
You
n
y
t
i
s
Periodic table position
y
∆
have
P
chloride, AlCl
χ
will
r
χ
aluminium
=
(Na)
∆
NaCl
χ
=
in
3
e
sodium
table
s
s
Compound
in
Compare the ionic character of bonding in the following pairs of compounds:
c aesium
b.
magnesium
uoride,
C sF,
oxide,
and
MgO,
c aesium iodide,
and
I
are
dierence
F,
in
in
distance
the
periodic
the
bond
0.8
∆χ(C sF)
=
in
∆χ
both
higher
4.0
and
χ(I)
1.9
v alues
are
=
are
between
them
is
more
Qualitative comparison:
In
the
from
periodic
e ach
than
ionic.
must
be
bond
ionic
so
the
character
than
bonds
C sF
C sI.
larger
Mg
C
and
and
dierence
than
between
χ(Mg)
=
has
a
Mg
1.3
(MgO)
χ(C)
1.8,
than
Mg
that
and
O
O
are
are.
between
between
O
further
must
C
be
away
Therefore,
Mg
and
more
and
O,
the
O
and
the
ionic.
Q uantitative comparison:
∆χ
However,
table,
other
electronegativity
=
2.6
∆χ(CO)
gre ater
compounds
percentage
2.7
a
Both
0.8
=
=
3.2
v
O
∆χ(C sI)
χ(F)
l
χ(C s)
=
and
Therefore,
u
f
x
=
other
larger
between
Q uantitative comparison:
χ(C s)
is
b.
e ach
table.
electronegativity
me aning
ionic.
from
t
a
and
gre ater
r
o
Cs
and
a
i
the
Cs
are
o
than
F
d
and
n
Qualitative comparison:
Cs
CO
U
a.
C sI
c arbon monoxide,
Solution
C
a.
=
and
than
1.8
ionic ally
O
and
=
χ(O)
=
3.4
2.1
and
χ(O)
=
3.4
0.8
bond
for
this
ionic ally
bec ause
compound.
bec ause
∆χ
is
C
lower
and
than
∆χ
O
is
do
gre ater
not
bond
1.8.
E
103
Structure
2
Models
of
bonding
and
structure
Activity
Determine
using
the
the
following
pairs
of
elements
are
likely
to
bond
ionic ally
s
s
i.
whether
following two methods:
look at their positions in the periodic table
ii.
refer
to
their
electronegativity
values in the data booklet.
d.
As and S
Rb
e.
P and Cl
c.
C a and I
f.
Ag
and
Ga
and
e
Li and F
b.
r
a.
Br
oen
incorrectly
said
element
,
that
do
not
only
ionic
together.
t
this
bonds
There
form when a metallic element and
are substances, such as aluminium
description. Aluminium is a metal and chlorine
3
a
non-metal,
Polarized
light
properties
that
would
are
expect
point
and
high
volatility.
+
two
elements
(1.6)
suggests
they
propellants
ionic
compounds
contain
+
which is made up of NH
polyatomic ions.
As
4
several atoms.
NO
You
are
expected
3
in table 4.
HCO
c arbonate
CO
3
2–
2–
sulfate
ATL
SO
4
4

Table 4
Common polyatomic ions
silver(I)
sulde
Cl,
4
+
,
are
4
polyatomic
ions
are ions that contain
and
formulas
of
the
polyatomic
ions
shown
engage
spend
in
some
table
What
with
4.
time
Some
strategies
memorizing
students
will
you
like
use?
the
to
names
use
How
and
formulas of the
ashc ards,
will
you
others make
make
sure
you
them?
patterns
the
in
list
their
of
ionic
compounds
shown
in
table
5.
C an
you
notice
any
names?
You should notice that, in the names of ionic compounds:
3
hydroxide
oxide
suggests,
names
instance, NH
anions. Ammonium ions, NH
C aCO
Ba(OH)
E
iron(III)
2
a
barium
Consider
KF
MgF
c arbonate
the
For
low
Naming ionic compounds
l
c alcium
uoride
to
ions
mnemonics.
actively
Formula
v
O
magnesium
need
t
a
uoride
up
u
potassium
will
polyatomic
i
r
o
f
x
Name
You
o
d
PO
elements.
a
Self-management skills
3–
phosphate
two
as
between these
n
3
know
U
hydrogenc arbonate
to
name
C
nitrate
i
n
OH
their
such
dierence
o
NH
hydroxide
than
But the compound
ionic ally.
4
+
ammonium
more
bond
c ations and Cl
v
Formula
ionic ally.
compounds,
p
Some
Name
bond
y
Polyatomic ions
rocket
not
r
e
Its uses include fertilizers and
3
to
covalent
electronegativity
do
4
.
of
The
nitrate crystals. Ammonium
nitrate contains two polyatomic ions: NH
and NO
them
characteristic
micrograph
melting
of ammonium
you
O
has

Figure 11
so
t
i
s
is
n
chloride, AlCl
that
bond
l
non-metallic
•
the
c ation
name
•
c ations
•
monatomic
is
given
rst
and
is
followed
by the anion
2
Fe
O
2
Ag
adopt
the
name
of
the
parent
atom
and
the
name
remains
unchanged
3
S
anions
adopt
the
rst
part
of
the
name
of
the
parent atom,
2
followed

Table 5
Names and
ionic compounds
104
by the sux -ide.
If
the
anion
is
polyatomic,
refer to table 4
formulas of some
•
the
name
of
the
compound
does
not
reect
the
number
of
ions
in
the
formula.
y
P
is
a
y
It
Structure
Practice questions
Anions
are
common
9.
State
the
name
of
each
of
the
2.1 The ionic model
conjugate bases
acids.
The
of
strength of
following compounds:
acids and stability of their anions
a.
RbF
d.
Sr(OH)
2
Al
S
2
c.
e.
c an
using
3
AlN
be
compared
quantitatively
BaCO
3
f.
NH
HCO
4
K
3
,
their
which
s
s
b.
dissociation constants,
will
be
introduced in
a
Reactivity 3.1
e
The formulas of ionic compounds
name
ratio
of
the
an
ionic
ions
in
charges
and
negative
the
c ation,
of
net
charge
charges
then
of
must
work
you what elements it contains, but not the
working
the
out
c ancel
out
how
the
formula of an ionic compound
compound
out.
many
First,
of
is
zero,
so
determine
each
ion
you
the
the
positive
charge of the
need
to
reach a
zero.
a.
c alcium
oxide
b.
c alcium nitride
c.
the
formula
of
c alcium
oxide,
work
through
The
the
second
method
charges
subscript,
and
ignoring
i
n
Step 1: Determine the charges of the c ation and
C alcium
2
1s
has
2
2s
2p
3s
6
3p
electrons,
has
Oxygen
atoms
.
C alcium
they
form
electron
have
a
of
2
six
atoms
ions
with
have
a
conguration
outer
shell
2+
of
two
outer
2
1s
2
2s
electrons,
4
2p
so
2+
.
they
charge.
Therefore, calcium ions = Ca
criss-cross
the
into
the
rule.
other
Swap
ion’s
sign:
Ca
charge.
o
with
4s
so
d
ions
an
conguration
2
the
them
n
Oxygen
form
electron
2
U
shell
an
6
C
the anion
is
turn
o
v
following steps.
aluminium nitrate
y
deduce
the
d.
p
To
c arbonate
r
e
Solution
a.
sodium
O
Deduce the formulas of the following ionic compounds:
t
i
s
Worked example 4
n
charge
the
tells
for
y
total
that
basis
l
anion
and
compound
The
y
remembering
it.
P
is
of
r
The
Ca
Then,
simplify
the
ratio:
2
and oxide ions = O
O
Ca
r
o
i
1
1
t
a
Step 2: Determine how many of each ion are needed
Step 3: Check that the net charge is zero
in order to achieve a net charge of zero
You
check
your
working
by
adding
up
the
charges
of
There are two methods you can use for this step. The rst
e ach
individual
ion.
If
you
did
Step
2
correctly,
the
f
x
u
is the bar diagram method. Write out the ions as blocks
charges
will
add
to
zero:
equal to the number of charges on each individual ion:
l
ion,
so
compound
is
a
oxide
diagram
E
bar
v
O
The
2+
Ca
+
2
Ca
Total
2+
2–
O
Total
of
=
2
contains
ratio
charge
O
one
c alcium
ion
and
c alcium
1:1.
to
oxide
in
negative
charge
=
2
one
Net
the
positive
charge
=
2
2
=
0
the
Step 4: Write the formula
Ca
This
O
1
is
a
straightforward
example
where
the
1
magnitude
anion,
and
of
charge
hence
the
is
equal
formula
for
is
the
c ation
and
C aO.
105
Structure
b.
2
Models
C alcium
ions
as
of
bonding
nitride
have
is
a
dierent
and
structure
more
complex
charges.
Work
example as the
c.
Sodium
through the steps
must
before.
c arbonate
not
split
polyatomic
and
draw
up
contains
or
cluster.
brackets
a
change
polyatomic
the
ion.
You
ratio of atoms in the
Treat it like an indivisible entity
around
it
if
the
formula contains
Step 1: Determine the charges of the cation and
s
s
more than one such ion.
the anion
2
C a:
2
1s
2s
6
2
2p
3s
6
3p
Step 1: Determine the charges of the cation and
2
4s
the anion
atoms
have
two
outer
shell
electrons,
so
2
N a:
they
form
ions
with
a
2+
1s
2
2s
Sodium
2
2s
have
ve
atoms
have
one
outer
shell
outer
shell
electrons,
form
ions
with
a
so
1+
charge.
2
ions,
CO
,
have
a
2
3
form
ions
with
a
3
charge.
+
Sodium
ions
=
Ca
=
Na
and
charge.
c arbonate
2
ions
=
CO
3
3
and
nitride
ions
=
N
needed in order to achieve a net charge of zero
Step 2: Determine how many of each ion are
t
i
s
needed in order to achieve a net charge of zero
Bar
diagram
method
+
+
Na
+
2
+
2
+
2
Ca
Ca
The
bar
diagram
c arbonate
in
the
ions,
contains
so
the
compound
three
ratio
is
of
its
simplest
the
ratio
here
form.
o
in
i
simplify
t
a
alre ady
N
in
the
contains
so
compound
Criss-cross
the
is
a
=
6
poly atomic
not
ratio
2:1
Na
3
CO
Na
you
ion
c an
to
change.
draw
remind
Again,
1
brackets
yourself
there
is
around
that
no
its
the
formula
need
to
simplify
here.
+
2
CO
+
Na
3
3
N
T
otal negative charge = 6
3
T
otal positive charge = 2+
Net
charge
=
2
2
=
T
otal negative charge = 2
0
Step 4: Write the formula
fo r mu l a
is
Na
CO
2
th e re
a re
no
.
Note
th a t
in
th e
fi n a l
3
bra c ke ts
a ro u n d
th e
0
po l y a t o m i c
one
Step 4: Write the formula
Ca
N
3
106
CO
2
Step 3: Check that the net charge is zero
Th e
=
one
it
a n sw e r
6
and
c arbonate
rule
Remember,
the
u
l
E
charge
ions
to
N
Ca
Net
sodium
sodium
Na
2+
T
otal positive charge = 6+
of
3
does
bec ause
2+
Ca
v
O
2+
two
ratio
Step 3: Check that the net charge is zero
Ca
3
n
U
to
d
r
o
need
f
x
is
no
2
rule
Ca
is
and
nitride
N
3
There
ions
to
3:2
Ca
Criss-cross
c alcium
c alcium
C
ions
diagram
ion,
o
nitride
v
bar
two
i
n
The
2–
CO
y
3
N
Na
p
r
e
3
N
O
Bar diagram method
Ca
n
y
Step 2: Determine how many of each ion are
l
2+
C alcium
ions
so
2
ion
c arbonate
be c a u s e
ion.
the
fo r mu l a
c o n ta i n s
only
y
C arbonate
they
electron,
3
2p
P
atoms
1
3s
r
1s
they
Nitrogen
6
2p
charge.
2
N:
e
C alcium
Structure
d.
The
nal
example,
polyatomic
ion.
aluminium
Follow
the
nitrate, also contains a
same
steps
as
2.1 The ionic model
Criss-cross rule
before.
1
Al
Step 1: Determine the charges of the cation and
the anion
1s
2
2s
6
2p
2
3s
1
3p
1
Aluminium
so
they
atoms
form
ions
have
with
three
a
3+
outer
shell
,
NO
have
a
1
there
is
no
need
to
simplify
r
=
Al
and
nitrate
ions
=
NO
3
do
not
remember
the
charge on a
3+
polyatomic
formulas
table
and
4.
M ake
charges
o
sure
by
that
Al
you
heart.
T
otal positive charge = 3+
charge
3
3
=
0
The
method
to
formula
indic ate
so
compound
is
the
one
ratio
aluminium
of
)
3
3
each
oxide
of
the
n
of
e.
lithium
f.
barium
c.
sodium sulde
g.
d.
lithium nitride
a
of
an
ionic
contain
the
written
as
are
used
nitrate
AlN
O
3
9
transition
elements
have
oxidation number of the
transition
hydrogenc arbonate
t
a
l
formation
be
one
ammonium
is
metal
covered in
ion
in
brackets. This
Structure 3.1
phosphate.
compound
from
its
elements
a
redox
(Reactivity 3.2)
is
formal
E
LHA
How
v
O
the
not
brackets
than
u
f
x
Linking questions
is
that
more
nitrate
i
magnesium
strontium chloride
r
o
a.
reaction?
should
of
following compounds:
b.
Why
Note
Names of ionic compounds that
o
formula
d
the
nitrate
U
Practice questions
formula
C
(NO
1
and
to
1:3.
Al
Deduce
ion
aluminium
.
3
o
the
contains
ions,
3
i
n
in
diagram
nitrate
NO
3
3
presence
y
bar
three
NO
3
The
v
The
ion.
)
Al(NO
p
NO
r
e
Al
is
the
O
t
i
s
Step 4: Write the formula
Bar
diagram
3
T
otal negative charge = 3
needed in order to achieve a net charge of zero
+
3
10.
=
3
NO
n
Net
Step 2: Determine how many of each ion are
3
NO
l
these
revise
y
learn
ion,
NO
y
you
here.
P
If
ratio
Step 3: Check that the net charge is zero
3+
ions
the
charge.
3
Al
e
ions,
3
electrons,
charge.
Again,
Nitrate
s
s
2
Al:
sulfate?
LHA
Polyatomic
the
charge
used
to
predict
the
preferred
structure of
(Structure 2.2)
anions
relationship
are
conjugate
between
dissociation constant,
K
their
?
bases
stability
(Reactivity
of
common
and
the
acids.
conjugate
What is
acid’
s
3.1)
a
107
Structure
2
Models
of
bonding
and
structure
Ionic lattices and properties of ionic
compounds (Structure 2.1.3)
L attices
ionic crystals,
continuous,
negative
ions.
The
exact
arranged in a
networks
arrangement
of
of
lattice structure.
repeating
ions
in
a
units
of
L attices
are
positive and
lattice depends on the size and
ratio of the ions.
Ionic compounds are made of
in a lattice structure
l
e asily
c o mp o u n d s
of
ion
in
contain
a
the
is
an
e mp i r i c a l
s tr u c tu re.
qu a dr i l l i o n
A
th a t
the
Na
ions
ra ti o.
a r ra n ge d
involves
reliability.
nding
n
and
ions
a re
in
it
of
a
i n d i c a te s
sodium
c o n ti n u o u s
p re s e n t
in
th e
c h l o r i d e,
t he
l a tti c e.
l a tt i c e
in
a
o
oen
U
Research
Cl
C
i
n
Thinking skills
ATL
and
g ra i n
p
i n d i c a te s
v
1:1
fo r mu l a
fo r m u l a :
single
+
Th e
O
c an
ionic
ty pe
y
N aCl,
of
e ach
r
e
of
n
y
fo r mu l a
ra ti o
t
i
s
Th e
information
but
also
evaluating
its
usefulness
Consider the statement “each grain of NaCl can easily contain a quadrillion ions”.
o
i
108
up
Ionic
you
with
compare
dierent?
bonds
Is
are
reasonable
would
reliably
your
it
the
to
nd
own
this
to
you
this
dierence
species
Bec ause
of
surrounding
to
How
of
do
between
it,
fact-check the statement?
information?
estimate
one.
non-directional.
charged
you?
need
This
with
the
number
they
the
two
means
the
of
ions
compare?
values
that
an
attraction
in
a
grain
of
salt
Why might they be
signic ant?
ion
will
being
attract
equal
in
all
all
oppositely
directions.
–
E ach ion attracts all oppositely
ions around it
sound
could
Come
and
Ionic bonding is non-
E
charged
+
How
u
directional.
–
•
•
+
a

Figure 13
–
information
–
l
+
+
v
O
–
+
f
x
–
+
–
this
What
t
a
+
Does
•
+
r
o
–
–
+
d
–
+
•
the
in
all
an
the
ionic
forces
in
this
non-directional
surrounding
lattice
an
ionic
are
anions,
very
lattice,
quality,
and
strong.
but
each
vice
Figure
remember,
c ation
versa.
13
This
shows
actual
in
the
means
a
2D
lattices
ionic
the
lattice
forces
attracts
of
representation
are
3D.
attraction
of
the
y
P
ions arranged
are
r
u
Figure 12
ions
e
charge
the
three-dimensional
s
s
Within
Structure
2.1 The ionic model
L attice enthalpy
L attice enthalpy
values
tell
us
how
strong
the
ionic
bonds
are in particular ionic

lattice.
L attice
enthalpy,
∆
,
is
the
standard
enthalpy
change
that
occurs on
lattice
formation
the
strength
all
the
need
of
of
gaseous
an
ionic
electrostatic
to
bond
forces
overcome.
A
of
from
one
bec ause,
attraction
general
mole
in
of
order
the
for
between
equation
for
solid
the
ions
c ations
the
lattice.
to
and
lattice
It
is
a
measure of
become
anions
in
dissociation
gaseous,
the
lattice
process is
below:
+
M

(g)
+
X
(g)
∆H
> 0
r
MX(s)
lattice
process is
quoted
forces
are
shall
is
enthalpy
of
and
energy
The
gure
formation
of
Two
strength
to
overcome
factors
of
the
the
aecting
increasing
lattice
ions
radius.
enthalpy
attraction

Figure 14
required
C
–12
Ionic radius/10
U

–1
/kJ mol
lattice
compound
m
C ation
Anion
+1
–1
n
Anion
133
NaF
930
102
133
+1
–1
C aF
2651
100
133
+2
–1
greater
The
lattice
enthalpy
smaller
C aF
is
KF
bec ause
between
considerably
2
larger
the
radius
of
Ca
in
NaF
ions and F
than
2+
ionic
c ations
are
smaller,
+
Na
that
of
ions
is
greater.
KF. This is in part
+
compared to K
.
However, it is mainly due
2+
greater
charge
between
on
the
the
Ca
a
attraction
of
than
attraction
l
the
enthalpy
electrostatic
c ations
E
v
O
to
the
the
u
therefore
to
lattice
f
x
and
due
i
a
Lattice enthalpies of selected compounds
t
a
has
r
o
NaF
o
138
d
829

Table 6
the lattice
Ionic charge
C ation
KF
2
Lattice enthalpy is the energy
to overcome the electrostatic
forces of attraction holding ions together in
enthalpy of the compounds in table 6.
Ionic
∆H
from
o
in
lattice
electrostatic
electrostatic
i
n
variations
the
lattice
charge
ionic
for
However, in
gaseous
charged ions:
ionic
of
14.
298 K
are oen
given in the data booklet.
required
increases.
ionic charge.
increasing
with
denition
formation
in
at
enthalpies
y
decreases
the
shown
endothermic
the
ions
exothermic
that
enthalpy
p
•
the
with
as
to
lattice
L attice
v
increases
the
with
between
oppositely
•
only
the
process
of
booklet.
r
e
between
at
increases
attraction
ionic radius
Look
consider
values
data
represent
opposite
consistent
the
O
L attice
we
which
that
in
n
book
lattice,
values
ions — the
Experimental
found
t
i
s
a
negative
gaseous
be
l
this
as
c an
y
from
endothermic.
compounds
y
some
P
The
e
shown
be
ions
s
s
the
c ation,
and
which
anions
in
results
in
greater
electrostatic
C aF
2
109
Structure
2
Models
of
bonding
and
structure
F actors aecting the lattice enthalpy of the group 1 chlorides
Charge
density
volume.
the
this
group
enthalpies
1
of
is
a
term
task,
you
c ations
the
used
will
and
to
describe
explore
relate
the
this
to
Part 3: Analysis
charge per unit
charge density
the
trend
in
5.
C alculate
the
volume
6.
C alculate
the
charge
7.
Relevant skills
Plot
two
graphs:
Tool 3: General mathematics
•
Inquiry 1:
•
Inquiry 2:
Identify,
explain
describe
and
between
Describe
explain
9.
patterns,
Discuss
the
relationship
enthalpy, and the
the
trends
differences
lattice
enthalpy.
shown in the
between the two
your
graphs
the
relationship
11.
prediction, including a comparison
you
obtained
in
7
and
the
sketched
O
predict
you
n
the
t
i
s
c ations,
the
and
l
Evaluate
graphs
1
explain
density
y
10.
group
and
charge
graphs.
relationships
Part 1: Prediction
the
lattice
graphs.
predictions
of
For
c ation.
P
and
Instructions
1.
and
each
showing
r
8.
State
and
radius
graph
spreadsheets to manipulate data
other,
trends
c ation.
of
obtained in 2.
Consider possible extensions to this investigation:
between:
what other aspects of ionic radius, charge density and
charge
y
ionic radius and lattice enthalpy of their chlorides
b.
r
e
a.
lattice enthalpy could you explore?
density
and
lattice
enthalpy of their
chlorides.
the
graphs
you
expect
to
obtain
for the
above.
3.
C
i
n
Part 2: Data collection
Collect the following data for the group 1 chlorides:
ionic charge, ionic radius, lattice enthalpy. Possible
U
sources of information include the data booklet and
Input
your
data
into
a
spreadsheet
table.
and
organize it into
Practice questions
1.
Write
equations,
u
enthalpies
of
including
KBr,
state
symbols,
that
represent
the
O
a
lattice
C aO and MgCl
2
2.
l
a
E
v
O
f
x
t
a
r
o
i
o
suitable
d
a
n
online databases. Cite each source appropriately.
4.
o
relationships
p
Sketch
v
2.
State
and
explain
whether
you
expect KF or K
to
have
lower
lattice
2
enthalpy
3.
State
have
value.
and
the
explain
which
of
greatest
lattice
enthalpy
the
following
value:
ionic
compounds
NaCl, MgCl
,
Na
2
4.
you
O
or
expect to
MgO.
2
Describe and explain the trend in lattice enthalpy of the group 1 chlorides
down the group from LiCl to CsCl.
Properties of ionic compounds
The
properties
contain
in
110
a
c ations
lattice.
of
ionic
and
compounds
anions
held
are
due
together
by
to
their
strong
structural
features: they
electrostatic
attractive
forces
y
•
ionic
one
density
e
between
Tool 2: Use
each
lattice
group 1 chlorides.
•
of
s
s
of
In
Structure
2.1 The ionic model
Volatility
Global impact
Volatility
(from
the
L atin
volare,
to
y)
refers to the tendency of a substance to
of science
vaporize
(turn
electrostatic
into
a
forces
gas).
of
For
an
attraction
ionic
compound
holding
the
ions
to
turn
into
together
a
gas,
must
be
the
strong
overcome.
Some
volatility
of
ionic
compounds
is
therefore
very
low:
they
are
ionic
compounds
said to be “nonuncharacteristic ally
volatile”.
This
also
means
they
points
and
sodium chloride is approximately 1 075 K. Magnesium oxide, frequently used in
be
furnaces due to its ability to withstand high temperatures, melts at around 3 098 K.
bec ause
c an
Such
described
they
they
as
and,
are
does
transportation
t
i
s
+
i
n
–
–
–
+
–
–
+
move. They
+
+
are fixed in the
–
+
U
–
+
regular
+
–
–
+
–
d
i
+
+
+
f
x
conduct
Electrolytes in batteries
electricity bec ause they contain
mobile ions.
separated
Used
batteries are frequently
from other types of waste to
prevent
them
from ending up in landll.
water
metals and
other substances,
be recycled.
which c an
Do you separate your used
+
batteries from
+
other household
waste?
+
+
+
+
l
+
a
v
+
+
+
+
water
+
+
+
molecule
+
+
+
+
E
+
+
+
O
+
(red) is for batteries.
This is bec ause batteries contain valuable
u
+
These are waste separation
dissolve in
t
a
r
o
+
o
+
heat

Figure 15
bins in Jakarta, Indonesia. The lemost bin
n
–
–
arrangement.
C
+
ions cannot
o
+
signic ant
O
+
–
–
v
+
electricity
have
y
compound
p
r
e
A solid ionic
necessarily
disposal and
c an
anions are free to move past one another
, allowing them to conduct electricity
when a potential dierence is applied.
not
environmental impacts.
ions in a solid lattice are not mobile. When molten or aqueous, both cations and
easily
recycled.
n
y
cannot change position. Solid ionic compounds do not conduct electricity because
more
oen,
harmless. Their
manufacturing,
a solid ionic lattice, cations and anions can vibrate around a xed point, but they
non-volatile. This
be
l
they
c an
solvents”
y
however,
mean
able to move. Ionic compounds contain charged particles, cations and anions. In
P
This,
In order to conduct electricity, substances must contain charged particles that are
“green
are
c an
Electrical conductivity
contained
used as
r
means
be
ionic liquids
e
solvents.
because the
low melting
have high boiling points.
Ionic compounds typically have high melting points too. The melting point of
cannot conduct
have
s
s
The
+
+
When an ionic compound is heated
When an ionic compound is
strongly and melts, the ions can
dissolved in water, it can conduct
move around and the molten
electricity because its ions can
compound conducts electricity.
move among the water molecules.
t
Figure 16
Molten and
aqueous ionic
compounds are electric al conductors
111
Structure
2
Models
of
bonding
and
structure
Solubility
Ionic
compounds
insoluble in
Water
is
discussed
in
in
a
polar
solvents
solvent.
The
such
as
dierence
polar
solvents
hydrogen
and
in
the
partial
atoms,
water
combined
molecule
positive
an
ionic
charges
with
in
the
electronegativity
bent
geometry
their
positive
the
the
water
oxygen
molecule,
a
lattice
and
solvent,
the
solvent
added
negative
point
to
towards
is
no
molecules,
water.
charges
become
there
negative
the
surrounded
attraction
so
the
of
the
remain
In
ionic
within
δ‒
H
δ+
ions
the
n
+
Na
+
Na
Cl
Cl
+
Na
U
+
Na
+
H
O
+
+
Na
Cl
Na
δ+
H
H
δ+
δ‒
Cl
i
u
l
a
E
v
O
f
x
ionic
competing
ionic
•
the
Ionic
O
ions
bonds
+
Na
δ‒
δ‒
δ+
O
H
and
and
of
O
H
H
δ+
δ+
are
water
in
the
insoluble
the
lattice
molecules.
c alcium
in
water.
This
is
bec ause
H
water molecules
there
are two
present:
c ations
between
anions
include
dissolve
attraction
between
compounds
water
δ+
H
δ+
compounds
forces
association
c ations
δ‒
O
The dissolution of ionic compounds in water involves interactions between ions and
•
O
δ‒
t
a
r
o
Cl
all
H
δ+
+
Na
+
+
Not
δ+
δ+
o
d
Na
Na
H
O
δ‒
H
Cl
Cl
+
H
+
Na
Cl
δ+
δ+
δ+
H
Na
Na
Cl
H
H
Cl
δ+
δ‒
O
Cl
H
n
Na
+

Figure 17
ions
anions
and
individual
O
H
Cl
the
position
c ations,
molecules.
δ+
δ+
C
i
n
H
+
and
the
O
p
δ+
Na
molecules
result,
water
between
o
δ+
O
v
δ‒
Cl
a
y
r
e
dissolved, but not when solid.
Cl
oxygen atom
δ‒
electricity when molten or
Cl
the
y
diagram
δ+
water
As
by
c ations
lattice.
why ionic compounds
H
on
towards
anions.
t
i
s
labelled
The
point
Activity
explaining
charge
l
and
being
partial
charges
the
non-polar
compound
the
of
their
partial
hydrogen atoms.
and
ions
anions
and
when
are
the
the
of
and
the
lattice
partial
charges
electrostatic
stronger
Examples
c arbonate
in
ionic
silver
than
the
water
attractions
association
compounds
chloride.
of
that
molecules
between the
between the
are insoluble in
y
of
out
that
a
the
P
pulled
c ase
112
between
of
r
so
partial
having
on
compound
themselves
are
conduct
water, and
e
Imagine
a
as
Structure 2.2.
result
Draw
such
hexane.
greater
and
detail
is
typic ally soluble in
s
s
Polarity
are
non-polar
Structure
2.1 The ionic model
Research skills
ATL
H e av y
Some
metal
h e av y
Use
the
and
their
as
le ad
t re a t m e n t
metals
internet
to
out
and
nickel,
p ro c e ss e s
of
industrial
re s e a rc h
other
often
take
form
insoluble
adv antage
effluents
ex a m p l e s
of
t h ro u g h
of
this
salts.
p ro p e r t y,
p re c i p i t a t i o n .
p re c i p i t a t i o n
re a c t i o n s
uses.
Describe
and
p re c i p i t a t e
is
ex p l a i n
the
changes
that
a re
o b s e r ve d
when
a
forme d.
r
data
demonstrate
the
physic al
properties of ionic
y
(Tool 1, Inquiry 2)
Close-up
photograph of the
O
t
i
s

Figure 18
How can lattice enthalpies and the bonding continuum explain the trend in
n
l
experimental
compounds?
y
P
Linking questions
What
e
•
such
s
s
re m o v i n g
•
ions,
w a s t ew a t e r
formation of a lead(II) chromate precipitate
melting points of metal chlorides across period 3? (Structure 3.1)
in the reaction between aqueous solutions
of lead(II) nitrate and
y
o
C
n
o
i
t
a
S alar de Uyuni in Bolivia,
chloride.
particularly lithium.
p
r
e
v
i
n
U
a
ats at
of sodium
u
l
ions,
S alt
v
E
mineral form
d
r
o
f
x
O

Figure 19
potassium chromate
which are mainly made of halite, the
The brine below the rock salt
Global demand
Lithium-ion batteries c an be used
for lithium
crust is rich in dissolved metal
is increasing due to its use in batteries.
to power mobile phones,
laptops and
electric vehicles
113
Structure
2
Models
of
bonding
and
structure
Solubility of ionic salts
The
patterns
ionic
in
aqueous
compounds
are
solubility
often
of
several common
referred to as
Instructions
solubility
Part 1: Solubility rules
us
We
c an
deduce
use
the
these
differences in solubility to help
identity
of
will
different solutions of ionic
an
unknown ionic compound.
General
In
this
task
you
mix
table
(known as a
and
observe
precipitate)
whether
is
an
insoluble
7.
solubility
For
rules
c an
be
inferred
from the data in
example:
product
•
All
•
Sulfates
nitrates
are
soluble.
produced.
generally
insoluble,
except
sulfates
issues
address
in
an
relevant
sulfate.
safety and
table
7
and
infer
at
least
three
more
investigation
solubility rules.
Inquiry
2:
Interpret
l
•
general
qualitative data
y
•
Dilute
eye
protection.
A,
will
B,
C,
sodium
c alcium
nitrate
and
silver
nitrate
solutions
be
provided
and
D.
chloride,
are
do
not
know
solution,
take
c are when handling all solutions
You
bec ause
you
do
not
know
exactly which is which.
of
are all potential irritants.
and
retain
any
precipitates
formed.
you
tubes.
i
n
•
materials
from
by
do
You
mixing
not
c an
a
of
Dispose of waste solutions and precipitates according
your
school’s guidelines.
Materials
Dilute acid solution
•
S amples
piece
of
on
a
copper
black
wire
solutions
–
COO
3
+
, Na
knowledge
of
other
areas
solution
observable,
small-sc ale
each
on
will
a
mixtures of solutions
plastic
become
particularly
if
sheet.
If
a
precipitate
opaque. This will be
you
lay
the
sheet on a
teacher
. If they approve it and if you have time, try it out!
C ations
ammonium,
barium,
lead,
2+
c alcium, C a
+
, K
)
+
NH
+
silver, Ag
2+
2+
Ba
Pb
4
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
soluble
insoluble
insoluble
soluble
soluble
soluble
slightly
soluble
insoluble
insoluble
soluble
soluble
insoluble
slightly
soluble
soluble
soluble
insoluble
insoluble
a
ethanoate,
you
each
background.
t
a
l
–
CH
u
+
(Li
v
O
f
x
group 1 c ations
3
but
Devise a method, present it clearly, and show it to your
cm long)
labelled A, B, C and D
nitrate, NO
black
nitrate,
identify
background
(~0.5
r
o
of
easily
the
to
n
Small
•
sheet
formed,
is
o
•
plastic
i
Pipettes
U
Clear
•
d
•
is
C
to
c alcium
job
labelled
c arbonate,
have to mix the solutions inside test
prepare
drop
solutions
potassium
listed and the solubility rules.
your
o
v
Collect
the
draw
and
Your
chemistry.
Note
•
using
also
nitrate
which.
p
They
may
is
of
are
y
should
r
e
You
samples
solutions
silver
which
irritants.
•
with
These
O
Wear
t
i
s
You
•
n
Part 2: Identification of ionic compounds
S afety
–
chloride, Cl
E
Anions
–
hydroxide, OH
2–
sulfate, SO
slightly
soluble
insoluble
4
2–
c arbonate, CO
3

Table 7
114
Aqueous solubility of common ionic compounds
insoluble
insoluble
y
and
Study
environmental
ammonium
P
Tool 1: Recognize
and
group 1
r
are
Relevant skills
•
e
compounds
s
s
rules.
Structure
2.1 The ionic model
End-of-topic questions
7
.
What
is
the
formula
of
sodium
nitrate?
A.
NaNO
B.
NaNO
C.
Na
D.
S
s
s
Topic review
2
1.
Using
your
answer
the
knowledge
guiding
from the
question
as
Structure 2.1 topic,
fully
as
possible:
3
e
What determines the ionic nature and properties of a
N
3
compound?
2.
Explain
why
ionic
substances
are
r
N
3
2
always compounds.
Which
compound
has
the
largest
enthalpy?
Exam-style questions
The
elements
group
17
generally
C aO
C.
K
form ions with
D.
A.
K
7+
1–
statement
A.
L attice
7–
B.
2
2s
1s
D.
1s
2
2
2s
2
Which
2
3s
6
2p
2
are
made
up
correct?
electrostatic
of
C aF
L attice
the
Which
of
equation
K
a
increases
ions
the
energy
needed to
c ation to an anion.
when
the
charge of the
correctly
when
the
charge density
increases.
represents
the
lattice
enthalpy
oxide?
+
A.
radii of the
increases.
component
potassium
from
decreases
ions
enthalpy
the
2–
O(s)
→
2K
(g)
O(s)
→
2K(s)
O(s)
→ K
+
O
(g)
2
B.
K
C.
K
+
½O
2
molecules.
(g)
2
2+
2
2–
(g)
+
O
(g)
2
2
contain
held
to
together
c ations.
name
of
c arbon sulte
c alcium sulte
C.
c arbon sulfate
D.
c alcium sulfate
E
B.
a
the
is
c ations and anions.
deloc alized
bec ause
l
A.
is
the
D.
represents
electron
enthalpy
when
increase.
D.
electrons when
K
O(s)
→
2K(g)
+
½O
2
(g)
2
dissolved, but not when solid.
v
O
What
from
u
Ions
or
is
compounds
an
component
10.
i
uoride
electrons
6.
between
f
x
D.
results
structures
molten
ionic
L attice
of
t
a
Ionic
2
3d
r
o
C.
2
4s
about
bonding
C alcium
2
3d
6
3p
statement
Ionic
2
4s
6
3p
3s
attraction
B.
6
3p
d
A.
6
2p
2
2s
2
3s
C.
o
5.
6
2p
conguration of this ion.
n
C.
2
2s
electron
6
3p
U
1s
full
2
3s
C
2
B.
the
6
2p
i
n
1s
enthalpy
transfer
o
Give
2
A.
L attice
v
protons.
charge and contains 20
increases
ions
p
D.
The ion of element X has a 2 +
correct?
enthalpy
component
4.
is
y
C.
Which
r
e
1+
O
2
9.
B.
S
2
O
charge?
t
i
s
which
in
n
B.
Multiple-choice questions
3.
lattice
l
C aS
of
y
A.
value
y
P
8.
C aSO
anions
transfer
11.
List
the
lithium
halides
in
order of
increasing
lattice
enthalpy.
A.
LiF,
LiCl,
LiBr,
LiBr,
LiCl,
LiI
B.
LiF,
C.
LiI,
LiBr,
LiCl,
LiF
D.
LiI,
LiCl,
LiBr,
LiF
?
4
LiI
115
Structure
12.
2
Models
Which
of
bonding
substance
has
Melting
and
an
structure
ionic
structure?
Electric al
Electric al
conductivity
conductivity
Solubility
point / °C
in water
when molten
when solid
high
none
none
B
186
low
none
none
C
1083
high
good
none
D
1710
low
good
good
r
Describe
applic ations in the eld of optics.
for
c alcium
uoride.
[1]
c.
Deduce
d.
Potassium
the
Chromium
the
structure and bonding in solid
charge
of
the
dichromate(VI)
dichromate(VI)
contains
is
a
uoride.
[3]
i.
c.
Explain
why
solid
form Cr
Write
c alcium uoride is a poor
c an
conduct
The
lattice
but
molten
c alcium uoride
electricity.
[2]
ii.
a
the
Copy
the
arrows
d.
enthalpy
of
c alcium uoride is
shows
lattice
the
process
enthalpy
Explain
why
including
of
the
state
uoride.
The
lattice
enthalpy
of
c alcium
oxide,
iii.
[1]
C aO,
15.
1
the
potassium
full
the
atom
v
E
116
Cr
2
[2]
orange
draw
The
[1]
4s
Write
the
3d
full
electron
conguration of a
3+
Cr
ion.
equation
uoride
Li(s)
ion.
+
from
F
(g)
[1]
below
its
→
shows
elements
the
formation of lithium
under
standard conditions.
LiF(s)
2
a.
Balance
b.
Identify
the
the
charge
equation.
c.
Identify
the
oxidized
[1]
O
2
7
of
the
lithium
ion.
[1]
in
mass
species
and
the
reduced
[1]
species
in
this
reaction.
[1]
between a
d.
Sketch
a
diagram
showing
the
structure of
potassium ion is
a
negligible.
a
bright
and
represent the
conguration of a
dierence
and
the
dichromate(VI), K
l
O
potassium
electron
contain
ion.
why
lattice
greater than
u
Explain
the
o
Write
is
uoride.
breathalyser
compound
potassium
b.
c alcium
f
x
a.
of
why
oxide
t
a
ionic
of
types
Explain
r
o
Certain
.
c alcium
i
that
14.
of
d
is3 401 kJ mol
enthalpy
n
iii.
[2]
process in part (i) is
endothermic.
to
2+
associated with the
c alcium
conguration
[1]
below
boxes
orbitals of a Cr
symbols,
U
ii.
equation,
diagram
the
electron
others.
electronconguration in the 3d and 4s
C
that
an
in
among
atom.
o
Write
i
n
i.
.
v
1
2 651 kJ mol
abbreviated
chromium
ions
y
conductor,
3+
ions and Cr
p
r
e
of
electric al
[1]
transition element that commonly
2+
forms
c alcium
ion.
chromium.
O
b.
has
formula
n
the
t
i
s
Deduce
l
uoride
a.
y
C alcium
lithiumuoride.
[1]
[2]
y
P
Extended-response questions
13.
e
36
s
s
A
Structure 2.2
The covalent model
s
s
What determines the covalent nature and properties of a substance?
Substances
From water to diamond to nitrogen gas, from oils to
characterized
plastics to polyatomic ions, these species contain atoms
low
held together by strong covalent bonds. Covalent bonds
molecular
lead to the formation of two dierent types of structure:
low melting points and boiling points. Their solubility and
covalent network structures (also known as giant covalent
volatility
structures) and molecular covalent structures
forces.
poor
electric al
Structure 2.2,
how
positively
shell
pair
with
a
two
total
of
8
electrons.
three
shared
pairs
of
the
same atom.
Structure 2.2.4 —
electron
domains
a
molecular
C arbon
v
E
molecules.
bond
Electron
from
the
repulsion of
central atom.
polarity
results
of
from the
the
bonded atoms.
<
You
structures.
dipole–dipole
forces <
is
more
in
a
than
Resonance
one
structures occur when
possible
position
Structure 2.2.12 —
Benzene, C
H
6
example
of
a
molecule
Structure 2.2.13 —
which
for a double
molecule.
they
have
an
Some
for
determine
which
has
each
Formal
atom
of
in
several
, is an important
6
resonance.
atoms
expanded
Structure 2.2.14 —
c alculated
that
c an
octet
charge
a
form
of
values
species
possible
molecules in
electrons.
and
c an be
used to
Lewis
formulas is
preferred.
The
is
Structure 2.2.15 —
geometry.
a
O
the
covalent
network
describe
Molecular polarity depends on both
molecules
(dispersion),
predicted
l
and
Structure 2.2.8 —
of
Bond
structures.
between
Shell
a
u
f
x
polarity
Structure 2.2.7 —
network
be
electronegativities
Structure 2.2.6 —
bond
to
around
Structure 2.2.5 —
in
Valence
is
shared pair
(VSEPR) model enables the
molecules
dierence
the
t
a
of
bond
of
we
forces.
mobile and stationary phases.
there
i
Repulsion
shapes
The
r
o
Pair
electrons
how
intermolecular
their relative attractions involving intermolecular forces to
Structure 2.2.11 —
o
from
coordination
the
d
originate
A
both
forces
covalent bond is,
as
LHA
which
covalent
a
well
intermolecular
used to separate the components of a mixture based on
n
in
and
their
have
Structure 2.2.10 — Chromatography is a technique
electrons
U
Structure 2.2.3 —
about
(dispersion)
Single, double and triple bonds
and
shapes
hydrogen bonding.
electrons
charged nuclei.
respectively.
bond
London
by the
of
learn
C
one,
formed
refers to the tendency of atoms to gain a
Structure 2.2.2 —
involve
is
shared
i
n
valence
rule
a
their
on
what
as
Substances with
generally
y
octet
bond
between
learn
water.
hand,
o
the
The
covalent
v
and
A
attraction
also
will
molecules
p
electrostatic
explain
depending
r
e
Structure 2.2.1 —
you
represent
in
other
O
will
we
the
t
i
s
and
Understandings
greatly
solubility
on
n
In
poor
y
are
are also
l
substances
structures
r
covalent
and
structures,
vary
network
P
general,
conductors.
volatility
covalent
by high melting points and boiling points,
y
In
with
e
Covalent bonds lead to a vast range of dierent substances.
and
Sigma bonds (σ) form by the head-on
combination of atomic orbitals where the electron density
silicon
form
covalent
is concentrated along the bond axis.
Pi
nature
of
the
force
that
bonds (π)
form
by
the
lateral combination of p orbitals
exists
where
determined
the
electron
density
is
concentrated on opposite
by the size and polarity
sides of the bond axis.
Intermolecular
dipole–induced
forces include London
dipole, dipole–dipole and
Structure 2.2.16 —
mixing
hydrogen bonding.
atomic
Hybridization is the concept of
orbitals
to
form
new
hybrid
orbitals
for
bonding.
Structure 2.2.9 —
relative
strengths
Given
of
comparable
intermolecular
molar
forces
mass, the
are
generally:
117
Structure
2
Models
of
bonding
and
structure
Covalent bonds and molecules
(Structure 2.2.1)
Covalent
covalent
bonds
bond
Diatomic
and
formed
the
from
positively
hydrogen, H
,
is
when
the
atoms
charged
the
share
electrostatic
nuclei
simplest
pairs
of
attraction
of
the
covalent
valence
electrons. A
between
atoms
a
shared pair of
s
s
electrons
are
results
involved in the bond.
molecule. It consists of two
2
from
hydrogen
electrons,
the
atom
forming
formation
together
(gure
single,
of
by
the
a
covalent
1).
double
double
Atoms
or
bond
c an
triple
covalent
made
share
bonds,
bond
in
diatomic
electrons gives both
arrangement and forms
a single covalent bond
This is a
double
covalent
bond (two
shared pairs of electrons).
Only the electrons in the
highest energy level
(outer
shell) are shown here
,
and
diatomic oxygen, O
2
2
o
i
t
a
u
l
a
v
Examples of substances that
contain covalent
bonds:
plastics,
graphite (in pencil leads),
zzy water,
and
the c arbon dioxide
y
n
A shared pair of
y
o
formation in diatomic hydrogen, H
d
r
o
118
E
bubbles in it
f
x
O

Figure 2
bond
O
.
2
Oxygen molecule
n
Covalent
U

Figure 1
O
C
i
n
Oxygen atoms
oxygen, O
atoms a stable
p
r
e
O
pairs
Figure 1 also
H
Hydrogen molecule
v
O
electron
three
O
Hydrogen atoms
or
l
H
H
one
two
respectively.
y
t
i
s
H
from
one,
P
shows
held
r
of
each
atoms
e
hydrogen
Structure
2.2
The
covalent model
Activity
Draw
a
diagram
to
show
the
triple
covalent
bond
in
diatomic
nitrogen, N
2
s
s
Electronegativity
bonds
generally
electronegativity,
1.8.
when
When
the
dierence
bond
in
dierence
them
is
relatively high
between
electronegativity
between
of
Structure 2.1,
between
predominantly
we
two
saw that ionic bonds
atoms
the
is
atoms
character
bond
together,
bond
bonds
they
form
are
covalent
bonds
periods
are
found
likely
with
to
be
each
the
referring
O
and
following
to
c arbon
b.
sodium
c.
c arbon
d.
chlorine
e.
iodine and iodine
f.
aluminium and uorine
to
so
the
they
same
form a
no
between
ionic
bonding.
a
covalent
borderline
and
Bonding
continuum,
character
Structure 2.1,
sharp
at
with
one
covalent
occurs along
greater ionic
end
character
at
and
the
greater
other.
groups. If two atoms of
Structure 2.4
addresses this
each other in the periodic
covalent.
For
continuum
example,
in
more detail with the
oxygen and
bonding triangle
pairs
of
elements
electronegativity
bond
covalently or
values:
oxygen
hydrogen
v
and
up
zero,
of
is
a
a.
and
oxygen
their
and
close
is
atoms
discussed in
there
t
a
whether
by
u
ionic ally,
l
f
x
Determine
dierence
other.
Practice questions
1.
non-metal
i
form
across
electronegativity
r
o
uorine
increases
two
o
the
electronegativity
d
table,
high
the
When
between them.
Electronegativity
relatively
compounds.
covalent substances
n
covalent
or
U
element
elements
compounds,
As
y
either
always
C
be
are
i
n
c an
which
covalent
o
that
p
Electronegativity dierences
are lower than 1.8 suggest
bonding is present
substances,
v
that
n
greater covalent
O
covalent
bonding
t
i
s
character
2.0
1.0
l
y
greater ionic
r
e
ecnereffid ytivitagenortcele
ionic
bonding
0
ionic
less than 1.8
covalent.
3.0

Figure 3
Unlike
greater than
is
y
3),
the
electronegativity
atoms
P
(gure
the
between
r
form
form
typic ally non-metals. In
e
Covalent
and
oxygen
E
119
Structure
2
Models
of
bonding
and
structure
Lewis formulas
Groups
way
of
atoms
valence
formulas.
of
are
Lewis
dots
or
covalently
are
formulas,
pairs
of
bonded
arranged
pairs
crosses,
in
of
or
a
valence
a
together
molecule
c an
are
be
electrons
combination
of
c alled
are
all
molecules. The
shown using
Lewis
represented as dashes,
three.
Consider the
F
various
Lewis
formulas
of
uoroamine, NH
F,
in
gure 4.
2
H
H
H
Here
are
the
dashes
rules
are
for
drawing
Lewis
formulas
regardless
of
whether
dots,
used:
F
Only
electrons
are
shown.
2.
Electrons
3.
Each pair of electrons shared between two atoms represents a covalent bond.
4.
Electrons
are
arranged
in
pairs.
These are all acceptable Lewis
formulas of uoroamine, NH
l
F
2
bond
two
(termed
atoms
region
positioned
in
the
(also
between
the
referred to as
two
atoms
lone pairs)
are
region
positioned
O
the
are
involved in the bond.
Non-bonding electrons
from
bonding electrons)
n
a
t
i
s
5.
in
the
y
between
away
involved in the bond.
Atoms in Lewis formulas generally have noble gas electron congurations. This is
r
e
known as the octet rule because noble gases oen have eight valence electrons.
y
Noble gases already have full octets and do not readily gain, lose or share
octet
rule
some
stable
of
1.
Work
c an
these
out
the
of
t
a
3.
least
always
O
u
l
a
5.
Bond
the
Assign
central
them.
help
there
by
of
the
of
than
and
limitations.
situations.
an
(this
draw
the
has
octet
is
seen
exceptions
Lewis
valence
valence
drawing
is
central
to
does
of
electrons.
at
AHL).
the
electrons
by
It
not
Sometimes
in
two
to
c an
Conversely,
rule.
most
for
explain
atoms
Species with odd
octet
formulas
electrons
their
usually
c ases:
each atom in the
work
out
how
many
single
pairs
symbols
on
found
the
in
the
page. The element with
centre.
Hydrogen atoms
atom(s).
peripheral
E ach
achieve
of
also
you
it
are.
atoms
non-bonding
they
but
certain
fewer
are
number
atoms
surround
the
will
number
number
between
until
total
in
expanded octets
total
electrons
Arrange
the
4.
N
Divide
pairs
N
v
O

Figure 5
f
x
+
O
2.
i
r
o
+
N
F
thumb,
with
o
d
F
of
bonds
electrons
steps
the
molecule.
even
form
valence
rule
form
n
U
Following
+
molecules
atoms
numbers
useful
gases
C
larger
a
i
n
form
is
noble
o
why
v
The
p
electrons. Hence, they are usually found in an unbonded, monatomic form.
atoms
bond
of
noble-gas
together
represents
electrons
to
the
an
by
drawing single bonds
electron
peripheral
pair.
atoms.
Keep going
congurations.
N
6.
Assign
any
7
.
Check
that
remaining
electron
pairs
to
the
central
atom(s).
By sharing a pair of electrons,
two uorine atoms c an obtain a full octet of
E
valence electrons.
the
central
atom
has
a
full
octet.
If
it
does
not,
try
the
following
Oxygen atoms c an do
two methods:
this by sharing two pairs of electrons, and
nitrogen atoms share three pairs
•
Reassign
non-bonding
additional
•
Check
octet
120
bonds
that
rule
the
(see
to
pairs
the
molecule
page
on
the
peripheral
atoms
to
become
central atom
123
you
for
are
looking
examples)
at
is
not
an
exception to the
y
1.
H
P

Figure 4
valence
r
H
crosses
e
or
s
s
pairs
In
that
electrons
Structure
2.2
The
covalent model
Worked example 1
Draw the Lewis formulas of each of the following molecules:
a.
water, H
O
b.
nitrogen trichloride, NCl
2
c.
c arbon dioxide,
CO
3
2
s
s
Solution
Follow
the
steps
above
for
e ach
molecule:
O
b. nitrogen trichloride, NCl
2
Step 1
Count
hydrogen: 1
2
nitrogen: 5
+
chlorine: 7
6
=
8
Total: 5
+
c arbon: 4
×
21
3
=
=
21
Total: 4
16
= 13
pairs
= 8
Cl
H
is
is
likely
only
the
to
one
be
in
periphery.
centre.
There
is
likely
only
to
so
far...
already
energy
need
more
full
electrons.
H
remaining
central atom
Check
the
central
full
remaining
assigned
Oxygen
the
full
Lewis
octet.
E
v
a
O
octet
has
to
4
electron
the
c arbon, so it
the
centre.
pairs
O
used
so
far...
O
The
oxygen
atoms
now
have
Cl
full
octets.
chlorine
atoms
now
pairs
used
so
far...
have
octets.
pairs
used
so
far...
We
have
used
all
8
available
Cl
electron
pairs
available
Cl
N
pairs.
for
None
the
are
central
c arbon.
Cl
oxygen.
electron
formula,
l
atom has a
are
2
u
that
f
x
Step 7
The
t
a
the
pairs on
r
o
electron
i
any
N
o
d
Put
far...
8
The
12
Step 6
2
Cl
Cl
Cl
in
level only holds up
electrons. They do not
any
so
one
be
O
centre.
N
used
n
two
bec ause the
U
to
the
have noble gas
congurations
rst
pairs
only
to
o
peripheral atoms
atoms
hydrogen
Cl
C
nitrogen, so it
C
pairs on
peripheral
v
bonding
The
i
n
Put non-
in
is
likely
y
used
3
Step 5
be
is
p
single bonds
pairs
one
O
There
Cl
H
2
pairs
Cl
r
e
is
Draw the
N
oxygen, so it
the
12
16
O
on
Cl
There
t
i
s
Hydrogens
=
=
n
O
2
2
pairs
H
×
12
l
2
Arrange
+
2
y
26
pairs
2
the atoms
Step 4
oxygen: 6
26
y
Step 3
=
8
= 4
electron
2
P
C alculate
the number of
×
oxygen: 6
Total: 2
Step 2
3
r
valence electrons
c. c arbon dioxide, CO
e
a. water, H
pairs in
The
remaining
assigned
to
Nitrogen
has
electron pair is
the
4
nitrogen.
electron
!
pairs in
O
therefore a
the
full
Lewis
octet.
formula,
therefore a
The
2
c arbon atom has only
electron
pairs
in
this
Lewis
formula.
We need to reassign two of the
non-bonding pairs on the oxygen
atoms to form double bonds:
O
Now
the
electron
full
c arbon atom has 4
pairs
and
therefore a
octet.
121
Structure
2
Models
of
bonding
and
structure
Polyatomic
magnitude
gained.
charge
the
valence
charged
charge
formulas
indic ated
subtract
an
of
you
a
you
add
electron.
of
how
poly atomic
with
electrons,
charge,
groups
indic ates
ions
superscript
also
an
need
to
electron,
This
is
covalently
many
are
factor
illustrated
in
for
in
the
in
been
square
brackets.
the
charge:
every
the
atoms.
have
enclosed
outside
and
bonded
electrons
for
set
of
every
2–
b.
c arbonate ion,
+
CO
c.
hydronium ion, H
O
3
Solution
steps
above
for
e ach
ion:
b c arbonate ion, CO
Count
c arbon: 4
hydrogen: 1
oxygen: 6
electrons
This
This
polyatomic ion has a
has
one
C alculate the
2
1
+
1
=
8
Total: 4
4
pairs
=
O
n
i
this
Total: 3
8
2
4
meaning that it
fewer
+
6
electron.
1
=
8
pairs
H
O
H
do
not
between
atoms
H
O
O
H
C
H
O
O
3
3
pairs
used
so
pairs
used
so
far...
far...
The peripheral hydrogen atoms
O
need
have
central
and
noble
gas
therefore
congurations
do
not
need
any
C
here.
more
O
electrons.
O
step.
The
full
12
oxygen
atoms
now
have
octets.
pairs
We
used
have
so
used
far...
all
12
available
O
H
electron
The
electron
you
a
remaining
Skip
E
any
Put
so
peripheral
v
O
Step 6
it,
distinguish
and
far...
l
atoms
to
in
24
3
C
hydroxide ion has only two
atoms
=
O
O
t
a
peripheral
so
u
pairs
used
r
o
on
The
f
x
Put non-
bonding
pair
2
one
=
o
single bonds
+
=
H
d
Step 4 Draw the
18
3
polyatomic ion has a
charge,
has
pairs
U
the atoms
Step 5
+
1+
electrons.
C
i
n
pairs
Arrange
1
additional
12
2
number of
electron
two
24
=
Step 3
has
This
meaning that it
o
8
+
electron.
polyatomic ion has a
charge,
×
oxygen: 6
v
Step 2
additional
2–
18
p
Total: 6
meaning that it
=
y
charge,
3
r
e
1
hydrogen: 1
×
O
3
O
oxygen: 6
valence
t
i
s
Step 1
+
c hydronium ion, H
3
n
2–
a hydroxide ion, OH
l
the
y
Follow
remaining
3
electron
pairs.
None
H
are
pairs
pairs
available
for
the
central
c arbon.
H
are
on
the
assigned
to
the
oxygen
central
atom.
The
hydrogen
already
The
fourth
electron pair is
atom
has
122
a
noble
gas
conguration.
assigned
to
the
oxygen atom.
y
P
3
charge,
e
r
–
hydroxide ion, OH
additional
examples.
Worked example 2
a.
with
counting
positive
worked
Draw the Lewis formulas of the following ions:
or
brackets,
When
additional
next
The
lost
s
s
negative
are
the
Lewis
the
you
ions
of
Structure
2.2
The
covalent model
+
2–
a hydroxide ion, OH
b c arbonate ion, CO
c hydronium ion, H
3
Step 7
O
3
+
Check
O
H
O
that
the
H
central
!
full
All
atoms
C
have noble-gas
H
octet
O
congurations.
O
All
Note
the
square
atoms
have noble gas
brackets and
The
c arbon atom has only
which
are
3pairs
in
this
Lewis
formula.
charge
Lewis
formulas
of
Brackets and
used in the
are
added to complete
polyatomic
We need to reassign one of the
Lewis
r
the
e
congurations.
charge,
formula.
ions.
non-bonding pairs on the oxygen
2
l
O
triuoride,
BF
in
Lewis
Consider
the
Lewis
atoms
that
formulas
the
do
that
arrangement
formula
shown
three
pairs
of
electrons
around
it.
The
follow
in
of
the
gure
the
atoms
12
6.
octet
with
Here,
boron atom is
rule.
At SL,
F
fewer than eight
electron
U
3
only
not
contain
p
electrons.
contain
of
formula.
o
aware
charge
C
valence
molecules
be
therefore a
and
v
should
i
n
you
and
added to complete the
Lewis
Sometimes
pairs
Brackets
y
octet.
are
O
c arbon atom has
r
e
full
t
i
s
the
4electron
n
y
C
y
P
atoms to form a double bond:
Now
s
s
atom has a
the
pairs
of
boron
boron atom has
B
electron decient.
n
F
F
Similar to boron, many other elements of groups 2 and 13 form stable electron-
o
d
decient molecules. These include beryllium, magnesium and aluminium. In Lewis
formulas of such molecules, group 2 elements (Be and Mg) have only two bonding

Figure 6
Lewis formula of boron
triuoride
t
a
r
o
i
electron pairs while group 13 elements (B and Al) have three bonding electron pairs.
Practice questions
Draw
the
Lewis
formula
of
each
BH
3
c.
BeCl
2
l
+
3
Draw
the
a.
HBr
c.
O
Lewis
2
a
3.
CH
formula
E
v
O
e.
of
of
the
u
a.
f
x
2.
each
of
the
following:
b.
BCl
d.
AlCl
3
3
following:
b.
OF
d.
O
2
3
e.
HCN
f.
CH
g.
CH
h.
N
2
4.
Draw
the
Lewis
Cl
3
O
H
2
formula
of
each
of
the
4
following:
+
a.
NH
b.
NO
d.
NO
4
3
+
c.
NO
2
e.
2
OCl
123
Structure
2
Models
of
bonding
and
structure
Linking questions
What
do
noble
some
gases
of
the
form
limitations
covalent
of
bonds
the
octet
less
rule?
readily
than
other
s
s
Why
are
elements?
(Structure 1.3)
do
bonds
ionic
c an
bonds
form
only
form
between
between
atoms
of
the
dierent
same
elements
element?
while
(Structure
e
Why
covalent
2.1)
r
hydrogen
molecule, H
,
shown
in
gure
7, contains a
molecule, O
, contains a
double bond.
Diatomic
2
triple bond
closer
this
order,
bonded
bonds
order
than
double
shows
p
8
o
C
n
than
o
u
a
E
124
(bond
stronger
atoms
of
Single
strength
dier
in
than
length and
pairs
order
bonds,
together
bond
electron
(bond
3)
single
triple
increasing length
between bond
number
bonds
l
are
The
order.
v
enthalpies
Reactivity 1.2
increasing
t
a
f
x
O
Bond
The relationship
order
double
i
d
r
o

Figure 8
nitrogen molecules with diering bond
v
i
n
U
single
O
oxygen and
, contains
2
y
Hydrogen,
r
e
H

Figure 7
nitrogen, N
t
i
s
a
single bond. The
n
y
2
oxygen
l
The
1),
their
and
Double
pattern
for
bonds
between
double
strength
triple
double
bonds.
bond
are,
in
two
bonds
and
bonds
bonds,
strength
length.
are
and
turn,
atoms
(bond
still.
triple
shorter
referred to as
Double
stronger
hence
is
bond
order 2) and triple
than
bonds
are
Triple bonds hold
bonds
single
are shorter
bonds.
Figure
c arbon–c arbon bonds.
discussed in
Bond
strength
bond
enthalpy.
and
length
c an
be
quantied.
The
stronger
a
bond
is,
the
larger its
y
P
Bond order (Structure 2.2.2)
Structure
2.2
The
covalent model
Bond length is dened as the average distance between two bonded nuclei. Table
1 shows bond enthalpy and bond length data for carbon–carbon single, double and
triple bonds. The carbon–carbon triple bond is the shortest but also the strongest.
–1
t
Table 1
–12
Bond enthalpy (at 298 K) / kJ mol
Bond length / 10
Bond enthalpies and bond
s
s
Bond
m
lengths of c arbon–c arbon bonds
346
154
614
134
C≡C
839
120
r
e
C–C
C=C
and its bond
m. Use these data and the information in table 1 to predict
the bond enthalpies and bond lengths of single and double carbon–nitrogen
why
two
hydrogen
between
atoms.
sketching
a
between
potential
Read
graph
of
it
and
try
potential
The
two
paragraph
hydrogen
by
energy and the distance
to
represent what it is
energy (y-axis)
hydrogen nuclei (x-axis).
below
atoms
vs distance
C
between
by
relationship
explanations.
forms
i
n
describing
written
bond
Two hydrogen atoms have no eect on each other if they are
o
between
the
covalent
v
describing
complement
a
O
oen
explains
p
Graphs
y
Communic ation skills
r
e
ATL
t
i
s
bonds. Then compare your predictions with the values in the data booklet.
n
y
12
length is 116 × 10
l
1
The bond enthalpy of a carbon–nitrogen triple bond is 890 kJ mol
y
P
Data-based question
U
separated by a suciently large distance. As the two atoms
n
approach each other, the electrostatic attraction between the
hydrogen nuclei and each other ’s electrons increases. This process
o
d
leads to a decrease in potential energy. As the atoms get closer,
they reach a point where the two nuclei attract the two electrons
i
r
o
in the pair, eectively sharing them. This arrangement occurs at a
t
a
potential energy minimum, meaning that the molecule is energetically
stable. If the atoms were to get any closer together, the resulting
repulsion between the two positively charged nuclei would outweigh
f
x
the attraction for the shared pair of electrons, leading to a rise in
What
c an
into
c an
search
graphic al
To
one
what
typing
your
sketch
“hydrogen
graph
to
molecule
a
potential
potential
energy
is
a
graphic al
Linking question
energy
engine.
extent
description?
compare
by
representations
v
above?
a
nished,
nd
a
curve”
have
l
You
O
you
curve.
u
potential energy.
When
How
add
to
explanations such as the one
representation
better
than
a
textual
and
does
triple
inuence
the
presence of double
bonds
their
in
molecules
reactivity?
(Reactivity 2.2)
E
125
Structure
The
2
Models
formation
bonds
is
a
of
of
bonding
and
structure
Coordination bonds (Structure 2.2.3)
coordination
feature
of
electron pair
We
sharing
reactions,
as
have
seen
that
a
covalent
bond
is
formed
when
each of the two atoms in
discussed in
the
bond
contributes
an
electron
to
the
bond.
Sometimes
both
the
electrons
Reactivity 3.4.
in
the
covalent
bond
come
water
molecule,
hydronium
the
electron
Dierent
is
direction
bond,
of
and
representations
the
which
hydrogen
formed,
bonds
arrow
atom
resulting
c ation
leading
are
shows
oen
the
the
atom
+
O
with
H
O
H
H
H
an
pair.
+
help
to
coordination
explain
bonds
how
are
bonding
Draw
the
Lewis
i
n
the
following
+
c arbon
4
3
monoxide,
ammonia
boron
CO
C
ozone, O
c.
triuoride, NH
molecules.
any other
molecules,
indic ating the
BF
3
n
U
LHA
of
clearly.
ammonium ion, NH
b.
d.
formulas
bonds
certain
o
coordination
a.
v
5.
in
from
y
Practice questions
p
r
e
covalentbond.
occurs
indistinguishable
n
bonds
formed,
O
Once
t
i
s
Coordination
H
y
coordination bond
l
H
H
arrow
donated the
electron
coordination bond
c alled
formation of a
+
showing the
H
are
encounters a
indic ated
which
accepted
to
bonds
3
Coordination bonds in transition metal complexes
their
of
complex ions
reactions
are
discussed in
cation and the surrounding atoms or groups of atoms, called
A
common
form
the
feature
of
all
ligands
is
a
lone
pair
of
electrons,
ligands.
which
c an
be
used to
coordination bond to the metal ion.
l
u
Why do Lewis acid–base
complex ions contain coordination bonds, which hold together the central metal
t
a
r
o
f
x
Linking question
Transition metals can form complex ions such as the one shown in Figure 10. These
i
Reactivity 3.4.
o
properties
and
d
The
reactions lead to the formation of
O
H
E
v
a
coordination bonds? (Reactivity 3.4)
H
+
2
O
2
OH
O
2
2
Cu
H
O
OH
2
H
2
O
2
+
2
[Cu(H

Figure 10
O)
2
]
6
Water molecules form coordination bonds with copper(II) c ations,
the formation of a blue hexaaquacopper(II) complex ion
126
leading to
y
ion,
the
bond
Coordination
a
The
P
of the hydronium
to
The
9).
when
atom.
r
u
Figure 9
coordination
(gure
bond.
pair
a
example,
same
e
along
ion
For
the
s
s
coordination bonds
from
Structure
The valence shell electron pair
2.2
The
covalent model
repulsion
model (VSEPR) (Structure 2.2.4)
You
might
have
hydrogen
come
atoms
This
an
c an
across
a
angle.
Such
also
be
Lewis
shown
formula
formulas
for
water
better
that
shows the two
illustrate
the
geometry of the
s
s
molecule.
at
by ball-and-stick and space-lling models
(gure 11).
reect
the
feature
antibiotic
of
a
molecules
geometry
properties.
be
shown
1.
Electron
pairs
is
based
repel
on
each
the
their
properties.
For
example,

Figure 11
gure 12.
Electron
following
other
three-
and
Space-lling model of a water
molecule
VSEPR model:
Pair
side chain
Repulsion
R
premises:
therefore
arrange
β-lactam ring
H
O
model
Shell
The
geometry, and it is an
t
i
s
VSEPR
in
to
formulas do not
molecule.
known as a beta-lactam ring is key to its
explored using the
Valence
The
molecular
contributes
structure
Lewis
a
N
C
n
c an
a
as
in
l
geometry
is
termed
which
within
This
is
such
atoms
y
Molecular
molecule
of
y
molecular
of
molecules
P
the
shape
of
arrangement
r
dimensional
essential
representations
three-dimensional
e
Two-dimensional
S
CH
3
themselves as far apart
O
Non-bonding
triple
region
of
domain
space
than
a
bonding
•
a
double
pair
or
pair
of
in
of
useful
in
density
electrons
electrons
bond
gure
these
due
discussions.
to
13
illustrates
which
one
is
a
CH
electron
of
electron
3
O
C
O
HO

Figure 12
The beta-lactam ring (in red
pairs.
above) is a common feature of antibiotic
molecules such as penicillins.
The geometry
lone pair)
multiple
this
N
of this structure is vital to its reactivity. The
(known as a
involve
C
amide bond
An
presence
point:
pairs
the
amide bond
of
hydrolyses readily due to the
strain c aused
by the 90° bond angles.
This means that
it
c an attach to enzymes
electrons).
responsible for building bacterial cell walls.
central sulfur atom has
i
of
the
(single bond)
(which
r
o
domains,
non-bonding
t
a
domains.
be
pairs
more space than single bonds.
c an be:
triple
formula
electron
will
electron
bonding
n
•
Lewis
more
o
non-bonding
high
occupy
U
a
bonding
bonds
d
•
three
occupy
C
a
electron
The
pairs
o
and
electron domain
is
electron
i
n
domain
lone)
bonds).
Double
The term
(or
possible.
v
3.
as
p
(single
An
other
y
2.
each
r
e
from
domain,
and
two
are
One of the bonding domains is a double bond, the other is a
single bond.
u
l
f
x
one non-bonding domain
composed of a lone
pair of electrons
S
one bonding domain
number
domain
Determine
how
one bonding domain
t
Figure 13
composed of a
The sulfur atom in sulfur
single bond
dioxide has three electron domains
geometries
Predicting
the
electron
2.
explore
VSEPR.
Count
a
1.
now
E
will
using
double bond
v
O
We
composed of a
the
of
involving
VSEPR
electron
shape
two,
of
domains
a
three
and
molecule
around
a
four
electron domains
involves two steps:
central
atom
to
deduce the
geometry.
many of these
are
bonding domains
and how many are
non-
bonding domains.
127
Structure
2
Models
of
bonding
and
structure
Two domains: Linear
If
there
other.
are
two
They
bonding
a
is
straight
Molecules
in
adopt
the
positions
molecule
is
linear to
electron
at
180°
known as the
illustrate
that
pairs
in
(gure
those
14).
The
domains
angle
bond angle. This
the
central
atom
this
electron
both
domains
domain
include
geometry
beryllium
also
have
chloride, BeCl
,
a
linear
c arbon
dioxide,
H
2

Figure 14
.
These
are
shown
in
gure 15.
r
ethyne, C
2
Linear geometry has two
Cl
180°
C
H
in which the central atom
has two electron
domains. Note that double and triple bonds count as one domain
there
each
Trigonal,
A
bec ause
trigonal
planar
all
domain
on
the
domains
only
triangle,
geometry
are
bonding
of
the
three
the
molecule has
type
i
o
each
are
pairs adopt positions at 120°
is
and
c alled
c an
trigonal planar.
planar,
have
two
domains
shown
in
domains,
the
bec ause the atoms lie
possible
molecular
molecule has
trigonal
are bonding domains, and one is a non-
bent
gure
(or
V-shaped) geometry.
17.
(a)
(b)
Cl
120°
N
B
Cl
O
Cl
<120°
O
u
A trigonal planar electron domain geometry c an lead to (a) trigonal planar
molecular geometry if all three domains are bonding domains,
l
a
E
v
O
f
x

Figure 17
geometry
domain,
t
a
r
o
d
of
two
a
geometry
electron
presence of non-bonding domains:
geometry.
bonding
Examples
form
n
U
When
electron
three
domain
domains
depending
the
C
When
planar
128
the
i
n
each other
•
electron
plane.
geometries,
•
domains,
o
Trigonal planar geometry has
three electron domains at 120° angles to
a
This
v

Figure 16
on
other.
bonding
p
at
three
y
from
are
r
e
If
O
t
i
s
Three domains: Trigonal planar
n
Examples of linear molecules,
O
l
O
y

Figure 15
Be
,
2
or (b) bent
(V-shaped)
molecular geometry if one of the domains is a lone pair of electrons
The bond angle in NO
is
smaller
than
the
predicted
120°.
This
is
bec ause the
2
non-bonding
domains,
and
pair
(or
lone
therefore
pair)
takes
exerts
up
a
more
stronger
space.
repulsion than the bonding
y
180°
other
Cl
CO
P
180°
electron domains at a 180° angle to each
are
molecular
e
Examples
each
electron domain
and
2
and
repel
between
line.
with
geometry.
a
c alled
geometry
domains,
s
s
on
therefore
pairs
geometry
electron
Structure
Four
If
the
are
each
ends
four
bonding
other.
of
might
the
This
domains
domains,
electron
domains
expect
conguration.
The
four
form
arrange
the
electron
However,
this
the
domain
electron
corners
domains
would
themselves
pairs
geometry
of
to
result
a
is
positions
at
tetrahedral
109.5°
bec ause
tetrahedron.
arrange
in
adopt
c alled
angles
themselves
of
90°
in
a
square
between the domains.
tetrahedrally to maximize the bond angles and
four
electron
Tetrahedral
geometry
depending
all
four
on
domains
geometry
the
are
gives
rise
to
three
possible
other
molecular
presence of non-bonding domains:
bonding
domains,
the
molecule has
109.5°
tetrahedral
l
geometry.
three
When
only
bonding
of
two
of
four
are bonding domains and one is a non-
molecule has
trigonal pyramidal
the
four
are
domains,
each
domains
the
type
the
are
domains
molecule has
shown
in
bonding
bent
gure
(or
geometry.
domains
V-shaped)
and
two
are non-
geometry.
20.
A
tetrahedral electron
P
trigonal
pairs
bond
occupy
angles.
more
space
Methane, CH
than
,
has
4
the
bond
angle
corresponds
to
the
pairs
and
therefore
or bent
(V-shaped)
molecular geometries
domains,
non-bonding
which
pairs,
109.5°. Ammonia, NH
an
stacked
and
revolutionized
packages
bec ause they c an be easily
packed.
the food
Their invention
packaging industry
in the mid-20th century
therefore
, has one
3
smaller bond angle (107°). Water, H
has
lled,
Tetrahedral food
leads to
and
o
d
non-bonding
trigonal pyramidal
V-shaped
bonding
no
predicted
non-bonding pair, and therefore has a
two
or
are convenient
n
decreased
bent
U
Non-bonding
pyramidal

Figure 19
leads to tetrahedral,
C
tetrahedral
domain geometry
o
F
i
n
F
S
v
Si
y
t
Figure 20
F
p
r
e
Examples
the
O
•
of
domain,
t
i
s
bonding
to
n
When
y
•
has
angles
y
When
domain
at
P
geometries,
electron
domains
r
e ach
tetrahedral
•
18
between them.
e

Figure
distances
A
covalent model
s
s
You
The
domains: Tetrahedral geometry
there
from
2.2
O, has
2
even smaller bond angle of 104.5°.
i
r
o
The angles and geometries are shown in gure 21. Note that some bonds are
wedges
(
) and
t
a
represented using
dashes
(
) to better convey their 3D
shape. Wedges represent bonds that are coming out of the plane of thepage at
u
f
x
an angle and dashes represent bonds that are going into the plane ofthe page.
H
l
a
H
v
O
C
H
N
O
H
H
H
H
109.5°
H
107°
H
104.5°
E
t
Figure 21
In methane, CH
, ammonia,
4
NH
,
and
water, H
3
all slightly dierent
repulsion exerted
O,
the bond
angles are
2
due to the stronger
by the lone pairs
129
Structure
2
Models
of
bonding
and
structure
Multiple bonds
Multiple bonds (double and triple bonds) count as one domain. A double bond
represents one domain, but it contains two electron pairs. Similarly, a triple bond
is one domain composed of three electron pairs. Since multiple bonds contain
s
s
more than one pair of electrons, they exert a greater repulsion than single bonds.
The increased repulsion causes the bond angles in the molecule to deviate from
predicted values.
121.5°
H
H
For
example,
hence
each
its
c arbon
molecular
is
ethene
is
has
trigonal
three
bonding
However, the H–C–H bond angle is 117° and the H–C =C bond
Multiple bonds generally
121.5°,
as
the
greater
repulsion
exerted
by the double C=C bond
a greater repulsion than single bonds
pushes
shows
a
C–H
bonding
summary
of
the
domains.
geometry
This
of
is
molecules
geometry
domains
of non-
Molecular
Bond
geometry
angle
0
U
4
1
u
domains using balloons. If
l
you hold several balloons
a
at a central point by their
knots, they will arrange
E
v
O
f
x
You can model electron
themselves as far away from
each other as possible. The
central point where the
balloons meet represents
the central atom in the
molecule. How else might
you model electron
130
o
i
Thinking skills
domains?
planar
V-shaped)
tetrahedral
2
four electron domains
t
a
r
o
ATL
three and
(or
and
four
BeCl
2
180°
CO
2
HCN
120°
BF
, NO
3
<120°
O
, SO
3
, SO
3
3
, NO
2
2
+
109.5°
CH
, NH
4
2
Summary of the geometries for two,
bent
0
3
d

Table 2
4
1
n
tetrahedral
trigonal
C
2
0
o
3
v
planar
i
n
3
trigonal
linear
p
2
three
Examples
domains
2
two,
y
r
e
domains
linear
with
bonding
bonding
gure 22.
O
t
i
s
Number
Number of
Number of
in
n
electron domains.
Electron domain
shown
l
2
both
y
Table
away
electron domains
planar. This would suggest 120°
, ClO
4
4
+
trigonal
pyramidal
<109.5°
NH
, H
3
bent
(or
V-shaped)
<109.5°
H
O
PBr
3
O, NH
2
,
3
, SF
2
2
y
angles.
angle
in
P
bond
atom
geometry
r
and
H

Figure 22
exert
117°
C
e
C
H
Structure
2.2
The
covalent model
VSEPR
Unlike
3D
a
formulas,
VSEPR
molecular
of
molecular
molecules.
database
to
You
develop
models
will
represent the
3.
your
Relevant skills
•
Tool
2:
Digital
molecular models
Identify
and
extract
data
•
molecular
geometry
•
molecular
formula
•
bond
•
3D
domain
geometry
from a database
angle
sketch
data
of
the
(if
available)
molecule, including bonding and
Write
list
of
the
molecular
geometries
that
you
need
4.
know.
Draw
the
Lewis
of
whether
in the Tools for chemistry chapter) and extract two
5.
Organize
6.
If
geometry. If possible, rotate the images to get a sense
bond
of their three-dimensional shape.
that
your
angle
data
data
the
into
is
a
and
molecular
numbers of bonding and
non-bonding domains.
the
geometry
a.
NF
c.
BF
multiple
bond
bonds
(e.g. double or triple
angle.
electron
of
each
of
domain
the
geometry
and
molecular
following:
b.
CH
d.
BeH
3
Cl
2
3
2
2
i
e.
HCN
f.
g.
SF
h.
SO
3
NO
t
a
2
9.
Predict
2
the
bond
angles
in
each
of
the
species in
question 8.
u
f
x
10.
Methanal
is
an
organic
compound
with
the
molecular
three bonding domains and one non-bonding
domain.
l
O
7
.
Deduce
of
on
two bonding domains and two non-bonding
domains
e.
r
o
d.
two bonding domains only
8.
o
two bonding domains and one non-bonding
domain
c.
d
b.
four bonding domains
effect
p
geometry
examples
o
domain
following
the
for
effect of non-bonding domains on the bond
bonds)
n
the
table.
search
y
the
C
electron
given
U
the
geometry
v
i
n
Practice questions
suitable
available,
angle
•
molecule is polar or non-
illustrate:
r
e
•
a.
deduce the
O
t
i
s
examples of species belonging to each molecular
Identify
Then
n
y
and
Access a VSEPR database (such as those suggested
polar.
6.
each.
number of bonding domains and non-bonding
domains,
2.
formula
l
to
a
y
P
non-bonding domains.
Instructions
1.
e
1:
electron
r
Tool
•
understanding of
geometry.
•
For each, write down the following:
explore the models
s
s
in
Lewis
geometry
formula CH
Identify the number of bonding and non-bonding
around
atoms
v
geometries.
a
domains
linear
b.
trigonal
c.
tetrahedral
d.
trigonal planar
e.
bent
E
a.
with
the
following
O.
2
a.
Draw
b.
Deduce
the
c.
Suggest
Lewis
the
formula
molecular
for methanal.
geometry of methanal.
molecular
Explain
values
your
for the bond angles in methanal.
reasoning.
pyramidal
(there
are
two
possible
answers
here).
131
Structure
2
Models
of
bonding
and
structure
Linking question
How
useful
is
the
VSEPR
model
at
predicting
molecular
geometry?
s
s
Bond polarity (Structure 2.2.5)
is
related
molecules.
the
dierence
c ase
in
of
pulls
the
the
atom
the
we
electrons
has
shared
a
in
much
partial
negative
positive
separation
charge,
of
This
polar
charge,
vector
the
less
is
δ+.
charge
The
shown
between
dipole
as
an
atom
δ–,
two
and
the
strongly
the
other
non-identic al
be
the
bond.
bond,
v
and
shared
two

Figure 23
in HF
. Note that the bond
dipole in HF c an be represented
as a vector
the
the
shied
the
involved
bond,
in
the
towards
dierence
formation
of
the
pure covalent bond. The
however,
hydrogen
hydrogen does. This
atom
(uorine)
(hydrogen) adopting
by
atoms
a
is
vector
more
towards
the
c alled a
(gure23).
covalent
is
more
electronegative atom in
electronegativity
charge
the
arrow starts as a plus sign
points
the
in
c ase,
than
bonded
The
there is an
2
the
one
atom
from
bond
between the
are
identic al, the
distributed
evenly about
meaning that it is non-polar and has no dipole moment.
R ank
the
Explain
following
bonds
in
order
C–H,
why
Br
is
non-polar,
of
C–C,
increasing polarity:
C–F, C–O
whereas
HBr
is
polar.
2
u
a
v
E
132
l
O
shared.
is
greater
than
with
,
results
o
f
x
12.
bonding pair of electrons is shied
electrons are transferred, not
The
greater this shi.
atoms
t
a
large (greater than 1.8), then the
are considered ionic because
electrons
said to be a
covalent
11.
between the two atoms is very
electronegative atom. Such bonds
is
necessarily
Practice questions
dierence of electronegativity
almost entirely towards the more
two
bond
the
i
r
o
As seen in Structure 2.1, if the
d
(arrow)
If
of
bond.
n
bond
U
covalent
atoms,
Representation of the
molecular electrostatic potential of the polar
pair
covalent
C
the
o
The
δ
i
n
F
+
δ
not
represented
electronegative atom in the bond.
H
is
electronegativity
more
c an
along
in
This
covalent bond,
moment
arrow
electronegative
pair
pair.
p
at
not
bonded atoms.
y
dipole moment.
greater
r
e
The
the
bonding
electron
describe as a
partial
the
H
a
of
is
O
a
electronegativities
bond
Bond polarity
t
i
s
what
adopting
covalent
between the two atoms in the bond.
uorine
it
a
n
so
distributed within bonds and
in
l
The
to
are
electrons
y
in
leads
electrons
of
identic al atoms, such as the two uorine atoms in F
sharing
atom,
F
of
equal
HF.
way
pair
Linking question
What properties of ionic compounds might be expected in compounds with
polar covalent bonding? (Structure 2.1)
y
the
the
P
In
equally
to
shared
r
shared
The
e
Polarity
Structure
2.2
The
covalent model
Molecular polarity (Structure 2.2.6)
Molecular
polarity
distribution
electron
and
a
μ. The
D.
value
is
an
to
of
the
partial
dipole
such
as
on
but
A
negative
charge
important
depends
are
a
polarity,
molecule.
on
the
moment
trait
of
it
describes
the
electron
molecule is polar when the
charge on one end of the
other.
is
oen
molecules
This
results in a
dipole
reported in the unit
bec ause
it
gives
rise
to
many
volatility, solubility and boiling point.
both
bond
polarity
and
molecular
geometry.
given in table 3.
e
polarity
bond
whole
positive
properties
examples
to
r
Molecular
leads
partial
Polarity
characteristic
the
s
s
moment,
Some
similar
distribution
molecule,
debye,
is
throughout
moment / D
bond dipole
O
bond dipole
polar
1.85
polar
1.01
O
2
net dipole
trichloromethane,
H
bond dipole
C
Cl
Cl
triuoride,
i
n
boron
non-
F
polar
bond dipole
3
B
dioxide,
non-
bond dipole
CO
polar
H
H
C
H
H
0
polar
H
u
C
non-
non-polar molecules
l
Examples of polar and
a
E
v
O
f
x

Table 3
H
0
o
6
i
H
2
t
a
r
o
ethane,
d
2
n
c arbon
F
U
F
C
0
C
BF
o
net dipole
v
Cl
y
3
p
r
e
CHCl
O
t
i
s
water, H
n
l
Polarity
y
Lewis formula
y
P
Dipole
Molecule
133
Structure
2
Models
of
bonding
and
structure
Molecules are polar
This
may
dierent
they
the
occur
due
polarities.
contain
polar
when their
to
the
Both
water
bonds
that
do
of
the
molecule,
happens
not
c ancel
out.
net
This
the
Examples
contain
polar
other
include
bonds
These
out.
bonds
means
the
that
of
the
geometry
bonds
dierence
around
each
and
triuoride
are
atoms
virtually
between
c arbon
atom
hexane, C
hydroc arbons
3
cyanide, HCN
So
State
why
we
have
CO
is
non-polar.
and
a
only
emulsiers,
soaps
and
2
and
explain whether PH
membranes.
is
or
polar
non-polar.
and
shows
regions
H
H
C
C
H
C
S
H
hydrophilic
head
A diagram
hydrophilic regions
l
a
the
way that they
a
net
to
well
bonds.
The
In
addition,
dipole
of
the
zero,
Long
molecules
interesting
features
of
a
tetrahedral
or
very close
the
phospholipid
detergent
c an
have a polar
applic ations including
molecule.
bilayer in cell
C an
you identify the
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
showing the hydrophobic and
molecule
make
it
“infrared
(IR)
H
H
C
hydrophobic tail
(Structure 3.2)
c arbon–c arbon
electronegativity.
non-polar.
many
as
are
Hydroc arbons
Linking question
What
dioxide
molecule?
H
of a synthetic detergent molecule,
LHA
E
v
O
134
H
u
f
x

Figure 24
i
H
H
t
a
r
o
d
O
C
o
O
in
as
typic al
H
+ –
Na
a
in
molecules.
leading
detergents,
24
a
c arbon
bec ause of the small
atoms.
generally
C
C
non-polar
small
region,
n
U
H
O
Figure
i
n
3
polar
discussed
non-polar
and
same
non-polar
two
results
are
the
o
Explain
15.
region
14
v
14.
far,
H
6
electronegativity of
y
d
Therefore,
the
have
p
hydrogen
zero.
r
e
c
to
in
such
bonds are non-polar.
c arbon–hydrogen
both
4
ammonia, NH
in
results
dioxide. Both
c arbon) such as ethane contain two types of
bonds
bec ause
c arbon–hydrogen
arrangement
n
non-polar
electronegativity
b
across
l
The
methane, CH
dipole
O
a
net
c arbon
dierence
y
molecules is polar or
c arbon–c arbon
and
presence of polar bonds.
when all their
are
their
positioned
boron
and
hydrogen
and
triuoride
are
(compounds
of
equal
boron
Molecules are non-polar
bonds
non-polar:
a
have
bec ause
bond dipoles c ancel each other out.
are
bec ause
polar
This
despite
dipoles
t
i
s
following
in
bonds
polar
active”?
H
H
y
non-polar,
bonds:
each of the
results
the
are
r
involved.
each
when their
bond
Practice questions
whether
3)
P
both
when
dipole.
atoms
c ancel
Determine
bec ause
(table
e
no
molecules
13.
or
trichloromethane
s
s
This
the
and
molecule.
Molecules are non-polar
in
bond dipoles do not c ancel each other out.
geometry
Structure
Covalent network structures
2.2
The
covalent model
Chemists talk about two types of
stability: kinetic (Reactivity 2.2)
(Structure 2.2.7)
and thermodynamic (Reactivity
Covalent bonds give rise to two dierent types of structure: molecular and
1), and diamond illustrates the
covalent network
(also known as giant covalent). The atoms in both cases are
s
s
dierence well. At high pressures,
joined by covalent bonds, but the two structures are very dierent in terms of their
such as those occurring deep
properties. Molecular substances consist of discrete groups of covalently bonded
inside the Earth’
s crust, diamond
atoms called molecules. By contrast, covalent network structures contain atoms
e
is the most stable allotrope.
that are held together by covalent bonds in a continuous three-dimensional
At standard temperature and
lattice. Examples of covalent network structures include silicon, silicon dioxide,
r
pressure, graphite is the most
and most of the allotropes of carbon, such as diamond and graphite.
thermodynamically stable allotrope
y
P
of carbon. Under these conditions,
Allotropes of c arbon
the transformation of diamond
group
of
but
substances
have
its
allotropes
known
dierent
as
forms
include
fullerenes.
chemic al
and
known as
diamond,
They
physic al
are
all
allotropes.
graphite,
due
into graphite is thermodynamically
graphene, and
composed
properties
C arbon is
to
of
spontaneous, but is so slow that
c arbon
their
diamonds eectively last forever.
dierent
arrangements.
Diamond is kinetically stable but
O
structural
and
structural
t
i
s
atoms
element,
dierent
n
a
such
have
y
one
elements
l
Some
thermodynamically unstable (we
describe it as being metastable).
(a)
(b)
You can read more about
p
o
C
n
i
d
t
a
r
o
u
l
f
x
a
v
O

Figure 25
(e)
o
(d)
y
r
e
v
i
n
U
(c)
spontaneity in Reactivity 1.4
The allotropes of c arbon:
(a) diamond,
(b) graphite,
(c) graphene,
(d) c arbon nanotubes and
(e) buckminsterfullerene, C
60
E
135
Structure
2
Models
of
bonding
and
structure
In
diamond,
each
arrangement.
oen
used
in
It
is
Diamond’
s
make
jewelry.
high
Diamond
through
the
is
cutting
is
a
tools
such
and
are
all
lattice
as
atoms
For
in
this
a
tetrahedral
reason, it is
saws, polishing tools and dental
mean
conductor
energy
and
other
known.
loc alized
vibrational
regular
four
durability
electric al
electrons
to
substances
index
poor
conductor:
highly
bonded
hardest
in
the
c arriers
strong
that
it
is
also
used to
bec ause it has no mobile
bonds.
c alled
It
is
s
s
thermal
atom
the
refractive
particles — the
excellent
well
of
heavy-duty
drills.
charged
c arbon
one
however an
phonons
travel
covalent bonds.
e
r
l
n
y
South Afric a
p
r
e
Kimberley,
O
The Big Hole:
a former diamond mine in
Graphite is composed of layers of sheets made of carbon atoms. Each carbon
v
atom is bonded to three other carbon atoms in a hexagonal arrangement where
o
i
n
the geometry around each carbon atom is trigonal planar. The carbon atoms in
graphite are bonded such that one electron per carbon atom is delocalized. These
C
delocalized electrons are free to move in the planes above and below each sheet
and therefore graphite is a good electrical conductor. While the covalent bonds
U
between the carbon atoms within the sheets are strong, the forces of attraction
n
(called London (dispersion) forces) between the sheets are weak. This means that
the sheets can be separated easily, making graphite a good lubricant, as well as an
Graphene
and
time,
to
conductor.
with
bendable
l
a
E
v
O
Fullerenes
hexagonal
these
are
are
vast
are
and
a
a
single
said
It
is
to
range
group
be
also
strong.
electronic
utilized in
less
a
sheet
exible,
isolated
Graphene is thus one-atom
Like
in
2004,
graphite,
transparent
graphene
applic ations,
from
it
is
and,
is
a
an
at
excellent
the
same
promising
desalination
new
technology
displays.
of
c arbon
rings.
allotropes
Some
with
with
atoms
fullerenes
c arbon nanotubes.
objects
graphite.
lightweight,
potential
nanotechnology.
and
of
two-dimensional.
First
of
pentagonal
known as
molecules
Due
to
their
Nanotechnology
dimensions
of
less
form
arranged in interlinking
long
size,
hollow
cylinders:
c arbon nanotubes
involves the use of atoms,
than
100 nm (about 1000 atoms or
across).
nanotubes
strength
C arbon
is
due
to
and
good
are
the
nanotubes
graphite
are
essentially
therefore
extraordinarily
C arbon
136
is
electric al
u
A microscope image of a
is
t
a
thick
material,
dentist’s drill showing the diamond chips
o
causes parts of the graphite sheets to come o, leaving a mark on the paper.
i
d
r
o
f
x

Figure 27
ideal material for pencil leads. When you use a pencil to write, the force applied
used
have
graphene,
electric al
to
strong
reinforce
covalent
potential
the
of
materials.
holding
applic ations
presence
conductors.
composite
bonds
in
the
electronics
deloc alized
Their
c arbon
exceptional
atoms
together.
bec ause, like
electrons
means that they
y
P
y
t
i
s
u
Figure 26
Structure
Buckminsterfullerenes, or
buckyballs,
have
a
covalent
molecular
formula C
,
the
atoms
in
buckminsterfullerene
are
arranged
in
The
covalent model
structure. With
Silicon
the
2.2
and
c arbon
are
found in
hexagons and
60
group
pentagons
to
suggest
a
very
familiar
shape:
a
football.
Along
with
other
14
similar
fullerenes,
buckyballs
could
have
exciting
new
roles
in
medicine
as
drug
of
its
molecular
structure, C
has
a
low
boiling
point:
they
have some
properties.
Other
patterns in
c arriers.
the
Bec ause
hence
spheric al
periodic
table
are
explored in
overcoming the
60
s
s
Structure 3.1.
weak
intermolecular
forces
of
attraction
does
not
require
much
thermal
energy.
Silicon and silicon dioxide
four
that
the
and
where
each
silicon
arrangement.
This
atom
bonds
in
boiling
The
Si–Si
the
lattice
point.
bond
is
4
we aker
Therefore,
in
result
Table
the
strength
than
Si–Si
also
in
high
contrasts
its
bond
helps
to
strength,
some
C–C
is
of
more
explain
re active
the
114
346
226
3265
between
E arth’s
O
in
1:2
to
insoluble
in
atoms.
of
silic a
of
poor
a
and
bond
of
it
is
table
group,
anticipate.
Diamond
Double
unusual
found
the
and
is
a
triple
For
the
properties
example,
poor
electric al
bonds
are
between silicon atoms.
and
sand
(table
4)
shows that single bonds
a
species containing Si–O bonds.
forms
.
a
covalent network containing
E ach
silicon
atom
is
bonded
2

Figure 28
each
oxygen
atom
is
bonded
Structure (le) and
silicon dioxide, SiO
form
of
silicon
dioxide
is
crystals of
covalently
quartz
,
known as quartz. The
2
(gure 28).
photograph also contains crystals of iron
and
glass.
disulde,
FeS
,
commonly known as
2
“fool’s gold”
dioxide
are
high
conductor
in
that
formula SiO
crystalline
has
metalloid.
strengths
mineral
atoms
include
silicon
water
a
is
hence
The
but
might
strong. Silicon is one of the most abundant elements
much
oxygen
E
is
ratio,
four
properties
it
C–O
a
forms
quartz,
and
v
The
a
atoms
very
a
semiconductor.
l
Other
crust
silicon
O
two
and
are
silicon dioxide,
covalently
to
O
is
periodic
one
u
and
Si–O
f
x
Silic a, or
Si
the
and
c arbon
silicon
same
as
i
the
Si
a
the
t
a
between
is
r
o
Comparing
while
silicon
to
similar
where X = Si or C
o
while
as
C
non-metal,
not
d
common
belonging
are
silicon,
n
a
conductor,
i
n
is
silicon
466
U
c arbon
despite
and
787
358
A selection of features of c arbon and
Interestingly,
in
1086
o
–1
p
1414
4827
enthalpy
the
melting
r
e
3500
Boiling point / °C
c arbon
than
dierent
of
silicon
v
Melting point / °C
of
as
O
Si
75
ionization energy / kJ mol
as
y
C (graphite)
m
–1

Table 4
well
fe atures
points.
enthalpy / kJ mol
X–O bond
as
the
counterpart
–12
First
bonded to
n
radius.
dierence
Atomic radius / 10
X–X bond
is
arrangement is similar
l
atomic
This
boiling
lattice
tetrahedral
t
i
s
and
cov alent
point
silicon.
larger
bond.
a
y
a
C–C
and
in
y
melting
c arbon
atoms
c arbon atoms in diamond.
extensive
high
has
three-dimensional
P
The
of
a
silicon
r
to
forms
other
e
Silicon
of
those
melting
heat
and
we
associate
point.
In
its
with
solid
sand:
it
crystalline
is
hard,
form,
electricity.
Linking question
Practice questions
Why
16.
Explain
why
graphite
and
graphene
conduct
electricity, but diamond
does not.
17
.
Explain
why
are silicon–silicon bonds
generally
c arbon
diamond
and
silicon
dioxide
weaker
bonds?
than
c arbon–
(Structure
3.1)
have high melting and
boiling points.
137
Structure
2
Models
of
bonding
and
structure
Activity
Summarize
the
properties
of
some
of
the
covalent
using
Graphite
a
table
like
the
Graphene
one
C
below.
fullerene
Silicon dioxide
s
s
Diamond
substances
60
Element(s)
Arrangement of atoms
e
Electric al conductivity
r
which
to
their
forces
name
substances.
The
and
of
polarity
collectively
type
the
c ause
water
some
that
of
are
weak
Figure
30
water
water
water
forces
such
as
δ
polar
l
covalent bond
a
intermolecular
of
through
to
ice
are
and
termed
than
strong
stay close to
What
break
answer
of
in
exactly
away
each
from
c ase is
chemic al bonds,
attraction that occur
dierence
+
force
forces
cubes?
in
weaker
forces
the
+
δ
ice
The
forces.
δ
London
volatility, solubility and boiling point of
electrostatic
highlights
atom
inside
are
intermolecular
molecular
molecules contain two
molecules
it?
now turn our
forces include
molecules
surrounding
Intermolecular
properties
the
to
intermolecular
oxygen
molecules
the
water
the
central
apply
y
intramolecular
a
three
will
dipole–dipole forces
p
o
i
and
to
rst
We
known as
forces depend on the size
intermolecular
o
forces.
forces
the
the
know
forces
liquid
physic al
molecules.
t
a
occur within molecules.
E
v
O
138
aect
between
u
bonds,
Intermolecular forces occur between
molecules
they
Intermolecular
Intramolecular forces, such
f
x
as covalent
d
Figure 30
but
about
when
molecular substances.
r
o
▸
intermolecular
what
the
You
bonded
n
the ice cubes
join
forces
intermolecular
of
C
Intermolecular forces hold
the molecules together in water and inside
But
How
of
Types
r
e

Figure 29
and
water.
each,
overcome
cube
U
the
bonds.
another?
being
of
i
n
covalent
glass
atoms
bonding.
molecules,
intermolecular
Collectively
v
a
hydrogen
attraction that hold atoms and ions
between
strength
Waals forces
Consider
is
suggests,
and
of
dipole–induced dipole forces,
hydrogen bonding.
one
attraction
molecules.
(dispersion) forces,
van der
of
forces
referred to as
O
forces. As
the
are
forces (Structure 2.2.8)
discussing
t
i
s
attention
been
n
together,
have
attraction
δ
between
intermolecular
y
we
l
far,
y
So
P
Intermolecular
Structure
When
molecular
between
oen
the
used
as
molecule
indic ators
do
Structure 1.5,
that
no
we
high
When
of
break
during
discussed
forces
in
deviates
account
the
the
or
sublime,
Therefore,
phase
the
are
ideal
for
from
gas
in
ideal
volume
the
which
gas.
the
gas
The
forces
boiling
points
weak intermolecular interactions
are
between I
2
molecules
break
on changing state
covalent bonds in
c arries
This
ideal
real gas
and
(gure31).
behaviour,
of
intermolecular
melting
strength.
law,
the
which
forces using a
actual
force
changes
present
temperatures,
gas
boil
intermolecular
intermolecular
into
negligible
a
of
melt,
overcome.
is
gas
we
assumption
equation
must
model.
particles,
the
generally
true
c an
correct
at
low
frequently
for the
Real gas models
which
is
r
take
and
are
covalent model
e
used.
presence
also
not
intermolecular
pressures
be
substances
molecules
The
s
s
In
the
2.2
considered
ideal gas model.
strong covalent bonds
500 K
2
molecule
between
do not
break on changing state
Intermolecular forces of
O
t
i
s

Figure 31
2
in I
n
y
atoms
l
200 K
y
P
3
attraction are overcome when a molecular
substance changes state,
but
the covalent
bonds remain intact
pV
1000 K
ideal gas
ideal gas behaviour at
lower temperatures. The
o
dierent
d
behaviour at
n
Gases deviate more from
horizontal black line represents an ideal gas.
o
900
p (atm)

Figure 32
C
600
U
300
v
i
n
0
p
1
y
r
e
RT
The curves show the deviations from ideal gas
temperatures for a real gas
i
t
a
r
o
London (dispersion) forces (LDFs)
All
molecules
experience
other
a
which
another
dipole
means
molecule.
electrons
when
negative
around
that
one
one
These
the
(dispersion)
from
charge.
a
of
a
forces
temporary
side
c arries
London
induced
a
partial
c auses
dipoles
(or
LDFs),
which
are
instantaneous dipoles. A
(dispersion)
molecule
molecule.
E
v
O
dipoles,
in
has
partial
London
resulting
u
molecule
forces
l
f
x
intermolecular
(or
occur
positive
forces
induces)
due
to
charge, and the
involve
a
the
induced
temporary dipole
random
movement

Figure 33
Intermolecular forces of
attraction are overcome when iodine
sublimes,
and
are formed when iodine is
deposited on a surface
139
Structure
2
Models
of
bonding
and
structure
In
a
simple
non-polar
molecule
such
as
hydrogen, H
,
the
electron distribution
2
is
on
average
around
of
within
electrons
perfectly
the
in
a
molecule.
molecule,
symmetric al.
with
one
in
the
If
the
of
the
There
molecule
34).
we
However,
could
electron
Instead,
region
negative (δ –).
elsewhere
(gure
we
somehow
would
a
also
see
a
be
a
would
somewhat
having
positive
are
more
region
partial
constantly
very unlikely be
unequal
electrons,
of
lower
electron
rendering it
electron density
charge (δ+).
e
r
H
H
is symmetric al
2
t
i
s
n
l
y
the electron distribution in H
This temporary dipole then induces further temporary dipoles in the molecules
O
around it. Due to electrostatic repulsion, a region of partial negative charge
(δ–) on one molecule will repel the electrons in neighbouring molecules.
This temporarily creates regions of partial positive charge (δ+). The resulting
y
r
e
electrostatic attraction between the δ– region on one molecule and the δ+ region
on the next is termed a London dispersion force. However, this arrangement
p
is temporary. In the next moment of time a dierent pattern of induced dipoles
+
δ
H
H
n
a
–
u
l
+
H
force
a temporary dipole is present
in a molecule,
A weak London dispersion force forms between the opposite
+
+
δ
δ
δ
δ
+
δ
δ
δ
+
δ
+
+
δ
–
δ
δ
δ
δ
+
δ
+
E
v
O
f
x
δ
δ
o
–
δ
+
δ
–
δ
dispersion
any given moment in time,
t
a
–
–
δ
partial charges
δ
δ
At
inducing a dipole in another.
i
d
r
o
+
δ
+
δ
H
London

Figure 35
o
U
C
i
n
δ
v
willemerge.
δ
+
δ
δ
–
δ
–
+
δ
δ
δ
+
+
δ
δ
+
δ
–
δ
δ
δ
+
δ
+
δ
+
δ
+
δ
+
δ

Figure 36
Dierent
arrangements of LDFs between molecules,
instantaneous dipole on one molecule and
140
an induced
which result
dipole on an adjacent
from
interactions between an
molecule
y
H
On average,
P
H

Figure 34
moving
freeze time and take a photo
distribution
would
molecule
with
electrons
s
s
density,
slightly
symmetric al
Structure
There
are
two
main
electrons
and
the
how
the
easily
electron
polarizability,
The
the
number
electron
elements,
higher
larger
more
boiling
is
all
distorted
of
less
you
an
form
down
LDFs: the number of
electric
of
eld.
non-polar
the
attracted
polarized
as
by
of
polarizability
the
The
molecule:
greater
forces.
which
going
and
easily
points
strength
aect the
dispersion
increases
become
become
the
the
They
and
descend
group.
to
the
the
the
diatomic
As
a
nuclei.
strength
molecules.
result, the
This
of
means that
LDFs
increases,
group (table 5).
r
Boiling point / °C
–188.1
uorine, F
2
Br
58.78
Increasing
iodine, I
184.4
2
LDF

Table 5
Boiling point
increases down group
17 due to an increasing number of
resulting in stronger LDFs
greater
electrons
polarizability

Figure 37
boiling
organic
point
also
increases
compounds:
each
for
successive
molecule
has
members of a
one
more CH
n
of
be
molecular
quantied
in
terms
in
of
and
mass,
terms
of
a
larger
so
we
their
oen
contradictory
not
c ausal.
each
is
the
responsible
requires
Therefore,
proportional
for
the
in
the
number
of
The
have
Homologous
one
series
and
isomers of
before
organic
compounds
are
discussed
Molecular size
in
molecular
to
larger
in
in
the
of
electrons.
the
Structure 3.2.
mass when
molecules is
number
the
number
of
of
This
of
increasing
may
But
here
electrons is
accompanied
protons and
mass.
overall
between
with
mass.
number
molecular
maintain
number
of
increases
negligible
and
greater
increase
increase
the
to
electrons.
relationships
strength
mass
mass
electron
any
increase
understand
LDF
electrons
molecular
electrons.
an
to
say
increasing
greater
of
try
oen
bec ause
number
proton
molecule.
E
neutrons
an
between
R ather,
greater
to
v
relationship
a
due
and
we
u
mass,
the
by
patterns
a
seem
for
example,
l
O
molecular
For
of
t
a
f
x
look
variables.
refer
LDFs.
Patterns and trends
Scientists
number
i
molecules
r
o
comparing
size
o
c an
greater
d
hence
point
IMFs
homologous series
than
2
it,
higher
boiling
overcome
LDFs
of molecular size on LDF strength and boiling point
U
The
The eect
energy
needed to
C
i
n
size
more
stronger
o
more
y
v
larger
molecular
p
r
e
electrons,
strength
O
t
i
s
2
n
2
bromine,
l
–34.04
y
chlorine, Cl
y
P
Substance
covalent model
e
to
stronger
17
increase
shape.
distribution
electrons
clouds
electrons
leading
the
group
of
that
The
s
s
Consider
the
the
factors
molecular
2.2
However,
neutrality of the
protons
always
leads to a
electrons.
141
Structure
2
Models
of
bonding
and
structure
Compare
the
boiling
points
of
the
two
isomers of C
H
5
CH
CH
3
CH
2
and
2,2-dimethylpropane
:
pentane,
12
(CH
3
)
3
C
(table
6
and
Space-lling models show
pentane (top) has a greater area of
contact
gure 38).
4
e
that
CH
2
s
s
u
Figure 38
CH
2
for LDFs than 2,2-dimethylpropane
r
(bottom)
contain
dierence
is
in
their
the
same
to
with
shapes.
number
Pentane
of
electrons
molecules
are
and
are
long
non-polar. The
and
are
therefore able
other
across
the
length
the
better
of
9.5
area
pentane.
As
a
interaction
result,
temperature.
bec ause
pentane
has
Molecules
of
of
a
so
they
pack
together
room
is
limited,
the
LDFs
less
in
eciently.
As
the
rst
point
H
6
H
propane, C
3
will
weaker, so it is a gas at
shown
below.
Describe
observe.
161
o
forces.
i
dipole
LDFs
more
42.1
of
2,2-dimethylpropane
respectively.
between
we
are
and
between the
88
CH
3
If
rounder
0.5
points
36.1 °C,
occur
electrons.
are
are
interaction
10
t
a
contrast London
forces and dipole-
8
2-methylbutane, CH
u
a
induced
l
and
(dispersion)
boiling
and
LDFs
v
E
Compare
d
r
o
f
x
O
Activity
4
The
you
H
butane, C
2.
alkanes
n
U
4
2
there is
Boiling point / °C
methane, CH
ethane, C
four
trend
C
Alkane
is,
molecules of
y
for
boiling
o
data
the
p
r
e
point
explain
i
n
and
v
Boiling
the
2,2-dimethylpropane
temperature.
Data-based questions
1.
that
between
relatively high boiling point and is a liquid at
isomers of pentane
molecules
contact
2,2-dimethylpropane
Boiling points of the two
compact,
molecule;
O
room
of
t
i
s
2,2-dimethylpropane
large
the
n
full
36.1
a

Table 6
each
y
pentane
interact
l
point / °C
have
all
a
Predict
CH(CH
2
types
a
of
value
)
3
.
made
pentane
are
9.5 °C
Explain
your
answer.
2
molecules
substance
and
for the boiling point of
up
bec ause
of
only
all
molecules contain
non-polar
molecules, only
occur.
Dipole–induced dipole forces
LDFs
are
forces
of
attraction
between
Dipole–induced dipole forces
occurring
presence
between
of
temporary
of
a
a
polar
permanent
dipole
intermolecular
in
the
force
are
a
molecule
dipole
in
temporary,
type
and
the
neighbouring
attracts
of
a
nearby
polar
instantaneous, dipoles.
intermolecular
non-polar
molecule
non-polar
non-polar
or
related
oxygen
induces
molecule.
force
molecule. The
For
the
formation of a
example, this type
molecules, O
,
to
polar
water
2
molecules.
aqueous
142
Dipole–induced
solubility
of
dipole
oxygen
is
forces
relatively
are
low.
weak,
which
explains
why the
y
isomers
P
Both
Boiling
Isomer
Structure
2.2
The
covalent model
Dipole–dipole forces
Dipole–dipole
temporary
therefore
forces
dipoles.
involve
When
experiences
a
permanent
molecule
dipole–dipole
is
dipoles,
whereas
LDFs
result
+
δ
from
forces
of
δ
Cl
H
attraction with neighbouring polar

Figure 39
Hydrogen chloride, HCl,
(gure 39).
molecules are polar and
attract
each other
due to dipole–dipole forces
Hydrogen chloride, HCl, and diatomic uorine, F
, have similar sizes and
2
comparable molecular masses (table 7). They both therefore experience LDFs of a
e
similar strength. However
, HCl molecules are polar and therefore experience dipole–
are stronger than those between F
r
dipole forces in addition to LDFs. The intermolecular forces between HCl molecules
molecules, so HCl has a higher boiling point.
2
38.00
Intermolecular forces
–188.1
–85.05
(dispersion)
(dispersion)
Table 7
Diatomic uorine and
hydrogen chloride have similar molecular masses but
dierent boiling points
of
the
following
species
B.
c arbon
dioxide,
C.
c arbon
monoxide,
experience
dipole–dipole
2
CO
2
c arbonate
ion,
CO
3
The
molecular
masses
of
ICl
and
Br
are
very
2
Draw
a
diagram
arrange
and
to
58.8 °C,
show
themselves
how
when
propanone
near
one
another.
a
are
strong
dipole
bonded
to
a
intermolecular
involving
highly
forces
hydrogen.
that
When
electronegative
u
f
x
covalently
bonds
strong
Label
t
a
contains
this
dierence.
COCH
3
i
r
o
Hydrogen bonding
Hydrogen
Explain
molecules, CH
a
atom
any
, might
3
dipole–dipole
forces
o
occur.
similar, while their boiling
respectively.
d
that
97.4 °C
n
20.
are
U
points
C
2–
D.
19.
i
n
CO
forces?
o
oxygen, O
v
A.
y
Which
p
r
e
Practice questions
18.
O

t
i
s
and dipole–dipole
y
36.46
n
London
London
2
HCl
l
F
Boiling point / °C
y
Molecular mass
P
Molecule
s
s
molecules
+
δ
δ
Cl
H
polar, it has a permanent dipole and
form
when
a
molecule
hydrogen atom is
such
as
oxygen,
nitrogen or
uorine, the bond between them is very polar. The electrons in the covalent bond
are
drawn
strong
another
more
electrostatic
electronegative
intramolecular
electronegative
on
a
a
the
charge (δ+)
v
O
form
towards
positive
l
partial
the
interaction — a
atom.
hydrogen
This
bonds
E
covalent
atom,
hydrogen.
is
c an
resulting
hydrogen
hydrogen
usually
also
This
found
in
a
bond — with
on
a
considerable
atom
c an then
the
dierent
electrons of
molecule, but
exist in certain situations.
bond
t
Figure 40
hydrogen bond
form
electronegative
electronegative
Hydrogen bonds
between the hydrogen in a
very polar bond
and
electrons on
hydrogen
atom
atom
a highly electronegative atom
atom
(e.g.
F,
O, N)
(e.g.
F,
O, N)
δ+
lone
electron pair
143
Structure
2
Models
of
bonding
and
structure
Hydrogen
1.
water
bonds
occur,
for
example,
between:
molecules
2.
ammonia (NH
3.
hydrogen
4.
water
)
molecules
3
and
molecules
dimethyl
ether
molecules,
(CH
)
3
These
hydrogen
bonds
are
depicted
with
blue
dashes
in
O
2
gure 41.
F
H
H
H
H
types
the
name
they
of
are
suggesting
otherwise,
intermolecular
intermolecular
force,
forces.
but
much
p
Enthalpies of hydrogen bonds
Single
single covalent
covalent
bonds compared
to
way
is
sucient
look
at
to
remember
what
to
overcome
1
(~20 kJ mol
the
),
but
i
in
water
happens
o
good
is
n
U
not
that
the
the
hydrogen
when
water
hydrogen
covalent
chemic al
covalent bonds.
–1
force enthalpy / kJ mol
20–40
150–600
bonds
boils.
At
bonds
bonds
are
weaker
100 °C,
between
between
the
than
water
oxygen
chemic al bonds
energy
available
molecules
and
hydrogen atoms
1
molecules
themselves
(463 kJ mol
).
Practice questions
u
l
a
E
v
O
f
x
21.
Which
their
two
of
the
following
substances
c an
form
hydrogen
bonds
between
molecules?
A.
hydrogen
B.
uorine, F
C.
methanol, CH
uoride, HF
D.
uoromethane, CH
2
OH
3
F
3
22.
Draw
a
diagram
molecules, CH
to
show
a
hydrogen
bond
between two methanol
a
hydrogen
bond
between
a
water
molecule and a
bond
between
a
water
molecule and a
OH.
3
23.
Draw
a
diagram
methanamine
to
show
molecule, CH
NH
3
24.
Draw
a
diagram
methanal
to
show
molecule, CH
a
2
hydrogen
O.
2
144
not
stronger than other
t
a
r
o
d
A
bonds
than
are
3
Typic al bond or intermolecular
C
Table 8
and
bonds
generally
o
v
i
n
u
bond
are
weaker
Bond or intermolecular force type
Hydrogen
hydrogen
They
C
3
CH
y
bonds:
r
e
Despite
H
H
n
H
O
O
t
i
s
H
Hydrogen bonds c an form
l
F
H
between a variety of dierent molecules
y
N
H
y
P
H
u
Figure 41
H
O
N
O
e
r
H
H
O
H
molecules
(HF)
s
s
uoride
Structure
2.2
The
covalent model
150
H
O
2
100
s
s
50
Te
r
2
period
2 3
4
nber
6
3
Se
2
H
S
2
3
y
SnH
H
4
3
Hr
3
HCl
GeH
SiH
4
4
4
v
boiling
points
of
the
group
14
hydrides
(CH
, SiH
4
cloud.
you
increasing
down
the
Descending
bottom
trend
up.
predicts
group
the
due
This
to
15
in
is
boiling
30 °C.
hydrogen
the
group
hydrides,
expected
16
Again,
this
the
molecule
boiling
E
therefore
precisely
and
what
of
and
, PH
rise
of
in
the
the
around
130 °C,
electron
data
show in the
H
O
AsH
point
mass.
is
and SbH
3
c an
O,
point
to
form
of
O
, in
3
H
from
when
in
H
phosphine,
Extrapolating
fact
it
the
is
trend
H
O
O
100 °C
H
H
H
H
higher-than-expected boiling point is due to
between
group
due
)
4
therefore an
3
boiling
molecular
,
and SnH
4
down the
17
and
ammonia
molecules.
O
H
hydrides
hydrogen
are
similar
uoride,
to
HF,
that
have
of
group 15:
very high
2
v
water
The
water, H
molecules.
single
point
bonds
a
O
points.
LDFs,
3
increasing
l
2
of
polarizability
hydrides, NH
u
trends
period
their
a
approximately
presence
boiling
A
the
see
,
f
x
the
expected.
and
stronger
3
higher,
The
of
We
stibane, SbH
ammonia
the
are
, GeH
4
increases
gure 42.
3
to
is
electrons
there
mass
o
curve
point
molecular
i
,
of
group,
The
t
a
next
PH
the
number
the
r
o
the
at
of
group?
d
look
to
boiling
curve
the
The boiling points of the
14 to 17 hydrides show the eect of
hydrogen bonding
n
increasing
Now
go
U
group,
as
C
the
i
n
do
Figure 42
group
o
–200
change
t
y
CH
p
r
e
–150
O
t
i
s
PH
–100
How
l
H
NH
–50
n
g n i l i ob
H
SbH
y
1
e
H
0
P
C° / tniop
HF
the
presence
of
hydrogen
bonds
H
O
water
is
to
four
very
hydrogen
high:
bonds
100 °C.
O
between
H
up
H
H
(gure 43) and

Figure 43
H
Water molecules c an form up
to four hydrogen bonds each
145
Structure
2
Models
of
bonding
and
structure
Hydrogen
higher
water
the
bonds
densities
bec ause
formation
this
ice
layer
The
to
is
that
between
dense
of
ice
lakes
notable
when
than
on
properties.
liquid.
liquid
lake
water.
surfaces
freezing
into
Most
Solid
This
the
have
(ice) oats on liquid
unusual
during
solid
substances
water
property
winter.
blocks
of
ice,
formation
of
a
leads to
Insulation
from
allowing aquatic
survive.
bonding
network
more
O–H
water
of
in
ice
results
molecules
random,
covalent
which
bonds
in
the
(gure
results
in
water
44).
in
a
are
In
the
higher
shorter
regular,
liquid
density
than
H
phase,
very
the
ordered
hydrogen
compared
-
-
-
r
Note
its
than
O
molecules.
e
bonding
solid
less
layers
prevents
hydrogen
open-c avity
is
water
s
s
ecosystems
it
of
give
when
to
ice.
hydrogen bonds
y
C
i
n
n
U
o
i
t
a
u
l
a
v
The hexagonal arrangement
of water molecules in
the ice lattice leads to the characteristic shape of snowakes

Figure 46
Extensive hydrogen bonding is responsible for
the high surface tension of water,
walk on the surface without
submerged
146
y
p
to be less dense than water (b) Ice oats on water bec ause of this
d
r
o
E
Figure 45
f
x
O

o
The open-c avity structure of ice c auses it
O
t
i
s
r
e
(a)
v

Figure 44
n
l
(b)
which allows small insects to
sinking or becoming even partly
y
P
(a)
Structure
2.2
The
covalent model
TOK
DNA
(deoxyribonucleic
accurate
along
composed
bonds.
two
DNA
two
is
generations.
strands
replic ation
strands,
molecules
of
allowing
DNA
organic
involves
copies
has
for
a
molecules
breaking
to
genetic
be
this
information. A highly
information to be
double-helix
held
these
structure
together
hydrogen
by
hydrogen
bonds
to
“unzip”
made.
r
two
store
essential
new DNA
A
A
G
G
“unzips” into two
T
n
forces
have
storage
systems
knowledge?
are
exist
electrostatic
discussed
the
four
o
of
t
a
table
which are
in
the
attractions
types
of
natural world, in
that
keep
intermolecular
9.
Type of intermolecular force
a
London
Waals
(dispersion)
v
dipole–induced
forces
dipole–dipole
forces,
u
in
We
together by hydrogen bonds,
i
areas
l
der
information
intermolecular
f
x
O
van
of
other
together.
summarised
U
summary,
molecules
in
r
o
In
types
and
pairs
Hydrogen bonds are shown by black dashes
d
other
science
T
The DNA double helix is held
broken during replic ation.
What
base
G
A

Figure 47
C
C
C
strands
i
n
different
G
o
v
double helix
A
y
T
p
A
O
t
i
s
r
e
G
C
n
y
C
G
C
T
T
A
l
T
T
G
A
T
C
A
y
form
C
A
P
C
G
molecules
e
the
between
of
acid)
mechanism
s
s
passed
DNA-copying
forces
dipole
(LDFs)
forces
Where does it occur?
between
all
molecules
E
between
a
between
polar
polar
between
a
molecule
and
a
non-polar
molecule
increasing
forces
hydrogen bonding
molecules
strength
and
a
highly
hydrogen
electronegative
covalently
atom
(F, O or N)
bonded to another highly
electronegative atom

Table 9
Summary of types of intermolecular forces
147
Structure
2
Models
of
bonding
and
structure
Figure
force
48
shows
present
in
the
a
overall
method
for
determining
the
strongest
intermolecular
substance.
Do the molecules contain a
hydrogen directly bonded to
s
s
a very electronegative atom
(F
, O, N)?
e
yes
no
r
Are the molecules
hydrogen bonding
some
molecules
l
are polar and some
no
t
i
s
London
(dispersion)
forces
forces

Figure 48
Figuring out
the strongest
intermolecular force present
dipole–induced
dipole
y
r
e
ethoxyethane, CH
CH
3
4
ethanamine, CH
CH
3
NH
2
uoroethane, CH
F
n
Cl,
and
methane, CH
3
water, H
O,
and
o
i
substances
u
l
a
E
v
O
f
x
t
a
r
o
d
Solution
in
this
Ethoxyethane is polar
, so it forms dipole–dipole forces.
Carbon tetrauoride is non-polar
, so it only forms LDFs.
d.
Ethanamine
is
also
polar,
form
it
forms
hydrogen
bonded
so
only
to
forms
London
London
c.
c an
it
form
b.
directly
so
example
Butane
e.
non-polar,
worked
a.
atom
is
4
hydrogen uoride, HF
2
the
3
2
chloromethane, CH
All
2
CH
3
CH
2
c arbon tetrauoride, CF
g.
bonds
nitrogen,
bec ause
which
dipole–dipole
is
(dispersion)
(dispersion)
a
it
very
forces.
forces.
contains
a
hydrogen
electronegative atom. It
forces.
Fluoroethane is polar, so it forms dipole–dipole forces. Note that it does not
form hydrogen bonds. The structural formula might suggest that uorine is
directly bonded to hydrogen, but this is not the case. If you draw the Lewis
formula, you will see that uorine is bonded to a carbon atom.
f.
Chloromethane
form
g.
is
polar
dipole–induced
Hydrogen
bonds
molecules
bec ause
that
is
directly
experience
148
3
OCH
2
C
i
n
f.
CH
2
CH
3
U
e.
CH
2
o
butane, CH
b.
d.
determine the intermolecular forces
the molecules to each other.
a.
c.
p
attract
v
that
forces
between a pair of interacting molecules
Worked example 3
For each of the following substances,
are non-polar
O
dipole–dipole
n
y
yes
are
while
dipole
formed
they
bonded
between
both
to
a
dipole–dipole
methane
is
non-polar.
They
would
therefore
forces.
contain
hydrogen.
forces.
water
a
and
highly
They
hydrogen uoride
electronegative atom
are
both
polar, so they also
y
P
polar?
Structure
2.2
The
covalent model
Practice questions
25.
For
each
that
of
attract
the
following
molecules
to
CH
substances,
a.
propane, CH
b.
hydrogen
bromide,
c.
hydrogen
uoride, HF
d.
buckminsterfullerene, C
e.
ammonia, NH
the
intermolecular
forces
CH
2
3
s
s
3
determine
each other:
HBr
and
diatomic
ethanal, CH
oxygen, O
2
r
f.
CHO
3
State
and
explain
water
which
of
the
following
species
c an
form
hydrogen bonds
molecules:
propane, CH
3
c.
y
CH
3
CH
2
ethanoic acid, CH
3
COOH
3
extent
c an
intermolecular
behaviour?
do
the
c an
of
“bonds”
Structure
advances
e.g.
in
the
Chemistry
2.1,
and
“forces”
technology
(IUPAC)
compare?
Structure 2.3)
updated
real gases
lead to changes in scientic
International
denition
of
Union
the
of
Pure and
hydrogen bond?
n
U
C
Applied
terms
1.1,
denitions,
deviation
i
n
How
the
o
(Structure
explain
v
How
forces
(Structure 1.5)
p
ideal
y
what
from
r
e
To
O
t
i
s
Linking questions
n
ammonia, NH
b.
l
a.
y
with
P
26.
e
60
,
3
The properties of covalent substances
generally:
(dispersion)
10
these
forces.
shows
typic al
The
in
forces
forces.
Intermolecular force type
E
London
(dispersion)
sipole–dipole
hydrogen

Table 10
forces
forces
bonds
relative
ranges
lies
forces
strengths
dipole–dipole
enthalpy
a
dipole
dipole–dipole
<
enthalpy
trend
v
O
dipole–induced
forces
the
u
Table
mass,
l
f
x
London
and
molar
the
somewhere
of
forces
associated
reects
o
comparable
t
a
Given
r
o
Relative strength of intermolecular
i
d
(Structure 2.2.9)
<
with
trend
in
intermolecular
forces
are
hydrogen bonding.
overcoming
strength.
in-between
The
London
each of
strength of
(dispersion)
–1
Typic al enthalpy / kJ mol
1 to 5
3 to 25
20 to 40
Comparison of the strength of various intermolecular forces
149
Structure
2
Models
of
bonding
and
structure
Data-based questions
The
relative
mole cular
of
Table
intermole cular
11
shows
forces
mole cules
are
best
with
illustrate d
mole cular
by
comparing
masses
in
the
mole cules
range
of
44.0
that
to
have
46.1
similar
and
wildly
points.
Name and
point
moment / D
/ °C
44.11
0.084
–42.1
46.08
1.30
–24.8
44.06
2.69
Intermolecular forces
London
Dipole–
Hydrogen
(dispersion)
dipole
bond
yes
no
mass
formula
H
C
C
C
H
H
H
H
3
8
H
H
2
O
6
C
H
O
O
4
H
H
H
C
H
H
down
are
four
shown
more
by
the
and
data
that
are
in
important
London
then
statements
data
78.2
yes
yes
yes
1.41
100.7
yes
yes
yes
intermolecular force data for organic compounds of similar molecular masses
experience
dipoles
1.69
u
temporary
E
Write
explaining
molecules
no
H
a
c ausing
and
all
C
O
boiling point
v
O
that
46.03
O
l
Polarity,
Interpreting
show
f
x

Table 11
H
i
2
H
t
a
O
2
O
yes
o
C
46.08
d
CH
150
H
r
o
methanoic acid
that
H
O
6
n
H
2
U
C
H
C
H
ethanol
i
n
C
yes
no
o
C
20.2
v
2
H
p
C
H
yes
y
H
ethanal
r
e
H
C
yes
O
H
O
C
l
y
H
H
t
i
s
methoxymethane
no
n
H
P
H
r
propane
induce
further
describing
table
11.
skills
in
science.
(dispersion)
the
Attempt
to
forces
temporary
relationship
explain
What
do
bec ause
dipoles
in
between
each
the
data
they
the
all
in
11
suggest?
electrons
that
For
example, they
move
around,
surroundingmolecules.
molecular
statement
table
contain
using
structure
your
and
intermolecular
knowledge
forces
from this topic.
y
Boiling
dipole
Structural formula
e
Molecular
Molecular
molecular
C
dierent
s
s
boiling
strengths
masses.
Structure
2.2
The
covalent model
Physic al properties of covalent substances
There
are
two
covalent.
properties
of
We
will
look
covalent
structures
molecular
with
properties
at
a
substances
covalent
are
physic al
by
are
covalent
dierent
governed
structure
their
properties
and
very
network
dictated
conductivity
structures:
have
lattice
of
are
network
and
molecular
properties. Most of the
by
intermolecular
not
comprised
of
forces.
molecules, so
features.
covalent
substances
including
volatility,
e
electric al
of
two
s
s
Substances
their
types
These
solubility.
r
Volatility

Figure 49
Molecular covalent
with
a
covalent
network
structure
are
solids
at
room
temperature and
intermolecular forces are overcome
pressure.
Vaporizing
holding
the
them
requires
structure
a
lot
together.
of
energy
They
are
bec ause
therefore
of
the
strong
non-volatile
covalent
and
have
to
vaporize
to
molecular
be
substances,
overcome.
Since
the
intermolecular
intermolecular
forces
forces
are
between the
relatively
27.
generally
a.
variety
variation
substances
in
we
,
the
have
CO
at
),
this
chapter
that
and
at
which
consist
boiling
each
at
much
points.
other.
500–600 °C,
boils
of
This
and
a
are
free
as
to
their
move.
and
a
Covalent
electrons
are
f
x
some
notable
exceptions,
due
to
O
meaning
conductors
insulators.
panels
consist
The
cells
E
electricity.
v
Solar
and
of
the
that
as
of
up”
cells,
larger
of
lower
of
CCl
c arbon
, is
4
uoromethane, CH
dierence
of
F, is
3
–78.2 °C.
volatility
molecules
3
point
76.7 °C. The boiling point
for
are gases or
Comment on this
using
your
intermolecular
knowledge
forces.
have
buckminsterfullerene
are
in
wax,
usually
to
(C
),
tetracosane
not
contain
electric al
charged
generally do not contain
loc alized
covalent bonds, and
graphite, which is a good
deloc alized
intermediate
photovoltaic
contain
such
presence
its
a
semiconductor,
l
conductor
have
or NH
boiling
tetrachloride,
molecular
molecules,
these
c andle
needs
substances
“locked
the
each
60
of
molecular,
u
are
electric al
include
substance
they do not contain ions.
There
small
of
The
however, a
of
in
o
that
particles,
network
electricity,
bec ause
t
a
such
both
conduct
is
i
d
particles
To
r
o
substances,
conductors.
of
M any
molecules
component
391 °C.
Electrical conductivity
larger
Examples
50
Covalent
consist
hydroc arbons.
M any
2
between
sublimes
H
24
in
some
is,
species
n
(C
and
substances.
U
which
O
there
temperature.
melting
LDFs
molecular
considered
, H
substances
higher
stronger
of
forces,
C
and
intermolecular
i
n
However,
volatility
2
room
and
o
2
liquids
sizes
v
example, N
of
CH
4
28.
p
large
the
He or Kr
b.
y
to
r
e
Due
volatile.
which
pair has the highest boiling point.
weak,
the energy required to overcome them is low and therefore molecular substances
are
Identify
O
need
t
i
s
order
molecules
n
y
Practice questions
very high melting and boiling points.
In
l
bonds
y
P
substances vaporize when the
Substances
electrons. Silicon is a
conductivity
which
convert
places
solar
it
between
energy into
semiconductors, such as silicon.

Figure 50
Solar panels containing
silicon provide this satellite with electric al
power
151
Structure
2
Models
of
bonding
and
structure
Solubility
When
a
substance
substance
network
A
structures
holding
molecular
between
is
oen
insoluble
atoms
substance
the
solute
said
is
and
in
attraction
the
most
are
solvent.
formed
between the
Substances
solvents
bec ause
of
with
the
covalent
strong
likely
to
solvent
dissolve
are
in
a
stronger
solvent
than
the
if
the
intermolecular
attraction
chemistry
in
to
polar
solvent
is
that
dissolve
in
solvents.
non-polar,
“like
dissolves
non-polar
like”.
solvents.
In
other
Similarly,
forces
between the
words, non-polar
polar
solutes
are likely
Dissolving is unlikely to occur if the solute is polar
or
vice
versa.
l
O
t
i
s
n
y
TOK
“Like dissolves like” is a useful rule of thumb, which is helpful in many cases
but not all, and it has no explanatory power. In a small group, discuss the
advantages and disadvantages of having rules like this. You may want to
y
r
e
consider: what are the alternatives? Do such rules have an “expiry date”? What
rules of thumb do you use in other areas of knowledge?
is
non-polar.
2
almost
molecules.
they
to
form
Methanol,
the
a
in
that
water
solution
ethanol
primary
and
alcohols,
CH
3
2

− O
−
l
H
readily
decreases
and
water
such
are
as
hexane.
as
with
ethanol,
said to be
are
water
more
miscible, as
proportion.
dissolve
with
in
water. The solubility of
increasing
c arbon chain length
Solubility g / 100 g of H
O
2
3
OH
CH
miscible
2
(CH
3
)
2
(CH
3
OH
6.8
OH
0.014
3
)
2
8
The aqueous solubility of primary alcohols decreases with increasing
Hydroc arbon
a
H
v
Ethanol and
F ats
and
easily
with
oils
chains
are
are
largely
non-polar
observable
oil
to
make
immiscible,
when
salad
which
non-polar
vinegar
dressing.
and
means
they
and
do
not
form
(an
aqueous
The
do
therefore
two
not
hydrogen
not
solution
liquids
mix,
do
bonds
due
form
to
dissolve
of
ethanoic
two
their
in
with
water.
layers
water. This is
acid)
and
is
are
mixed
said to be
diering polarities.
water c an form
Therefore, ethanol is
Long
hydroc arbon
chains,
such
hydrophobic (water-hating).
hydrophobic tail and a
they
associate
greasy
are
152
any
bonds,
such
strongly
O − H
−
hydrogen bonds.
water soluble
in
solvents
associate
hydroc arbon chain length
between ethanol and water
Figure 51
Table 12
c annot
hydrogen
Ethanol
mixed
CH
CH
nonan-1-ol
51).
non-polar
it
Structural formula
CH
butan-1-ol
u
2
E

f
x
CH
3
O
(c)
2
between ethanol molecules
CH
o
(b)
CH
3
i
CH
form
(gure
in
bec ause
propan-1-ol
Primary alcohol
ethanol
t
a
r
o
d
O − H
−
c an
however,
− O
−
H
readily
water
when
(table12).
CH
in
n
U
(a) between water molecules
Substances
dissolve
c an
dissolves
C
likely
O − H
−
H
It
insoluble
i
n
H
is
o
Iodine
p
,
v
Iodine, I
H − O
−
stains
with
those
and
in
fats
and
detergent
hydrophilic (water-loving)
both
(gure
as
Soap
52).
water
and
the
Molecules
non-polar
with
c alled surface-active agents, or
a
oils,
head.
fats
said to be
oen
They
and
hydrophobic
surfactants
are
molecules
oils
tail
work
have a
bec ause
frequently
and
found in
hydrophilic
head
y
the
in
likely
covalent
together.
P
are
dissolve
and
their
of
and
molecules.
solutes
to
are
forces
solute)
r
It
the
e
solute
dissolves,
as
s
s
bonds
(known
Structure
2.2
The
covalent model
Water
Water
s
s
Water
e
Grease
(a)
Surface of material
Detergent
'heads'
attracted
by water
(c)
y
The
eect
a
high
they
known
as
potential
opiates,
In
are
diamorphine,
groups
for misuse and
are
reduces
have on the body depends on
2.
their ability to bind to opioid receptors in the brain.
ability
to
cross
the
barrier.
blood-brain barrier
lipids
the
morphine,
This
mechanism
i
n
times
more
same
time,
c an
heroin,
easily
action
an
tolerance,
to
cross
makes
has
is
the
analgesic
hydroxyl
which
the
blood-brain
opioid
metabolized
receptor.
diamorphine
and
greatly
Diamorphine is
quickly
than
more
addiction
both
groups,
molecule.
binds
diamorphine
about
morphine.
severe
side
central
ve
At the
eects,
nervous
system
depression. In most countries the use of diamorphine
they are permeable to lipid molecules. The presence
U
C
including
of
potent
the
as
ester
diamorphine
which
prevent polar molecules from entering the central nervous
system. These membranes are lipophilic, which means
and
brain,
into
of
o
membranes that coat the blood vessels in the brain and
v
The blood-brain barrier consists of a series of cell
in
In
polarity
with
y
their
known
p
factors:
1.
the
r
e
soluble
two
also
substituted
n
addiction.
derivatives,
with
O
its
l
y
and
painkillers
t
i
s
strong
water
In a surfactant molecule, the hydrophilic (water-loving) head and a hydrophobic (water-hating) tail li greasy stains o materials
Global impact of science
Morphine
Grease washed aw ay by
P
(b)
Surface of material

Figure 52
r
Grease
is
either
banned
or
restricted to terminally ill patients
of one amino and two hydroxyl groups in the morphine
CH
OH
2
u
f
x
l
c ancer
or
central
nervous
hydroxyl
(b)
CH
system
2
CH
3
C
H
C
3
N
CH
2
O
groups
O
morphine derivatives)
E
v
a
O
amino group
of
O
(can be substituted in
O
tertiary

Figure 53
i
2
t
a
N
C
3
d
r
o
H
CH
forms
o
in lipids and therefore limits its ability to reach the opioid
(a)
certain
disorders.
soluble in water but at the same time reduces its solubility
receptors in the brain.
n
with
molecule (gure 53) makes it suciently polar to be
O
C
O
CH
3
OH
The structures of (a) morphine and
(b) diamorphine (heroin)
153
Structure
2
Models
of
bonding
and
structure
Physic al properties of ionic and covalent substances
You
its
c an
figure
physic al
out
the
structure
properties
such
as
of
a
substance
electric al
by testing
conductivity,
you
do
that
could
not
know
be
which
is
which.
Design
a
safe method
used to identify A, B and C.
s
s
aqueous solubility and melting point.
Possible
Tool
1:
Recognize
or
Inquiry
1:
and
address
environmental
the
issues
in
relevant
an
spatulas,
safety,
leads,
investigation.
beakers,
ammeter,
crucible,
•
substances
you could use
distilled
lamp,
Bunsen
water,
pipettes,
conductivity
burner,
pipeclay
probe,
triangle,
Demonstrate independent thinking,
mat,
mortar
and
pestle,
melting
point
Inquiry
2:
Identify
and
record
relevant
qualitative
a
proposed
method
risk
assessment
and
show
this
and
your
presented
bottle
with
three
contains
a
bottles
crystalline
labelled A, B and
solid.
Extension
One is sodium
What
chloride,
another
is
sugar,
and
the
third
is
white
if
would
bec ause
electric al
•
insoluble
strong
experimental data
the
physic al
of
covalent
substances?
•
(Tool 1, Inquiry 2)
group
determine
the
•
and
compounds
for
explain
graphite
the
of
their
molecules
conductors
form
solubility,
with
Explain
why
and
although
32.
Explain
E
volatile,
covalent
are
and
not
electric al
33.
graphene.
relationship
they
both
and
are non-
are
would
require
breaking the
typic ally:
together
by
intermolecular
forces,
they
do
not
contain
mobile
charged
on
the
strength
of
the
intermolecular
forces
Which
two
in
of
the
following
substances
are
readily
water?
A.
ethanoic acid, CH
B.
octane, CH
C.
octan-1-ol, CH
(CH
COOH
)
2
CH
6
3
(CH
)
2
CH
6
OH
2
volatile,
D.
propanone, CH
COCH
3
3
covalent bonds.
34.
CH
Explain
why
ionic
substances,
such
as
sodium
chloride,
OH, is miscible
do
2
not
dissolve in oil.
water.
35.
Explain
why
hexane
than
diatomic
bromine,
Br
,
2
154
graphene)
molecules.
soluble
between
substances
substances
3
with
held
are
3
contain
typic ally:
covalent bonds
graphite
dissolution
3
why ethanol, CH
strong
are
3
network
molecular
bec ause
depending
solvent
volatility and boiling point.
31.
by
include
structures
are
structures
weak
electric al
varying
they
a
Describe
except
v
O
30.
covalent
bec ause
are
soluble? How
charged particles
their
covalent
water
you
particles
l
why
conductors,
poor
bec ause
molecular
together
mobile
not
If
u
f
x
Explain
•
forces?
Practice questions
29.
which
t
a
(Structure 3.2)
nature
volatile
with
i
theintermolecular
r
o
of
extent does a functional
solvents
network
(exceptions
contain
o
what
d
To
all
not
n
Substances
properties
in
do
covalent bonds.
U
demonstrate
they
held
C
Linking questions
What
conductors
i
n
bec ause
covalent
are
were
your method?
o
poor
they
v
between atoms
with
substance
y
substances
non-volatile
•
ionic
modify
p
•
summarize,
you
r
e
To
the
sand, but
approval.
n
are
E ach
time, try it out.
for
O
C.
teacher
t
i
s
You
have
your
y
Instructions
to
l
Complete
observations.
in
water.
is
more soluble in
y
hood, combustion spoon.
•
heat-proof
apparatus, fume
P
initiative or insight.
power pack,
electrodes,
r
ethic al
and
e
•
equipment
mightinclude:
Relevant skills
Structure
2.2
The
covalent model
Chromatography (Structure 2.2.10)
The
components
of
chromatography.
underlying
most
based
involve
of
their
forms
the
phase
oen
be
separated
chromatography
existence of a
chromatography,
dierent
intermolecular
mobile
c an
of
anities
the
for
and
(gure
mobile phase
components
of
identied using
54)
have
and a
the
the
same
stationary phase.
mixture
are
separated
each of the two phases. These anities
forces.
moves
through
the
stationary
phase,
c arrying the components
Methods
the
mixture.
The
components
all
start
at
the
same
point
but
are
are
through
the
the
stationary
two
phase
phases.
The
at
dierent
rates
components
due
to
their
eventually
diering
anities
for
separated.
(b) liquid column chromatography
y
sample
O
t
i
s
solvent front
move down the
column at
different
rates,
forming bands
n
o
i
t
a
1
2
3
Sample is placed
Solvent is
Different components in the sample
on the top of the
poured down
become separated into bands and are
packed column
the column
collected at the bottom of the column
a
E
v
O
l
u
chromatography
o
2
C
i
n
SiO
U
d
r
o
f
x
gas–liquid
p
of the sample
with
adsorbent
(c)
components
packed
e
lu
b
v
oran
ge
y
r
e
column
water (the solvent)
solvent
n
solvent
paper clip
l
(a) paper chromatography
are
mixtures
Structure 1.1.
y
of
separating
discussed in
P
each
for
transported
r
of
e
The
principle:
forms
on
mixture
s
s
In
a
All
detector
heated
chamber
recorder
sample
injected
t
Figure
54
All types of
here
column
–
c apillary tube
chromatography
involve a mobile
c arrier gas
with
liquid-coated
walls
phase and a
stationary phase
155
Structure
2
Models
of
bonding
and
structure
Research skills
ATL
Chromatographic
be
grouped
methods
according
adsorption,
we
will
the
focus
former
latter
is
on
the
two
on
ion
of
c ategories.
separation
planar
(for
based
gas
on
principle
at
notes
in
example, partition,
a
paper
of
In
this
and
chapter,
thin-layer.
partition,
whereas
least one other method of
chromatography)
your
c an
column
chromatography).
the
Research
They
vs
and
Venn
compare
r
as
of
into
example,
planar chromatography:
mixtures
Organize
(for
exchange
adsorption.
(such
methods.
or
types
separates
based
classied
mechanism
exclusion
chromatography
dierent
or
be
format
e
The
size
c an
their
s
s
chromatography),
on
and
diagram.
contrast the
solvent.
paper
The
55).
(made
mixture
is
The
of
to
a
hydrated
be
of
the
paper.
The
paper
is
then
chromatography experiment using dierent
containing
a
small
amount
You c an see that the purple ink at the
right-hand
the
vapour
solvent
atmosphere
side is a mixture containing red,
solvent
yellow and
allows
and
prevents
blue dyes
in
the
solvent
at
is
placed
the
i
n
coloured
ink
are
dropped onto
piece of
to
onto
chamber
start
line
as
a
near the
beaker,
Placing a lid on top of the
become
through
a
such
saturated with the
evaporation.
After
Different components of
the inks rise up the paper to
different
extents
and
are
separated
o
i
d
u
l
a
E
v
O
f
x
t
a
r
o
solvent
paper
rectangular
n
U
paper
C
chromatography
paper
a
p
different
a
dotted
in
o
v
of
is
bottom.
chamber
loss
demonstrate
cellulose) and the mobile phase is a
Before
Spots
to
phase
y
beaker
of
r
e
inks.
way
separated
Simple paper
bottom
simple
stationary
O
suitable

Figure 55
pigments
t
i
s
chromatography
of
(gure
n
mixtures
chromatography
y
Separating
l
chromatography
The
solvent
The
components
to
a
their
is
allowed
relative
greater
to
the
anities
anities
transported
known as
of
for
further
for
the
up
move
up
mixture
the
paper
will
paper.
until
dierent
stationary
solvent
the
the
move
and
dissolve
The
results
it
reaches
distances
mobile
more
of
a
up
a
phases.
readily
point
the
in
near
paper
the
top.
according
Components with
it
and
chromatography
are
therefore
experiment
are
chromatogram.
Chromatography paper is composed of hydrated cellulose, which contains many –OH
groups. These groups are very polar and attract water molecules tightly, saturating the
surface of cellulose with water
. When coupled with a less polar organic solvent, this
technique can be used to investigate mixtures such as leaf pigments or amino acid
mixtures. The mixture components are partitioned between the water layer and the
solvent: less polar components dissolve in the solvent and get carried further up the
paper
, whereas more polar components stay predominantly in the water layer
. The
separation can be optimized by using dierent solvents, or mixtures of solvents.
156
y
P
Paper
Structure
Thin layer
2.2
The
covalent model
chromatography (TLC)
Thin layer chromatography (TLC) is more expensive than its paper counterpart, but
it oers greater sensitivity. In TLC, the stationary phase is a rectangular plate made of
glass or metal coated with silica (silicon dioxide, SiO
) or alumina (aluminium oxide,
2
O
2
s
s
Al
). The silica or alumina surfaces are very polar and contain many hydroxyl
3
groups. The mixture is spotted onto the plate and placed into an eluting chamber
containing a non-polar organic solvent. The polar substances in the mixture travel
surface is greater than their tendency to dissolve in the solvent. Conversely, non-
r
polar components dissolve in the solvent and travel further up the plate.
e
up the plate slowly because their tendency to adsorb onto the silica or alumina
n
l
y
reacts with the spots to form coloured products. Other
purposes. Coloured substances are easily located on
colourless compounds can be located in dierent ways.
a chromatogram. When a substance is colourless, an
For example, some might show up under UV light. C an
additional step is needed. For example, when a mixture
you think of any other examples of how existing scientic
contains amino acids, the chromatogram can be sprayed
methods were adapted to solve a problem?
chromatography
R
distance
travelled
for
a
c an
spot
be
on
by the spot (b)
to
the
=
F
a
quantied
by
travelled
by
of
R
,
are
56
Retardation
c alculated
factors,
from the distances
F
measured
on
a
chromatogram
o
paper
type
i
on
and
front
t
a
depending
travel
ratio of
t
Figure
solvent
u
in
the
solvent (a):
mixture on
baseline
a
c annot
always
a
l
spots
are
v
E
the
is
the
n
U
d
r
o
f
x
O
Since
values
c alculating the
chromatogram
distance
b
R
a
C
the
experiment
value
F
i
n
). The
F
y
a
o
of
v
results
retardation factor ( R
p
)
F
The
r
e
Retardation factor (R
O
t
i
s
Experimental techniques are adapted to suit dierent
with a locating agent. A locating agent is a substance that
y
P
Experiments
pencil baseline
b
further
than
the
solvent
that
transports them,
R
F
the
factors
range
such
used
(in
of
as
0
the
paper
to
1.
The
distance
composition
of
travelled
the
chromatography).
solvent,
R
values
by
a
spot
c an
vary
temperature, pH
are
reproducible
F
and
therefore
values
to
c an
be
accepted
used
values,
to
identify
provided
substances
that
the
by
comparing
conditions
are
the
experimental
same.
157
Structure
2
Models
of
bonding
and
structure
Worked example 4
Consider the thin layer chromatogram shown in gure 57
.
Given that the stationary phase is silica and the solvent is a
Calculate the retardation factor for the red and blue spots.
hydrocarbon, determine which component is more polar
.
s
s
cm
6.0
cm
r
8.0
maximum
solvent
distance the
front
distance
cm
level
of
the
solvent
baseline:
2.0 cm
and
the
(blue).
solvent
c alculate
the
at
The
front
contains
distinct
8.0 cm
(red),
distance
is
three
spots
factors
(R
)
F
blue
spots
(top
bottom)
=
by
the
red
R
(2
signic ant
gures)
l
a
v
E
This experimental
between substances than the previous
158
do
a
each
substances
blue
phase
is
for
is
blends
signic ant
than
R
spot
solvent
gures)
(blue).
of
into
the
clear
the
red
spot
(non-polar
This
so
the
than
me ans
dots
you
next.
you
are
will
If
c an
phase.
spot
The
that
the
frequently
you
obtain
want
using
a
soluble
solvent)
that
in
in
the
and
a
the
gre ater
phase
substances
the
shown
a
have
stationary
contains
substances
oen
try
more
substances
blue
the
are
hydroc arbon
of
stationary
polar
phase,
the
composed
experiment,
mobile
(2
of
polar,
more
composed
the
0.17
larger
spot
anity
colour
blue
the
F
mobile
are
in
is
component(s)
chromatography
which
=
(red)
(silic a)
Chromatograms
chromatogram has less clear separation
examples we looked at
spot
solvent
u
f
x
O
u
Figure 58
0.67
red
the
i
=
follows:
t
a
12.0 cm
by
r
o
8.0 cm
=
travelled
the
by
12.0 cm
the
the
by
travelled
2.0 cm
6.0 cm
for
travelled
distance
o
travelled
(red) =
F
distance
as
d
distance
R
and
=
F
F
and
distance
(blue)
R
dierent
between
12.0 cm.
retardation
at
o
c an
the
as
n
baseline
You
from
and
mixture
U
(green)
the
appe ar
C
distances
that
which
the resulting chromatogram (right)
v
shows
i
n
58
components,
p
The apparatus for thin layer chromatography (le) and
Solution
Figure
solvent
y
the
r
e

Figure 57
above
O
t
i
s
solvent
mixture
n
cm
0.0
travelled
y
2.0
l
the spot
12.0 cm
in
red
that
spot.
books.
If
you
chromatogram in
clearer
dierent
separation
solvents.
between
y
P
solvent
travelled
baseline
e
lid
Structure
2.2
The
covalent model
Thin-layer chromatography of plant pigments
The
pigments
found
chromatography,
in
plants
c an
depending
on
be
separated
their
by
Part 2: Chromatography
relative affinities to
3
4.
mobile
and
stationary
phases.
The
separation
Place
by
changing
the
phases.
In
this
few
cm
tank.
Place
will
compare
the
chromatograms
stationary
phases
are
on
top,
the
chamber
to
the
in
solvent
vapours.
becomes
You
saturated
c an place a paper
changed.
towel
soaked
in
the
solvent
vertic ally inside the tank
Relevant skills
Measuring length
1:
Paper
5.
or
thin
layer
Then,
Record
3:
Propagate
uncertainties
in
uncertainties
in
the
starting
sides
of
you
must
make
attaching
top of the
Wear
•
Flammable
eye
Keep
away
from flames and
solvents
a
are
the
toxic
well-ventilated
to
the
space
environment.
and
dispense
Chromatography paper
•
TLC plate
•
Pencil
•
green
beaker
the
solvents
ether:propanone
mixture)
Method
using
splint
the
before
pigment
it
a
the
once
line
that
paper,
the
placed
in
the
tank
by
across the
solvent
pencil
evaporates.
spots
Allow
tank
draw
in
Draw
over
circles
are visible in the
plate
to
dry.
the
distance
travelled
by
the
Measure
the
distance
travelled
by
each pigment
the
base
Determine
a
line
to
the
reasonable
you
have
solvent.
centre of the spot.
uncertainty
for the
measured.
value
C alculate the R
of
each pigment, and the
F
uncertainty
Compare
of
your
(9:1
values
of
each.
your
class.
results
to
Compare
those
your
of
other
results
to
members
literature
for the R
values,
if
available.
Search online
F
for
information
that
will
help
you
identify
each of
thepigments.
petroleum
O
l
u
Chromatography
10.
a
are
upright
chromatography tank
leaves
f
x
•
Fresh
or
for
o
Propanone
tank
foil
i
•
aluminium
t
a
Lid
U
Chromatography
•
d
Ruler
•
r
o
•
from
Q uickly
front
the
distances
9.
to
is
Measure
from
8.
or
o
•
spotter
solvent
C
Chromatography
n
Pestle and mortar
•
v
i
n
Materials
plate
top.
chromatogram.
7
.
lid
it
tank.
the
the
around
solvents
cupboard.
•
near
the
Ifyou
that
y
in
fume
or
is
sure
p
Work
Remove
r
e
Organic
•
a
solvents.
6.
sources of ignition.
•
in
protection.
to
solvent, taking
O
S afety
it
the
Ensure the plate does not touch
beaker.
t
i
s
conclusions
other
line.
the
Discuss the impact of uncertainties on the
•
into
below the pigment spot
n
3:
plate
solvent
processed data
the
Inquiry
TLC
the
measurements
on
•
the
keep
l
3:
Tool
to
y
Tool
•
process.
chromatography
c are
•
lower
the
y
1:
Tool
up
P
Tool
•
speed
r
to
•
solvent into
ensure that the
e
and
chromatography
lid
obtained when the
with
mobile
the
the
investigation,
atmosphere
you
of
c an be
the
optimized
a
s
s
the
11.
Comment
accuracy,
of
on
of
the
your
measurement
precision
class’
s
and,
results.
uncertainty
on
if
possible,
Discuss the impact
the
validity
of
your
Part 1: Preparation of the chromatography plate
a
1.
conclusions.
3
Grind the leaves with a few cm
of propanone until
v
Part 3: Changing the phases
you have obtained a concentrated pigment extract.
Handle
the
a
pencil,
line
in
bottom.
TLC
E
2.
12.
plate
c arefully
by
the
edges.
the
plate,
1
to
2
cm
Spot
the
plant
spot
taking
of
the
solvent
results
would
system,
or
a
differ
if
you
used
different stationary
phase. Then, try it out.
pigment
Clear
c are
to
pigment.
create
Allow
up
as
instructed
by
your
teacher,
taking
c are
extract onto the starting
not
line,
how
different
from the
13.
3.
Consider
Draw
a
across
it
a
to
small,
to
dispose
of
the
solvents
down the sink.
concentrated
dry.
159
Structure
2
Models
of
bonding
and
structure
Practice questions
36.
C alculate
the
retardation
factor,
R
,
for
the
green spot
F
at
37
.
6.0 cm
Suggest
in
gure 58.
why
the
baseline
in
paper
chromatograms is
s
s
drawn in pencil.
38.
Chromatographic
and
analysis
individual
of
a
pigments
mixture
(B,
C
of
and
dyes (A)
D)
produced
chromatogram on the right.
and
explain
which
pigments
are
present in A.
A
B
C
r
Identify
D
e
the
three
l
LHA
a
two
molecule
or
molecule.
double
be
drawn
Resonance
or
for
triple
oen
bond
ozone, O
alternative
in
a
described
happens
resonance
by
when
molecule.
(gure
3
show
be
59).
Note
For
that
a
bonds
identic al
•
intermediate
of
bond
have
a
Lewis
formulas
arrow
is
c an
used to
O
that
ozone
the
double.
In
Bond
two
oxygen–oxygen
reality,
length
the
and
ozone
bonds
molecule
strength
data
in
is
(table
ozone
a
13)
show
are:
each other
bond
distributed
between a single O–O bond and a double O =O bond in terms
order
of
1.5,
bec ause
there
are
3
bonding
electron
pairs
across two domains.
l
u
Average bond
–12
Bond
Length / 10
m
Bond order
–1
enthalpy / kJ mol
E
v
a
O–O
O
are
hybrid of
strength and length
t
a
•
to
two
o
•
one
structures.
in
Instead,
represent
more than one position
double-headed
p
resonance
two
i
d
r
o
f
x
O
160
two
the
suggest
and
is
formula.
collectively
o
that
might
single
n
U
these
one
Lewis
that
Ozone has two possible resonance structures
C
structures
dierent:
there
structures.
v
i
n
These
single
example,
O

Figure 59
a
resonance structures
y
a
c annot
possible
r
e
for
more
n
the
are
O
there
t
i
s
Sometimes,
single
O
O=O
bond
bond
148
144
1
in
128
362
1.5
121
498
2
double

Table 13
ozone
bond
Bond
data for oxygen–oxygen bonds
y
P
y
Resonance (Structure 2.2.11)
Structure
2.2
The
covalent model
LHA
Global impact of science
Diatomic
oxygen,
O
,
and
ozone,
O
2
allotropes
that
ultraviolet
in
O
,
(UV)
with
dierent
radiation
section
bond
are
oxygen
of
order
corresponding
the
2,
wavelengths
UV-C
is
spectrum.
stronger
to
The
and
of
and
resulting
UV-B
the
Ozone
double
therefore
bond
has
a
(CFC s)
c an
in
radiation
skin
layer
in
depletion
the
reach
c ancer
c an
c ause DNA mutations,
(melanomas)
c aused
atmosphere
the
E arth’
s
by
means
surface.
and
c ataracts.
s
s
electromagnetic
,
3
absorb
chlorouoroc arbons
that
more
International
UV
radiation
collaboration
2
O
,
bond
with
enthalpy
bond
order
than
1.5.
the
oxygen–oxygen
Breaking
the
bond
in
involving
of
oxygen–oxygen
the
1987
Montreal
ozone-depleting
e
higher
Protocol and the banning
substances
has
slowed
down the
3
bond
requires
be
of
the
double
of
depletion.
concentration
oxygen–oxygen bond in O
of
Evidence
ozone
in
in
the
recent
stratosphere
2
brought
wavelengths
be
correspond
electromagnetic
in
the
The
bonds
330 nm
to
the
and
UV-C
spectrum,
stratosphere
in
UV-C
surface
and
UV-B
radiation
from
and
UV-B
prevents
the
nearly all
E arth’s
(gure 60).
the
ozone
layer
has
increased,
(gure 61).
y
p
r
e
C
o
v
i
n
n
U
o
i
d

Figure 61
S atellite images of a hole in the ozone layer over
Antarctic a from
1979 to 2011.
The largest
ozone hole to date was
recorded in 2006. Since then, data suggests that the hole has
The ozone layer in the stratosphere absorbs
radiation and
most
UV-B radiation
been getting smaller again
u
l
a
E
v
O
f
x
t
a
UV-C
r
o

Figure 60
of
ozone
respectively.
reaching
recovery
below.
3
solar
the
O
presence of O
of
in
n
the
below.
resulting
radiation with
t
i
s
of
and
wavelengths
wavelengths
regions
The
240 nm
by
electromagnetic
y
These
of
broken
by
l
c an
about
years suggests the
y
dissociation
c an
rate
energy.
P
The
higher
r
double
t
Figure 62
Former US president
Barack
Obama presents the Presidential Medal
of Freedom
to Mexic an chemist and
environmental scientist
helped
and
M ario Molina. Molina
to uncover the link between CFC s
ozone layer depletion
161
2
Models
of
bonding
and
structure
LHA
Structure
Deloc alization
Resonance
of
structures
deloc alization.
loc alized
are
shared
loc alized
c an
between
by
more
be
described
covalent
the
than
bonds,
two
two
atoms.
atoms
as
two
in
a
single
atoms
structure using the concept
share
Deloc alization
a
molecule
or
a
pair
occurs
ion,
as
of
electrons, which
when
electrons
s
s
are
In
opposed to being
between a pair of atoms.
Consider the ion NO
.
It
is
represented
by
two
resonance
structures
(gure63a).
e
2
The
nitrogen–oxygen
bond
has
bond
order
1.5,
and
its
length
electrons
is
deloc alized
c an
be
the
deloc alization
across
represented
the
with
a
two
N–O
structure
bonding
domains.
containing
a
Practice questions
Draw
the
structure
of
electron
(a)
deloc alized
structure
N
the
c arbonate
ion,
CO
why
the
C–O
bond
O
order
O
in this ion is 1.33.
–

Figure 63
(a) The ion NO
c an be represented
by two Lewis formulas (b) An alternative
2
and
to
those
in
in
the
c arbonate
c arbon
dioxide
OH.
3
Note
that
the
does
oxygen
dissociated
by
and
Lewis
ozone
different
wavelengths of light?
actual
molecule
which
is
best
is
to
a
the
or
single
the
63b
and
a
Lewis
of
that
“in-between”
the
single
between
be
bec ause it does not
pairs.
Instead,
c annot
electron
formula
electron
movement.
molecule
somewhere
by
not
interconversion
bond
presence
represented
is
non-bonding
the
represented
deloc alization.
the
various
deloc alized
various
resonance
resonance
structures
by a single
Therefore, the
resonance
structures,
structure.
o
i
Benzene and resonance (Structure 2.2.12)
Benzene, C
H
6
,
is
an
important
example
of
a
molecule
that
demonstrates
6
resonance.
u
l
a
E
v
O
f
x
t
a
r
o
d
(Structure1.3)
due
involve
rotation
describe
formula
Figure
n
are
U
collectively
Why
not
through
in
bonding
C
structures
of
i
n
Resonance
Linking question
structure
loc ations
electron pair
o
specify
v
and methanol, CH
representation of the ion shows the deloc alized
p
ion
strengths
y
Compare the C–O bond lengths
r
e
41.
N
N
, and
3
explain
O
the
2–
of
(b)
t
i
s
Draw
shows
ozone that
deloc alized
pair.
40.
reason,
that
n
the
y
shows
63b).
this
line
l
39.
(gure
For
dashed

Figure 64
The ball-and-stick structure and
electron cloud
of benzene, C
H
6
162
are
One pair of
6
y
ion
P
the
bond.
strength
r
between those of a single N–O bond and a double N =O
and
Structure
Lewis
formula
connected
structures
for
through
are
benzene
contains
alternating
possible
single
a
six-membered
and
double
ring
bonds.
of
Two
c arbon atoms
resonance
H
H
H
H
H
H
H
electrons
in
the
c arbon–c arbon
bonds
in
of benzene
benzene
are
deloc alized
this
reason,
these
deloc alized
electrons
are oen
skeletal
formulas,
straight lines
n
y
For
l
In
ring.
y
the
the
Resonance structures
P
of
Figure 65
r
t
H
e
H
s
s
H
H
around
covalent model
(gure 65).
H
Some
The
LHA
The
2.2
represent bonds. The ends and
represented
with
a
circle:
vertices
(corners) of the bonds
hydrogen
Detailed
skeletal
be
found
are
is
in
a
planar
the
and
simplest
large
range
stabilized
aromatic
of
due
substances
(gure 66).
structures
of
1,2-dichlorobenzene
structure.
Cl
benzene
the
1,2-dichlorobenzene
structures
rings
in
of
the
l
E
v
a
O
curcumin

Figure 66
(a) Curcumin,
that
contains
following
below.
drawn.
drawing
molecules
are
Structure 3.2.
TOK
Skeletal
of
formulas
“chemistry
extensively
what
extent
using
are
and
language?
are
an
example
shorthand”
used
formulas
a
of
for
by
that
chemists.
chemic al
are
To
symbols,
equations like a
What
chemic al
is
gained or lost
notation?
circle to
molecules and identify the
(b)
(c)
Kevlar
found
shown
formulas
not
electrons.
each:
(a)
is
o
Research
of
deloc alized
u
f
x
3.
the
i
structure
represent
Cl
t
a
a
d
Draw
r
o
2.
deloc alized
structures similar
n
resonance
of
given in
describe
are
o
resonance
other
and
to
C
the
the
presence
i
n
of
Draw
used
U
One
the
hydroc arbon,
Activity
1.
to
is
p
c an
that
Benzene
hydroc arbon. The term “aromatic”
v
it
aromatic
molecules
electrons.
to
an
atoms
instructions
y
cyclic
is
r
e
Benzene
c arbon atoms, and
O
t
i
s
represent
in the yellow food
estradiol
spice turmeric (b) Kevlar is used
for making personal armour (c) Synthetic estradiol is used
as medic ation in menopausal
hormone therapy
163
2
Models
of
bonding
and
structure
LHA
Structure
Communic ation skills
ATL
In
ordinary
conversation,
something
a
to
having
indic ate
compound.
a
a
the
term
pleasant
benzene
However,
“aromatic”
smell.
derivative
some
In
or
aromatic
c an
be
chemistry,
the
used
For
presence of a benzene ring in
compounds
example,
azulene — a
component
of
some
blue
are non-benzenoids.
pigments
two
planar
rings
that
are
joined
r
together:
C an
you
think
that
of
have
other
examples
dierent
of
chemic al
meanings in
C
C
everydaylife?
C
H
benzene,
The
deloc alized
electrons in the p orbitals of benzene
six
plane
p
of
one
electron
orbitals
the
in
overlap,
molecule.
each
of
forming
The
a
of
electron
the
plane
density
of
the
above
c arbon
atoms
are sp
hybridized
step
H
to
c arbon
(gure
have
hexagon
of
69).
come
bec ause
to
of
August
him
the
(b)
structure of
68
above
deloc alized
and
(gure
below the
67). The
molecular
with
(gure
68).
multiple
alternating
bonds.
single
Kekulé
was
credited
dream.
The
structure
diering
formula
The
and
with
1:1
for
benzene,
ratio
of
Eventually
double
the
c arbon
a
cyclic
bonds
was
structure, which is
proposed
was
a
lop-sided
c arbon-c arbon bond lengths.
CH
2
o
i
stamp
the
t
a
Kekulé on
u
for
a
August
postage
suggestion
benzene
l
(a)
E
Kekulé’
s
164
69
German
(a)
v
E ast
d
r
o
an
f
x
O
u
Figure

Figure
a
the
of
n
CH
2
in
was
6
structure
presence
atoms
U
H
the
its
C
i
n
said
deduce
suggested
arrangement
proposed
to
o
hydrogen
was
v
to
next
are
the structure of benzene
established that C
6
the
ring
y
was
electrondensity
this
p
it
in
r
e
Once
c arbon atoms occupies a p orbital.
of
(seeStructure 2.2.16).
molecule
Physic al evidence for
six
ring
electrons
2
ring
below
the
O
The
t
i
s
67
n
H
In

Figure
y
C
C
H
l
C
H
Some
of
the
19th
century
suggestions
for
the
(b)
structure of benzene
y
terms
P
H
H
in
e
bonds
H
H
a
found in
nature — contains alternating single and double
H
and
describe
H
H
form
to
however, this term is
s
s
used
as
Structure
diraction
c arbon
atoms
suggesting
The
patterns
in
they
were
c arbon–c arbon
that
and
they
a
double
are
also
of
all
showed
where
all
equivalent
bond
length
regular
in
(gure
hexagonal
bond
bonds
arrangement of
had
the
same length,
70).
benzene
c arbon–c arbon
intermediate
a
c arbon–c arbon
covalent model
is
intermediate
(table
14).
Bond
between that of a
enthalpy
data
s
s
single
later
benzene
The
LHA
X-ray
2.2
shows
strength.
0.140 nm
e
Average bond
–12
Bond
Bond length / 10
m
–1
enthalpy / kJ mol
single
C=C
in
bond
benzene
double
bond
154
346
140
507
134
614
length and
strength data for the c arbon–c arbon bond in benzene
diraction
compared
to single and
double c arbon–c arbon bonds
c arbon
The
chemic al
does
not
end
form
readily
where
bonds
the
of
addition
double
new
the
Instead,
diers
alkenes,
bond
species.
benzene
is
reactions.
breaks
In
deloc alized
from what would normally be
benzene
and
the
tends
to
ring,
which
undergo
value
increases
c alled
cyclic
comparing
the
thermochemistry
bond,
contains
change
in
is
This
with
hydrogenation
two
double
is
data
152 kJ mol
for
c arbon–
0.140 nm
are
and
substitution
discussed in
reactions
Reactivity 3.4
quantied
bond.
.
benzene
When
120 kJ mol
This
we
and
add
value is
those
for
hydrogen
.
1
ΔH
hyd
=
–120 kJ mol
cyclohexane
reaction,
enthalpy of hydrogenation,
molecule
is
1
u
a
a
c alled
v
similar
nearly double:
l
is
double
energy is
2
f
x
O
reaction
referred to as the
one
E ach
length
1
it
t
a
r
o
H
the
cyclohexene
a
molecule.
benzene
o
cyclohexene
double
+
This
benzene
For
reactions.
six-membered alkenes.
example,
this
the
i
to
of
d
For
by
stability
molecule.
bond
Addition
energetic ally
n
established
the
resonance energy.
be
C
i
n
a
U
by
would
substitution
Resonance energy of benzene
Resonance
undergo addition
c arbon atoms on either
benzene, addition would disrupt the
electron
X-ray
unsaturated, but unlike alkenes,
Alkenes
o
unfavourable.
undergo
with
eect
benzene
v
stabilizing
of
Like
p
reactions,
alkenes.
c alled
y
it
behaviour
from
r
e
expected
technique
electron density in an individual
benzene
the structure of benzene
A
shows a contour map of
O
t
i
s
the
Chemic al evidence for
70
n
Bond
l
y

Figure

Table 14
y
C
P
C
r
C–C
and
∆H
the
.
change
in
energy is
1,3-cyclohexadiene is
hyd
bonds,
so
its
enthalpy
of
hydrogenation is
1
232 kJ mol
E
1
ΔH
+
1,3-cyclohexadiene
2H
hyd
=
–232 kJ mol
2
cyclohexane
165
LHA
Structure
2
Models
of
bonding
and
structure
The
Changes
in
energy
next
molecule
theoretic al
reaction
are
known
as
Enthalpy
the
molecule,
series
matches
Kekulé’
s
1,3,5-cyclohexatriene.
suggested
structure and is the
1,3,5-cyclohexatriene
contains
three
enthalpy
double
changes.
in
across a
changes
bonds
and
its
enthalpy
of
hydrogenation
is
predicted to be
120
×
3
=
are
1
1
360 kJ mol
discussed in
Reactivity 1.1.
is
However,
the
actual
value
for benzene is
208 kJ mol
is
due
to
the
added
stability
provided
by
.
resonance in the benzene ring. The
s
s
compounds
71).
Naming
This
cyclic
(gure
covered in
1
dierence
between
these
enthalpy
values
is
152 kJ mol
,
which
is
the
resonance
Structure 3.2.
energy.
e
r
lom Jk / yplahtne
(152 kJ mol
theoretic al
1,3,5-cyclohexatriene
of
p
an
benzene
is
less
exothermic
C
has
additional
o
v
i
n
Hydrogenation
reaction
)
energetic
cost
to
than
break
n
compound
the
208
y
to
l
benzene
r
e
of
compared
O
resonance
–1
energy
y
The
t
i
s
71
360
expected.
the
This
deloc alized
is
bec ause the
electron rings.
1,2-disubstituted benzene compounds
and
the
single
Kekulé
bonds.
structure
When
i
whereas
c arbon–c arbon
evenly
the
in
the
you
two
distributed,
,
reacts
with
benzene
to
form
2
expect
chlorine
atoms
that
at
two
possible
either
end
of
a
isomers
c an be
c arbon–c arbon single
isomer, the chlorines would be at either end of a
bond
these
benzene, containing alternating double
would
other
double
of
chlorine, Cl
two
(gure
72a).
isomers
However,
represent
the
bec ause
same
u
l
all
electrons
molecule
(gure
are
72b).
(b)
Cl
Cl
Cl
Cl
Cl
a
E
v
O
f
x
(a)
Cl

Figure
Kekulé
only
166
with
t
a
r
o
d
bond,
one
o
1,2-dichlorobenzene,
formed:
n
U
Consider
72
(a)
structure
one
isomer
The
of
two
theoretic al
benzene
bec ause
(b)
the
The
isomers
actual
c arbon
of
1,2-dichlorobenzene
structure
atoms
are
all
of
for the
1,2-dichlorobenzene has
chemic ally
equivalent
y
P
1
u
Figure
152
Structure
2.2
The
covalent model
LHA
Practice questions
42.
Describe
structure
43.
How
many
pieces
ring
State
bond
the
the
evidence — one
deloc alized
benzene
of
of
physic al
and
one
chemic al — for the
benzene.
electrons
contribute
angle
in
to
the
does
each
deloc alized
benzene
and
the
of
the
c arbon atoms in a
s
s
44.
two
of
electron cloud?
molecular
geometry
around
each
c arbon atoms.
are
resonance
energy
in
benzene
explain
its
relative
unreactivity?
the
structural
features
of
benzene
reactions?
that
favour
it
undergoing
(Reactivity 3.4)
n
substitution
y
electrophilic
l
What
the
2.1, 2.2)
(a)
eight
or
in
12
these
electrons.
molecules
electrons,
,
This
c an
the
in
the
molecules
is
have
respectively.
and
discussed
with
SL
known as an
ve
or
Examples
six
of
pairs
such
of
same
containing
do
not
(b)
central
electrons, amounting to
,
shown
in
gure
73.
steps
used
expanded
check
that
the
expanded octets
for
drawing
octets.
Lewis
However,
this
at
SL
for
Cl
Cl
I
Cl

Figure
73
Cl
Two
species with
expanded octets: PF
and ICl
5
c ase, we omit the nal
central atom has a full octet.
ve
the
and
six
central
pairs
of
atom,
electrons
, with
4
around
respectively
o
d
Worked example 5
in
formulas
n
step — we
The
F
F
At
more
4
U
molecules
the
chapter.
with
C
follow
this
species include phosphorus
tetrachloroiodide ion, ICl
Drawing Lewis formulas for
c an
of
atoms
expanded octet.
5
You
section
containing
i
n
pentauoride, PF
were
familiar
o
10
valence
VSEPR
be
v
atoms
and
also
p
than
should
P
F
y
formulas
you
r
e
HL
O
t
i
s
Expanded octets (Structure 2.2.13)
Lewis
F
F
y
does
(Reactivity
P
How
e
r
Linking questions
Draw the Lewis formulas of each of the following species with expanded octets:
sulfur hexauoride, SF
b.
Count
above
for
e ach
Fluorine: 7
electrons
Total: 6
E
C alculate the
v
valence
Step 2
c.
+
×
42
molecule:
b triiodide ion, I
6
6
=
=
42
48
48
This
×
3
pairs
21
meaning that it has an
3
Xenon: 8
Oxygen: 6
Total: 8
+
×
18
3
=
=
18
26
electron.
+
1
=
22
26
=
2
=
polyatomic ion has a –1
charge
22
24
c xenon trioxide, XeO
3
Iodine: 7
additional
=
XeO
3
Total: 21
2
xenon trioxide,
3
a sulfur hexauoride, SF
Sulfur: 6
a
O
Step 1
steps
u
the
l
f
x
Follow
triiodide ion, I
t
a
6
Solution
i
r
o
a.
11
pairs
=
13
pairs
2
number of
electron
pairs
167
2
Models
of
bonding
LHA
Structure
and
structure
a sulfur hexauoride, SF
b triiodide ion, I
6
Step 3
Arrange
c xenon trioxide, XeO
3
3
O
F
the atoms
F
F
I
I
Xe
O
F
F
s
s
I
S
O
F
Draw
e
Step 4
F
O
the single
F
I
bonds
I
r
F
I
S
P
Xe
O
F
used
so
far...
2
pairs
used
3
t
i
s
F
peripheral
S
now
have noble gas
congurations
do
F
not
need
any
atoms
now
have
8
full
We
remaining
have
pairs
used
pairs,
above
nal.
24
is
so
the
There
structure
are
6
pairs
electrons
on
the
central
of
electrons
around
the
central
are
three
pairs of
C
atom.
It
has
an
expanded
octet.
U
use
is
for
n
we
far...
O
Xe
All
assign
to
13
have
the
O
the
nal
available
been
electron
xenon atom.
electron
pairs
used.
iodine atom has an expanded
XeO
have
,
each
3
an
atom
expanded
incorrect,
deducing
and
an
Lewis
around it.
has
L ater
octet.
in
formal
exception
this
chapter,
charge:
structure
of
this
XeO
we
will
and
will
discuss
help
explain
3
formulas.
to
have
an
expanded
octet.
a
E
v
O
Practice questions
45.
Draw
the
a.
PCl
b.
BrF
Lewis
formula
of
each
d.
of
the
following:
ICl
5
2
+
e.
3
c.
IF
ICl
2
f.
5
168
so
o
structure
method
for
used
needed.
i
this
none
pairs
We
species is an ion, so
have
We can see that the central
u
the
and
are
now
octet, with 5 pairs of electrons
l
f
x
However,
to
deduced
I
atoms
O
square brackets and the charge
t
a
structure
electrons
This
oxygen
octets.
pair
d
the
r
o
In
eight
I
I
sulfur
12
assigned to the
central iodine atom.
atom
full
far...
I
remaining
i
n
pairs
so
I
The
electron
used
available
The
O
o
electron
all
pairs
used.
v
any
Put
24
more
O
p
Step 6
octets.
O
y
uorine
far...
Xe
therefore
r
e
electrons
The
and
I
F
so
peripheral iodine atoms
atoms
F
used
O
The
pairs
n
I
pairs
F
on
far...
y
F
bonding
so
I
I
pairs
Put non-
l
6
Step 5
O
SOF
4
us
the
concept
deduce
why
it
is
the
of
true
considered
y
F
F
Structure
In
the
SL
atoms
occur
for
covalent model
geometry of expanded octets
section
have
The
LHA
Molecular
2.2
of
two,
this
three
expanded
chapter,
or
four
octets
we
discussed
electron
containing
the
domains.
ve
or
six
geometries that arise when
Several
additional
electron
geometries
pairs.
s
s
Five domains: trigonal bipyramidal geometry
If
there
are
and
electron
shape
electron
domain
domains,
adopt
geometry
consisting
of
two
c alled
a
trigonal
120°
in
and
those
90°
domains
from
each
trigonal bipyramidal.
that
share
the
bipyramidal
same
The
repel
ve
triangular
molecule
fall
each
other. This
into
domains
form
base.
The
electron
two
c ategories:
1.
Equatorial: the bonds forming the triangular base of the pyramids, at 120° from
the
bonds
triangular
that
form
the
axis
of
the
molecule,
at
90°

Figure
from the plane of
trigonal
base.
74
A balloon model of the
bipyramidal
geometry
A
trigonal
molecular
When
geometries,
all
ve
four
When
of
only
of
domains
the
the
two
the
molecule has
ve
of
the
the
domains
are
domains
the
two
are non-
and
three
geometry.

Figure
75
75.
The
four types of
are
trigonal
bipyramidal
geometry:
seesaw
(c)
(a)
electron domain
trigonal
T-shaped
(c)
bipyramidal
(d)
(b)
linear
(d)
F
F
I
F
Cl
F
I
S
F
F
a
E
e
domains,
i
l
B
linear
(b)
u
F
v
position
bonding
gure
and
geometry.
t
a
r
o
f
x
O
equatorial
are
F
P
domains,
T-shaped
o
F
F
in
d
(a)
F
bonding
molecular has
shown
geometry.
n
type
are
seesaw
molecule has
ve
trigonal
are bonding domains, and one is a non-
molecule has
domains,
each
B
presence of non-bonding domains:
domains,
U
of
ve
domains,
non-bonding
Examples
the
possible
C
•
four
geometry.
domain,
three
bonding
the
to
o
When
on
bonding
rise
i
n
•
are
gives
v
bonding
depending
domains
geometry
p
When
domain
y
bipyramidal
•
electron
r
e
•
bipyramidal
electron domain
O
t
i
s
the
n
Axial:
l
2.
y
each other
.
y
in
is
pyramids
electrons
at
P
domains
the
positions
r
a
ve
therefore
e
other
I
F
B
a
B
a
a
B
B
B
e
a
e
120°
A
A
B
B
e
90°
B
less than 120°
B
e
A
A
e
B
a
B
a
less than 90°
180°
a
less than
B
a
90°
axial position
169
LHA
Structure
2
Models
of
bonding
and
structure
Like
other
geometries,
the
120°
distorted
the
presence
and
90°
bond
angles
in
a
trigonal
bipyramidal
Practice question
shape
46.
Consider
the
T-shaped
molecular
Why
do
seesaw and
of
non-bonding
pairs,
due
to
their
greater
geometries.
are
Six domains: octahedral geometry
s
s
domains
by
repulsion.
you think the non-
bonding
are
loc ated in
If
there
are
six
domains,
the
electron
pairs in those domains adopt positions at
equatorial positions, not in axial
90°
from
each
other.
This
electron
domain
geometry
is
c alled
octahedral. This is
positions?
bec ause
the
electron
is
no
electron
six
octahedral
domains
the
are
geometry
bonding
four
domains,
the
each
type
90°
bond
geometry
presence
F
angles
in
pairs,
o
170
77
i
E

Figure
t
a
a
B
each
other, so
to
three
common
molecular
the
molecule has
octahedral
in
gure
to
domains,
their
and
two
are non-bonding
geometry.
77.
molecular
due
geometry.
geometries
greater
are
distorted
by the
repulsion.
(c)
F
F
F
Br
F
Xe
F
F
F
u
l
v
O
f
x
B
A
(c)
F
F
B
F
d
r
o
F
to
F
(b)
F
S
shown
octahedral
non-bonding
bonding
C
(a)
are
square planar
n
U
F
of
domains
are
90°
are bonding domains, and one is a non-bonding
o
domain
i
n
electron
six
domains,
square pyramidal
molecule has
A balloon model of the
The
octahedral
of
domains
v
76
of
the
rise
p
Examples

Figure
six
at
y
When
the
molecule has
r
e
•
of
the
gives
all
equatorial domains.
O
ve
domain,
are
and
t
i
s
When
axial
presence of non-bonding domains:
geometry.
•
molecule
between
domain
on
polyhedron
equilateral triangles.
distinguish
depending
eight-sided
n
all
an
an
l
When
in
is
B
B
B
B
B
B
A
90°
A
B
B
90°
<90°
B
B
B
B
The
square planar
three
types
of
y
octahedral
to
which
y
•
domains
need
geometries,
identic al
octahedron,
P
there
An
eight
an
r
The
of
form
e
consisting
domains
octahedral
electron
domain
geometry:
(a)
octahedral
(b)
square
pyramidal
Structure
The
covalent model
LHA
Electron
2.2
Number of
Number of
Number of bonding
domain
Molecular
Bond
geometry
angle
non-bonding
domains
Example
domains
geometry
domains
trigonal
5
5
0
trigonal
90°,
120°
PF
5
s
s
bipyramidal
bipyramidal
4
1
seesaw
<90°,
SF
4
<120°
T-shaped
3
linear
1
octahedral
90°
square
<90°
pyramidal
six electron domains
and
molecular
geometry of a
two
e.
four bonding domains and two non-bonding domains
the
number
following
seesaw
f.
square
g.
linear
the
are
l
PCl
3
IF
5
d.
ICl
e.
ICl
2
+
2
E
Predict
the
bond
a
BrF
v
O
c.
possible
domain
following:
5
b.
two
electron
answers
geometry
u
the
a.
pyramidal
f
x
Deduce
bipyramidal
(there
around atoms
n
square planar
e.
domains
o
octahedral
d.
non-bonding
i
c.
and
geometries:
t
a
T-shaped
three non-bonding domains
d
trigonal
b.
bonding
molecular
r
o
a.
of
and
U
the
domains
C
Identify
bonding
p
ve bonding domains
d.
4
o
four bonding domains and one non-bonding domain
c.
v
six bonding domains
XeF
O
geometry
b.
of
4.
domain
following bonding and non-bonding domains:
a.
with
3.
the
i
n
2.
electron
with
6
y
the
species
r
e
Identify
SF
BrF
5
90°
t
i
s
Activity
1.
planar
3
n
Summary of molecular geometries with ve and
square
3
I
l

Table 15
2
y
4
ClF
y
0
5
180°
P
6
<90°
e
6
2
2
r
octahedral
3
f.
SOF
g.
SiF
here)
and
molecular
geometry
of
each
4
2
6
+
h.
IF
4
j.
SCl
4
angles
in
molecules
a
to
j
above.
Linking question
How
does
position
in
the
ability
the
periodic
of
some
atoms
table?
to
expand
(Structure
their
octet
relate to their
3.1)
171
2
Models
of
bonding
and
structure
LHA
Structure
Formal charge (Structure 2.2.14)
Formal charge
in
the
at
all.
Lewis
looking
at
the
charge
an
were
shared
equally,
We
c an
formula
the
assign
when
electrons
charge
formal
several
an
=
have
its
to
of
if
all
the
bonding
non-bonding
atoms
possible.
has
number
if
charge
are
atom
would
and
in
Formal
a
electrons
electrons
molecule
charge
is
to
were not
select the
c alculated
electrons
before bonding – number of
electrons aer bonding
=
VE
– (NBE
+
r
FC
½BE)
VE
=
number
of
number
valence
of
electrons
non-bonding
of
sum
of
overall
the
bonding
formal
charge
of
,
all
atoms
Cl
=
5
of
the
follows:
(bec ause
nitrogen
nitrogen
is
in
atom
c an
be
f
x
t
a
VE
charge
as
The
u
6
(bec ause
FC(N)
Therefore,
the
Lewis
which
electrons)
5
the
formula
(2
+
0.5
formal
E
Lewis
=
in
bonds,
is
zero.
×
formula
are
the
6)
charge
=
of
BE
nitrogen
=
atom
equivalent to six
7
=
atom
or
ion
Lewis
Lewis
formula
should
be
equal to
c an
e ach
(bec ause
6
has
2
of
in
the
one
in
the
bond,
to
determine
the
atom:
is
in
group 17)
Lewis
non-bonding
(bec ause
has
repe ated
chlorine
(bec ause
six
be
chlorine
formula
each chlorine
electrons)
Lewis
formula
each chlorine
therefore two bonding
electrons)
FC(Cl)
0
the
nitrogen
atom
in
this
Therefore,
this
The
Lewis
sum
overall
172
=
NBE
a
bonding
three
v
O
has
l
=
process
charge
VE
group 15)
atom has two non-bonding electrons)
atom
same
formal
NBE = 2 (because in the Lewis formula the nitrogen
BE
molecule
the
the
o
formal
c alculated
r
o
The
i
d
Solution
C
N
in
in
n
U
Cl
atom
is zero by c alculating the formal charges of the atoms in the
3
Cl
a
atom
molecule or ion.
is shown below.
3
in
the
p
the overall charge of nitrogen trichloride, NCl
The Lewis formula of NCl
of
to
the
o
i
n
Show that
molecule.
charges
that
assigned
v
Worked example 6
electrons
to
y
the
of
r
e
The
number
before bonding
assigned
O
=
atom
electrons
formula
BE
the
t
i
s
=
charge
n
formal
l
=
y
FC
=
7
the
(6
formula
of
the
charge
+
0.5
formal
is
2)
=
of
0
e ach
chlorine
atom
in
zero.
formal
of
×
charge
the
charges
is
molecule.
zero,
which
is
also
the
y
P
Where:
NBE
by
before and aer bonding:
e
Formal
atom
s
s
shared
best
is
molecule
Structure
1.
formal
2.
a
3.
negative
charges
dierence
When
these
conditions
in
close
to
formal
formal
presented
are
generally
favourable:
zero
charges
charge(s)
with
covalent model
LHA
following
The
in
the
assigned
alternative
Lewis
molecule
to
the
close
more
formulas,
to
zero
electronegative
we
look
for
the
s
s
The
2.2
atom(s).
one
that
satises
requirements most.
2–
Assign formal charges to the atoms in the sulfate ion, SO
,
e
r
Worked example 7
to deduce its Lewis formula.
Solution
l
2–
We
start
by
drawing
the
Lewis
formula
of
SO
in
a
We
way
know
that
sulfur
is
prone
the
octet
octets,
rule:
so
the
FC(O) = 6
+
0.5
×
8)
Lewis
the
sulfur
atoms
in
the
this
FC(S)
=
FC(O
(6 + 0.5 × 2) = –1 (all the oxygen atoms
formal
Lewis
formula:
= +2
S
O
y
(0
this
of
2
6
(0
singly
p
6
on
charges
charges
formula
o
=
based
formal
C
FC(S)
atoms
the
v
oxygen
c alculate
i
n
and
c an
proposed:
O
C alculating
we
expanded
is
O
O
r
e
O
forming
formula
O
O
to
Lewis
t
i
s
2
O
Now
following
n
follows
y
4
that
y
P
4
+
0.5
×
bonded)
U
doubly
more
electronegative
the
sulfur
and
oxygen
12) = 0
=
FC(O
of
gives:
bonded)
6
=
(6
6
+
(4
0.5
+
×
0.5
2)
×
= –1
4)
=
0
are equivalent in this Lewis formula and therefore each
The
and
conditions
the
are
in
this
Lewis
dierence
not
formula
between
O
2
2
OH
E
c.
these
draw
charge
the
i
CO
3:
same
structure.
second
Lewis
a
of
range
Therefore,
In
both
which
as
1:
in
these
the
equal
to
Lewis
sums
of
to
are
v alues
more
formal
of
the
negative
suggested
zero
formula
the
charge
the
charge
closer
conditions
the
has
previously
formal
are
second
the
the
the
formula
c ases,
is
that
However,
oxygen
is
and
in
this
have
favourable.
preferred.
charges
are
poly atomic
2,
ion.
u
H
b.
formal
to
v
a.
the
need
a
will
l
You
is
close
t
a
r
o
f
x
O
Determine
them
charge,
not
favourable.
Practice questions
47
.
are
o
zero,
charges
d
to
formal
1)
n
have the same formal charge of
The
of
Lewis
every
atom
formulas
in
each
of
the
molecules
below.
rst.
d.
NO
e.
SF
3
6
f.
BH
3
173
2
Models
of
bonding
and
structure
LHA
Structure
Practice questions
48.
Based
on
formal
charge
considerations,
determine
which
Lewis
formula is
preferred:
a.
are
in
the
the
dierent
assumptions
c alculation
of
r
made
b.
formal
49.
in
a
of
oxidation
species?
states
the
below
suggest
formal
charge
of
each
atom
in
the
XeO
(Structure
for
and
an
alternative
Lewis
formula
in
molecule
P
atoms
and
which
3.1,
charges
are
all
zero.
Reactivity 3.2)
the
shown
formal
l
t
i
s
n
y
O
Xe
Draw
two
alternative
of
the
Lewis
two
formulas
is
for ClO
.
Determine,
using
formal
3
y
which
r
e
charge,
O
50.
favourable.
p
o
v
Sigma bonds () and pi bonds (π)
(Structure 2.2.15 )
this
attraction
will
on
delve
that
overlap
with
that
i
with that of a
the
double
This
is
nuclei
into
are
other,
bond
covalent
and
how
electrons
each
bond
dened
atoms
o
know
average
l
u
than
double
a
In
a
the
the
triple
other
bonding
shared
of
pair
electrons
shared using
allowing
the
the
electrostatic
electrons.
In
this
section,
occupying atomic orbitals
valence bond theory. This is the
electrons to pair their
increases
with
greater
bond
spins
order.
Consider the
1
for a single C
C
bond
(346 kJ mol
)
and
contrast it
1
double C=C
bond
is
the
bond
(614 kJ mol
signic antly higher but
two
bonds.
This
bonds
in
the
is
why
the
).
The
double
bond
pi bond (π).
strength
average
not double
of
Pi
the
are
bond
enthalpy
for
that of the single bond.
not
bonds
double
the
are
same: one is a
generally
bond
is
not
weaker
exactly
strength of the single bond.
c arbon–c arbon
two
are
pi
σ
bond,
bonds
one
(gure
of
the
three bonds is a sigma bond and
78).
σ
E
π
bond.
as
of
form bonds when orbitals in neighbouring atoms
strength
enthalpy
bec ause
sigma
a
pairs
sigma bond () and the other is a
v
O
f
x
t
a
r
o
d
We
valence

Figure 78
174
deeper
neighbouring
idea
we
two
n
U
we
topic,
between
C
in
i
n
E arlier
σ
π
π
Single bonds are sigma bonds. Double bonds contain one sigma and one pi
Triple bonds contain one sigma and two pi bonds
y
Determine
3
charge
e
What
s
s
Linking question
Structure
2.2
The
covalent model
LHA
Sigma bonds ()
Imagine
two
unpaired
they
do
so,
is
an
the
this
axis
to
form
a
along
of
pair
each
Eventually
up
forming
other.
their
a
1s
They
each
orbitals
will
have one
overlap.
When
covalent bond.
s
between the nuclei of the two bonding atoms. This line
or
internuclear axis. The
sigma
the
orbitals
bond.
bond
of
This
axis.
means
Figure
79
two
that
1s
orbitals
there
shows
that
is
a
will
s
s
p
region of high
sigma
bonds
c an
form
dierent types.

Figure 79
bonds
are
always
sigma
bonds.
p
Sigma bonds are formed
y
P
Single
p
overlap
r
pairs
line
approaching
orbital.
will
bond axis
density
between
1s
e
electron
atoms
the
electrons
invisible
known as the
along
in
s
s
Imagine
hydrogen
electron,
when orbitals overlap along the bond axis
The
shapes
of
discussed in
Pi
bonds
get
lobes
above
other
name
from
the
Greek
letter
π,
and
things
in
are
named
aer
things
bonds,
the
the
in
two
bond
electron
neighbouring
axis.
density
This
is
type
pi
bond
two lobes
contains
two
at
opposite
electrons,
sigma
Therefore,
count
double
the
bond
bonds
there
pi
and
(gure80).
of
the
bond
overlap
above and
below bond axis
axis,
as
shown in

Figure 80
Pi bonds are formed when
occupy both lobes.
porbitals overlap
above and
below the
bond axis
C=CH
2
.
2
in
the
H
structural
formula.
Remember that single
each of the multiple bonds contains one sigma
are 7 sigma bonds.
bonds.
contains
sideways,
C
H
bonds
bond
u
the
sigma
π
p
C
a
v
Then,
E
bond.
are
l
f
x
O
count
bond
pi bonds in the molecule (NC)
The structural formula is given below.
bonds
sides
which
i
Determine the number of sigma and
First,
overlap
pi
t
a
r
o
Worked example 8
Solution
a
o
A
c an
forms
n
bond consists of
80.
d
gure
they
overlap
p
concentrated at opposite sides of the
bondaxis.
One pi
atoms,
of
o
present
C
pi
are
below
U
In
and
resemble?
v
orbitals
they
i
n
p
above
corresponds to the
below the axis.
science
Pi bonds (π)
If
which
are
you look along the bond axis, the pi bond looks like a p orbital,
p
What
their
When
orbitals
y
with
p.
r
e
letter
p
O
came from in the Latin alphabet. When you look directly along the bond axis, the
sigma bond looks spherical, like an s orbital.
and
Structure 1.3.
t
i
s
Sigma bonds get their name from the Greek letter , which is where the letter s
s
n
l
y
Thinking skills
ATL
E ach
one
pi
triple
bond.
bond
There
contains
two
pi
bonds.
E ach
are 5 pi bonds in total.
175
2
Models
of
bonding
and
structure
LHA
Structure
Worked example 9
Determine the number of sigma and pi bonds in the phosphate anion,
3–
PO
.
The Lewis formula is given below.
4
s
s
3
O
r
O
Practice questions
and
contrast sigma
bonds () and pi
Solution
bonds (π).
why
two
orbitals
c annot
overlapping s
y
Explain
l
The three single bonds and one of the bonds in the double bond are all sigma
52.
the double bond.
O
t
i
s
Activity
n
bonds, giving a total of 4 sigma bonds. There is one pi bond: the second bond in
form a pi bond.
r
e
number of sigma
and pi
ethanoic acid, CH
bonds.
p
a.
COOH
3
v
propyne, CH
c.
ethanenitrile, CH
CCH
d.
c arbonate
e.
ammonium ion, NH
o
b.
3
CN
3
i
n
phosphorus
such
be
being
phenomenon
example,
explained
the
explain
on
M any
ion,
HCOO
•
Valence
sigma
energies
bond
bond
theory
describes
framework
the
strong,
hexagonal
benzene.
on
a
•
Molecular
orbital
theory
shows
that
deloc alization
more
gives
benzene
a
distinct
energetic
advantage.
orbital theory. This
are
spread
best
in
across
the
Molecular
entire
orbital
mechanics.
specic bonds.
we
use
the
C an
theory
you
other
draws
think
sciences
of
to
heavily
any
other
enhance
from quantum
examples
our
where
understanding
described using both
of chemistry?
aromatic stability of benzene
by two models:
E
Hybridization (Structure 2.2.16)
By
now
you
covalent
from
should
bonds.
tetrahedral
one
If
know
these
structure
another.
that
around
For
c arbon is
bonds
are
the
example,
tetravalent,
single
c arbon
bonds,
atom
at
methane, CH
,
4
176
of
discussions
based
loc alized
is
3
molecular
electron
u
c an
For
molecular
electrons
than
are
a
theories.
a
that
to
l
Sometimes
magnetism.
v
O
rather
helps
information
chemists
theory,
assumes
molecule
as
amongst
sophistic ated
theory
theory
give
PCl
o
properties
bonding
to
f
x
of
bond
fails
methanoate
i
or
it
trichloride,
3
t
a
valence
geometry,
r
o
While
+
4
azide ion, N
d
Theories
3
n
h.
U
g.
CO
C
f.
2–
ion,
y
Look up the Lewis formula of each of the following molecules. Determine their
meaning
they
are
it
approximately
has
a
c an
usually
form
four
arranged in a
109.5° angles
tetrahedral
structure.
y
Compare
P
51.
e
O
Structure
2
the
ground
state
electron
conguration
of
c arbon, 1s
2
these
(gure
and
observations.
C arbon
contains
two
unpaired
2p
2p
,
c an
deduce
therefore
should
form
two
bonds
(one
involving
each
not
four.
Furthermore,
the
two
occupied
2p
orbitals
are
at
of
an
electron
element
from its
unpaired
atomic
electron),
the
electrons
conguration
81a)
covalent model
2
2s
You
contradicts
The
LHA
However,
2.2
90°
number.
This
is
covered in
from
Structure 1.3.
one
another,
changes
Hybridization
c arbon
for
when
is
the
There
An
means
that
the
atomic
orbitals
must
undergo
form bonds.
of
mixing
example
of
atomic
this
orbitals
process
is
to
form
shown
in
new
hybrid
gure
81
for the
are two steps:
2s
electron
is
r
A
they
concept
bonding.
atom.
This
promoted to one of the 2p orbitals.
2.
Hybridization: The singly occupied 2s and 2p atomic orbitals are hybridized,
meaning they combine and give rise to orbitals of new shapes. The resulting
l
orbitals are called hybrid orbitals and they all have the same energy.
(c)
2s
2p
state electron conguration of c arbon
(b) One of the 2s electrons is promoted to a 2p orbital
The four atomic orbitals are hybridized
to form
four sp
Promotion
the
repulsion
is
energy
does
it
energetic ally
released
not
by
require
experiences
favourable.
the
much
when
energy
of
combined
in
orbitals
therefore
has
s
character
c an
2s
electron
hybrid
2s
and
of
orbitals
one
75%
p
part
character
are
2s
three
2p)
are
produced.
and
three
E ach of
parts
2p,
(gure 82).
H
central
109.5°
+
+
+
carbon atom
3
sp
3
sp
H
3
atomic orbitals combine to form four equivalent sp
H
H
1s atomic
orbitals
orbitals
angles.
therefore
three 2p
v
E
hybrid
a
One 2s and
bond
and
+
(one
3
sp
the
equal to the number of atomic orbitals
orbitals
u
+
is
composed
l
O
four sp
atomic
mixture
2p
3
and
a
25%
hybrid orbitals
109.5°
is
2s

Figure 82
orbitals
four
equivalent
f
x
+
The
four
If
t
a
and
c arbon,
hybrid
hybrid
them.
r
o
these
resulting
make
i
to
formation outweighs
relieves
o
number
combined
it
n
hybrid orbitals
d
The
promotion step absorbs
bond
bec ause
paired.
3
sp
The
subsequent
U
of
the
o
this.
hybridization
but
C
energy,
i
n
O verall,
hybrid orbitals
v
of identic al energy
p
3
(c)
y
(a) The ground
hybrid orbitals
r
e

Figure 81
sp
2p
O
3
2s
n
y
t
i
s
(b)
(a)
y
Promotion:
P
1.
e
orbitals
109.5°.
s
s
certain
not
In
arrange
c arbon,
themselves
these
orbitals
tetrahedrally,
are
each
leading to the
occupied
by
one
electron
3

Figure 83
form
four
sigma
bonds.
For
example,
in
methane, CH
,
each
4
In methane,
each c arbon sp
orbital overlaps with a hydrogen 1s orbital.
3
of the sp
forming
hybrid
four
orbitals
covalent
overlaps
bonds
with
a
(gure 83).
1s
atomic
orbital
on
a
hydrogen atom,
The geometry around
the c arbon atom is
tetrahedral
177
Structure
2
Models
of
bonding
and
structure
LHA
2
sp
hybrid orbitals
2
p
Carbon can also become sp
z
hybridized. The combination of one 2s and two 2p
2
atomic orbitals produces three sp
2
hybrid orbitals. These sp
hybrid orbitals arrange
themselves in a trigonal planar fashion, at 120° from one another (gure 84).
2
sp
The
hybridized
remaining
orbital
contains
unhybridized p
C
one
electron,
orbital
c an
so
then
they
go
on
c an
to
s
s
E ach
2
form sigma bonds.
form a pi bond with a
z
sp
parallel p
orbital
on
a
dierent
atom
(gure 85).
e
z
r
2
sp
2
2
sp
sp
pi
2
2

Figure 84
E ach of the three sp
hybrid
2
sp
sp
H
l
orbitals (blue) contains one electron. The
y
C
atomic orbital (white) also
C
z
contains one electron
t
i
s
n
unhybridized p
sigma
2
sp
p
H
2
orbitals (white) overlap
z
side-by-side and
v
sp
hybrid
of
one
orbitals.
2s
U
them
and
They
C
combination
one
adopt
2p
a
hybrid orbitals
form a pi bond,
The remaining
resulting in a
o
i
n
sp hybrid orbitals
two
2
two hydrogen atoms.
c arbon–c arbon double bond
The
z
the two c arbon atoms each have three sp
p
unhybridized p
,
p
4
sigma bonds with each other and
H
2
sp
y
In ethene, C
(blue) which form
r
e

Figure 85
z
O
H
atomic
linear
orbital
leads
arrangement,
to
with
the
formation of
180°
between
(gure 86).
n
The hybrid orbitals can form sigma bonds. The remaining unhybridized p
sp
C
sp
E ach of the two sp
hybrid
orbitals (blue) contains one electron. The
two unhybridized p
and p
y
atomic orbitals
z
l
u
f
x

Figure 86
z
pi
t
a
p
(gure 87).
i
y
d
r
o
p
sp
sp
H
H
sigma
(white) also contain one electron each
a
E
v
O
178
and
y
orbitals can form two pi bonds with parallel p orbitals on a neighbouring atom
z
o
p
p
pi
p

Figure 87
In ethyne, C
p
,
the one sp–sp
overlap
and
z
the two 1s–sp
overlaps form
three
2
The four unhybridized
c arbon triple bond
y
z
H
2
sigma bonds.
p
y
p
orbitals form two pi bonds,
resulting in a c arbon–
y
P
H
Structure
2.2
The
covalent model
LHA
Hybridization in other atoms
Hybridization
in
the
p
also
orbitals
occurs
will
in
dier,
atoms
but
the
other
than
general
c arbon.
principles
The
are
number
the
same.
of
electrons
For
example,
3
consider
the
oxygen
atom
in
water, H
O, which is sp
hybridized:
2s
2
•
The
•
We
oxygen
c an
distinguish
contains
2
1s
2
2s
ground
2
a
of
the
electrons
electron
three
and
conguration is 1s
2p
the
orbitals
other
to
two
2s
2p
.
show that one of
are
singly
occupied:
1
2p
2p
y
z
3
The
2s
and
2p
orbitals
hybridize
to
form
four
equivalent sp
3
hybrid orbitals
sp
six
electrons
in
total
(gure
88).
Two of the sp
hybrid orbitals

Figure 88
contain
a
pair
of
electrons
each
and
therefore
do
not
form bonds: they
hybrid orbitals
The 2s and 2p atomic orbitals
3
the
two
lone
pairs
on
oxygen
that
you
to form
four sp
are familiar with. The
y
hybrid orbitals
close
the
to
oxygen
atom
orbitals
However,
has
the
on
two
lone
bond
are
are
two
is
occupied
tetrahedrally
of
pairs.
angle
singly
and
go
on
the
This
not
hybrid
arranged,
orbitals
corresponds
exactly
109.5°
are
hybridization
is
close
to
to
the
bent
but
bond
but
not
molecular
exactly sp
diagrams
described
to
for
show
the
the
above.
and
number
its
shape.
remember
This
have
a
as
is
from
single
large
the
eect
pi
E
v
a
O
2
sp
by
to
an
the
of
changes
nitrogen
involve
sp
by the atomic
to
hydrogen to
3
electron domain
atom
of
a
when
is
equal to the
determining
molecular
molecule.
Number
Number of
Electron domain
of electron
non-bonding
Molecular geometry
geometry
domains
4
4
domains
tetrahedral
0
1
2
Hybridization,
bonds
form ammonia, NH
3
trigonal
planar
2
electron domain geometry and
linear
tetrahedral
trigonal
pyramidal
2
bent
0
trigonal planar
1

Table 16
it
hybridized orbitals, so they do
orbitals
3
undergone
when
VSEPR that double and triple bonds
domains
u
l
of hybrid
3
not
geometry
Number
Hybridization
sp
the
do
linked
formed
of
electron
bonds
on
closely
discussion
additional
bec ause
f
x
not
electron domains.
are
orbitals
i
will
treated
hybrid
t
a
are
bonding
of
r
o
You
of
sigma
number
Describe
orbitals
o
The
d
geometry.
54.
n
U
Hybridization and geometry
Hybridization
promotion
oxygen atom in
C
water,
process
geometry.
p
box
are
o
orbital
hybridization
i
n
and
v
Draw
angles
rather 104.5° suggests that
Practice questions
53.
form sigma
already full, so the
3
the
to
hydrogen atoms.
y
that
orbitals
orbitals
hybrid
109.5°.
fact
s
r
e
The
four
hybrid
the
O
Since
two
with
t
i
s
bonds
n
remaining
l
in oxygen are hybridized
constitute
y
3
containing
P
•
r
x
between
pair
1
2p
state
4
e
them
atom’
s
2
s
s
2p
2
0
bent
linear
molecular geometry
179
LHA
Structure
2
Models
of
bonding
and
structure
Practice questions
55.
State
the
56.
hybridization of:
Consider
the
following
species: CF
, HCN, N
4
an
atom
with
b.
the
c arbon
c.
the
oxygen
d.
the
nitrogen
e.
an
tetrahedral
For
electron domain
each
species,
geometry
atom
in
ethene, C
H
2
atom
in
oxygen
of
the
c arbon
deduce
of
the
a.
number
b.
electron
c.
hybridization.
and
H
2
following:
electron domains
4
diuoride, OF
domain
geometry
2
in
molecular
nitrogen, N
2
atom
with
trigonal
pyramidal
57.
molecular
The
Lewis
domains,
f.
an
atom
with
bent
molecular
formula
Deduce
electron
domain
for atoms A, B and C.
H
H
O
H
O
H
H
B
C
r
e
Hybridization and deloc alization
Consider
the
ethanoate
ion,
CH
COO
.
It
forms
when
3
hydrogen
H
bond
orders
bond
become
change
in
gure
from
the
2
c arbon–oxygen
and
1
across
to
1.5
the
89.
In the C=O bond of ethanoic acid, CH
O
2
0.128 nm
sp
hybridized.
U
bond
H
C
3
C
unhybridized
the
them.
–OH
The
When ethanoic acid loses
a hydrogen ion,
atom
orbitals
shown
in
the
c arbon
loses
in
the
double
domains.
and
a
the
This
oxygen
is
orbital
electron
in
each
atom
are both
forms the pi
figure
is
oxygen atom, in –OH, has sp
lost
to
have
thre e
form
As
a
ele ctron
ethanoate,
result,
the
domains
the
remaining
oxygen
around
hybridization.
atoms
them.
oxygen
and
Thre e
the
ele ctron
2
to
sp
overlap
hy bridization
allowing
the
p
and
one
unhy bridize d
ele ctrons
in
them
to
2p
be
orbital
deloc alize d
90.
ethanoate ion
shown as
O
E
v
H
C
CH
C
3
3
— C
—
O
O

Figure 90
180
COOH,
acid
change:
O
a
O
of the ethanoate ion.
2p
hy bridization.
correspond
The
i
structures
as
electrons
c arbon–oxygen
u
resonance
l
two
e ach.
t
a
r
o
f
x
Activity
domains
sp
e ach
the C–O bonds become
equivalent
the
c arbon
an
o
d

Figure 89
adopts
other
hydrogen
2
atom
The
ethanoic
lengths
3
between
When
O
bond
3
The
n
0.128 nm
e ach.
two
C
(b)
so,
deloc alized
i
n
shown
doing
o
O
By
v
0.133 nm
ion.
p
C
C
3
y
O
0.123 nm
Draw
H
O
t
i
s
H
electron
molecular
n
C
A
(a)
and
shown
of
l
y
H
H
geometry
here).
is
number
The deloc alization of electrons due to resonance in the ethanoate anion
y
possible
propanoate
P
are
methyl
hybridization,
geometry (two
geometry
answers
of
the
r
below.
geometry
e
atom
.
2
nitrogen atoms in these
s
s
a.
Structure
2.2
The
covalent model
End-of-topic questions
7
.
The
hydronium
ion
is
formed
when
a
water
molecule,
s
s
Topic review
+
H
O,
reacts
with
a
hydrogen ion, H
.
Which type of
2
1.
Using
your
answer
knowledge
the
guiding
from the
question
as
Structure 2.2
fully
as
topic,
bond
is
formed
in
this
reaction?
possible:
coordination bond
B.
ionic bond
C.
hydrogen bond
e
A.
What determines the covalent nature and properties of a
Compare
and
contrast
ionic
bonding,
covalent bonding
D.
intermolecularforces.
and
contrast
molecular
and
8.
covalent network
intermolecular
The
force
electronegativities,
Multiple-choice questions
Which
of
the
following
species
is
molecular?
A.
2
NO
4
N
4
CH
3
)
)
CO
What
is
the
electron
domain
Cl–P–Cl
trichloride,
Electron domain
A.
3.4
3.2
C–O
Which
of
the
10.
following
allotropes
A.
graphite
B.
graphene
C.
buckminsterfullerene
D.
diamond
What
are
the
intermolecular
of
forces
c arbon is
molecules of CH
present
between
Cl–P–Cl bond
angle / °
F?
3
molecule
A.
London
B.
London (dispersion) forces and dipole–dipole forces
C.
London
and
109.5
109.5
D.
(dispersion)
(dispersion)
forces
forces,
dipole–dipole
forces
hydrogen bonding
hydrogen bonding
pyramidal
11.
trigonal
What
are
the
formal
charges
on
P
and
O
in
the
Lewis
100.3
formula
of
the
phosphate
oxoanion?
LHA
trigonal pyramidal
2.6
molecular
the
geometry
trigonal
a
D.
tetrahedral
the
for
Molecular
tetrahedral
tetrahedral
v
O
C.
?
3
l
B.
tetrahedral
PCl
angle
u
f
x
geometry
geometry,
bond
H–Cl
D.
molecular?
t
a
phosphorus
the
i
and
r
o
geometry
of
CO
2
o
6.
O
d
D.
OH
2
3
9.
C.
c arbon–oxygen bond?
n
(CH
shortest
2
3
C.
the
U
(CH
has
O–H
C
molecule
CH
B.
i
n
A.
3
O
2
Which
Cl
C–H
B.
o
D.
5.
O
p
NH
v
KBr
C.
are:
C
Which bond is the most polar?
O
B.
elements
y
Na
2.2
r
e
A.
four
O
t
i
s
Exam-style questions
4.
H
χ
of
n
Element
χ,
y
substances.
y
3.
and
l
Compare
P
2.
r
substance?
pyramidal
trigonal
100.3
A.
P is
3
1 and O is 0
E
O
pyramidal
B.
P is
+5 and O is
C.
P is 0 and O is 0 and
D.
Both
are
2
1
O
3
O
181
Structure
LHA
12.
2
What
Models
is
the
of
bonding
electron
and
structure
domain
geometry of the sulte
17
.
What
is
the
hybridization
of
the
oxygen atom in ethanol,
of
the
c arbon atom in
2
oxoanion, [SO
]
?
CH
3
CH
3
OH?
2
A.
trigonal planar
A.
sp
B.
trigonal
B.
sp
C.
tetrahedral
C.
sp
D.
bent
D.
It
2
pyramidal
s
s
3
What
is
the
molecular
geometry
of
BrF
18.
?
What
is
not
is
the
hybridized.
hybridization
hydrogen
C.
T-shaped
D.
square
A.
sp
B.
sp
C.
sp
D.
It
2
is
the
molecular
geometry of [PF
]
is
not
?
6
19.
A.
How
many
trigonal planar
molecule
octahedral
following
molecules is non-polar?
4
B.
ClF
C.
BrCl
3
U
D.
SeF
20.
21.
atomic
orbitals
results in a sigma bond?
I and II only
I and III only
I, II and III
pi
3
1
2
2
3
1
Describe
the
meaning of the term
covalent bond.
[2]
+
The phosphonium ion, PH
,
is
formed when a
4
+
hydrogen ion, H
PH
.
For
each
,
reacts
with
a
molecule
of
phosphine,
species, phosphine and phosphonium:
a.
Draw
the
Lewis
formula.
Identify
any
coordination
bonds
[3]
Deduce
c.
Suggest
d.
Deduce
the
molecular
a
v
O
b.
electron
domain
geometry and
geometry.
the
bond
[2]
angle.
[2]
+
I.
Explain
whether
your
it
is
polar
or
non-polar.
reasoning.
[2]
+
II.
E
III.
22.
Methane, CH
, ammonia, NH
4
have
tetrahedral
angles
of
,
and
water, H
3
electron
109.5°,
107°
domain
and
O, all
2
geometry with bond
104.5°,
respectively.
+
Explain
23.
these
Diamond
Describe
these
182
present in a
3
u
II and III only
D.
l
f
x
C.
shown
o
of
are
Extended-response questions
i
B.
le
combinations
2
D.
t
a
A.
the
r
o
below
of
d
Which
B.
C.
6
16.
1
n
5
sigma
A.
C
SF
bonds
cyanide, HCN?
o
the
v
A.
of
pyramidal
pi
y
square
D.
Which
bipyramidal
and
hydrogen
p
C.
sigma
of
r
e
trigonal
i
n
15.
B.
hybridized.
O
What
t
i
s
14.
l
y
3
pyramidal
and
and
two
dierences
graphite
explain
materials.
in
are
the
bond
angle
allotropes
of
values.
[2]
c arbon.
electric al conductivity of
[3]
y
square planar
n
octahedral
B.
cyanide, HCN?
P
A.
r
5
e
13.
(V-shaped)
Structure
24.
Household
acid, CH
vinegar
is
made
from
aqueous ethanoic
25.
The
following
substances
all
have
2.2
The
similar
covalent model
molar
masses.
COOH.
3
i
ethanoic acid, CH
ii
propan-1-ol, CH
COOH
iii
methoxyethane, CH
iv
butane, CH
3
a.
Draw
the
Lewis
formula
of
ethanoic
acid.
[2]
CH
3
b.
Draw
a
diagram
to
represent
the
intermolecular
CH
2
3
between
a
and
a
water
molecule.
CH
acid
molecules
when
c an
gaseous
form
or
Use
your
and
structure
solvent.
acid
Dimerization
molecules
c an
as
of
your
Explain
why,
molar
26.
Explain
27.
Water
O
why
increases
is
mass of an ethanoic
dimer.
[1]
a.
the
down
an
Oxygen
ethanoic
acid
is
more likely to
of
of
the
solvent.
in
aquatic
[1]
group 17 elements
[2]
water bodies supports the
y
why
point
strength of
have similar molar
group.
dissolved
r
e
Suggest
the
excellent
presence
ii.
boiling
that
O
the
H
3
the
forces, it is helpful to
n
CH
C
[4]
l
y
acid
comparing
substances
t
i
s
C alculate
when
intermolecular
masses.
i.
reasoning.
O
C
forces
four substances in
y
H
O
intermolecular
increasing boiling point.
Explain
compare
C
of
these
P
b.
3
list
shown
dierent
H
to
means that two
associate
below.
O
knowledge
dissolved in a
r
ethanoic
3
dimers,
order
non-polar
3
CH
2
e
particularly
CH
2
[2]
a.
Ethanoic
2
molecule of ethanoic acid
3
c.
OCH
s
s
interaction
OH
2
CH
organisms.
Describe the type
of intermolecular forces that occur between oxygen
form
dimers
when
dissolved in non-polar
hexane) than in polar
as
water).
COOH,
[2]
dissolves
readily in
3
water.
The
CH
3
Explain
CH
2
this
solubility
CH
2
dierence
in
low.
in
ethanoic
c arbon and
acid.
a.
[2]
b.
as
a
of
and
higher
dissolve
Explain
dissolve
group
shown
Identify
has
in
cross
from
brain
barrier
two
of
in
water but not
strong
substances
morphine
solvents.
[2]
painkillers
known as opiates.
produces diamorphine
below:
explain which of these two opiates
aqueous
solubility.
[3]
the
is
blood
Identify
opiates
into
the
composed
and
dissolves
of
brain. The blood–
lipids
which
are
explain which of these
more
readily in non-polar
environments.
CH
[1]
2
O
CH
3
a
CH
[1]
why ionic compounds
hydroc arbon
are
OH
N
O.
The potency of opiates depends on their ability to
non-polar.
u
l
2
the
a
molecules, H
2
diamorphine
esteric ation
(heroin),
t
a
r
o
CH
of
[1]
n
hybridization
atoms
f
x
E
v
O
C
acid.
to
water
oen
solvents.
readily
and
belonging
The
i
oxygen
the
ethanoic
not
Morphine
[2]
and
compounds
organic
do
28.
o
Deduce
of
in
hexanoic acid,
very
solubility.
d
f.
3
is
State the number of pi and sigma bonds in a
molecule
H
of
COOH,
2
U
LHA
e.
aqueous
CH
2
Ionic
C
CH
b.
,
2
o
Ethanoic acid, CH
v
as
(such
i
n
d.
(such
solvents
p
molecules, O
solvents
C
H
2
3
CH
2
O
O
2H
O
C
CH
3
C
C
O
O
3
O
2CH
3
COOH
O
C
OH
morphine
O
CH
3
diamorphine (heroin)
183
Structure
A
Models
paper
of
bonding
and
chromatogram
obtained
using
a
structure
for
a
non-polar
mixture of amino acids is
31.
The
c arbonate
ion
contains
deloc alized
LHA
29.
2
electrons.
solvent.
a.
Draw
the
three
resonance
structures of the
2
a.
C alculate
the
retardation
factor,
R
,
for the spot
c arbonate
ion,
CO
F
X
in
the
Deduce
which
chromatogram.
b.
b.
to
have
the
of
the
lowest
[2]
[2]
three components is likely
polarity.
State
the
bond
in
bond
the
order
of
c arbonate
the
c arbon–oxygen
s
s
labelled
.
3
ion.
[1]
[1]
32.
Benzene, C
H
,
contains
deloc alized
electrons and is
6
represented
as
e
6
oen
follows.
r
X
l
a.
Explain,
with
reference
Y
deloc alized
Z
b.
and
Outline
The
diagram
a
the
a
chromatogram
of
sauce
food
chromatogram.
of
(labelled
colourings
Outline
an
S).
have
extract
further
Four
also
three conclusions
33.
c an
be
derived
from
the
information in the
chromatogram.
Sulfur
leading
a.
n
U
a
S
E122
E124
o
i
t
a
u
l
v
E
184
d
r
o
f
x
O
E110
to
Draw
the
front
the
molecule.
[2]
of
the
,
is
an
[1]
chemic al evidence that
structure
of
benzene
shown
[2]
acidic
gas.
If
present in the
3
b.
E102
trioxide, SO
atmosphere,
[3]
solvent
supports
an
acid
two
octet
Lewis
Deduce,
which
of
Explain
c an
dissolve
in
rainwater
formulas
for SO
,
one
that
follows
3
rule,
Determine
in
trioxide
rain.
expanded
atoms
c.
sulfur
C
i
n
that
piece
above.
been
of
why the
form a ring
benzene.
one
o
on
samples
is
curry
v
reference
run
below
supermarket
of
Describe
p
from
plane
y
r
e
30.
the
hybridization,
benzene
how bond length data conrms the
structure
c.
below
to
in
O
t
i
s
above
1.70 cm
0.75 cm
electrons
n
y
4.10 cm
and
your
formal
Lewis
using
the
where the sulfur atom has
octet.
the
the
one
the
two
[3]
charges
formulas
concept
Lewis
of
you
of
formal
formulas
reasoning.
each of the
drew
is
in
(a).
[2]
charge,
preferred.
[1]
y
P
4.90 cm
5.40 cm
Structure
LHA
34.
Ethene, C
H
2
,
belongs
to
a
group
of
substances
known
36.
The
condensed
structural
formula
2.2
of
The
covalent model
phenylamine is
4
as the alkenes.
C
H
6
NH
5
.
A
molecular
model
of
phenylamine
is
shown
2
below.
a.
Draw
the
b.
Deduce
Lewis
formula
for
ethene.
[1]
H
c.
the
molecular
atoms
Suggest
in
values
geometry
of
each of the
ethene.
for
the
[1]
H
following bond angles in
H
ii.
HCC
r
HCH
N
C
Deduce
f.
Explain
why
around
the
of
ethene.
[1]
H
C
of
the
H
c arbon atoms in
ethene.
main
type
between
your
of
in
ethene.
intermolecular
ethene
[2]
force
molecules.
reasoning.
[2]
the
hybridization
in
graphite.
electric al
conductivity
of
c arbon atoms in
Using
each
these
of
the
data,
two
explain the
materials.
Deduce
the
molecular
atoms
in
[4]
b.
State
the
U
C
A
domain
of
the
geometry and
c arbon
and
nitrogen
phenylamine.
nitrogen
c.
electron
geometry
o
and
the
i
n
diamond
of
v
State
H
H
p
a.
35.
C
C
y
Explain
the
bond
restricted
r
e
present
c arbon
is
O
Identify
c arbon
rotation
t
i
s
g.
[1]
molecular
n
hybridization
y
C
the
hybridization
atoms
theoretic al
y
e.
l
State the number of sigma and pi bonds in a
P
d.
molecule
C
[2]
e
ethene:
i.
s
s
c arbon
in
[2]
of
the
c arbon and
phenylamine.
study
of
the
[2]
electron
structure of
phenylamine suggests that the H–N–H bond
n
angle
is
in
very
o
i
u
l
you
to
may
geometry
the
should
conclude
around
be
experimental
the
about
112.79°, which
value.
the
Discuss
molecular
nitrogen in the –NH
group
2
in
the
structure
hybridization
37
.
C arbon
bonds
c arbon
with
atoms
with
atoms
each
are
other.
phenylamine,
state
c an
other
of
of
form
nitrogen
single,
c arbon
to
on
deduce the
this
double
atoms.
unlikely
and
or
Suggest
basis.
triple
[3]
covalent
why two
form quadruple bonds
[2]
a
E
v
O
f
x
t
a
r
o
d
what
phenylamine
close
185
The metallic model
Structure 2.3
s
s
What determines the metallic nature and properties of an element?
A
large
proportion
elements,
millennia.
into
very
ability
wires,
useful
of
and
and
E arth’
s
humans
metals
to
conduct
versatile.
crust
have
be
is
The
composed of
been
using
metals
metals
moulded into shapes,
electricity
These
and
are
a
are
Metal
result
c an
many
found as elementary substances in
are
found
in
ores,
in
their
oxidized
reduced
be
and
state.
posing
treated
is
the
c ations
electrostatic
and
provided
safely
Structure 2.3.3
deloc alized
require
and
—
that
waste
many metals
is
collected
Transition
elements
have
deloc alized
o
C
i
n
v
on the charge of the ions and the radius of the metal ion.
However,
metallic
correctly.
p
Structure 2.3.2 — The strength of a metallic bond depends
large amounts of
risks.
d-electrons.
electrons.
reactivity
of their molten
Metallic structures (Structure 2.3.1)
100
described
electrons.
related
to
6 ,0 0 0
metal
y e a rs
artefacts
a
v
E
186
be
oldest
a
Jordan
o
the
example,
known
elements
of
electron
groups.
the
they
Therefore,
of
metals.
between
deloc alization
which
increase
properties
are
attraction
electronegativity,
trend:
i
of
the
For
118
electrostatic
degree
down
opposite
t
a
estimated
O
k n ow n
is
one
aw l ,
The
the
gradually
the
the
their
decreases
u
It
and
in
l
old,
copper
discovere d
d
Va l l e y.
Ancient
tool,
r
o
piercing
f
x

Figure 1
to
of
as
n
U
Nearly
period
3
in
generally
metallic
down
Bonding
metal
pure
and
elements,
from metallic to non-metallic (table 1).
and
of
is
c an be
deloc alized
inversely
across periods and
elements
decrease
from
metals
elements
increases
properties
groups
in
c ations
demonstrate
across periods.
sodium
to
argon, change
LHA
bond
of
achieved with the
Higher
n
lattice
is
O
metallic
ore
y
a
processes
t
i
s
A
between
its
furnace.
electrolysis
r
e
attraction
by
from
blast
environmental
recycled,
Understandings
Structure 2.3.1 —
a
y
but
in
y
Some
nature,
iron(II)
l
energy,
are
of
c arbon
extraction
of metallic bonding.
metals
of
compounds.
heat makes
properties
reduction
addition
for
P
them
the
r
drawn
The
of
and
e
metallic
Structure 2.3 The metallic model
Na
Mg
Al
Si
P
S
Cl
Ar
metal
metal
metal
metalloid
non-metal
non-metal
non-metal
non-metal
metallic
metallic
metallic
covalent
molecular
molecular
molecular
monatomic
network
covalent
covalent
covalent
semi-
low
low
low
2.2
2.6
3.2
Metal, metalloid,
or non-metal?
Structure
high
high
high
conductivity
3
1.6
to
the
notes
to
way
a
properties
eort
their
peer ’s
to
particles
and
is
a
connect
see
if
central
each
of
behave.
there
any
are
non-metallic
highly
are
shown
values
electronegative
somewhere
in
the
properties.
deloc alized.
rather
move
negatively
of
results
electrons
lattice
electrons
from
the
of
the
greater
greater
metallic
covalent
character
character
metallic
covalent
bonding
bonding
electrons
c ations.
This
creates a
1.0
2.0
3.0
4.0
mean electronegativity
between
metal
c ations
(gure 3).
+
p
Figure 2
bonding
The
metallic–cov alent
continuum
+
+
+
-
+
+
+
+
+
+
+
+
+
+
+
an
form
c ations.
+
s h ow i n g
that
have both
attached to individual ions,
attraction
them
On the le,
form metallic
valence
packed
t
a
metal
not
their
surrounds
around
u
a
are
closely
that
+
a
of
deloc alized
atoms,
electrostatic
electrons
l
Structure
of
the
+
v
“s e a ”
of
deloc alized
E
a
“sea”
f
x
O
p
Figure 3
by
within
metallic
2.
to
they
o
bonding
sea
Deloc alized
tend
non-metals
i
the
charged
other
r
o
and
freely
to
d
Metallic
bond
gure
compare
dierences.
n
become
atoms
U
metallic
nished,
and/or
middle:
read,
C
The metallic bonding model
When
in
that
you
o
and
there
Metalloids
have
As
characteristics of
y
right,
is
electronegativity
you
similarities
i
n
metallic
the
bonds.
low
chapter.
p
On
covalent
with
this
v
bonds.
elements
in
observable
aects
between
r
e
are
behaviour
connection
idea
the
When
are
metallic–covalent bonding continuum
there
microscopic
The
y
your
physic al
how
materials.
n
metals
and
conscious
about
and
l
a
thinking
matter
O
make
of
t
i
s
structure
involves
features
N/A
P
oen
observable
but
1.9
elements
e
1.3
period
r
of
y
Chemistry
The
0.9
Properties
Thinking skills
ATL
low
conductor
Electronegativity
p
Table 1
s
s
Electric al
a r ray
of
positive
ions
(c a t i o n s)
surrounde d
ele ctrons
187
Structure
2
Models
of
bonding
and
structure
+
–
The properties of metallic structures
+
−
−
The
presence
of
deloc alized
electrons
gives
rise
to
many
characteristic
properties
−
of
+
+
metals.
Metals
are good
electric al conductors, thermal conductors, and they
+
are
malleable
and
ductile.
−
+
+
Substances
+
−
Consider
−
a
c an
conduct
sample
of
electricity
metal.
It
when
contains
a
mobile
lattice
charged
of
c ations
particles
are
surrounded
s
s
−
−
present.
by
−
free-moving,
applied,
donated
deloc alized
the
negative
the
difference
ele ctrons
terminal
positive
are
randomly
applied,
and
and
negatively
throughout
there
towards
is
the
a
net
charged
the
electrons.
metallic
movement
positive
structure.
of
terminal
The
deloc alized
When
electrons
away
electrons
potential
from
the
move
dierence is
negative terminal
(gure 4).
terminal
objects.
In
some
They
Mirrors
have
have
metal
types
of
and
to
knowledge
some
appeared
a
most
glass
might
in
why
they
are
used to make
believed
as
to
have
magic al
valuable, status-signalling
literature, both as instruments of clarity and
we
know
the
household
surface.
approach
knowledge
were
perceived
is
sparked human fascination
more
These
the
deloc alized
mirrors
are
are
electrons in metals
made
examples
by applying a
of
how
dierent
same topic.
open
to
interpretation
than
others?
n
U
C
i
n
o
Are
of
light,
mirrors
been
Nowadays,
v
areas
silver
also
which
y
of
cultures,
well,
historic ally
p
layer
incident
light
have
r
e
misrepresentation.
reect
reect
n
curiosity.
properties.
and
surfaces
O
and
lustrous
Reective
l
are
mirrors.
t
i
s
Metals
y
TOK
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
p
Figure 5
p e o p l e,
188
a
Anish
buildings
K a p o o r ’s
and
sky
Cloud
around
G ate
it
on
in
its
Millennium
curved,
Park,
s t a i n l e ss
C h i c a g o,
steel
refle cts
surface
the
y
t ow a r d s
potential
P
move
the
by
a
r
is
When
e
p
Figure 4
Structure 2.3 The metallic model
One
of
the
reasons
conductivity,
substance
is
heated,
become
is
its
particles
more
a
vibrations
vibrations
passed
vibrations
and
movement
this
metals
collisions
each
which
collision.
increases
reduces
path.
In
this
in
electrons
the
other
lattice
ions.
collisions
the
in
energy to the
metals
parts
of
vibrate
Some
are
temperature
frequency
c ase,
transfer
electrons
to
magnitude.
of
are
the
lattice.
heat.
the
and
These
with
of
in
bec ause of their
the
energy is
c ause of
(gure
collisions,
resistance
more, so
kinetic
6).
electric al
Conversely,
so
the
electrons
decreases.

Figure 6
400
Re s i s t a n c e
line arly
t e m p e ra t u r e
300
Ω / e c n at s i s e r
300
600
900
temperature / K
without
wide
to
array
this
ability
This
of
of
oer
very
no
low
resistance
a
solid
useful
shapes,
to
be
property
ranging
anchors.
to
temperatures
property.
pressed
is
the
l
u
a
zippers
the
breaking.
f
x
such
is
exhibit
that
At
t
a
M alleability
materials
i
materials
are
temperature.
r
o
other
“critic al”
from
o
certain
d
Superconductors
or
–
+
–
–
+
+
+
+
+
+
+
+
+
+
+
+
+
+
–
push
–
current
–
–
below
many metals and some
+
–
pounded
reason
sewing
electric
–
–
1200
n
0
C
U
–100
o
v
i
n
100
with
most
p
metals
200
0
a
for
y
r
e
inc re ases
O
direct
ions
also
the
energy
conductors
increase
c ations
c ations
Since
kinetic
increases,
between
temperature
more
vibrating
good
c ations
other
n
with
its
to
collisions.
increased
are
of
along
y
in
The
through
metals
metals,
in
their
energy of a
When a substance
y
a
thermal
t
i
s
in
pass
heat
in
the
easily
arrangement.
why
to
decrease
move
are
energy
resistance
a
heated,
c an
more
converted
so
The
particles.
l
are
energy,
well.
its
e
thermal
there
of
P
As
explains
kinetic
cookware is their thermal
heat
kinetic energy
gain
electrons
they
make
r
This
is
packed
surrounding
to
conduct
s
s
metal
used
they
vigorous.
These
mobile,
are
that
related to the
When
closely
metals
meaning
into
–
dierent shapes
electron ‘sea’
why
metallic
needles
to
objects
have
aeroplanes to
–
–
+
+
+
–
+
–
+
a
O
Metallic bonding is non-directional. When a force is applied to a metallic structure,
+
+
+
+
the layers of cations can slide past each other without breaking the electrostatic
–
–
–
attraction to the surrounding delocalized electrons. The cations can therefore
v
+
+
+
+
+
+
be rearranged, allowing the metal to take on a new shape (gure 7). Aluminium
foil, oen used to wrap food, demonstrates that the metal aluminium is readily
E
malleable.
A
related
Ductility
that
property,
and
allows
+
–
ductility,
electric al
electric al
is
the
ability
conductivity
wires
to
be
are
a
of
a
solid
very
produced
at
to
useful
the
be
stretched
combination
industrial
sc ale.
into
of
wires.
properties
p
Figure 7
Metallic
intact
even
into
sheet
a
after
or
a
bonding
metal
other
–
is
remains
hammere d
shape
189
Structure
2
Models
of
bonding
and
structure
To
summarize,
metals
are
typic ally:
•
good electrical conductors because they contain mobile delocalized electrons
•
good
and
malleable
without
conductors
packed
and
bec ause
they
contain
mobile
deloc alized
electrons
c ations
ductile,
bec ause
the
layers
of
c ations
c an
slide
past
s
s
•
thermal
closely
each other
breaking the metallic bonds.
e
r
l
blast
furnace
p
Figure 9
Aluminium
aluminium
c ans
retains
C
i
n
o
metals,
n
a
O
f rom
y
metal
p
iron
its
in
a
re cycling
properties
plant.
after
Like
being
other
re cycle d
Database investigation of the properties of metals
this
skills
task,
you
will
Relevant skills
Inquiry
2: Identify
•
Inquiry
1:
and
Formulate
2:
a
Collect
database
extract
o
Tool
•
i
d
data
to
explore
the
properties of metals.
from databases
research questions
and
record
sufficient
relevant
quantitative data
Instructions
Using
that
u
l
a
E
v
O
f
x
t
a
r
o
•
use
n
U
In
This
at
will
least
may
an
you
involve
conductivity
are
one
allow
AHL
of
database
to
looking
a
at
number
student,
of
your
investigate
you
choice,
the
trend
electric al
of
metals
may
wish
extract
to
the
and
analyse data
properties of metals.
conductivity, melting point or thermal
across
to
identify,
relating
a
period
explore
a
or
down
property
of
a
group.
the
(If
you
transition
elements.)
State
a
research
question
that
is
addressed
by
your
data
collection and
processing.
Before
starting
this
task,
create
a
plan
of
the
steps
you think it will
ATL
require
do
the
and
task
estimate
and
keep
the
a
time
log
of
needed
the
to
actual
complete
time
each
step. Then,
you spent completing the
various steps.
When you have finished, compare the two sets of steps and times. How
do they compare? How can these observations help you plan extended
tasks in the future?
190
y
P
y
t
i
s
r
e
Liquid
v
p
Figure 8
Structure 2.3 The metallic model
Practice questions
1.
State
the
property
(or
properties)
of
metals
that
makes
them
suitable
for
manufacturing:
kitchen pans
b.
electric
c.
radiators
d.
c ars
e.
articial hip joints
f.
guitar strings.
power
labelled
labelled
diagram
diagrams
electric al
good
thermal
bonding
metals
occurs in metals.
are:
conductors
conductors.
O
t
i
s
Linking questions
n
good
c.
how
why
l
malleable
b.
explain
explain
y
a.
to
to
y
Draw
e
a
3.
r
Draw
c ables
P
2.
s
s
a.
r
e
What experimental data demonstrate the physical properties of metals, and
trends
in
reactivity
of
metals
c an
be
predicted
from
the
periodic
table?
C
i
n
The strength of metallic bonds
(Structure 2.3.2)
stronger
the
attractions
between
the
U
The
c ations
strength
decreases
greater ionic charge
factor,
point
the
group
and
boiling
bonding
1
metals
data
a
melting
and
A
the
trends
c an
be
density
used
to
of
the
with
“sea” of
compare
the
strength
Consider the melting and boiling points of
and
potassium (table 2).

Table 2
Melting
data
10
point / °C
point / °C
181
1342
1+
102
98
883
1+
138
63
759
points
ionic
m
decrease
radius
deloc alized
Melting
and
boiling
point
Boiling
for
group
1
metals
–12
76
boiling
greater
electron
increases
1+
v
E
potassium
and
the
Ionic radius /
the c ation
increases.
is
metals.
sodium
Charge of
sodium
attraction
point
dierent
lithium,
lithium
c ations
in
l
O
Element
The
charge,
electrons, the
u
metallic
ionic
electrons.
f
x
of
to
t
a
Melting
related
greater ionic radius, and
i
third
deloc alized
r
o
A
with
deloc alized
o
bond
d
Metallic
and
n
stronger the metallic bond.
o
v
(Reactivity 3.2)
p
What
y
trends in these properties, in the periodic table? (Tool 1, Inquiry 2, Structure 3.1)
means
electrons.
a
down
the
greater
This
group
average
weakens
the
as
the
ionic
distance
radius
between the
electrostatic
forces of
between them.
191
Structure
2
Models
of
bonding
and
structure
Now
compare sodium against two other period 3 metals, magnesium and
aluminium.

Table 3
Melting
and
boiling
point
Element
data
for
period
3
Charge of
Ionic radius /
Melting
Boiling
point / °C
point / °C
metals
–12
the c ation
m
s
s
sodium
10
102
98
883
2+
72
650
1090
aluminium
3+
54
660
melting
and
boiling
The
points
Decrease in ionic radius:
Greater
number
contributes
one
the
distance
attraction
increases
and
valence
the
valence
electron
that
to
the
between
strength
deloc alized
electrons
increase
sea
becomes
the
c ations and
between them
of
the
electrostatic
electrons
of deloc alized electrons per ion:
three
aluminium
of
E ach aluminium atom
electrons. Sodium atoms
y
only
this
c ations
r
e
have
the
and
O
3.
between
reduces
electrostatic
Increase in ionic charge:
attraction
this
magnesium
are:
n
the
this
t
i
s
2.
increasing
sodium,
for
y
electrons,
of
reasons
deloc alized.
p
v
Using trends in data to predict properties
skills
the
task,
you
will
properties
examine
of
U
•
that
result,
e asily
in
cut
1.
Examine
of
is
with
2: C arry
Instructions
sodium
sodium
these
trends
a
2.
u
l
a
E
v
O
192
be
f
x
knife
c an
bonding
a
Inquiry
3.
the
Look
your
down
Briefly
5.
Draw
6.
Reflect
of
that
in
your
second
on
relevant
point
will
help
how
second
graph
group
values
evaluate
a
interpret
charts
and
graphs
and
accurate
data
processing
and
you
boiling
point
data
in
table
2.
Plot
a
graph
analyse the melting and boiling point
group 1.
predicted
4.
group 1 melting points to
graphs
out
your
metals
the
and
melting
data
Extrapolate
other
in
o
metal
As
3: Extrapolate
i
soft
Metallic
we ak.
3: Construct
Tool
t
a
a
relatively
d
is
r
o
p
Figure 10
Tool
n
•
trends
unknown elements.
Relevant skills
•
the
C
predict
o
this
i
n
In
up
in
predict the melting and boiling points of the
(rubidium
the
values.
the
to
1
chemistry
C alculate
validity
graph
of
plotting
graph.
and
of
the
these
data
data
the
booklet
and
percentage
compare them to
errors.
extrapolation.
data,
that
c aesium).
including
covers
a
Rb
larger
and
Cs
range
this
time.
affects the quality
y
period.
l
1.
the
P
across
2519
r
The
e
1+
magnesium
Structure 2.3 The metallic model
Practice questions
4.
The
melting
points
respectively.
than
List
of
A.
Be,
C.
C a,
Which
C a,
of
and
c alcium
why
the
are
63 °C
melting
and
point
of
842 °C,
c alcium is higher
Sr,
in
order
of
increasing melting point.
Ba
Ba, Be
the
following
explains
B.
Be,
Ba,
D.
Ba,
Sr,
C a, Sr
C a, Be
why aluminium has a higher melting point
r
than
reasons
potassium.
metals
Sr,
potassium
magnesium?
The
contains
ionic
more
III.
The
charge
deloc alized
electrons per ion than magnesium
3+
radius of Al
of
an
2+
is smaller than that of Mg
aluminium
ion
is
greater
than
that
of
a
magnesium
B.
II only
II and III only
D.
I, II and III
features
of
metallic
bonding
that
make
it
possible
for metals to
(Structure 2.4)
p
The
position
transition
table
are
of
the
elements
discussed
d-block
in
the
in
LHA
o
C
(Structure 2.3.3)
v
i
n
Bonding in transition elements
O
the
y
are
alloys?
r
e
form
n
I only
C.
t
i
s
A.
Linking question
What
l
y
ion.
y
Al
II.
P
I.
e
6.
that
these
of
three
s
s
5.
State
and
periodic
Structure
3.1.
Transition elements are found in the d-block in the periodic table, and they have
U
many properties typically associated with metals: hardness, strength and high
n
density. They exhibit variable oxidation states and, frequently, catalytic properties.
Transition elements are dened as those that have atoms, or give rise to ions, with
o
d
incomplete d sublevels. Most of the d-block elements are transition elements, but
there are some exceptions. For example, zinc is not a transition element because its
i
r
o
2+
3d sublevel is full, both when it is an atom and when it forms the only stable ion Zn
t
a
Observations
obtain
are
alludes
by
analysed
precipitating
new
to
the
French
a
lead
out
element
colourful
the
E arth
and
all
The
other
name
compounds
Nicolas-Louis
sample
lead,
aer
at
phenomena.
chemist
ore
the
looking
natural
a
discovered
this
about
v
O
chromium
and
observers,
data
u
to
l
f
x
Scientists
it
parts
of
the
forms.
Vauquelin,
of
the
universe,
transition element
Chromium
who
was
observed
p
Figure 11
made
containing
chromium
from
Siberia.
f rom
palladium.
Vauquelin
noted
the
dierent
colours
and
word
for
colour,
c atalytic
c o nv e r t e r
rhodium
Tra n s i t i o n
metals
and
often
h av e
named
c atalytic
Greek
C ar
platinum,
Aer
properties
due
to
their
v ariable
chromos.
E
ox i d a t i o n
states
193
2
Models
of
bonding
and
structure
LHA
Structure
In
transition
well
the
as
the
electrostatic
that
surrounds
compared
chemic al
properties
to
the
of
forces
them.
group
of
outer
results
and
in
the
level.
attraction
This
1
electrons
the
in
d
The
sublevel
greater
between
higher
the
become
electron
c ations
melting
and
points
deloc alized, as
density
for
the
strengthens
electron
sea
transition elements
group 2 metals.
s
s
The
elements,
electrons
of
The large number of delocalized electrons in transition elements also allows for
transition
elements
are
discussed
good electrical conductivity. This is because there are more delocalized electrons
in
Reactivity
3.4.
e
that can move along the metal sample when a potential dierence is applied.
The
inc andescent
Tu n g s t e n ’s
melting
point
high
t e m p e ra t u r e
stone:
suitable
and
f rom
“ tung
It
d e n s e.
the
is
In
Swedish
for
also
fact,
for
sten”
o
of
the
presence
following
properties
deloc alized
d
coloured compounds
B.
high density
Which
the
Transition
M any
i
B.
oen
C.
elements
due
contain
Transition
u
D.
the
element
Transition
of
are
elements
C.
good
D.
c atalytic
is
a
result of the
d
d
oen
electric al conductivity
properties
incorrect?
have high melting points and boiling
unpaired
d
electrons.
are
coloured
bec ause their ions
sublevels.
good
deloc alized
elements
transition
compounds
incomplete
elements
of
generally
presence
transition
presence
electric al
conductors due to the
electrons.
have
multiple
oxidation states.
2+
9.
l
a
v
O
f
x
t
a
r
o
d
points
of
electrons?
following statements is
o
A.
of
n
A.
U
8.
of
C
i
n
Which
y
p
r
e
v
Practice questions
7
.
O
t
i
s
h e av y
it
applic ations.
strong
derives
ex t ra o r d i n a r i l y
makes
n
name
in
contains
y
exc e p t i o n a l l y
filament
bulb
l
high
its
light
Explain,
zinc
is
with
not
a
reference
to
the
electron
congurations of Zn and Zn
,
why
transition element.
Thinking skills
ATL
As
you
know,
there
are
various linking questions in this book that aim to
Why
is
the
E
Linking question
interconnect
networked
of
metals
evident
that
194
explores
least
one
other
3.1)
the
part
across the d-block?
compare
(Structure
content
understanding
of
and
encourage
chemistry.
Write
a
the
development of a
linking
connection
between
the
question
content
in
this
of
your
section and
across a period
at
less
course
trend in melting
own,
points
the
answers.
of
the
DP
chemistry
course.
Share it with a partner and
y
tungsten.
g l ow i n g
P
this
r

Figure 12
Structure 2.3 The metallic model
End-of-topic questions
1.
Using
your
question
knowledge
as
fully
as
from the
Structure 2.3
topic,
s
s
Topic review
answer the guiding
possible:
and
contrast
ionic,
metallic
Compare
and
contrast
ionic,
metallic,
and
r
Compare
3.
covalent bonding.
covalent
network,
and
molecular
structures.
Electrostatic
deloc alized
attractions
a
lattice
of
positive ions and
Electrostatic
attractions
between
a
lattice
of
negative ions and
attractions
between
protons
Electrostatic
attractions
between
oppositely
has
a
metallic
structure?
Solubility in
Electric al
water
conductivity when
A
36
high
B
186
insoluble
low
C
1083
high
high
D
1710
molten
when solid
low
t
a
high
o
i
d
r
o
insoluble
Electric al
conductivity
n
U
Melting
point / °C
C
substance
electrons.
charged ions.
i
n
Which
and
o
Electrostatic
D.
v
C.
p
protons.
y
between
electrons.
deloc alized
5.
describes metallic bonding?
r
e
B.
best
n
A.
statement
O
Which
t
i
s
Multiple-choice questions
4.
l
y
Exam-style questions
y
P
2.
e
What determines the metallic nature and properties of an element?
low
low
low
high
u
l
f
x
a
E
v
O
195
Structure
2
Models
of
bonding
and
structure
Extended-response questions
6.
Metals
have
many
useful
properties
including
malleability
and
electric al
conductivity.
s
s
sodium,
magnesium,
aluminium,
copper,
Na
Mg
Al
Cu
2.1
2.3
3.8
5.9
Element
Electric al
7
× 10
e
conductivity /
–1
S m
r
why
metals
are
malleable.
Include
a
labelled
answer.
conductivity
Describe
Explain
why
increasing
of
metals
decreases with
[2]
y
data
for
the
period
3
elements
Si
Identify
b.
Plot
c.
Explain
d.
Describe
a
of
trend
and
of
boiling
o
the
3
structure
graph
i
t
a
the
the
519
n
U
d
a.
2
in
C
090
each
point
the
of
vs
shown
P
the
S
atomic
Cl
in
period
2
–34
3.
number.
across
in
Chlorine,
8
445
elements
[1]
below.
Sulfur,
4
281
point
trend
is
Phosphorus,
265
boiling
explain
o
v
Silicon,
Al
i
n
Aluminium,
1
p
point
Mg
the
electric al
Argon,
Ar
–186
[1]
[1]
period.
conductivity
[3]
across
period.
[3]
u
elements
8.
Explain
of
196
conductivity
[2]
a
3
electric al
M agnesium,
l
v
E
period
electric al conductivity of
temperature.
Na
883
r
o
f
x
O
The
the
below:
Suggest why the electrical conductivity of copper is signicantly higher
Boiling
point / °C
Figure 13
in
[3]
shown
aluminium.
Sodium,
Boiling
▴
trend
is
than that of the other three elements shown in the table.
7
.
Element
the
the
and
r
e
LHA
iii.
explain
elements
O
ii.
and
magnesium
several
t
i
s
sodium,
of
your
n
i.
electric al
in
l
The
y
b.
diagram
why
metallic
melting
bond
point
strength.
and
boiling
point
are
good
indic ators
[1]
y
Explain
P
a.
From models to materials
Structure 2.4
s
s
What role do bonding and structure have in the design of materials?
So far in
Structure 2,
classied
and
metallic.
use,
were
chemistry,
civilizations
Bronze
developed
before
development
according to
Age,
based
explanations
on
for
Another
by the materials
and
Iron
Age.
Uses
into
for
these
observations of their
those
From
metals
materials
to
nanotechnology,
sometimes
are
discovered
and
by
the
uses
sometimes
into
their
structures,
of
Polymers
while
molecules
chance.
and
polymers.
will
help
Alloys
polymers
very
are
p
macromolecules,
represented by a bonding triangle.
monomers.
C
Structure 2.4.4 —
between the ionic, covalent and metallic models, and can be
Polymers
made
grouping them
ceramics.
Two of
Structure 2.4: metals
Your
you
are
the
from
knowledge of
understand
the
nature
predominantly metallic
covalent compounds.
structure, although their
large.
o
i
n
Structure 2.4.1 — Bonding is best described as a continuum
and
and
and
heavily
involves
polymers
molecular
v
Understandings
are
draws
considered in
polymers.
have
It
materials
metals,
bonding
structure
y
properties
research
deliberate,
and
r
e
unique
is
alloys
the
applic ation,
engineering.
classify
alloys)
and
of
their
materials.
and
are
study
and
O
of
to
c ategories
structure
the
new
types:
(specic ally,
properties had
proposed.
of
physics
way
three
is
materials
t
i
s
been
Age,
of
n
properties,
classied
science
properties
y
materials
be
M aterials
c an
covalent
properties.
characterized
Stone
also
materials
ionic,
y
has
use:
and
c an
how
bonding:
l
they
origin,
discussed
their
P
History
M aterials
have
to
r
their
we
according
e
be
are
from
large
molecules, or
repeating
subunits
c alled
Structure 2.4.5 — Addition polymers form by the breaking
bonding triangle is determined by the relative contributions
of
Structure 2.4.3 —
non-metals.
They
mixtures of a metal and other
have
in
each
monomer.
Condensation
reaction
functional
with
the
between
release
of
a
small
polymers
groups
in
form
by the
each monomer
molecule.
i
u
printer
bond
a
3D
l
A
v
E
1
properties.
double
t
a
r
o
f
x
O
p
Figure
enhanced
o
or
are
d
metals
Alloys
a
Structure 2.4.6 —
LHA
of the three bonding types to the overall bond.
n
U
Structure 2.4.2 — The position of a compound in the
cre ating
a
model
from
a
polymer
material,
and
t ra d i t i o n a l
J av a n e s e
gamelan
metallophones
197
Structure
2
Models
of
bonding
and
structure
The bonding continuum (Structure 2.4.1)
strongly ionic,
Some materials cannot be categorized easily into substances with metallic, ionic and
e.g. CsF
covalent bonding. We saw this in Structure 2.1, with compounds such as aluminium
chloride, AlCl
, having a hybrid of covalent and ionic bonding. We also discussed
3
s
s
metalloids in Structure 2.3, which exhibit covalent and metallic properties.
Therefore, bonding is best described as a continuum between these models.
chemists
diagram
to
Anton
represent
Eduard
the
van
Arkel
continuum
of
and
the
Jan
Ketelaar
three
constructed a triangular
bonding
types.
The
three
strongly
corners
represent
sides
the
“pure”
metallic,
ionic
and
covalent
bonding
r
strongly
(gure 2). The
covalent,
metallic
of
bonding
example,
an
the
Al–F
intermediates
bond
c an
be
between one bonding type and
positioned
roughly
ionic–covalent continuum.
Arkel-Ketelaar
halfway along the
d i a g ra m
classic ation
This
anomalies.
M aterials
of
us
c an
properties,
disadvantages
science
each
to
be
to
organize
recognize
classied
structure,
and
objects,
trends
and
according
bonding.
to
information and
patterns as well as
dierent
What
system?
are
v
p
y
origin,
in
help
r
e
use,
c an
O
use
concepts.
t
i
s
We
n
y
Patterns and trends
criteria such as
the
advantages and
solubility
bond
type,
conductivity
U
particles.
and
volatility
structure
of
Table
a
1
and,
if
of
a
substance
C
i
n
The
applic able,
substance
depends
summarizes
o
The properties of materials with dierent bonding types
these
on
and
c an
be
explained in terms of
intermolecular
the
presence
other
forces. The
of
mobile
electric al
charged
properties of substances with
Metallic
soluble
conductor
corrosion
Ty p i c a l
where
properties
substances
Molecular covalent
Covalent network
low
high
low
high
low
varies
varies
no
no when solid
no
no
yes when liquid/
(except
and
for
graphite
graphene)
aqueous
a
v
exc e p t i o n s
to
E
p
Table 1
198
yes
conductor
susceptible
no
l
O
thermal
low
u
electric al
brittle
f
x
water
high
i
volatility
varies
Bonding type
t
a
boiling point
r
o
melting point and
Ionic
o
d
Property
n
dierent bonding types.
yes
no
no
no
yes
no,
varies
except some
yes
polymers
yes
of
varies
metallic,
do
not
cov alent
ex h i b i t
and
typic al
varies
ionic
substances.
b e h av i o u r
This
table
no
is
me ant
as
a
g e n e ra l
guide
and
there
are
y
v an
For
l
triangular
The
represent
2
another.
p
Figure 2
triangle
P
e.g. F
e.g. Cs
e
Dutch
Structure
Brittleness
malleable.
they
c annot
unable
to
is
and
coil
up
again.
property is
even
aer
with
describe
the
force
the
oxidation of a metal, such as
more
surrounding
a
manner
Graphite
molecular
more
than
This
size
and
one
to
its
is
generally
refers to a
environment, which
bonding
some
bonding
explain
are
vary
have
widely.
molecular
type
the
For
covalent network
substances
polarity
type.
volatile, despite the fact
graphene
covalent
and
of
mercury
whereas
helps
bonding
typic al
while
variable
Waxes
are
covalent plastics
c an contribute to the
variations
c ategory.
of
properties
C
i
n
n
U
o
i
d
t
a
u
a
means that
removed.
a
is
is
O
in
density
l
v
Rusting
observe this
n
its
external
c an
which
plasticity.
Corrosion
and
You
plasticity,
l
water.
Molecular
r
o
E
and
return to
y
to
force,
contrast, rubber is elastic due to its
and
material
and
a
e
shape
a
to
way.
low
f
x
O
p
Figure 3
are
y
given
substance
are
bec ause
p
a
metals
force
o
within
In
opposite
metals.
substance.
subjected
uncoil
substances,
c ases,
the
while
a
y
their
within
to
v
observed
a
a
are
ions
brittle
removed. In a metal spring, the metallic
material
behave
covalent
some
of
a
when
is
to
used
electricity.
bec ause
In
properties
of
some
has
or
are
subjected
t
i
s
hard.
not
crystals
when
r
e
are
do
The
oxygen
between
conduct
molecular
oen
in
atoms
elasticity.
able
deformed
of
substances
that
properties
so
is
aluminium
both
the
band.
material
materials
for
shape
force
example
presence
the
example,
solids
an
reaction
damages
that
is
corrosion
the
chemic al
M any
clay
the
being
its
The
Ionic
break
other.
change
aer
elastic
retains
easily.
and
r
in
will
malleability.
snap
P
The term
each
chains
an
material
iron,
deformed
responsible
in
of
substances
shape
polymer
Modelling
opposite
past
materials
property
the
be
slide
original
bonding
long
the
From models to materials
s
s
Elastic
their
is
Brittle
2.4
type
of
corrosion.
Iron
and
steel
rust
in
the
presence
of
water
ox y g e n
199
Structure
2
Models
of
bonding
and
structure
Worked example 1
Explain
the
given
that
those
in
trends
the
in
bonds
potassium
melting
in
points
silver
halides.
for
halides
Use
the
silver
have
data
halides
greater
given
in
and
potassium
covalent
table
2
halides,
character
and
table
than
3.
s
s
Melting
point (˚C)
e
point
AgBr
432
AgI
558
generally
increases
points
of
the
silver
halides
KBr
734
O
t
i
s
point
773
n
y
Melting
point (˚C)
KCl
l
Melting
P
p
Table 2
681
points
of
the
p o t a ss i u m
of
the
electrons
the
going
London
ions
u
l
a
E
v
O
f
x
You
ionic
and
try
2.
to
The
region
of
the
loc ate
silver
section.
so
In
halides
the
these
halides
contrast,
from
is
the
the
all
in
to
the
The
AgI
that
their
melting
increasing
increases
melting
point
the
to
strength
increase
one
of
the
another.
attraction
melting
the
in
also
halides
depends
on
how well
region
are
oppositely
charged
decreases (table 3).
triangular
covalent
potassium
between
point
the
bonds
From KF to KI, the size of the
all
bonding
of
the
rmly
diagram in
diagram,
near the
loc ated in the ionic
triangle.
Data-based questions
Plot a graph of the data given in tables 2 and 3. Compare your graph with the
tables and suggest which is a better method for illustrating the data.
200
points
number
predominantly ionic, with minimal
compounds
are
AgF
strength
with
means
forces.
c ausing
electrostatic
therefore
halides
Waals
group
interact
the
and
silver
der
forces,
Therefore,
anions
of
van
the
potassium
increases,
decreases
c an
gure
o
c ations
anion
in
i
the
down
character.
t
a
r
o
d
bonding
covalent
of
(dispersion)
(table2).
The
character
strength
n
U
of
covalent
on
halides
o
partial
depend
C
i
n
Solution
The
p
Melting
v
p
Table 3
y
KI
r
e
decreases
y
455
r
AgCl
Structure
Use of triangular
2.4
From models to materials
bonding diagrams
(Structure 2.4.2)
O ver
time,
chemists
have
further
developed
van
Arkel
and
Ketelaar ’
s original
You
work.
Eventually,
electronegativity
was
incorporated
in
a
way
that
gures
4a
and
encountered
Structure
resulting
triangular
and
bonding
diagram
(or
“bonding
triangle”)
is
2.3.
shown in
greater
covalent
character
character
greater ionic
covalent
greater covalent
bonding
character
2.0
2.0
3.0
O
1.0
l
covalent
bonding
y
1.0
metallic
bonding
4.0
mean electronegativity
0
Ionic
to
bond.
of
a
of
know
We
E
position
•
dierence
•
mean
and
three
the
to
out
χ,
50
75
25
OF
3.0
which
triangle
to
values
two
(b)
diagram
The
3.5
the
of
is
two
dierent
Examples
of
determined
overall
the
metallic–covalent
incorporates
electronegativity.
types
electronegativities,
electronegativity,
in
This
bonding
bonding
work
50
NO
2.5
continuum
diagram.
dierence
in
75
0
.0
Covalent
electronegativity
need
in
the
25
Mean electronegativity (휒
bonding
bonding
substance
the
the
then
a
ionic–covalent
triangular
electronegativity
contributions
need
The
v
O
The
(a)
resulting
92
100
2.0
l
Metallic
p
Figure 4
u
1.5
8
o
i
Mg
1.0
mean
n
AIN
t
a
f
x
0.0
o
d
1.0
r
o
ytivitagenortcelE
1.5
AILi
The
C
2.0
0.5
(c)
v
MgO
U
ecnereffid
i
n
)휒∆ (
2.5
%Ionic
p
CsF
3.0
y
3.5
%Covalent
r
e
(c)
n
character
P
bonding
in
y
greater
metallic
r
ionic
t
i
s
ecnereffid ytivitagenortcele
(b)
3.0
4a
4b
e
gure 4c.
(a)
gure
gure
4b.
Structure
The
2.2,
s
s
in
rst
merges the two
in
continua
bond.
To
elements
bonding
continuum
bonding
c ategories,
several
by
the
substances
are
the
included
relative
loc ate it, we
involved in the
parameters:
∆χ,
is
which
the
is
the
ionic–covalent
metallic–covalent
parameter
parameter.
201
Structure
2
Models
of
bonding
and
structure
We
do
more
not
need
dierent
c arbon–oxygen
diagram
as
to
consider
bonds
the
may
bond
triple
in
molecular
have
formula
identic al
c arbon
dioxide
c arbon–oxygen
or
bond
parameters.
is
in
bond
the
in
order.
For
As
a
result, two or
example, the double
same place in the bonding
c arbon
monoxide.
s
s
Worked example 2
Determine the position of the following substances in the triangular bonding diagram:
pure silicon, Si
c.
barium
iodide,
BaI
4
2
example,
between C and F in CF
=
c alculate
4.0
of
2.6
the
and
4
.
In
C arbon has an
uorine
is
electronegativity
4.0.
From this,
dierence,
∆χ:
this
we
atoms
are
all
the
same
electronegativity.
dierence
2.6
consider
atoms. In elemental
mean
=
example,
two Si
type
1.4
and
Therefore,
between
two
electronegativity
Si
is
the
bond
between
substances like this, the
thus
the
the
same
electronegative
atoms
equal
have
O
∆χ
c an
pure silicon, Si
t
i
s
electronegativity
we
b.
4
need to look at the
n
bond
we
l
tetrauoride, CF
this
y
c arbon
For
is
to
zero, and the
the
Pauling
r
e
electronegativity of one Si atom (1.9).
c alculate
the
mean
electronegativity,
χ:
This
2.6 + 4.0
χ
=
places
3.3
these
the
triangle
between
C
in
the
(gure
and
F
is
we
upper
5).
c an
part
This
polar
loc ate
of
the
the
c arbon–
covalent
∆χ
=
atoms
2.7
2.7
χ
U
=
=
+
in
BaI
2
in
region
the
bond
0.9
=
these
ionic
region
n
With
parameters,
of
the
we
c an
diagram
%Ionic
Ionic
l
a
v
E
ytivitagenortcelE
Bal
8
92
25
75
50
50
75
25
CF
1.0
0.5
Si
100 0
1.5
p
Figure 5
2.0
2.5
3.0
ean electronegativity (휒 )
Metallic
The
loc ation
of
0.9), and
certain
bonds
in
the
triangular
3.5
4.0
Covalent
bonding
d i a g ra m
loc ate the bond in the
(gure 5).
%Covalent
1.5
1.0
=
= 1.8
o
2.0
are barium ( χ
1.8
u
ecnereffid
O
f
x
2.5
(gure 5).
0.9
t
a
3.0
between metallic and
triangle
2.7).
0.0
202
the
bonding
2
i
d
)휒∆ (
r
o
3.5
two
iodine ( χ
region
means that the bond
covalent.
The
iodide,
C
of
parameters,
bond
i
n
uorine
barium
the
o
With
c.
v
=
silicon
in
p
covalent
2
y
Then,
y
P
Solution
a.
e
c arbon tetrauoride, CF
b.
r
a.
Structure
2.4
From models to materials
Practice questions
1.
Using
electronegativity
of
following
the
values,
substances
deduce
the
main
type
of
bonding
in
each
by determining their position in the bonding
triangle.
c.
lithium
d.
brass,
e.
aluminium
f.
silicon
be
discussed
NO,
and
bonding
typic al
the
three
nitrogen
dioxide, NO
of
pure ionic
look
at
a
selection
of
substances
with
signic ant
contributions
covalent
Silicon
network
substances,
is
brittle
conductivity,
(like
most
Semiconductors
is
and
an
covalent
are
forms
atoms
a
poor
and
in
the
is
regions in the bonding
both
the
covalent and metallic
other
electrons.
acidic
between
electric al
or
On
share
(weakly)
substances)
illuminated,
covalent
exhibits
metal.
intermediate
generally
heated,
which
a
it
hand,
Like
it
forms
covalent
oxide. In terms of
conductors
classied as a
(like
metals) and
semiconductor.
conductors, but their conductivity
o
when
it
in
like
presence of certain impurities.
i

Figure 6
hundre ds
of
A
silicon
t i ny
wafer
containing
mic rochips
for
use
in
ele ctronics.
u
l
a
E
v
O
f
x
t
a
r
o
d
increases
structure
(shiny)
and
that
n
insulators
silicon
lustrous
metallic
means
U
electric al
is
the
which
C
a
between
metalloid,
p
properties.
a
i
n
is
o
positioned
It
from
v
is
, are
covalent (Structure 2.2), and metallic (Structure 2.3) substances.
Silicon: metallic and covalent character
triangle.
2
by its position in the bonding
properties
more than one type of bonding.
Silicon
dierent types
diagram.
predicted
the
of
previous question.
n
c an
the
the
l
in
in
y
contributions
e
will
monoxide,
point
substance
already
relative
substances
y
we
a
same
the
the
r
e
Now
of
of
nitrogen
the
have
(Structure 2.1),
each
3
2
O
We
describe
in
r
properties
triangle.
chloride, AlCl
t
i
s
The
at
C sF
O
dioxide, SiO
why
loc ated
2
composed of Cu and Zn
bonding
Explain
Li
y
3.
2
uoride,
oxide,
Qualitatively
of
s
s
uorine, F
c aesium
P
2.
a.
b.
203
Structure
2
Models
of
bonding
and
structure
M agnesium iodide: ionic and covalent character
L attice
enthalpy
Structure
at
2.1,
additional
is
and
discussed
in
higher
in
Reactivity
1.2
In
level.
magnesium
character
are
unusually
high
iodide,
both
∆χ = 1.4 and
about
lattice
50%.
As
χ
a
=
2.0.
The
percentage
ionic
and
covalent
result of this, magnesium iodide has an
enthalpy.
s
s
Data-based questions
c an
test
the
compound
ions
are
signic antly
larger
ions,
lattice
electron
the
enthalpy
sodium
Compound
are
with
lower
between
chloride
silver
Theoretic al lattice
in
2
the
Cl
6
of
between
and
734
the oretic al
theoretic al
compare
the
and
ionic
and
l a tt i c e
e n t h a l py
experimental
covalent
lattice
character of
Aluminium chloride: ionic and covalent character
compound
the
aluminium
chloride
unusually
c annot
true
In
if
which
a
at
the
is
ionic
melting
as
used
in
classied
which
compound,
to
as
∆χ
predict
ionic
but
solid.
very
the
low
aluminium
ionic
χ
=
for
high
an
a
has
a
signic ant
covalent
appears
in
the
of
is
region of
a
substance. This is particularly
character.
covalent
covalent
in
and
the
bonding
ionic
pressure, it melts into Al
ionic
same
Therefore, aluminium chloride is
polar
both
it
shows that the bonding triangle
mixed
2.4.
and
chloride
This
properties
has
exhibits
At
though
iodide.
= 1.6 and
chloride
when
is
even
substance
between
aluminium
lattices
point,
magnesium
that
chloride,
boundary
7) at 190 °C,
ionic
be
bonding
result,
forms
low
triangle
always
the
aluminium
right
As
representing
an
bonding
trichloromethane.
204
y
p
o
Al
w ays
o
d i m e r,
Cl
Cl
Tw o
ex p e r i m e n t a l
and
–1
each compound.
i
bonding
Al
E
p
Figure 7
a
Cl
v
O
Cl
in
has
Cl
Cl
Al
bonding
assess
values of
contribution is aluminium chloride. Unlike magnesium iodide, aluminium chloride
Cl
Cl
dierences
values,
Another
Al
l
Cl
the
enthalpy
u
Al
Cl
of
compounds
t
a
Cl
d
r
o
f
x
Cl
on
Comparison
two
theoretic al
enthalpy / kJ mol
918
n
U
Based
for
and
750
C
AgCl
values,
allows a certain
chloride.
–1
chloride,
p
Table 4
This
experimental
and
r
e
v
i
n
silver
enthalpy tends to
electronegativity
790
NaCl
v alues
lattice
polarisable.
enthalpy / kJ mol
chloride,
for an ionic
prediction. This oen happens
Experimental lattice
sodium
enthalpy
However, if the bonds
experimental
more
predicted and
lattice
(covalent bonding) in addition to ionic bonding.
dierence
for
behaviour.
theoretic al
those
clouds
sharing
the
comparing
O
shows
the
by
theoretic al
t
i
s
of
Table
ionic
covalent,
than
particularly
electron
degree
4
partly
the
n
their
completely
greater
model
y
bec ause
a
l
with
of
example,
compound.
soluble
in
Also,
non-polar
triangle.
properties. AlCl
2
Cl
6
unlike
dimers
a
3
(gure
typic al
solvents such as
y
be
assumes
For
P
between
assumptions
values.
r
experimental
e
You
Structure
2.4
From models to materials
Practice questions
Activity
4.
Use
your
knowledge
of
the
behaviour
of
compounds
containing
a
Mean
electronegativity and
hydrogen–
dierence
oxygen
bond
to
predict
where
in
the
bonding
triangle
you
c alculate
the
∆χ
and
χ
values
to
see
if
your
predictions
for
electronegativity
dierent
bonds
were
shown
below.
O utline the
close.
properties
are
s
s
Then
in
expect to nd this
values
bond.
you
would
expect a
compound with these bonds to
e
have.
Linking question
composites
bonded
reinforced
concretes,
components
and
steel
which
bars,
are
made
have
c.
χ = 2.3
from ionic
unique
and
∆χ
properties?
2.1, 2.2, 2.3)
= 1.4
∆χ
and
=
∆χ
l
metallic
elements
c an
be
mixed with other elements, metallic or non-
Structure
a
metallic,
therefore
maintaining
the
retain
metallic
by
of
The
identity
from
their
are
held
electrons
such
as
of
traces
original
ratio
the
to
of
the
components
substance.
about
properties.
2%.
All
In
For
to
and
those
the
of
melting
an
example,
addition,
these
in
alloy
the
c an
characteristics
are
the
c ations
layers
of
in
the
metallic
c ations
slide
lattice
past
are
the
i
the
c an be
of
of
typic al
discussed
c ations
in
a
contain
sea
of
bonding
between
and
electrons
This
accounts
is
and
the
the
fact
c ations
non-directional.
for
the
malleability
ofmetals.
c arbon in
alloys
of
electrons,
retain
mixtures.
n
all
applied,
that
vary without
components
of
deloc alized
pure metals.
proportion
each
o
metals,
same
other
size.
When a
easily.
t
a
r
o
is
d
pure
force
lattice
we
structures
structure,
Alloys
point,
Properties of alloys
In
the
electric al conductivity and
dierent
resistance
within
throughout.
U
many
alloys
be
C
ranges
of
corrosion
c an
alloying a metal with other elements.
mixtures.
the
of
characteristics
properties
hardness,
sea
ions
i
n
are
changing
deloc alized
or
o
enhanced
the
as
atoms
v
such
steel
many
However,
Properties
Alloys
Dierent
p
lustre.
alloys.
2.3,
metallic
y
still
form
r
e
while
to
O
In
how
Pure
n
y
t
i
s
Alloys (Structure 2.4.3)
0.3
= 1.9
y
covalently
(Structure
like
χ = 1.9 and
r
and
do
χ = 2.9
b.
P
Why
a.
While malleability is a useful property of metals, sometimes we require a metallic
structure to be stronger. Alloying involves the addition of atoms or ions with a
u
f
x
dierent radius to the cations of the pure metal, which disrupts the regular structure
of the lattice. When an alloy is struck with a force, the layers of cations do not slide
past each other as easily. Therefore, alloys are usually stronger than pure metals.
l
a
E
v
O

Figure 8
iron
alloy
are
packed
where as
element
In
in
the
to
pure
a
metals,
re gular
addition
form
a r ra n g e m e n t ,
an
of
ions
another
a l l oy
re ducing
the
p a tt e r n ,
disrupts
the
this
malle ability
205
Structure
2
Models
of
bonding
and
structure
Examples of alloys
NaK is a sodium–potassium alloy that has a lower melting point than its constituent
elements, so it is a liquid at room temperature. Because of this, NaK is used as a
nuclear reactor coolant, which is non-volatile and can be pumped as any other liquid.
metals
are
to
used
spectacle
is
an
abundant
are
harder
from
original
being
shape
upon
deformed
than
heating. They
through
use, such as
tools.
presence
not
the
objects
from
rusting
have
methods
(such as
to
barrier
be
deformed
c arbon
them
to
ideal
many
varieties
relatively
create
for
of
a
steel.
variety of
steel,
each
and
into
from
(for
preserve
methods
the
of
valuable
metal
iron
and
and
instance,
their
(such
galvanising).
elemental
hydrated
properties
exposing
steel
ships
physic al
as
oxygen, O
iron(III)
has
and
and
of
metals.
the
iron
economic
bridges)
integrity.
painting
Rust
oiling)
2
.
Rusting
oxide. This is an
Rust also
underneath
implic ations.
need to be
protection
and
sacricial
i
n
o
include
made
structures
water
iron
v
methods
are
p
protected
making
There
c an
with
y
of
of
volume
corrosion.
steel-containing
iron,
it
c arbon and other elements.
transforms
does
iron
n
it
pure
Pure
alloying
O
so
it
to
by
l
the
of
crust.
y
in
lowering
degradation
L arge
amounts
rusts
E arth’
s
overcome
r
e
The
the
stronger
bec ause
easily,
further
to
t
i
s
to
o
in
be
construction
dierent
steel
c an
and
compound,
akes
their
Stainless steels are iron alloys that contain at least 11% chromium. The chromium
C
reacts with oxygen in the air or water to form a thin layer of chromium oxide on
the surface of the steel. This chromium oxide layer prevents rusting, and therefore
U
stainless steels have useful applications in cooking and medical-grade equipment.
o
t
a
u
Tow e r
a
Eiffel
several
l
The
lasts
v
(a)
i
d
r
o
206
itself
E
process
f
x
O
p
Figure 9
(b)
n
(a)
in
months
Paris,
(b)
F ra n c e,
Adding
is
made
chromium
to
of
iron.
steel
It
is
painted
cre ates
every
stainless
f ew
steel,
a
y e a rs
to
p ro t e c t
rust-resistant
alloy
it
f ro m
rusting.
suitable
for
The
kitchen
painting
equipment
y
ranging
problematic
ionic
metal
limitation
containing
iron,
to
prone
r
Steels
is
return
are
P
This
Like
that
that
frames.
easily.
uses
alloys
objects
e
Iron
are
make
s
s
Memory
Structure
2.4
From models to materials
How does the c arbon content aect the hardness of steel?
Steels
are
that
are
known
as
made
up
medium
of
0.3%
steels.
to
The
0.6%
c arbon
hardness
of
by
mass
Relevant skills
medium•
samples
with
varying
c arbon
contents
were
Inquiry
1:
Demonstrate independent thinking,
s
s
steel
tested,
initiative or insight
giving
the
results
shown
in
figure
10.
•
Hardness
known
materials
have
(units:
as
the
HV)
were
Vickers
greater
hardness
Vickers
Inquiry
determined using a
hardness
test.
1:
Formulate
research questions and
hypotheses
Harder
•
Inquiry
2:
Interpret
graphs
•
Inquiry
3:
Interpret
processed
values.
data
research
information
0.45
0.50
0.55
0.60
Medium
of
of
data:
C alik,
C arbon
C arbon
A.,
Akin,
Content
Steels”,
Z.
on
D. ,
the
of
S ahin,
Natur forsch.
O.
and
65a,
U c a r,
of
copper
and
in
tin.
It
is
harder
common,
6.
than
bronze
pure
i
bec ame
What
4 6 8 – 472
was
t
a
r
o
steel
5.
to
system
copper
do
steels
of
you
your
O utline
an
and
least
the
performed.
compose
a
answered
possible
by the
you include the:
variable
studied
trend
one
might
c arbon
is
(in
this
c ase,
steels).
explain
at
think
with
Outline
your
being
c arbon
and
think
N.
n
alloy
Before
Try
o
an
Describe
4.
Properties
U
is
corrosion.
d
Bronze
specific
3.
sure
independent
–
c arbon-containing
Mechanic al
the
o
“Effect
h a r d n e ss
of
methodology
C
Source
(2010)
the
i
n
steels.
s h ow i n g
v
G ra p h
range
–
medium
Carbon content (wt. %)
p
Figure 10
–
M ake
be
y
0.40
above.
variables
given,
test
O
shown
–
hardness
would
p
0.25
that
t
i
s
0.30
question
r
e
600
and
the
1200
800
of
on
Vickers
y
Based
the
n
2.
how
l
Research
y
VH ,ssendraH
1.
data
analysis to
P
Instructions
1400
1000
and
r
draw and justify conclusions
1600
0.25
e
method
values
shown
alternative
happen
content
to
in
the
graph.
interpretation.
the
greater
hardness
than
0.55%?
ideas.
experiment
answers
to
that
questions
could
4
and
be
5
done
to
explore
above.
resistant to
used in shipbuilding, tools
various household artefacts including coins.
Brass
is
a
highly
malleable
properties,
due
to
which
its
is
alloy
acoustic
why
it
is
of
copper
properties.
oen
such as hospitals.
and
The
zinc.
It
is
copper
used
in
to
make
brass has
music al
antimicrobial
a
E
v
O
l
u
f
x
instruments
used to make door handles in public buildings

Figure 11
coins
f rom
Old
bronze
China
p
Figure 12
made
for
of
An
b ra ss .
me asuring
19th
This
century
theodolite
instrument
was
used
angles
207
Structure
2
Models
of
bonding
and
structure
Research skills and social skills
ATL
The
life
cycle
of
a
product
present
properties,
determines
involves
undergo
the
various
many
of
the
process
and
materials
transformations
processes
involved
throughout
involved
in
a
in
(figure
product
life
its
production,
13).
Structure,
usage
as
well
and
as
disposal. The
chemic al
and
physic al
s
s
substances
cycle.
crushing the ore
e
r
mineral
pure mineral
recycle
end of
waste tip
waste metal
useful life
v
U
products from
task,
you
will
work
particular
skills
involved.
people.
Collaboratively,
plastic
the
life
4.
Plan
to
research
E
of
the
a.
extraction
b.
manufacturing
c.
use
d.
disposal.
Consider
reflect on the
–
if
and
how
you
will
divide
up
the
work
within
your
to
of
the
team
–
how
you
with
each
and
your
team
will
share
your findings
other.
following
investigate:
5.
C arry
6.
Prepare
out
your
a
research.
visual
depicting
the
representation
life
cycle
of
your
of
your
research,
product.
Before
bottle.
you
various
to
stages
raw materials
of
the
start,
take
everyone’
s
Briefly reflect on how you collectively reached a choice.
lifecycle:
208
a
research
a
3.
in
one
aircraft
v
2.
c arbon
to
also
ingot
Work with one or two other
products
an
will
l
–
in
iron in a ship
O
–
and
aluminium
group
u
substances
–
choose
f
x
1.
a
You
t
a
collaboration
with
product.
i
a
r
o
of
o
this
cycle
the metal
metal
The life cycle of a metal
d
In
extracting
n
the metal
p
Figure 13
scrap metal
C
manufacturing
i
n
the metal
o
recycle
shaping
y
scrap metal
in use
p
r
e
recycle
metal
O
t
i
s
rubbish to
n
y
l
crushed ore
a
moment
opinions
represent
into
to
reflect
account
on
how to take
when
deciding
how
your work.
product’
s
7
.
Share
8.
Individually,
and
your
work
identify
with
reflect
one
IB
your
on
class.
your
learner
role
in
profile
developed during this task.
your
group,
attribute
that
you
y
P
purifying the
mining the ore
Structure
2.4
From models to materials
Practice questions
5.
Explain
6.
Suggest
of
a
why
alloys
why
the
such
as
steel
electric al
are
harder
conductivity
than
of
an
pure metals.
alloy
is
oen
lower than that
pure metal.
s
s
are
alloys
compounds?
more
correctly
(Structure
described
as
mixtures
rather than as
r
Why
1.1)
e
Linking question
and
as
Polymers
synthetic
N atural
many
of
covalently
macromolecules,
such
as
nylon
include
biologic al
which
are
brous
molecules
characterized
density
compared
are
very
examples
materials
involved
in
of
like
life
bonded
large
polymer
chains
are
end-to-end
(gure
15),
This
process
and
is
c alled
like
when
small
polymerization.
There
m o n o m e rs
be ads
represent
repeating
brackets,
subscript
the
n
polymer.
n
aer
A
atoms
that
appears
polymers
c an
vary,
with
unit
is
the
the
represents
a
structural
shown
indic ating
typic al
E
units, so
is
of
that
closing
large
so
p
Figure 14
polyethene
A
digital
model
terephthalate
of
(PET),
a
polymer
to
make
and
a
chain
the
c alled
resulting
repeatedly along a polymer
the
formula
enclosed
the
in
polymer
bracket
polymer
very
up
i
polymer
of
are two types of
t
a
the
the
group
a
The
a
a
l
across
of
is
lengths
v
O
formula,
exact
u
The
structure
in
f
x
repeating unit
chain.
join
m o n o m e rs
polymer
Repeating units
A
as
as
n
small
silk,
such
sy n t h e t i c
o
the
and
molecules
U
the
polymerization,
r
o
chain,
In
Here
molecules.
synthetic
lignin
low
types
beads strung together to
condensation
d
p o l y m e r.
their
C
addition
made
by
other
o
join
necklace.
p
Figure 15
a
to
processes,
i
n
a
polymerization:
materials
low
proteins.
monomers
create
and
p
Long,
c alled
and
bonded
v
starch
are
textiles
polymers
the
DNA,
cov alently
conductivity,
y
well
of
r
e
polymers.
class
n
materials.
Plastics
a
electric al
l
of
are
and
O
thermal
t
i
s
Polymers
y
P
y
Polymers (Structure 2.4.4)
the
large
chain
represents
contains
best
of
to
brackets
or
describe the
with
continues
the
hundreds
way
repeating unit. In the
number
on
of
thousands
bonds
either
drawn
side.
repeating units
of
repeating
number.
209
Structure
2
Models
of
bonding
and
structure
Worked example 3
A
section
Draw
of
the
the
polypropene
repeating
unit
of
CH
H
CH
H
CH
H
H
H
H
H
3
3
molecule
is
shown
below.
polypropene.
H
3
CH
H
H
H
3
s
s
r
Solution
the
section,
pattern
you
every
c an
two
see
that
When drawing a repeating unit, make sure that the
there is a
connecting bonds on both sides cross the brackets. If you
c arbon atoms along the
l
need to represent the whole polymer chain, do not forget
polymer chain.
n
y
to include the subscript n to signify that the polymer is
Therefore
the
repeating unit is
composed of many repeating units:
3
H
H
3
H
H
n
the
repeating
unit
for
each
of
the
polymer
H
H
H
H
H
H
—
—
—
—
—
—
C
— C — C — C — C — C — C —
H
Cl
H
H
C
C
C
H
2
H
H
CH
5
H
O
CH
structure
n
9.
The
a
E
Draw
section
structure
of
of
the
also
(CH
polymer
nylon-6,6
)
2
6
H
2
OH
H
O
H
CH
5
3
OH
OH
H
H
H
H
as
C
O
H
H
O
CH
OH
polystyrene,
is
showing
shown
three
is
shown
2
OH
below.
below.
repeating units.
Draw
a
section
O
(CH
)
2
C
4
n
210
H
3
O
OH
known
O
NH
2
H
2
H
O
CH
H
a
—
H
H
poly(phenylethene),
v
2
OH
OH
l
O
— CH — CH
H
H
—
of
O
u
The
OH
CH
OH
H
H
f
x
8.
2
H
t
a
H
OH
i
OH
r
o
O
2
O
H
o
CH
H
d
OH
H
C
c.
H
H
n
H
Cl
U
—
H
—
Cl
—
—
—
—
H
shown.
C
H
sections
i
n
b.
p
Draw
a.
o
7
.
v
Practice questions
y
r
e
H
O
CH
H
t
i
s
CH
of
the
polymer
showing
two
repeating units.
y
at
repeating
P
Looking
e
H
Structure
2.4
From models to materials
The structure of the repeating unit originates from the monomer used to make the
polymer.
the
As
prex
a
result,
poly-.
the
some
However,
refer
to
their
commercial
monomer.
polymer
many
Similarly,
brand
names.
names
natural
many
C an
are
monomer
polymers
synthetic
you
think
polymers
of
name
in
brackets, with
have unique names that do not
any
names
are
referred
to
by
examples?
s
s
Polymers can be classied according to their source (natural or synthetic), or type of
polymerization reaction undergone by the monomers (addition or condensation).
examples
are
e
Some
shown in table 5.
Repeating unit
synthetic?
Type of
balloons, elastic
polymer
natural
addition
2
(natural rubber)
c ar
tyres
(vulc anized
t
i
s
n
bands,
y
poly(isoprene)
rubber)
polyethene
C
H
3
H
c arrier bags,
H
packaging,
H
OH
walls,
toys
paper,
U
natural
condensation
synthetic
condensation
o
HO
cell
cellophane
C
i
n
O
O
v
cellulose
addition
y
H
insulation,
synthetic
c able
p
C
r
e
C
O
H
y
CH
2
Natural or
uses
l
CH
Examples of
P
cis-1,4-
r
Name
OH
H
equipment,
C
personal armour
t
a
of
sports
O
O
H
Examples
C
o
N
i
r
o
d
N
p
Table 5
n
Kevlar
p o l y m e rs
u
l
a
E
v
O
f
x

Figure 16
chemist
dire ctly
Kev lar
Stephanie
linked
to
the
anti-stab
clothing.
a ny t h i n g
like
and
was
f i rs t
Kw o l e k .
use
of
Kw o l e k
s av i n g
sy n t h e s i z e d
O ver
3 ,0 0 0
Kev lar
in
life
Po l i s h - A m e r i c a n
lives
bulletproof
remarke d,
s o m e o n e’s
by
s av e d
to
“I
d o n ’t
bring
think
you
h av e
vests
been
and
t h e r e’s
satisfaction
h a p p i n e s s .”
211
Structure
2
Models
of
bonding
and
structure
Properties of polymers
In
general,
most
thermal
and
points.
length
is
of
due
the
each
strong,
they
the
and
have
number
polymer
other
not
are
their
macromolecular
moving
also
intermolecular
The
long
entangled,
forces that occur along
polymer
requiring
strands
c an also wind
energy to disentangle and
the
which
houses
a
Starch
a
collagen,
an
natural
which
organism’s
c arbohydrate
through
is
and
a
photosynthesis.
structures
such
as
proteins,
structure
gives
genetic
natural
as
well
our
linear
as
is
n a t u ra l
of
plants.
polymer
way
t
a
tightly
as
u
l
a
unit
of
starch
in
polymers
stalks
polymers
by
and
oil-based
other
are
important
store
of
as
by
therefore
humanmade.
coating
These
and
M any
crude
on
some
from
have
oil.
pans
of
for
a
an
The
them
as
and
important
v
Explain
why
polymers
have
11.
Explain
why
polymers
are
The
see the
resulting
structural
bres
solvents. The
integrity of
in
starch
starch
are
polymer, like
so
the
chains
c an
dierence lies in the
have
an
irregular
c annot pack as
bonds
with
one
another.
petroleum-based, which
include
PET
rubber
biomedic al
bottles,
produced
alternative
applic ations.
insulators.
nylon
tyres. Semi-
materials
sustainable
electric al
c an
by using cellulose to
relatively high melting points.
generally
Plants make
You
common
Practice questions
10.
store in
elasticity, and
means that the
strong,
biologic al
a
trunks.
main
synthetic
provide
others
cellophane.
chains
in
while
energy
glucose.
glucose
not
Examples
and
the
hydrogen
renewable
polymers
is
chains
fewer
many
like
also
it
polymer
form
tree
properties
humans.
the
the
from
is
way these
walls. It supports
another.
as
and
properties:
of
cell
cellulose
material
polymer
derived
well
these
a
result,
in
in
digested
in
one
materials
plants
as
structure
and
polymer
dierent
derived
bacteria.
polymers,
an
water
with
connected:
Teon
synthetic
it
in
skin
acts
polymer
groups)
bonds
advantage
and
and
are
are
non-stick
plants
As
cellulose
they
has
are
structure.
that
textiles,
OH
it
easily
monomers
branched
means
repe ating
E
212
the
down
and
energy
However
broken
taken
o
be
main
insoluble
makes
have
textiles,
i
d
r
o
f
x
O
The
the
Synthetic
HO
p
Figure 18
is
cellulose.
O
O
cellulose
paper,
Starch
and
(hydroxyl
hydrogen
n
make
OH
of
strong
Humans
groups
form
organisms,
that
found
o
a
up
c an
C
is
made
U
c e l l u l o s e,
Co tt o n ,
chains
uniquely
strength
p
Figure 17
–OH
i
n
are
of
v
presence
polymer
the
information.
Cellulose
cotton,
living
polymer
repeating unit of cellulose in table 5.
The
and
c arbohydrates and nucleic acids,
to
p
brous
is
monomers
y
glucose
include
including
provide
r
e
Cellulose
of
n
DNA,
Some
processes.
Others
type
O
plants.
molecules,
polymers.
life
the
t
i
s
support
together,
l
biologic al
natural
chains
arranged.
to
crude
y
are
y
are
r
monomers
M any
properties of a polymer depend on its chain length, the type of
holding
P
specic
forces
Like
low.
separatethem.
The
nature.
charged particles, so their
relatively high melting points and boiling
of
molecules.
become
of
contain
conductivities
polymers
to
bec ause
do
e
around
electric al
molecular,
This
are
materials,
s
s
Though
the
polymers
covalent
Structure
2.4

Figure 19
This
c re ate d
a
f rom
From models to materials
artificial
mixture
of
s e m i - sy n t h e t i c
p o l y m e r.
c an
organ
be
used
in
t i ss u e
living
T i ss u e s
was
cells
like
and
this
t ra n s p l a n t a t i o n
s
s
e
r
mainly
although
of
they
c arbon
c an
and
c atch
hydrogen.
re and sustain
in
packing
chains
be
it
is
it
chains
widely
overcome
On
It
is
a
forces
by
non-polar
new
hydroc arbon
(LDFs) hold the polymer
heating,
cooling,
used plastic
so
LDFs
polystyrene
c an
form
CH — CH
2
n
c an
between
p
Figure 20
used
plastics
inert
biodegrade
are
covalent
network
The
structure
of
p o l ys t y r e n e
is
as
that,
and
by
phenyl
insulating
due
to
As
groups
(C
6
H
5
),
Polymers
crosslinking
covalent
many
contain
sites.
which
London
dispersion
and
other
is
In
provides
very
of
the
(LDFs),
discussed
intermolecular
in
Structure
forces
2.2.
bonds,
plastics
do
materials,
reactive
some
known
forces
for the
breakdown
rarely
binding
together,
strong
result,
the
panels
are
their
a
involves
bacterial
bonded
formed
bulky
insulator, particularly in its
durable.
molecules.
act
the
plastics,
as
the
crosslinking.
few
entry
points
a
v
E
Activity
small
covalently
microorganisms.
useful
Biodegradation
could
to
coolers
therefore
l
chains
into
that
make
due
thermal
u
groups
are
and
easily
good
o
is
a
industry.
f
x
O
dense
for
to
it
microorganisms,
polymer
A
so
reasons
typic ally
rotate
is
chemic ally
functional
20.
i
the
be
gure
dispersion
t
a
are
c annot
Polystyrene
r
o
of
they
in
easily.
d
form,
construction
One
c an
a
molecules, so the polymer solidies again.
brittle.
expanded
LDFs
shown
London
reshaped
is
n
making
and
is
water.
U
Polystyrene
in
These
down
polystyrene
by
polystyrene
together.
melted
not
of
insoluble
polystyrene,
containers.
o
so
known as
food
C
structure
and
and
i
n
The
commonly
peanuts
v
found
p
Poly(phenylethene),
y
combustion.
—
composed
unreactive,
n
l
polymers
chemic ally
O
synthetic
oen
y
P
y
are
are
r
e
They
t
i
s
Plastics
Compare and contrast two of the polymers discussed in this topic.
213
Structure
2
Models
of
bonding
and
structure
Investigating hydrogels
Smart
materials
response
to
undergo
a
surrounding
change
in
conditions
Instructions
properties in
(such as pH,
Part 1: The effect of pH variations on the hydrogel
materials
or
used
applic ations.
to
swell.
21).
basic
water,
investigate
surrounding conditions.
numerous
groups
the
environments,
ions,
–COO
c arboxylate
.
dishes,
c ausing the gel
the
c ausing
Prepare
the
response of a
c arboxyl
form
groups.
hydrogen
hydrogel
to
become
ionized,
The
negative
charges
groups
repel
each
–
weak acid solution
–
weak base solution
Place
the
3.
Place
a
hydrogel
1:
on
a
lens
is
the
following:
white
in
the
dicult
to
surface.
rst
dish
and
measure its
see, it might be inside out.
Reverse it and try again.)
5.
Swap the contact lenses in the acidic and basic
other,
changes
swell.
the
it
of
Place contact lenses in the other two dishes and note
solutions
Justify
(If
one
buer solution
4.
changes
Inquiry
dishes
contact
diameter.
Relevant skills
•
lens
in
and
diameter.
observe
what
happens.
O
the
shallow dishes, such as petri
contact
any
making
clear
containing
–
2.
contract
they
three
each
n
neighbouring
hold
–COOH
1.
biomedic al
will
in
another,
c arboxylate
and
crosslinked polymer
t
i
s
on
In
of
Are the
reversible?
range and quantity of
y
r
e
Part 2: The effect of salinity variations on the hydrogel
measurements
•
Inquiry
2:
Interpret
qualitative
and
quantitative data
6.
Repeat
the
Part
1
experiment, using these solutions
p
instead:
S afety
protection.
Commercially
available soft contact lenses
(manufactured
from
poly(2-hydroxyethyl
•
Contact
lens
buffer solution
Sodium chloride
base
solution
(for
–3
)
equipment including petri
pipettes and balances.
u
l
a
E
v
O
f
x
Ruler
laboratory
beakers,
0.06 mol dm
i
Standard
dishes,
•
water
t
a
r
o
•
Distilled
example, sodium
o
Weak
hydrogenc arbonate
214
d
Weak acid solution (for example, household vinegar)
reuse
Describe
Propose
of
9.
lens
buer solution
water
saturated sodium chloride solution
c an
damaged.
8.
•
•
–
You
7
.
•
distilled
n
U
methacrylate-co-methacrylic acid))
•
contact
–
C
•
i
n
Materials
–
o
eye
v
Wear
the
the
your
a
Part
Devise
Part
an
1
lenses,
they
are not
results.
hypothesis
2
provided
to
explain
the
observed
results
experiment.
experiment
that
you
could
do
to
test
your
hypothesis. Identify the independent, dependent
and
control
you
make?
variables.
Justify
measurements.
the
What
measurements would
range and quantity of these
y
(figure
one
and
contain
these
up
l
with
you
chains
industrial
y
bonds
forming
task,
media,
of
made
absorb
variations
hydrogel
acidic
range
are
are smart
r
In
c an
this
to
a
Hydrogels
P
The
that
In
hydrogel
in
They
fields).
e
networks
magnetic
s
s
temperature
Structure
acid form
CO
2.4
From models to materials
base form
COO
H
2
O
O
O
OH
s
s
OH
O
+
H
O
O
e
+
H
HO
O
O
O
O
O
r
OH
O
COO
H
electrostatic repulsion
groups causes hydrogel
to contract
to expand
between
– CO O H
groups
and
a
result,
disposed
challenge,
durable
the
a
and
longer
term.
petroleum-based
risk
an
enough
Much
to
area
be
Work
light-sensitive.
must
be
taken
to
For
action.
c an
down
slow
or
gas
also
are
quality
is
bacterial
functional
oxygen
in
action
up
time
ways
by
groups
largely
unaected
by
end
for
to
in
the
but
in
the
D isc arde d
oce an.
plastics
The
cold
c an
and
oce an
oceans,
s l ows
d ow n
dark
plastic
d e g ra d a t i o n
biodegradable
make
e nv i r o n m e n t
where it
Greenhouse
at
AHL
in
gases
Structure
are
discussed
3.2.
existing
microorganisms.
having
that
up
incinerated or
are
increased surface
attractive to bacteria
hydro-biodegradable plastics
down
in
conditions
environments
produce
nd
of
feedstock
plant-based
them
is
instead
ends
period
to
adjustments
recycled.
interest, is designing plastic materials that
done
broken
cleaned
prevents some
p
Figure 22
and
plastic
certain
appetizing
or
are
of
a
being
more
that
even
E
greenhouse
low
for
facilitate
dispose
In
growing
example,
v
bacterial
of
chains,
c arbohydrates
disc arded
pH
release atmospheric pollutants
the
wildlife.
a
O
contain
recycled
u
polymer
not
l
f
x
or
shorter
of
used
is
plastics
Biodegradable plastics
area,
to
is
they
and
which
o
poses
waste
when
sorted
melted,
they
where
be
for mitigating the
several challenges.
i
in
and
landlls.
deteriorate
must
and
t
a
are
in
plastic
up
bec ause
aluminium,
r
o
The
of
oen
and
d
persists
most
recycled
glass
waste
poses
n
As
plastics
be
U
recycling,
being
Unlike
to
approach
C
from
needs
third
recycling
i
n
plastic
broken
plastic
types
the
then
It
heating.
is
Plastic
energy-intensive:
beforehand.
on
recycling
plastics.
the
y
is
of
with
o
process
reuse,
v
The
and
impact
a ss o c i a t e d
p
reduction
groups
r
e
Polymers and the environment
environmental
– CO O
O
t
i
s
–
I n t e r c o nv e rs i o n
n
hydrogel
Aer
l
between carboxylate
groups cause
y
between carboxyl
y
2
hydrogen bonds
p
Figure 21
P
CO
a
process
that
found
environmentally
c alled
provide
in
landlls,
harmful
hydrolysis.
enough
C are
oxygen
for
decomposition
products, such as the
methane.
p
Figure 23
m i c r o g ra p h
embedded
c arrier
and
are a
bag.
sw e l l s
for
A
of
in
sc anning
starch
a
b i o d e g ra d a b l e
The
u p,
ele ctron
g ra n u l e s
starch
absorbs
providing
bacterial
plastic
water
gre ater
surface
action
215
Structure
2
Models
of
bonding
and
structure
Global impact of science
The 12
the
green chemistry
benets
ecient
reduction
use
of
of
waste,
of
The
adopted
choice
by
many
of
and
reactants,
renewable materials,
materials
philosophy
and
of
and
prevention of
green chemistry has
educ ational
eventually
and
commercial
r
organisations
chemic als and
energy
passed into national and
internationallaws.
of
renewable
their
Even
c arbon
c arbon-neutral
biosynthesis
is
of
renewable, such as
polymers
bec ause
removed
from
the
the
releases
are sometimes
same quantity
atmosphere
by
monomers in plants.
evaluate
have
for
require
factors.
For
a
product,
household
example,
used
fertilizers
or
to
grow
other
pesticides,
crops,
which
which
have
their
c an
12.
Outline two challenges
remain
recycling plastic
waste.
two
structural
features of
and
Their
impact
be auty
banned
in
m i c r o g ra p h
on
are
a
single
sometimes
products,
m a ny
particles,
but
countries
re d
of
this
s h ow i n g
k n ow n
blood
added
use
to
is
is
into
polyethene
t
a
Ele ctron
polyethene
mic robe ads,
Mic robe ads
216
health
small
way
a
v
E
mic roplastic
such
as
cell.
he alth
n ow
food
samples
banana
shortages.
environmental
fragments
not
unchanged.
their
u
l
f
x
O

Figure 25
and
nd
polystyrene
biodegradable plastics.
as
very
process
chemic ally
and
i
Outline
r
o
13.
chain
are
This
o
with
d
associated
5 mm.
c ause
associated
n
U
under
potentially
own
Microplastics
Practice questions
bioplastic
waste
p i n e a p p l e  c r ow n s
plant-based plastics uses land that
be
These
food
C
may
maize
otherwise
many
of
i
n
could
consider
impact
o
growing
to
environmental
v
you
the
p
p
Figure 24
To
the
of
same
Bec ause
were
as
is
still
made
onion
f rom
skins
and
even
Additionally, farming maize
broken
down
biodegradation
their
size,
places.
discovered
microplastics
were
peel,
implic ations.
plastic,
of
unexpected
y
the
are
biodegradation
plant-based
dioxide
petroleum-
r
e
c arbon
plastic
are
non-renewable. The
bioplastics
though
dioxide,
considered
of
for
plastics
is
O
materials
maize.
Most
feedstock
n
so
t
i
s
based,
raw
feedstocks and designing materials
biodegradable.
y
be
bioplastics,
green chemistry:
l
use
to
biodegradable plastics, or
aligned with two principles of
in
In
they
2022,
human
unknown.
over time to sizes
bec ause the plastics
c an
enter
the
microplastics
blood
for
the
food
of
rst
PET,
time.
y
are
P
Plant-based
e
been
use
biodegradable
accidents.
principles emphasize
non-hazardous
s
s
solvents,
of
Structure
2.4
From models to materials
Addition polymers (Structure 2.4.5)
During
way
polymerization,
this
contain
c an happen is
a
double
bond
due
to
and
chemic al
their
2
H
4
properties
,
is
the
polymerizes
Consider
c arbon-3
that
involves
very
large number of ethene
o
i
d
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
a
H
H
H
H
E
v
O
l
u
f
x
t
a
r
o
H
H
H
a
n
U
process
form a
o
4
H
to
C
i
n
H
similar
and
O
c arbon-2
y
between
p
shared
n
up…
monomers together:
H
H
physic al
r
e
are
the
3
a
open
H
H
polyethene.
H
H
H
H
form
4
H
imagine
bonds
monomers:
H
to
v
joining
2
H
dierent
monomers.
H
electrons
H
H
form a
order, and
y
double
H
H
1
quite
bond
t
i
s
their
3
these
H
have
in
H
H
bond,
H
Now
in
2
and
electrons
y
4
H
H
single
molecules
two
change
One
monomers that
l
C
3
H
H
It
the
the
polymer.
H
C
1
and
of
corresponding
alkene.
2
electrons
H
simplest
breaks
Bec ause
polymer
their
H
H
The
length,
than
C
H
bond
a
P
1
double
monomer.
form
involves
monomers:
H
C
…
extraordinary
ethene
H
The
another
to
which
r
two
together
e
Ethene, C
bond.
with
combine
s
s
covalent
monomers
addition polymerization,
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
217
Structure
2
Models
of
bonding
and
structure
We
c an
convey
simplify
that
the
there
H
equation
are
a
very
for
the
large
addition
number
H
H
H
H
of
polymerization
reaction using
n
to
monomers:
H
s
s
n
H
H
n
n
H
also
use
structural
CH
2
formulas:
CH
2
2
n
Any monomer that contains a c arbon–c arbon double bond c an undergo addition
non-polar
therefore
in
chlorine
where
stronger
of
the
with
in
past,
People
u
l
f
x
a
E
v
O
the
and
and
of
why
required
is
CH
is
polar,
are
present.
PVC is
H
CH
H
H
3
3
H
p o l y (c h l o r o e t h e n e)
the
and
the
and
they
worked.
materials
of
practic al
materials
the
( b)
polypropene
are
a
the
centuries
the
particular
to
in
scientic
scientic
time. Science is
and
scientic
natural world.
knowledge.
knowledge.
understanding the
properties.
scientic
Empiric al
understanding of
research
are
scientic
science
technology, and
over
before
applic ation
using
observation
of
observable
contributed
and
physic al
preceded
materials’
engineered
empiric al
science
disciplines
applic ation
for
Nowadays,
for
two
understanding
technologies
properties
between
these
developments
behind
early
role
dierence
developed
reasons
Desirable
What
is
technologic al
observation
rst.
H
between
observing
technology
used
underlying
how
forces
y
focus
the
relationship
i
concerned
In
dispersion
p
o
illustrates
changing
t
a
r
o
d
Structure 2.4
The
(a)
bond
n
U
TOK
the
London
n
form
chloroethene
c arbon in the
polymers.
n
to
from
other
c arbon–chlorine
H
H
re actions
The
made
Every
o
polymerization
only
two
C
Ad d i t i o n
polymer
between the polymer chains, unlike in
r
e
H
a
substituent.
forces
v
i
n
H
H
is
(b)
Cl
n
218
the
Cl
H
p
Figure 26
a
PVC,
poly(chloroethene).
dipole–dipole
polyethene
the
(a)
H
has
is
n
polymer
results
chloride,
name
O
which
Polyvinyl
IUPAC
t
i
s
resulting
Its
l
molecules.
y
polymerization.
oen
occurs
identied, and then
knowledge.
and
technology?
n
y
P
2
condensed
e
c an
r
We
Structure
2.4
From models to materials
Practice questions
14.
For
each
of
draw
ii.
deduce
following
section
the
of
monomers:
the
polymer
structure
of
the
F
F
F
polymer.
H
H
H
—
—
—
—
structure
of
of
the
the
shown:
repeating unit
monomer.
CH
−
2
CH
−
2
CH−
CH
3
H
H
and
C
CH
−
…
3
C
C
Linking questions
What
C
functional
molecules
act
H
CH
as
groups in
c an enable them to
monomers
for addition
3
reactions?
(Structure 3.2)
Why
atom
C
3
is
the
economy 100%
for an addition polymerization
ethene
point
H
i
contrast
3
u
melting
a
and
bonding,
C
l
Compare
CH
3
t
a
r
o
f
x
O
16.
H
CH
o
H
d
C
−
2
3
n
b.
CH−
CH
3
−
− CH
CH
−
− CH
2
3
U
− CH
CH
3
−
−
…
C
a.
o
the
sections
structure
i
n
draw
polymer
the
p
the
v
of
deduce
H
y
—
—
H
each
r
e
— C — C = C — C — H
O
t
i
s
but-2-ene
y
H
n
H
l
C
r
C
(prop-2-enenitrile)
P
CN
H
ii.
repeating units
the
y
H
H
i.
of
e
F
acrylonitrile
c.
For
three
unit
tetrauoroethene
b.
15.
containing
repeating
s
s
a.
a
the
i.
and
polyethene
in
terms
of
their
structure,
reaction?
electric al
(Reactivity
2.1)
conductivity.
v
E
219
2
Models
of
bonding
and
structure
LHA
Structure
Condensation polymers (Structure 2.4.6)
Condensation
reaction
For
every
balance
webbing
typic ally
tensioned
Nylon
is
a
is
a
between
have
two
nylon
and
reactive
cellulose,
functional
monomers,
a
small
are
produced in a
groups on either end.
molecule
is
released
water or HCl).
p ra c t i c e
uses
ny l o n
between
two
condensation
anchor
r
points.
Slacklining
that
formed
include
that
(for
polymer
e
in
bond
which
monomers
s
s
instance,

Figure 27
polymers,
between
l
and
between an alcohol and
the
a
hydroxyl
water
H
O
H
functional
used
ethyl
formed
to
make
in
the
condensation
condensation
H
H
ethanoate
re action
polymers
H−O
+
−
H
−
reactions
two
H
−
The
released. This
−
heat
H
is
is
H−C−C−O−C−C−H
−
e s t e r,
c arboxyl
−
−
4
H
=
SO
2
ethanol
an
and
molecule
esteric ation
conc. H
−
H
n
reaction
reaction,
linkage,
known as
O
ester
y
is
this
H
−
t
a
3.2.
further
e t h a n o a t e,
u
properties
E
220
a
Structure
and
discussed
v
are
l
structure
esters
ethanol
p
i
−
f
x
O
The
and
O lost
2
E t hy l
an
In
H−O−C−C−H
+
acid
28).
−
−
−
=
r
o
ethanoic
p
Figure 28
form
reaction
H
H
H
condensation
o
O
d
H
H−C−C−O−H
a
o
groups
condensation
in
(gure
n
U
functional
acid
C
formed
c arboxylic
y
P
y
t
i
s
r
e
are
v
a
i
n
Esters
are
between
similar
to
water
ethanoic
acid
esteric ation:
of
groups
react
to
form
a
linkage,
and
a
small
molecule
is
released.
in
However,
groups,
We
will
in
polymerization,
one
at
each
look
at
two
end
of
the
the
examples
reacting
monomers
molecule,
of
how
this
which
have
allows
c an happen.
a
two reactive functional
polymer
chain
to
form.
Structure
2.4
From models to materials
Condensation between two dierent monomers
alternating
monomer
a
between
to
form
=
=
HO − C
an
ester
pattern
acid
linkage
is
for
ABABABAB
in
and
gure
H O−
−
form
−
+
OH
H
29
based
on
two
each
ends
structure,
polymer
product
OH
+
H O
n
−
−
O
+
−
OH
2n H
number
the
length
condensation
=
31).
formed
4
during
polymer
the
is
and
chain
is
known
as
a
condensation
H
−
N − (CH
)
2
6
O
=
=
− N
−
H
−
H
HO
−
C − (CH
)
2
4
− C
HO
−
−
H
−
− N
(CH
−
)
2
N − (CH
4
−
−
6
−
H
−
…
C
(CH
−
O
+
O
2
repe ating
unit
is
H
)
2
…
4
−
−
C
−
C
=
The
− N
=
ny l o n - 6 , 6 .
6
=
−
O
H
)
2
C
=
E
)
2
H
−
of
resulting
n
i nv o l v e d
between
product
− C
−
HO
N − (CH
formation
where
p a i rs
−
)
2
+
The
the
condensation
p o l y e s t e r,
=
C − (CH
H
p
Figure 31
a
monomer
re action
The
O
O
linkages
g e n e ra l i z e d
−
−
(gure
u
− N
amide
polyester.
o
multiple
of
of
O
i
the
A
make
the
t
a
in
to
2
d
the
−
produced
a
6
is
l
)
2
each monomer
c alled a
n
−
hexane-1,6-diamine
v
O
N − (CH
H
and
to
is
U
C− O−
subsequently
o
−
f
x
−
H
due
with
re action
nylon-6,6
acid
c an

Figure 30
r
o
polymer
hexanedioic
that
forms a long polymer chain with
monomer
resulting
=
=
−
reactive
This
p
The
has
30).
y
(gure
LHA
gure
O
in
O
−
O
2
n
−
y
(diol)
n
process.
m o n o m e rs
ester
C
C
poly amide
an
t
i
s
C
O
The
two
OH
v
O
C
condensation
linkage
linkage.
O
−
A
between
i
n
−
pattern
ester
=
HO
an
O
monomers
=
n
by
(diol)
l
−
molecule
more
alternating
joined
general
alcohol
r
e
an
reacts
=
=
C
product
with
A
The
O
C
Ester
react
formed.
y
O
Alcohol
creates
P
acid
This
29.
re action
Dic arboxylic
The
is
react with
monomer

Figure 29
C − OH
−
if
c an
end.
difunctional
to
HO
end
either
example,
monomer
shown
at
e
O
dic arboxylic
structure:
the
either
r
O
a
with
at
group
H
monomer
monomer
polymer
group
functional
H
reaction
B,
functional
dierent
s
s
with
of
same
a
H
polymer
the
with
−
a
with
monomer
HO
monomer
another
O
One
O
+
O
2
s h ow n
in
H
O
O
2
+ H
O
2
gre en
221
2
Models
of
bonding
and
structure
LHA
Structure
Condensation of the same monomer
If
there
is
a
dierent
polymerize
OH
O
acid
with
(gure
32)
of
by
contains
group
on
condensation.
both
a
each
For
hydroxyl
3-hydroxypentanoic
acid
end
group
c an
of
a
example,
and
react
a
monomer,
then
it
c an
3-hydroxypentanoic
c arboxyl
with
one
group.
another
to
Multiple
form
apolymer.
OH
CH
Proteins
The
structure
of
biopolymers
that
play
structural
a
central
integrity,
role
and
in
life.
Proteins
are
2-amino
processes,
polypeptides,
which
are
polymers
a
as
acids.
building
–COOH
group,
There
are
blocks
so
they
of
20
c an
polymerization.
types
proteins.
In
form
a
of
2-amino
2-amino
amide
polypeptide,
H
v
COOH
R
is
group
also
through
known as a
COOH
O
NH
2
R
C
H
1
H
N
C
R
H
2
2
NH
o
O
n
3
2
H
2
C
R
i
H
H
2
H
N
R
C
H
C
H
3
COOH
NH
2
C
R
H
R
4
4
COOH
O
2
NH
u
l
2
C
R
H
H
N
C
R
5
H
5
COOH
H
O
2
NH
a
E
v
O
f
x
t
a
r
o
d
H
2
C
C
i
n
U
H
NH
o
O
link
2
other
n
2
p
C
1
2
amide
each
O
NH
H
by living
y
r
e
a
R
with
aspects
monomers
have an –NH
t
i
s
H
of
the
used
of
y
O
formation
acids
acids
linkages
peptide bond:
The
up
l
condensation
chain
made
transport
molecular
2
C
R
H
6
H
N
C
R
H
6
COOH
COOH
amino acid molecules
(where R
1
, R
2
, R
3
etc. represent
any of the 20 or so groups
found in naturally occurring amino
acids)
polypeptide
(part of)
y
and
polypeptide
other
P
organisms

Figure 33
all
monomer
c alled
222
metabolic
virtually
r
acid
functions,
the
of
3 - hy d r ox y p e n t a n o i c
sensory
e
and
p
Figure 32
are
3
s
s
monomers
functional
itself
Structure
2.4
From models to materials
LHA
Activity
Figure 34 shows the 3D structure of lysozyme, the natural antibacterial enzyme
contained in saliva, milk, mucus, tears and egg whites. The carbon atoms are
s
s
shown as white, the oxygen atoms are red and the nitrogen atoms are blue.
C an you nd any amide linkages?
e
r
Identify
the
type
of
v
monomers
a
between
polymer
containing
condensation
and
functional
hydroxyl
group,
–OH,
A
c arboxyl
group,
–COOH,
c.
Two
d.
An
E
A
b.
acyl
addition
made
these
a.
hydroxyl
polymer
made
from
(gure 32).
a
19.
n
l
dierences
O
two
polymerization.
of
acid
l
Outline
O
18.
y
unit
p
repeating
u
the
3-hydroxypentanoic
o
t
a
r
o
f
x
Draw
C
l ys oz y m e
Practice questions
17
.
n
of
o
structure
i
3D
y
P
y
t
i
s
r
e
v
i
n
U
The
d
p
Figure 34
and
a
the
and
small
groups
c arboxyl
and
an
condensation
molecule
form
group,
amino
released when
polymers:
–COOH
group, –NH
2
groups, –OH
chloride
group,
–COCl,
and
an
amino
group, NH
2
223
2
Models
of
bonding
and
structure
LHA
Structure
Hydrolysis
The
reverse
formed
by
of
condensation
condensation
macromolecules
35
formed
shows
again
by
form
how
the
monosaccharides
energy
hydrolysis,
of
condensation
reactions
and
hydrolysis
hydroxyl
reserves
up
in
Note
form
c arbohydrates,
living
CH
down
via
the
two
are
by
hydrolysis.
reactive ends of
with
used
a
glycosidic
as
energy
organisms.
OH
linkage
condensation and
polymers
which
the
biologic al
r
OH
2
which
where
All
break
polymerized
reaction.
groups,
make
are
water.
2
H
monosaccharides
l
y
(water added)
OH
CH
2
H
O
HO
OH
i
n
glycosidic
C
bond
condensation
OH
t
a
Some
—OH
lactose
CH
OH
2
O
O
H
polysaccharide
e.g., starch, glycogen
O
O
OH
Condens ation
c a r b o hy d ra t e s .
OH
2
o
i
r
o
O
p
Figure 35
CH
2
O
HO
e.g., maltose, sucrose,
hydrolysis
n
U
CH
O
d
H
OH
2
disaccharide
o
v
p
H
OH
2
O
O
y
r
e
CH
n
hydrolysis
O
2
(water removed)
CH
O
H
galactose
t
i
s
condensation
e.g., glucose, fructose,
and
hy d r o l ys i s
groups
are
of
monosaccharides,
o m i tt e d
for
the
p o l y m e rs
that
make
up
clarity
u
l
a
E
v
O
f
x
Evidence
Water
is
oen
biologic al
for
water
forms
to
referred
reactions.
on
M ars,
as
to
For
as
the
this
the
‘molecule
reason,
presence
of
NASA
of
water
life’
due
scientists
is
to
its
have
thought
to
vital
role in
been
be
searching
required
for
life
exist.
Linking question
What
functional
condensation
224
groups
in
reactions?
molecules
c an
(Structure 3.2)
enable
them
to
act
as
monomers
for
y
P
CH
and
c alled
molecule
monosaccharides
Polysaccharides
sources
by
is
a
reverse
are
up
e
linkage.
by
reactions
split
s
s
Figure
is
Structure
2.4
From models to materials
End-of-topic questions
Which
of
1.
Using
your
knowledge
from the
Structure 2.4
to
the
following
pairs
of
molecules will
form a polymer?
topic,
A.
answer
the
guiding
question
as
fully
as
OH
O
OH
possible:
OH
H
H
(CH
mean
electronegativity
χ
PbBr
(Pb) = 1.8 and
χ
2
∆χ,
for
6
the
bonds
H
found in
.
(Br)
=
O
O
3.0.
∆χ
)
2
4
HO
4.2
C
2.4
1.2
D
1.2
2.4
Which
of
the
ionic
values
diagram
and
the
properties of
electronegativity
Cs
F
Zn
S
0.8
4.0
1.6
2.6
a
sodium-potassium
C.
Zinc
D.
Bromine,
uoride,
sulde, ZnS
the
iron
and
mixtures
and
is
an
2
Br
and
H
H
C
N
COOH
2
CH
2
alloy?
SH
a
B.
following
chloride
water
D.
H
H
C
C
H
H
O
H
vanadium
v
sodium
2
CH
COOH
3.0
u
of
A.
Br
l
Which
alloy
C sF
1.9
t
a
NaK,
C aesium
Ag
i
K
0.8
A.
C
N
2
o
d
Na
0.9
r
o
Element
B.
H
Refer to the
below.
f
x
C.
c arbon
D.
magnesium and chlorine
bre and a polymer
E
5.
bonding
has
compounds?
Electronegativity
O
4.
substances
covalent
H
OH
n
triangular
following
and
U
both
C.
p
2.6
o
B
C
2.6
v
4.2
i
n
A
y
Electronegativity
dierence,
(CH
r
e
Mean
electronegativity, χ
3.
H
and
O

dierence,
bromide,
)
2
χ, and
t
i
s
lead(II)
electronegativity,
n
the
y
C alculate
l
Multiple-choice questions
2.
OH
3
y
B.
CH
3
O
P
Exam-style questions
r
ofmaterials?
e
and
What role do bonding and structure have in the design
CH
not
s
s
react
LHA
6.
Topic review
Which of these substances does
H
not
H
and
H
C
H
have the ability to
polymerize?
A.
ethane, CH
B.
tetrauoroethene, CF
C.
propene, CH
D.
propenenitrile, CH
3
CH
2
3
CHCH
2
2
CF
2
3
CHCN
225
Structure
2
Models
of
bonding
and
structure
Extended-response questions
7
.
Tin(II)
chloride
lead(II)
used
in
the
a
melting
has
a
production
glassware.
point
melting
One
of
of
of
point
aurene
these
10.
247 °C while
of
500 °C.
Both
Chloroethene
shown
glass, an iridescent
two
substances
in
the
vapour
substance
and
explain
C= C
—
H
Cl
[3]
chloride, AlCl
3
,
forms
white
Identify
the
structural
feature
of
it
to
form
a
polymer.
the
mean
State
values
type
of
[2]
The
the
a
mean
repeating
molecular
sample
reference to the triangular
of
unit
of
the
polymer.
poly(chloroethene)
is
O
diagram,
why aluminium chloride
characteristics
of
both
covalent and
d.
ionic
substances.
Deduce
[1]
per
Using
your
2
Cl
6
VSEPR
chloride.
theory,
deduce
[2]
is
oen
11.
U
Al
The dimer contains
coordination bonds.
Outline
the
meaning
of the term
the
is
broken
point
Explain
of
–154 °C,
poly(chloroethene)
this
dierence in
(PLA)
[2]
is
a
feedstocks
down
by
bioplastic
such
as
made
from
sugarc ane
and
maize.
microorganisms under high
O
nHO
C
CH
OH
CH
3
2-hydroxypropanoic (lactic) acid
[1]
formal
charges
l
O
of the atoms inthe aluminium
Explain
why
metals.
Include
dimer
alloys
a
are
a
chloride
above.
generally
v
O
LHA
9.
Determine
It
[2]
is
points.
acid
renewable
answer to two
chloroethene
melting
210 °C.
repeating units
u
f
x
coordination bond.
ii.
Polylactic
t
a
two
melting
o
i
d
r
o
i.
the
around
of
of
your
temperature and high humidity conditions.
Al
Cl
is
point
n
Cl
Cl
melting
found as the dimer
Cl
Cl
The
whereas
chloride.
.
Cl
e.
geometry and bond angle of
chloride
signic antgures.
[2]
C
Al
of
aluminium
State
o
Aluminium
for
i
n
d.
knowledge
molecular
aluminium
formula
v
the
Lewis
number
chain.
y
the
mean
p
c.
Draw
r
e
b.
the
polymer
[2]
found to be
–1
displays
[1]
mass of the polymer chains
69,000 g mol
bonding
[1]
polymerization.
n
with
Draw
for
t
i
s
Explain,
this
l
dierence
chloride.
in
ii.
of
y
aluminium
name
electronegativity and
c.
electronegativity
the
diagram
[2]
O
stronger
than
CH
+
C
nH
O
2
pure
CH
3
in
your
answer.
n
[3]
E
polylactic acid (PLA)
a.
Deduce
action
why
than
PLA
is
more susceptible to bacterial
petroleum-based plastics such as
poly(propene).
b.
Outline
of
226
two
[2]
advantages
bioplastics
such
as
and
PLA.
two
disadvantages
[4]
y
b.
P
low melting point and boiling point.
C alculate
chloroethene
crystals with a
thatallows
i.
r
a.
Aluminium
a.
known as
chloroethene is
your
reasoning.
8.
of
e
that
polymer
—
identify
a
H
phase. Using
electronegativity tables and the triangular bonding
diagram,
form
structure
below.
—
molecules
to
The
—
discrete
react
exists
H
as
c an
poly(chloroethene).
are
s
s
artisan
has
chloride
Structure
Kevlar
is
a
high
synthesized
The
structure
…
tensile-strength
in
1964
of
by
Kevlar
polymer
chemist
is
shown
13.
rst
Stephanie
The
repeating
units
of
polymers
A
From models to materials
and
B
LHA
LHA
12.
2.4
are
shownbelow.
Kwolek.
below.
−
= O
s
s
C
−
CH —
CH
2
n
−
H
r
N
−
H
C
C
C
H
H
H
O
O
C
−
O
C
C
C
H
H
N
= O
C
−
polymer B
O
−
For
H
or
−
For
each
−
N
14.
−
…
linkage
State the formula of the other compound produced
is
the
monomers
made.
Cl
[2]
name of this type of polymerization.
chains
hydrogen
Draw
a
bonding
diagram
two
showing
adjacent
to
one
another
which
occur.
Draw
b.
Deduce
the
monomers
formed
in
structural
[3]
are
shown
below:
O
C
H
N
NH
2
2
Cl
repeating
the
[1]
the
unit
formula
the
of
of
the
the
resulting
inorganic
polymerization
polymer.
[2]
product
process.
[1]
through
features
of
i
hydrogen
to
Identify
Kevlar
[1]
hydrogen bonds
Kevlar
polymer
chains.
[2]
u
l
a
E
v
O
f
x
t
a
allow
between
attracted
r
o
e.
are
bonding.
a.
draw
o
Kevlar
d
d.
[2]
n
U
when Kevlar monomers polymerize and state the
C
c.
Kevlar
of
C
and
monomer(s).
condensation
O
[1]
Draw
which
formulas
Kevlar.
i
n
structural
in
b.
from
the
amide
its
o
the
v
Identify
Two
of
deduce
p
−
polymer.
polymer,
y
r
e
formulas
H
polymer, state whether it is an addition
condensation
N
b.
each
O
a.
−
t
i
s
−
= C
n
n
y
H
a.
O
−
l
−
H
H
y
H
P
H
e
−
polymer A
−
= C
O
227
s
s
e
r
n
l
O
of
y
p
o
v
i
n
C l a ss i f i c a t i o n
m a tt e r
y
P
y
t
i
s
3
r
e
Structure
C
n
U
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
Structure 3.1
The periodic table:
Classication of elements
s
s
How does the periodic table help us to predict patterns and trends in the properties of elements?
of
use
the
it
according
to
we
their
we
classify
the
oils
bec ause
depending
classify
matter
motor
viscosity,
purposes
c an
c an
example,
on
is
based on what
c an
be
they
how
are
thick
The
classied
used
they
for
are.
increase
correlates
or
So,
down
and
elements?
the
trends
with
and
with
decrease
the
groups
in
the
of
position
of
the
periodic
certain
of
periodic
table
number
shows the outer
energy
r
e
Structure 3.1.8 — Transition
occupied
by
electrons. Elements in a
d-sublevels
have
a
common
number
of
valence
electrons.
that
give
them
Structure 3.1.3 — Periodicity
elements
across
a
period
include
and
in
properties
increasing
decreasing
metallic
a
in
properties
group.
of
elements
character
non-metallic
of
character
metal
a
continuum.
oxides
This
includes
if
forming
the
a
show
the
bond.
compound
It
number
is
the
were
of
a
v
E
nd
a
detailed
to
the
between
be
ionization
the
of
orbitals
oxidation
explained
energies
element
absorption
properties.
variable
complexes
light
in
by the
are close
when
the
split
an
are
electron
colour
d-sublevels.
absorbed is complementary to the colour
observed.
electrons
charge that atom
composed of ions.
u
l
f
x
O
c an
promoted
The
from basic
Periods, groups and blocks in the periodic
table (Structure 3.1.1 and
Structure 3.1.2)
The periodic table consists of horizontal periods and vertical groups. The
periodic table is also divided into four blocks: s, p, d and f. Elements are
organized into these blocks according to the arrangement of their outermost
valence electrons. For example, elements in groups 1 and 2 have their outermost
valence electrons in the s sublevel, so these are in the s-block. The blocks are
shown in gure 1 on the next page.
We
You
is
due
of
o
have
coloured
have incomplete
properties
oxidation state is a number
to
value.
group
i
in
atom
successive
c an
trend of
t
a
would
an
r
o
transferred
trend
d
Structure 3.1.6 — The
to
the
in
elements
Structure 3.1.10 — Transition
through amphoteric to acidic non-metal
oxides.
assigned
non-metallic
their
formation
the
n
show
and
that
transition
down a
U
Structure 3.1.5 — Metallic
fact
in
group 1
of
17 elements.
states
C
elements
the
trends
i
n
group
to
down
o
Structure 3.1.4 — Trends
refers
and
v
of
elements
characteristic
p
Structure 3.1.9 — The
in
y
is
occur
across a period.
LHA
period
ionization
n
Structure 3.1.2 — The
rst
patterns
l
increasing
group
These
described visually
O
Structure 3.1.7 — Discontinuities
groups and blocks.
that
be
y
t
i
s
Structure 3.1.1 — The periodic table consists of periods,
level
across the periods
table.
c an
graphs.
Understandings
energy
properties closely
elements
y
how
For
P
dierent
ways
for.
r
we
e
One
c an
also
classify elements in the periodic table as
metals,
non-metals
and
version of
metalloids.
Metals
generally
have
three
or
fewer
valence
electrons,
which
are
the periodic table at the back of
deloc alized and thus contribute to metallic bonding ( Structure 2.3). Non-metals
this book.
usually
have
four
or
more
valence
electrons
which
are
not
deloc alized.
229
Structure
3
Classic ation
of
matter
s-block
1
18
transition elements
1s
1s
2s
14
15
16
2p
d-block
e
3s
3p
3
4
5
6
7
8
4f
7p
y
r
e
5f
p, d and f blocks of the periodic table
ionic
and
show
number(s)
directly
1
either
alkali metals
transition
Groups
elements
groups
18
noble gases
Periods
periodic
given
are
the
a
red
zig-zag
in
table
are
which
to
7
.
The
n,
of
the
this
period.
are
Elements
in
u
groups
For
l
a
E
v
O
f
x
t
a
electrons
1
to
the
ve
You
This
3
it
the
and
is
use
group
valence
this
all
2p
They
metalloids
Metals
c an
form both
are usually
are to the le of the
have
to
period
summarized in table 1.
number
outermost
in
the
is
period
18,
electrons.
same
equal
group 1, the
13
from 1 to 18. Elements in some
are
2
corresponds to the
electron
have
n
=
sublevels
2,
and
for elements
their
valence
sublevels.
number
information
n
=
3
electrons
gas
in
or
groups
straightforward
ve
noble
of
For
elements
2s
group
1.
be
to
number
the
alkali metals,
the
For
last
digit
example,
of
valence
number
all
of
of
have
the
one
group
elements
electrons.
valence
in
For
electrons.
electron in their
number
group
15
is
all
equal
have
electrons.
means that
has
the
the
elements
number
c an
in
same
2,
sublevel.
valence
example,
found
example,
valence
For
from
numbered
name,
1
in
gure
behaviour.
to
are on the right.
collective
numbered
non-metallic
considered
principal quantum number,
Names of groups in the periodic
table
the
i
r
o

Table 1
in
are
of
and
Elements
non-metals
o
halogens
d
17
side
and
metallic
bonds.
n
U
metalloids,
3 – 11
hybrid
covalent
o
Group name
C
i
n
Metalloids
Group
p
The s,
v

Figure 1
6p
to
for
in
the
2
[Ne] 3s
3
3p
deduce
valence
these
represent
conguration:
to
the
electron
congurations of elements.
for periods 2 and 3. Consider phosphorus: being in period
the
sublevels,
sublevels.
inner
You
electrons
and
c an
to
being
use
give
the
the
in
group
symbol
of
15
means that
the
condensed
previous
electron
y
6d
n
7s
l
5d
4p
O
6s
r
4d
12
P
5s
11
y
3d
10
t
i
s
4s
9
f-block
230
17
s
s
13
2
Structure
With
period
4,
we
need
to
take
the
3d
3.1
The
periodic
table:
sublevel into account, which lls aer the
Remember
4s
sublevel
and
before
the
4p
sublevel.
The
group
number
is
of
electrons
occupying
the
4s,
3d
and
4p
sublevels.
Consider
all
in
group
17
means
that
17
electrons
ll
the
outer
sublevels.
go
into
3d
and
the
remaining
ve
go
into
4p,
so
the
electron
dierent
of
electrons.
on
how to ll
conguration of
with
electrons
is
5
3d
4p
Structure1.3
of
a
properties
By
of
the
also
found
certain
table
as
the
we
know
elements
time,
but
table
also
depicted
their
very
chemist
Dmitri
periodic
which
properties.
mass and
Mendeleev
system
eventually
today.
like
be
this
not
only
based
helped
presence of gaps in
signalled
elements.
the
possible
G allium,
for
existence of
example,
was
not known to Mendeleev, but he had assigned an element
close
to
was
of
For
you
the
have
periodic
of
this
table.
positions
book
E
a.
element
to
not
name of the element
name
the
Sc,
you
think
Y
,
La
table
and
do
it
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231
Structure
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Classic ation
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233
Structure
3
Classic ation
of
matter
Figure
5
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ionic
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+
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across
234
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Classic ation of elements
2500
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235
Structure
3
Classic ation
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237
Structure
3
Classic ation
of
matter
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whereas carbon atoms contain six. Fluorine and carbon
Why?
atoms have two electrons in the rst energy level, and their
Because the electrons in the bond are not distributed
r
valence electrons are in the second energy level.
equally between the two atoms.
Because the rst energy level holds up to two electrons.
Because the dierence in electronegativity is large.
l
Why?
Why?
And
so
on.
use
to
How
build
Because uorine has a stronger tendency to attract a
good
about
the
Why?
explanations
Why?
acquire
the
such
knowledge that
as
this?
What makes a
“explain”
How
c an
should
questions
you
know
in
your IB
how
detailed
your
y
r
e
examinations.
Because uorine has a smaller atomic radius than carbon.
you
explanation?
bonded pair of electrons towards itself than carbon.
Think
did
explanations
O
you
t
i
s
Why?
n
y
Because the rst energy level only has one orbital.
Because uorine is more electronegative than carbon.
be?
p
o
v
i
n
Metallic character and periodicity
C
(Structure 3.1.4)
character
value
it.
to
ionization
electrons
not
in
energy.
their
a
the
quantitative
metallic
Elements
structure.
For
example,
the
that
greater
u
The
in
they
group
of
l
character
17
is
a
E
v
to
accept
We
c an
some
the
an
more
donate
linked
of
are
to
the
group
high
that
an
elements
is,
we
element
metallic
in
these
chemic al
known
metallic
reactive
their
do
is
character
with
known
as
as
lower
the
not
assign a
closely
have
linked to
deloc alized
ionization
decrease
chemic al
energies
trends
in
the
reactions.
alkali
the
We
have
down
group,
in
the
as
decreasing
group. This
they
chemic al
have a
reactions.
describe the halogens
property is the opposite to metallic
The
non-metallic
going
reactive,
reactions
metals,
going
electron
halogens.
electronegativity.
more
alkali
down
valence
character — this
therefore
the
character
going
single
halogens
are
electron
illustrate
typic al
also
to
elements,
non-metallic
electronegativity
up
1
increasing
elements
their
and
and
are
tendency
terms
higher
group
energy
t
a
means
with
of
o
more likely to be metallic.
ionization
property,
character
Therefore,
are
i
d
r
o
f
x
O
238
is
However,
n
its
U
Metallic
and
down
as
ll
metals
the
they
their
and
character and
group.
have
a
valence
the
Halogens
greater tendency
sublevel.
halogens
by looking at
y
P
Why?
Why?
Structure
3.1
The
periodic
table:
Classic ation of elements
Reactions of alkali metals with water
+
A
Brønsted–Lowry acid
is
a
substance
that
donates
protons, H
. A
Brønsted–
Brønsted–Lowry acids and bases,
+
Lowry base
is
a
substance
that
accepts
protons, H
.
The
alkali
metals
all
form
and
aqueous
Brønsted–Lowry
bases
when
reacted
with
water.
The
general
the
alkaline
the
reaction
of
a
group
1
metal
with
water
is
as
and
basic
between
substances
follows:
covered in
Reactivity 3.1.
1
M(s)
+
H
2
O(l)
H
→
2
(g)
+
MOH(aq)
2
example,
lithium
reacts
with
water
as
e
For
follows:
1
+
H
2
O(l)
→
H
2
(g)
+
r
Li(s)
LiOH(aq)
2
going down the group. As a result, reactivity increases and the reactions of alkali
character
octet,
of
the
react
with
becoming
halogens
other
species
anions.
decrease
The
to
gain
electrons,
and
to
O
halogens
complete
achieve
electronegativity and non-metallic
going
down
the
group, and hence their

Figure 10
c an
illustrate
elemental
and
halogens
a
elemental
as
other
uorine
+
Cl
→
F
Cl
+
2
is
bec ause
not
not
happen,
uorine
is
more
however:
chlorine
electronegative
favourable.
an
c annot
than
When elemental
electron,
This eect
increases going down group 1
reactions
forming a
In
this
reaction, chloride loses
electrons
and
(undergoes
uorine
gains
oxidation)
electrons
oxidize uoride
(undergoes
reduction). Oxidation
chlorine, so the
and
n
is
will
the
halogens.
follows:
2
reaction
at
gains
U
This
chlorine
of
looking
Elemental sodium metal
reacts vigorously with water.
C
2
reverse
reaction
anions
elemental
by
1
F
ions.
halogens
the
ion,
group.
i
n
1
the
and
chloride
2
The
of
the
o
ion
with
reactivity
down
v
reacts
uoride
the
going
p
between
uorine
decreases
y
We
also
r
e
reactivity
n
y
a
t
i
s
Elemental
l
metals with water occur faster and more vigorously going down the group.
y
P
Ionization energy decreases and the metallic character of the alkali metals increases
Reactions of halide ions with halogens
are
s
s
for
distinction
equation
reduction
are
discussed in
Reactivity 3.2.
The greater the electronegativity dierence between the two reacting species, the
2
o
d
more readily the reaction occurs. Hence, F
will react faster with I
than with Cl
This is also the case for the reactions between halogens and alkali metals: F
2
.
reacts
i
r
o
faster with K than with Li. The equation for the former reaction is shown below:
t
a
1
+
F
2
+
K → K
2
+
F
u
f
x
Activity
the
row
table
has
to
been
show
a
rst
l
The
done
v
O
Complete
which
for
reactions
F
F
Cl
Br
I
2
2
2
No
reaction
E
2
c an
occur
between
halogen
Cl
F
species.
you.
2
+
No
2Cl
→
2F
Br
+
Cl
2
F
2
+
2Br
→
2F
I
+
Br
2
F
2
+
2I
→
2F
+
I
2
reaction
No
reaction
No
reaction
239
Structure
3
Classic ation
of
matter
Practice questions
1.
Write
the
between
2.
List
the
word
alkali
symbol
in
metals
terms
of
and
equations
from
the
their
reactions
most
with
1
17
least
b.
Ba
c.
K
d.
F
I
+
most
or
2
I
2
Li
or
reactive
+
Ba
F
pair
of
reactants in
to
+
Cl
2
water.
ATL
+
2
Cl
+
2
or
Li
+
KCl or F
2
d
Cl
+
2
KI
Thinking skills
The modern periodic table is a very useful way of organizing the elements.
Conduct an online search for alternative representations of the periodic table.
exploring
An example is shown in gure 11.
reactivity of
elements?
Tool 2)
O
t
i
s
y
p
r
e
C
o
v
i
n
n
U
t
a

Figure 11
o
i
d
r
o
This periodic table has a “bicycle wheel”
structure,
with each group
being a spoke of the wheel
u
l
f
x
a
E
v
O
240
a
2
n
2,
group
the
the
+
y
(Inquiry
and
chemic al
to
Li
l
group
in
in
Identify
a.
Select two or three alternative periodic tables and consider:
•
What
•
To
features
•
What are the advantages and disadvantages of the dierent periodic tables?
what
of
extent
the
are
elements
they
are
essentially
being
the
highlighted
same
as
the
in
each
standard
c ase?
periodic
table?
y
trends
used
3.
r
often
reaction
P
the
reactive
are simulations and online
reactions
the
e
Linking question
Why
for
water.
s
s
reactive,
and
potassium
Structure
3.1
The
periodic
table:
Classic ation of elements
Metal oxides and non-metal oxides
(Structure 3.1.5)
Another way to describe acids and bases is through their ability to accept or donate
a pair of electrons: a Lewis acid can accept an electron pair, and a Lewis base can
s
s
donate an electron pair. Many metal oxides are Lewis bases. They react with water
to form hydroxides, also bases, by donating an electron pair to hydrogen in water.
Reactions of alkali metal oxides with water have the general equation:
are
O(s)
some
2
O(l)
→
forms
that
H
a
2
2
6H
O(l)
O(l)
or
donate
an
acids
by
CO
3
(aq)
H
2
2
SO
3
(aq)
sulfurous acid
H
2
SO
4
(aq)
sulfuric acid
O(l)
4H
oxides
a
of
the
elements
stronger
base
stronger acid than P
as
both
a
Lewis
For example, aluminium oxide, Al
pair
other
examples:
3
PO
4
(aq)
phosphoric acid
forms
behaves
2
p
species
(l)
form
c arbonic acid
u
3
the
O(s)
to
some
O(l)
a
v
period,
2
are
(aq)
hydroxide
o
H
water
Here
i
H
l
+
oxide
Na
with
water.
2
o
2
react
in
(s)
n
H
+
(s)
They
oxygen
Ba(OH)
barium
2
hydroxide
C
O(l)
t
a
(l)
any
form
v
acids.
from
+
water and SO
electron
2
trioxide
10
to
Mg(OH)
i
n
(g)
3
water
O(l)
d
2
example,
chemic al
H
+
dioxide
amphoteric.
an
Lewis
pair
r
o
A
are
(g)
E
with
For
2
U
2
with
(aq)
H
+
dioxide
across
react
magnesium
f
x
O
acidic.
2
also
equation:
oxide
phosphorus(V)
Going
M(OH)
+
electron
SO
They
general
oxide
oxides
an
bases.
hydroxide
examples:
accepting
O
2LiOH(aq)
n
H
Lewis
following
Non-metallic
4
O(l)
l
also
the
BaO(s)
barium
P
2
hydroxide
O
are
+
magnesium
sulfur
H
lithium
MgO(s)
SO
2NaOH(aq)
y
some
sulfur
O(l)
sodium
+
with
CO
2
oxide
oxides
c arbon
H
r
e
are
+
O(s)
MO(s)
Here
examples:
t
i
s
2
hydroxides,
2MOH(aq)
oxide
lithium
Group
→
y
2
O(l)
O(s)
sodium
Li
2
y
2
H
P
Na
+
r
Here
2
e
M
electron
pair
2
4
become
than
O
10
acid
O
3
less
MgO(s)
basic
in
its
and
more
reaction
(s).
and
a
Lewis
base
is
termed
(s), is amphoteric: it c an accept
depending
on
what
it
is
reacting with.
241
Structure
3
Classic ation
of
matter
For
example,
oxide
Al
2
acts
O
3
as
when
a
reacting
(s)
+
with
(s)
+
3H
2
→
O(l)
2Na[Al(OH)
4
acid,
aluminium
→
6HCl(aq)
2AlCl
(aq)
as
a
Lewis base:
+
3H
2
r
chloride
neutralization
reactions:
an
acid
each other to form a salt.
and
a
O(l)
base
react
standard
Na
level of DP
2
O(s)
MgO(s)
Al
know
and
do
not
denition
bases.
of
need to
Lewis acids
Acid or
Lewis acids and bases
in
basic
3
(s)
Trend
O(l)
o
dioxide
and
atmosphere
by
2
is
of
nitrogen
to
P
4
O
10
(s)
SO
acidic
SO
3
2
(l) and
(g)
acidic
3
corresponds
naturally
weak
acidic
to
a
due
neutral solution: it
to
the
presence of
c arbonic acid:
(aq)
5.6.
We
know
the
right
rain
is
in
2
,
the
more
that
are
other
more
periodic
acidic
oxides, such as
acidic
than
than
CO
2
bec ause
table. If these gases
normal,
and
is
known as
less than 5.6.
oxides
vegetation.
industrial
is
dioxide, SO
ensuing
pH
which
forms
CO
sulfur
further
the
a
H
rainwater
are
has
7
.0,
which
→
and
rainwater,
decomposing
are
These
processes,
produced
pollutants
such
as
naturally
c an
the
also
by
be
volc anic eruptions
released into the
combustion
of
fossil fuels with
levels of sulfur impurities.
u
You have seen the reaction of SO
l
2
with water that produces sulfurous acid. The
reaction between water and nitrogen dioxide produces nitric acid and nitrous acid:
a
v
O
f
x
high
2
for
sulfur
i
and
H
value
which
t
a
r
o
d
in
acid rain,
Sulfur
+
of
R ainwater
dioxide,
oxides, NO
and
pH
C
pH
(g)
a
basic.
x
nitrogen
dissolve
2
has
n
U
typic al
nitrogen
nor
c arbon
CO
A
298 K
i
n
dissolved
(s)
acidic
p
at
acidic
o
water
neither
v
Acid rain and ocean acidic ation
is
2
in the acid–base properties of the oxides of some period 3 elements
Reactivity 3.4 (AHL )
Pure
SiO
amphoteric
base?
more detail in

Table 2
O
y
discussed
basic
2
r
e
are
the
you
O
oxide
chemistry,
n
the
t
i
s
Formula of
At
l
y
The acid–base properties of some period 3 elements are shown in table 2.
with
2NO
nitrogen
2
(g)
+
H
2
→
O(l)
dioxide
HNO
nitric
3
(aq)
+
acid
HNO
2
(aq)
nitrous acid
Linking question
do
explain
E
How
Ultimately,
to
nitric
nitrous
acid
and
acid
and
sulfuric
sulfurous
acid,
acid
are
oxidized
by
atmospheric
oxygen
respectively:
differences in bonding
the
differences in the
2HNO
2
(aq)
+
O
2
(g)
→
2HNO
3
(aq)
properties of metal and non2H
metal
oxides?
Structure 2.2)
(Structure
2
SO
(aq)
+
O
2
(g)
→
2H
2
SO
4
(aq)
2.1 and
Oceans
absorb
atmosphere.
acidity
242
3
c an
As
a
large
a
aect
proportion
result,
the
c arbonic
ability
of
of
the
acid
coral
is
reefs
c arbon
dioxide
formed
in
and
the
shellsh
released into the
ocean.
to
Increased
ocean
c alcify their skeletons.
y
are
3
acts
P
reactions
oxide
aluminium
acid
these
](aq)
sodium aluminate
hydrochloric
oxide
of
+
hydrochloric
aluminium
Both
base, aluminium
e
3
a
s
s
O
hydroxide,
hydroxide
reacting
2
sodium
sodium
oxide
Al
aqueous
2NaOH(aq)
aluminium
When
with
Lewis acid:
Structure
3.1
The
periodic
table:

Figure 12
reacts
water and
Classic ation of elements
Sulfur dioxide reacts with
oxygen in the atmosphere to
in the air
form
acid
sulfur
dioxide
sulfur
sulfuric acid,
c ausing acid
rain
rain
trioxide
H
SO
2
4
s
s
e
r
corrodes
metals
burning
kills
erodes
trees
wet
acidifies
limestone
other
surfaces
lakes
(fish
t
i
s
buildings
fuels
die)
O
Oxidation states (Structure 3.1.6)
purely ionic or covalent, but rather occurs on a continuum with unequal sharing
v
represent the charge on an atom in a compound if it were composed of ions. It also
C
i
n
Water
, for example, is a predominantly covalent molecule: H
o
describes the number of electrons shared or transferred when forming a bond.
p
of electrons. The concept of oxidation state, or oxidation number, is used to
y
r
e
The bonding triangle introduced in Structure 2.4 shows that bonding is oen not
n
and
y
erodes
coal
l
power
stations
y
P
volc anoes
2
O
O. If it were an ionic
2
H
+1
H
+1
compound, then oxygen would gain two electrons and each hydrogen atom would

Figure 13
assigned an oxidation state of –2 and each hydrogen has an oxidation state of +1.
elements in water
a
more
electronegative
electronegative
number
of
oxidation
useful
deducing
oxidation states of atoms:
u
of
all
zero.
atoms
in
a
polyatomic
Rule 4
The
oxidation state of uorine is –1 in all compounds.
Rule 5
The
oxidation
the
E
The
of
group
oxidation
more
state
v
state
Rule 6
of
2
state
The
oxidation
state
electronegative
with
state
the
less
is
of
2
so
O
of
a
is
is
equal to
2
,
1
metals
is
always
+1
and
the
oxidation
always +2.
oxygen
is
–2
oxygen
except in OF
has
an
2
,
where uorine is
oxidation state of +2, and in
where it is –1.
hydrogen
elements
electronegative
c alled
group
metals
electronegative,
peroxides such as H
Rule 7
ion
charge of the ion.
a
O
l
f
x
The
states
is
oxidations
oxidation states.
The sum of the oxidation states of all atoms in a neutral compound is zero.
oxidation
element
the
negative
positive
Rule 3
the
free
assigned
Rule 2
of
any
are
assigned
The
sum
of
for
atoms
are
Rule 1
the
state
rules
atoms
i
are
the
less
t
a
There
the
r
o
and
o
compounds,
states
d
Deducing oxidation states in compounds
In
The oxidation states of
n
U
lose one, as oxygen is more electronegative than hydrogen. Therefore, oxygen is
(most
is
+1
when
it
non-metals),
elements
(metals).
is
combined
and
–1
when
Hydrogen
in
with
it
a
is
–1
more
combined
oxidation
hydride ion.
243
Structure
3
Classic ation
of
matter
The
is
rst
rule
bec ause
states
atoms
that
of
electrons
are
example,
consider
shared
electronegativity
is
oxidation
same
evenly
across
elemental
of
the
state
element
bonds
oxygen, O
atoms
is
of
have
2
in
:
identic al.
an
element
equal
elemental
the
is
always
zero. This
electronegativity, so the
compounds.
molecule
Therefore,
is
the
For
symmetric al, and the
oxidation
state
zero.
Deduce the oxidation state of
the atoms in H
c.
P in the phosphate ion,
d.
all the elements in LiAlF
4
Cu,
Zn,
Xe,
and
Au
P
,
3–
PO
4
states
of
its
(oxidation state of K)
0
multiply
the
4
is
a
neutral compound, the sum of the
equal
to
zero:
(oxidation state of Mn)
oxidation
state
value
for
rule
the
+
the
above
to
rule
5,
oxidation
equation
the
b.
the
a.
2
Cr
2
O
CO
d.
HClO
2
4
2
2
3
Fe
j.
CH
k.
CH
l.
Na
2
O
3
2
2
3
OH
O
O
2
E
h.
v
NaH
a
SO
4
of
its
to
3,
is
as
+
rule
6,
above
oxidation state of P
Using
rule
state of F
2:
According
4,
the
above
1
+
to
is
of
–2.
and
then
states
nd
there
the
are
value
four
for Mn
zero.
potassium is
Substituting
(–2
×
4)
=
+1.
According to
these
values into
0
so
each
atom
is
4
to
has
an
oxidation state of 0
polyatomic
(–2
×
state
above
4)
ion,
the
sum
of
the
oxidation
charge on the ion:
oxidation
equation
+
a
the
(oxidation state of O
=
of
×
4)
=
oxygen
–3
is
–2.
Substituting this
gives:
–3
= +5
(oxidation
× 4)
K,
as
gives:
+
PO
equal
the
the
state
oxygen
species,
According
into
and
four,
3
rule
atoms
to
O
oxidation
by
= +7
elemental
(oxidation state of P)
d.
of
above
(oxidation state of P)
value
l
O
OF
f.
According
states
u
CH
c.
e.
7
c.
following
f
x
b.
g.
244
K
all
o
in
are
(rule 1).
i
atom
compounds:
These
to
the
oxidation
(oxidation state of Mn)
t
a
each
oxidation states to
d
Assign
r
o
4.
of
state
oxidation state of Mn
Practice questions
states
sum
n
U
1
6,
the
C
i
n
According
oxidation
make
p
assign
o
would
v
c an
that
(oxidation state of O
oxygen
oxygen atoms in the compound.
You
+
y
We
=
+
is
r
e
4)
atoms
O
oxidation
t
i
s
According to rule 2, as KMnO
n
y
Solution
×
l
4
state
of
Li)
+
(oxidation state of Al)
+
(oxidation
= 0
rule
oxidation
equation
5,
the
state
of
above
oxidation state of lithium is
uorine
–1.
Substituting
gives:
(oxidation state of Al)
oxidation state of Al
is
+
= +3
(–1
×
4)
=
0
+1.
According to rule
these
values into the
y
Mn in KMnO
b.
r
a.
2
each
e
Worked example 1
a.
of
s
s
atom
the
the
Structure
3.1
The
periodic
table:
Classic ation of elements
Naming of oxyanions
Oxyanions
are
oxyanions,
you
a
Roman
with
no
at
anions
include
the
space
end
of
that
include
the
oxidation
the
name.
between
the
ion
example, consider the anion MnO
Mn
has
an
potassium
This
.
4
The
Roman
name
The
and
atoms.
the
numerals
the
anion
is
When naming
non-oxygen atom as
are
named manganate(VII),
neutral compound KMnO
compound
is
also
enclosed in
bracket.
known
under
4
its
is
c alled
traditional
Practice questions
permanganate.
5.
the
KClO
,
metal,
for
non-oxygen
is
c alled
Roman
used
atom
in
potassium
numeral
is
to
an
refer
to
the
oxyanion
c an also be a non-metal.
omitted
for
some
common
oxyanions. These
nitrite
nitrate(III)
nitrate
nitrate(V)
2
3
2
SO
sulte
3
systematic names of some oxyanions.
have
nomenclature
the
to
Your
data
IUPAC
and
study
booklet,
brackets
by
of
in
the
periodic
Linking question
How
c an
used
to
will
chemic al
oxidation states be
analyse
redox
reactions?
for a long time
table, and the
International
chemistry
writing
where
elements
the
i
subscripts
dierent
symbols
bromate(III)
Union
involve
formulas
of
Pure and
using
symbols
correctly
necessary) and naming substances
t
a
according
in
identify
the
encouraged
(IUPAC).
r
o
(including
stated
to
use
conventions
Chemistry
are
we
o
they
symbols
d
Applied
used
Nowadays,
n
14).
U
(gure
Both are
C
Communic ation skills
Scientists
as
i
n
acceptable in examinations
ATL
magnesium
potassium sulte
o
The common names and
c.
(Reactivity 3.2)
sulfate(VI)
v

Table 3
sulfate
4
for
nitrate
d.
p
SO
hydrogen
sodium phosphate(V)
sulfate(IV)
2
formulas
a.
O
Systematic name
chemic al
b.
y
Common name
r
e
NO
t
i
s
NO
the
following compounds:
chlorate(VII).
shown in table 3.
Ionic formula
Write
the
oxidation state of a
n
the
oen
l
practice,
are
the
example,
are
y
In
4
numerals
y
transition
Roman
P
While
r
name,
manganate(VII).
+7
.
The
of
e
potassium
oxidation state of
oxygen
state
s
s
as
should
numeral
brackets
For
polyatomic
guidance.
T
est your chemical communication skills by nding the error in each of the
following and writing the correct answer:
3.
O3
iron,
4.
MgOH2
5.
cuprum(II)
oxide
a
E
v
O

Figure 14
u
NE
l
HCL
2.
f
x
1.
The symbols used
by alchemists for copper and
respectively
245
3
Classic ation
of
matter
LHA
Structure
Discontinuities in ionization energy trends
(Structure 3.1.7)
E arlier
going
in
this
topic,
across
13,
you
period.
and
from
saw
You
group
that
may
15
there
have
to
was
a
trend
noticed
group
16
two
of
increasing
ionization
discontinuities
from
(gure 15).
e
2

Ne
l
N
Ar
n o it a  i n o i
Be

t
i
s
B
t s r if
O
Mg

Al
Na
the
for
o
u
between
l
a
E
v
O
f
x
t
a
come
is
2s
the
in
therefore
the
between
two
of
to
higher
This
of
c an
be
s
remove
energy
drop
and
nitrogen
in
p
explained
conguration
2
beryllium is 1s
easier
a
boron,
electron
an
2s
by
boron
1
2p
. Despite the
electron
sublevel,
ionization
of
from this
which
energy
is
shielded
therefore
sublevels.
and
oxygen,
c an
be
explained
by
2
close
space
in
existence
2
into
of
is
is
sublevel.
orbitals.
oxygen it is 1s
region
electron
electron
it
and
The
third period elements
2
conguration
boron,
discontinuity,
repulsion
for
not
same
by
evidence
i
d
r
o
while
in
valence
nucleus
second
electron
p
the
the
electron
charge
n
U
as
provides
The
the
nuclear
beryllium
congurations.
Z
and
o
and
between
electron
C
2
element
from
their
y
r
e
discontinuity,
at
2s
greater
v
2
is 1s
The ionization energies of the second
i
n
rst
looking
3rd period
Si
Li
The
Cl
S
P
n
y
O
C
The
2
2s
.
have
easier
conguration
of
nitrogen is 1s
In
to
nitrogen,
The
paired
increased
remove.
the
three
3
2p
electrons
repulsion.
This
electrons in the 2p orbitals
explains
in
The
the
oxygen
occupy the
higher-energy
drop
in
+
N
N
3
2
2p
2p
2
2
2s
2s
+
O
O
4
3
2p
2p
2
2
2s
2s
A half-lled
electron repulsion,
p
sublevel is stable due to decreased
as none of the electrons are paired
paired
ionization
groups.

Figure 16
2
2s
4
2p
proximity.
and
electron
energy
y
P
  grene
F


Figure 15
246
r
lom
2nd period
2
atomic number,
do
energy
group 2 to
s
s
group
a
Structure
3.1
The
periodic
table:
Classic ation of elements
LHA
Properties of the transition elements
(Structure 3.1.8)
A
transition
or
c an
element
form
gives
example,
the
metallic
ions
elements
is
elements
have
This
not
which
considered
a
not
also
c atalyst
occur,
a
type
solid
oen
of
that
the
2
incomplete
exist
as
up
in
a
rhodium
in
the
2CO
of
d-sublevel.
electrons
the
phase
to
the
required
the
reactants.
Gas
are
transition
exhaust
of
elements
converters
vehicle,
2
used
adsorb
a
reaction
and
Transition
chemic al
as
C atalysts
desorb
activation
energy
are
Reactivity 2.2
monoxide
c arbon
dioxide:
Linking question
n
What are the arguments for and
against including scandium as a
transition element? (Structure 2.3)
adsorbs
onto surface of solid
catalyst at active site
u
E
v
a
tsylatac
l
tsylatac
tcudorp
O
f
x
t
a
r
o
i
o
d
c
u
d
A
and
discussed further in
heterogeneous
c arbon
reactant A
catalyst
for
undergo
t
o
r
p
from catalyst
electron
molecules adsorb onto
molecules
heterogeneous catalyst
product desorbs
Its
form 2+ ions,
electrons in elemental
(gure 17).
C atalytic
to
Heterogeneous c atalysts
c atalysts.
and
d
energy
reaction.
separate
element,
4s
means that zinc is not magnetic.
activation
that
heterogeneous
desorb
its
All
which
the
in
loses
intact.
state,
reduces
converters.
→
an
o
O
with
readily
remains
used
then
and
it
U
+
so
transition
molecules
2CO
,
C
c atalytic
oxygen
that
used
palladium
in
c ations
neutral
being
products
creates a
v
c atalysts
The
result they
external magnetic eld.
i
n
Platinum,
not
c atalyst
surface
changes.
and
of
are
its
species
a
p
elements
a
a
as
electron
y
are
is
while
d
r
e
to
in
unpaired
have high melting
and
10
3d
d-sublevel
paired
elements
electrons,
throughout
electrons and the
O
A
complete
are
an
d
an
deloc alized
t
i
s
the
[Ar] 4s
with
of
properties.
deloc alized
these
d-sublevel
transition element despite being in the d-block. This
form
2
is
align
spin
are
lled
Having an incomplete
n
does
c an
and
the
unpaired
the
partially
y
it
bec ause
so
a
characteristic
energy
strong,
has
d-sublevel.
y
moment,
is
some
between
transition
conguration
zinc
high
attraction
transition
bec ause
but
elements
have
that
l
is
The
element
incomplete
in
paramagnetic.
Zinc
electrons
an
an
Most
magnetic
is
d
transition
structure.
as
with
P
are
the
the
dened
r
points.
is
c ation
e
For
metal
stable
s
s
d-sublevel
a
A
B
reaction occurs
reactant B adsorbs onto surface
on catalyst
catalyst
of solid catalyst at active site
B

Figure 17
c atalysts.
The action of heterogeneous
Transition elements are oen
A
good
heterogeneous c atalysts
247
3
Classic ation
of
matter
LHA
Structure
This
reduces
when
fuel
H
2
cells.
In
emission
fuel.
this
of
the
Platinum
process, H
2
is
products
also
used
(g) and O
2
of
as
(g)
incomplete
a
combustion
heterogeneous
adsorb
onto
the
c atalyst
created
in
hydrogen
c atalyst surface and
O(l) desorbs.
some
of
elements
which
are
have
also
able
distinctive
to
form
compounds
known as
s
s
Transition
is
the
burning
complex ions,
colours. The chemistry behind this phenomenon
detailed later in this topic.
e
r
Variable oxidation states in transition
elements (Structure 3.1.9)
other
table,
element,
you
c an
if
you
know
determine
its
3d.
c ase
sublevels.
into
4s
2
t
a
argon: 1s
The
two
u
sublevel
2
and
to
in
achieve
to
a
Iron
3d
valence
8,
electron
the
is
in
period
comes
aer
electrons.
so
it
six
has
previous noble gas in
go
into
4,
4s
This
eight
6
2
3p
4s
the
In
so
its
c an
be
valence
the
valence
bec ause it is higher
3d
deduced
electrons.
sublevel.
the
of
iron
is
electron
[Ar] 4s
6
3d
.
If
you
conguration of
6
method
described
c ases,
stable
l
[Ar] 4s
a
4s
electron
1
is
expand
from
Two go
3d
both
more
copper
conguration
conguration,
2
3s
copper.
and
with
argon: [Ar]
remaining
5
3d
starts
Fe.
2
6
2p
of
group
the
condensed
exceptions
1
[Ar] 4s
and
electron
2s
is
o
the
full
number
Iron
sublevel,
i
the
the
number.
Therefore,
need
in
the
electron
worked
is
example
are
promoted into the 3d
conguration.
Therefore,
chromium is
10
3d
.
a
Variable oxidation states
Another
property
of
the
transition
elements
is
their
ability
to
form
a
variety of
oxidation states of
elements
stable
are
ions
in
dierent
oxidation
states.
These
are
known as
variable oxidation
given in
states.
Some
of
the
booklet.
given
248
the
chromium
v
transition
the data
d
r
o
common
E
the
f
x
O
The
determine
group
n
U
the
is
Remember,
inenergy.
Then,
O
valence
and
transition element on the
y
the
4s
this
C
are
conguration
in
o
identify
sublevels
which
a
electron conguration of iron,
p
electron
brackets,
i
n
Then,
v
condensed
square
of
conguration.
r
e
Solution
The
position
t
i
s
Worked example 2
Determine the condensed
the
electron
n
any
l
with
periodic
y
As
in
gure 18.
most
common
oxidation
states
of
transition
elements
are
y
P
Deducing electron congurations of transition elements
Structure
Ti
V
Cr
Mn
Fe
Co
Ni
The
periodic
table:
Classic ation of elements
LHA
Sc
3.1
Cu
7
6
6
6
5
5
5
5
5
4
4
4
4
4
4
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
oxidation
that occur
in
s
s
states
4
compounds
+2,
c an
+4
of
formation
most
manganese(IV)
variable
oxidation
common
means
that
oxide,
states
in
and
it
oxidation states of
forms
the
following
manganese(VII)
transition
elements
oxides:
oxide.
c an
be
explained
Methods
by
the
fact
that
their
successive
ionization
energies
are
close
in
value. The
lling
successive
ionization
energies
of
chromium
and
copper
are
shown
in
gure
for
drawing orbital
diagrams
are
19,
Structure 1.3
along
with
orbital
lling
diagrams.
0
(b) copper
om
30000
l
10000
5000
v
E
n o it a z i n o i
15000
10
12
14
10
12
14
[Ar]
4s
3d
[Ar]
4s
3d
a
O
k / ygrene
20000
8
ionization
35000
25000
6
u
1–
f
x
40000
4
i
2
t
a
r
o
0
o
5000
n
10000
d
n o it a z i n o i
15000
o
20000
C
25000
U
k / y g r e n e
i
n
om
30000
v
1–
35000
described in
y
chromium
40000
p
r
e
(a)
their
n
The
the
This
O
oxide,
that
+7.
t
i
s
manganese(II)
see
and
y
you
are
y
18,
4 transition elements,
common oxidation states are shown in a larger font
l
gure
manganese
The most
e
From
Common oxidation states of the period
to zinc.
P
sc andium
r

Figure 18
1
1
0
0
2
4
6
8
ionization

Figure 19
The rst
13 successive ionization energies and
orbital lling diagrams for (a) chromium
(b) copper
249
Structure
3
Classic ation
of
matter
LHA
1
Chromium
forms
ions
with
oxidation
states
ranging
from +1 to +6, losing the 4s
3
electron
rst
and
then
the
d
electrons.
The
most
common
an
electron
chromium
energy
has
conguration of [Ar] 3d
an
between
electron
losing
, and CrO
conguration
the
rst
of
electron
[Ar].
to
, which
2
and Cr
4
There
through
+
are Cr
2
3
has
ions
is
the
a
2
O
very
sixth
, in which
7
small
increase in
electron, so all six
s
s
Activity
oxidation
Find
data
for
the
rst 13 ionization
The
states
increments
are
possible.
between
successive
ionization
energies
for
copper
are
larger
+
4
graph
for
your
which
has
an
electron
forms
fewer
conguration
conguration of [Ar] 3d
for
the
it
stable
ions.
of
[Ar]
2
3d
the
.
Cu(III)
superconductors,
but
higher
Sc andium
electron
and
Cu(IV)
oxidation
the
up
until
the
1920s,
are
conguration [Ar] 4s
and
it
was
not
has
an
are Cu
not
stable.
1
3d
3
+
. Sc
considered
a
was
the
only
known
transition element, as a
energies,
for
element
has
to
be
able
to
form
ions
with
each
However,
sc andium
c an
also
exist in the
+2
t
i
s
1
conguration
of
[Ar] 3d
.
An
example
both
lling
here:
c an
exist in
diagrams
v
atom/ion
of
d-sublevels.
having
compound
an
is
electron
C sScCl
+2 and
these
+3
3
oxidation states. The
elements
and
their
ions
are
electron structure
o
4s
3d
C
i
n
Mn
iron
sc andium(II)
p
given
and
orbital
a
y
condensed
r
e
M anganese
of
lled
state,
O
Practice questions
6.
partially
oxidation
transition element.
n
ions
[Ar]
U
2+
Mn
n
[Ar]
3+
Mn
Fe
o
i
d
t
a
r
o
[Ar]
2+
Fe
u
l
f
x
a
E
v
O
250
[Ar]
[Ar]
3+
Fe
[Ar]
2
Explain
why Mn
3
oxidized
7
.
to
Vanadium
+
3
is
+
more common than Mn
2
and
why
Fe
+
c an
be
easily
+
Fe
commonly
a.
Determine
b.
Explain
the
forms
compounds
condensed
electron
where
it
has
a
+5
oxidation
state.
conguration of elemental
vanadium.
why
the
most
common
,
electron
l
ionization
common
has
ions
exist in some compounds, such as in
states
y
the
which
dierence in
transition
and
,
electron
ion
successive
common
+
and Cu
2
congurations,
The
oxidation
state
of
vanadium is +5.
y
between
so
P
connections
look
chromium,
9
each element.
graphs,
for
10
transition elements and plot
Using
those
r
a
than
e
energies of the other period
Structure
3.1
The
periodic
table:
Classic ation of elements
LHA
Transition element complexes
(Structure 3.1.10)
Transition
ions
elements
when
form
bonded with
coloured
ligands.
compounds
Ligands
are
known as
complexes
or
complex
electrons
that
coordination
c an
bond
be
donated
formed
to
a
between
transition
the
element
ligands
and
c ation.
the
This
central
s
s
OH
of
3+
molecules or ions with a lone pair
2
results in a
c ation
H
O
OH
2
(gure 20).
When
ligands
the
form
Normally,
visible
gap
light
all
ve
d
between
region
of
are
bonds
split
orbitals
into
have
with
two
the
orbitals
transition
sets
same
the
split
the
electromagnetic
with
energy:
corresponds
element
dierent
to
they
a
c ations, the
energies
are
H
(gure
O
2
degenerate. The
OH
wavelength within the
spectrum.
2
2
3
The complex ion [Cr(H
O)
y
l

Figure 20
OH
y
energy
coordination
d-sublevel
P
21).
in
r
orbitals
2
e
Cr
2
6
+
]
has six coordination bonds from the lone
+
2
t
i
s
ygrene
The
formation
detailed in
(b)
gaseous ion
with
non-degenerate
in
wavelength
orbitals.
absorbed.
gure
22.
the
The
the
split
higher
corresponding
colour
colour
wheel
is
colours
also
electrons
happens
the
energy
in
when
gap
the
the
lower
complex
between the
are
opposites
on
the
colour
wheel
shown in the data booklet.
i
585 nm
orange
u
l
yellow
575 nm
a
400 nm
E
v
O
f
x
red
to
t
a
r
o
647 nm
700 nm
allows
This
observed is the complementary colour to
Complementary
The
orbitals
orbitals.
o
colour
given
a
between
into
d
the
energy
complex ions is
o
light
in
promoted
C
absorbs
be
orbitals of the transition element ion
n
to
of ligands on the d
i
n
dierence
orbitals
degenerate orbitals
U
The
The eect
ion — non-
v
— degenerate orbitals

Figure 21
five d orbitals in a
complex
of
Reactivity 3.4
y
isolated
p
five d orbitals in an
r
e
(a)
O molecules to the central
ion. It is violet in colour
O
level
3
Cr
n
pairs on the H
green
violet
blue
424 nm

Figure 22
491 nm
The colour wheel
251
Structure
3
Classic ation
of
matter
LHA
3
For
example,
[Cr(H
between
the
opposite
colour
The
larger
shorter
the
an
d
on
O)
+
]
absorbs
orbitals.
the
splitting
electron.
wavelength
wavelength,
6
As
colour
of
the
and
frequency
light
(3.00 × 10
light
when
electrons
the
complex
the
greater
orbitals,
for
a
larger
frequency.
the
splitting,
energy
the
Remember
light
the
1
m s
frequency (in s
)
f
have a
between
×
wavelength (in m)
×
λ
r
=
constant
×
J s)
(6.63 × 10
1
frequency (in s
×
)
f
2
The colour of [CuCl
]
ions is equivalent to visible light with a wavelength of
4
2
647 nm. Calculate the energy gap of the split d orbitals in [CuCl
of
the
of
light
complex
the
wavelength
metres.
U
splitting
l
stronger
a
gap
×
10
to
determine
of
d
the
of
491
×
6.11 × 10
f × λ.
wheel
647 nm, the
(gure 22):
Remember
to
convert
10
m
×
f
1
s
energy
gap
between the orbitals:
34
E = 6.63 × 10
ligand
the
the
split
and
the
complex
orbitals
coordination
1
14
J s
×
6.11 × 10
depending
bonds
d
transition
observed.
with
orbitals
higher
on
the
s
the
J
element
The
same
central
central
ion
c ation
ligand
metal
will
are
will
c ation.
produce
a
two
factors that
c ause
dierent
Ligands
greater
that
form
energy
(gure 23).
greater
splitting of
of light
bonds
orbitals
absorbed
strength and
wavelength
colour
9
=
the
stronger
between ligand
=
14
f =
coordination
The relationship
the
19
the
between
c
E = 4.05 × 10
colour
in
× f
to
the
1
m s
n
u
the
o
i
t
a
identities
aect
v

Figure 23
E
strength
d
r
o
f
x
O
greater ligand
8
E = h
on
frequency of the light, use
to
3
Then use
The
opposite
C
i
n
determine
the
corresponds
is
o
491 nm.
To
ion
absorbed
y
the
v
colour
wavelength
]
4
p
r
e
Solution
If
O
t
i
s
Worked example 3
n
y
l
h
energy
smaller
greater
wavelength of
wavelength of
light
absorbed
complex colour
the colour of complexes
The third factor aecting the colour observed is the charge on the central transitional
+
3
element cation. A Cr
+
2
complex will have a dierent splitting of d orbitals than a Cr
complex with the same ligands, and they will therefore be of dierent colours.
252
y
Planck
to
1
=
)
P
=
required to
needs
relationships
34
=
E
promoted
energy:
c
energy (in J)
are
appears violet, as this is the
e
of
d
higher
and
yellow
result,
wheel.
Therefore,
8
speed
a
s
s
promote
split
2
Structure
3.1
The
periodic
table:
Classic ation of elements
LHA
Practice question
2
8.
A
complex
Chloride
colour
[Cu(H
ions
2
are
change
O)
6
+
]
(aq)
changes
considered
to
be
colour
weaker
when
excess Cl
ligands
than
(aq)
is
added.
water. Explain the
observed.
s
s
Colorimetric analysis of a solution of unknown concentration
this
of
a
task,
you
transition
curve.
app
that
in
You
c an
the
solution
will
need
measure
unknown
Method
concentration
by interpolating a
a
1.
Review
2.
Read
the
colorimetry
section of the
colorimeter, or a mobile
RGB
through
the
procedure
values.
equipment
you
will
chapter.
need.
Relevant skills
•
Tool
•
Tool 1:
Prepare
•
Tool 1:
C arry out dilutions
•
Tool 3:
Construct
•
Tool 3:
Interpolate
1: Use and apply colorimetry
standard solution
Prepare
a
interpret
C alibrate
uncertainty
measuring
bars
apparatus, including
•
Inquiry 1:
research questions
transition metal solutions include
U
Possible
Formulate
those containing manganate(VII) ions, MnO
and
complex
ions,
pentaaquathiocyanatoiron(III)
O)
5
(SCN)]
is
also
of
a
of
concentration
task
of
a
You
in
,
complex ions is not
you will be
to
obtain
concentrations.
five
colorimeter.
the
absorbances
a
c alibration
curve
that
of
the
shows
five
standard
absorbance
vs
Check
9.
for
errors
Propagate
and
by
verifying
that
your line of best fit
through the origin.
uncertainties
graph.
and
include
error
bars on
Draw lines to determine the maximum
minimum
gradients.
Part 2: Analysis of solution of unknown concentration
10.
Obtain
a
small
sample
of
the
solution
of
unknown
technique, so
for SL students.
this
complex
technician.
of
analytic al
Plot
goes
concentration. Determine the absorbance of the
analysed solution.
involves determining the
solution
prepared
by
11.
Use
your
c alibration
curve to determine the
your
concentration
of
the
analysed ion in the solution.
c an also use it to determine
particular
u
f
x
the
or
important
suitable
version
concentration
teacher
Knowledge
]
i
task
simplied
.
this
6
t
a
A
use
r
o
this
to
O)
ion
dilution
concentration.
8.
o
2
necessary
2
7
.
your
complex ions,
d
2+
[Fe(H
[Cu(H
Determine
,
2+
hexaaquacopper(II)
6.
n
4
the
C
i
n
conclusions
C alibrate
o
Discuss the impact of uncertainties on the
v
Inquiry 3:
the
wavelength of incident light.
5.
solutions.
sensors
of
serial
y
Inquiry 1:
and
different
p
•
Draw
graphs
r
e
Tool 3:
of
a
Depending on the colour of the solution, determine a
graphs
suitable
•
solution
perform
O
4.
interpret
and
t
i
s
solutions
and
standard
analysing,
n
a
y
3.
•
l
Part 1: Calibration curve preparation
y
the
Skills
below and make a list of
P
phone
determine
r
c alibration
will
metal
e
In
coloured component in
everyday substances.
12.
Determine
uncertainty
the
of
absolute,
the
relative
and
percentage
concentration.
l
a
E
v
O
Part 3: Developing a research question
13.
Brainstorm
involve
possible
using
this
research questions that could
analytic al
technique.
Linking question
What
is
ligands
How
the
in
c an
nature
forming
of
colorimetry
concentration
of
a
the
reaction
complex
or
ions?
between
transition element ions and
(Reactivity 3.4)
spectrophotometry
solution
of
coloured
be
ions?
used
to
c alculate the
(Tool 1, Inquiry 2)
253
Structure
3
Classic ation
of
matter
End of topic questions
A.
–141
B.
–141
C.
(–141
D.
–141
Topic review
23
Using
your
knowledge
from the
Structure 3.1
×
2
×
s
s
1.
6.02 × 10
topic,
23
answer
the
guiding
question
as
fully
as
possible:
+
753)
×
6.02 × 10
753
e
How does the periodic table help us to predict patterns
+
and trends in the properties of elements?
7
.
Which
of
Li and I
of
the
following does
C.
C s and I
D.
Li and F
transition element
D.
element
Which
the
following
added
SiO
forms an acidic solution
water?
2
9.
SO
II and III only
Which
of
the
following
charge
constant shielding
following
C.
X
D.
X
(g)
+
e
→
+
3
(g)
→
2
of
energy,
from
→
2e
the
in
one
kJ,
(g)
(g)
+
X
(g)
E
(g)
X
+
X
+
Which
+
2
(g)
2
+
X
+
+
EA
of
oxygen
254
N
A
=
a
tendency
lower
reduce I
the
+1
oxidation
B.
of
the
state
+2
C.
following is
to
be
of
reduced
D.
a
form
variable
low
oxidation states
D.
complex
electric al conductivity
of
the
ion
formation
following
metal
Electrons
statements
compounds
absorb
Electrons
release
lower-energy
one
mole
of
=
oxide ions
atoms?
–141 kJ mol
6.02 x 10
; 2nd
1
mol
)
EA
of
oxygen
2
property of
are
energy
explains
why
coloured?
when
they
move to a
C.
Transition
D.
Electrons
What
is
the
energy
when
they
drop
elements
are
4
]
=
A.
2
have
variable
deloc alized
oxidation
state
of
in
oxidation states
transition elements
cobalt
in
the
?
B.
0
down to a
level
represents the total amount of
oxygen
Br
+6
typic al
2
to
than
nitrogen in sodium
+5
not
2
properties
C.
Which
correct?
elements?
c atalytic
[CoCl
23
;
Br
will
is
reducing agent than F
oxidize
B.
B.
12.
1
753 kJ mol
2
weaker
halogens
higher-energy d-orbital
1
(1st
has
2
Which
A.
e
following
of
a
nitrate.
second ionization
e
required
mole
a
Cl
transition
a
X
→
v
O
B.
X(g)
shows
l
2
A.
increases
11.
u
the
electrons
o
D.
of
valence
is
transition
why
t
a
nuclear
Which
of
explain
periods?
decreases
C.
energy?
6.
number
radius
not
D.
A.
i
atomic
across
r
o
increasing
B.
f
x
5.
A.
trends does
increases
d
electronegativity
I
n
I and II only
D.
U
C.
2
C.
A.
10.
2
will
the
C.
+2
D.
+4
complex ion
LHA
III only
I
C
i
n
B.
4.
SrO
II only
Cl
B.
about
o
III.
A.
2
A.
Deduce
v
II.
statement
p
I.
to
oxides
Which
y
when
of
8.
2
2
n
d-block element
C.
zinc?
O
B.
to
vigorously?
y
metal
apply
most
2
2
t
i
s
A.
not
r
e
3.
Which
react
l
2.
would
y
B.
Multiple-choice questions
elements
P
C s and F
r
Exam-style questions
pair
A.
Structure
LHA
13.
What
[NH
4
]
is
2
the
ligand
[Fe(H
2
O)
6
in
the
][SO
4
]
2
17
.
complex
The
periodic
table:
Classic ation of elements
This question is about acid deposition.
?
a.
2
A.
3.1
Explain,
using
relevant
chemic al
equation(s),
+
why
Fe
rain
is
naturally
acidic.
[2]
2
[SO
C.
H
D.
[NH
4
b.
]
Explain
why
production
2
burning
of
acid
coal
leads to the
s
s
B.
rain.
[2]
O
c.
+
4
Figure
12
on
page
243
shows some of the
]
consequences of acid deposition.
e
x
14.
[Ni(NH
3
)
6
]
is
a
complex ion with nickel in the
+2
i.
oxidation
state.
What
is
the
overall
charge,
x, of
List
the
ve
consequences of acid deposition
the
in
gure
r
shown
12.
complex?
0
B.
1+
rain.
Suggest
graph
energies
how
State
of
c.
State
the
c an
the
be
copper.
abbreviated
electron
15.
This
question
is
Explain
why
the
has
a
greater
the
chloride
chlorine
ion
has
atom.
gas
is
bubbled
an
through
balanced
symbols,
metals
State
Explain
alkali
why
metals
the
Describe
E
of
Suggest
the
Write
a
a
added
indic ator
water.
to
and
the
a
for
the
pink
with
a
split
b.
[2]
d
colourless.
to
reaction
solution
to
coloured
[2]
aqueous
takes
change
place,
from pink
2
+
4Cl
⇌
[CoCl
4
]
+
6H
2
O
blue
[Co(H
the
2
O)
how
that
6
+
]
complex ion absorbs light
of
approximately
energy
orbitals
Explain
is
added
+
]
wavelength
C alculate
nitrate
is
a
of
the
dierence
cobalt
colour
chloride
in
this
complex
change
ions
are
540 nm.
between the
in
this
ion.
[2]
reaction
weaker ligands than
to
small
a
molecules.
[2]
[1]
21.
Explain,
why
[2]
by
referring
titanium
c alcium
only
c an
to
exist
successive
in
variable
occurs in the
+2
ionization
energies,
oxidation states, but
oxidation
state.
[2]
large
piece of
water.
explain the change in the
value
balanced
The
water
metals.
group.
added
a
6
forms
exchange
the
2
a.
[2]
energy of the
is
O)
of
nitrate
water.
alkali
the
ligand
2
shows
indic ator.
container
symbols,
with
the
Then,
the
above.
occurs.
down
l
full
ii.
of
of
ionization
sodium
colour
vigorously
decreases
container
is
reaction
number
Phenolphthalein
i.
d.
react
group
v
O
c.
the
this
described
u
b.
alkali
f
x
a.
reaction
i
The
why
aqueous
t
a
16.
Explain
the
r
o
ii.
for
a
zinc
acid
pink
orange-brown liquid is
equation, including state
aqueous
colour
2
copper(II)
hydrochloric
ions,
the
[Co(H
o
a
whereas
to blue:
radius
n
bromide,
c ausing
[1]
aqueous
excess
cobalt(II)
[2]
d
Write
greater
why
solution,
electronegativity
a
produced.
i.
When
[2]
[2]
chlorine
potassium
20.
radius of the halogens
U
When
Explain
p
Explain
19.
o
uorine
[1]
conguration of a copper(II)
ion.
are in
group.
oxygen.
than
d.
why
the
halogens
C
c.
atomic
down
The
i
n
than
the
halogens.
table.
v
Explain
the
periodic
why
increases
b.
about
the
electron
[2]
LHA
of
full
y
a.
17
r
e
group
the
[2]
conguration of
copper.
Extended-response questions
oset.
rst 10 ionization
O
b.
it
showing
n
3+
a
[1]
consequences of acid
y
2+
D.
Sketch
the
t
i
s
C.
a.
of
l
18.
one
y
A.
Choose
P
ii.
[1]
for the pH of the solution in
once
the
reaction
has
ended.
[1]
equation, including state
reaction
of
rubidium
with
water.
[2]
255
Functional groups:
Structure 3.2
Classic ation of
organic compounds
s
s
e
How does the classication of organic molecules help us to predict their properties?
r
a
common
on
compounds
the
that
human
basis
share
of
approach
shared
important
utilized
in
many
characteristics,
features
of
elds
chemists
structure
and
of
science.
utilize
a
Just
unique
as
biology
system
of
uses
scientic
nomenclature
group and
reactivity.
(atom
but
dierent
Structure 3.2.8 — M ass
c an
c ause
spectrometry
o
v
type
of
U
C
i
n
the
bond
(IR)
spectra
present
Structure 3.2.10 — Proton
the
same
arrangements of atoms.
fragmentation
Structure 3.2.9 — Infrared
identify
spatial
p
compounds
have
connectivities and bond
y
r
e
multiplicities)
identities,
in
nuclear
a
(MS)
of
of
organic
molecules.
c an
be
used to
molecule.
magnetic
resonance
1
spectroscopy (
n
chemic al
H
NMR)
environments
gives
of
hydrogen
Structure 3.2.11 — Individual
o
i
u
l
a
E
v
O
f
x
t
a
r
o
d
clusters
of
signals
on
atoms
the
in
a
dierent
molecule.
c an be split into
peaks.
Structure 3.2.12 — D ata
oen
information
combined
in
from
structural
dierent
analysis.
techniques
are
LHA
constitution
O
Structure 3.2.7 — Stereoisomers
n
y
t
i
s
Understandings
256
taxonomy to
to
l
name
is
organisms
y
classify
P
Classic ation
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Formulas of organic compounds
(Structure 3.2.1)
Introduction to organic chemistry
are
chemistry
studie d.
four
to
process
other
by
This
spe cial
fe ature
alcohol,
plastics,
fertilizers.
are
why
drugs
mole cules,
considere d
to
be
or
together
and
fuels,
me dicines,
such
as
DNA,
by
which
is
This
such
paints,
foods,
contain
a
wide
dyes,
pesticides
c arbon
organic.
represent
empiric al
of
present
the
6
H
12
the
6
and
glucose
an
the
formula
Both
of
the
be
types
structures
of
of
formula
ratio
the
of
several
formula
of
deduced.
by using
oer
2
O.
you
or
no
Empiric al
molecules.
has
the
molecular
From
this
information,
the
were
and
molecular
introduced
in
formulas
Structure
1.4.
C
n
o
i
12
a
t
a
H
little
complex
U
6
of
in
atoms
u
C
of
a
structure
l
G l u c o s e,
v
The
E
groups.
d
r
o
f
x
O
p
Figure 1
ways
present
number
more
glucose
CH
atoms
actual
larger,
c arbohydrate
empiric al
c annot
simplest
describes
these
possible
molecule
O
in
i
n
of
compound
p
C
structure
molecule.
a
organic
o
formula
represents
molecular
about
example,
an
v
For
of
y
The
in
structure
formula
molecule.
information
the
formulas.
r
e
An
types
O
c an
dierent
t
i
s
Types of formulas
You
is
cov alent
structures.
chemistry
include
form
c atenation ,
cyclic
organic
compounds
solvents,
biologic al
joine d
branche d,
re ason
O rganic
industrial
also
the
are
compounds
c an
n
and
study.
Most
atoms
undergo
they
y
atoms
of
is
c an
so
l
and
field
identic al
c arbon-base d
ele ctrons,
y
v arie d
atoms
straight-chain,
c arbon
where
v alence
P
and
chemistry
four
C arbon
many
produces
of
have
atoms.
which
of
field
r
bonds.
the
atoms
e
the
bonds
is
C arbon
s
s
O rganic
O
the
6
,
is
ring
a
is
simple
not
s i x- c a r b o n
refle cte d
by
sugar
the
with
hy d r ox y l
molecular
functional
formula
257
Structure
3
Classic ation
of
matter
Structural formulas, unlike molecular formulas, describe the structure of a
compound. There are three types of structural formula: full, condensed
and skeletal.
•
Full structural formulas
to one another in a
•
In a
A
the
but
some
skeletal formula
and
functional
is
all
the
the
most
basic
included
c arbon
H
C
C
CH
H
o
CH
H
C
H
H
C
C
structural
The
atoms
Condensed
H
H
the
formula,
present in
3
3
Skeletal
formula
CH
2
CH
formula
3
y
H
H
O
atom.
p
H
H
of
shown but the end of
shown in table 1.
r
e
H
OH
CH(OH)CH
3
O
n
o
i
H
C
H
O
H
C
C
C
t
a
u
l
a
E
v
structural
line
CH
2
CHO
O
H
H
CH
3
C(O)CH
3
H
H
H
C
CH
C
3
CH=CH
2
or
H
Full,
of
3
H
H
condensed
organic
and
a
a
and
skeletal
covalent
describes
for
3
CHCH
2
formulas
c an
all
be
used
to
represent
compounds
formulas,
that
CH
H
H
p
Table 1
CH
H
C
used,
not
n
C
relative positions
l
C
their
O
C
H
propene
are
a
as
are
t
i
s
H
H
i
n
U
d
r
o
f
x
O
258
H
and
relative
O
propanone
single
atoms
structural
H
C
H
propanal
structures
positions
omitted.
representation
structural
v
H
propan-2-ol
the
are
atoms
represents
also
formula
H
all
bonds
hydrogen
vertex
are
Full
propane
their
y
Name
or
and
each
groups
and
triple
two
bond
bonding
bond,
three
between
two
electrons.
lines
are
atoms
is
represented
by a
For a double bond, two lines
used.
y
line
c arbon
are two-dimensional
bonds,
P
each
the
and
compound.
condensed structural formula,
shown,
where
In
atoms
r
•
displayed formulas
all
e
are
or
showing
s
s
representations
Structure
Constructing
technique,
of
individual
from
the
optic al
3D
models
enhancing
atoms.
naming
isomers)
of
to
of
your
organic
ability
Models
organic
c an
compounds
to
visualize
enhance
molecules
complex
reaction
of
is
an
excellent
Functional
groups:
Classic ation
of
organic compounds
interactive
molecules and mutual orientation
understanding
and
3.2
visualizing
of
a
variety of concepts,
stereoisomers (including
mechanisms.
s
s
Practice questions
Draw
the
full
structural
formula
and
skeletal
formula
for
each
of
the
following
organic compounds.
C
III.
C
2
2
5
H
H
H
6
4
O
formulas
are
compounds
C
4
3
H
H
H
4
8
62
O
O
88
72
t
a
r
o
Linking question
the
skeletal
advantages
compound
formula,
3D
u
are
organic
that
elements’
a
an
c arbon
other
l
of
about
the
E
v
O
f
x
unique
90
O
o
D
2
–1
i
C
formula and
Molar mass / g mol
d
C
2
empiric al
C
CH
What
same
n
CO
B
all
the
U
A
of
has
y
following
formula?
p
I, II and III
is
formulas?
o
D.
sum
empiric al
r
e
II and III only
Empiric al formula
the
also
v
I and III only
What
3
2
C.
the
CH
12
B.
of
2
2
OH
O
I and II only
molecular
CH
C=CHCl
molecular
A.
Which
2
2
CH
O
C
II.
2
CH
t
i
s
I.
2
2
CH
n
these
CH
i
n
3.
of
3
2
CH
y
chloroethene, H
Which
CH
l
c.
3
r
butan-1-ol, CH
P
hexane, CH
b.
y
2.
a.
e
1.
enables
it
to
form
compounds?
and
disadvantages
(structural
models,
more
etc.)?
compounds
(Structure
formula,
of
dierent
depictions
stereochemic al
(Structure
than
2.2)
formula,
2.2)
259
Structure
3
Classic ation
of
matter
Models
Models
are
useful
representations
of
the
physic al
more
and
complex
computer
3D models in the
direct
molecules
Skills
observation
(such
as
features
models.
that
You
is
dicult.
structural
Two-dimensional
formulas)
become
clear
explore
some
c an
in
of
do
not
three-dimensional
the
dierent types of
chapter.
laboratories
animals
and
are
group
of
atoms,
synthesized
research
synthesized
in
by
living
found
fe atures
n a t u ra l l y
hy d r ox y l ,
O
O
ester
o
C
n
It
y
p
r
e
steroid
fibrillation.
found in
OH
H
OH
O
O
H
o
i
c ardioactive
O
OH
hydroxyl
in
the
a l k ox y
f oxg l o v e
and
plant.
ester
It
is
functional
used
to
tre at
congestive
he art
failure
and
groups
u
f
x
All organic compounds are divided into classes of organic compounds
l
a
E
v
O
260
a
atrial
D-glucose-
t
a
is
or
Natural compounds
organisms.
H
alkoxy
organic
pharmaceutic al companies,
universities.
v
i
n
U
D igitalin
he artbe ats
d
irre gular
r
o
p
Figure 2
gives
organic compounds, and this number is constantly
are
OMe
β
that
characteristics.
O
and
of
a
dependent on the specic functional group found in their molecules. Compounds
of the same class have similar chemical properties. Some functional groups and
their corresponding classes are given in table 2. R
symbolize
formulas
other
any
for
c arbon-containing
classes
functionalities
of
organic
are
groups.
They
compounds.
c alled alkyl groups
groups
are
useful
are
for
used in table 2 to
writing
C arbon-containing
general
groups with no
y
millions
compounds
or
chemic al
n
of
and
t
i
s
industrial
plants
tens
new
atom,
l
are
as
an
e
rising
is
physic al
y
There
their
P
functional group
compounds
r
Functional groups (Structure 3.2.2)
A
represent
s
s
some
when
of
Structure
Functional
3.2
Functional
Name
chloro,
of
organic compounds
Sux
Formula
Class
Example
chloromethane
bromo
R
X
—
halogenoalkanes
H
C
Cl
3
s
s
or
Classic ation
group
halogeno
(uoro,
groups:
iodo)
ethanol
R
OH
-ol
H
alcohols
C
e
hydroxyl
H
OH
C
3
r
H
H
aldehydes
H
c arbonyl
-one
ketones
C
-oic
O
R′
—
U
methoxymethane
H
C
O
3
CH
3
H
H
C
C
i
o
3
NH
2
H
t
a
ethanamide
O
H
amides
C
C
3
u
f
x
l
a
E
v
O
R
C
OH
primary amines
-amide
NH
NH
2
2
methyl ethanoate
O
O
C
-oate
esters
H
C
C
3
O
O
R’
CH
3
methylbenzene
R
phenyl
S ummary
C
C
3
ethanamine
-amine
2
O
R
ester
p
Table 2
NH
d
r
o
amido
R
H
n
amino
ethers
3
O
o
R
ethanoic acid
c arboxylic acids
C
i
n
OH
alkoxy
acid
v
R
c arboxyl
CH
O
y
O
C
3
C
p
r
e
O
H
O
propanone
R’
C
C
H
t
i
s
R
C
3
n
y
-al
y
C
O
l
R
c arbonyl
P
ethanal
O
of
c l a ss e s
—
of
organic
compounds
and
aromatics
their
functional
H
C
3
groups
261
Structure
3
Classic ation
of
matter
Self-management skills
ATL
How
c an
names,
build
the
of
and
various
develop
familiarity
structures?
maintain
functional
You
your
may
with
uency.
groups
the
wish
will
to
As
various
create
you
help
will
you
to
a
functional
ashc ard
see in
groups, their
deck
Reactivity
predict
the
help
you
recognizing
chemic al
behaviour
various substances.
or
c an
unsaturated.
, a
single.
hydroc arbon
must
C=CH
saturated
double
2
.
Alkanes
no
or
and
are
c arbon
methane,
unsaturated
c arbon bonds.
alkenes. The
are both
aliphatic.
H
p
c arbon
is
H
o
a
an
y
H
H
with
in
c arbon
c alled
alkenes
C
H
(left)
chain
are
c arbon
hydroc arbon
triple
H
they
c arbon
aromatic rings.
C
C
s a t u ra t e d
whether
the
c arbon
bonds
r
e
is
2
a
primary
double
v
i
n
Ethene
(right)
more
contain
of
The
H
C
ethane
or
c arbon
ethene, H
molecules
H
p
Figure 3
one
c arbon
H
example
family.
to
all
O
is
their
alkane
contain
with
alkene
means
simplest
according
compound,
t
i
s
Hydroc arbons
simplest
The
classied
n
member of the
4
be
saturated
l
are
CH
a
y
bonds
also
In
double
bond,
is
u n s a t u ra t e d ,
where as
U
TOK
d
the
organic
word
i
context:
What
is
c an
organic.
the
role
dier
This
food,
organic
t
a
r
o
u
to
of
and
from
word
takes
organic
signify
context
a
in
chemical
their
have
meanings
on
a
variety
chemistry
and
in
of
precise
meanings in
everyday
language.
meanings depending on its
organic
household
waste all use
dierent thing.
the
choice
and
interpretation
of
language?
Linking questions
E
v
a
LHA
l
f
x
O
262
Consider
natural, organic
which
o
chemistry,
n
Terms such as
What
form
is
a
How
a
the
nature
dipeptide?
c an
the
functional
reaction
ethene
of
pathway
into
reaction
(Structure
group
reactivity
between
ethanoic
that
occurs
when
two
amino
2.4)
acid?
be
used
compounds,
(Reactivity
3.2,
to
e.g.,
3.4)
determine
converting
acids
y
compounds
saturated
P
Organic
e
r
S aturated and unsaturated hydroc arbons
This
to
3,
s
s
to
you
and
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Homologous series (Structure 3.2.3)
A
homologous series
based
the
on
same
similarities
general
you
its
c an
family
formula,
group.
identify
a
their
If
you
which
of
compounds
structure
which
know
group
a
and
varies
the
from
general
particular
that
c an
be
grouped together
reactivity. A homologous series has
one
member
formula
for
compound
to
another
by one CH
2
s
s
(methylene)
is
in
each homologous series,
belongs
to
without
drawing
structure.
contain
series
homologous
series.
general
formula C
hydrogen
You
should
member
diers
be
able
from
to
the
n
H
2n+2
only.
.
Alkanes
Table
identify
previous
are
shows
how
the
hydroc arbons
the
alkane
structural
by a single CH
2
formula of
group.
structural
formula
structural
formula
formula
t
i
s
H
CH
CH
4
O
methane
C
4
H
3
CH
2
H
3
CH
2
CH
2
of
3
CH
CH
2
2
CH
CH
2
2
CH
CH
2
2
CH
CH
2
H
o
series
3
CH
i
a
homologous
CH
t
a
CH
14
3
u
H
CH
12
l
The
6
H
3
d
C
5
CH
10
E
v
O
p
Table 3
f
x
hexane
C
H
r
o
pentane
4
H
n
C
C
H
H
3
U
butane
C
H
H
H
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
y
CH
8
CH
o
H
3
C
3
CH
6
v
C
H
i
n
propane
2
H
p
C
r
e
ethane
H
n
Full
Name
l
Condensed
3
y
Molecular
atoms
y
successive
the
and
P
each
has
c arbon
e
alkane
that
r
The
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
3
CH
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
H
3
alkanes
263
Structure
3
Classic ation
of
matter
The
You
c an
alkenes
are
c arbon–c arbon
halogenoalkanes
from
a
homologous
process
known
as
This
form
alkyne
series
The
of
unsaturated
alkene
series
has
hydroc arbons that contain a
the
family have
c arbon–c arbon
triple
bonds,
general
with
the
formula of C
general
what
c arbon–halogen
was
a
detailed
R adic al
in
a
n
H
2n–2
.
Halogenoalkanes
substituted
with
a
are
alkanes
halogen
atom.
where
They
one
have
of
the
the
formula C
n
H
2n+1
bond
into
alkane
substitution
is
3.3.
r
formula
C
n
H
alkynes
C
2n
H
C
2n–2
C
C
2
H
H
ethyne
ethene
r
e
3
C
H
CH
3
U
H
C
C
C
3
CH
3
3
i
of
s t ra i g h t - c h a i n
alkenes,
Members
u
f
x
group (
of
H
3
H
H
H
C
C
C
H
H
H
Cl
1-chloropropane
H
H
H
H
l
Another
a
contain
the
as
aldehydes,
c arbon
the
and
H
3
C
C
C
C
H
H
H
Cl
halogenoalkanes
have
a
series
general
of
chain.
but
ends.
series,
the
Ketones
the
the
c arbonyl
These
and
aldehydes,
the
group
homologous
H
contain
2n+1
have
group
same
is
all
n
oxidation
functional
have
c arbonyl
alcohols
formula of C
reactions (Reactivity 1.3)
of
except
CH
1-chlorobutane
homologous
They
homologous
Aldehydes
C
H
alkynes
the
OH).
combustion
E
v
O
264
series
t
a
homologous
CH
o
d
r
o
The
H
but-2-yne
but-2-ene
p
Table 4
n
C
H
Cl
H
C
4
C
propyne
C
i
n
propene
C
H
o
v
C
C
H
chloroethane
y
H
X
2n+1
C
H
p
H
H
H
O
C
n
H
n
y
H
C
t
i
s
H
n
halogenoalkanes
hydroxyl functional
c an
undergo
general
formula of C
n
H
2n
O.
(C=O) in the terminal position
formula
anywhere
are
the
Alcohols
reactions (Reactivity 3.2).
general
found
series
a
OH.
on
and
the
functional
group
c arbon chain
summarized in table 5.
y
alkenes
l
General
series
X,
P
Homologous
.
hydrogen atoms has
general
highly
inert
Reactivity
2n
e
molecule.
relatively
H
s
s
introduces
where X is the halogen atom.
polar
n
formula
of
been
substitution
bond.
radic al
C
substitution.
double
alkanes
The
by
a
synthesize
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Homologous
alcohols
aldehydes
ketones
series
General
C
n
H
2n+1
OH
C
n
H
2n
O
C
n
H
2n
O
formula
H
s
s
H
H
O
C
C
H
OH
C
C
no C
2
e
H
C
ketones
2
H
H
H
ethanol
ethanal
H
H
H
C
H
H
H
H
OH
C
C
H
H
C
3
H
t
i
s
C
C
C
H
O
H
n
C
C
H
H
H
H
H
H
H
H
H
C
C
C
O
H
H
H
O
C
C
OH
H
H
H
H
H
C
C
C
H
H
H
H
H
C
C
C
C
C
H
H
H
H
OH
C
have
OR
c arbon
atom:
of
is
formula
c alled
OCH
3
.
where
an
This
is
ethers:
methoxymethane
3
=
1,
ROR’,
is
n
a l d e hy d e s
H
the
where
group.
known
a
H
3
formula C
n
alkoxy
l
3
general
E
v
O
examples
H
the
group
general
u
The
family.
f
x
Ethers
the
as
a
2n
O
2
R
and
CH
and
and
R’
C
C
C
H
H
H
H
butanone
H
H
C
C
H
H
H
C
C
C
O
H
H
O
C
H
H
pentanal
H
pentan-3-one
ketones
have
a
c arboxyl
group.
c arboxylic acid of this
represent
simplest
methoxy
2
,
simplest
The
H
C
n
homologous
have
HCOOH,
alcohols,
i
acid,
of
H
H
H
H
t
a
acids
r
o
C arboxylic
Methanoic
series
H
C
o
homologous
d
The
H
H
pentan-1-ol
p
Table 5
H
U
5
butanal
o
H
H
H
C
H
i
n
H
H
v
butan-1-ol
C
y
C
4
p
C
r
e
H
C
H
propanone
propanal
H
O
propan-1-ol
l
C
y
H
H
H
O
y
H
P
H
r
H
alkoxy
group.
alkyl
group
Here
are
groups.
has
one
some
3
methoxyethane
265
Structure
3
Classic ation
of
matter
Members
There
are
amines.
atom.
and
in
of
In
In
the
three
amine
types
primary
amines:
amines,
secondary
tertiary
homologous
of
the
amines,
amines,
it
is
series
primary
nitrogen
the
atom
nitrogen
bonded
to
all
have
amines,
is
three
is
the
general
formula C
n
H
bonded
bonded
c arbon
to
to
two
atoms.
only
one
c arbon
This
is
summarized
have
the
have
medic ations
of
c arbon
also
derived
They
atoms
and
bonded
the
to
are
c arboxylic
amido
acids
and
functional
nitrogen
H
H
H
C
C
C
H
H
H
p
C
o
H
n
H
o
i
tertiary
l
a
E
v
O
266
and
formula of
.
Example
H
NH
2
H
C
N
C
H
H
H
H
3
2
H
dimethylamine
H
C
N
C
H
H
H
H
t
a
secondary
general
C
H
H
trimethylamine
u
P r i m a r y,
a
–C(O)NH
y
r
e
v
i
n
U
d
r
o
f
x
p
Table 6
have
propanamine
2
tertiary
c arboxylic
O
t
i
s
1
secondary
from
avouring agents and
group,
H
primary
derived
from
explosives.
from
contain
They
ranging
amines
Linking question
How
useful
are
3D
models
(real
or
virtual)
to
visualize
the
invisible?
(Tool
2)
y
Number
are
NO.
R’.
n
amine
2n+1
2
l
of
H
RCO
applic ations
y
Type
n
solvents
of
P
C
formula
variety
r
Amides
to
general
a
e
and
s
s
acids
N.
c arbon atoms,
in table 6.
Esters
2n+3
secondary amines and tertiary
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Practice questions
4.
Which
functional
group
is
circled
in
this
molecule?
7
.
Which
of
the
following
functional
groups
is
present in
paracetamol?
A.
hydroxyl
B.
c arboxyl
s
s
O
OH
C
C.
c arbonyl
D.
ether
H
Which
functional
groups
are
present
in
this
molecule?
r
O
CO
3
N
following
II.
c arboxyl
are
present
i
H
O
H
drug
used
Its
to
treat
structure
is
symptoms of
shown
below.
OH
O
O
3
N
CH
3
H
Which
of
the
following
functional
groups
are
present
C
C
C
in
this
molecule?
A.
amido,
hydroxyl, ester
B.
amino,
hydroxide, ether
C.
amino,
hydroxyl, ether
D.
amido,
H
O
a
E
C
l
C
v
C
ester
t
a
C
III.
u
f
x
O
H
r
o
O
H
O
groups
a
o
ether
functional
molecule?
O
the
d
I.
of
aspirin
CH
is
disease.
y
the
G alantamine
n
in
hydroxyl
p
ester, amino
Which
phenyl
phenyl
n
y
ether, amino
c arboxyl,
amino
D.
o
c arboxyl,
D.
B.
alkene
C
C.
c arboxyl
v
ester,
i
n
ether,
c arbonyl,
U
6.
c arbonyl,
3
C.
Alzheimer ’s
CN
CH
A.
r
e
8.
CN
C
t
i
s
O
B.
O
3
H
A.
l
CH
y
P
H
e
5.
C
hydroxide, ester
H
C
H
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
267
Structure
3
Classic ation
of
matter
Research skills
ATL
Our
atmosphere
oxygen, O
.
precursors
formed
is
mostly
Organic
of
billions
made
biomolecules,
of
up
compounds,
of
nitrogen, N
which
are
may
thought
2
, and
have
to
ammonia
water vapour
been
(NH
was
these
as the Miller
very
dierent.
conditions
Urey
in
An
the
attempt
mid-20th
was made to
century,
4
electrode
)
2
)
known
e
recreate
)
hydrogen
years ago when the composition of the
(H
atmosphere
3
methane (CH
have
experiment.
condenser
(or
two
another
many
in
of
your
own
possible
document
the
Urey
experiment
biomolecules), and
to
words,
your
drawing
knowledge of
information
sources, and
cold
water in
reliability.
cooled water containing
organic compounds
y
sample taken for
chemical analysis
a p p a ra t u s
for
the
M i l l e r– U r e y
ex p e r i m e n t
o
i
n
v
p
The
O
t
i
s
r
e
p
Figure 4
n
y
their
as
origin
l
Fully
the
ndings
connections
chemistry.
evaluate
your
of
Physic al trends in homologous series
C
(Structure 3.2.4)
of
they
the
have
types
given
the
forces
The
c arbon
and
u
a
7.
points
These
the
and
melting
as
point
chemic al
group
is
properties, as
responsible
for the
undergoes.
member. This
time,
altering
is
the
bec ause the length
intermolecular
properties.
points
increase
boiling
each
2
similar
properties such as melting and boiling
successive
by CH
physic al
have
functional
compound
physic al
with each
increases
values
consequence,
series,
the
The
for
the
and
the
straight-chain
number
melting
of
c arbon
point
alkanes
atoms
increase
are
shown
increases.
with
increasing
molarmass.
l
a
Name
Formula
methane
CH
E
ethane
C
propane
C
butane
C
pentane
C
hexane
C
p
Table 7
268
chain
therefore
boiling
table
As
v
O
f
x
in
gradually
series
group.
reactions
homologous
change
i
of
homologous
functional
o
a
points
same
same
characteristic
t
a
r
o
d
For
of
the
n
U
Members
The
2
3
4
5
6
H
H
H
H
6
8
10
12
14
and
point / °C
Melting
161
4
H
melting
Boiling
boiling
points
point / °C
183
89
172
42
188
0.5
135
36
130
69
95
of
the
s t ra i g h t - c h a i n
alkanes
y
as
theory
P
summarize
sources on the Miller
r
Research
s
s
the
2
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
The close correlation between c arbon chain length and boiling point enables you
to
predict
The
the
properties
of
neighbouring
members of the homologous series.
trend in melting points is not as smooth as that of boiling points and would
have
less
certainty
in
terms
of
the
prediction
of
properties.
You
c an
boiling
use
simple
points
for
distillation
successive
apparatus
and
a
temperature
probe
to
measure the
learned
techniques
exist
between
London
non-polar
polar
results
in
forces,
(for
is
HCl).
or
H
London
are
2
).
molecule
polar.
(dispersion)
forces
intermolecular
covered
in
Structure
and
forces
2.2.
were
Which
of
will
the
so
homologous
far
series
experience
discussed
hydrogen
bonding?
n
number of
carbon atoms
5
6
7
8
9
10
alkane
alkene
150
other
o
E
v
50
3
Every
non-polar
i
2
a
1
forces
hydrogen
u
l
f
x
O
C° / erutarepmet
0
(dispersion)
t
a
r
o
150
d
200
100
example,
example,
it
London
(dispersion)
U
250
50
(for
whether
300
100
stronger
London
molecules
molecules
(dispersion)
from
length.
C
experience
between
point
increases
y
forces
boiling
chain
i
n
also
c arbon
p
increasing
probe
o
attractive
They
in
the
t e m p e ra t u r e
v
as
a
r
e
trend
forces
i n c o r p o ra t i n g
n
distillate
from faucet
a p p a ra t u s
l
water
O
t
i
s
to sink
y
P
y
water
The
1.1.
e
r
probe
condenser
D istillation
distillation
Structure
members of a homologous series.
temperature
p
Figure 5
about
in
s
s
You
alcohol
carboxlc ac

Figure 6
inc re ases
The
with
member
of
series
the
g r ows
as
in
a
boiling
e ach
point
s u c c e ss i v e
homologous
c arbon
chain
length
200
269
Structure
3
Classic ation
of
matter
Constructing a simple graphic al model
In
this
task
model
will
construct
and
test
a
simple
Part 3: Graph
graphic al
homologous series of
7
.
Create
8.
Using
a
sc atter
graph of the data.
choice.
software,
draw
a
line
or
curve of best fit that
describes the data well.
Relevant skills
2
9.
Inquiry
1:
State
and
explain
Determine
the
R
value
and
the
equation
of
predictions using scientific
or
curve.
understanding
Use
•
Tool
2:
Represent
•
Tool
3:
Sketch
and
extract
data
from databases
Part 4: Test the model
spreadsheets to manipulate data
data
graphs
in
to
graphic al
You have now created a simple graphical model of the
relationship between boiling point and carbon chain length.
form
qualitatively
describe
You can test your model by using it to make predictions and
trends
Tool 3: Apply the coefficient of determination ( R
•
Tool
3:
Construct
graphs
and
draw
lines
or
)
10.
curves of
3:
Tool
•
Inquiry
3:
C alculate
2:
possible, the 20th members of the homologous series.
graphs
and
Describe
interpret
and
percentage
explain
patterns,
error
trends and
11.
Look
12.
C alculate
up
their
actual
the
boiling
percentage
point
errors
predictions.
relationships
p
Instructions
values.
of
each of the
y
r
e
•
Extrapolate
O
Tool
t
i
s
•
Use the equation of the line or curve of best fit to
predict the boiling point of the 7th, 10th, and if
best fit
n
•
y
then checking the accuracy of these predictions.
2
Part 5: Interpret the results
2.
Predict
your
first
the
chain
length.
knowledge
a
your
that
qualitatively
have
i
t
a
r
o
for
a
variety
Chemistry
data
on
the
you
boiling
have
do
This
their
the
the
accuracies
is
accuracy
c arbon
an
data.
a
value
How
errors
for the
following questions:
compare?
Why might
and
of
the
extrapolation
that
the
explain
predictions
change
for
lengths?
might
interpolated
Describe
of
chain
example
predicted
had
14.
percentage
be?
Does
the
the
Consider
was
the
accuracy
data
the
bec ause
outside
the
you
range of
change
if
you
instead?
number
data:
of
point
enter
c arbon
the
a
the
v
O
and
instance,
for
the
six
you
trend
shown
in
the
graph
plotted.
NIST
15.
If uncertainties were reported along with the boiling
16.
Briefly
point data, what impact did these have on your analysis?
organic
evaluate
your
model,
giving
examples of its
selected and enter the data into
spreadsheet.
Organize
(for
u
a
substances
l
f
x
Collect
of
Web Book).
compounds
6.
•
describes the
contrast
Identify an online database that contains boiling point
data
5.
How
longer
predicted.
Part 2: Data collection
•
referring
and
predictions.
this
o
you
•
forces and
d
relationship
4.
prediction,
intermolecular
groups.
graph
three
between boiling point and
Justify
of
Compare
n
Sketch
relationship
U
your
functional
3.
members of a homologous series of
choice.
c arbon
to
six
o
the
C
List
i
n
1.
v
13.
Part 1: Prediction
17
.
name,
atoms,
specific
How
strengths
might
the
and
weaknesses.
accuracy of this simple model be
formula,
improved?
and
convert all
temperatures into a consistent unit.
E
270
Extension
Choose
task.
a
different
Compare
homologous
and
contrast
the
series
two
and
sets
of
repeat the
results.
y
Identify
2:
l
2:
Tool
r
Tool
•
line
P
•
the
e
•
s
s
your
you
of the boiling points of a
Structure
The
dierent
members
this
principle
mixture
of
is
passed
Therefore,
the
as
a
through
length
of
a
their
similar
will
column.
of
mean
that
petrochemic al
example,
oil
dierent
industry,
crude oil is a
c arbon chain.
crude
organic compounds
is
Fractional
vaporized and
through the column, it cools.
levels in the column depending
Compounds with a similar boiling point,
condense
down
the
where
members
of
at
the
column,
have
a
same
as
they
level. Long-chain
have a higher
lower boiling point, so will
it
is
cooler.
a
homologous
series
also
increase with
y
What
the
is
the
inuence
functional
of
groups
the
on
c arbon
chain
intermolecular
length,
forces?
branching
(Structure
and
the
nature
2.2)
O
t
i
s
Linking question
of
existence
of
an
example,
is
substituents
organic
any
class
alkyl
group
a
in
of
and
compound
considered
dierent
an
the
is
compound
compound,
functional
that
an
chemic al substances, names help us to
organic
not
the
groups
part
of
the
needs
length
present.
longest
to
of
A
provide enough
its
longest
substituent
c arbon
chain.
c arbon
is
any
For
alkane compound that is not part of the main
substituent.
An
U
chain
any
the
of
of
o
part
and
identify
name
C
chain,
to
millions
The
i
n
information
of
them.
p
the
dierentiate
v
With
y
r
e
Naming organic compounds (Structure 3.2.5)
n
l
c arbon chain length.
y
viscosity
series
the
their
rises
dierent
compounds
column,
For
of
where
vapour
at
will
further
Shorter-chain
and
the
chain.
length,
length
process
condense
c arbon
the
As
the
In
of
P
density
will
homologous
technique.
in
separation
chain
up
vary
of
temperatures.
separation
that
condense
further
members
Classic ation
r
point.
condense
physic al
compounds
compounds
The
a
therefore
boiling
used
of
dierent
groups:
e
and
is
at
Functional
s
s
then
points
condense
hydroc arbons
distillation
on
boiling
will
3.2
alkane
with
alkyl
substituents
is
known as a
n
branched-chain alkane
d
o
The International Union of Pure and Applied Chemistry (IUPAC) has developed
a universal chemical naming system for both organic and inorganic compounds:
r
o
i
IUPAC nomenclature. IUPAC nomenclature rules helps chemists to avoid
confusion, eliminate language barriers, and freely communicate knowledge. An
t
a
overview of the IUPAC naming system for organic compounds is shown in gure 7:
l
longest
includes
or
the
or
a
E
v
b
u
f
x
O
a
the
c arbon chain that
the
principal
most
complex
heterocyclic
alphabetic al
cyclic
suffix
system
the
principal
(parent) chain
O utline
of
the
group
-ene
-yne
suffixes
order
saturation
indic ating
or
unsaturation
of the principal
p
Figure 7
indic ating
principal
-ane
c
prefixes
in
group,
nomenclature
of
organic
chain
compounds
271
Structure
3
Classic ation
of
matter
Naming alkanes
Alkanes are saturated organic molecules that only contain
In the example below, numbering from le to right results in the
carbon
methyl substituent being on carbon2. Numbering from right to
carbon single bonds. You can use the following
steps to deduce the IUPAC name of an alkane:
le would incorrectly have the substituent on carbon 5.
s
s
H
1
Examine
the
structure of the compound and determine
H
the
longest
root name
atoms
the
c arbon
chain.
This
(table8).
If
an
alkane
is
a
C
H
C
C
1
name
is
the
same
as
its
root
have
the
same
sux:
H
C
C
H
2
3
4
H
C
5
6
name. All
H
alkanes
H
C
H
H
H
H
r
its
H
straight-chain
H
molecule,
H
provides the
alkane depending on the number of
H
-ane.
name
name
as
of
the
alkyl
prex,
number
and
and
substituent
a
the
hyphen
prex.
goes
is
used
before
to
Therefore,
the
root
separate the
the
l
loc ant
1
a
above
methane
3
propane
4
butane
5
pentane
6
hexane
5.
is
c alled
2-methylhexane.
n
ethane
y
compound
2
If there are multiple dierent substituents, they are arranged
t
i
s
in alphabetical order prior to the root name. For example,
O
the below compound is called 3-ethyl-2-methylhexane.
H
ethyl
alkyl
substituents
of
a
are
substituent
for
the
alkane
series
determined
by the number of
H
Some
with
the
examples
sux
are
changing
shown
in
from
table
“-ane” to
9.
H
methyl
H
H
C
C
H
H
H
H
specify
the
numbers,
E
c arbon
the
chain.
The
substituent
is
the
c arbon
given
are
substituent
multiple
is
identic al substituents, the number
indic ated
name,
as
by
a
in
before the
table10.
Multiplier name
2
di
3
tri
4
tetra
5
penta
CH
3
2
2
2
CH
CH
CH
3
2
2
CH
CH
p
Table 10
3
2
CH
3
The
Numeric al
loc ants
of
For
m u l t i p l i e rs
each
in
the
substituent
example,
the
I U PAC
are
below
nomenclature
separated
compound
substituents
a
substituent
H
by
loc ants, to the main
H
chain
is
C
H
numbered such that
H
the
multiplier
summarized
Number of identic al substituents
lowest
value
H
H
loc ant.
H
C
C
H
H
H
C
C
H
272
3-ethyl-2-methylhexane
2,3-dimethylhexane.
of
known as
H
6
H
methyl
substituents
comma.
alkyl
position
there
C
5
H
H
t
a
c an
assigning
for
u
You
names
CH
a
3.
I U PAC
CH
propyl
v
O
p
Table 9
CH
ethyl
butyl
If
4
H
formula
l
4
3
H
C
H
C
Condensed
name
methyl
f
x
3
substituent
H
2
H
of
i
Substituent
1
C
substituent
r
o
Number of c arbon
C
1
H
o
C
C
H
6.
d
C
atoms in alkyl group
2
substituent
H
C
n
H
C
U
H
H
C
i
n
“-yl”.
atoms,
C
o
c arbon
H
H
present in the compound, the
is
v
name
names
H
y
If
root
C
p
2.
I U PAC
r
e
p
Table 8
H
C
C
C
H
H
H
H
H
by a
is
c alled
y
The
Name
P
4.
Number of c arbon atoms
e
c arbon
continuous
for
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Structural isomers are compounds that have the same molecular formula but
dierent structural formulas. Dierent isomers have unique physical and chemical
properties. In the following worked example, we will look at the structural isomers of
the hydrocarbon hexane, and practise using the rules above to name these isomers.
s
s
Worked example 1
Three dierent
structural isomers of hexane, C
6
H
12
,
are shown below.
Using the IUPAC
nomenclature,
e
name these isomers.
a.
H
H
H
H
C
C
C
C
C
b.
c.
H
H
H
C
H
H
H
H
H
C
C
C
C
H
H
H
H
C
H
H
H
H
C
C
H
H
H
H
H
H
Solution
a.
First,
count
straight
so
the
c arbon
of
There
c arbon atoms in the
chain.
parent chain is
There
are
ve
c arbon
bonded
pentane
Since
number
has
CH
)
the
lowest
possible
substituent
name
is
loc ant. The
bonded
to
c arbon 2.
for this compound is
2-methylpentane.
There
are
In
this
chain,
c ase,
the
is
so
the
irrelevant
for
the
longest
parent chain is
direction
both
the
of
numbering
as
the
would
3
CH
2
CH
2
a
CH
3
CH
E
CH
2
C
CH
CH
CH
3
CH
CH
2
3
)
are
substituents,
to
c arbon
identic al,
one
3.
a
numeric al
indic ate the number of
this
c ase,
di indic ates two
name
for this compound is
between
Like
(
b,
CH
The
3
)
a
the
Remember
numbers,
number
and
and
a
parent chain is
substituents
IUPAC
name
are
for
commas
are
are
used
letter.
butane.
bonded
this
that
hyphens
to
Two
methyl
c arbon 2.
compound
is
therefore
2,2-dimethylbutane.
10.
Draw
the
structural
formulas
of
the
following
molecules:
2
CH
3
a.
2,2,3-trimethylpentane
b.
5-ethyl-4,4-dimethyloctane
3
3
CH
2
3
CH
3
CH
c.
the
nomenclature rules, state the name of
following compounds:
v
O
b.
CH
IUPAC
the
l
a.
of
u
each
is
give
to
In
3
one
t
a
r
o
f
x
Applying
the
substituents.
Practice questions
9.
IUPAC
used
c.
of
compound
approaches
substituents.
between
butane
CH
and
substituents
required
2,3-dimethylbutane.
straight
i
loc ants
so
in
The
o
chain
symmetric al,
same
and
atoms
d
c arbon
c arbon
is
2
(
substituents.
n
c arbon
four
U
b.
these
c arbon
two
C
IUPAC
3
i
n
The
(
the
H
H
methyl
o
methyl
to
multiplier
c arbon chain so that the
v
substituent
the
two
p
Then,
are
C
y
atoms,
number
r
e
longest
the
H
H
O
t
i
s
H
C
n
C
y
H
C
l
H
y
C
P
H
H
H
r
H
H
2
CH
3
CH
3
3
3
273
Structure
3
Classic ation
of
matter
Naming alkenes
Alkenes
bond.
same
as
unsaturated
IUPAC
those
of
for
-ane
following
organic
nomenclature
the
used
alkane
for
molecules
rules
series.
for
All
that
the
contain
naming
alkene
names
of
a
c arbon
alkene
end
with
c arbon double
molecules
the
sux
are the
-ene,
s
s
instead
The
are
The
alkanes.
compound
has
both
a
c arbon
c arbon
double
bond,
and
an
side chain.
CH
3
1
2
3
methyl (
on
from
the
position
taking
right
c arbon-4.
to
3
)
substituent
le
at
follows:
loc ant
4,
the
of
the
over
would
Therefore,
example.
CH
priority
c arbon
incorrectly place the
numbering
name
of
this
occurs
from
molecule is
double
substituent
c arbon
any substituents
bond
y
r
e
loc ant
name
p
4-methylhex-2-ene
C
C
i
C
r
o
H
C
H
H
molecule
so
the
Numbering
the
atoms
the
possible
in
the
a
c arbon
straight
from
C
C
H
right
le
There
The
so
-ene.
to
b.
are
gives
sux
double
c arbon
loc ant.
bonded
double
H
C
C
H
H
C
bond
follows
c arbon
molecule
the
The
six
chain,
c arbon–c arbon double bond the
The
H
H
H
c arbon
butane.
from -ane to
chain
H
C
H
H
H
H
c arbon–c arbon double
changes
H
H
longest
parent chain is
contains
unsaturated
lowest
the
sux
E
bond,
c arbon
so
indic ates
unsaturation
H
H
u
The
four
chain,
a
are
c arbon
l
There
v
O
a.
H
f
x
Solution
b.
t
a
H
o
H
o
H
H
suffix
n
name the following alkenes.
d
a.
C
chain
U
Worked example 2
Using the IUPAC rules,
v
parent
loc ant
i
n
methyl substituent
atoms
the
contains
bond
the
longest
a
-ene.
c arbon-2 in the longest
methyl
(–CH
3
)
substituents
c arbon atoms number 4 and 5.
c arbon-1.
The
The
274
IUPAC
name
for
this
compound
is
but-1-ene.
IUPAC
straight
hexane.
c arbon-c arbon double bond,
from -ane to
follows
Two
in
parent chain is
changes
chain.
to
so
name
for
this
4,5-dimethylhex-2-ene.
molecule is
are
y
the
this
Numbering
bond
chain,
loc ant,
O
as
in
double
c arbon
lowest
t
i
s
With
molecule.
c arbon
right
the
n
to
longest
y
le
the
the
have
3
l
in
c arbon
must
r
present
number
bond
3
6
P
you
double
CH
2
5
CH
When
CH
4
e
H
alkyl
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Naming halogenoalkanes
When
naming
identied
each
halogenoalkanes, the position of the halogen substituent is
the
halogeno
derivatives
loc ants
for
the
loc ant
on
the
parent
chain.
Table
11
lists
the
prexes
used
for
H alogeno
Substituent
group
name
group.
of
methane
substituent
and
are
ethane
featuring
equivalent,
so
only
they
are
one
halogeno
omitted.
s
s
For
by
F
uoro
Cl
chloro
group, all
Consider the
following compound:
C
H
H
I
H
Br
N ames
substituents
this
atom
halogenoalkane
alphabetic al
order.
to
other
molecule
with
For
is
positions
will
bromoethane,
more
than
example,
the
one
form
not
halogeno
following
the
same
molecule. The
1-bromoethane.
group, list the halogens in
compound
is
named
2-chloro-
1,1,1-triuoroethane.
C
loc ants.
So
compound
takes
the
also
priority
following
features
over
the
molecule
c arbon
is
C
C
C
i
H
C
C
t
a
C
E
b.
u
H
H
a
H
H
rules
assigning
o
Cl
nomenclature
of
2-chlorobut-3-ene.
to
deduce the
c.
Cl
following compounds.
v
O
H
IUPAC
the
l
a.
of
not
H
2
d
r
o
f
x
the
names
terms
n
H
Practice questions
Apply
in
3-chlorobut-1-ene,
H
H
11.
c arbon double bond, the
group
U
H
a
halogeno
o
bond
C
halogeno
i
n
a
double
v
Cl
F
If
H
y
C
p
H
r
e
F
F
n
for
O
a
bromine
t
i
s
For
the
name
halogeno
l
IUPAC
of
y
Moving
P
p
Table 11
iodo
y
C
bromo
r
H
e
Br
H
C
Cl
H
C
Cl
H
C
Br
H
12.
C
Deduce
the
structural
formulas
for
the
following
H
compounds.
a.
2,3,4-tribromohexane
b.
1,3-dibromo-2-chlorobutane
c.
2-uorobut-2-ene
H
H
H
C
C
Cl
H
H
H
l
275
Structure
3
Classic ation
of
matter
Naming alcohols
All
alcohols
you
is
need
contain
to
numbered
The
over
group
ethanol
the
in
this
any
or
functional
sux
group
c arbon
series
position
do
results
from
has
the
c arbon chain
loc ant, taking
methanol.
loc ants.
The
This
molecule,
as
is
rst two
bec ause the
seen in methanol
more
The
H
2
C
H
H
H
1
H
H
H
propan-1-ol
loc ants
to
specify
H
2
C
p
C
require
before the -ol sux,
H
3
C
o
v
H
atoms
comes
y
r
e
1
HO
H
c arbon
loc ant
n
or
group.
example:
H
H
ethanol
3
C
C
OH
H
H
propan-2-ol
C
i
n
Naming compounds with c arbonyl groups
c arbonyl
aldehydes
at
the
end
of
to
the
c arbon
the
chain.
atoms
aldehyde
t
a
When
the
primary
u
is
priority
l
have
over
the
other
the
chain.
that
is
group
for
c arbonyl
known as the
group
in
two
classes of compounds:
group
is
terminal
aldehydes,
as
it
always
positioned
position.
always
There is
occurs at the
aldehyde includes a substituent, the numbering of
c arbon
the
the
the
so
This
the
the
sux
-al,
attached
so
group
c arbonyl
or
propane
to
ethane
is
compound
substituents
-one,
functional
the
functional
chain,
such
sux
at
have
is
aldehydes,
c arbonyl
When
c arbonyl
c arbon
numbered
the
begins
names
In
c arbon
number
i
d
r
o
f
x
a
E
v
O
276
need
of
(C=O)
ketones.
o
no
end
group
and
n
U
The
not
is
a
the
has
the
c arbon–c arbon
becomes
group.
For
example,
loc ated at the terminal position in the
ketone.
group
c arbonyl
becomes ethanal.
In
ketones,
lowest
double
propanone.
the
c arbon chain
possible
bonds.
loc ant, taking
Ketone names
y
H
three
l
with
hydroxyl
H
O
the
C
y
of
C
t
i
s
compounds
H
r
HO
H
P
H
C
H
for
is
same
methanol
position
The
possible
below.
HO
the
When naming alcohols,
-anol.
lowest
require
H
Alcohol
the
series
not
in
group.
-ane to
c arbon double bonds.
homologous
homologous
in
OH)
chain
hydroxyl
substituents
molecule
this
parent
e
hydroxyl
of
that
other
simplest
members
and
such
hydroxyl (
the
s
s
priority
the
change
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Worked example 3
Using the IUPAC rules,
name the following aldehydes.
There
are
no
substituents,
so
the
IUPAC name is
methanal.
a.
b.
O
CH
H
H
3
H
O
b.
parent
and
the
chain
H
C
C
C
C
H
H
H
H
The
c arbon
sux
comes
from the C
5
alkane,
pentane
becomes -al.
C
H
numbering
starts
from
the
c arbonyl
H
and
attached
c.
H
F
therefore
to
the
methyl substituent is
c arbon-4.
H
r
O
The
C
C
C
is
4-methylpentanal.
This is an example of an aldehyde featuring a halogeno
substituent. The parent chain comes from the C
H
H
4
l
H
name
alkane, butane, and the sux becomes -al.
The
The
parent
chain
comes
from the C
alkane,
group,
molecule
terminal
contains
position,
so
a
the
c arbonyl
sux
becomes -al.
The
O
H
molecule
the
terminal
ketone
bec ause
that
c an
Therefore,
parent
and
the
so
the
becomes
required,
as
c an
exist.
So
The
parent
and
the
or
sux
propanone.
propanone
exist;
if
the
c arbon-3,
it
is
The
the
all
the
information
in
the
name
-one.
b.
chain
sux
H
CH
H
H
3
C
C
C
H
H
H
O
H
the
nal
attached
C
comes
is
the
is
C
becomes
this
c arbonyl
from the C
-one.
only
The
4
alkane,
H
butane,
loc ant is not
four-c arbon ketone that
IUPAC
name
will
be
butanone.
loc ant is not
only
c arbonyl
would
propane.
be
three-c arbon
group
an
required
were
aldehyde.
to
deduce the
c.
numbered
group
chain
sux
has
at
Therefore,
from
the
c arbon-2,
comes
becomes
right
lowest
and
the
the
to
from the C
-one.
le
such
loc ant.
methyl
IUPAC
name
The
The
alkane,
that
the
is
at
pentane,
atoms
are
c arbonyl
c arbonyl
group
is
5
c arbon
group is
c arbon-4.
4-methylpentan-2-one.
a
v
O
‘propanone’.
l
is
The
alkane,
3
the
substituent
3-uorobutanal.
H
group that is not in
u
f
x
c arbon-1
structure
is
H
i
required
position,
name
O
H
c arbonyl
from the C
a
H
t
a
IUPAC
comes
contains
r
o
The
at
chain
C
o
parent
The
d
The
H
C
n
Solution
a.
is
c.
H
C
from
uoro
y
H
C
H
H
starts
the
p
C
H
C
C
H
U
C
name
o
b.
i
n
H
name the following ketones.
H
H
v
a.
IUPAC
r
e
Worked example 4
Using the IUPAC rules,
numbering
therefore
c arbon-3.
group in the
-ane
and
O
to
The
c arbon
methane.
t
i
s
1
n
y
Solution
a.
y
c.
H
IUPAC
C
P
H
e
group,
s
s
The
H
Naming c arboxylic acids
E
C arboxylic
acids
are
C arboxylic
acids
have the sux -oic acid.
parent
chain
is
a
class
of
compound
featuring
So
a
a
c arboxyl
group (–COOH).
c arboxylic acid with an ethane
c alled ethanoic acid.
277
Structure
3
Classic ation
of
matter
In
c arboxylic
group,
aldehydes,
always
acids,
including
in
there
the
is
chains
c arbon
no
terminal
need
from
to
are
the
assign
position.
Here
H
numbered starting with the functional
c arboxyl
loc ants
are
two
H
group
to
in
examples
CH
the
c arboxyl
3
of
H
H
chain.
groups,
C
H
H
O
C
H
C
C
C
C
H
H
H
H
C
OH
OH
acid
4-methylpentanoic acid
the
structures
of
the
following
molecules
2-chloropropan-1-ol
d.
5-ethyl-2-methylhept-3-ene
e.
2-methylpentan-2-ol
CH
O
the
following
c.
CH
molecules:
CH
3
CCH
3
CH
e.
OH
2
3
O
H
3
i
u
l
a
E
v
O
f
x
t
a
r
o
Science as a shared endeavour
Scientists
collaborate
facilitates
unambiguous
in
this
chapter
structures
bottle
to
of
are
that
still
foll ow
u se d
aci d,
four m i,
or
of
c ases,
name
the
of
and
the
phenyl)propanoic
of
and
you
such
stems
ant:
will
names
naming
chemic al
as
and
bu t
to
ac id
for mic a .
re co gnize
you
have
compounds
listed
on
you
the
will
learnt
using
label
now
both
of
be
a
able
suxes.
enc ourages
cer tain
a llude
terminology
rules
shampoo,
n omenc lature
these
for
agreed-upon
The
ingredients
me thano ic
hor miga,
acid.
the
guideline s,
L atin
mild
at
product
naming
M any
IUPAC
for
look
example,
the
world,
communic ate
orga nic
Span ish,
for miga
some
IUPAC
after
you
IUPAC
Fo r
to
hygiene
widel y.
su bstan ces.
formi c
In
some
If
the
communic ation.
you
names.
s tanda rdization
names
across
enable
personal
recognize
The
of
a
and
Por tugue se
278
CH
2
name
3
o
in
3
3
CCHCH
to
CH
3
CH
3
2
C
n a t u ra l l y
rules
CH
n
o c c u rs
2
o
acid,
CH
v
formic
ants
3
i
n
as
U
some
3
CH
d
k n ow n
nomenclature
CH=CHCH
p
CH
d.
commonly
2
y
3
O
IUPAC
CH
r
e
Apply
n
b.
3,3-dimethylpentane
t
i
s
hexan-3-one
CH
acid,
IUPAC name
c.
b.
Methanoic
the
a.
a.
p
Figure 8
on
y
given:
14.
based
l
Draw
If
the
is
us es
an d
commonly
you
this
the
u se
non-systemati c
as
spe ak
the
cha ra cteri stics
known
Italian ,
roo t
of
names
of
as
Fren ch,
for mic a,
respe ctively.
are
far
painkiller
too
long
ibuprofen
to
is
be
useful.
For
example,
the
(RS)-2-(4-(2-methylpropyl)
y
P
Practice questions
13.
r
3-bromopropanoic
are
e
C
they
c arboxylic acids:
O
Br
Like in
as
s
s
H
c arbon
the
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Structural isomers (Structure 3.2.6)
Structural
molecular
of
or
constitutional isomers
formula
structural
but
dierent
are
compounds
connectivity
of
the
that
atoms.
have
the
There
same
are
several types
isomers.
s
s
Chain isomers
Chain isomers
Butane
is
a
structural
and
straight-chain
However,
isomers
with
methylpropane
molecule
methylpropane
is
dierent
both
have
bec ause
a
it
lengths
the
has
of
c arbon
molecular
no
branched-chain
chains.
formula C
substituents
molecule,
on
the
4
H
10
For
.
c arbon
r
chain.
are
butane
bec ause it has an
methylpropane
C
C
C
H
H
H
CH
3
CH
2
CH
m e t hy l p r o p a n e
are
chain
i s o m e rs
Positional isomers
changes.
isomers,
all
with
the
molecular
isomers
formula C
5
H
11
has
C
H
H
Skeletal
E
p
Table 13
v
formula
Po s i t i o n a l
H
i s o m e rs
H
C
H
H
CH(CH
3
)
3
H
H
C
Br
H
2-bromopentane
C
H
C
dierent positional
o
i
C
t
a
C
a
O
H
u
H
H
l
f
x
structural
H
three
Br.
d
r
o
formula
H
where the position of the functional
bromopentane
1-bromopentane
Molecular
formula
structural
example,
Isomer
Full
are
For
C
H
n
group
U
Positional isomers
C
H
o
and
3
C
Butane
i
n
p
Table 12
CH
C
H
H
v
Skeletal formula
2
H
p
Condensed structural formula
H
H
n
C
H
O
H
y
H
10
r
e
H
H
t
i
s
H
Full structural formula
H
4
y
C
y
butane
Molecular formula
l
Isomer
P
alkyl substituent.
e
example,
H
5
H
11
3-bromopentane
Br
H
H
H
H
H
C
C
C
C
C
H
H
H
Br
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
Br
H
H
H
Br
Br
Br
of
bromopentane
279
Structure
3
Classic ation
of
matter
Worked example 5
Draw the structural formulas of all possible structural isomers of a compound with
the molecular formula C
4
H
9
F
.
Name the isomers using IUPAC nomenclature.
First,
draw
the
molecule,
longest
it
is
a
possible
straight
c arbon
chain.
Having
For
you
four-c arbon chain.
identied
need
longest
C
C
look
chain
by
and
at
named
possible
one
c arbon,
and
chain
in
place
as
the
many
uorine
positions
share
add
the
hydrogen
atoms.
Remember
numbering
of
the
c arbon
chain
c an
occur
in
to
same
add
molecule,
the
either
c arbon
in
the
else
H
H
3
4
C
C
H
H
H
H
H
2
C
3
C
4
C
H
H
in
H
are
to
to
the
ways
2-uorobutane,
H
as
is
methyl
on
C
C
uorine
to
the
group
For
central
anywhere
H
C
substituent
possible
above,
chain.
create
on
the
unique
c arbon chain
molecules.
H
C
H
3
isomers
H
H
H
H
2
1
H
C
C
C
H
H
H
C
3
F
H
H
2
1
H
C
C
H
F
H
1-uorobutane, and
favoured.
u
l
f
x
a
E
v
O
280
equivalent
possible
o
loc ants
is
equivalent
not
i
lower
is
are
t
a
the
3-uorobutane
4-uorobutane
d
and
4-uorobutane
r
o
isomers:
and
position
many
the
propane
H
2-fluorobutane
3-uorobutane
as
n
F
C
C
Then,
U
H
H
as
the
position
the
H
H
C
1
H
i
n
H
adding
to
four-c arbon butane chain.
o
1-fluorobutane
H
as
possible
p
C
a
v
F
create
formula
group
y
2
C
only
r
e
1
H
H
will
methyl
chain,
direction.
H
molecular
a
C
O
this
the
need
that
t
i
s
then
the
as
you
possible,
C
1-fluoro-2-methylpropane
2-fluoro-2-methylpropane
Activity
Draw and name all the structural isomers of chloropentane, C
H
5
Cl.
11
y
chain
propane chain.
n
c arbon
a
l
c arbon
the
isomers,
P
the
To
on
butane
isomers. Shorten the
creating
y
number
substituent
the
C
C
Next,
all
chain
r
C
to
e
this
s
s
Solution
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Primary, secondary and tertiary compounds
Changing
the
the
chemic al
alcohols,
the
You
two
to
halogeno
the
c arbon
atoms
other
group
is
is
to
bonded
this
secondary
to
a
c arbon
secondary
or
tertiary (3°)
H
H
H
2°
H
c arbon
atoms
class
of
H
C
C
C
H
Cl
H
secondary
in
attached
the
H
H
H
alcohol
the
to
(2°)
same
the
and
way
c arbon
as
H
c arbon
H
C
C
C
H
Cl
H
H
2-chloro-2-methylpropane
tertiary
(3°)
atom
next
to
the
functional
group
C
C
C
H
H
H
OH
H
of
to
the
primary,
and
groups
functional
OH
used
secondary
halogenoalkanes
Reactivity
to
and
are
form
tertiary
discussed
in
3.3.
c arbon
H
Primary,
secondary
and
tertiary
H
alcohols
form
dierent
products
2-methylpropan-2-ol
and
tertiary
secondary
or
in
(3°)
directly to the
and
is
Amine
slightly
reactions.
in
Reactivity
This
is
3.2.
dierent to the
classic ation depends
nitrogen
alcohols
oxidation
discussed
alcohols
tertiary
halogenoalkanes.
bonded
3°
C
C
H
halogenoalkanes
a
atom
alkyl
unlike
E
v
O
group,
as
alcohols
C
H
H
(2°)
H
H
i
number
secondary
C
H
t
a
amines
with
functional
next
C
u
the
of
(1°),
l
on
used
H
OH
r
o
f
x
method
C
c arbon
propan-2-ol
Primary
Classic ation
C
H
propan-1-ol
p
Figure 10
H
H
primary,
mechanisms
n
C
H
H
o
H
U
H
2°
d
H
H
c arbon
conditions
indic ates
H
1°
reaction
and
halogenoalkanes
halogenoalkanes: the number of
(gure 10).
The
p
classied
bonded to
o
the
(1°),
is
H
i
n
c arbon
are
H
3°
C
v
Alcohols
H
c arbon
2-chloropropane
Primary
atom
y
p
Figure 9
c arbon
c arbon atom is
r
e
1-chloropropane
primary (1°)
(gure 9).
c arbon
Cl
order to
y
C
In
n
C
atom.
l
C
tertiary,
O
H
and
y
H
aects
t
i
s
H
H
chain
tertiary, we count
secondary (2°)
H
1°
c arbon
c arbon atom. A
c arbon atom, a
c arbon atoms, and a
atoms
parent
group.
primary,
bonded
other
primary,
functional
compound
one
c arbon
the
the
classify compounds such as
e
to
other
a
of
on
c an
P
three
of
bonded
bonded
group
r
is
functional
compounds.
position
whether
number
atom
the
of
s
s
on
halogenoalkanes,
determine
the
of
halogenoalkanes and amines as
depending
In
position
properties
that
atom of the
consider
the
c arbon
group.
281
Structure
3
Classic ation
of
matter
When
naming
“amine”.
amines,
When
the
substituents
are
methyl
are
parent
name
loses
the
bonded
to
the
nitrogen
atom.
2°
is
replaced
H
C
H
C
H
gure 11, the substituents
H
C
C
C
H
H
H
H
H
H
C
tertiary
(3°)
amines
Classify
a
useful
skeletal
organic
formula
propan-2-ol
of
the
as
a
propan-2-ol.
primary,
following
is
a
secondary
H
H
H
H
H
H
H
i
t
a
H
C
H
H
a
OH
E
v
O
l
C
H
u
H
H
a
C
H
H
C
CH
reason.
H
C
3
H
H
D.
H
C
C
H
d
H
f
x
C
H
OH
n
C
o
C
U
C
r
o
H
C.
282
C
H
giving
C
i
n
H
H
alcohol,
o
B.
OH
H
tertiary
v
A.
H
or
secondary alcohol.
p
which
solvent.
of
H
CH
3
O
b.
Deduce
is
the
H
H
C
C
C
C
H
OH
H
H
H
y
Draw
n
y
a.
r
e
16.
Propan-2-ol
3
3
H
1-amine
and
t
i
s
Practice questions
CH
y
(2°)
CH
N
l
secondary
C
N,N-dimethylpropan-
1-amine
(1°),
H
H
amine
P
N-methylpropan-
Primary
C
H
H
p
Figure 11
H
3
N
r
H
3°
H
CH
N
propan-1-amine
15.
by
used to signify that
amine
H
H
which
N is
e
C
In
-e,
s
s
H
amine
H
H
sux
secondary and tertiary amines,
groups.
1°
H
the
naming
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
Functional group isomers
Functional group isomers
dierently
such
Compounds
each
group
and
with
other.
For
isomers
of
they
are
have
hydroxyl
and
example,
this
an
alkoxy
consider
molecule
methoxymethane,
C
H
H
H
of
propanal.
C
C
H
H
group,
H
O
C
H
H
H
be
functional
Propanone,
isomers
group
the
isomers
simplest
have
the
of
each
other, as they
ketone, is a functional
molecular
H
formula C
3
H
6
O.
H
O
C
H
C
C
H
O
H
H
H
simplest
The
ester
and
c arboxyl
ester,
groups
c an
also
methyl methanoate C
2
4
be
O
2
functional
group
, is an isomer of the
C
c arboxylic acid, ethanoic acid.
H
H
U
O
H
C
C
O
O
H
O
H
i
r
o
ethanoic acid
t
a
Molecular models
H
8
and
O.
obtained
In
u
information
3
methoxyethane
this
from
task,
you
different
all
will
types
have the
3.
of
Tool
v
•
skills
1:
Physic al
and
digital
models
molecular
MolView).
formulas
of
the
three
visualization
Explore
the
isomers using
software
(for
example,
different visualization
possibilities, including: ball and stick model, space-
and models of these substances.
Relevant
Create
online
reflect on
a
O
the
propan-2-ol
formula C
l
f
x
molecular
H
o
d
H
methyl methanoate
Propan-1-ol,
C
C
H
n
H
o
with
isomers.
propanone
i
n
Compounds
v
propanal
p
C
r
e
H
y
H
hydroxyl
methoxymethane
Both
H
a
isomers
Functional
n
isomer
c an
groups.
O.
l
ketones
with
6
O
group
and
c arbonyl
structural
H
y
have
alcohol
2
t
i
s
both
be
group.
C
ethanol
Aldehydes
an
c an
formula C
H
OH
arranged
y
C
alkoxy
atoms
e
H
groups
molecular
ethanol,
an
the
groups.
P
H
with
with
functional
the
include
ether
isomers
functional
r
H
structural
dierent
s
s
of
that
filling
model,
molecular
electrostatic potential
(MEP)surface.
digital
molecular modelling
E
4.
Compare
and
contrast
the
strengths
and
uses
of
each
Instructions
1.
State
the
structural
full
of
structural
formula
formula,
skeletal
various
formula
for
e ach
thre e
and
molecular models.
How
does
the
molecular
structure
of
these
three
of
compounds
the
formulas
condense d
5.
and
the
relate
to
their
physic al
properties and
isomers.
reactivity?
2.
Using
three
a
molecular model kit, construct models of the
isomers.
283
LHA
Structure
3
Classic ation
of
matter
Linking question
LHA
How
does
there
fact
In the Molecular models task, you gathered information about three
that
three
compounds, including: types of formulas, types of models, discussion of
isomers
strengths and properties. Reect on how this information could be organized.
dibromobenzene
support
of
the
only
the
current
benzene’s
Would you opt for a table, Venn diagram, concept map, or anything dierent?
model
Do you have a preferred method? Why does eective organization of
structure?
information support learning?
2.2)
r
e
(Structure
s
s
of
are
Thinking skills
ATL
Another
type
of
isomerism
arrangements
of
the
multiplicity.
atoms.
organic compounds is
have
However,
Stereoisomers
c an
the
they
same
be
have
and
formula,
congurational
isomers
(gure 12).
O
conformational isomers
molecular
dierent spatial
subdivided into two major
t
i
s
classes:
bond
some
n
and
by
stereoisomers
y
connectivities
exhibited
pair of
l
stereoisomerism. A
isomerism
H
H
H
H
staggered
of
e t h a n e,
eclipsed
The
two
v i ew e d
configurational
isomerism
along
isomerism
the
bond
p
Figure 12
c an
For
about
the
f
x
This
process
u
separate
the
groups
the
in
bond
isomers or
does
not
a
O
l
Congurational
bond
ethane, CH
c arbon
3
CH
breaking
together
to
3
resulting
conformers
involve
individual
bonded
single
form
by a single bond, these
dierent
, the two CH
in
two
3
conformational
groups
possible
c an
rotate
orientations,
known as
(gure 13).
any
bonds,
and
it
is
virtually
impossible to
conformers.
isomers
c an
only
be
interconverted
by
breaking
and
reforming
optic al
cis–trans
bonds.
Congurational
isomers
c an
be
subdivided into
cis
trans isomers
and
isomerism
of
v
Ty p e s
optic al isomers
(gure 14).
c o n f i g u ra t i o n a l
isomerism
Cis–trans isomers
E
In
aliphatic
able
to
on
the
two
alkenes,
rotate
dierent,
two
same
“this
about
groups
the
isomers
side
identic al
means
284
c arbon
two
about
example,
conformational
isomerism
isomerism
contains
rotate
i
isomers.
configurational
p
Figure 14
molecule
of
o
a
groups
t
a
r
o
d
If
Ty p e s
n
c arbon–c arbon
conformational
c o n f o r m e rs
U
p
Figure 13
isomerism
C
i
n
H
isomerism
o
H
structural
v
stereo-
H
y
H
H
H
p
r
e
H
H
of
c an
the
of ” and
are
the
cis
unsaturated
double
isomer
bond.
where
c arbon
If
two
these
opposite
means
“the
sides
other
of
the
side
atoms
two
identic al
c arbon double bond, and a
on
trans
to
c arbon
exist: a
c arbon
substituents
side
bonded
c arbon
are
substituents
trans
double
of ”.
are not
groups
isomer
bond.
In
are
where
L atin,
cis
y
P
Stereoisomers (Structure 3.2.7)
Structure
example,
isomers
bond is the
trans,
the
about
symmetric al
the
double
reference plane:
depending
on
molecule of but-2-ene has
bond
which
it
trans
groups:
Classic ation
of
organic compounds
congurational
(gure 15). The horizontal axis of the double
is
used to determine whether an isomer is
groups
are
either
H
H
side
of
the
cis
or
plane.
C
s
s
H
cis
Functional
LHA
For
3.2
H
3
reference
C
C
C
C
plane
H
C
3
CH
H
CH
3
3
trans-but-2-ene
r
e
cis-but-2-ene
an
C=CHCH
atoms
has
be
3
sides
of
cis–trans
c arbon
atom
the
, will not experience
two
identic al
identic al
groups
reference
isomerism,
have
two
to
be
the
two
dierent.
cis–trans
(hydrogen
groups
For
the
atoms),
H
plane
H
C
so
the
potential
C
C
o
also propene
not
ex h i b i t
i
does
cis
trans
isomerism
t
a
Cis
Propene
r
o
p
Figure 16
d
propene
isomers will
3
H
3
propene,
n
C
methyl
attached to
example,
reference
C
two
isomerism bec ause one of the c arbon
U
H
has
plane.
(gure 16).
H
methyl substituents on the
C
also
exhibit
the
trans-isomer
o
2
to
has
i
n
H
alkene
unsaturated
but-2-ene
plane, while the
v
each
opposite
of
p
For
on
reference
but-2-ene
n
l
substituents
cis-isomer
the
of
y
of
i s o m e rs
r
e
side
trans
O
gure 15, the
same
cis
y
P
y
In
The
t
i
s
p
Figure 15
trans isomerism also occurs in disubstituted cycloalkane molecules. In this case,
the reference plane is the at face of the ring structure. For example, the compound
trans congurational isomers (gure 17).
u
f
x
1,2-dimethylcyclobutane has cis
The cis-isomer has both methyl substituents on the same side of the plane of the ring.
The trans-isomer has methyl substituents on opposite sides of the plane of the ring.
l
O
Disubstituted three-carbon rings also experience cis
CH
2
H
E
2
C
3
CH
v
C
CH
a
H
H
H
CH
2
H
2
H
The
cis–trans
3
CH
2
H
C
C
i s o m e rs
C
C
CH
3
trans
1 , 2 - d i m e t hy l c y c l o b u t a n e
CH
H
2
C
CH
3
cis
p
Figure 17
3
trans isomerism (gure 18).
of
H
H
C
3
CH
CH
3
cis
p
Figure 18
3
trans
The
cis–trans
i s o m e rs
of
1 , 2 - d i m e t hy l c y c l o p r o p a n e
285
3
Classic ation
of
matter
LHA
Structure
Practice questions
17
.
Deduce
and
draw the
cis-
and
trans-isomers of
hex-2-ene.
For each of the following molecules, state whether they exhibit
cis
s
s
18.
trans
isomerism. For the molecules that do, deduce whether they are in the
or trans conguration. Name each molecule using IUPAC rules.
b.
CH
H
CH
2
3
CH
H
d.
2
H
H
structural
c an
form
formula
cis
trans
of
e.
1-ethyl-2-methylcyclopropane
f.
2,3-dimethylpent-2-ene
is
dened
atoms.
one
c alled
is
or
C
2
Chiral
t
a
r
o
Pe n i c i l l a m i n e
2 - a m i n o - 3 - m e t hy l
3 - s u l f a ny l b u t a n o i c
molecular
the
C arbon-1
-
and
acid
l
a
E
v
O
286
has
Penicillamine
u
name
H
f
x
p
Figure 19
I U PAC
C
3
c arbon
have
more
the
atom
chiral
c arbon
ability
to
A
bonded
to
stereocentre
atoms
pair
rotate
of
four
or
exhibit
optic al
dierent atoms
asymmetric
a
type
isomers
plane-polarized
of
are
light;
centre.
congurational
c alled
this
is
known as
optic al activity.
OH
H
molecules
1
C
a
known as a
optic al isomerism.
o
C
as
also
n
with
It
C
of
o
v
c arbon
groups
p
1,1-dibromobut-1-ene
i
HS
y
r
e
pent-2-ene
d.
i
n
O
3
d
2
deduce
b.
U
NH
then
1-bromo-2-chloroethene
enantiomers.
C
and
2-bromoprop-1-ene
isomerism
3
compounds,
c.
Molecules
H
following
a.
chiral
or
the
H
C
3
isomers.
Optic al isomers
A
H
C
O
the
they
t
i
s
if
C
3
C
H
Draw
H
3
C
C
19.
CH
2
n
3
CH
C
l
CH
y
H
is
a
drug
formula C
is
only
c arbon-3
bonded
achiral.
C arbon-2
chiral.
used
H
11
NO
bonded
each
groups
therefore
5
to
have
to
in
2
is
Its
three
four
c arbon-3
bonded
The
the
S.
treatment of rheumatoid arthritis. It has the
structural
atoms,
bonded
are
to
presence
so
four
it
atoms
identic al
of
formula
(the
dierent
this
chiral
is
or
not
is
shown
chiral, or
groups.
two
or
gure
achiral.
However,
methyl
atoms
c arbon
in
19.
C arbon-2
two
groups),
so
of
the
four
c arbon-3 is
groups of atoms, and it is
atom
means that penicillamine
is optic ally active and c an exist as a pair of two dierent enantiomers. Enantiomers
are
non-superimposable
ofsymmetry.
mirror
images
of
each
other,
and
they
have no plane
y
H
H
c.
3
P
H
C
CH
2
C
C
3
CH
2
r
C
C
H
e
a.
cis
Structure
atoms
formulas,
and
groups
like
the
present
one
in
a
in
gure
19,
molecule
are
provide
information
connected.
on
However,
Functional
groups:
Classic ation
how the
these
of
organic compounds
LHA
Structural
3.2
H
formulas
*
do
not
allow
centre.
As
a
you
to
result,
dierentiate
enantiomers
between
are
enantiomers
drawn
as
in
a
molecule
three-dimensional
with
a
structures
N
chiral
known
CH
N
as
stereochemic al formulas
the
20),
where
specic
symbols
are
3
used to
s
s
designate
(gure
direction of a bond (table 14).
OH
Name
Symbol
Directionality
*
bond
C
—
aligned with the plane of paper
c
coming
dash bond
c
going
out
of
the
plane
(towards
the
viewer)
r
wedge bond
HO
behind
the
plane
(away
from
the
viewer)
NH
2
e
line
P
adopts
tetrahedral
and
as
line
dash
are
central
arrangement
around
bonds,
bonds
a
a
one
chiral
as
a
in
stere ochemic al
atom
with
with
bond
c arbon
four
angles
atom,
two
regions
of
of
formulas
of
109.5°.
the
electron
For the
bonds
are
tapered,
meaning
that
they
start
o
narrow
enantiomer
are
of
a
example,
shown
in
stereochemic al
the
formula,
stereochemic al
you
need
gure 21.
C
2
HOOC
COOH
H
were
to
the
to
align
same
enantiomers,
the
in
gure
models
the
models
direction.
mirror
21,
of
are
them
such
that
However,
image
non-superimposable.
using
would
if
the
a
you
show
SH
Nicotine
the
All
c arbon
( t o p)
is
tobacco
( m i d d l e)
is
t hy r ox i n e
t hy r o i d
a
of
these
atom,
molecules
designated
n a t u ra l l y
plant,
is
h av e
*.
sy n t h e s i z e d
norepinephrine
n e u r o t ra n s m i t t e r,
( b o tt o m )
by
a
and
hormone
f rom
the
gland
2
This
means that if
molecular model kit, it would be
i
in
3D
penicillamine
functional
u
l
were
the
to
groups
hold
structure
a
of
of
the
same type
mirror to one of the
the
other
enantiomer.

Figure 22
e n a n t i o m e rs
HOCH
the
2
Molecular
of
C H (O H ) C H O.
c h i ra l
models
of
the
2 , 3 - d i hy d r ox y p r o p a n a l ,
C an
you
identify
c e n t r e s?
a
E
v
O
f
x
t
a
point
those
r
o
impossible
like
create
)
3
c h i ra l
by
I
H
o
you
of
d
Enantiomers,
E n a n t i o m e rs
C(CH
a
2
n
p
Figure 21
2
C
U
C)
3
I
p
Figure 20
o
NH
C
i
n
N
2
C
HS(H
draw its
v
mirror plane
H
to
formulas of the penicillamine
p
enantiomers
For
O
HO
NH
y
the
image.
I
I
from the
r
e
draw
mirror
OH
*
wedge bond, and one as a dash bond.
c arbon atom and widen.
To
O
O
drawn
Wedge
a
arrangement
that
used
t
i
s
tetrahedral
usually
from Structure 2.2
bonds
n
density
rec all
dire ctional
l
will
of
y
You
S ummary
y
OH
p
Table 14
287
3
Classic ation
of
matter
LHA
Structure
Worked example 6
Draw
the
stereochemic al
2-aminopropanoic
chiral
c arbon
acid
formulas
showing
of
a
the
two
enantiomers
tetrahedral
of
arrangement
around
the
atom.
s
s
Solution
draw
the
structural
formula
to
deduce
C
C
H
NH
atom
O
chiral.
P
*
H
is
C
the
to
mirror
and
the
draw
of
is
the
attached
and
atoms
to
four
methyl, amino and
and
groups
around
projected out of the plane of the
behind
atoms
the
plane
of
the
paper
(dash
are in the plane of the paper (line
enantiomer.
H
which
bond
bonds
dashes
draw
t
a
a
is
It
atom
C
o
Finally,
and
rst
H
the
i
d
r
o
are
in
just
other
the
C
3
is
the
assigned
2
to
tetrahedral
used
to
give
enantiomer,
reected
COOH
NH
you
draw
image
of
which
group
structure
a
a
sense
dashed
your
are
of
for
all
the
3D
vertic al
existing
this
rst
equivalent. The
structure.
line
to
represent
enantiomer.
H
H
u
l
f
x
a
E
v
O
288
wedges
as
your
arrange
projected
groups
asterisk.
o
matter
one
C
not
is
an
hydrogen
p
you
that
n
does
enantiomer,
or
v
gives
formula,
such
one
atoms
a
y
bond),
two
This
atom
i
n
U
It
with
atoms:
r
e
and
bonds).
of
stereochemic al
c arbon
(wedge
bond),
marked
groups.
chiral
paper
is
groups
n
draw
the
atom
or
O
c arboxyl
To
c arbon
atoms
y
chiral
dierent
t
i
s
The
l
OH
2
C
C
H
C
COOH
3
NH
M ake
are
sure
that
correct.
le-hand
the
For
image
to
the
correctly
H
2
connectivities
example,
HOOC
between
methyl
show
a
the
group
c arbon
central
has
to
CH
3
N
2
been
atom
and
the
reversed (H
c arbon bond.
3
groups
C) in the
y
H
c arbon
r
H
which
e
First,
we
discussed the
cis
trans
1,2-dimethylcyclobutane. The
isomerism
trans-isomer
of
of
a
3.2
Functional
exhibits
optic al
groups:
cycloalkane,
this
molecule
Dierent
This
is
bec ause
enantiomers
taste
c an
of
also
the
be
mirror
same
images
are
compound
aected
by
a
non-superimposable
have
specic
dierent
odours.
enantiomer
CH
C
3
dierent
chemic al
environments,
Our
present
pair
light
of
by
molecules
enantiomers
the
other
plane
under
angle,
of
the
in
rotates
the
human
same
in
foods.
plane
of
other
chiral compounds,
and
is
will
rotate
(gure
p
Figure 23
polarization
enantiomer.
anti-clockwise and is
a
explain
E
Draw
v
2-hydroxypropanoic acid, C
b.
4
H
the
Construct
chiral
3D
10
(or
racemate),
the
tasks
(a)-(c):
O
which
stereochemic al
around
racemic mixture
complete
ii.
and
O
molecules,
butan-2-ol, C
Identify
y
following
i.
a.
c.
c alled a
viewer
enantiomer
u
the
l
of
is
o
enantiomers
f
x
O
each
pure
plane-polarized light.
Activity
For
a
of
o
two
by
i
of
rotate
light
isomerism
polarizer
t
a
not
plane-polarized
r
o
mixture
does
of
d
Ro t a t i o n
n
U
C
i
n
pure enantiomer
v
polarizer
p
r
e
polarized light
p
Figure 24
3
n
y
t
i
s
light
it
optic al
enantiomer.
unpolarized
50:50
The
One enantiomer
designated the (+)
light
and
CH
plane-polarized
24).
source
A
C
3
t r a n s - 1 , 2 - d i m e t hy l c y c l o b u t a n e
directions
clockwise
of
non-chiral
body.
conditions
opposite
the
presence
in
l
)
the
polarization
enantiomer
designated the (
but
in
in
identic al
y
the
same
dierent
are
P
rotates
The
biologic al
enantiomers
r
A
as
become
of
e
such
but
properties
3
(gure 23).
H
The
organic compounds
s
s
perceived
its
of
mirror plane
also
H
isomerism.
Classic ation
LHA
E arlier,
Structure
c arbon
formulas
3
H
6
O
atom
showing
3
is
chiral.
the
tetrahedral
arrangement
c arbon atom.
models
for
the
pair
of
enantiomers.
289
LHA
Structure
3
Classic ation
of
matter
Rotation of plane-polarized light by optic ally active compounds
Optic ally
active
Here
will
you
rotate
plane-polarized light.
5.
Place
6.
Dim
the
filters,
and
consider
ways
to
the
look
filter
the
above
light
the
source
tube.
through the
polarizing
filter.
further.
skills
Inquiry
at
1:
polarizing filter
e
•
polarizing
and
develop the
second
investigation
Relevant
second
lights
s
s
polarizing
compounds
explore this phenomenon using simple
Demonstrate independent thinking and
Identify
the
design
•
Inquiry
1:
Design
•
Inquiry
2:
of
an
control
investigation
variables
sample
and
Identify
explain
and
record
a
valid methodology
relevant
qualitative
l
observations
S afety
n
y
polarizing filter
light source
eye
protection.
teacher
will
provide
you
with
further
Rotate
the
safety
changes
8.
Repe at
analysed.
M aterials
•
Light
source,
filters
for
example, a mobile phone flashlight
glass
a
measuring
clear
cover
at
one
end,
cylinder
Clamp and stand
S amples
optic ally
glucose
or
in
on
2.
to
the
Place
a
Pour
of
flat
surface,
the
water
in
light
the
l
of
a
details
are
a
over
pointing
method
in
on
of
the
in
time
the
the
variables
observe the
placing
you
of
one
Explore
rotation
that
effect
tube.
of
must
of
the
plane-
keep constant
different compounds
plane-polarized light.
explain
measure
investigate
•
the
of
the
order
changes
to
do
you would make
one
or
more of the
•
of
Show
your
on
of
effect
rotation
of
concentration on the
the
effect of path length on the
plane-polarized light
investigate
mixture
angle
the
plane-polarized light
investigate
rotation
3.
the
the
ideas
effect of the composition of a
rotation
to
your
of
plane-polarized light.
teacher,
and
if
you
have
time,
try them out.
M ass spectrometry (MS) (Structure 3.2.8)
(MS)
compounds
into
is
an
analytic al
fragments,
technique
some
of
that
which
will
c an
be
be
used
ions.
to
This
is
break up
known
mass
as
fragmentation.
The
molecular
masses
of
these
fragments
will
correspond to
in
the
masses
MS
c an
of
certain
functional
groups
or
parts
of
the
c arbon
chain.
As
a
result,
1.2.
be
fragments
The
a
290
and
rotation
up.
tube.
organic
discussed
rotation
•
source.
M ass spectrometry
the
the
explore
•
plane-polarized light.
v
Structure
E
spectrometer
of
height
operational
on
filter
u
the
aside.
you
water into the tube and clamp it in place
source
f
x
O
The
the
Record
source
polarizing
distilled
above
4.
light
them
substance(s)
i
Place
Set
the
t
a
3.
test.
of
r
o
2.
to
solutions
o
wish
aqueous
d
Prepare
this
of
to
and
following:
Instructions
1.
the
Identify
active substances, such as
fructose.
some
order
n
sucrose,
of
List
for
U
•
a
with
solution
light.
C
example,
•
tube
i
n
Clear
this
Q uestions
torch
1.
•
of
polarize d
this
solutions
o
or
effe ct
polarizing
experiment,
prepare d
filter
intensity.
p
Two
v
•
polarizing
light
y
your
the
the
r
e
being
second
to
prec autions, depending on the identity of the substances
O
7
.
Your
t
i
s
Wear
used
to
present
molecular
obtain
in
an
masses
mass spectrum.
The
information
about
the
functional
groups
and
structural
organic compound.
of
the
dierent
dierent
fragments
peaks
within
a
are
mass
recorded
spectrum
in
a
are
graph
c alled
known
y
1:
r
in
Inquiry
P
creativity
•
Structure
charge
is
fragmentation pattern.
ratio (m/z)
known as the
in
a
mass
molecular
spectrum
The
peak
with
corresponds
to
the
the
highest
Functional
groups:
Classic ation
of
organic compounds
LHA
collectively as a
3.2
mass to
parent compound. This
ion peak.
The mass spectrum
s
s
Worked example 7
for propan-1-ol is shown in gure 25.
r
80
15
20
25
30
35
40
45
50
55
60
m/z
spectrum
of
p r o p a n -1 - o l
to the structure of propan-1-ol.
you
expected
structural
formula
the
will
have
analysed
help
more
molecules.
this
option,
c ase,
C
C
C
H
H
and
u
Molecular fragment
the
only
fragmentation
structural
one
formulas
possible
OH
H
identify
for
drawing its
o
H
is
by
possible
i
H
begin
the
draw
there
H
structures
out
t
a
spectrum
propose
the
each
peaks and their
of
the
m/z
values. Use
fragments:
Explanation
l
a
Molecular
ion
(the
original
molecule that
+
[CH
3
CH
2
CH
2
v
O
60
the
to
In
gure
one
d
m/z
values
r
o
examine
m/z
f
x
the
than
compound,
to
(propan-1-ol):
H
Then,
you
o
the
When
of
This
n
all
name
U
patterns.
the
formula.
C
know
v
you
structural
i
n
Solution
p
to show that the fragmentation pattern corresponds
y
M a ss
Use the mass spectrum
r
e
p
Figure 25
O
12
n
y
t
i
s
0
of
l
evitaler
40
y
P
ytisnetni
60
20
If
e
100
OH]
has
lost
one
electron)
+
31
[CH
29
[CH
2
OH]
Loss of CH
3
CH
2
from the
m/z
60
+
CH
E
3
2
]
Loss of CH
H
H
H
H
C
C
C
H
H
H
M
r
=
29
2
M
OH
from the
m/z
60
OH
r
=
31
291
3
Classic ation
of
matter
LHA
Structure
D ata
Here
on
specic
are
some
MS
fragments
are
provided
in
section 21 of the data booklet.
examples:
peak with
m/z
=
15
corresponds to a CH
•
A
peak with
m/z
=
17
corresponds
•
A
peak with
m/z
=
29
•
A
peak with
m/z
=
31
•
A
peak with
m/z
=
45
to
an
fragment
3
OH
fragment
corresponds to a CHO or CH
corresponds to a CH
corresponds
to
a
3
2
O or CH
COOH
CH
2
3
OH
fragment
fragment
fragment
r
straight-chain
molecular
formula C
the
structural
spectra
Explain
A
which
B
formulas
of
spectrum
booklet.
the
is
of
two
the
produced
C
29
60
o
v
40
n
U
evitaler
i
n
ytisnetni
80
20
o
i
0.0
20
40
60
80
100
m/z
B
43
100
evitaler
a
ytisnetni
u
l
E
v
O
f
x
t
a
r
o
d
0.0
86
80
60
40
86
20
0.0
0.0
20
40
60
m/z
292
80
100
10
O has two
isomers.
given.
each
p
57
100
by
H
possible
are
compound
y
A
two
isomers
r
e
22 of the data
and
5
O
b.
the
n
Deduce
M ass
with
isomers.
t
i
s
a.
ketone
y
structural
l
A
using
section
y
P
Practice questions
20.
e
A
s
s
•
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
LHA
Infrared (IR) spectroscopy (Structure 3.2.9)
Infrared (IR) spectroscopy
the
types
of
functional
IR
groups.
radiation.
present
IR
is
an
analytic al
in
an
organic
spectroscopy
Molecules
spectroscopy is a type of
absorb
technique
exposes
IR
that
compound,
a
and
sample
radiation,
of
c ausing
c an
be
used to identify
therefore determine its
an
organic compound
certain
bonds
to
s
s
to
bonds
vibrate. IR
vibrational spectroscopy
e
There are two basic types of vibration that are possible in molecules: stretching/
compression and bending. The type of movement observed depends on whether
r
the molecule is diatomic or polyatomic. For example, the diatomic molecule of
hydrogen chloride can only undergo stretching and compression (gure 26).
l
a
molecule
the
absorbs
bond
IR
radiation,
enthalpy
and
the
the
HCl
molecule
frequency
of
the
a
shorter
of
the
enthalpy,
iodide
so
contains
vibration
halide
lower
a
the
large
occurs
at
frequency
frequency
iodine
a
of
atom
lower
of
the
vibration.
the
and
molecules (table 15).
has
frequency
a
relatively
when
Frequency of
Br
I
Bond
Wavenumber
the
fourth
column
1
=
λ
is
of
–1
/ cm
2886
usually
table
of
halogen
The
described
1.3,
in
frequency
Structure
(f )
is
related
to
2230
wavelength
halide
(λ)
via
the
relationship
c
molecules
f
=
,
where
c
is
the
speed
of
λ
described in terms of its
14.
As
2559
wavenumber,
ν,
light.
is
wavenumber
is
the
This
related
means
to
that
frequency
the
wavenumber
c
f.
as
reciproc al of the
follows:
ν
=
×
a
ν
E
v
O
wavelength:
absorption
w av e n u m b e rs
l
in
the
and
–1
s
6.69
u
spectra,
shown
e n t h a l py
f
x
IR
7.68
298
p
Table 15
In
8.66
t
a
H
431
366
i
Cl
H
r
o
H
vibration / 10
o
11
d
–1
low bond
n
Bond
in turn result
larger the
compared to the other
Molecule
enthalpy / kJ mol
The
vibration.
U
hydrogen
its
the
higher
C
Hydrogen
atoms,
and
i
n
masses
wavelength,
o
v
bond. The higher the bond enthalpy, the stronger the bond. This will
in
resultant
masses of the atoms in the
p
on
in
n
in
depends
c o m p re ss i o n
O
bond
and
y
a
vibration
Stretching
y
P
y
When
Cl
r
e
p
Figure 26
t
i
s
H
293
3
Classic ation
of
matter
LHA
Structure
In
polyatomic
Polyatomic
molecules,
molecules
bond
angle
changes.
bend
when
they
spectroscopy.
In
the
c an
Both
absorb
other
stretching
also
IR
water
and
radiation
words,
c an
undergo
such
be
symmetric
bending,
c arbon
(gure
a
dioxide
27),
molecules
so
are
they
c an
be
stretch and
analysed
O
H
H
H
H
r
IR
All
of
water
thre e
C
C
and
modes
of
active
compound
dipole,
such
hydrogen, H
performing
change
in
the
molecule
homonuclear
heteronuclear
of
the
vibration;
2
,
dipole
molecules
oxygen, O
molecules,
the
such
stretching
moment,
must
otherwise,
2
as
and
such
,
and
no
that
experience a change in
peak
do
will
not
hydrogen
2
,
are
observed in the
are
uoride,
compression
molecules
be
have a permanent
chlorine, Cl
p
a
as
i
n
is
Diatomic
active,
the
o
c apable
IR
v
Diatomic
be
y
spectrum.
to
during
r
e
IR
moment
O
a
dipole
t
i
s
For
C
n
are
v i b ra t i o n
bending
l
v i b ra t i o n
of
molecules.
stretching
IR
HF,
vibrations.
IR
inactive.
are only
Bec ause
there
active.
C
Plotting the wavenumbers of the molecular vibrations against their intensity gives a
graph known as an IR spectrum. Certain wavenumbers are characteristic of specic
U
functional groups, which are listed in section 20 of the data booklet. We can
n
therefore analyse the IR spectrum to determine the types of functional group present
in a molecule. For example, the IR spectrum of butanoic acid is shown in gure 28.
o
i
u
ecnattimsnart
l
50
a
E
v
O
f
x
t
a
)% (
r
o
d
100
0
4000
3000
2000
1500
1000
500
1
wavenumber / cm
p
Figure 28
IR
spectrum
of
butanoic
acid
–1
There
is
a
strong,
very
broad
peak
in
the
range 2500
3000 cm
characteristic of
–1
the O
is
294
H
bond
in
c arboxylic
acids.
characteristic of the C=O bond.
The
strong
peak
in
the
range
1700
1750 cm
y
Modes
d i ox i d e
asymmetric
stretching
y
c arbon
symmetric
P

Figure 27
by IR
IR-active.
e
H
where the
s
s
H
asymmetric.
molecules
O
O
or
vibration
Structure
spectroscopy
in
a
of
molecule,
an
must
unknown
be
used
techniques,
only
so
in
with
about
c annot
be
information
information
spectroscopy,
the
functional
used
groups
determine
gathered
derived
nuclear
to
from
from
IR
other
groups:
Classic ation
of
organic compounds
present
the
structure
spectroscopy
analytic al
magnetic resonance (NMR)
s
s
or
mass
alone
Therefore,
conjunction
as
information
technique
molecule.
such
spectroscopy
provides
this
Functional
LHA
IR
3.2
combustion analysis.
composed
hydrogen and
oxygen is
10.43% hydrogen by mass.
showing your
spectrum of the compound
functional group
present
is shown below.
Deduce the
in the molecule.
3000
2500
2000
1500
1000
o
3500
U
C
0.0
i
n
0.2
v
0.4
y
0.6
p
r
e
% / e c n att im s n a rt
0.8
O
t
i
s
1.0
n
The infrared
y
b.
l
working.
y
Determine the empiric al formula of the compound,
P
a.
containing only c arbon,
of 62.02% c arbon and
r
An organic compound
e
Worked example 8
500
1
wavenumber / cm
The mass spectrum
of the compound
is shown below.
n
c.
Deduce the
o
d
relative molecular mass of the compound, and hence determine the
molecular formula.
16
E
15
u
14
0
v
15
20
a
40
20
d.
l
O
evitaler
f
x
ytisnetni
60
i
80
43
t
a
r
o
100
58
42
26 27 29
25
28
25
37
31
30
Through experimental analysis,
did
not
36
35
39
38
41
44
57
40
40
45
50
55
60
m/z
it
was determined that the compound
contain a terminal c arbonyl group.
Deduce the structural
formula of the compound.
295
3
Classic ation
of
matter
LHA
Structure
Solution
a.
The
method
for
determining
an
empiric al
formula
C
1:
O
62.02
10.43
=
5.16
=
12.01
Step
Divide
result
=
by the
value
(1.72)
5.16
10.33
=
3.00
1.72
=
6.01
to determine the
rounding
if
ratio,
necessary
is
therefore C
3
H
6
O.
1.00
n
l
formula
y
empiric al
=
1.72
1
b.
A
strong
peak
c arbonyl
to
the
c arboxylic
deduce
range
the
of
1700
Possible
acids
functional
and
molecule
atom
terminal
middle
of
in
equal
not
the
an
is
58,
or
c arbon
of
the
so
up
the
formula.
so
and
the
its
is
molecular
acid,
As
the
the
masses
molecule
c arbonyl
is
a
by the
the
3
empiric al
H
6
O.
there is only one
c arbonyl
structural
of
formula is C
as
you
answer is
represented
atomic
c arboxylic
molecule,
chain,
the
the
aldehydes,
asked
group is not in
group must be in the
ketone.
Therefore, the
formula is CH
3
COCH
3
.
n
U
magnetic resonance
o
i
t
a
1
Proton nuclear
the
u
a
spe ctrometer
ester
compound
so
has
and Structure 3.2.11)
in
magnetic
the
add
presence of
H NMR) spectroscopy (Structure 3.2.10
most
useful
determine
l
(NMR)
1
molecules.
v
nucle ar
d
r
o
296
A
E
resonance
f
x
O
p
Figure 29
to
propanone,
Proton nuclear
(
of
you
molecular
position
the
compound
is
If
C
the
i
n
oxygen
also
mass
58.
the
include
question
compound,
o
The
is
v
d.
this
peak:
the
the
p
formula,
molecular
ion
of
but
shows
molecules
y
relative
molecular
r
e
The
esters,
group
c arbonyl/C=O.
c.
1750 cm
organic
O
ketones,
in
group, C=O.
t
i
s
a
the
magnetic resonance (
analytic al
technique
for
H NMR) spectroscopy
structural
is perhaps
determination
of
organic
1
The
the
spectra
number
molecule.
An
produced
of
dierent
atom's
by
H
NMR
chemic al
chemic al
spectroscopy
environments
environment
is
help
of
dened
you to
hydrogen atoms
by the other atoms
attached to it.
1
H
NMR
spectroscopy
is
an
example of
spin resonance spectroscopy. The
1
nucleus
of
a
nucleus
has
hydrogen atom
two
possible
quantum number (m
I
),
H
spin
with
contains
a
single
orientations,
possible
as
positively
described
values of
m
I
charged
by the
= +½ or
m
I
proton. This
nuclear spin
= –½.
y
number
1.72
P
1.72
The
1.72
16.00
r
lowest
10.33
1.01
mass
e
atomic
2:
27.55
by the
relative
whole
below:
Divide the
percentage
each
shown
s
s
Step
is
H
Structure
a
hydrogen
small
nucleus
magnet.
behaves
Similar
to
the
as
a
spinning
electron
spin,
charged
the
two
particle
that
Functional
groups:
Classic ation
of
organic compounds
LHA
The
3.2
c an act as
possible spin states of a
1
H
nucleus
are
degenerate:
they
is
the
exist
at
the
same
energy
level.
When
a
strong
1
magnetic
become
eld
applied
non-equivalent
to
and
sample containing
split
into
two
H nuclei, the two spin states
dierent
energy
levels
(gure 30).
s
s
the magnetic field generated by
the higher energy spin state is
e
high energy
opposite to the applied external
1
spin
=
magnetic field
–
r
2
1
spin
+
=
l
=
–
2
n
y
2
the magnetic field generated by
low energy
=
+
O
t
i
s
the lower energy spin state is
1
spin
aligned with the applied external
2
magnetic field
no magnetic field
magnetic field applied
magnetic
magnetic
equal
required
in
for
the
to
dierence
process
of
Chemic al
is
radiation
molecule
with
four
1
to
and
E
spectrum
small
different
a
be
two
hydrogen
by
as
be
spin
of
dierent
signals.
when
CH
H
3
Si
C
CH
3
CH
promoted
to
from
radio
the
p
Figure 31
3
states.
The
waves of
exact
environment
in
dierent
dierent
is
3
the
accepted
reference
u n i v e rs a l l y
standard
used
in
1
NMR
s p e c t r o s c o py
energy
hydrogen
chemic al
A
summary
of
chemic al
shi
frequencies.
hydrogen
These
of
T MS
lower
H
atoms
radiation
energy
increases along with the
exposed
chemic al
relative
is
values
are
dierent
environments
environments
signals
for
of
given in
the
data
is
hydrogen
given
in
section
21
booklet.
sc ale,
assigned
a
where
chemic al
methyl
the
signal
shi
groups
of
0
produced
ppm.
by a
TMS is a
(gure 31).
1
resolution
H
NMR
spectrum,
the
area
under
each signal
1
dicult
indic ate
the
to
in
H NMR
low
a
is
must
equivalent
proportional to the number of
signal
are
expressed in units of parts per million (ppm).
using
standard
v
a
a
δ,
it
the
absorbed
l
O
is
examining
So,
states
nucleus
u
(TMS)
Low-resolution
When
molecule.
on
H NMR spectrum
f
x
symmetric al
between
electromagnetic
measured
tetramethylsilane
state,
depends
chemic al shi,
shi
the
spin
t
a
displayed in a
terms of their
nucleus
i
frequencies
1
are
the
d
absorb
For
energy
r
o
The
the
between
strength.
higher
analysed
environments
ΔE,
eld
the
this
hy d r o g e n
o
atoms
to
the
n
energy
dierence,
state
of
U
energy
states
o
external
spin
applied
C
energy
two
is
i
n
The
The
field
v
a
y
p
Figure 30
p
r
e
applied
y
P
훥 E
1
spin
to
H
nuclei
generating
measure, so an
this
signal.
integration trace
The
is
area under
added to the
1
the
relative number of
H
nuclei
represented
by
each signal.
297
Structure
3
Classic ation
of
matter
LHA
1
a
Cl
b
CH
CH
2
Consider the
H
two
the
an
signals
in
integration
NMR
spectrum
spectrum
trace
of
2,
of
chloroethane,
indic ating
and
one
with
two
an
shown
hydrogen
in
gure
32.
There
integration
trace of 3. These signals
3
a
b
therefore
correspond to the CH
and CH
2
groups
3
in
chloroethane,
respectively.
Figure
33
groups
shows the
with
spectrum.
This
identic al
is
and
NMR
spectrum
atoms
bec ause
in
the
therefore
the
of
molecule
form
a
2-bromopropane.
There
are
is
single
symmetric al,
hydrogen
so
the
two
methyl
an
integration
hydrogen
chemical
shift
trace
of
6
corresponds
to
these
methyl
groups,
as
1
atoms.
The
other
signal
corresponds
to
the
CH
they
have six
r
2
groups
environment. The signal with
3
3
three
molecule, but only two signals in the
e
are
2
H
hydrogen
s
s
1
4
are
environments, one with
group.
δ/ppm
l
a
c h l o ro e t h a n e,
C
2
H
H
5
NMR
spectrum
of
CH
CH
3
CH
1
3
2
1
v
area
under
U
hydrogen
i
interactions
t
a
The
radiation
These
u
l
a
R aman,
interactions
bending
the
c ow o r k e r,
The
his
R aman
Sir
K ariamanikkam
298
many
result
When
the
2
in
the
of
2 - b ro m o p ro p a n e,
a
For
and
the
4,
actual
or
3
H
7
Br
instance, imagine a
one
signal
correspond
so
C
1
ratio.
under
3
in
this
sc attering
to
for
sc attering
to
the
is
6,
or
H NMR
equal to 1, and the
relative number of
numbers
and
dierent
spin
radiation
of
R aman
is
of
any
hydrogen atoms
other
values in a
(NMR)
radiation
known
who
of
of
a
electromagnetic
we
or
have
bond
certain
looked at.
stretching
wavelength
sc attered.
incident
is
types
methods
changes
electromagnetic
was
and
as
sc attered
“R aman
awarded
radiation
are
sc attering” aer Sir
the
1930
Nobel
Physics
area.
eect
formed
determine
was
experiments,
and
spectroscopic
nuclear
the
Venkata
work
particles
of
phenomenon
used
R aman
as
area
These
be
wavelengths
This
nowadays
the
structural
very
R aman
basis
dicult
to
and
team
his
of
features
obtain
R aman
of
at
spectroscopy, which is
various
rst.
arrived
at
materials.
Evidence
Eventually, with
conclusive
evidence
h e re,
existence
of
the
eect.
his
Sriniv asa
How
Krishnan
(IR).
same.
for
the
2.
could
molecule,
Chandrasekhara
the
with
a
Occ asionally,
not
to
given
environment,
between
underpin
encounters
Prize
picture d
effect
equal
environments
TOK
and
v
E
V.
R aman
is
each
are
where
spectrum
o
d
r
o
f
x
O
C.
the
other
in
traces
signals
ratio.
ongoing
discovere d
the
atoms
dierent
1:2
two
NMR
n
in
integration
with
H
o
spectrum
1
The
C
i
n
Sometimes
δ / ppm
p
chemic al shift
p
Figure 33
6
y
r
e
4
O
t
i
s
b
p
Figure 34
3
b
Cl
n
y
a
The
much
evidence
is
required
in
order
to
support
a
robust
conclusion?
for
y
P
Br
a
1
p
Figure 32
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
LHA
Worked example 9
1
Sketch the
H
NMR spectra for the following molecules:
propanone, CH
b.
ethanoic acid, CH
3
C(O)CH
3
3
COOH
s
s
a.
Solution
Propanone, CH
The
structural
and
C(O)CH
two
is
a
symmetric al
shows
a
c arbonyl functional
methyl
3
,
groups.
The
ketone.
b.
Ethanoic acid, CH
in
two
the
hydrogen
atoms
is
O
H
signal
methy l
C
H
refle ct
the
21).
Integration
there
is
in
propanone
range of 2.2
only
traces
one
will
2.7 ppm
are
not
hydrogen
produce
(data booklet,
useful
here
environment. A
H
NMR
spectrum
is
shown
below:
formulas
E
v
COOH
the
two
organic
CH
number
1
two
NMR
environments,
spe ctrum
is
10
8
6
chemic al shift
4
2
0
–2
δ / ppm
0
compounds
are
COOH
3
H
3
H
of
H
of
should
shown
below.
b.
C
OH
Deduce
of
a
a.
H
l
structural
the
signals
1
the
the
re gion
u
f
x
O
The
2
in
and
the
δ / ppm
Practice questions
21.
3
of
the
in
n
4
under
protons
produce
o
5
i
6
chemic al shift
t
a
7
C
i
n
12
U
8
d
9
r
o
10
of
sketch
signal
protons
o
1
sketch of the
A
a
will
13.0 ppm,
below:
v
bec ause
protons
the
ratio
1:3.
are as
p
section
in
group
9.0
produce
The
four
O
signal
will
of
of
y
equivalent
c arboxy l
re gion
r
e
one
the
t
i
s
chemic al
environments
The
group
is
shown
H
identic al
of
the
2.5 ppm.
which
H
total
n
C
in
y
C
a
l
2.0
H
proton
identic al:
a
H
has
environments.
y
all
COOH,
P
for
3
chemic al
chemic al
The
environment
dierent
r
group
3
formula
e
a.
signals
the
ratio
3
C
3
H
and
CH
of
OH
areas
under
the
signals
in
the
1
H
NMR
spectra of these compounds.
299
Structure
3
Classic ation
of
matter
LHA
1
High-resolution
H NMR
Advances in technology have led to improvements in the accuracy of analytical
1
instruments like
H NMR. There are distinct dierences between low resolution and
1
high resolution
H NMR spectra and the information that you can extract from them.
s
s
1
We
have
already
information
The
demonstrated
about
number
of
signals
while
that
chemic al
corresponds
the
low
resolution
environment
to
integration
the
trace
of
H
NMR
hydrogen
number
shows
of
the
provides us with
atoms
dierent
number
in
a
molecule.
chemic al
of
e
environments,
the
hydrogen atoms
1
signals
are
split
each
into
atoms
in
a
clusters
about
the
In
a
of
high
resolution
peaks.
number
The
of
H
splitting
hydrogen
NMR
spectrum, individual
pattern
atoms
of
a
signal
provides us
attached to neighbouring
molecule.
H
The
signals
a
signal
spectrum
split
for the H
into
a
atom
b
is
into
a
peaks
and
cluster
of
(gure 35). The
is
referred to as a
three
peaks and is
a
noitargetni
noitargetni
7
two
=
=
n
o
i
u
l
a
E
v
O
f
x
t
a
8
of
2
1
C
i
n
U
d
r
o
9
1,1,2-trichloroethane
split
triplet.
b
6
5
4
chemic al shift
3
2
1
0
δ / ppm
1
p
Figure 35
are
dierent
c aused
discussed
with
or
The
patterns
previously,
against
close
other,
splitting
by the spin
the
proximity,
resulting
The
number
c an
be
of
in
external
the
high
resolution
of
in
proton
cluster
has
has
H
N
eld.
elds
for
NMR
spectrum
for
the
hydrogen
neighbouring
If
spin
This
that
hydrogen
generated
an
N + 1
of
nuclear
splitting.
by using the
x
signals
magnetic
signal
a
the
interactions
magnetic
NMR
peaks
c alculated
spin
each
If the proton H
of
is
by
c an
known as
individual
either
atoms
their
in
nuclei
spin
signal,
environments
hydrogen
a
nuclei.
be
molecule
c an
x
aect
N
spin coupling
known as
+
are
each
multiplicity,
rule:
will be split into
As
aligned
protons attached to its nearest neighbours,
the NMR signal of H
300
of
cluster
1 ,1 , 2 - t r i c h l o r o e t h a n e
The
in
the
is
o
referred to as a
in
atoms
v
doublet.
two
1 peaks.
:
b
b
y
for the H
H
a
Cl
p
are
signal
r
e
There
C
and H
n
C
a
O
t
i
s
H
Cl
a
l
Cl
y
Consider 1,1,2-trichloroethane, which has two hydrogen environments: H
y
information
signal.
P
with
with
r
associated
Structure
the
are
given
peak ratio,
is
in
table
deduced
16.
The
using
ratio
of
Pasc al’s
the
heights
of
the
peaks
in
a
Functional
groups:
Classic ation
Number of
Number
of peaks in
hydrogen atoms (N)
cluster (N + 1)
Splitting pattern
cluster,
singlet
doublet
(s)
2
3
triplet
3
4
quartet
1
(d)
1:1
(t)
1:2:1
(q)
1:3:3:1
1,1,2-trichloroethane, the spins of the nuclei in the H
The H
a
b
environment
doublet
a
environment
triplet
(t),
and
couple
the
with
has
+
two
peak
proton,
1
rule.
protons,
ratio
each
and
The
will
be
therefore the H
heights
and
of
the
signal
a
two
peaks
therefore the H
1:2:1.
The
b
signal
magnetic ally
In
to
new
form
the
is
the
row
row.
of
of
the
parts
high-resolution
pattern
triangle,
Evidence
dierent
role
in
of
exists
the
mathematics
in
known
adjacent
that
world
this
H
as
numbers
pattern
including
scientic
NMR
spectra
correspond
Pasc al’
s triangle
was
China
are
added together
known
and
by ancient
Persia.
n
What
in
observed
mathematic al
U
civilizations
each
nuclei
p
(gure36).
a
a
C
in
equal.
be split into
o
patterns
numbers
i
n
splitting
the
will
v
1
to
be split into
be
equivalent H
other.
TOK
The
will
will
y
not
one
N
r
e
do
has
(d) as per the
O
a
a
environment.
t
i
s
The H
b
n
in
environment will couple with the spins in the adjacent H
l
example,
NMR spectra.
y
For
H
more peaks are oen c alled “multiplets”
y
P
1
Splitting patterns due to neighbouring hydrogen atoms in
e
2
s
s
1
1
Peak ratio
r
0
Splitting patterns of ve and
organic compounds
triangle.
neighbouring
p
Table 16
of
LHA
Examples
3.2
knowledge-building?
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
etc
p
Figure 36
etc
P a s c a l ’s
etc
triangle
301
3
Classic ation
of
matter
LHA
Structure
Worked example 10
1
Show that
the splitting patterns in the signals of the
gure 37 correspond
H
to the structure of butanone, H
3
NMR spectrum in
CC(O)CH
2
CH
3
s
s
e
r
9
8
7
6
5
4
3
draw
the
structure
of
butanone
v
three
Hydrogen
the
to
Environment
B
i
in
The
A
0.
is
adjacent
to
not
adjacent
in
to
the
of
hydrogen
3
any
N
+
1
which
other
rule,
correspond to the
hydrogen
this
environment
spectrum.
B
environment
ratio
The
u
l
a
E
v
O
f
x
predicted
in
each
cluster
chemic al
values
from
shis
the
C
C,
which
has
three
hydrogen
corresponds to the quartet. The signal
is
split
into
corresponds
of
data
the
signals
booklet.
Hydrogen
a
triplet
by
the
two
hydrogen
to
that
predicted
by
Pasc al’
s
for B and C also closely match the
This
is
summarized in table 17:
Splitting
Peak
Predicted
Actual
pattern
ratio
δ / ppm
δ / ppm
1
2.2–2.7
2.1
1:3:3:1
2.2–2.7
2.4
1:2:1
0.9–1.0
1.1
N + 1 rule
environment
N =
0
A
singlet
0
+
1
=
N =
(s)
1
3
B
quartet (q)
3
+
1
=
N =
4
2
C
triplet
2
p
Table 17
Yo u
c an
+ 1
(t)
= 3
d i s p l ay
data
like
this
when
1
compound
302
to
environment
environment
number
C
environments,
According to the
singlet
CH
2
butanone
environment B.
peak
triangle.
=
o
is
corresponding
atoms
the
Therefore,
t
a
r
o
d
atoms.
hydrogen
n
U
corresponds
N
CH
B
spectrum.
environment
environments, so
C
A
dierent
in
the
o
are
signals
deduce
O
C
three
i
n
There
C
3
and
of
y
H
spectrum
p
r
e
environments.
NMR
O
Solution
First,
H
0
n
resolution
t
i
s
High
1
δ / ppm
1
p
Figure 37
2
y
chemic al shift
l
10
to
signals
in
a
H
NMR
spectrum
matching
the
s t r u c t u ra l
details
of
a
y
P
11
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
LHA
Combining analytic al techniques
(Structure 3.2.12)
To
determine
the
in
a
variety
types
of
the
of
data
It
is
the
of
a
molecule,
example,
group
you
dierent
present
combination
and
of
the
data
may
be
methods
required
are
arrangement
that
will
to
needed
enable
of
the
you
to
interpret a
to
deduce
c arbon atoms
deduce the
structure.
may include:
analysis
•
the
•
the
•
information
from the
•
description
of
spectroscopy
formula)
(MS) data
spectrum
1
H
NMR
spectrum
reactions
that
the
compound
c an
undergo
section of DP chemistry).
(covered in the
y
i
n
propan-1-ol, propan-2-ol and methoxyethane. Now you will examine these
C
o
v
In the skills task on page 283, you were asked to create molecular models of
p
Predict spectra from molecular models
O
t
i
s
data booklet
r
e
Reactivity
l
(IR)
empiric al
y
infrared
c alculate
y
mass
(to
n
•
data
r
combustion
P
•
e
This
For
functional
molecule.
unknown
structure
data.
s
s
wide
molecular models and consider the spectroscopic data of these substances.
Tool
2:
Identify
and
extract
data
from databases
n
•
U
Skills
Instructions
using
models
molecular
the
spectrum
of
each
patterns
the
done
determine
molecule.
number
of
so
the
and
methoxyethane
already.
number
of
different
1
Then,
predict the
signals,
area
H NMR
under
each signal,
shifts).
u
For each compound, predict some of the peaks found in its mass spectrum.
spectra
a
Spectral
l
the
compare
to
in
your
D atabase
E
v
O
AIST
for
spectrum
propan-2-ol
not
Predict
spectra
IR
and
each
chemic al
have
4.
Search
major
you
3.
5.
the
in
(consider
and
if
models
environments
f
x
splitting
molecular
propan-1-ol,
t
a
hydrogen
of
kit,
i
Examine
model
r
o
2.
molecular
a
o
Create
d
1.
a
signals
suitable
expected
database
for
and
each compound.
see
how the actual
predictions. Examples of databases include the
for
Organic
Compounds.
303
LHA
Structure
3
Classic ation
of
matter
Worked example 11
Ethane reacts with chlorine to form
Combustion analysis was used
product X.
to determine that
a.
Determine the empiric al formula of product X.
b.
The mass and
product
1
X has the following composition by mass:
H NMR spectra of product X are shown
c arbon
s
s
below. Deduce, giving your reasons, its structural
24.27%,
hydrogen 4.08%, chlorine 71.65%.
formula, and hence the name of the compound.
mass
spectrum
1
NMR
spectrum
e
H
100
r
10
20
30
40
50
60
70
80
90
100
110
10
percentages
by
the
relative
atomic
mass
for
=
2.021
hydrogen:
=
71.65
=
2.021
35.45
When
analysing
containing
abundant
chlorine-35
of
isotopes
and
of
dierent
chlorine-37.
product
f
x
X
will
H
4
Cl and C
2
H
4
.
the
We
c an
to
molecular
c an
elements
Therefore,
3
35
also
occur
the
2
.
[C
27
[C
H
2
]
+
Cl]
4
35
2
H
chlorine-37
and
deduce that the
in
We
304
formula of
c an conrm
1
H
NMR
spectroscopy.
The
H
NMR
a
mixture of
dierent
spectrum
The
hydrogen
has
two
signals,
indic ating two
integration
there
are
three
environment
environments.
trace
times
than
in
shows
as
The
a
signal
at
a
many
the
ratio
of
3:1,
indic ating
hydrogen atoms in one
other.
2.0 ppm is a doublet (H
neighbouring
hydrogen
proton.
a
At
5.8 ppm
neighbouring
A
),
indic ating
environment with one
there is a quartet (H
hydrogen
protons.
We
c an
environment
therefore
conclude
1,1-dichloroethane:
loss
chlorine-35
molecular
ion
atom
peak (m/z
3
A
Cl
from
100)
H
A
C
C
+
Cl]
4
H
of
loss
of
chlorine-35
ion
atom
peak (m/z
from
98)
H
p
Table 18
molecule:
1
molecular ion containing two
molecular
H
the
B
),
indic ating
chlorine-35
+
2
m/z
mass of the
structural
CHCl
formula
peak
molecular ion containing
+
Cl
4
E
63
2
0
with
three
isotopes
37
[C
ion
molecular
chlorine-35 isotopes
65
1
deduce that the
the
molecular
relative
(99).
structural
2
relative atomic
a
2
+
Cl]
H
the
formula
following
2
to
we
identic al
of
3
δ / ppm
Reason
l
[C
4
37
Cl
v
O
98
2
H
empiric al
is
average
equal
this using
Therefore, the
contain
is
4
product X is likely to be CH
u
35
[C
2
need to consider that chlorine
fragment
100
Cl
chlorine-37 atoms (table 18).
Molecular
m/z
4
t
a
and
H
i
ions
chlorine-35
2
spectra of compounds
we
r
o
fragment
therefore C
values
C
o
two
mass:
mass
chlorine,
is
d
has
formula
the
n
b.
empiric al
U
The
bec ause
5
spectrum,
formula
4.04
1.01
chlorine:
mass
C
12.01
the
empiric al
6
chemic al shift
o
4.08
From
v
c arbon:
i
n
24.27
7
p
the
each element.
8
y
Solution
Divide
9
r
e
m/z
n
0
O
t
i
s
20
a.
l
40
]
F ra g m e n t
loss of HCl and Cl
ions
of
product
X
A
Cl
H
B
that
product X is
y
60
y
evitaler
P
ytisnetni
80
Structure
3.2
Functional
groups:
Classic ation
of
organic compounds
End of topic questions
Which
of
the
following
analytic al
LHA
4.
techniques would
Topic review
1.
Using
your
answer
the
knowledge
guiding
from the
question
as
Structure 3.2
fully
as
topic,
CH
3
the
dierence
CH(OH)CH
3
,
between
and
propan-2-ol,
propanal, CH
3
CH
2
mass
II.
infrared
e
I.
to predict their properties?
r
1
III.
H NMR
I and II only
Multiple-choice questions
name
of
the
I and III only
II and III only
D.
I, II and III
rules?
5.
H
Which
C
2
H
H
3
2-ethylpent-3-ene
D.
3-methylhex-4-ene
Which
of
the
3
3
3
COOCH
CH
CH
2
2
CH
CH
2
2
the
same homologous
3
i
CH
COCH
in
OH
OCH
3
H
NMR
spectrum?
4
3
2
1
0
PPM
A.
propane
B.
propanal
C.
ethyl ethanoate
D.
butanoic acid
Extended-response questions
6.
a.
State,
giving
a
reason,
if
but-1-ene
exhibits
cis–trans
isomerism.
b.
The
a
the
a
[1]
reaction
hydrogen
which
iodide
at
occurs
room
halogenoalkane,
reason,
if
LHA
u
l
E
v
O
f
x
t
a
CH
D.
3
is
?
r
o
C.
3
3
o
CH
CH
OCH
d
A.
B.
following
3
n
series as CH
5
following
o
C.
C
4-ethylpent-2-ene
v
4-methylhex-2-ene
B.
i
n
A.
p
H
U
3.
CH
C
3
the
y
C
C
1
has
r
e
HC
substance
O
3
t
i
s
CH
n
B.
C.
following compound, applying
l
IUPAC
the
y
is
y
A.
P
Exam-style questions
What
CHO?
possible:
How does the classication of organic molecules help us
2.
s
s
show
the
2-iodobutane.
product
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State, giving
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305
Structure
3
Classic ation
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306
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[3]
307
To o l s
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mass
example
The
value,
than
a
p
in
balance
a
four
In
number
y
Another
to
the
r
e
decrease
down
recorded
precise
precision.
0.5 g).
Gravimetric analyses
chemic al
writing
mass
of
greater
y
with
When
a
the
O
place
to.
levels
balance,
t
i
s
measured
mass
example,
dierent
a
n
For
of
paper.
l
measure
zeroes.
have
precision
y
will
balances
the
are
values
P
greater
pH
valid?
dierence.
Dierent
pH
experimental outcomes be
reliable
electric potential
pH
precise
determined
5.
mass,
measurements
e
volume,
will
teams
the
out
s
s
when
c arried
Comparing
research
One
but
photo
was
balance?
really
massless?
313
Tools
for Chemistry
Volume
The
its
suitability
intended
designed
to
of
a
particular
function
and
measure
chemic al
reaction.
asks
needed.
the
volume
approximate
involve
For
and
are
the
and
level
volume
for
a
asks
holding
equipment depends on
precision.
are
analyses,
conic al
designed
of
measurement
volumetric
Beakers
volume-measuring
not
of
very
Measuring
cylinders
precise.
volume changes during a
pipette,
give
an
liquids,
burette
and/or
approximate
not
volumetric
measurement
measuring
e
of
are
of
desired
s
s
Volumetric analyses
type
the
volume.
r
– burette,
including
vessel,
measurements
such
should
a
be
solutions,
burette
read
or
from
the
the
form
neck
of
bottom
a
a
U-shaped
volumetric
of
the
ask.
meniscus
o
i
u
l
a
v
E
Figure 10 Water and
form
314
aqueous solutions
a U-shaped meniscus

Figure 11 Mercury forms an upside-down meniscus,
where the centre bulges upwards above the sides
inside
Volume
(gure 10).
tap

B – volumetric
meniscus
t
a
r
o
f
x
O
burette
aqueous
n
U
d
read from the bottom
of the meniscus
as
C
liquids,
narrow
D – volumetric ask
i
n
M any
C
D
A – measuring cylinder,
o
pipette,
a
C
Figure 9 Various volume-measurement devices.
n
l
volumetric pipette into a conic al ask
B
O

y
from a
p
being released
v
Figure 8 A liquid
y
P
y
t
i
s
r
e
A

Tool
1:
Experimental
techniques
Practice questions
8.
Which
of
the
following
volumetric
•
volumetric ask
appropriate
for
precise
volume
measurements?
pipette
s
s
•
are
cm
100
•
burette
•
beaker
•
conic al ask
•
measuring
90
80
e
70
60
60
r
cylinder
50
P
What
glass
container
would
you use to:
30
determine
b.
prepare
c.
c arry
the
volume
of
liquid
added
during
a
titration
l
a.
50
20
you
need
for
from
a
pH
a
solid
reagent
experiment
from
10
the
reagent
bottle

10.
approximately
measure
25 cm
at
of liquid?
Figure 12 The meniscus should
eye-level
be read
What is the volume of liquid contained in the measuring cylinder in gure12?
for
time
is
the
example,
measurements
measured.
certain
The
volume
uncertainty
arising
from
of
usually
of
the
gas
a
to
be
depends
the
measuring
stopwatch
to
the
the
rate
record
produced.
on
The
of
a
chemic al
how long it takes, in
uncertainty
associated
context in which time is being
stopwatch itself is oen far smaller than the
experimenter ’
s
reaction
time.
n
U
uncertainty
a
When
use
C
for
with
(s).
might
i
n
seconds,
time
second
you
o
for
y
unit
v
SI
reaction,
p
r
e
Time
The
O
your lab bench
3
d.
n
acid
solution
y
the
standard
t
i
s
to
a
y
40
9.
For example, if you are recording how long it takes for a clock reaction to produce
a change in colour, there will be a short delay between the instant the colour
o
d
changes and your nger stopping the stopwatch (gure 13). The stopwatch display
might indicate a time of 16.45 seconds, which might suggest an uncertainty of
i
r
o
± 0.01 s indicated by the two-decimal-places display (gure 13). A more realistic
u
l
f
x
ATL
Figure 13 In the “iodine clock”
Thinking skills
Some
a
E
v
O

t
a
uncertainty, given your reaction time, would be around 0.5 to 1second.
over
as
hours,
these
to
experiments take place
long
periods
days,
c ases,
record
is
or
it
times
of
time, such
even
always
to
the
weeks. In
necessary
nearest
second?
reaction,
the clear and
colourless reaction mixture
suddenly turns blue-black aer a certain period of time
315
Tools
for Chemistry
Temperature
You
c an
use
required
digital
accuracy
dependent
on
analogue
the
context
monitoring
temperature
probe
to
thermometers
precision
a
of
of
of
the
a
measure
temperature. The
measurement is highly
experiment.
temperature
data
to
temperature
logger
as
c an
be
shown
achieved
in
by
connecting a
gure 14
e
interface
r
O
dissolving in water
n
l
y
t
i
s
ammonium chloride
temperature
sensor
y
p
r
e
o
v
i
n
Figure 14 Temperature sensors and
data loggers c an be used to
C
monitor the temperature changes over time
U
Length
gures
are
measurements
discussed in
in
Tool 3: Mathematics.
involves
loc ation
the
be
usually
used.
estimating
between
on
i
impact
u
2
l
a
v
Figure 15
usually
also
two
number
As
with
the
sc ale
of
involve
last
rulers although other devices, such as
other
digit
divisions.
signic ant
analogue
in
the
Sc ale
gures
devices,
resolution
in
reading
the
sc ale
measurement, depending on its
the
will
therefore
measurement
have an
(gure 15).
t
a
r
o
f
x
O
E
316
1
This ruler gives a

d
detail
c an
o
c alipers,
n
Length
Signic ant
3
4
1
1.7 cm
reading to two significant figures
The relationship
between sc ale resolution and
2
3
4
1.75 cm
This ruler gives a
signic ant
gures
reading to three significant figures
y
P
.pmet
time

s
s
Continuous
or
and
Tool
1:
Experimental
techniques
Practice questions
11.
A
student
conducted
concentrations
results
and
obtained,
in
trials
of
minutes,
an
iodine
were
seconds
the
and
clock
same
reaction.
in
all
The
reagent
three trials. The
milliseconds
(min:s.ms)
are
s
s
shown
three
temperatures
below.
State
the
b.
Suggest
times
a
to
the
suitable
nearest
second.
uncertainty
for
the
time
measurements, in
seconds.
C alculate
the
mean
seconds.
for
the
colour
change
the
limitations
and
of
each
propose
an
of
the
following
3
A
student
heats
temperature
student
from
the
wishing
down
ice
to
bath,
to
of
the
and
a
beaker
in
and
the
checks the
water, with the bulb
beaker.
perform
this
in
thermometer
a
reaction
temperature
adds
the
in
a
at
2 °C cools one of the
beaker,
second
removes
beaker
reagent.
U
c.
the
C
reagents
bottom
water
a
o
A
of
i
n
b.
the
5 cm
placing
v
touching
by
p
a.
temperature
alternative method.
y
scenarios
r
e
measurement
n
01:12.985
explain
O
01:15.120
01:10.400
t
i
s
1
and
l
Mean / s
1 ms
2
3
Identify
occur
y
/ s
±
12.
to
/ min:s.ms
y
taken
in
P
Time
Trial
time,
r
c.
e
a.
In a study of an enzyme action at 40 °C, a student places a test tube
n
containing a sample of the enzyme solution into a 40 °C water bath. The
student then immediately adds the substrate to the enzyme solution.
is
16
orange bar?
a
14
device
you
15
c alled
a
16
c aliper.
u
might
obtain
l
using
a
17
What
18
19
20
chemistry-related
c aliper?
a
E
v
O

13
shows
measurements
the
i
Figure
of
t
a
12
f
x
14.
length
r
o
11
the
o
What
d
13.
Figure 16 A vernier c aliper
317
Tools
for Chemistry
pH of solution
The
pH
of
solution.
c an
be
is
related
sc ale.
be
estimated
pH
of
probe
a
The
red
colour
concentration
indic ators,
c abbage.
shown
is
of
such
These
hydrogen ions in that
as
universal
indic ators
compared
to
a
indic ator, or the
vary
key,
in
colour
allowing
a
(gure 17).
solution
c an
be
more
reliably
determined
with
a
properly
(gure 18).
paper turns dierent
r
colours depending on the pH of the

substance being tested
across
pH
value
c alibrated
e
pH
pH
from
the
with
s
s
the
entire
to
estimated
extracted
The
Figure 17 pH
solution
anthocyanins
to

a
pH


l




t
i
s




Figure 18 A pH
use
the
indic ator
juice
c abbage as a
anthocyanins
(HA),
which
U
amphiprotic
Materials
•
red
•
two
are
2
A
.
In
basic
accept
with
protons
form anions, A
high
:
+
and
form
•
c ations,
H
H
vinegar or
neutral
visible
producing
Relevant skills
•
Tool
•
Inquiry 1: Designing
Measuring
at
solid
household
products, such as baking soda or
•
five
•
cloth or filter paper
•
standard
clear
glasses
laboratory
glassware
Method
1.
Prepare
the
indic ator
by
blending
several
leaves
anionic) of
different
of
red
c abbage
with
a
glass
of
water and filter the
wavelengths
resulting
mixture
through
cloth
or
paper.
characteristic colour
a
c abbage
2.
juice.
v
O
red
and
light
the
two
anionic form
u
(c ationic,
absorb
1.3),
1:
as
(green to yellow)
l
f
x
forms
anthocyanins
of
such
+
H
form
(blue)
(Structure
products,
A
+
neutral
t
a
+
(red)
changes
r
o
+
three
+
HA
A
2
cationic form
All
household
detergent
antacid tablets
anthocyanins lose
i
H
+
H
pH,
o
and
solutions
d
protons
liquid
organic compounds. In acidic solutions with
anthocyanins
+
H
c abbage
liquid
n
pH,
red
products.
c abbage juice is rich in
low
of
for determining the pH of
y
will
C
household
Red
you
acid–base
p
task,
to the colour of the universal indic ator
o
this
natural
i
n
In
probe reading compared
v
Using a natural acid–base indic ator
r
e


O

To
test
one
of
your
of
the
liquid
household
products, pour
3
20–30 cm
filtered
c abbage
juice
into
and
add
2–3 cm
of
the
E
gloves
and
3.
To
test
one
of
the
solid
and
the
c abbage
eye
Compare
your
solid
dissolve
it
household
in
water,
products,
318
Do
not
use
products
colours
of
your solutions with
protection
with
a
corrosive
safety
warning
in
figure19 to determine their
approximate pH.
crush
before mixing with
juice.
the
thoseshown
•
clear
analysed solution.
variables
4.
Wear
a
3
glass
S afety
•
n
y

y
P

Tool
1:
Experimental
techniques
s
s
e
r
l
Figure19 Red
c abbage juice at pH 1 (le) to 10 (right)
What
18?
the
pH
might
values
a
of
suitable
indic ators
advantage
b.
pH
probes.
substances
and
one
be
in
each
the
a
conductive
liquid
made
ensure
on
the
of
Electric al conductivity,
is
a
related
be
used
E
measured
property,
directly
to
monitor
produced
bec ause
identity
mobile
of
such
and
charge
the
as
not
so
be
in
A
amperes
connected in
where all the components
path,
the
with
no
ammeter
branches.
c an
be
solid to be tested
placed
circuit 2
switch
samples in a series
solid
into
known
a
circuit is usually
touch
as
electrode
materials
the
A
as
graphite
the
within
one
measured
must
leads to either end of the solid. Inserting
such
do
is
liquid
a
is
electrolytes)
or
each
surface
platinum.
other.
area,
involved.
circuit,
requires
C are must be
distance
These
known
as
beaker
between the
factors
all
aect the
carbon
electrodes
resistance.
oen
solution
a
given
in
microsiemens
per
centimetre,
µS cm
to be
,
tested
is
the
probe.
ability
to
Electric al
conduct
charge
conductivity
and
which
measurements
c an be
c an

the
progress
conductivity
ions
switch
1
which
with
and
It
ammeter
single
circuit,
(collectively
electrodes
factors
a
Inserting
materials,
applied
ow
solid
pack.
a
to
the
series
connect
solution
v
O
opposition
battery
inert
that
voltage
connect
simply
or
of
depends
electrodes,
a
l
Current
a
you
to
circuit
following
u
to
by
f
x
electrodes
taken
how
of
charge.
the
t
a
powered
straightforward:
throughout
electric al
current,
type
another,
circuit.
shows
of
o
the
simplest
aer
same
ow
measure
r
o
circuit
20
in
The
one
of
i
Figure
the
To
d
anywhere
rate
n
is
the
circuit.
connected
Current
is
ammeter.
circuit 1
o
in
an
U
series
are
current
using
gures 17 and
C
i
n
(A),
in
disadvantage of
Electric current
Electric
tested
c ase?
y
pH
the
p
one
a.
of
uncertainty
v
O utline
each
r
e
16.
are
What
O
15.
n
y
t
i
s
Practice questions
y
P

present.
of
of
a
a
reaction
solution
in
which
depends
ions
on
are
the
consumed or
concentration and
Figure 20
series.
solid
Circuit
Ammeters are connected in
1 shows how to connect a
sample while circuit
2 shows how to
connect a solution
319
Tools
for Chemistry
Electric potential dierence
Potential difference
in
a
circuit.
measure
It
the
6
8
potential
in
In
chemistry,
in
the
is
the
expressed
parallel
you
volts
dierence
to
will
energy
in
the
most
dierence
(V)
and
across
circuit
component
likely
per
unit
measured
a
measured
potential
between two points
voltmeter.
components
being
encounter
charge
using
and
are
Voltmeters
therefore
s
s
connected
4
2
0
is
(gure 21).
dierence
measurements
10
V
context of
the
known
to
as
the
cell
potential
potential.
two
half-cells
dierence
Measuring
and
cell
completing
between the anode and
potential
the
involves
circuit
with
Figure 21 Voltmeters are connected
shown
in parallel to a component
in
a
connecting
salt bridge as
r

is
voltmeter
The
e
c athode
voltaic cells.
gure 22.
to measure
V
potential dierence of the lamp is being
voltmeter
l
measured)
salt bridge
how
current
passes
through solid materials, such
as
a
copper
wire
or
graphite
O utline
how
through
current
passes
electrolytes, such as
solution containing
solution containing
lead(II)
v
molten
bromide or
2+
A
sodium
chloride.
i
n

(aq)
(aq)
Figure 22 The potential dierence across the two half-cells is measured
Another
term
you
value
This
reason
is
the
potential.
Their
why
high
the
come
potential
across
high-resistance
resistance
cell
is
dierence
i
o
hence
potential
means
electromotive
across
a
cell
voltmeters
that
measured
are
virtually
is
close
force
when
used
no
to
(emf ), which is the
no
to
current
current
the
is
owing.
measure cell
ows
through
emf.
Practice questions
19.
Explain
the
u
l
a
E
v
O
f
x
t
a
r
o
d
them,
might
of
by connecting a
B represent metals
n
U
maximum
A and
C
voltmeter to the electrodes.
320
2+
B
o
aqueous
y
18.
p
r
e
electrode.
O
O utline
t
i
s
17
.
B
n
y
A
Practice questions
a.
In
the
AgNO
this
mass,
For
3
+
+
you
volume,
each
NaOH(aq)
(aq)
section
involved
following
mixtures
undergo a change in conductivity as
proceeds.
HCl(aq)
b.
20.
why
reaction
NaBr(aq)
have
time,
variable,
→
→
of
that
an
+
AgBr(s)
considered
temperature,
think
measuring
NaCl(aq)
+
how
pH,
H
and
O(l)
NaNO
to
you
state
3
(aq)
measure
current
experiment
variable,
2
and
key
types
potential
of
variables:
dierence.
have done in the past which
the
instrument
used.
y
P
its potential dierence (in this c ase, the
Tool
1:
Experimental
techniques
Applying experimental techniques
In
this
and
section,
analysing
we
will
discuss
chemic al
experimental
techniques
for
preparing, isolating
species and phenomena.
s
s
Preparation
Preparation of a standard solution
volumes
dilution
high
of
standard
precision.
volumetric glassware,
This
such
is
as
solutions
requires
achieved
burettes,
by
measuring
using
pipettes
an
masses and
analytic al balance
and
r
and
and
with
volumetric asks
(gure23).
The
solid
volume
with
a
solution
4.
The
beaker,
each
rod
water
solution
formula
of
the
or
is
is
added
of
and
who
the
volumetric
funnel
into
to
are
the
the
to
ask
beaker
rinsed
the
beaker,
and
over
at
using
three
a
to
a
reagent
substance,
the
its
bottle
mixture is

glass funnel.
times
ask
ask
least
the
completely.
volumetric
volumetric
turned
prepared
to
with
deionized
until
ten
with
water,
ensure that the solute
its
times
a
Figure23
level
to
label
reaches the
Volumetric ask and
pipette
ensure that the
showing the
concentration, the date and initials
solution,
as
well
as
any
hazard labels, if
o
i
d
f
x
t
a
r
o
stirring
rod
paper
u
wedge
stirring
rod
l
wash
bottle
18/08/2023,
P.M.D., 1.51
–3
mol
a
O
1
2
3
4
5

dm

v
Figure 24 Preparation of a standard solution
E
Measuring
for
added
dissolves
thoroughly.
relevant.

a
is
solid
on the ask.
transferred
name
person
to
water
stoppered
mixed
dry
n
The
and
this
water
the
U
7
.
is
is
and
C
ask
solution
until
i
n
The
clean
completely.
graduation mark
6.
a
o
Deionized
in
v
5.
glass
transferred
rod
transferred
adding
weighed
p
is
time
deionized
glass
is
is
y
The
solute
balance.
r
e
3.
of
liquid)
follows:
n
small
stirred
sometimes
as
O
A
(or
analytic al
prepared
t
i
s
2.
an
usually
y
using
is
l
1.
solution
y
standard
P
A
e
Preparation
cylinders
preparing
volumes
sample
with
with
or
and
diluting
sucient
at
least
beakers
standard
precision.
four
with
graduation
solutions,
Similarly,
signic ant
as
the
marks
they
do
balance
should
not
never
allow
must
be
used
measurement of
show
the
mass of a
gures.
321
Tools
for Chemistry
Practice questions
3
22.
21.
These
steps
outline
how
to
prepare
a
A
student
solution
solution.
List
them
in
the
correct
prepares
100 cm
3
of
a
0.50 mol dm
standard
of
copper(II)
sulfate
by
c arrying out the
order.
following steps. Identify the mistakes.
Add
deionized
s
s
A.
water up to the mark on the
A.
A
sample
of
7.98 g
of
hydrated copper(II) sulfate
volumetric ask.
was
B.
Dissolve
the
solute
in
a
small
amount
of
weighed
and
transferred
into
a
volumetric
deionized
ask using a funnel.
in
a
beaker.
B.
the
solution
to
a
labelled
reagent
Tap
it
for
was
solution
into
a
water
volumetric ask.
to
mark
the
on
volumetric ask until
the
desired
mass
of
the
was
neck
beaker,
invert
and
the
the
mark.
ask.
stirring
until
the
meniscus
rod with
C.
The
D.
A
ask
was
label
several times.
for
23.
was
stoppered
placed
on
water
was
c arefully
was back at the mark.
and
the
inverted
hydroxide
air
property
sodium
is
hygroscopic,
moisture.
for
the
Suggest
preparation
hydroxide.
the
meaning that it
implic ations of this
of
standard solutions of
o
v
p
y
r
e
absorbs
twice.
volumetric ask
storage.
Sodium
At
added, so the meniscus
Excess
solute.
volumetric ask.
ask
the
Too much
O
and
funnel
over
the
t
i
s
Stopper
water
accidentally
of
dropwise.
n
the
deionized
added
y
Rinse
was
l
the
water
above
removed
G.
added
the
Preparation of solutions by dilution
dilute
burette
water
is
a
standard
or
added
U
several
solution,
volumetric
times
to
to
the
certain
and
graduation
achieve
a
volume
transferred
mark,
and
of
the
to
the
a
solution
is
volumetric
stoppered
measured using
ask.
ask
is
Deionized
turned
over
uniform composition of the nal solution.
n
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
bottle

Figure 25
A
The solution in bottle B c an be prepared
stock solution in bottle A
322
a
pipette
C
a
i
n
To
bottle
B
by performing a simple dilution of the
y
the
Weigh
F.
to
P
Pour
point
went
E.
was
close
storage.
this
D.
water
bottle
r
Transfer
e
water
C.
Tool
1:
Experimental
techniques
Worked example 1
3
3
A teacher wishes to prepare 0.500 dm
of 0.400 mol dm
HCl for a class
3
practic al.
A 11.0 mol dm
HCl stock solution is available. Determine the
volume of stock solution required
to prepare the desired solution.
s
s
Solution
We
know that:
×
V
1
1
c
indic ates
V
stock
known
solution
values
and
2
indic ates
3
(11.0
mol dm
the
desired solution.
gives:
3
3
)
×
V
=
(0.400 mol dm
)
×
(0.500 dm
)
l
1
V
gives:
V
=
1
3
3
0.400 mol dm
×
0.500 dm
3
11.0
mol dm
3
V
=
0.01818182… dm
1
3
to
cm
,
which
are
more
easily
measured
in
a
lab,
and
round
3
=
18.1818182…
3
cm
≈
18.2 cm
(3
sf )
1
as
extremely
c arry
times
out
more
a
accurately
low
serial
dilute
for
prepare
c alibration
concentrations
dilution
than
that
where
in
the
are
the
prepared
water.
This
1 cm
of
procedure
is
often
previous
the
required.
one.
in
each
The
Figure
solution
until
solution
the
and
desired
3
+9 cm
in
1
C alibration
in
are
introduced
1.4
shows
final
9 cm
of
Practice questions
concentration
24.
A
technician
the
wishes
following
to
solutions
prepare
by
3
diluting
HCl
3
a
5.00 mol dm
solution.
What
stock
volume of
cm
the
stock
each
3
+9 cm
curves
Structure
each test tube
solution
is
needed in
c ase?
3
3
a.
+9 cm
750 cm
u
HCl
3
of
1.00 mol dm
(aq)
3
b.
3.00 dm
of
3
0.150 mol dm
a
25.
Suppose
of
the
that
stock
the
HCl
(aq)
concentration
solution
in
gure
3
26 is
1
0.1
0.01
0.001
10
10
10
10
of
1
(
100
1
(
1000
0.50 mol dm
. Determine
3
the
(
v
E

l
f
x
O
stock
cm
3
+9 cm
solution
3
1
26
adding
o
3
cm
t
a
r
o
1
i
d
3
cm
increasingly
successive tube is
obtained.
1
of
prepared using this
3
preceding
repeated
solutions
often
concentration,
each
of
the
in
mol dm
,
diluted solutions.
1
(
is
transferring
of
are
n
deionized
by
series
solution
3
is
a
curves
p
to
to
Solutions
U
ten
used
C
method,
how
are
concentration.
o
dilutions
i
n
lower
v
Serial
three
y
gures:
V
to
r
e
signic ant
(
10000
(
Convert
O
t
i
s
1
n
for
y
Solving
y
the
2
the
P
Substituting
×
2
r
where
=
1
e
c
Figure 26 A serial dilution
323
Tools
for Chemistry
Reux and distillation
Reux
and
(gure
is
a
A
used
piece
condenser
a
we
in
will
equipment
essentially
ow
the
both
reaction
of
inner
cold
the
known as a
tube
within
As
a
they
very
whereas
between
similar
is
used
The
is
in
them
to
a
of
c ases.
tube
the
condense.
to
Reux
separation
both
outer
temperature
granules
equipment
purposes.
distillation
tube.
the
c ausing
involve
dierent
condenser
another
anti-bumping
dierences
for
result,
decreases,
use
is
cooled
gaseous
You will also
ensure smooth boiling.
reux and distillation.
(b)
l
thermometer
y
C
purified product
heat
i
Figure 27 (a) Reux apparatus (b) simple distillation apparatus
is
heating
u
l
a
E
v
O
f
x
t
a
Reflux
that
any
into
the
used
of
a
boiling
ask
still
is
of
liquid
27(b)
a
to
is
The
in
that
shown
on
volatile substances during continuous
reaction
the
is
placed
mixture
components
until
substance
the
the
the
with
condenser,
temperature
above
condense
being
the
reaction ask, so
and
drip
back
down
two
solution
of
a
mixture
point
according to their
of
the
more
volatile
lower boiling point rises into
and
is
collected
at
the
bottom of the
stays constant during state changes, so
will
distilled
liquids
remain constant at the boiling
o.
simple distillation,
(or
of
boiling
the
thermometer
is
solid
aqueous
and
of
condenser
heated
substance
non-volatile
sulfate
the
separate
mixture
condenses
an
loss
A
esc aping.
above depicts
example,
copper(II)
of
Remember
whichever
from
from
reached.
temperature
point
For
The
the
mixture.
rising
used
is
head,
condenser.
the
minimize
instead
points.
component
the
to
reaction
vapours
Distillation
Figure
324
anti-bumping granules
o
d
r
o

water in
impure product
n
granules
o
v
i
n
anti-bumping
condenser
heat
U
reaction mixture
water out
p
r
e
water in
O
t
i
s
out
n
y
water
condenser
which
with
copper(II)
water using this method.
separates
very
sulfate
dierent
c an
be
a
volatile
boiling
points).
separated into
y
P
(a)
bec ause
used
mixtures,
water.
tube
techniques
discuss
a
are
r
Next,
that
heat
confused
methods
e
substances
of
is
continuous
notice
to
easily
these
s
s
The
are
However,
method
method.
by
distillation
27).
Tool
Fractional distillation
is
a
related
technique,
used
to
separate
1:
Experimental
techniques
mixtures of two
Practice questions
or
more
as
an
miscible
and
ethanol–water
mixture
is
column,
heated,
volatile
mixture.
the
condense
liquids
A
the
similar
fractionating
substances
on
with
glass
in
the
beads
column
mixture
and
boiling
drip
is
points
tted
continually
back
to
(gure 28), such
the
ask.
As the
26.
vaporize, rise into the
cycle
repeats
many
times.
The
vapours
the
lower-boiling-point
column
and
a
greater
Compare
collect
bath,
each
the
component,
condenser.
particularly
if
the
The
which
eventually
receiving
distillate
is
ask
volatile.
A
is
and
fractional distillation.
reaches the top of the
sometimes
fresh
28.
placed in an
receiver
is
Primary alcohols such
as
used to
ethanol
by
distillate.
are
acidied
distillation
is
used
air.
to
separate
o
fractional
liquid
i
and
d
29)
r
o
(gure
y
Figure 28 Fractional distillation apparatus
Industrial-sc ale
C

if
you
obtain
(b)
a
(a)
technique,
reux,
you would
were looking to
an
aldehyde and
c arboxylic acid.
n
U
heat
use
which
or
y
receiver
oxidized
o
i
n
v
distillation flask
outline
distillation
p
r
e
cold
water in
further
n
condenser
large surface area
are
O
t
i
s
column with a
aldehydes.
there is
c arboxylic acids. Identify
and
water out
to
l
y
to
if
oxidizing agent, the
aldehydes
granules
r
enough
thermometer
anti-bumping
rst
and
P
Eventually,
oxidized
potassium
dichromate,
fractionating
contrast simple
proportion
e
ice-water
enters
reux
down into the distillation
contain
and
of
contrast
s
s
The
and
and simple distillation.
27.
ask.
Compare
mixtures
such
as
crude oil
Figure 29 An industrial fractional
distillation facility,
used
to separate the
components of crude oil
u
l
a
E
v
O
f
x
t
a
t
325
Tools
for Chemistry
Isolation
Drying to a constant mass
When a solid is obtained in the lab, it may contain trace amounts of water, even if it
dried to a constant mass
to ensure that they do
s
s
looks dry. Solid samples are oen
not contain any water (or other volatile impurities). Drying to a constant mass involves:
allowing
3.
weighing it
4.
repeating
cool
for
the
until
method
drying
two
should
Common
be
mass
before
the
2
when
of
aer
hydration
the
salt
to
a
hydrated
heating:
heat
O(s)
xH
determining
magnesium
of
remove
p
why,
sulfate
water
be
obtained.
desired solid
Bunsen
burners,
of
2
the
O(g)
salt, such as
water
and
+ MgSO
4
measuring
(s)
hydration of magnesium
heated
to
a
constant
mass.
n
U
Separation of mixtures
the
should
C
i
n
sulfate,
•xH
water
heating
o
Explain
4
and
v
MgSO
by
the
includes
y
the
the
sulfate,
that
are
O
determine
masses
ensure
equipment
t
i
s
c an
magnesium
to
equal
ovens.
r
e
You
chosen
heating
Practice question
29.
consecutive
n
and
few minutes
l
heating
process
decompose.
hotplates
a
y
not
to
r
suitable
does
it
sample
The method used to separate a mixture depends on the properties of the mixture
components. For example, if one of the components is magnetic, it can be removed
o
crystallization or simple distillation. Mixtures of liquids can be separated with
i
fractional distillation or using a piece of equipment known as a separating funnel.
t
a
Filtration
medium,
used
to
u
usually
the
l
a
v
Figure 30 Gravity ltration
E
326
d
r
o
f
x
O

using a magnet. Mixtures of solids and liquids can be separated using ltration,
In
the
as
separate
much
lter
and
separates
such
paper.
liquid
paper
due
The
that
to
paper,
insoluble
smaller
than
solid
(gure
gravity.
according
which
solids
solid
le
passes
gravity filtration
the
particles
lter
to
from
behind
in
as
its
so
the
sizes
tiny
liquids.
particles,
through is the
30),
their
contains
Liquid
they
lter
by
holes
or
easily
paper
passing
or
them
pores.
aqueous
pass
is
through a
Filtration is oen
particles
are
through the holes in
known as the
residue
filtrate.
name
suggests,
the
ltrate
passes
through
y
2.
A
the
e
heating
P
1.
Tool
This
process
involves
c an
folding
be
very
the
lter
slow,
so
paper
it
is
into
sped
an
up
by
fluting
accordion
shape
the
lter
(gure
1:
Experimental
techniques
paper. Fluting
31)
before placing
it inside the funnel.
s
s
fold to crease the
this is the
open filter
quarter sections
eighth sections
“accordion” shape
cone
Figure 31 Figure showing the steps for making a uted
funnel
to
usually
a
the
three
pump
solid
or
four
mixture.
on
the
perforated
draws
(gure
from
the
32).
slipping
times
and
the
with
liquid
C are
it
is
oen
mixture
plate
used when the
is
loaded onto a
inside
a
Buchner funnel.
through the lter paper and into
must
under
the
The
the
solvent
be
taken
lter
to
not
paper.
to
The
Buchner
residue is
funnel
remove soluble impurities.
out of a solution by removing the solvent. Most of
the solvent is evaporated o by heating it over boiling water until the solution
p
flask
v
drop of it onto a cold tile and watching for the formation of crystals as it cools. The
i
n
evaporating
r
o
boiling
burner
l
u
f
x
a
O
ltration apparatus
solution from
evaporating basin
cold tile
leave for a
few
days to crystallize
Figure 33 Crystallization apparatus
E
v

Bunsen
t
a
water
Figure 32 Vacuum
o
i
d
basin

n
U
solution
C
the crystals are allowed to dry to a constant mass on a watchglass.
o
bulk of the remaining solution is le to cool slowly to allow crystals to form. Finally,
any remaining solvent is removed from the crystals by ltration (if necessary), and
vacuum pump
conical
becomes saturated. The solution c an be checked for saturation by spotting a
copper(II) sulfate
filter paper
overll the
y
crystallized
at
ltration,
of
r
e
A solid can be
laid
underneath
prevent
rinsed
paper
vacuum
ask
gravity
O
receiving
lter
than
component
t
i
s
a
moist
from
quicker
y
Suction
is
desired
n
of
the
y
sheet
is
l
residue
cone from circular lter paper
P
Vacuum filtration
r

e
fold to crease the
327
Tools
for Chemistry
A
solvent
c an
evaporation,
the
also
be
which
is
removed
a
type
of
from
a
solute
distillation
through
performed
simple
at
low
distillation,
pressures
or
to
solvent’
s boiling point.
s
s
Practice questions
30.
Figure
base.
34
shows
Copper(II)
the
preparation
sulfate
crystals
and
c an
isolation
be
made
of
salt
this
crystals
way,
from
from
the
the
reaction
reaction
between an acid and an insoluble
between
base
and
filter
l
crystals
funnel
form
O
t
i
s
r
e
y
filter to remove
heat
evaporate the filtrate until
unreacted base
the
chapter
solution
is
not
mixture
isolation of an inorganic salt
ltration
by
and
evaporating
the
evaporating dish
disadvantages of doing
dish
over
a
(gure 33).
evaporated
by reacting an insoluble base with an acid
step.
heating
advantages
p
the
rst
o
heating
a
the
the
l
why
E
v
O
Suggest
in
the
evaporated
Discuss
with
earlier
are
in
C
solvents
shown.
compared
discussed
removed
o
i
as
is
cool
crystals
filter the cooled
t
a
Sometimes
so,
what
u
b.
directly
328
d
Identify
f
x
a.
c.
r
o
Figure 34 Summary of the preparation and
leave to
salt
n
constant mass
crystals start to form
v
i
n
U
dry crystals to
to
dryness.
hot
n
y
start to
water bath as
y
base
and acid
oxide.
P
mixture of
excess base
copper(II)
e
unreacted
acid
r
salt
solution
sulfuric
warmed

rotary
reduce
Tool
Miscible
liquids
liquids
(such
as
funnel
(gure
c an
oil
be
and
35).
separated
water)
Aer
c an
ensuring
by
be
fractional distillation.
separated
that
the
is
Experimental
techniques
Two immiscible
by placing them in a
stopcock
1:
closed,
the
separating
mixture is
oil
poured
into
denser,
lower
narrower
layer
slowing
neck
Separating
at
is
then
down
the
funnels
Soon
aer,
the
drained
as
the
liquids
into
an
interface
form
two
underlying
between
distinct
beaker
the
two
liquid
layers. The
layers
interface
water
by opening the
bottom end of the funnel.
are
also
used
to
selectively
extract
solutes
from
one
solvent
another.
r
Recrystallization
used
to
the
is
a
puric ation
desired
organic
solid
synthesis
selectively
process
from
a
based
mixture
products.
dissolves
dierent
of
This
on
selective
solids.
solubility
that
c an be
Recrystallization is oen
isolation
components
technique
of
the
requires a
mixture depending
temperature.
mixture
of
solids
typic ally
contains
three
types
of
components,
of funnel
Figure 35 Separating an oil–water
O
t
i
s

The
water runs out
n
on
purify
that
a
y
solvent
isolate
y
to
l
used
P
Recrystallization
oil
approaches the
e
into
funnel.
s
s
stopcock,
the
each with
mixture with a separating funnel
dierent
solubilities
in
the
identied
•
impurities
that
are
soluble
•
the
The
recrystallization
desired
solid
that
we
in
in
the
the
wish
solvent
solvent
to
at
at
isolate,
all
all
temperatures
temperatures
which
must
be
when hot, but insoluble when cold.
process
is
detailed
in
gure 36.
filter
t
a
r
o
Figure 36 Recrystallization process
the
solid
to
insoluble
The
ltrate
the
a
more
starts
o
dissolves
desired
usual,
the
solid
are
form
but
to
hot
solvent.
liquids,
molecular
intermolecular
discussed in
forces
Structure
2.2.
pure crystals
vacuum
pump
pure crystals forming
This
c auses
the
desired
any soluble impurities.
removed
cool
slowly.
follows,
then
of
aer
remaining
are
in
all
of
and
by
hot
Crystals
ltration.
of
the
desired
solid
form while
remain in solution.
ltration
which
complex
as
to
impurities
vacuum
crystals
v
Cold
allowed
dissolved
with
impurities
E
4.
is
soluble
is
along
a
Any
the
solid
l
3.
impure
dissolve,
u
First,
O
2.
f
x
1.
In
i
d
water
are
cold
o
hot
polarity
solvent
n
U
solvent
the
C
i
n
solid
heat
in
insoluble impurities
impure

soluble
Miscibility
y
insoluble
p
are
o
that
r
e
impurities
v
•
solvent:
removing the soluble impurities and isolating
dried.
recrystallization,
the
hot
impurities
ltration
but
not
two
solvents
step,
the
a
are
second
desired
used.
solvent
solid.
This
The
is
process
added that
c auses the
precipitate out.
329
Tools
for Chemistry
Analysis
Practice questions
Melting point determination
31.
What
use
method
to
would
separate
the
you
following
The
mixtures?
purity
solids
a
solid
is
placed
in
c an
melt
a
be
at
assessed
low
c apillary
to
by
measuring
moderate
tube.
The
its
melting
temperatures.
sample
is
heated,
A
point.
small
Organic
sample of the
observed
s
s
solid
a.
of
frequently
closely, noting
a suspension containing
down
the
temperature(s)
at
which
it
starts
and
nishes
melting.
Pure substances
solid barium sulfate and an
have
sharp
melting
points
that
agree
with
published
values.
If
impurities
present,
the
melting
point
is
usually
lowered
and
the
solid
melts
over
a
solution
temperatures.
ethanol–water
mixture
a sodium chloride solution
Figure
The
d.
a
mixture
of
37
lowered
water and
as
into
the
from
a
oil
bath.
expected
rst,
Thiele
The
melting
when
the
tiny
tube
the
oil
c an
bath
point
yet
be
sample
is
is
is
used in a melting point determination.
attached to a thermometer and
then
heated,
approached.
visible
droplets
sample
has
just
melted.
down
and
O
impurities
heating
are
second, when the
t
i
s
of soluble and insoluble
the
temperatures
appear,
mixture
entire
slowing
Two
thermometer
y
r
e
opening to allow air in
n
aspirin,
an
a
containing
y
recorded:
e.
how
tube
l
rate
cyclohexane
shows
c apillary
range of
and out of the apparatus
p
v
o
or water bath
C
i
n
rubber band
oil
substance being tested
n
U
o
i
d
heat

Figure 37 Melting point determination
with a Thiele tube
melting
point
observation
of
apparatus
the
may
contain
a
built-in
magnifying
glass that
sample.
u
l
O
f
x
t
a
r
o
Electronic
facilitates
capillary tube
Data-based question
following
melting
about
the
point
E
v
concluded
purity
a
The
data
of
were
the
collected
aer
performing
certain
organic
syntheses.
O utline
what
c an be
products.
Experimental
melting
point
of
Theoretic al
melting
Experiment
330
Synthesis
of
aspirin
Synthesis
of
methyl
3-nitrobenzoate
product / °C ± 1 °C
product / °C
122–129
136
76–77
78
point
of
y
c.
r
an
P
b.
are
e
aqueous sodium chloride
Tool
1:
Experimental
techniques
Chromatography
Chromatography
analysed
to
the
the
mobile
likely
to
(gure38)
is
a
collective
separation
and
stationary
encounter
and
of
as
thin-layer
a
term
its
phases.
DP
for
a
group
components
The
chemistry
two
of
types
student
chromatography
methods
according
(TLC)
of
are
to
in
which
their
a
chromatography
paper
mixture is
relative anities
you
s
s
are
by
chromatography
(gure 39).
front
B
0
t
i
s
(stationary
phase)
L
B
A
O
A
L
C
L
start line
sample
A
A
B
phase)
v
Figure 38 The set-up for paper chromatography, and the resulting chromatogram
C
the tank is saturated
adsorbent
coating – SiO
with solvent vapour
2
U
components
separate out
on plate
solvent
3
amount of the
The end of the TLC
l
sample in solution is
plate is immersed in
spotted onto the plate
a pool of solvent
a
O
Figure 39 Thin-layer chromatography (TLC)
38,
the
c alculate
R
lengths
A,
E
components
v
gure
by
to
n
thin layer of
o
u
A small
adsorbent coating

i
f
x
2
TLC plate is prepared
with a
front
t
a
(plastic or metal)
solvent
d
r
o
plate
1
o
i
n
atmosphere within
thin layer of
In
p
(mobile

y
r
e
start line
solvent
n
l
L
paper
y
C
chromatographic
y
P
with lid
B
e
r
solvent
beaker
B,
C
marked
L
,
L
A
and
the
,
L
B
solvent,
and
L
C
indic ate
the
distances
travelled
0
respectively.
These
lengths
c an
be
used
values.
F
The
c alculation of
R
values
was
discussed in
Structure
2.2.
Review the content of
F
Structure
2.2
and
answer
the
practice
questions
on
the
next
page.
331
Tools
for Chemistry
Practice questions
32.
40
shows
a.
O utline
b.
Explain
A
student
a
why
a
which
chromatography
lid
has
investigating
chromatography
been
component
the
experiment
placed
in
the
on
components
experiment
as
the
mixture
of
a
involving
leaf pigments.
container.
has
the
black
greatest
anity
for
the
s
s
33.
Figure
solvent.
food colouring sets up a
shownbelow:
e
beaker with lid
r
C
D
E
F
drawn in ink
black
food
colouring
explain
errors
in
the
separated
chromatography.
rotated 90°,
and
run
set-up
using
The
again
in
a
a
shown.
technique
second
origin
n
i
paper
acid /
90°
and
which
answer
the
amino
acid
questions
has
the
run the chromatogram
(phenol / water then add
locating agent
v
Between
E
attached.
glutamic
these
H
The
acid
amino
these
identity
(Glu)
acids
CH
aspartic
have
COOH
2
acid,
of
and
COOH
2
two
Asp
)
R
group
leucine
(Leu)
dierent
anity
and
functional
the
H
2
is
for
is
a
dierent
are
shown
for
CH
2
two
glutamic
COOH
acid,
Glu
to
With
which
which
amino
a
acids.
reference
solvent
H
groups,
atom,
dierent
below.
the
2
(–COOH)
c arbon
for
COOH
2
solvent 1.
c arboxyl
groups
anities
CH
mixture
below:
greatest
Amino acids all contain amino (–NH
Asp
in solvent 2
u
41
a
State
bonds.
332
the
l
gure
O
b.
turn
Figure 41 Two-dimensional chromatography of an amino acid
Study
a.
f
x

miture
in
t
a
1
(butanol / ethanoic
water
d
r
o
chromatogram
Glu
o
Asp
the
Leu
U
Glu
solvent
(gure 41).
C
Leu
solvent
i
n
=
run
known as
chromatogram is run in one
o
solvent,
be
v
two-dimensional
two
c an
p
mixtures
Figure 40 Paper
chromatogram of leaf pigments
y
and
Complex
r
e
Describe
34.

O
t
i
s
B
n
y
solvent
start line
A
l
paper
to
CH(CH
2
leucine, Leu
and
their
structural
)
3
2
form
known
structures
COOH
CH
polar
The
systems.
2
are
substituent
of
hydrogen
as
an
aspartic
features,
R
group is
acid
(Asp),
suggest
why
y
P
chromatography
Tool
1:
Experimental
techniques
Practice questions
35.
Remember that the
R
value
is
c alculated
as
follows:
F
distance
R
travelled
by the spot
=
F
distance
to
the
travelled
by
the
chromatogram
solvent
s
s
Refer
B
below and:
A
a.
c alculate the
b.
estimate the
R
value
of
the
top
spot.
Assume that length
A
=
3.95 cm and
B
=
5.02 cm.
F
value
of
the
middle
and
bottom black spots.
F
r
Calorimetry
measure
the
amount
of
heat
released
(or
absorbed) during a
thermometer
reaction
measurable
heat
c an
be
is
a
no
illustrates
out
in
an
lost
of
to
the
error
in
of
a
c alorimetry is that the
known
surroundings,
c alorimetry
a
c alorimetry
and
dilute
insulated
loss
to
the
experiment
involving
container,
such
as
a
polystyrene
a
excess
is
that
rise
explain
Determine
e.
Identify
v
C alculate
d.
E
c.
the
whether
amount
the
two
equation
a
O
and
of
to
was
for
the
further
heat
in
an
that
will
minimize
determine
and
reaction,
reaction
hole)
insulated
container
magnesium
powder
hydrochloric
acid

Figure 42 C alorimetry of the
reaction between magnesium and dilute
hydrochloric acid
the
enthalpy of
lower
heat
the
heat
loss.
insulated container using
to
11.0 °C
this
determine
diagram
acid
(with
hydrochloric acid.
reaction.
would
performed
temperature
balanced
State
the
the
hydrochloric
200 g.
b.
cup, to
o
Write
experiment
in
during
l
a.
modic ation
The
was
shown
used
dilute
i
solution
measures
is
and
t
a
a
42
u
reaction.
gure
surroundings
and
f
x
of
zinc
in
magnesium
dilute
C
two
the
c alorimetry
of
set-up
between
Suggest
5.0 g
loss
answer these questions.
r
o
A
to
practice
Heat
reaction
o
lost
37
.
and
experimental
reaction
Identify
b.
1.1
the
surroundings.
d
a.
in
lid
n
the
although
i
n
Reactivity
The
c auses
water. It is
hydrochloric acid. Note that it is
U
36.
of
experiments.
Practice questions
Review
directly)
mass
v
heat
temperature
of
measure
by insulation.
magnesium
minimize
c annot
p
c arried
is
source
principle
we
y
42
the
energy
large
minimized
between
in
basic
(which
r
e
Figure
change
that
loss
The
c alorimetry, which is
O
assumed
1.1.
change
known as
t
i
s
a
Reactivity
enthalpy
technique
n
reaction’
s
a
y
discussed in
using
l
chemic al
y
c an
P
We
e
R
is
the
the
enthalpy
mass of the acid
including
state
symbols.
exothermic or endothermic.
released, in kJ.
1
enthalpy
change
assumptions
made
for
in
this
the
reaction,
c alculation
in
kJ mol
above.
333
Tools
for Chemistry
Enthalpies
gure43.
a
copper
of
A
combustion
measured
c alorimeter
c an
mass
be
of
determined
the
containing
a
fuel
is
using
the
combusted
known
mass
of
apparatus
in
a
water.
spirit
In
shown in
burner
this
below
c ase, the
thermometer
c alorimeter
is
made
of
a
thermally
conductive
material
to
facilitate
the
energy
calorimeter
transfer
from
is
also
combustion
transferred
reaction
to
the
to
the
water.
c alorimeter,
and
Heat
this
from the combustion
should
be
considered
water
when
processing
this
dicult
is
to
the
results.
quantify
Heat
and
is
should
also
transferred
therefore
be
to
the
surrounding
e
wick
fuel
r
Data-based question
Figure 43 Determining the combustion
student
the
c arried
enthalpy
of
out
the
experiment
combustion
of
shown
butan-1-ol, C
4
in
H
9
gure 43 to determine
OH.
The
following
results
l
wereobtained:
A
student
of
copper
c alorimeter / g ± 0.01 g
M ass
of
copper
c alorimeter
performed the
took
shown
the
in
gure 43.
and
water / g ± 0.01 g
following steps:
Temperature
of
water
before
poured
20 cm
of
water
the
butan-1-ol
spirit
c alorimeter
g ± 0.01 g
weighed and lit the spirit
Temperature
burner
without
the
temperature
for one minute
Use
these
extinguished the spirit
that
the
burner,
c apacity
keeping
replenished
the
c ap o
to
of
c alculate
heat
water in
c alorimeter
copper
weighed the spirit burner
mass
fuel
reliability
the
titration
(gure 44).
measures
experiment.
0.1 °C
60.1
with
a
standard
of
a
is
enthalpy of combustion. Note
4.18 kJ kg
1
K
and
the
specic
heat
1
K
substance
solution
74.38
of
in
known
use a
to fill
solution
c an
be
concentration
determined
and
by
volume in a
beaker and funnel
burette
u
l
when filling
0
a
E
v
O
f
x
leave air gap
the burette reading is
taken from the bottom
of the meniscus
use left
swirl
the flask
hand to
with right hand while
control the
the drops are being added
flow rate

Figure 44 Titration apparatus.
le hand
334
concentration
it
improve the
of
unknown
reacting
i
would
consumed.
practic al
±
–1
water
t
a
that
of
two
r
o
Suggest
The
experimental
of
0.385 kJ kg
o
second time to nd the
d
a
the
–1
is
Titration
•
combustion / °C
c apacity
n
the
the
data
specic
U
•
the
water
aer
C
•
the
i
n
of
water
21.7
75.47
butan-1-ol spirit burner aer combustion / g ± 0.01 g
o
measured
of
v
M ass
•
of
c ap
before combustion /
p
•
burner
y
into
of
r
e
M ass
99.92
combustion / °C ± 0.1 °C
3
•
24.03
O
experiment
They
M ass
t
i
s
38.
n
y
Practice question
and
Note:
le-handed
control the ow rate with their right
chemists usually swirl the ask with their
y
P
enthalpy of a fuel
A
air, but
minimized.
spirit burner

s
s
reaction
clamp
the
Tool
The
equivalence point,
stoichiometric
by
observing
an
indic ator
at
or
near
a
is
the
added
the
solution
to
the
a
the
burette
of
In
point.
the
acid–base
This
redox
oxidation
present in
this
and
is
oen
many
achieved
redox
titrations,
indic ator changes colour
titrations
state
are
and
titrations,
mixture.
Some
in
reagents
identiable,
of
at
are
said to be self-
least
one
of
the
reagents is
change.
holds
unknown
which
reaction
change
colour
at
easily
the
standard
concentration
solution,
(or
and
analyte).
the
The
ask
below contains
titration
is
performed
r
asfollows:
burette,
into
contact
through
from
by
mark
the
tip
taking
a
With
or
by
just
a
solution.
and
at
the
funnel,
below.
draining
reading
the
standard
to
The
process
the
The
some
the
of
rotated
of
the
is
that
solution
is
burette
funnel
solution
ensure
is
is
lled
titrant,
and
drained
twice
with
the
added to the
entire inner surface
then
repeated
removed,
is
its
from the
more.
standard solution to
bubbles
burette
are
removed
volume
is
noted
bottom of the meniscus.

Figure 45 A student
redox titration.
is potassium
Rinse the volumetric pipette:
analyte
and
white
ask
the
conic al
the
you
end
(when
a
point.
burette.
The
is
end
is
approached
until
at
will
the
conic al
with
not
two
is
and
allow
is
indic ator
be read
are
may choose to place it
the
delivered
solution
noted,
ask
distilled
be
titration
volume
least
3
of
of
In this c ase the top of the meniscus should
volume of
colour
from
change.
the
burette
are
and
is
required
the
burette
disc arded
water
and
it
is
in
a
to
is
reach
relled if
suitable
placed
back
waste
below
aected if it is slightly wet, as long as it
repeated.
the
This
addition
addition
an
exact
of
titrant
of
(to
is
concordant
the
the
is
time,
standard solution
down
within
a
few
standard solution as the end
nearest
recorded,
results
the
slowed
are
drop)
and
the
volume
of
accurate
obtained
titrant
titration
(typic ally within
a
cm
You
, which
swirled, until the end point is
standard
volume
Dropwise
The
each other).
E
v
O
repeated
0.1
much
swirling,
point.
established.
is
will
The
with
to
be
of
rinsed
results
point
quickly
known
drops
distinguish
i
the
is
continuously
u
of
how
l
f
x
3
cm
swily,
burette.
t
a
added
the
better
solution
burette
contents
Accurate titrations:
is
of
nal
ask
clean.
is
to
few
4
obscures the meniscus.
indic ator changes colour). The purpose of this step is to
r
o
is
under
able
d
the
The
The
which
idea
The
be
standard
ask,
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to
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o
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container.
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rough
placed
ask.
a
n
give
is
tile
conic al
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reached
or
clean
pipette,
is dark purple and
C
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into
•
the
paper
a
volumetric
permanganate, KMnO
rinsed with a small
p
a
into
the
i
n
on
transferred
Using
performing a
The solution in the burette
o
added,
is
v
is
pipette
three times.
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analyte
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of
volumetric
y
amount
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r
e
•
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zero
tap
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t
i
s
the
amount
tilted
with
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Fill the burette:
small
n
•
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then
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burette
is
l
comes
which
y
Rinse the burette:
P
•
techniques
e
the
by
be
change.
equivalence
bec ause
accompanied
Normally
colour
point
must
Experimental
s
s
indic ating
the
amounts,
1:
335
Tools
for Chemistry
Titrations
changes
do
in
not
temperature,
titrations
always
certain
are
need
properties
which
c an
known
as
be
indic ators.
of
the
The
end
solution,
measured
over
conductometric
the
and
point
such
as
c an
be
determined
course
of
the
thermometric
titration. These
titrations,
respectively.
a.
Explain
why
of
the
burette
reagents,
a
rough
c.
Explain
why
it
does
titration
not
is
matter
need
to
be
rinsed with the
purewater.
needed.
if
the
conic al ask holding the analyte
is wet.
at
group
in
their
lab
c arbonate,
is
a
water
The
or
is
balanced
and
sodium
students
dioxide
equation
that
by
Determine
solution
from
c arbonate
in
a
the
the
.
a
They
from
for
the
the
0.300 g
a
titration.
hydrochloric
nd
the
exact
a
primary
reaction
of
standard:
form and does not
atmosphere.
between
sodium
hydrochloric
c arbonate
was
3
of
the
hydrochloric acid
3
concentration,
students’
is
high-purity
23.83 cm
U
C
i
n
solution.
Sodium
available
found
of
3
0.2 mol dm
c arbonate.
exactly
performing
titrating it against a solution of sodium
o
neutralized
the
(aq).
c arbon
v
ii.
a
3
that
by
p
acid
CO
when
concentration
y
Write
2
avoid
the
approximately
HCl(aq)
solid
to
that
r
e
absorb
i.
Na
stable
is
of
mistakes
know
O
concentration
it
three
students
t
i
s
acid
least
of
n
A
l
Identify
e.
y
d.
in
mol dm
, of HCl(aq) in
results.
In
chemistry
you
need
and
o
electrolytic
complete
circuit,
electrode
(reduction)
i
d
t
a
r
o
Electrolytic cells
Electrolysis
energy
u
f
x
are
is
input,
usually
l
electrodes
a
a
site
for
a
allowing
and
convert
usually
of
such
that
to
know
voltaic.
oxidation
an
as
are
in
at
the
electric al
the
inert
form
to
both
electrons
lost
to
be
a
electroplating,
cell
reduction
to
into
and
as
or
occur.
by
of
electrochemic al
electrolytes, and a
chemic al
chemic al
therefore
battery
involve
oxidized
types
species at one
(oxidation).
or
such
two
electrodes,
gained
energy
process
of
construct
involve
other
material,
themselves
or
how
They
non-spontaneous
made
applic ations,
E
v
O
336
DP
cells:
n
Construction of electrochemical cells
energy (Reactivity3.2).
it
pack.
requires a continuous
The
electrodes
graphite or platinum. Some
active
electrodes.
reduced,
These
are
rather than simply acting as
y
why
of
r
Explain
pipette
P
b.
and
instead
e
solutions
s
s
Practice questions
39.
from
electric al conductivity or
Tool
The
silver
anode
in
gure
46
is
an
example
of
an
active
1:
Experimental
techniques
electrode.
+
s
s
+
Ag→ Ag
+ e
silver
nickel
Ag
+ e
spoon
to be plated
→ Ag
(cathode)
r
nitrate solution
Figure 46 An electrolytic cell used
to electroplate an object
placed
at
the c athode
electrodes. Gaseous products can be collected in inverted tubes. Figure47
O
t
i
s
showsa Hofmann voltameter, in which slightly acidied water is being
electrolysed, producing hydrogen gas and oxygen gas. The electrodes are located
at the bottom of the tube to allow the gas bubbles to rise for collection at the top.
i
the top
readily
requires
collecting
of
a
C an you explain the
of the tubes? How could the identity of the
solid
the diagram
product, it will usually plate the
observable.
measuring
from
The
the
quantic ation of solids
dierence
in
mass of the
any solid particles or akes that might fall to the
electrolytic cell.
electrolytic
electrode
The
t
a
as
E
applied.
well
the
Quantitative
including
electrolysis
as
formation
be
for electrolysing water.
u
of
the
a
electrode,
in
to
therefore
)
the power supply has been omitted
l
produced
bottom
leads
and
v
O
electrolysis
electrode
at
o
d
r
o
f
x
gases be conrmed? Note that
If
cathode (
Figure 47 A Hofmann voltameter is used
relative volumes of the gases collected
y
foils

p
platinum
anode (+)
o
n
U
C
i
n
v
hydrogen
r
e
oxygen
n
l
y
Electrode placement depends on the nature of the products released at the
y
P
silver

e
+
anode
investigations
surface
electrodes
area,
should
require
c areful
temperature,
be
thoroughly
time
control
and
cleaned
of
many
potential
and
dried
variables,
dierence
before the
experiment.
337
Tools
for Chemistry
In
voltaic
energy.
arises,
Electrolytic
and
voltaic
cells
Reactivity
chemic al
two
c ausing
energy
half-cells
current
to
are
ow
is
spontaneously
connected
as
oxidation
in
a
converted
circuit,
occurs
in
a
into
electric al
potential
one
half-cell,
dierence
and
reduction
are
occurs
discussed in
cells,
When
in
the
other
(see
Reactivity
3.2).
The
potential
dierence
c an
be
measured
3.2.
by
connecting
a
high-resistance
voltmeter
to
the
electrodes
(gure 48).
s
s
e
e
V
+
Na
3
e
+
NO
Zn anode
r
Cu cathode
3
NO
t
i
s
→
Zn
+
2
(aq)
+
2e
Cu
(aq)
+
2e
→
Cu(s)
O
movement of cations
movement of anions
bridge
wool.
A
solution
lling
chloride
simpler
of
the
a
or
It
a
half-cells
source
U-tube
sodium
alternative
inert
two
is
is
with
of
agar
sulfate)
using
a
to
complete
mobile
inert
mixed
and
C
potassium
by
the
charge.
i
n
constructed
connects
of
o
build-up
Electrons ow in the external circuit
from anode to c athode. In
anions ow towards the c athode and
strip
p
salt
the
c ations and
v
A
bridge,
y
Figure 48 A voltaic cell.
the salt
r
e

n
+
2
Cu
3
+
2
Zn(s)
3
y
NO
+
l
+
2
Zn
3
with
plugging
of
lter
the
ions.
an
it
anode,
circuit
A
salt
inert
at
paper
respectively
and
prevent
bridge
c an be
electrolyte
either
end
soaked
in
(e.g.
with
a
cotton
saturated
electrolyte and dipping its ends in the two half-cells
U
(gure49).
n
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O


Figure 49 The zinc and
copper half-cells in this voltaic cell are connected
by a paper salt bridge
Electrode
factors
identity,
aecting
electrolyte
cell
how
concentration
potentials.
placement
and
long
the
electrolyte
concentrations).
Other
cell
As
has
with
and
factors
been
temperature
include
running
electrolytic
(bec ause
cells,
the
surface
this
cleaned
and
dried
before
connecting
them
electrodes should be
in
LED. It is a series of alternating copper coins,
paper soaked
in vinegar and
similar to the rst
zinc washers,
battery constructed
by
Practice question
Alessandro Volta in the 18th century
40.
Compare
and
contrast
electrolytic
and
voltaic cells.
area,
aects the
Figure 50 A voltaic pile powering an
thoroughly
338
are among the
electrode
the
circuit.
y
NO
P
NO
Tool
1:
Experimental
techniques
Colorimetry and spectrophotometry
Spectrophotometry
interaction
of
UV,
with
visible
or
light
IR
is
of
an
a
analytic al
certain
radiation.
technique
wavelength.
Colorimetry
is
a
that
is
based
on
a
sample’
s
Spectrophotometry
similar
technique
uses
that
is
a
range
limited to
s
s
wavelengths of visible light.
In both spectrophotometry and colorimetry, light is passed through the sample,
then detected (gure 51). The incident and transmitted intensities of light are
r
and concentration is quantiable, and it is oen presented in the form of a
calibration curve. The calibration curve can be used to determine the concentration
e
compared to determine the absorbance. The relationship between absorbance
l
280 nm
t
i
s
n
y
wavelength
absorbance
O
0.347
detector
with
and
sample
amplifier
source
Figure 51 A single-beam
UV-vis spectrophotometer
c alculated
=
0
log
(
)
I
t
where
I
is the intensity of the incident light and
I
0
the
depends
the
used,
solvent
same
the
the
vs
product of
low
in
cm, and
2.0
ε
the
temperature
experimental
and
l
c an
also
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measurement
concentration
non-linear
a
with
and
and
concentration
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length,
extinction coecient)
nature
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absorbance
absorbance
solutions
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molar
u
an
relationship
concentrations
v
precise
the
the
l
of
O
this
the
as
εcl
to
becomes a
determined
a
c alibration
(gure 52). Note that
at
high
should
ecnabrosba
are
If
Therefore,
comparing
curve
so
on
f
x
constant.
(sample
(known
solution.
conditions
by
cuvette
=
o
of
is
A
follows:
concentration of the solute (c):
t
a
that
l
constant
the
i
a
to
as
n
U
is
proportional
r
o
ε
also
d
where
is
intensity
is the intensity of the
t
transmitted light.
Absorbance
light
C
A
from
i
n
I
o
is
v
M athematic ally, the absorbance (A)
device
p

digital
output
y
cuvette
monochromator
r
e
light
y
P
of a solution by measuring its absorbance of light of a particular wavelength.
absorbance of
unknown sample
1.0
concentrations,
be
used
for
concentration of
measurements.
unknown sample
E
0
10
20
30
40
concentration/µg cm

50
3
Figure 52 Use of a c alibration curve to determine the
concentration of an unknown sample
339
Tools
for Chemistry
While
spectrophotometers
wavelengths,
at
a
specic
should
light
a
simple
wavelength
correspond
absorbed
by
(gure
53).
For
wavelengths
a
of
The
that
coloured solution is
example,
between
are
a
placed
blue
585
therefore
and
use
is
determine
over
the
colorimeter ’
s
strongly
solution
each
absorbs
A
other
orange
colorimetry
wavelength
that
falls
range of
wavelength
to
in
by
the
the
light
the
colour
colour
this
585 nm
involving this
range.
l
yellow
t
i
s
575 nm
O
400 nm
green
violet
n
y
red
700 nm
r
e
blue
y
424 nm
491 nm
p
o
C
i
n
v
Figure 53 The colour wheel
Data-based question
Practice questions
A
student
constructed a
0.285.
the
c alibration
absorbance
at
recording
490 nm of
containing
varying
2+
[FeSCN]
.
0.40
Suggest the colour of the
2+
[FeSCN]
linear
the
A
unknown
below,
concentration
determine
the
was
found to
concentration of
0.20
is absorbance
490 nm and
the
c
of
curve
a
at
l
where
equation:
14 000 ×
c
is
v
O
=
solution
0.30
gave
relationship with
following
A
curve
u
a
c alibration
ecnabrosba
The
solution.
f
x
b.
protein
c alibration
t
a
r
o
a.
of
i
concentrations
a
the
protein in this solution.
d
solutions
by
of
o
the
curve
Using
n
be
41.
absorbance
U
The
0.10
concentration of
2+
E
[FeSCN]
,
in
3
mol dm
.
Determine the
0.00
concentration
of
a
sample
0.00
0.10
0.20
0.30
0.40
2+
of
[FeSCN]
(aq) with an
protein
absorbance
340
of
0.225.
concentration / mmol dm
3
0.50
y
P
orange

observed.
wheel
corresponding to
experiment
within
setting
sample. The
r
647 nm
a
sample’s absorbance
absorbed
complementary
opposite
647 nm.
a
absorbance
only
light.
wavelength
colours
should
visible
c an
e
substance
a
measure
s
s
Complementary
to
oen
colorimeter
Tool
1:
Experimental
techniques
Molecular models
Certain
from
features
Lewis
structure
of
observe
allowing
Models
c an
be
be
might
interact
kits
or
are
the
not
apparent
three-dimensional
rotated (manually or digitally)
molecular
model
kits
c an
visualize
geometry, as well as
with
one
another.
constructed using digital
are usually ball-and-stick models, digital
based
(MEP)
on
van
surfaces.
der
Waals
Digital
radii)
models
and
c an
molecular
easily
be
saved as les
communic ation.

Figure 54 A molecular model of ice
H
H
O
O
H
O
O
H
O
O
O
H
H
H
O
H
a.
in
The
interactions
biologic al
of
digital
biomolecules,
E
other
research.
computer-aided
are
computationally (in
laboratory (in
technology
and
vast
silico),
of
If
they
Drug
computing
numbers
of
very
drug
explored
design.
vitro).
of
digital
molecular models
following scenarios:
large
revolutionized
potential
now
drug
models
has
between
targets
the
a
scientic
construction
screening
disadvantages
school science labs
and
a
and
models
l
The
physic al
v
O
b.
advantages
to
y
the
compared
u
f
x
Discuss
o
C
n
i
t
a
r
o
Practice questions
in
o
MEP surface
Figure 55 Various representations of sucrose
42.
p
r
e
v
i
n
U
d
space-filling model

O
t
i
s
O
O
ball-and-stick model
n
l
y
O
structural formula
y
P
or
and
geometry,
us
between this and other types of visualizations, including
(oen
potential
molecules
model
help
r
models
molecular
models
features
molecular
While
toggled
reference
addition,
how
from
as
Models
e
be
electrostatic
future
built
soware.
c an
In
structural
predict
such
formulas.
molecule.
to
space-lling
for
molecules,
dierent
us
modelling
models
a
of
structural
s
s
to
or
a
c andidate
using
then
be
development
power
c an
dierent
group
drug
This
of
help
extremely
to
molecules
proteins
and
is
is
known as
identied
investigated
resource-intensive;
streamline
and
as
process
molecules
synthesized
is
such
development.
molecules and their
soware.
promising
c an
molecules,
modern
the
targets.
process
by quickly

Figure 56 Molecular model of the
protein rhodopsin,
which is involved in
vision
341
Tool 2: Technology
Sensors
are
digital
transmit
labs
include
probes.
the
devices
results
temperature
Digital
that
c an
be
electronic ally.
sensors
probes,
and
other
pH
used
to
measure
Common
probes,
devices,
physic al
light
such
as
sensors and conductivity
colorimeters,
practic al
and
easy
way
to
measure
certain
properties that would
otherwise require bulky equipment or time-consuming analyses (for example,
produce
at
be
probes
are
a
Jacobson
c ardboard
3
cuvette (top) and
a 0.50 mol dm
copper(II)
app
and
devised
values
were
used as
I
Bec ause
the
phones
laboratory,
are
c are
and
create
a
paper
c ase
is
I ,
of
along
lab.
chamber,
a
that
US
blue
absorbed,
use.
was
yet
respectively,
compared to
you
are going to use
Colorimeters and
frequent
apps
Thomas
c alibration.
they
c an make
Kuntzleman and
eective colorimeter using
red
green
blue
(RGB)
published in the Journal of
sulfate
incident
to
as
large amounts of data
rst.
certain
copper(II)
the
If
require
smartphone
that
save
manual
chemists
simple
a
and
with
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sensors
and
c alculate
solution,
where the
transmitted
absorbance
R
(red)
(gure
57).
t
items
must
i
cuvette
and
the
pH
that
be
people
taken
holder
to
could
frequently
keep
be
them
designed
touch
well
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and
away
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sturdier support of the
analysed
u
l
a
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v
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f
x
t
a
r
o
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sample.
Activity
1.
M ake
2.
If
a
list
of
available,
Investigate
app.
the
sensors
download
how
the
Remember
to
an
available
your school.
RGB app onto a smartphone or tablet.
following
get
in
aect
approval
the
from
RGB
your
values
teacher
registered
before
by the
trying
any of
these out.
a.
Colour
of
dierent
a
solution
food
Concentration
c.
Colour
d.
Path
of
the
length
of
a
could
dyes,
coloured
background
(i.e.
the
(you
colouring
b.
reaching
342
the
red
o
d
for the copper(II)
In
before
devices
sensors,
to
collect
checking
chemistry
way
0
the analysed solution (53), an absorbance
sulfate sample
a
colour
Given the R values for the blank (190) and
of 0.554 was obtained
a
construction
Educ ation.
complementary
B values of the pixels within the circle.
in
cuvette-holding
coloured
Chemic al
applic ation records the average R, G, and
contain
c alibration
by
digital
n
U
sulfate solution (bottom). The phone
of
it
C
Erik
also
that
o
Figure 57 The view through the
examples
measurements
smartphone RGB analyser on a blank
require
c alibrate
v

to
i
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Figure 58 Modern-day scientists
What
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research being done in the Antarctic
use
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quickly
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the
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Figure 59 An Excel spreadsheet
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1
a
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v
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l
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2
mass / g
F
Molar mass of
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mass / g
to c alculate the enthalpy of combustion of primary alcohols from
lower images,
experimental data.
The values
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343
Tools
for Chemistry
While
symbols
c alculators),
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and
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subtraction
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report.
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Remember to include
section
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Tool
3
for
details).
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spreadsheets
sc atter
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344
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Tool
2:
Technology
Practice questions
43.
State
the
formulas
you
would
enter
into
a
spreadsheet
to
c alculate the
following:
0.400
s
s
a.
5
×
4.18
The
mean of 42, 32, 45, 46 and 48.
c.
−log(0.0034)
d.
10
e.
the
12.5
values in cells A5 to A8.
operation
performed
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each
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following
spreadsheet
formulas:
investigation
gave
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into
following
the
eect
of
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impurity
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45
−5.3
into
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of
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results.
information,
chemistry
WebElements,
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SDBSWeb,
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Chemspider,
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National
Science
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properties
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information on:
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345
Tools
for Chemistry
You
may
you
will
of
data.
use
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up
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three
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entries
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346
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databases
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Tool
2:
Technology
Modelling
You have already come across models in your study of chemistry, for instance the
ideal gas model, or atomic models. Scatter graphs help us model the relationship
between two variables. As described in the NOS section of the subject guide,
s
s
“models are simplications of complex systems”. With the aid of technology,
scientists can model the structure and reactivity of intricate chemical phenomena.
The best way to develop a model is to start with a simple relationship or concept
are facilitated by technology: spreadsheet modelling and molecular mechanics.
r
Spreadsheet modelling
set
disturbances
M arzzacco
to
on
transform
such
as
equilibrium
proposed
an
input
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equilibria.
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concentrations,
introductory
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for
output
explore
variables in
the
example.
eect of
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k
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r
v
We can now explore the eects of changing the initial concentrations and the rate
spreadsheet
scrollbars
c alculation
of
the
to
values
the
and
formulas
spreadsheet
equilibrium constant,
K,
k
k
and
f
k ,
where
K
=
r
k
r
C
D
[H]
an
=C5/100
[T]
8
0
9
k
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12
14
19
=J31
v
20
a
0.1
0.05
Here we
and
reverse
rate
G
coinfliin
H
eqilirim
I
J
K
tem
time /
[H]
[T]
0
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=B
= 3+1
=J3B10J3+B13K3
=K3B13K3+B10J3
= 4+1
=J4B10J4+B13K4
=K4B13K4+B10J4
= 5+1
=J5B10J5+B13K5
=K5B13K5+B10J5
= +1
=JB10J+B13K
=KB13K+B10J
= +1
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= 8+1
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= +1
=JB10J+B13K
=KB13K+B10J
= 10+1
=J10B10J10+B13K10
=K10B13K10+B10J10
= 11+1
=J11B10J11+B13K11
=K11B13K11+B10J11
= 12+1
=J12B10J12+B13K12
=K12B13K12+B10J12
= 13+1
=J13B10J13+B13K13
=K13B13K13+B10J13
= 14+1
=J14B10J14+B13K14
=K14B13K14+B10J14
= 15+1
=J15B10J15+B13K15
=K15B13K15+B10J15
= 1+1
=J1B10J1+B13K1
=K1B13K1+B10J1
= 1+1
=J1B10J1+B13K1
=K1B13K1+B10J1
0
Outputs
[H]eq
0.15
a
18
O
17
220
0.2
l
15
16
145
=C14/1000
k
for
60.
u
11
13
f
x
10
forward
F
time
gure
t
a
7
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lom / noitartnecnoC
6
m
23
the
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i
[H]
5
E
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3
4
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o
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3
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d
A
1
2
shown
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f
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U
a
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C
resulting
have
i
n
The
o
constants on the equilibrium concentrations and the equilibrium constant, K
p
f
concentrations over time, from which a concentration vs time graph is generated.
y
r
e
The input parameters are the initial concentrations of H and T
, and the forward and
reverse rate constants, k
n
Charles
be
given
y
external
of
0
5
10
15
20
25
30
35
40
Time / 
= 18+1
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=K18B13K18+B10J18
= 1+1
=J1B10J1+B13K1
=K1B13K1+B10J1
= 20+1
=J20B10J20+B13K20
=K20B13K20+B10J20
= 21+1
=J21B10J21+B13K21
=K21B13K21+B10J21
= 22+1
=J22B10J22+B13K22
=K22B13K22+B10J22
=J23B10J23+B13K23
=K23B13K23+B10J23
= 24+1
=J24B10J24+B13K24
=K24B13K24+B10J24
= 25+1
=J25B10J25+B13K25
=K25B13K25+B10J25
27
= 2+1
=J2B10J2+B13K2
=K2B13K2+B10J2
28
= 2+1
=J2B10J2+B13K2
=K2B13K2+B10J2
29
= 28+1
=J28B10J28+B13K28
=K28B13K28+B10J28
30
= 2+1
=J2B10J2+B13K2
=K2B13K2+B10J2
31
= 30+1
=J30B10J30+B13K30
=K30B13K30+B10J30
21
[T]eq
=K31
22
24
25
26

Keq
E
23
y
a
context
l
c an
the
P
Spreadsheets
e
and build up from there. Here we will look at two examples of modelling tasks that
=B10/B13
Figure 60 Spreadsheet
= 23+1
model of a simple equilibrium
system.
Source of data:
C.
M arzzacco, J.
Chem.
Ed. , 1993, 70(12), p.993
347
Tools
for Chemistry
Activity
Replic ate
bars,
the
leave
Adjust
out
the
following
and
input
ii.
rate constants.
study.
molecular
Here
beyond
we
modelling
mechanics.
the
mechanic al
by
scope
of
principles
by
a
A
the
to
will
more
DP
study
common
modelling
greatly
that
three
has
H
species,
species,
profoundly
for
or
for
T.
example,
example,
impacted our
extending the scope of what chemists
introduce
treats
molecular
atoms
complex
chemistry
the
of
removing
O
classic al
related
molecular
chemistry
values).
equilibrium
y
of
composed
C
of
of
investigate:
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n
with
system
eect
p
⇌
to
the
o
B
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of
C
mechanics
i
substituents)
the
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currently in that cell.
explore
r
e
and
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equilibria
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development
“see”
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v
i
n
3
number
spreadsheet
as
area
balls
of
course,
distribution
of
mechanics, which is a
joined
by springs, in line
molecular modelling, which
involves the use of quantum
electron
density
in
reacting
species.
Figure
61
shows
c an
be
molecular
analysed
oxygen on the R–O–R′
model
are
to
models
explore
bond
of
the
angle.
water,
eect
The
methanol
of
bond
and
substituents
angle
data
methoxymethane,
bonded to the
derived
from
each
shown in table 2.
a
E
v
O
l
H or CH
to
equilibrium
which
R’ represent either
H
very
by manually entering a number into cell J18
i.
two
you
of
and
the
an
how
the
addition
low
Extend
u
methoxymethane (R and
than
sudden
you input both
(very
e.
t
a
angles in molecular
models of water, methanol and
a
when
extremes
Determine some of the limitations of this model.
U
d
Table 2 Bond
f
x

111.5°
of
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happens
the
n
larger
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r
o
methoxymethane
eect
15
what
and
l
the
at
check
values
*1
*1
*1
*2
*3
*2
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*2

348
Figure 61 Molecular models of water,
*3
methanol and
methoxymethane
y
to
scroll
manually.)
eect of changing the
Suggest
understanding
is
the
d.
Molecular
107.1°
include
equilibrium:
c.
A
104.0°
to
P
is
ii.
methanol
explore
able
parameters
concentrations
A + B
water
not
t
i
s
that
c an
and
the
are
input
y
Explore
mixture
angle
on
you
the
e
initial
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r
LHA
b.
R–O–R′
60.
change
parameters
i.
intermediate
Molecule
gure
parameters
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The
in
simply
s
s
a.
spreadsheet
them
Tool
2:
Technology
Activity
a.
Outline
bond
Using
in
eect
of
replacing
the
H
atoms
with
methyl
groups on the
a
molecular
editor
of
your
choice,
replic ate
the
molecules
s
s
b.
the
angle.
shown
gure 61.
Compare
the
bond
angle
values
in
your
models
with
those
given in
e
i.
table 2.
62
each
shows
a
model
distance,
is
two
water
this
activity.
molecules
energy
between
approaching
the
two
one
water
another.
molecules
was
summarized in table 3.
*1
*2
Potential
−0.15
−0.45
5.07
−3.88
U
1.78
Table 3 Distances between two water molecules and
the
hydrogen
atom
10
other,
formed.
the
The
the
1.78 × 10
by
and
and
an
model
Explore
c.
Add
d.
Using
a
the
these
the
e.
Find
out
the
two
what
is
loc ated
15.5 Å
bond
to form
(or
molecule, and the potential
–0.15 kJ mol
.
As
the
molecules
TOK
approach
We
decrease
hydrogen
of
until
a
.
observe
will
have
a
length
of
our
validity
of
the
model
allows
with
naked
at
a
which
eye.
are
To
extended
a
observe
molecular
molecular
editor
of
our
invisible to the
what
advancements
using
Molecular
to
experimental data.
sc ale,
molecule
us
c an
interactions
values
natural world
senses.
1.78 Å
modelling
The
the
hydrogen bond is
through
bond
–27.7 kJ mol
predicted
water
and
water
molecular
the
c ase, the
in
extent
have
technology
senses?
your
Avogadro.
tools
second
between
of
as
E
b.
such
other
1
a
a
choice,
molecule
the
both
the
l
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v
O
a.
is
energy
that
Activity
in
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1
energy
comparing
one
u
veried
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in
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molecules
predicts
f
x
be
two
distance
model
10
(or
oxygen
i
between
the
t
a
each
from
r
o
energy
m)
1.78 Å apart.
molecules are close enough for a hydrogen
the corresponding potential
o
with,
d
begin
15.5 × 10
5.07 and
−27.66
energy values
To
Figure 62 A molecular model of two
water molecules at distances of 15.5, 11.2,
n

C
11.2
o
i
n
15.5

1
energy / kJ mol
p
molecules / Å
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*2
v
between
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r
e
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n
*1
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t
i
s
*1
Distance
l
This
of
potential
to
y
c alculated.
the
extensions
y
For
possible
P
Figure
Suggest
r
ii.
visualizations
available
in
that
editor.
molecule.
editor,
water
happens
c alculate
the
energy
of
the
interaction
molecules.
when
you
add
more
water
molecules and run an
energy optimization.
f.
Build
a
dierent
molecule
and
explore
its
interactions with other
molecules.
349
Tool 3: M athematics
The
purpose
of
this
reliability
of
your
communic ation.
graphs
is
to
M athematic al
relevant
opportunity
to
to
help
quantitative
nearly
skills,
every
reinforce
you
ensure,
measurements
units,
topic
these
in
understand
and
signic ant
DP
they
will
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you
take
International
Bureau
•
Dierent
numeric al
should
be
ounce,
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foot, 
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cubic
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degree
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millimetre
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kelvin, K
•
metre
for
each
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a
energy
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amount of substance
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pressure
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o
unit
l
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second, m/s
u
SI
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a
per
the
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time
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350
is
of
d
a.
cm
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r
o
What
centimetre,
f
x
47
.
•
a
space.
correct
the
following:
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time
b.
temperature
c.
enthalpy
d.
mass
n
U
3
and
by
for
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o
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Identify
reaction to happen
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i
n
metre, m
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separated
y
48.
are
value and its unit.
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units
a
reaction
e.
melting point
f.
density
sec
= −77
.5 kJmol
=
277 °F
g.
volume
h.
amount
=
.5 mol
i.
amount
=
3.0 g
j.
R
3
=
16
= 304 °K
0.33g
=
=
each of the
1
of
=
in
0.78 g / dm
500 cc
8.31 J/(Kmol)
n
space
y
between
Decimal markers should be preceded by a number, even if this number is zero.
l
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units?
•
used, and these
•
p
following
be
y
the
Measures (Bureau
should
t
i
s
of
and
•
v
SI
SI units
followed:
r
e
Which
Weights
conventions
Practice questions
46.
of
BIPM),
r
the
P
to
International des Poids et Mesures,
advantage of
become habitual.
Units
According
evaluate
gures, uncertainties and
chemistry.
concepts,
and
analyses as well as good
e
every
are
section
s
s
the
Tool
3:
M athematics
Uncertainties
Measurements
Repeating
a
instrument
during
produces
known as
readings
uncertainty.
randomly
human
uncertainty
the
and
0.1) °C
varying
range
should
unit.
of
For
gure
36.5 °C
them)
values.
time.
R andom
example,
the
This
tells
inexact.
with
the
inuencing
instrument
be
to
always
and
F actors
therefore
63.
are
conditions
include
reaction
in
from
same
same
this
random
imprecision, uctuations
error
is
described
by the
recorded along with their
temperature
us
that
the
is
given
measured
as
36.6 °C
value is
r
in
the
random error,
and
±
derived
under
Measurements
0.1 °C or (36.6
somewhere
values
36.7 °C.
e
associated
±
the
s
s
variation,
(and
measurement
rise
to
measurements
uncertainty
of
3
around
±
0.5 cm
Measured
burette
burette
are
with
is
therefore
the
correct
±
more
level
may
3
typic ally
0.05 cm
. The
precise.
of
precision and the
o
involves
one
the
the
based
following:
on
the
manufacturer
over
i
by
range
of
display
or
sc ale on the instrument
t
a
stated
the
a
lower
of
the
measuring instrument
u
estimating
the
recorded
uncertainty
f
x
uncertainty
•
be
r
o
the
are using
•
using
of
have
cylinder
stated. The determination of the uncertainty of a particular
usually
estimating
you
should
be
uncertainty
d
must
measurement
•
obtained
values
uncertainty
the
that
measuring
n
measurements
,
U
be
a
n
l
give
the
O
while
y
precision
instance,
p
greater
For
o
with
uncertainties.
C
Instruments
y
P
y
t
i
s
r
e
v
Figure 63 The temperature shown here is 36.6 °C ± 0.1 °C or (36.6 ± 0.1) °C
i
n

which
a
value uctuates.
way
to
estimate
analogue
(has
a
least
that
decimal
count
recorded
the
as
is
0.7 g
v
lowest
The
E
the
zero
instrument’
s uncertainty depends on whether it is digital
sc ale).
The uncertainty of a
above
an
a
or
O
The
l
Estimating the uncertainty of an instrument
digital instrument is its
instrument
place
on
c an
the
least count,
register.
display.
0.1 g.
is
Consider
0.1 g, so the uncertainty is
±
This
±
that
usually
the
0.1 g
mass
and
is,
the
the
the
lowest
value
reading of “1” in
shown
mass
in
of
gure 64.
the
powder is

Figure 64 The uncertainty of the
measurement is ± 0.1 g
351
Tools
for Chemistry
Analogue
instruments
as
smallest
half
the
divisions
at
Therefore,
that
the
last
The
in
that
the
the
marked
The
should
be
temperature
reading
is
on
on
them.
thermometer
uncertainty
temperature
digit
means
1 °C.
sc ales
division.
is
half
the
recorded
value
level
is
The
in
smallest
as
the
In
37 °C
65
has
is
±
0.5 °C.
this
tick
c ase,
mark
it
on
the
uncertainty
of
the
measuring
cylinder
shown
below.
is
tick
marks
at
every
1 cm
zero,
the
The
r
3
has
0.5 °C.
sc ale.
e
State
±
Remember
Practice questions
49.
estimated
sc ale
sc ale division,
37
.0 °C
estimated.
with
uncertainty
gure
s
s
which
every
the
have
sc ale
sc ale
n
l
y
o
C
n
the
measurement,
i
three
o
State
p
r
e
v
i
n
U
images
including
units
and
uncertainties,
shown in the
below.
3
2
4
1
5
0
u
l
V
a
E
v
O
f
x
t
a
r
o
d
50.
O
Figure 65 The temperature is
37
.0 °C ± 0.5 °C
80 °C
90 °C
Uncertainty given by the manufacturer

Figure 66 The uncertainty of this
burette is stated in the label: ± 0.05 ml
Sometimes
3
(or ± 0.05 cm
352
)
(gure 66).
the
uncertainty
is
stated
on
a
label
somewhere on the instrument
y
P
y
t
i
s

Tool
3:
M athematics
Values that uctuate
Sometimes
certain
uncertainty
3
)
and
dioxide
gas
volume
the
of
the
at
acid
using
specic
meniscus.
smallest
sc ale
of
on
their
to
use
a
is
the
the
gas
chips
by
(c alcium
measuring
gure
67.
dicult
pause
syringe
at
make
dierent
instead
c arbonate,
the
rate
uncertainty
cylinder.
uncertainties,
could
it
c ase of
of
c arbon
bec ause the bubbles
measuring
you
the
However, determining
measurement
associated
Alternatively,
subsequently
is
time
c ase
in
of
a
To
greater
you could try
video
points
the
is
better estimate
recording
to
examine the
measuring
cylinder.
time
include
with
cross
is
human
no
to
the
by
quantify. Examples
appearance of the
the
cloudy
experimenter.
solution
The
it
takes
the
making
kinetics
for
of
a
this
the
picture
precipitate
reaction
reaction
cross
c an be
placed
white paper
below
(gure 68).
seconds, and the uncertainty of the
cross invisible
is
count).
However, the actual uncertainty
probably
much
higher, as it must

reaction
experimenter ’
s
becomes
by
42
least
measurement
the
the
(the
noted
precipitate,
long
obscured
E
the
how
reading
± 1 s
dicult
in
o
be
stopwatch
instrument is
of
i
The
to
a
opaque.
a
mixture
be
l
measuring
be
change
clear
solution
between sodium thiosulfate and
produces
v
O
by
or
should
reaction
progressively
studied
the
the
acid
c an
u
example,
These
time
from above
t
a
mixture.
hydrochloric
mixture
uncertainty
reaction
f
x
For
of
human
reaction
d
include
increase the uncertainty
sources of uncertainty
sources
time
and
the
uncertainty
associated
perception of the moment when the
longer
y
the bubbles interfere with the
r
o
Some
o
for measuring the rate of a
volume measurements and
Further
C
gas-producing reaction,
look down at cross
n
Figure 67 In this set-up
p
r
e
and marble chips

v
U
water
i
n
hydrochloric acid
O
t
i
s
gas collecting in
measuring cylinder
y
option
this
in
instance
c an try to estimate the
observe.
marble
explored
for
you
n
Another
and
you
apparatus
points
and
reaction.
reaction
be
division
measurements
the
between
c an
the
In
values
time,
c ases,
y
sc ale.
gas
the
down
the
reaction
in
over
these
l
during
of
variation
In
P
volume
slowing
rate
uctuates
displays.
r
than
the
the
production
with
half
on
hydrochloric
interfere
value
probe
e
the
based
example,
C aCO
measured
s
s
For
the
conductivity
visible.
Figure 68 Studying the kinetics of the reaction between
hydrochloric acid
and
sodium thiosulfate involves measuring
the time taken for the reaction mixture to obscure a cross on a
piece of paper under the ask
353
Tools
for Chemistry
In
TOK
the
c ase
(gure
of
68),
determining
M athematic al
reaction
random
the
range
between
error
over
c an
sodium
be
which
thiosulfate
estimated
they
vary.
by
and
hydrochloric acid
repeating
Suppose
that
measurements and
the
measurement
techniques
is
help
the
the
scientists
repeated
ve
times
under
the
same
conditions
and
the
following
data
are
communic ate
obtained
for
the
time
taken
for
the
cross
to
become
invisible:
s
s
measurements and their
uncertainties,
process
and
make
42 s
predictions,
analyse
data
The
What
is
the
in
the
mean
this
knowledge?
compare
to
42.6 s
ways:
≈
43 s.
one
The
based
uncertainty
on
the
of
range
a
of
set
of
replic ate
values
other
value.
role
Estimating the uncertainty by halving the range of readings
areas of
We
knowledge?
c an be
readings, and the other on the
c an
estimate
the
uncertainty
by
halving
the
range
of
the
readings.
Here, the
l
6
is
45 s
−
39 s
=
6 s.
Half
the
range is
=
3 s.
invisible
is
43 s
±
3 s.
time
for the
Estimating the uncertainty by finding the furthest reading from the
meanvalue
The
values
range
s).
mean
larger
time
=
of
s)
to
and
two
for
45 s.
the
4 s.
4
is
s
The
away
used
cross
as
to
mean
from
an
value
the
is
2 s
bottom
estimate
be
obscured
of
away
of
the
the
under
from the top of the
range (43 s
uncertainty.
these
−
39 s
Therefore,
conditions
c an be
o
C
i
n
v
±
39 s
the
taken
recorded as 43 s
2
p
the
The
from
43 s
y
4
−
r
e
range (45 s
=
the
O
t
i
s
2.
become
n
2
to
Therefore,
y
range
cross
Data-based question
A
student
measured
the
magnesium
following
volume
and
results.
i
u
l
a
E
v
O
f
x
t
a
r
o
d
o
Volume
354
dilute
n
U
between
of
H
2
of
hydrogen
sulfuric
(g)
produced
aer
produced
cm
2
in
the
minutes,
aer
3
/
47
.7
the
acid
2
minutes
3
±
0.5
cm
43.5
mean
reaction
producing the
a.
C alculate
b.
Estimate
the
uncertainty
by
halving
c.
Estimate
the
uncertainty
by
nding
44.0
volume.
the
the
range
of
furthest
readings.
reading
from the
meanvalue.
d.
Which
uncertainty
value
would
you
report
with
the
data?
Why?
y
in
is
two
P
mathematics
in
How
the
1.
of
time
r
scientic
does
45 s
production
mean
of
44 s
role of
estimated
mathematics
43 s
e
experiments.
39 s
from
Tool
3:
M athematics
Measurement
Measurements
science
are
in
the
Dening
SI
in
have
system.
precision
no
These
constant
and
uncertainty,
constants
Symbol
Some
instance,
are
shown
Numeric al
Δv
transition
accuracy.
for
the
values in
seven dening
below:
value
Unit
9 192 631 770
Hz
Cs
frequency
of
C s–133
r
1
speed
of
light
in
c
vacuum
299 792 458
m s
34
6.626 × 10
elementary
e
1.602 176 634 × 10
J s
19
charge
C
Boltzmann constant
1
1.380 649 × 10
J K
1
6.022 140 76 × 10
mol
A
K
ec acy
683
lm W
cd
we
4 The seven dening constants of the SI
have
seen,
division)
additional
includes
it
noting
(Was
based
it
on
this,
you
least
the
might
estimated
count?
uctuations
understand
you
communic ation
how
the
do
impact
in
of
Was
the
it
be
0.2 cm
.
the
values?)
3
±
0.2 cm
is
the
the
The
value
absolute
of
the
value
is
The
it
is
x,
in
this
associated
associated
absolute
with.
In
l
a
v
O
the
example
Doing
of
the
with
dierent
manufacturer?
so
will
uncertainty
help
on
you
your
relative or
measurement:
c ase
the
uncertainty
for
a
3
15.0 cm
given
example
, and the
value. It has
above,
expressed as a
associated with.
absolute uncertainty of
percentage uncertainty =
Using
the
volume
3
uncertainty
it
by
absolute,
0.2 cm
u
f
x
percentage
value be
±
uncertainty.
Percentage uncertainty:
formats:
following
i
as
the
uncertainty
measurement
t
a
units
measured
r
o
same
dierent
u(x), which is
Absolute uncertainty:
the
•
Let
uncertainty,
in
consider
3
±
associated
•
this,
d
3
15.0 cm
expressed
illustrate
account
n
c an
To
to
collected during
o
percentage.
need
data
given
U
Uncertainties
of
the
conclusions.
Expressing uncertainties
measurement
(least count or half the smallest
C
reader)
to
determining
o
estimate
your
way
Eective
of
uncertainty
v
you
(and
common
measured.
ways
i
n
Did
you
a
several
equipment
uncertainties.
experiments
data
is
are
the
p
sc ale
there
While
y
uncertainties.
r
e
As
Table
O

t
i
s
1
luminous
n
N
y
23
Avogadro constant
l
23
k
y
h
P
Planck constant
e
hyperne
limited
and
s
s
constants
are
exact
x
× 100%
x
u(x)
=
× 100%
x
above:
E
3
0.2 cm
percentage uncertainty =
3
× 100%
15.0 cm
=
1.3%
(2
sf )
355
Tools
for Chemistry
•
Relative (or
the
fractional) uncertainty:
absolute
uncertainty
to
the
The
ratio comparing the magnitude of
magnitude
of
the
associated
absolute uncertainty of
value:
x
relative uncertainty =
x
u(x)
s
s
=
x
Using
the
example
above:
3
0.2 cm
e
relative uncertainty =
3
15.0 cm
the
oen
estimate
as
given
is
3
so
only
±
that
to
1
the
In
as
the
(which
10.30 g
±
the
is
decimal
not
±
be
Sometimes
two
signic ant
very small. Either
precision
other
experimenter
should
is
consistently.
of
words,
a
uncertainty
gives
place.
0.1 g,
associated with it.
in
terms
of
to
2
precision
decimal
Perhaps the uncertainty
have
omitted
value
value itself should
readings
giving
might
measured
the
inconsistent
balance
down and it is in fact
the
it
0.01 g
U
C
mass
the
use
o
aer
you
places
10.3 g
suggests
gure.
uncertainty
uncertainty.
decimal
Alternatively,
i
n
correct
is
written
as
precision,
its
v
0.1 g.
zero
of
long
signic ant
the
y
mass
of
of
given
uncertainty
the
as
if
p
but
mass
one
r
e
the
that
number
a
incorrectly
trailing
the
same
to
to
particularly
expressed as a
n
an
the
example,
±
uncertainty
O
is
have
bec ause
stated
acceptable,
correspond
10.3 g
percentage
t
i
s
is
must
was
the
l
are
acceptable,
Uncertainty
places,
is
y
are
convention
For
uncertainty
fraction.
Uncertainties
gures
relative
sf )
a
correct
failed
bec ause
it
to
is
mass
write
down a
signic ant), so
0.01 g.
Activity
n
Copy and complete the table:
Absolute
Percentage
Relative
uncertainty
uncertainty
uncertainty
3
3
19.96 cm
±
0.04 cm
1.08 V
±
2%
±
0.01
u
l
a
E
v
O
f
x
78.5 °C
Decimal places and signic ant gures
Any
has
measurement
a

involves
limited number of
measured
0
356
o
i
value
t
a
r
o
d
Measured
with
1
a
ruler
uncertainty,
so
the
signific ant figures
typic ally
2
has
no
more
3
result
(sf ).
The
than
of
the
length
three
4
Figure 69 Measurement uncertainty when using a ruler
sf
measurement
of
as
a
small
shown
5
in
always
object
gure69.
y
that
decimal
(2
P
Note
0.013
r
=
Tool
In
gure
certain,
69,
as
uncertain,
way
of
third
as
measured
length
the
getting
actual
the
measurements
these
of
fourth
3.67 cm.
3.6
could
gure
be
using
but
The
less
3.66
this
or
ruler,
rst
than
two
3.68 cm.
as
we
gures,
3.7 cm.
are
The
3
and
last
6,
are
gure,
7
, is
There is absolutely no
not
even
sure about the
never
being
subtracting
number
of
are
the
two
results
rules
of
of
the
c alculations
thumb, which depend on
c arried out:
measured
decimal
dividing
number
are
values:
places
measured
signic ant
the
present
values:
gures
in
answer
the
the
present
should
values
answer
in
the
be
recorded to
used.
should
values
be
recorded to
used.
the initial burette reading was 1.03 cm
.
At
3
point,
of the titrant
the nal burette reading was 24.13 cm
.
C alculate the volume
used.
used
=
nal
volume
−
initial
volume
24.13 cm
=
23.10 cm
3
−
1.03 cm
3
values
answer
used
have
should
two
also
be
decimal
stated
to
places.
two
24.31
=
M
0.500 g
=
−1
40.31 g mol
least
therefore
rounded
E
M any
involves
number
c alculations
incorrect
required
answers
number
of
to
a
three
of
to
several
decimal
intermediate
steps.
Always
intermediate
c alculations.
gures
signic ant
steps.
rounding
16.00
in a 0.500 g
1
=
40.31 g mol
or
two
mol
(3
so
the
answer
in
the
values
Rounding
Round
signic ant
or
sf )
should
used.
be
The
rounded
answer is
gures.
errors.
places
c arry
+
0.0124
multiplic ation,
signic ant
require
due
≈
a
operation
the
l
to
0.012404… mol
v
O
This
=
o
=
therefore
u
f
x
n(MgO)
in mol,
i
M(MgO)
m
n
MgO,
t
a
=
and
n
U
mass of MgO
subtracted
oxide.
r
o
Molar
d
of magnesium oxide,
sample of pure magnesium
Solution
are
decimal places.
Worked example 3
C alculate the amount
They
C
the
dp)
i
n
Both
(2
o
=
v
3
y
Volume
p
r
e
Solution
n
the beginning of a titration,
the end
O
t
i
s
3
At
l
y
Worked example 2
y
and
of
neither
There
P
least
exact,
r
and
least
Multiplying
the
are
measurements.
c alculation
Adding
the
2.
length
is
than
e
type
1.
length
greater
gure.
involving
the
is
M athematics
s
s
As
the
the
3:
three
too
the
early
nal
gures,
extra
could
lead to
answer to the
but
not
signic ant
the
answers to
gures
through
357
Tools
for Chemistry
Propagating uncertainties
Processed
also
have
data
are
result
by
the
of
c alculation
c an
by
be
with
the
raw
The
data
2.
Multiplic ation and division
3.
Exponents (AHL only)
variables.
be
and
as
an
to
consider
these
are
only
involving
addition
of
t
a
u
l
a
E
v
O
f
x
m
=
0.621 g
propagate
the
associated
of
speed of
purposes
you
c an
uncertainty.
uncertainty
propagation
uncertainty
we
propagation will not
propagate uncertainties
that
are
being
added or
before and aer
respectively.
The uncertainty for
C alculate the mass of ethanol combusted and
–
in
mass:
combusted
=
initial
the
uncertainties.
= ± 0.001 g
u(m)
= ± 0.002 g
the
0.621 g
+
overall
±
mass
–
nal
mass
70.350 g
u(m)
=
of
values
70.350 g,
You
must add the absolute uncertainties:
write
the
C
change
ethanol
70.971 g
Finally,
no
most
n
i
o
the
=
Then
values or constants
uncertainty.
m
m
358
of
three types:
o
i
n
U
d
r
o
mass
±0.001 g.
of
include
For
methods
burner is measured
combustion giving 70.971 g and
both mass values is
have
subtraction,
uncertainties
The mass of an ethanol spirit
c alculate
into
experiments with a small number of
methods
or
uncertainty
Examples
and
simplied
statistic al
Worked example 4
First,
c ategorized
c apacities.
exact
v
subtracted.
heat
laboratory
course.
absolute
Solution
be
the
uncertainty.
specic
discuss
typic al
advanced
the associated
c an
p
the
will
in
this
c alculation
adding
it
y
a
by
in
such
need
with
r
e
In
More
assessed
rarely
you
values
we
used
introduced
n
be
to
values
section,
c an
uncertainty
O
this
that
and
t
i
s
In
that
the
subtraction
will
mass
propagating
l
molar
assume
you
given
therefore
way this is done depends on the type of
y
light,
not
and
(± 0.001 g)
result:
0.002 g
are
subtracting
the
mass
values,
so
you
y
level,
are
measurements
overall uncertainty of a
P
this
that
raw
The
e
Addition
by
them.
r
1.
At
and
with
estimated
measurements.
done
manipulating
associated
s
s
c alculated
each
obtained
uncertainties
Tool
3:
M athematics
Worked example 5
3
A student
uses a volumetric pipette to transfer 25.00 cm
3
the ask,
3
±
0.05 cm
of dilute acid
into a conic al ask and then adds a
3
further 10.0 cm
±
0.5 cm
of the acid
along with its associated
solution using a measuring cylinder.
C alculate the total volume of acid
added to
uncertainty.
s
s
Solution
First,
c alculate
the
total
volume
added:
You
are
adding
total
volume of acid
=
rst
volume
+
second
the
the
volume
values
uncertainties
and
therefore to
e
propagate
you must add the absolute
volume
uncertainties:
V
=
3
25.00 cm
+
10.0 cm
3
= ± 0.05 cm
3
+
(± 0.5 cm
=
35.0 cm
(1
dp)
3
= ± 0.55 cm
the
result
consistent
rounded
to
one
decimal place
3
≈
be
least
number
of
the
second
decimal
± 0.6 cm
(1
sf )
volume, which has
places
in
the
raw data.
This
gives
the
overall
result:
3
V
The
=
35.0 cm
large
or
divided.
in
Since
decimal
and
c alculations
form,
percentage
you
expressing
c an
them
involving
uncertainties
think
f
x
least
c alculation
is
given
number
of
of
the
the
relative
measuring
of
the
largely
pipette.
sucrose in water to produce
–3
in g dm
,
and
the associated
sucrose:
This
gives
the
overall
uncertainty.
result:
–3
c
This
=
103.5 g dm
percentage
± 0.3%
uncertainty
c an
be
converted to an
absolute uncertainty:
0.3
–3
u(c)
=
×
103.5 g dm
100
four
signic ant
gures,
–3
=
signic ant
0.3105 g dm
gures in the
–3
a
division,
therefore
to
0.3 g dm
(1
sf )
propagate
–3
c
you
cylinder
volumetric
c alculation.
≈
involves
uncertainties
to
of
u
the
result
a
is
the
E
the
3
103.5 g dm
rawdata.
The
dm
l
this
3
3
× 10
v
O
as
end
n
concentration
10.35 g
100.00
that
adding
the
o
mass
i
V
Note
as
the
t
a
=
=
at
are being
relative
C alculate the concentration of sucrose,
d
the
m
=
this
that
the
by dissolving 10.35 g ± 0.02 g of solid
of solution.
r
o
c alculate
c
just
3
± 0.10 cm
Solution
First
values
are
U
3
100.00 cm
of
percentages
Worked example 6
A sucrose solution is prepared
the
uncertainties
also
as
multiplic ation or division
of
C
uncertainties
in
percentage
i
n
uncertainty
the
of
precision
o
multiplied
add
high
y
to
you
the
p
uncertainties
requires
v
Propagating
uncertainty
r
e
eclipses
3
± 0.6 cm
O
t
i
s
the
with
n
to
is
l
that
y
Note
)
y
V
P
u(V)
3
r
3
must
compute
the
=
103.5 g dm
–3
±
0.3 g dm
percentage
uncertainties and then add them together:
3
± 0.02 g
u(c)
=
(
± 0.10 cm
+
10.35 g
3
100.00 cm
= ± 0.29324…%
≈
)
± 0.3%
× 100%
(1
sf )
359
Tools
for Chemistry
Worked example 7
The temperature of a 50.00 g
spirit
burner until it
–1
0.01 g sample of water is 23.0 °C
absorbed
±
±
0.1 °C.
The sample is then heated with an ethanol
0.1 °C.
by the water,
in J,
and
its associated
uncertainty.
The specic heat
s
s
C alculate the heat
±
reaches 33.0 °C
c apacity of water is
–1
4.18 J g
K
First,
nd
=
temperature
T
–
–
=
2090 J
×
(3
4.18 J g
10.0 K
is
rounded
the
to
three
a
subtraction,
so
to
propagate
we
are
(
± 0.2 K
+
50.00 g
10.0 K
heat
have
no
×
2090 J
100
≈
Q
360
=
E
=
v
=
42.8 J
40 J
a
uncertainty
2
(1
value
uncertainty.
2090 J ± 2%
percentage
u(Q)
c apacity
sf )
2090 J ± 40 J
i
to
therefore
100%
c an
is
The
l
O
This
sf )
×
u
specic
assumed
=
(1
f
x
the
± 2%
and
uncertainties
in
the
we
must
add
the
least
number
temperature
of
signic ant
gures in
you add them together:
percentage uncertainties:
t
a
r
o
= ± 2.02%
≈
)
values
with
o
=
multiplying
d
± 0.01 g
u(Q)
the
n
mc∆T,
consistent
C
+ (± 0.1 °C)
U
=
be
o
involves
= ± 0.2 K
Q
to
i
n
= ± 0.1 °C
gures
v
c alculation
signic ant
c alculation.
p
in
×
sf )
result
used
–1
K
O
50.00 g
y
=
the
Q:
t
i
s
mc∆T
that
Q
absorbed,
23.0 °C
= ± 0.2 °C
is
heat
be
assumed
overall
to
be
result
is
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as
(bec ause
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was
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converted to an absolute uncertainty:
not
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y
10.0 K
u(T)
Note
the
n
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Then, in
determine
l
10.0 °C
rst
to
P
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values
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this
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percentage
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the
value
third
percentage
of
c alculations
The
the
power),
exponent.
then
uncertainty
involving
percentage
by
exponents
uncertainty
For
instance,
propagating
its
of
if
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the
requires
raw data is
raw
uncertainty
data
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involves multiplying
three.
s
s
Worked example 8
e
The rate equation for the decomposition of hydrogen iodide is found to be:
2
=
k[HI]
k is the rate constant and
[HI] is the concentration of hydrogen
C alculate the rate of the decomposition of hydrogen iodide when
–3
[HI]
=
–3
0.60 mol dm
–1
±
0.03 mol dm
.
The value of
k
at
this temperature is
–1
s
2
rate
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3
–1
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s
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)
–3
result
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s
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two
signic ant
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an
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the
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exponent
to
therefore
percentage uncertainty:
=
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3
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2
overall
also
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s
± 10%
100
=
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0.57 mol dm
notice
processed,
–3
0.06 mol dm
E
will
a
–3
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that
the
2
in
this
c ase:
ATL
–3
of
is
good
mathematic al
communic ation?
important?
Why
Why is it
are
you
–1
s
encouraged
–1
out
for
all
to
your
show
your working
c alculations?
–1
s
(1
sf )
–1
s
Communic ation skills
What
s
–3
± 0.06 mol dm
propagation
range
–3
0.57 mol dm
v
O
You
×
l
=
≈
rate
is
express this as an absolute uncertainty:
10
u(rate)
which
u
c an
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f
x
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the
= ±10%
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o
the
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it
d
multiply
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with
n
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consistent
U
3
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be
used.
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c alculation
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i
n
The
is
of
–1
s
o
number
2
)
v
The
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(0.60 mol dm
p
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≈
×
y
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r
e
=
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t
i
s
Solution
n
y
mol
l
3
1.58 dm
y
iodide.
P
where
r
rate
M athematics
LHA
Propagating
3:
values
increases
they
c an
–1
s
the
cover
overall
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becomes
As
raw
data
are
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361
Tools
for Chemistry
Uncertainties and means
Trials
are
oen
Uncertainties
of
repeated
do
not
to
need
check
to
be
for
repeatability
propagated
and
when
minimize
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consider
the
following
set
of
results
from
a
titration:
trial
2
0.00
0.00
11.00
11.25
11.00
21.95
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3
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.
follows:
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+
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3
10.966... cm
the
the
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n
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to
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3
× ±
C
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n
answer
given
initial
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o
volume
U
The
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v
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Subtracting
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t
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s
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two
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places.
uncertainty
of
Note
each
places
that
of
the
bec ause
the
the
uncertainty
measured
of
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values
mean
are
volume is
three trials.
o
i
d
A
student
results
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l
are
performed
given
Volume
of
an
experiment
to
nd
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density
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below.
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cm
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3
cylinder /
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a
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v
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f
x
t
a
r
o
Data-based question
8.1
Determine
b.
C alculate
the
density
c.
C alculate
the
uncertainty
absolute
g ± 0.01 g
42.82
a.
the
mass
uncertainty
of
the
of
butan-1-ol
the
and
of
(ii)
a
sample.
density.
relative
49.38
sample.
butan-1-ol
the
cylinder /
Express
your
uncertainty.
answer as (i) an
y
1
e
Rough
r
Initial
s
s
example,
Titration
362
error.
mean of a set
values.
For
all
random
the
Tool
3:
M athematics
Graphs and tables
Graphs
show
variable.
how
changes
Graphic al
relationship
in
an
techniques
between
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independent
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variables
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s
s
Sketching graphs
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graphs
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labelled
such
as
Examples
but
variables
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that
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are
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qualitative
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70.
r
(b)
3
2 mol dm
acid (highest concentration)
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n
c
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g
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m
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gas volume is directly proportional to temperature (c) two graphs showing that
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show:
concentration
system
r
a
to
v
graphs
E
a.
E
l
O
Sketch
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energy,
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Activity
taht selucelom fo
t
a
graphs:
(K)
3
u
f
x
Figure 70 Examples of sketched
T
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1/V / dm
metal c arbonate (b) graph showing that
o
C
n
i
o
,erusserp
r
o
d
,erusserp
V/dm
temperature,
(d)
U
aP/P
aP/P
3
volume,
v
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n
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y
acid
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p
r
e
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OC fo
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0.5 mol dm
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V
t
i
s
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acid
time
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y
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P
(a)
of
reactants
point
changes
of
the
when
and
products
elements
a
strong
over
time
in
an
equilibrium
across period 3
acid
is
added
to
a
strong
base.
363
Tools
for Chemistry
D ata,
particularly
processed
into
quantitative
charts.
There
data,
are
c an
be
dierent
organized into tables and then
types
of
charts,
including
graphs.
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following
•
data
c an
be
presented
as
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table. It is customary to include the
features in data tables:
Descriptive title to aid communic ation.
each
should
be
If
more than one table is being
numbered as well.
listed
in
order
value.
increasing
Dependent variable(s) in the columns to the right of the independent
variable.
List
the
results
from
repeating
trials
next
to
each
other, and the
l
mean in the rightmost column.
Units and uncertainties in column headings.
•
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r
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U
M ass of NaCl
Figure 71 Common features of a data table
places for
values
obtained using te
anoalies identified
if applic able
sae equipent
y
P
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of
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s
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Control
tables
variables
bec ause
sometimes
qualitative
include
data
you
information
data
in
that
a
c an
you
for
need
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found
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of
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are
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way
to
usually
the
or
applic able.
below
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databases).
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c an
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table.
record additional
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examples
of
how to do
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r
Activity
an
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data
looking
dierent
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the
at
how
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combinations
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potential
of
dierence of a
half-cells
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wrote
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Organize the data into a suitable
table.
magnesium – copper: 2.05 V, 2.00 V, 1.34 V
each
and
and
used
graphs
represent
patterns
to
in
present
the
all
the
data.
information
are
c ategoric al
compare
used
to
variables,
group
such
data
as
straight
means
•
that
a
Sc atter
data
a
they
c an
graphs
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line
bec ause
are
is
be
are
usually
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data,
along
to
0–1 cm,
but
a
explore
graphs,
known as
forces, or identity of a
range
in
1–2 cm,
they
useful
a
to
and
expressed as
c ategories
intermolecular
particularly
organized
represent
added
what
data
length
us
interpretation. Examples
whole,
distinct
charts
the
value of the
2–3 cm).
contain
points
joined
by
for continuous data, which
numbered
line
and
have
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value
range.
plotted
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continuous
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Unlike
show
Line
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within
also
lines.
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Line graphs
grouping
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O
•
involve
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quantitative
of
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f
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a
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simultaneously,
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U
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i
n
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v
magnesium – zinc: 1.54 V, 1.43 V, 1.23 V
y
zinc – silver: 1.30 V, 1.25 V, 1.27 V
p
r
e
copper – silver: 0.50 V, 0.53 V, 0.51 V
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t
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s
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n
following
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y
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did
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y
student
voltaic
P
A
e
this in scientic journal
directly
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throughout
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think
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included
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have
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be
will
your
be
should
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s
s
Sometimes
for
should
they
3:
to
the
matters
the
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points
is
relationship
then
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a
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sc atter
overall
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relationship
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variables. The
t
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the
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two
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365
Tools
for Chemistry
(2)
(1)
carbon dioxide
argon
40
)%(
and methane
(1%)
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21%
s
s
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30
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1920
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100
200
300
400
500
600
time in seconds
types of charts and
graphs.
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showing the present
composition of the E arth’s atmosphere (2) Histogram
showing the distribution in the size of microplastics in a sample (3) Line graph showing global temperature variation and
concentration obtained
1980
n
U
(5)
Figure 72 Dierent
0.4
o
v
parts per million by volume
p
0
10
i
n
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=
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i
s
CO
temperature variation

y
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(3)
atmospheric CO
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2
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graph showing the linear increase of c arbon dioxide concentration over time.
Note that
366
the line of best
t
does not
go through all the points bec ause it
represents the overall relationship
between the two variables
y
P
under
Tool
3:
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Plotting graphs
All
graphs,
regardless
technology
•
should
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title
to
each
whether
the
aid
they
are
drawn
communic ation.
should
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If
more
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•
Axis labels, including units and uncertainties
•
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6
8
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logarithmic sc ale on the y-axis in second
10
graph are linear.
graph gives a straight
Plotting the same data with a
line,
conrming that
the trend in the
data is exponential
367
Tools
for Chemistry
Data-based questions
1.
Identify
the
outlier
in
the
graph
below:
120
s
s
100
r
1
2
3
4
5
6
7
a
the
logarithmic
logarithm
of
sc ale on the
the
successive
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for
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useful
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3
4
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5
6
7
8
9
10
11
12
number of the ionization
13
14
15
16
17
18
19
LHA
why
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v
Explain
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p
r
e
time (minutes)
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0
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t
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s
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l
y
40
y
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20
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Error
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that
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represents
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s
s
error
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bars
bars
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3:
variable.
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e
1
s
r
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Figure 74 Error bars show the uncertainty
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0.1
0.2
0.3
0.4
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for this point
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associated with each point, the soware might not accommodate this. In this case
p
you can either draw the error bars for each point by hand or have the program draw
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v
this eect). You should determine which option is best in the context of your analysis.
o
curves of best t
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t
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Figure 75 A generalized
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a line of best t
independent variable
369
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371
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for Chemistry
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372
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so the gradient
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373
Tools
for Chemistry
We can determine the gradient at a specic point on the curve by drawing a
tangent line
and working out its gradient. In gure 82, two tangent lines are
shown in green. The one at
t = 0 is clearly steeper because the rate is higher at
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s
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changes
ClO•,
18 km
gure87.
E
The
in
The
concentrations
in
the
minimum
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latitudes
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375
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0.8
3.0
O
s
s
2.0
72
70
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curve
x-axis.
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under
x-axis
the
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88
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to compare the proportion of particles that
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t
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1.
Students
investigating
the
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temperature
on
air
volume
plotted
the
following
graph
of
their
results:
140
s
s
120
40
0
5
10
15
20
25
30
35
40
temperature (°C)
following
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concentration.
377
Tools
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for Chemistry
The
graph
hydrogen
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gas
experiments.
drawn
at
25
shows
the
collected
A
tangent
seconds
volume of
over
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time
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for two
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45
s
s
3
a.
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for
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experiment 1 at
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rate of
t =
experiment 2
40
25 s.
(high temperature)
Describe
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explain whether the
than
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30
experiment 1
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less than that in
20
10
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dinitrogen
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concentration of
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,
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reaction.
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3
concentration of
N
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3.40
10
2.60
20
1.95
30
1.50
40
1.15
×
×
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10
–8
10
–8
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10
×
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–8
× 10
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5
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Tangent
2
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l
u
/ ])g(
dierent.
f
x
at
t = 0
1
a
E
v
O
378
reaction
35
×
why
are
of
01
above
rate
t
a
Explain
average
30
–3
/s
8–
the
r
o
c.
C alculate
i
b.
25
time
o
3–
Give
has
2
20
time / s
n
a.
A
the
U
time.
how
pentoxide, N
15
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The
i
n
4.
10
y
5
o
v
0
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r
e
5
O
t
i
s
15
0
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y
experiment 1.
10
20
30
time/s
40
–8
y
greater
experiment 1.
explain whether the
35
l
is
for
greater or
em u l ov
2
rate
is
fo
overall
of
and
reaction
25 s
P
Describe
of
=
r
less
c.
t
e
rate
experiment 2 at
mc / negordyh
instantaneous
3
b.
Tool
3:
M athematics
2
Coecient of determination, R
We
have
se en
associate d
likely
to
of
fall
it.
all
me asurements,
dire ctly
One
of
on
the
As
the
tools
a
result
line
we
of
of
and
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t,
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the
relationship
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the
v ariables
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graph
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R
,
the
coefficient
of
determination .
e
2
In
DP
chemistry,
you
will
use
technology to nd
R
values,
and
you will be
2
to
interpret
them.
However,
you
will
not
be
required
to
c alculate
R
r
expected
2
yourself.
data
R
points
lies
between
match
the
line
0
and
of
1.
best
The
t
closer
(gure
it
89)
is
to
and
1,
the
the
more
more
2
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model
is
at
predicting the changes in the
y-axis variable.
R
is only
linear models.
0.7
fo
0.4
0.2
0.1
0.3
0.4
0.6
0.5
of
ethanol / cm
U
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0.2
0.3
of
R
0.4
0.5
0.6
0.7
0.8
3
of
ethanol / cm
2
R
= 0.250
0.6
0.6
0.5
0.2
0.1
0
0
0.7
0.1
0.3
0.4
0.5
0.6
0.7
0.8
3
3
volume
value is 1 when the points are all on the line of best
l
O
points in graphs 2 to 4 become increasingly more spread
0.2
0.8
ethanol / cm
2
Figure 89 The
0.5
u
volume
0.4
0.3
volume
0.3
s s am
0.1
f
x
0
0.2
0.4
fo
0
g / l o n a ht e
n
0.1
t
a
r
o
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0.2
i
fo
0.3
o
0.4
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0.5
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g / l o n a ht e
0.6
0
C
2
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0.7
0
0.8
i
n
3
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0.7
0.1
o
0.2
0.2
v
0.1
0.3
p
s s am
s s am
fo
0.3
0

0.5
r
e
0.4
0.6
O
0.5
= 0.989
y
g / l o n a ht e
g / l o n a ht e
0.6
2
R
t
i
s
= 1.000
n
y
2
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0.7
0
l
suitable
the
y
the
value of
P
closely
The
t
of
ethanol / cm
(graph 1). The
2
out,
c ausing the R
to decrease
a
v
E
379
Tools
for Chemistry
Interpolation and extrapolation
Lines
and
involves
curves
of
predicting
extrapolation
best
t
values
involves
c an
be
within
interpolated
the
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s
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interpolation
l
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line
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the
concentration
obtained
concentration.
by
graph
shown
in
gure 91.
measuring the absorbance of six
plot of
A
versus
c
0.9
n
U
C
solutions
absorbance
on
x
extrapolation using a straight line
o
the
points
i
n
The
p
Figure 90 Interpolation and
Consider
0.8
(x
,
2
y
)
2
o
0.7
d
0.6
i
0.5
t
a
r
o
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l
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,
1
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0.1
0
a
E
v
O
380
v

intercept
r
e
x
n
y
extrapolation
0
0.1
0.2
0.3
c/mol

0.4
dm
0.5
0.6
0.7
3
Figure 91 Interpolation involves nding values within the range of the experimental data
y
P
extrapolation
Tool
The
Its
absorbance
of
concentration
a
sample
c an
then
of
be
unknown
found
by
concentration
interpolation,
is
as
found
to
be
3:
M athematics
0.460.
shown, giving
3
0.34 mol dm
.
concentration,
a
c alorimetry
copper(II)
are
c alculate
∆T
∆H).
lowered
due
92).
time)
metal
The
the
curve
the
is
could
solving
of
the
of
be
over
mixture containing
time
temperature
heat,
measured
Q
used to nd the
x
reaction
monitored
amount
also
for
(and
during
the
and
value
later,
the
is
the
results
needed
enthalpy
experiment is
surroundings.
(the
backwards
solution,
line
maximum
temperature
to
the
and
allows
part
to
us
of
the
to
the
time
graph
at
predict
a
that
which
shows
the
corrected
zinc
a
decreasing
was
value of
added to the
T
. Doing so
a
constant
gure
For
91
is
involves
same
example,
linear,
the
but
the
beyond
during
assumption
the
range
relationship
the
the
linearity
is
of
reaction.
that
between
lost
at
the
data.
In
relationship
practice,
absorbance
higher
and
temperature of
0
highest temperature
=
temperature that would have
actually reached
2
T
been reached if there were no
2
1
rate of cooling
i
ΔT for reaction
0
l
a
O

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T
T
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u
reactants mixed
o
extrapolation at same
t
a
r
o
f
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T
d
erutarepmet
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n
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T
C
i
n
T
o
v
=
1
p
reactants
T
always
y
initial
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r
e
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is
concentration
concentrations.
compensating for heat lost
T
between
this
n
in
c ase.
the
loss
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the
always
stays
heat
t
i
s
variables
of
y
Extrapolation
rate
l
max
assumes
y
over
sulfate
loss
cooling
y
temperature
zinc
subsequently
highest
heat
the
(gure
of
for
P
temperature
and
The
to
Extrapolating
graph
the
and
in
r
change,
copper(II)
a
solution
equation
0.460
e
to
on
the
entering
experiment,
sulfate
plotted
known,
s
s
In
If
by
time
Figure 92 Extrapolation of temperature data during a zinc metal and copper(II)
v
sulfate reaction to determine a corrected
value of T
, the maximum
temperature
2
E
381
Tools
for Chemistry
Data-based questions
1.
A
student
metal
performed
and
copper(II)
a
c alorimetry
sulfate.
A
experiment
graph
of
the
to
determine
student’s
results
the
is
enthalpy
shown
change
for
the
reaction
between zinc
below.
s
s
68
66
60
r
58
56
52
C
50
44.5
19.5
º
C
42
38
34
28
26
24
20
0
1
2
3
4
5
6
time / min
the
Explain
d.
Outline
why
a
the
what
temperature
reason,
whether
student
other
collected
hydrochloric
NaOH(aq).
D ata
f
x
Plot
c.
The
maximum
cooling
volume
need
data
every
minute
to
c alculate
while
the
titrating
t
=
2.5 min.
heat,
Q,
and
enthalpy
Volume
of
change,
N aOH
∆H,
for this
volume
0.600 mol dm
of
sodium
sodium
hydroxide
hydroxide,
Temperature / °C
3
(aq) / cm
added and
0.00
22.4
shown in the table on the right.
graph of these data.
the
rate of cooling back to
would
temperature
by
extrapolation
a
and
the
with
l
Find
for
exothermic or endothermic.
3
HCl(aq),
of
the
heating
5.00
23.1
10.00
23.9
15.00
24.6
20.00
25.4
curves.
v
O
b.
a
are
temperature
the
is
u
temperature
a.
acid,
you
t
a
student
r
o
A
reaction
i
reaction.
measured.
the
extrapolated
information
8
n
c.
minimum
giving
o
State,
U
State
b.
d
a.
7
C
i
n
18
o
v
22
y
30
p
r
e
32
O
40
of
hydrochloric
3
acid
was
25.00 cm
. Determine the
25.00
25.6
30.00
25.5
concentration of the acid.
E
d.
C alculate
reaction,
382
⦵
the
in
enthalpy
change
of
n
=
y
ΔT = 64.0
44
t
i
s
/ e r u t a r ep m e t
46
36
2.
l
º
48
neutralization,
∆H
,
for this
35.00
25.4
40.00
25.3
45.00
25.2
50.00
25.1
1
kJ mol
y
P
54
e
62
Tool
3:
M athematics
Sources of experimental error
Experimental
always
have
systematic
has
the
or
in
dierent
everyday
meaning.
consideration
outcome
of
the
fact
that
with
measurements
them.
are
Experimental
never
errors
exact: they
c an be
random.
“error ”
very
from
associated
an
of
language
One
sources
investigation,
of
of
c an
the
mean
features
experimental
and
possible
“mistake”.
of
a
In
thorough
science,
“error ”
scientic
research
errors, their impact on the
ways to minimize them.
r
R andom and systematic error
be
minimized,
random
an
include
associated
variations
in
the
be
by
bec ause
provides
errors
uncertainty
random
eliminated,
precision
uncertainty
to
in
too
measured
high
as
repeating
the
estimate
values
of
too
values,
low.
L arge
are
a
random
spread
over
readings that uctuate
with
reading
a
sc ale,
measurements made with a
in
the
same
direction
y
results
p
results
greater
error.
r
e
aect
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errors
measurements.
particular instrument.
Systematic errors
leading to
R andom
O
and
of
time,
reduce
not
variations
likely
t
i
s
Sources
over
errors
The
natural
n
range.
but
to
equally
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random
due
are
l
c an
arise
that
y
measurements
P
R andom errors
e
is
word
a
result
s
s
The
errors
uncertainties
meniscus
correct reading of volume
making
meniscus
(gure
the
and
c an be
values
errors
background
c an
then
include
radiation,
during
a
be
corrected
misc alibrated
leaks
in
c alorimetry
random
uncertainties,
a
basis
where
evaluating
error
how
(not
far
to
an
available,
the
be
that
literature
the
=
E
Note
error
theoretic al
and
with
value
value
collection
uncertainty
−
propagation,
values
accuracy
and
of
percentage
is
from
the
theoretic al
value
is
and
Figure 93 Examples of systematic error
when taking burette readings
heat
a
are useful in
between
percentage
percentage
processed
uncertainty)
theoretic al
is
errors
c an
result.
a
way of
value:
value
×
theoretic al

probe,
apparatus,
Comparisons
theoretic al
confused
experimental
experimental
Percentage
by
results.
precision
a
quantifying
for
obtained
experimental
v
O
Percentage
and,
errors,
in
l
give
errors
u
of
temperature
experiments.
t
a
of
analysis
gas
for it.
i
surroundings
f
x
the
at bottom of meniscus
by making changes to the
Errors and processed results
Estimates
are too
o
systematic
of
the
that
methodology. If a particular systematic error
r
o
to
accuracy
eliminated,
d
of
presence
loss
reduce
not
quantied,
Sources
readings
from
n
be
in
For
reading
U
but
experimental
c an
burette
low).
93).
errors
minimized,
result
too
C
Systematic
will
a
or
i
n
high
the
taking
high
o
above
consistently
too
v
example,
the
100%
value
sometimes
c alled
the
accepted
value or
value.
383
Tools
for Chemistry
For
example,
the
theoretic al
value
for
the
activation
energy
of
a
reaction is
1
reported
in
activation
the
literature
energy
as
65 kJ mol
experimentally
± 1%. Students A and B determine this
and
obtain
the
data
shown in table 5. Student A
1
nds
the
activation
energy
to
be
60 kJ mol
± 10%. The
± 10% part is an estimate
1
of
the
c an
value,
see
that
(10%)
that
theoretic al
that
the
two
indic ates
the
value
are
low
result
lies
is
in
within
aligned.
precision,
the
the
range
range
However,
the
particularly
of
of
54–66 kJ mol
the
high
when
percentage
compared to that of
value (1%).
Uncertainty
Absolute
r
Experimental E
R ange
Percentage
a
1
1
value / kJ mol
/
%
uncertainty
/ kJ mol
6
4
2

Table 5
Student
results for the experimentally determined
the
result
obtained
by
1
± 4%.
not
result
lie
in
is
this
be
(though
the
attributed
the
random
the
is
errors.
loss
of
to
following
the
precise
This
reaction:
obtained
as
the
by student A,
theoretic al
time,
the
theoretic al
value).
value does
inaccurate. In addition, since the
This
C
use
the
same
errors
surroundings
suggests
in
between
as
systematic
an
that
student
or
systematic
B’s
errors
result and the
random:
investigation
involving
c alorimetry
o
heat
b.
a
misc alibrated colorimeter to determine the absorbance of
i
asample
c.
taking
a
burette
reading
by
positioning
the
viewer
slightly
below the
meniscus rather than positioning the eyes on a level with themeniscus
d.
u
52.
l
a
E
v
O
f
x
t
a
r
o
d
a.
of
as
the
that
error is 23%, only up to 4% of the 23%
dierence
n
U
each
for
o
to
.
value.
Classify
A
use
of
student
and
pH
probe
c alculated
obtained
C alculate
53.
a
that
the
reports
enthalpy
values
change
to
the
for
nearest
the
whole
combustion
number.
of
butan-1-
1
ol
the
Two
students
with
a
strong
−1 430 kJ mol
percentage
error
investigated
base.
Their
1
.
the
The
in
theoretic al
the
student
C:
−35 kJ mol
student
D:
−50 kJ mol
student’s
enthalpy
results
of
value
is
−2 676 kJ mol
neutralization
of
a
strong acid
1
± 4 kJ mol
1
1
± 10 kJ mol
1
theoretic al
a.
C alculate
b.
Comment
value
the
on
theresults.
for
the
enthalpy
percentage
the
relative
errors
impact
of
of
of
neutralization
the
.
result.
were:
1
The
384
not
B
than
1
result
Practice questions
51.
still
percentage
to
signic antly
smaller
48–52 kJ mol
therefore
and
i
n
theoretic al
4%
of
v
c an
range
range,
is
contributed
the
student
is
p
uncertainty
error
in
precision
error
y
The
greater
random
r
e
indic ating
The
23
O
consider
50 kJ mol
7.7
48–52
activation energy of a reaction
t
i
s
Now
54–66
students’
random
is
−57 kJ mol
results.
errors
on
each of
.
y
10
50
n
60
B
l
A
student
y
student
error / %
P
1
/ ± kJ mol
.
experimental
e
theoretic al
indic ating
the
suggesting
uncertainty
the
error,
s
s
We
random
Tool
3:
M athematics
Errors and graphs
A ytitnauq
Graphs c an also provide information regarding systematic and random
error
line
(gure
of
best
R andom
Systematic
(either
up
or
errors
errors
down).
c ause
c ause
On
data
all
points
the
graphs,
data
this
to
to
deviate
shi
in
from the
the
same
means that the points
perfect results
random error
s
s
direction
94).
t.
systematic error
are
systematic ally
especially
loc ated
higher
noticeable in the
(or
lower)
y-intercept: a
than
expected. This is
y-intercept
that
is
not
where
quantity B
would
expect
it
to
be
suggests
the
presence
of
a
systematic
error.

Figure 94 The presence of systematic and
errors c an be inferred
sodium hydrogencarbonate on the rate of its reaction
n
experimentally (table 6).
y
A student investigated the eect of concentration of
y
determined
l
55.
The combustion of primary alcohols methanol to
was
graphic al data
P
Practice questions
octan-1-ol
from
r
random
54.
e
you
with citric acid. The reaction produces carbon dioxide,
atoms in primary
of combustion
alcohol
/ kJ mol
-
–400
¹
–814
–1300
–1440
8
–1700
n
Table 6
U

Experimental enthalpy of combustion of the
primary alcohols

dioxide / s ± 1 s
1
2
3
210
277
166
0.50
195
135
118
1.00
66
93
53
1.50
21
27
35
2.00
15
25
15
0.25
C
7
³ ± 5 %
o
–990
6
i
n
5
-
p
4
v
–680
25 cm³ of c arbon
hydrogenc arbonate /
mol dm
3
Time taken to produce
Concentration of sodium
y
2
-
r
e
–230
of
this gas. The student’
s data collected is shown in table 7
.
¹ ± 40 kJ mol
1
3
so the student timed how long it took to collect 25 cm
O
Experimental enthalpy
t
i
s
Number of c arbon
3
Table 7
Time taken to produce 25 cm
of c arbon dioxide
in the reaction of sodium hydrogenc arbonate and citric acid at
a
line
your
or
and
published
dierent
concentrations of sodium
hydrogenc arbonate
error
curve of best t.
graph,
o
and
including
i
Using
graph of the data in table 6,
a.
enthalpy of
t
a
b.
bars
a
d
Plot
r
o
a.
C alculate
the
mean
time
hydrogenc arbonate
for
each of the sodium
concentrations.
combustion data, comment on the impact of
systematic
Using
your
random
on
the
results.
u
f
x
c.
error
b.
Determine
halving
an
the
estimate
range
of
of
each
the
set
time
of
uncertainty
by
three trials.
graph, comment on the impact of
error
on
the
c.
Plot
a
graph
to
show
the
relationship
between
results.
concentration
l
a
E
v
O
time,
of
including
sodium
error
hydrogenc arbonate and
bars
and
a
line
or
curve of
best t.
d.
Using
your
random
graph, comment on the impact of
error
on
the
analysis.
1
e.
C alculate the rate by nding
for each
T
concentration.
f.
Plot
a
graph
to
concentration
and
g.
show
of
the
sodium
relationship
between
hydrogenc arbonate
rate.
Using
your
systematic
graph, comment on the impact of
and
random
error
on
the
analysis.
385
s
s
e
r
n
l
O
y
p
chemic al
C
n
U
r e a c t i o n s?
o
drives
y
P
y
t
i
s
v
i
n
What
1
r
e
Re a c t i v i t y
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
Reactivity 1.1
Measuring enthalpy
changes
s
s
What c an be deduced from the temperature change that accompanies chemic al or
Chemistry
involves
the
study
of
chemic al
An
reactions and
understanding
energy
principle
of
Conservation
science,
which
of
is
energy is a fundamental
examined
empiric al
experimental data, the language of
or
scientic
understanding
of
progress
energy
changes
associated with
reactions.
is
reactions
system
and
the
involve
a
Reactivity 1.1.4 — The
transfer of
chemic al
surroundings, while
conserved.
pressure
It
determined
c an
be
the
system
Reactivity 1.1.3 — The
whether
the
of
reactions
In
l
a
E
v
O
surroundings
a
chemic al
stored
of
the
in
the
control
enthalpy
over the
refers
to
the
change
heat
for a
transferred at
standard conditions and states.
from
the
change
in
temperature of a
in chemic al reactions
All
reaction,
energy
mixture
is
a
is
conserved.
reactants
function
of
the
and
Chemic al
products,
kinetic
potential
while
the
energy is
temperature
energy of the atoms, ions and
present.
chemic al
reactions
surroundings
system
total
chemic al bonds of
reaction
molecules
from
the
heat,
involve
from the
energy
surroundings.
form
of
but
it
In an
open system,
changes.
reaction system
may
also
Most
be
in
or
it
Energy
may
commonly,
the
form
of
be
the
may
be
absorbed
energy
is
released into the
by
the
reaction
transferred in the
electricity, sound or light.
(contents
the
transfer
of
matter
and
energy
is
possible
across its
of flask)
boundary

Figure 1
better
(Reactivity 1.1.1)
surroundings
system
used
o
i
t
a
+
Energy transfer
u
system
substance.
reactants and
be
relative stability of
are endothermic
d
r
o
f
x
=
pure
energy
surroundings.
stability
exothermic.
universe
of
to
system and
turn
n
or
determines
and
relative
direction
U
products
the
,
under
o
between
on
the
in
studied.
standard
ΔH
constant
C
transfer
depending
i
n
exothermic,
being
c an
⦵
reaction,
Reactivity 1.1.2 — Reactions are described as endothermic
or
of
leading
reaction
p
energy
the
v
total
between
products,
the
chemic al
y
Reactivity 1.1.1 — Chemic al
understanding
r
e
Understandings
energy
and
of
a
information
between
how
O
chemic al
reactants
terminology, all contribute to
an
exist
n
and
develop
between
This
that
understanding
t
i
s
our
to
experimentation. The use of models,
transferred
surroundings.
relationships
involves
l
and
is
the
y
observation
mathematics
the
through
of
energy
y
compounds.
and
r
chemistry
P
physic al changes of state of the elements and their
e
physic alchange?
The universe is the
(for
transferred
combination of the system and its
though
surroundings
matter
example,
through
energy
and
may
energy
its
c an
sides). A
be
c an
matter
added
to
closed system
transferred
neither
be
across
enter
nor
the
exit
a
beaker,
allows
no
and
transfer
boundary. In an
the
energy
of
c an be
matter,
isolated system,
system.
387
Reactivity
1
What
drives
chemic al
reactions?
s
s
e
r
In each of the above scenarios,
energy is transferred.
In hot
springs,
energy is transferred
and
in reworks,
as heat,
industrial
during
reaction,
an
but
processes
industrial
also
contributes
place
in
chemic al
c an
a
be
industries
in
not
loss
used
as
a
open
only
of
or
useful
to
closed
aects
map,
systems.
eciency
energy,
model
heat
the
p
Thermography
structures
to
heat
an
ow
where
The
of
loss of
the
chemic al
increase in thermal
and
loss
from
red is hot and purple
n
U
C
i
n
o
v
pollution.
iscold.
take
process
y
heat
r
e
Most
O
t
i
s
Models
o
i
d
t
a
r
o
u
l
f
x
a
E
v
O
388
as heat,
Both scenarios are opensystems
n
light.
l
and
y
sound

Figure 3
What
c an
are
Thermograph of industrial engineering system
the
chemic al
industrial
advantages
engineers
processes?
of
use
How
modelling
the
does
data
this
heat
distribution
collected
help
our
to
and
transfer?
How
improve the eciency of
environment?
y
P

Figure 2
Reactivity
1.1
Measuring
enthalpy changes
What is the dierence between heat and temperature?
Temperature, T,
change
in
value
is
is
an
example of a
independent
of
state function.
the
pathway
For
a
state
function,
any
between the initial and nal
measurements.
in
it
example,
the
will
you
take
(the
initial
not
tell
you
occurred
the
the
temperature
value)
complete
throughout
the
and
then
story
day.
of
of
any
The
the
again
water
in
the
in
temperature
c alculation
of
a
swimming
aernoon
uctuations
the
pool
(nal
early
value),
that
may
temperature change
=
T
nal
−
T
initial
of
energy
thermal energy.
convection
and
It
c an
absolute
a
E
v
and
temperature
a
(−273.15
entropy,
S,
(in
When
from
of
a
phase
°C),
all
system
kelvin)
is
by
heat
kinetic energy
potentially
absolute zero, 0 K
stops and the
transferred
work.
average
from liquid to gas.
At
be
radiation.
the
temperature
transferred
l
O
in
in
is
temperature gradient.
Heat has the ability to do
increase
that
volume,
is
of
enthalpy
a
Heat
the
motion
reaches
for
of
its
proportional
is
sometimes
to
an
This
referred
object,
c auses
the
an
result is
increase
example, a change of state
the
particles
minimum
to
pressure.
processes of conduction,
particles.
change,
of a day,
throughout the
warmer body to a cooler
transferred
its
and
O
form
f
x
to as
a
result of the
include
y
a
i
is
as
functions
t
a
q,
body,
state
u
Heat,
of
r
o
examples
the end
has occurred
p
only the overall temperature change
Other
an
the beginning and
cooling that
o
C
the temperature of a pool at
n
day,
If you record
give you an indic ation of the heating and
o
will not
d
it
the
theoretic ally
possible
average
value. The
kinetic
energy of
Entropy
the
particles
of
matter.
As
the
temperature
n
l
y
t
i
s
r
e
v
i
n
U

Figure 4
y
P
ΔT
r
issimple:
e
have
if
morning
s
s
For
increases,
the
kinetic
is
dened
and
explored in
energy of the
Reactivity 1.4
particles
also
increases.
389
Reactivity
1
What
drives
reactions?
Communic ation skills
ATL
Your
communic ation
throughout
the
terms
of
“heat”
solution
When
you
describing
particles
within
→
MgCl
released
surroundings,
considering
rises.
the
refer
the
that
between
When
the
(aq)
from
and
transfer
to
2
the
the
you
of
+
H
reaction
think
thermal
kinetic
(g)
temperature of
about
heat,
energy in the
temperature
average
2
of
a
system,
energy of the
system.
y
p
r
e
M agnesium
ribbon reacting with hydrochloric acid
n
U
C
o
v
i
n

Figure 5
Thermochemistry is the study of heat changes that occur during chemical
reactions. Heat changes are oen described in terms of enthalpy. At constant
o
a closed system to the surroundings during a chemical reaction. The terms
i
“enthalpy change” and “heat of reaction” are commonly used when describing
the thermodynamics of a reaction. The most common unit of enthalpy change is kJ.
u
l
a
E
v
O
f
x
t
a
r
o
d
pressure, the enthalpy change, ∆H, is dened as the heat transferred from
Activity
Imagine
a
undergo
glass
a
the
of
•
Identify
the
system
•
Explain
the
movement
the
What
of
containing
ice
cubes
sitting in the summer sun. It will
enthalpy.
Is
•
glass
in
•
and
water
an
open,
and
the
of
closed
or
isolated
system?
surroundings.
energy
in
the
form
of
heat,
between
the
surroundings.
would
observation
390
water
change
you
in
observe
terms
of
a
on
the
change
outside
of
state
of
the
and
glass? Explain this
transfer
of
energy.
system
y
are
is
reaction
n
you
2HCl(aq)
heat
simple
l
system.
the
hydrochloric acid.
O
are
and
e
aqueous
explanations will
t
i
s
you
the
accurate
y
the
into
in
and
Your use of the
r
reaction,
system
consistent
terminology.
P
this
+
incrementally
programme.
understanding of theseconcepts.
consider
metal
Mg(s)
develop
“temperature”
your
example,
magnesium
include
scientic
and
demonstrate
For
skills
will
chemistry
s
s
applic ation
skills
entire
Communic ation
In
chemic al
Reactivity
1.1
Measuring
enthalpy changes
Exothermic and endothermic reactions
(Reactivity 1.1.2)
A
chemic al
reaction
surroundings
in
dened as
to
a
to
it
c an
be
an
the
chemic al
surroundings
are
place,
to
form
chemic al
This
energy
each
process. The
part
the
bond.
are
the
reactants
Energy
and
is
are
rearranged
broken, and
absorbed
bond
by the
breaking is
bond dissociation energy
Energy
made,
of energy
are
therefore
termed the
bonds
your
of
reactants
products.
of
transfer
of
atoms
in
bonds,
is
type
chemic al
important
the
bonds
is
released into the
and
therefore bond making
between
understanding
of
the
the
surroundings and
energy changes in
Exothermic
reactions
example,
so
determine
is
the
reaction
in
is
between
experiments
which
most
any
the
focus
c ases
the
reaction,
the
change
or
to
heat
is
exothermic,
measure
For
system and so the
the
absorbed
we
by
c an use a
amount
of
heat
surroundings. In the school
in
temperature,
ΔT,
of
the
reaction
n
collect
used
enthalpy change
surroundings.
o
ofchemistry.
to
as
by
i
technology
such
and
water.
observations
instruments
the
the
c an be made using human senses, or with
data-logging
data
is
one
of
equipment.
the
The
applic ation of
essential skills in the study
u
l
f
x
t
a
digital
of
lost
d
laboratory,
aid
r
o
the
the
being
endothermic
the
of
not
positive.
apparatus
on
is
is
is
endothermic
is
system
Observations
In
an
heat
sign
and
U
solvent,
a
For
change
system
C
exchanged
laboratory,
negative.
c alorimeter
the
reaction,
enthalpy
whether
A
of
change, and endothermic
The
i
n
c alorimeter.
being
exothermic
change
system,
perspective
enthalpy
change.
o
To
an
negative
v
the
in
the
a
enthalpy
p
enthalpy
from
have
positive
y
dened
a
r
e
is
reactions
have
O
areaction.
n
is
from
y
exothermic
system
made
for
new
system
system to the
contrast,
t
i
s
the
takes
the
process.
when
the
the
In
l
surroundings
is an
are
quantied
into
from
reaction.
reactions.
Chemic al
break
transferred
y
endothermic
and
absorbed
reaction
bonds
system
is
exothermic
P
an
is
products.
chemic al
reaction
heat
endothermic
new
heat
r
new
which
chemic al
create
which
e
When
in
dened as an
s
s
reactions
is
a
O
Energy proles (Reactivity 1.1.3)
Energy proles are a visual representation of the enthalpy change during a reaction.
Activation energy, E
, is the
a
From an energy prole, you can determine the enthalpy of the reactants and the
v
minimum energy required for the
products, the activation energy (E
from
E
M any
chemic al
the
energy
reactions
system
level
reactions
to
and
are
at
a
the
), and the enthalpy change for the reaction.
a
reaction to take place. You will study
are
exothermic.
surroundings.
considered
lower
to
energy
be
The
lower
level
and
In
these
reactions,
reactants
in
of
stability.
this
energy
reaction
Products
considered
to
be
for
more
is
released
activation energy in Reactivity 2.2.
are at a higher
exothermic
energetic ally
favourable.
391
Reactivity
1
What
drives
chemic al
reactions?
Consider
the
reaction
displacement
Zn(s)
Measured
+ CuSO
quantities
The
is
using
measured
is
(aq)
of
a
→
the
Cu(s)
copper(II)
is
the
the
heat
+
ZnSO
and
or
and
change
data-logging
As
the
copper(II)
a
transferred
in
are
the
the
r
NO
i
4
This
heat
is
apparatus
example
the
3
(s)
3
,
to
is
are
an
form
the
less
products
stable
important
aqueous
than
of
the
the
component
ammonium
reaction
of
and
fertilizers.
When
nitrate ions, the
decreases.
NH
by
4
the
the
(aq)
+
reaction
reaction
endothermic
NO
3
(aq)
system
mixture
reaction
from
will
the
surroundings. The
feel cold to touch. This is an
(gure 7).
endothermic reaction
a
ygrene
products
E
a
l a i t n e t op
ΔH
reactants
reaction

Figure 7
coordinate
Using a thermometer or a temperature probe,
you would
in the temperature of the reaction mixture in an endothermic reaction.
the products is greater than that of the reactants.
energetic ally less stable than the reactants
392
are at a
reactants.
+
→
absorbed
an
NO
solution
containing
of
4
water
o
of
in
energy prole
reaction,
therefore
u
l
E
v
O
f
x
t
a
r
o
d
NH
and
nitrate, NH
dissolves
temperature
example of an
endothermic
level
describe the products as being
o
solid
an
C
Ammonium
an
is
n
U
energy
6
n
p
You would
observe an increase
The enthalpy of the
observe a decrease
The enthalpy of
The products are described as being
y
l
you would
O
products
y
r
e
v
consider
higher
the
gure
reaction is
y
t
i
s
laitnetop
i
n
you
in
The
P
ygrene
of the reactants.
energetic ally more stable than the reactants
graph
reaction,
ΔH
Using a thermometer or a temperature probe,
products is lower than that
this
a
reactants
in the temperature of the reaction mixture in an exothermic reaction.
If
In
temperature of the
surroundings.
reaction coordinate
The
mixed in a
equipment.
result,
to
zinc
exothermic reaction

Figure 6
by zinc:
temperature of the solution
exothermic.
E
ion
(aq)
solution
the
system.
is
4
of
e
and
reaction
displacement
sulfate
stirred,
thermometer
by
increases,
therefore
4
mixture
generated
solution
involving
s
s
c alorimeter.
heat
between zinc and copper(II) sulfate solution. It is a single
reaction
Reactivity
1.1
Measuring
enthalpy changes
Global impact of science
Developments in science may have ethical, environmental, political, social,
cultural and economic consequences, which must be considered during
s
s
decision making. The pursuit of science may have unintended consequences.
German chemist Fritz Haber was awarded the Nobel Prize in Chemistry in
1918 for developing a method to chemically extract nitrogen from the air
production of fertilizers that began during the green revolution and continues
r
today. However, his process also provided Germany with a source of
ammonia that was used for the production of explosives during the First World
oen
and
you
would
graphs.
exothermic
to
in
able
writing
sketch
C an
and
be
you
read
texts
write
writing
reports
graphs
and
and
endothermic
thinking
communic ation
of
to
are
intended
formal
styles.
and
also
for
answering
extract
analyse
data
the
reactions?
involves
form
to
of
meaningful
that
of
the
4
Cl(s)
→
BaCl
following
is
2
(aq)
correct
l
a
enthalpy
decreases
v
D
enthalpy
products
E
C
products
decreases
for
3
(g)
+
2H
this
chloride, NH
products
enthalpy
products
have
than
have
than
reactants
lower
the
reactants
have higher
than
the
Cl:
O(l)
ΔH
= +164 kJ mol
reaction?
lower
the
4
1
2
Enthalpy
increases
O
B
2NH
ammonium
u
f
x
Temperature
A
+
with
represent
n
2NH
reacts
specic
information
o
Which
+
,
using
communic ation
proles
i
(s)
2
t
a
2
Ba(OH)
r
o
Ba(OH)
hydroxide,
your ability to
reports
U
Barium
d
1.
written
accurately sketch these
diagrams, including all of the components.
Practice question
in
examination questions,
and
energy
Try
and
improving
dierent audiences. In
laboratory
Another
verbally
about
C
from
utilize
ability
to
accepted
you
i
n
your
need
forms
forms of
o
you
However,
dierent
and
when
and
v
terminology
skills.
mind
skills
communic ate
p
science,
write
of
y
read
to
range
eectively
r
e
communic ation
comes
wide
to
n
most
a
ability
l
form
cover
Your
O
skills
communic ation.
t
i
s
Communic ation
is
y
Communic ation skills
y
P
War. The global impact of science is evident in Haber ’
s research.
ATL
e
by reacting it with hydrogen. Haber ’
s discovery allowed for the large-sc ale
reactants
have higher
Stability
products
than
the
products
stable
are
than
products
than
are
the
less stable
reactants
more
the
are
reactants
less stable
reactants
products
are
more
increases
enthalpy
than
the
reactants
stable
than
the
reactants
393
Reactivity
1
What
drives
chemic al
reactions?
⦵
Standard
temperature
and
pressure
(STP) conditions are denoted by the
symbol
⦵.
STP
is
a
Standard enthalpy change, ΔH
(Reactivity1.1.4)
temperature of
⦵
The
273.15
K
and
a
pressure
of
standard enthalpy change for
transferred
ambient
at
pressure
(SATP)
refer
to
from
⦵
,
refers
conditions
to
the
and
heat
states.
the
change
in
temperature
of
a
pure
It
c an be
substance. The units of
1
are kJ mol
reaction conditions of
K
and
100
kPa.
STP and
⦵
T
o calculate ∆H
SATP
ΔH
standard
conditions
are
e
298.15
under
more
ΔH
practic al
pressure
temperature
determined
and
constant
s
s
Standard
a reaction,
100kPa.
for a reaction, you therefore need to determine the amount of heat
given in the
released or absorbed in the course of that reaction. This can be done by measuring
r
section 2 of the data booklet.
the change in temperature of a pure substance to or from which this heat is
such as water
, you need to know the specic heat capacity, c, of thatsubstance.
Specic heat
l
Substance
1
c apacity / kJ kg
1
The
specic
example,
a
pure
kJ
to
the
specic
raise
the
heat
of
copper
0.385
of
1
c apacity
same
amount
Specic
heat
with
size
the
is
heat
up
has
a
given
heat
is
substance,
transferred
c apacity is an
of
the
the
pure
system
same
dened
as
of
the
to
heat
the
rise
higher
the
the
by
1
amount
c apacity
in
1
the
•
the
amount
supplied.
K.
rise
of
that
For
as
a
heat
does
ton
For
, so it takes
lower
the
specic
temperature when the
not
example,
1
K.
1
K
The
in
1
vary in magnitude
3
a
block
10
of
cm
sample of
copper.
When
you
o
mass of that substance
1
or
temperature depends on:
C
the
by
°C
sample.
v
•
i
n
the identity of that substance
heat
ethanol
described.
•
of
substance
intensive property
being
specic
substance,
that
p
copper
of
a
of
y
c apacities of
water, ethanol and copper
of
kg
kg
r
e
The specic heat
substance
1
c apacity of ethanol is 2.44 kJ kg
temperature
2.44
heat

Table 1
of
temperature
O
ethanol
c apacity
the
t
i
s
2.44
raise
n
4.18
to
y
needed
water
heat
K
n
U
Practice questions
2.
Using
table
1,
o
1000 kg of copper
When
u
l
a
E
v
O
f
x
Which
of
masses
energy,
energy
is
the
of
two
their
following
dierent
substances,
temperatures
is
rise
by
specic
heat
c apacity
of
X
is
twice
b.
The
specic
heat
c apacity
of
X
is
half
c.
The
specic
heat
c apacity
of
X
is
one
d.
The
specic
heat
c apacity
of
X
is
the
a.
table
More
50
b.
If
°C
the
c.
If
1,
state
heat
than
same
equal
is
If
equal
which
of
the
needed
to
increase
50
g
heat
of
masses
together,
d.
to
raise the
the
is
of
of
ethanol
supplied
the
nal
masses
temperature,
of
5
X
°C
and
and
Y
,
10
absorb
°C,
the
same
respectively.
correct?
The
Using
required
by 1 K.
a.
temperature
394
much
water
equal
of
how
following
1
amount
4.
of
the
b.
i
3.
kg
c alculate
of
a.
t
a
r
o
d
temperature
water
water
h
the
of
of
Y
.
Y
.
that
same
following
as
of
Y
.
that
statements
of
is
temperature
Y
.
correct.
of
50
g
of
water
to
equal
masses
increase
20
°C
and
of
ethanol
and
water, the
more.
ethanol
at
50
°C
are
mixed
temperature will be 35 °C.
water
ethanol
and
gives
by
by 50 °C.
will
at
that
that
ethanol
out
at
more
50
°C
heat.
cool
down
to
room
y
P
transferred. When calculating the amount of heat lost or gained by a pure substance
Reactivity
Specic
a
pure
heat
Q
where
c apacity,
substance
m
=
is
c,
using
is
the
used
to
c alculate
the
amount
of
heat,
Q,
absorbed
1.1
Measuring
enthalpy changes
by
relationship:
mcΔT
mass
of
the
pure substance in kg and
∆T
is
the
change
in
temperature
s
s
of that substance in K.
Heat,
Q,
is
related
to
enthalpy
change,
∆H,
by
the
following
equation:
e
Q
ΔH = −
n
n
is the number of moles of the
the
reacting
therefore
substances
how
are
much
limiting reactant.
with
the
product
said to be in
least
c an
In
a
reaction, the limiting
stoichiometric
be
formed.
In
amount
present,
contrast, the other
excess
experimental
that
directionality
the
are
and
on
heat
be
direction.
With
change
the
experimental
in
the
coee-cup
enthalpy
be
transferred
surroundings
in
in solution
A coee-cup
c alorimeter
n
2.
to 25.0 g of water in a
a temperature rise of 3.80K
is transferred
to a 100.0 g sample
resulting in a temperature rise from 22.0°C
to 26.0 °C.
C alculate the specic heat
c apacity
ofiron.
Assume that
u
f
x
chloride.
180.0 J of heat
ofiron,
C alculate the enthalpy change of
dissolution for 1 mol of lithium
containing reactants

Figure 8
t
a
c alorimeter,
was recorded.
i
was added
o
d
coee-cup
r
o
LiCl,
nested together
experiment.
When a 1.15 g sample of anhydrous lithium
chloride,
cork stopper
for a
between the
every
glass stirrer
two polystyrene cups
lower in magnitude than the actual
will
Worked example 1
1.
of
empiric al data is constant
measured
the
eect of
U
contents
the
C
always
some
consequence
same
reaction
will
a
eect
the
as
introduces
and
assessed.
c alorimeter,
value,
method
analysed
thermometer
o
in
be
coee cup
convenient
i
n
always
Their
This
a
p
errors
procedure.
c an
is
v
Systematic
and
procedure.
errors
polystyrene
change
y
their
a
r
e
systematic
in
enthalpy
O
the
measure
n
y
reactions
to
t
i
s
Measurement
Performing
l
reacting
limits
substance
y
which
is
P
reactant
r
where
the heat capacity of lithium chloride itself is negligible.
=
mcΔT
=
0.025 kg
=
0.397 kJ
1
mol
E
Now
you
×
need
a
Q
2.
1
4.18 kJ kg
v
O
1.
l
Solution
to
First,
determine
the
change
in
temperature,
∆T:
1
K
× 3.80 K
∆T
=
Substitute
determine
the
energy
gained
for
(299
the
0.180 kJ
=
−
295) K
=
values into
0.100 kg
4 K.
Q
×
c
=
mcΔT:
×
4 K
ofLiCl.
M ake
c
the
subject
of
the
equation
and
solve:
1.15 g
n(LiCl)
=
=
0.180kJ
0.0271 mol
1
c
42.39gmol
−
=
=
0.450 kJ kg
−
1
1
K
0.100kg×4K
−0.397kJ
Q
1
=
ΔH = −
n
= −14.6 kJ mol
0.0271mol
395
Reactivity
1
What
drives
chemic al
reactions?
Practice questions
5.
C alculate
the
energy
absorbed
by
water
when
the
temperature of 30 g of
1
water
6.
0.675
is
raised
kJ
of
by 30
heat
is
°C.
The
specic
transferred
to
125
heat
g
copper
of
water
metal.
is
4.18 J g
1
specic heat c apacity of 385 J kg
K
. C alculate the change in temperature
of the copper metal.
task,
the
we
will
enthalpy
at
reaction
the
for
method
the
3.
used to
At
of
exothermic metal
three
zinc
minutes,
powder,
introduce
record
the
4
(aq)
→
Cu(s)
+
ZnSO
4
Continue
minutes
(aq)
to
after
5.
Tool
1:
Measuring
Produce
a
variables
Applying
techniques
6.
Use
Q,
2:
Applying
•
Tool
3:
Processing uncertainties
•
Tool
3:
Graphing
•
Inquiry
2:
Inquiry
3:
Processing data
Evaluating
Materials
•
The
•
the
•
measuring
•
thermometer
c alorimeter
r
o
or
temperature
zinc
copper(II) sulfate solution
powder
l
O
Method
Using
the
a
the
25
cm
of
solution
E
Using
of
temperature
The
the
or
the
to
threeminutes,
or
c apacity
but
The
Loss
of
measure
specific
same
heat
396
is
to
as
graph to
c alculate
for
the
the
heat
released,
reaction,
∆H
are made when using this
from
the
reaction is completely
as
an
insulator
However,
and
heat
much
is
the
heat
the
against
heat
loss to
coffee cup also has
transferred
smaller
temperature
of
five
water.
acts
heat
that
from
source
of
from
a
c apacity
of
the
error
quantify.
c alculated
CuSO
coffee-cup
4
solution.
c alorimeter.
This
a
temperature
probe,
loss
solution
every
30
of
recorded
until
a
constant
heat
than
that
reached
to
of
is
it
from the
water.
an
accurate
evolved during the
system
in
of
an
aqueous solution is
water.
this
The
to
the
change
graph
surroundings is the
experiment and one that is
will
in
temperature,
include
a
∆T,
systematic
will
means
be
c alculated
lower
value of
that
than
Q
the
the
lower
maximum
error.
temperature
theoretic al
than
the
value, making
actual
value. The
record
seconds
of
errors
on
the
result
of
subsequent
c alculations is
for
in
temperature is
procedures.
achieved.
it
maximum
important
up
cup
to
been
reaction.
•
effect
of
the
representation
the
thermometer
to
time
up
has
3
1.0 mol dm
to
•
the
accurately
coffee
heat
difficult
balance,
released
surroundings.
water,
main
3
mass
Transfer
2.
electronic
v
the
an
a
1.
probe
u
•
f
x
3
1.0 mol dm
t
a
cylinder
o
coffee-cup
i
•
•
d
electronic balance
assumptions
heat
The
a
•
of
transferred
n
U
•
variables
number
method:
for
temperature.
change
Assumptions and errors
A
Controlling
∆T to
enthalpy
C
1:
process data
v
Inquiry
to
i
n
•
technology
versus
in
p
Tool
value of
the
o
•
your
and
change
readings
temperature
y
1:
the
r
e
Tool
maximum
temperature
determine
•
temperature
the
O
reached.
Relevant skills
•
take
t
i
s
CuSO
n
y
4.
+
mass of zinc and
commence stirring.
between zinc and copper(II)
sulfate:
Zn(s)
between 1.3 g and 1.4g
exact
l
displacement
look
change
considering
improvements
in
experimental
y
skills
P
this
c alculate
e
r
Investigation to nd the enthalpy change for a reaction
In
1
K
Copper metal has a
s
s
1
of
c apacity
Reactivity
An
accepted
temperature
is
to
look
is
in
figure
cooling
complete,
when
9.
zinc
Amore
is
Measuring
enthalpy changes
c alculating the maximum
for
section
and
systematic
of
the
errors in data
curve after the
extrapolate this back to the
introduced
accurate
at
value
3
for
minutes,
ΔT
as
shown
c an then be
determined.
s
s
moment
the
of
compensate
C° / erutarepmet
reaction
at
method
to
1.1
ΔT
e
r
4
6
8
10
time / min
to measure the temperature change for the
a 1.0 mol dm
solution of copper(II) sulfate.
The following results were recorded:
determine
=
4.69 kJ
the
4.18 kJ kg
limiting
heat
released:
1
K
×
reactant
for
In
39.0 K
the
f
x
n(Zn)
12
These
truth,
u
l
E
in
and
Scientists
certainty,
power,
explanation,
responsibility.
perform
experiments
r
and
process
the
raw data to
1.37g
=
enable
us
to
draw conclusions.
1
We
=
compare
experimental
0.0210 mol
and
theoretic al
values.
What
concepts do we utilize when
n(CuSO
4
) = c × v
justifying
3
=
0.0288 dm
How
so
it
is
the
do
our
we
conclusions?
use
evidence?
judgments
subjective
0.0288 mol
or
amount,
our
3
×
Are
smaller
evidence,
interpretation,
values
= 1.00 mol dm
a
there
focus.
=
M
a
v
O
present
are:
in
objectivity, perspective, culture,
objective?
and
is
knowledge,
65.38gmol
Number of moles of copper(II) sulfate,
Zinc
of
concepts
justic ation,
reaction.
m
Number of moles of zinc,
theory
are
t
a
determine
×
of
o
0.0288 kg
amount
1
i
=
the
o
to
r
o
Then,
mcΔT
d
Q
=
TOK
n
Q
the enthalpy change for this
U
Solution
First, use
and
39.0
C
i
n
released
reaction.
v
1.37
M ass of zinc / g
∆T / °C
p
28.8
M ass of copper(II) sulfate solution / g
y
r
e
3
reaction between zinc powder and
of heat
n
c alorimeter was used
O
t
i
s
Worked example 2
Determine the amount
l
Example of a temperature vs time graph for a
c alorimetry experiment
A coee-cup
y

Figure 9
y
2
P
0
limiting
reactant.
You
c an
When
appraising
analysing
experimental
c alculate
limitations,
how do
Q
the
enthalpy
change
of
reaction
from
ΔH = −
:
assumptions
have an impact on
n
4.69kJ
1
∆H
=
–
=
our
perceptions?
−223 kJ mol
0.0210mol
397
Reactivity
1
What
drives
chemic al
reactions?
Combustion of primary alcohols
You
c an
determine
common
experiment
the
in
a
enthalpy change of combustion of
school
laboratory.
After
2.
Determine
an
repeating
alcohols,
you
c an
subsequently
mass
of
the
spirit
burners using
3
Accurately
determine
the
analyse this data and
mass
of
30
beaker
or
metal
cm
in
a
250
cm
Using
either
determine
a
temperature
and
record
the
probe
initial
or
a
temperature of
1:
Recognise
ethic al
or
and
address
environmental
the
issues
in
relevant
an
safety,
investigation
Measuring
temperature
and
Ignite
a
period
1:
C alorimetry
Inquiry
2:
when
and
how to insulate
a.
loss or gain
Identify
observations
and
and
record
relevant
relevant
qualitative
b.
quantitative data
which
each
allow
of
Determine
the
burners,
each containing one of the
c are
alcohols:
methanol,
ethanol,
propan-1-ol,
and
pentan-1-ol
beaker
•
tripod
•
temperature
or
metal
c alorimeter
S afety
they
be
are
handled
generally
suitable
your
level of
v
•
identify suitable
the
this
a
final
to
mass
the
temperature
burn
for a period of two
of
flame
each spirit burner
is
extinguished.
Take
burner to calculate the amount of heat released, Q,
and the enthalpy change of combustion, ∆H, for
eachalcohol.
thermometer
beaker
clamp
water
spirit burner
wick
methanol

Figure 10
E
aligned with
safety policies.
extra
bec ause the burner will be hot.
A typic al arrangement
of experimental
apparatus for an enthalpy of combustion determination
disposal methods
and
until
hazards and
control measures
health
burn
experiment. In
risk
relevant
school’s
hazardous and
a
the
determine
for
you should:
hazards
•
your
398
assessment,
assess
identify
assessment
l
•
sources,
risk
identify the
O
•
risk
a
c are
u
complete
with
i
Using
f
x
1.
of
t
a
Instructions
disposed
flammable,
r
o
volatile.
and
to
reached
o
bec ause
should
d
Alcohols
water. The
c an be set in one of two
Use your values of ∆T of the water and ∆m of the
n
U
probe or thermometer
c alorimeter
o
•
alcohol
after
C
electronic balance
i
n
•
v
7
.
butan-1-ol
is
p
following
or
the
y
immediately
spirit
burns
alcohol
30°C
each
r
e
6.
five
it
heat
minutes.
Materials
•
beaker
to
ways:
allow
change
sufficient
the
burn
O
•
different
Appreciate
heat
over
t
i
s
against
1:
to
n
Inquiry
under
alcohol
l
Tool
•
burner
the
y
•
spirit
allow
mass
y
1:
P
5.
and
Tool
r
thewater.
Tool
thermometer,
e
4.
•
water
c alorimeter.
patterns.
Relevant skills
•
of
3
contained
identify
initial
balance.
several times with a homologous series
3.
of
the
electronic
s
s
the
alcohols
Reactivity
1.1
Measuring
enthalpy changes
Research skills
ATL
Cite
your
sources
fully,
according
to
your
school’s
citing
and
referencing
system.
s
s
Worked example 3
to measure the amount
of heat
by
The following results were recorded:
the enthalpy change of
Solution
=
mcΔT
to
determine
the
amount
1
Q
3.91 kJ
with
for
oxygen
this
in
a
reaction
K
×
30.0 K
combustion
bec ause
reaction. Methanol is the
oxygen
Number of moles of methanol,
n(CH
3
OH)
=
Mr
is
present
in
U
1
32.05gmol
the
=
−
change
of
1
−359 kJ mol
=
0.0109mol
experimental
identic ation
to
measure.
environment
discrepancies
the
give
values.
E
the
of
typic ally
theoretic al
dicult
to
aspects
This
smaller
is
the
a
useful
and
result
TOK
experimental
set
from
of
for
Q
:
ΔH = −
n
of
helps
and
data,
examination
in
the
us
results.
loss
from
involve
systematic
their
errors.
directionality
C alorimetry
than
the
assumption
to
and
and
of
temperature
heat
make
raw
random
experimental
change
usually
negligible.
between
of
reaction
evaluated
errors
analysis
a
Scientists
is
be
systematic
v
from
the
provide
c an
a
essential
experiments
of
that
l
are
O
The
experiments
procedures
u
f
x
Thermochemistry
air.
t
a
r
o
3.91kJ
ΔH
enthalpy
o
c alculate
the
0.0109 mol
i
c an
d
You
in
n
0.348g
=
=
excess
C
m
o
reacts
reactant
4.18 kJ kg
p
=
×
released:
v
limiting
0.0312 kg
heat
i
n
Methanol
=
of
1
y
Q
r
e
First, use
O
combustion for methanol.
n
and
t
i
s
released
l
y
30.0
∆T / °C
y
0.348
Change in mass of methanol / g
of heat
P
31.2
M ass of water / g
Determine the amount
released
r
the combustion of methanol.
e
A metal c alorimeter was used
is
predicted
system, which is
that
the
heat lost
understand our judgments of
theoretic al
values.
399
Reactivity
1
What
drives
chemic al
reactions?
Thermometric titration
The
neutralization
base
reaction
exothermic.
unknown
In
this
2.
between an acid and a
skills
concentration
of
task,
Review
hydrochloric
acid
the
change
in
is
added
to
percentage
sections in the
error and
Skills
chapter.
Rinse
and
fill
the
burette
with
sodium
hydroxide
temperature while sodium
solution.
hydroxide
titration,
by
3.
measuring
the
uncertainties
you will determine
the
acid.
The
Record
its
concentration.
temperature will
3
a
maximum
when
the
acid
and
base
are
Add 25 cm
of acid solution to the cup and place it
e
4.
reach
s
s
the
is
mixed
under the burette. Nest it inside a second cup, for
together in stoichiometric amounts.
r
additional thermal insulation. For safety,these cups
should be placed inside a beaker to avoid tipping over.
Relevant skills
•
Tool
2:
Use
•
Tool
3:
C alculate
C alorimetry
and
acid–base
titration
5.
Position
the
the
initial
temperature
temperature
probe
of
the
in
acid
the
in
acid
the
and
record
cup.
sensors
percentage
Add
a
small
solution
Tool
3:
Understand
and
the
signific ance of uncertainties in
temperature
processed data
7
.
•
Tool
3:
Continue
Propagate uncertainties and state them to an
appropriate
level
of
the
(~5
solution
precision
acid,
cm
stirring
and
small
the
graphs
decreases
8.
Inquiry
3:
when
and
Identify
random
and
discuss
sources
of
systematic
error
•
hydroxide
Hydrochloric
acid
solution
is
is
thermometer
•
graduated
•
burette
•
~50.0 cm
or
temperature
probe
pipette and filler
l
O
over
v
this
a
through
the
information,
risk
your
400
E
Read
by
showing
hydroxide
Extrapolate
the
two
Determine
the
with
absolute
sure
you
consecutive
teacher.
temperature
vs
volume of
added.
sections
of
the
reached
graph to find the
during
the
titration.
concentration of the acid, along
and
state
several
your
solution
temperature
hydroxide
all
percentage
values
to
an
uncertainties.
appropriate
M ake
level of
precision.
4.
C alculate
acid
5.
the
percentage
error
of
your
experimental
concentration.
Determine
absolute
values
the
and
to
an
C alculate
enthalpy
of
neutralization, along with
percentage uncertainties. State all
appropriate
the
level
percentage
of
error
precision.
enthalpy
7
.
of
of
your
experimental
Comment
neutralization.
on
the
relative
impacts
of
systematic and
hydrochloric acid of unknown concentration
Method
1.
graph
maximum
6.
a
3
30.0 cm
instructed
sodium
sodium hydroxide solution of known
concentration
•
u
f
x
3
t
a
•
polystyrene cups
o
cm
2.
3.
i
250
d
r
o
3
two
corrosive.
a
sodium
corrosive.
Materials
•
as
C
Sodium
protection.
Plot
n
•
eye
1.
U
Wear
up
Questions
S afety
•
Clear
how to insulate
loss or gain
i
n
and
Appreciate
heat
o
•
1:
v
against
of
y
Inquiry
Extrapolate
hydroxide
Record the highest
temperature until the
p
•
3:
sodium
volumes
recording
temperature
r
e
Tool
of
reached with this addition.
adding
readings.
•
)
gently.
O
raw
to
t
i
s
•
volume
error
n
interpret
y
3
6.
and
teacher.
for
8.
relevant
this
error
on
concentration
safety, materials and method. Use
and
assessment
random
safety data, to complete
practic al
work
and
show it to
Suggest
and
the
and
values
obtained
enthalpy
explain
two
made to this method.
of
for the acid
neutralization.
improvements that could be
y
1:
l
Tool
P
•
Reactivity
1.1
Measuring
enthalpy changes
Measurement
Experimental
precision.
errors
systematic
extent
are
c ause
errors
these
values
errors
c an
be
an
assessed
measurement
inaccuracy.
in
errors
in
What
enthalpy
of
in
lead
are
terms
to
of
their
accuracy and
imprecision,
some
of
neutralization
the
whereas
sources
experiment?
of
s
s
systematic
and
enthalpy
R andom
random
To what
quantiable?
water
is
a
bomb c alorimeter
“bomb”),
and
laboratories
described
the
in
(gure
this
11).
resulting
utilize
the
same
chapter. The
A
sample
is
burned
temperature change of
measured.
ignition
O
surrounding
c alled a
(c alled
research
p
oxygen
of
o
list
all
the
in research laboratories
features
that
are
labelled.
feature.
measurements
and
properties
i
the
and
each
a
why
accurate
of
l
What
c arefully
purpose
v
O
Consider
highly
diagram
the
t
a
the
c alorimeter used
u
Deduce
d
r
o
Study
2.
4.
of a bomb
f
x
1.
3.
Diagram
o
n
water bath
bomb

Figure 11
temperature
probe
U
calorimeter
supply
C
i
n
stirrer
v
jacket
y
r
e
device
y
is
chamber
in
experiments
n
used
conducted
c alorimetry
t
i
s
the
a
the
y
instrument
inside
experiments
as
l
principles
e
r
C alorimetry
P
Thinking skills
ATL
obtained
with
a
bomb
c alorimeter
are
precise.
water
make
it
suitable
for
c alorimetry
experiments?
E
401
Reactivity
1
What
drives
chemic al
reactions?
End of topic questions
Topic review
Using
your
knowledge
from the
Reactivity 1.1
topic,
answer
the
guiding
question
as
fully
as
s
s
1.
possible:
What can be deduced from the temperature change that accompanies chemical or physical change?
r
Multiple-choice questions
is
correct
for
+ 6HCl(aq)
the
→
following
2AlCl
3
(aq)
reaction?
+
3H
2
(g)
Reactants
are
less
B
Reactants
are
more
stable
than
products
and
the
reaction is endothermic.
C
Reactants
are
more
stable
than
products
and
the
reaction
D
Reactants
are
less
exothermic
reaction,
B
In
an
exothermic
reversible
the
reaction,
endothermic
reversible
the
products
reaction,
are
the
is
correct
→
for
2Fe(s)
+
this
more
stable
activation
2
(g)
1
26.6
kJ
are
absorbed
C
53.2
kJ
are
released
for
every
mole
of
Fe
produced.
D
26.6
kJ
are
released
for
every
mole
of
Fe
produced.
4
reaction
(g)
2O
→
do
2
(g)
H(g)
+
the
→
(g)
+ Cl
NaOH(aq)
(g)
+
→
+
have
2H
2
Fe
of
→
forward
produced.
a
lower
O(g)
NaCl(aq)
reactants.
the
produced.
Fe
NaCl(s)
HCl(aq)
E
402
2
(g)
of
mole
a
Na
v
D
mole
every
reactants
CO
Br(g)
+
C
every
for
l
HBr(g)
+
for
+
H
2
O(l)
energy
than
the
reaction
the
t
a
CH
O
B
which
released
u
A
are
i
13.3
B
r
o
A
In
kJ
of
reactants.
forward
o
−26.6 kJ mol
f
x
5.
=
d
ΔH
the
the
than
energy
reaction?
3CO
than
of
n
3
energy
energy
U
2
more
activation
reaction.
3CO(g)
exothermic.
exothermic.
products?
is
greater than that of the
o
endothermic
an
statement
is
is
C
an
In
+
have
the
i
n
In
(s)
reaction
reaction.
D
O
products
reaction,
C
Fe
the
reaction is endothermic.
p
an
Which
and
the
v
In
reverse
products
and
correct?
A
reverse
4.
is
than
products
O
statement
stable
than
y
Which
stable
t
i
s
A
r
e
3.
= −1049 kJ mol
n
y
1
ΔH
l
2Al(s)
reaction
is
greater than that of the
y
Which
P
2.
e
Exam-style questions
Reactivity
1.1
Measuring
enthalpy changes
Extended-response questions
6.
Nitrogen
NO
2
(g)
dioxide
+
CO(g)
and
→
c arbon
NO(g)
+
monoxide
CO
2
react
according
to
the
following
equation:
(g)
1
ΔH
=
−226 kJ mol
C alculate
b.
State
the
enthalpy
acid
for
change
the
for
the
reverse
reaction of NO
in
2
reaction.
the
[1]
atmosphere to
deposition.
[1]
3
3
Powdered
an
zinc
insulated
was
reacted
beaker.
with
25.00
Temperature
was
cm
of
plotted
1.000 mol dm
against
copper(II) sulfate solution in
r
7
.
time.
40
Y
30
25
2
4
6
8
10
time / min
a
point
State
the
d.
To
in
f.
Predict,
giving
a
experiment
chip
each
made
reason,
would
(crisp)
experiment
copper(II)
c alculate
were
when
how
m
the
using
nal
with
ignited,
=
and
E
the
the
was
[1]
c arried
was
used
out
ve times.
but
the
mass of
of
[1]
energy
released.
1
values
enthalpy
of
c
=
for
4.18 J g
m
and
reaction
the
theoretic al
the
ame
was
1
K
.
c.
[1]
c alculated
from
value.
used
to
[1]
heat a test
7.8
1.2
21.3
Final temperature / °C
in
assumption
water.
Initial temperature / °C
data
one
was chosen
line.
sulfate
amount
these
M ass of chip / g
C alculate
the
reaction
[1]
[1]
reason, if zinc or copper(II) sulfate
25.00 g and
M ass of water / g
a.
a
of
beaker.
used, and outline
the
with
the
change
extrapolated
trial.
to
compare
was
be
in
a
containing
used
enthalpy
the
of
Suggest,
l
potato
tube
for
time.
c alculation
v
O
A
the
reaction,
concentration
was
assumption
f
x
this
an
in
should
on
placed
i
State
used
that
was
t
a
e.
excess
of
the
line.
u
values
and
each
q = mcΔT
c alculate
(red)
temperature
enthalpy
volume
in
to
zinc
represents.
temperature
this
r
o
The
the
dierent
be
formula
used
extrapolated
maximum
powdered
o
should
the
graph
d
was
The
8.
the
choosing
same
zinc
the
temperature
on
determine
The
which
on
n
made
at
Y
C
maximum
time
U
c.
the
o
State what point
v
Estimate
b.
i
n
a.
y
0
p
r
e
20
O
t
i
s
ΔT
35
n
y
C° / erutarepmet
45
at
l
50
y
P
55
The
e
produce
the
equation
s
s
a.
heat
table
22.6
required,
above
and
in
kJ,
from
to
raise
section
the
2
of
temperature
the
data
of
the
water, using
booklet.
[1]
1
b.
Determine
the
enthalpy
of
combustion
of
the
potato
chip, in kJ g
.
[1]
403
Reactivity 1.2
Energy cycles
in reactions
The
utilization
of
energy
energy
to
its
consume
lives.
Thermodynamics
interconversions.
states
another
that
and
energy
that
the
c an
total
rst
nor
of
to
law of
converted
amount
law
is
oen
According
vast
is the
The
be
This
Agricultural,
energy
another.
of
for
the
for
most
c alled the
this
destroyed;
account
from
to
it
This
law,
c an
only
means
and
law of conservation of energy.
energy
be
that
quantify
fundamental
c an
for
all
enthalpy
change
for
a
reaction
is
independent
of
the
pathway
in
and
thermodynamic
uses
enthalpy
of
a
data
to
reaction
U
changes
in
of
law,
the
used
of
or
,
f
data
are
Hess’
s
law
enthalpy of
enthalpy change
Born–Haber
Hess’
s
the
data
c alculate
Reactivity 1.2.5 — A
applic ation
applic ation
formation
C
i
n
combustion
of
⦵
ΔH
cycle is an
to
show
energy
formation of an ionic compound.
n
o
i
(Reactivity 1.2.1)
When
to
a
chemic al
create
chemic al
u
chemic al
when
l
a
E
v
O
f
x
t
a
r
o
d
Bond-breaking and bond-forming
reaction
products.
bonds
of
made
to
place,
bonds
form
in
the
bond-breaking is an
chemic al
energy
understanding
takes
Chemic al
are
bonds:
new
transfer
bonds
between
the
energy
the
atoms
the
reactants
products.
of
the
system
and
the
involved
in
reactants
are
Energy
endothermic
form: bond-making is an
changes
the
broken
is
required
process.
chemic al
is
an
rearranged
new
to
Energy
exothermic
surroundings
a
are
and
break the
is
released
process. This
essential part of
reaction (Reactivity 1.1).
Laws and theories
The
nor
law of conservation of energy
destroyed;
predictions
power.
404
science.
c alculations.
o
v
Reactivity 1.2.4 — An
between the initial and nal states.
c an
enthalpy changes of
formation,
p
used
,
form
we
LHA
the
c
one
reaction
O
that
of
y
r
e
states
ΔH
created
from
energy changes. This is one
⦵
combustion,
law
given
principles
Reactivity 1.2.3 — Standard
energy.
Reactivity 1.2.2 — Hess’s
neither
t
i
s
Reactivity 1.2.1 — Bond-breaking absorbs and bond-
releases
a
the
remains constant.
Understandings
forming
be
converted
n
system
our
l
form
given
to
all
y
a
central
activities
daily.
and
thermodynamics
one
is
to
What
it
c an
be
only
made
other
laws
be
but,
unlike
have
states
converted
you
that
theories,
come
energy
between
laws
across
in
c an
dierent
do
not
be
neither
forms.
have
created
Laws
allow
explanatory
your study of chemistry?
y
study
of
energy
domestic
P
amounts
of
and
r
industrial,
e
during reactions?
s
s
How does application of the law of conservation of energy help us to predict energy changes
Reactivity
1.2
Energy
cycles
in
reactions
Bond enthalpy
Imagine
a
molecule
(BE) is
The
dened
in
process
between
the
of
energy
of
new
required
gaseous
homolytic
two
to
covalent
ssion
species.
.
The
break
results
one
the
in
of
mole
of
under
electrons
the
the
hydrogen
energy. The
molecules
distributes
This
breaking
requires
bond enthalpy
bonds
by
homolytic
standard conditions.
from
formation
of
the
bond
radic als,
equally
indic ated
symbol.
example,
the
mole
of
the
the
2
atoms
the
following
bond
enthalpy
of
the
H–H
bond
is
equal
to
the
enthalpy change
r
For
•
as
molecule, H
hydrogen
reaction:
2
(g)
→
is
an
∆H =
endothermic
+436 kJ mol
process,
and
therefore it has a
positive
the
breaking
the
C–H
values
booklet.
of
bond
are
provided
Bond
the
enthalpy
same
enthalpy
bond
will
in
in
vary
table
values
a
1
(on
are
wide
the
variety
through
next
derived
the
of
page)
from
and
experimental data on
compounds.
alkane
series
For
bec ause
molecule
methane
This
is
bec ause
3
series
chemic al
of
steps
enthalpy for
environment
•CH
3
(g)
in
which
each
→
•CH
(g)
→
•CH(g)
→
•C(g)
+
+
+
•H(g)
BE
•H(g)
BE
•H(g)
BE
dierences
changes
(C‒H)
2
(C
in
the
changes
3
4
the
removal of
1
1
H)
= +462 kJ mol
(C
H)
= +424 kJ mol
(C
H)
= +338 kJ mol
1
products
for
upon
= +439 kJ mol
each
of
each
equation
1
and
explain
successivestep?
t
a
r
o
i
environment
1
hydrogen
o
the
(g)
BE
d
recognize
chemic al
2
•H(g)
one
removal would be
n
(g)
+
U
2
→
•CH(g)
the
a
bond
C
(g)
•CH
you
the
the
i
n
4
•CH
how
underwent
time,
hydrogen atoms.
CH
C an
a
o
successive
at
v
dierent.
of
removed
p
was
chemic al
y
a
atom
example,
the
r
e
environment of the individual bonds changes.
If
section 12 of
O
data
t
i
s
BE
the
n
y
enthalpyvalue.
Selected
l
Bond-breaking
2H•(g)
y
P
1
H
e
by
one
hydrogen
individual
s
s
ssion
simple
into
Bond enthalpies are average values and are therefore only an approximation. The
average bond enthalpy value of the C
u
l
f
x
H
H
H
H
C
H
a
C
H
C
H
v
O
H
1
H bond is +414 kJ mol
C
C
H
C
E
H
H
H
H

Figure 1
H
When you examine the structure of this hexane molecule,
chemic al environment
of the C–H
c an you see how the
bonds diers throughout the molecule?
405
Reactivity
1
What
drives
chemic al
reactions?
Average bond
Average bond
Bond
Bond
1
1
enthalpy / kJ mol
C =O
804
144
C =N
615
O=O
498
C ≡N
890
O−H
463
C−H
414
N =N
346
N ≡N
C−O
358
I
natural
What
sciences?
used
in
do
further
do
play
we
areas
of
n
U
H
data
models
How
other
chemic al
bonds
reactions
formed.
c an
in
be
the
c an
These
obtained
whether
be
explained
explanations
to
acquisition
determine
C
models
and
role
i
n
model.
data
in
and
o
empiric al
changes
broken
y
energy
bonds
p
of
v
model
151
298 K
r
e
Measured
I
O
enthalpies at
TOK
193
t
i
s
Average bond
242
a
of
modify
by the
are
and
based
rene the
knowledge in the
model
is
useful?
How
are
knowledge?
You can use bond enthalpy data to calculate the enthalpy change of reaction, ΔH:
C
H
ΔH = Σ(BE
H
Figure 2
H
2
and
A molecule of bromoethane,
Br, has one C
5
ve C
C bond, one C
Br bond
When
data
value,
do
when
you
of
as
not
formulas
c an
bonds
a
take
bond
of
– Σ(BE
c alculated
enthalpies
are
intermolecular
in
a
reaction
enthalpy
the
determine
broken)
reaction
bond
substances
performing
structural
type
and
and
bonds
from
formed)
bond
average
enthalpies
values.
will
dier
Additionally, bond
forces into account, which is particularly
are solid or liquid.
c alculations,
reactants
the
of
it
is
products
number
of
useful
to
involved
each
draw out the full
in
the
chemic al
reaction.
bond
From
present.
l
a
E
v
O
Models
The
model
of
bond
changes
taking
to
this
which
theoretic al
evidence.
central
406
change
actual
important
this,
H bonds
its
enthalpy
u
C
f
x

enthalpy
from
o
H
The
i
Br
t
a
r
o
d
C
of
breaking
place
model
values
Using
within
is
in
empiric al
in
formation
reaction
agreement
generated
methodology
and
a
from
data
to
science.
is
used
system.
with
bond
We
empiric al
energy
replace
or
to
represent
c an
assess
data,
data
modify
the
the
energy
extent
by comparing the
with
experimental
proposed models is a
y
Br
n
Cl
Br
945
l
Cl
839
470
r
614
C≡C
158
y
C=C

Table 1
H
N
P
C
N
e
436
s
s
H−H
O−O
C
on
enthalpy / kJ mol
Reactivity
1.2
Energy
cycles
in
reactions
Worked example 1
Using bond
C
2
enthalpy data,
H
4
(g)
+
HBr(g)
determine the enthalpy change of reaction for the following process:
→ C
2
H
5
Br(g)
s
s
Solution
Start
by
drawing
the
full
structural
formulas
of
all
of
the
H
H
bond.
values
Br)] − [5BE(C
H) +
2701
gas
to
data
formed.
changes
(g)
→
this
gas
2HCl(g)
c alculate
on
c an
CH
be
the
4
(g)
+
H
enthalpy
2
O(g)
change,
−185 kJ mol
bond
number
formed
→
O
l
Determine
the
the
steam:
there
as
are
ve C
follows:
Br)]
H
is
combined with
1
reaction is
u
with
gas
hydrogen chloride:
depend
Hydrogen
f
x
gas
2
for
is
used to make chlorine
chlorine
3H
2
+
Using
section
Linking question
1
in
kJ mol
,
for the
How
and
type
of
bonds
by
the
reaction
of
expect bond
to
relate to bond
length and polarity? (Structure 2.2)
How
CO(g)
this
you
data
natural
reaction,
in
kJ mol
does
the
strength of a
c arbon–halogen
1
for
would
broken and
enthalpy
industrially
(g)
ΔH,
.
enthalpy,
t
a
Enthalpy
H
booklet,
H–Clbond.
2.
+
process
the
o
the
(g)
of
i
of
2
change
r
o
12
enthalpy
produce
industrial
Much
d
Cl
The
an
product,
reaction
n
hydrogen
is
hydroxide.
the
o
process
sodium
U
chloralkali
BE(C
C
i
n
Practice questions
and
+
v
1
In
the
y
2636
gas
C)
for
p
=
The
BE(C
bond.
r
e
[(4 × 414) + 614 + 366] − [(5 × 414) + 346 + 285]
= −65 kJ mol
1.
Br
change
from the data booklet:
=
−
enthalpy
n
BE
the
O
Then, substitute in the
Therefore,
Σ(BE of bonds formed)
H) + BE(C=C) + BE(H
Br
y
= [4BE(C
H
t
i
s
ΔH = Σ(BE of bonds broken) −
Br
C
H
H
H bonds, one C=C bond and one H
C bond, and one C
H
C
y
four C
H
l
are
Br
e
there
H
+ H
C
r
reactants,
products:
P
the
bonds, one C
and
H
C
In
reactants
bond
aect the
, using
rate
of
a
nucleophilic substitution
3.
Methane
and
chlorine
Using
the
c.
Deduce
E
Write
b.
v
chloride.
a.
reaction? (Reactivity 3.4)
react
balanced
bond
enthalpy
a
section 12 of the data booklet.
change
produce
chemic al
enthalpy
and
to
for
explain
values
this
chloromethane
equation
for
this
and
hydrogen
reaction.
from the data booklet, determine the
reaction.
whether
this
reaction
is
exothermic or
endothermic.
d.
State
which
are
more
energetic ally
stable,
the
reactants or
theproducts.
407
Reactivity
1
What
drives
chemic al
reactions?
Hess’s law (Reactivity 1.2.2)
If
you
have
Delhi,
and
travelled
have
transport
one
the
millions
way
fun
is
to
New
York
experienced
to
of
travel
oen
the
people
between
working
out
City,
Shanghai,
subways
every
point
the
that
day.
A
In
and
fastest
Paris,
Seoul,
London,
M adrid or
criss-cross these enormous cities
any
B.
If
transport
you
are
network
an
there
adventurous
is
more
s
s
than
half
you
traveller,
route.
e
r
l
in Seoul provides many alternative
y
r
e
same
net
added
true
c an
together,
example,
the
3
(l),
a
is
chemistry.
summary
result
overall
trioxide, SO
for
be
in
an
reaction
as
for
follows:
1:
S(s)
(l)
usually
dierent
shows
reactions, which,
reaction.
oxidation
of
elemental
sulfur,
S(s), to
dicult
to
+
O
2
(g)
2
c arry
(g),
→ SO
2
out
and
in
one
then
step.
sulfur
Typic ally, sulfur is
dioxide
is
oxidized further
(g)
SO
2
(g)
O
+
2
(g)
→ SO
3
(l)
2
up
of
is
1
2:
the
would
basis
enthalpy
be
Hess’s
the
changes
same
as
the
that
occur
enthalpy
in
each of these individual steps,
change
of
the
overall
reaction. This is
law:
Regardless of the route by which a chemic al reaction proceeds, the
statesof the system are the same.
z
Hess’s
law
Hess’s
law:
is
an
the
applic ation
enthalpy
of
the
change
conservation
for
reaction A
of
energy
→ B (ΔH
x
)
law.
is
Figure
4
illustrates
equal to the sum of
Hess’s law
the
enthalpy
Hess’s
law
reaction
408
3
dioxide, SO
o
E
Figure 4
t
a
∆H
trioxide:
add
total
SO
equation
of
enthalpychange will always be the same, as long as the initial and nal
a
C
B

l
x
v
O
∆H
u
the
y
sulfur
i
d
r
o
f
x
∆H
you
→
process
to
Step
the
A
this
rst
Step
If
(g)
n
U
oxidized
sulfur
2
the
chemic al
number
2
However,
to
O
+
A
a
overall
3
S(s)
of
C
sulfur
be
it
i
n
For
c an
o
when
idea
reaction –
v
the
p
The
n
The extensive metro system
O
Figure 3
pathways for navigating the urban sprawl
changes
c an
by
be
for
the
applied
combining
to
other
reactions A
nd
the
→ C and C → B (ΔH
unknown
reactions
with
enthalpy
known
y
+ ΔH
change
for
z
a
).
given
enthalpy changes.
y
P
y
t
i
s

Reactivity
1.2
Energy
cycles
in
reactions
Worked example 2
The reaction for the formation of methanol is shown below:
1
C(s) + 2H
2
(g)
O
+
2
→ CH
(g)
3
OH(l)
Use Hess’s law and
s
s
2
equations 1–3 to c alculate the enthalpy change for the reaction above.
3
1
3
OH(l)
O
+
2
(g)
→
CO
2
(g)
+
2H
2
1
O(l)
=
−726 kJ mol
∆H
=
−394 kJ mol
∆H
=
−286 kJ mol
2
1
C(s)
+
O
2
(g)
→
O
(g)
CO
2
(g)
r
2
P
1
3
1
H
2
(g)
+
2
→
H
2
O(l)
2
are
two
methods
for
using
Hess’s
law
for a question like this: the
summation of equations method, and the
Summation of equations method
inspect
the
the
reactants.
methanol
that
reaction
involve
C arbon
is
a
c arbon,
reactant
two
moles
of
hydrogen
as
coecients.
a
reactant.
This
2H
2
→ CO
(g)
means
(g)
+
O
2
2
(g)
the
→ 2H
equation
ΔH
=
equation
enthalpy
2
you
need
reactants,
c an
be
used
−394 kJ mol
3
value
O(l)
ΔH
c an
be
must
=
used
also
to
and
1
(g)
Therefore,
that
this
as
U
C
stoichiometric
2
so
c alculation,
oxygen
o
require
doubled
O
your
and
equation2,
i
n
You
+
In
v
C(s)
in
scheme.
hydrogen
be
use
in
enthalpy
y
with
of
reactions
methanol
as
p
Start
formation
to
r
e
corresponding
O
t
i
s
enthalpy cycle diagram method
First,
n
There
l
y
Solution
y
∆H
e
CH
the
as
a
values
product.
written:
direction
as
written, but with
doubled:
1
−572 kJ mol
For
the
product
a
methanol,
reactant.
you
When
need
2
O
2
2
(g)
(g)
+
together,
O
+
+
+ 2H
2
2
2H
O(l)
→ CH
c ancelling
(g)
O
2
chemic al
u
CO
2
3
However,
equation,
OH(l)
the
the
equation
sign
of
must
the
v
the
common
are
reversed to make methanol a
change
must
also
be
changed:
3
1
O
+
2
(g)
ΔH
=
+726 kJ mol
2
the
species
common
to
both
sides
of
the
reaction, and add the
1
→
CO
2
(g)
ΔH
=
−394 kJ mol
1
→
(g)
2H
2
O(l)
ΔH
=
−572 kJ mol
ΔH
=
+726 kJ mol
ΔH
=
−240 kJ mol
3
1
2
→
O(l)
CH
3
OH(l)
+
O
2
(g)
2
3
1
2H
2
+
2
+
2
O
→
2
+
2
O
+ CH
3
OH
+
2
2
1
species
gives
the
overall
equation C
+
2H
2
O
+
2
→ CH
3
OH,
which
is
identic al
to
the
reaction
2
–1
Therefore,
omitted
be
enthalpy
2
question.
states
2H
+
a
+
E
the
C(s)
1
C
C ancelling
the
equations
(g)
l
f
x
O
Total
in
three
2
equation1.
the
t
a
the
values:
r
o
add
enthalpy
use
i
CO
Now
to
reversing
o
not
d
product,
n
We also need half a mole of oxygen. Oxygen is present asa reactant in two of the above equations, so ignore this for now.
to
the
save
total
enthalpy
change
for
the
reaction
is
equal
to
–240
kJ mol
.
Note
that
here
and
below
space.
409
Reactivity
1
What
drives
chemic al
reactions?
Enthalpy cycle diagram method
You
c an
cycle
represent
the
c alculations
needed
to
determine
the
enthalpy
change
of
reaction
in
the
form
of
an
enthalpy
diagram.
1
write
the
equation
you
are
trying
to
nd
the
enthalpy
change
for: C
+
2H
2
O
+
2
→ CH
3
OH.
Below this, write the
s
s
First,
2
other
species
included
draw
equations
enthalpy
two
cycle
arrows
1,
2
and
3
not
included
in
the
original
reaction:
CO
+
2
2H
2
O.
Oxygen, O
2
, is usually not
diagrams.
from
the
reactants
to
the
species
at
the
bottom
to
represent
1
equations
2
and
3.
Draw
an
arrow
from
the
products
to
the
species
at
the
C
bottom to
+
2H
2
+
O
equation 1.
the
arrows
method,
the
with
the
enthalpy
enthalpy
change
change
for
values.
equation
3
As with the summation of
needs
to
be
doubled
to
give
394
value
for
two
moles
of
hydrogen.
The
resulting
enthalpy
cycle
diagram
is
286
726
shown in
l
the
× 2
Then, calculate the sum of the enthalpy change values, following the alternative
and 2H
2
t
i
s
pathway from the reactants to the products, via the CO
O intermediates. If the
2
2
+
O
CO
2H
pathway you are following is in the opposite direction of an arrow (in this case, the last
arrow), reverse the sign of the enthalpy change. This calculation is summarized below:
(−286
×
2)
+
(+726)
= −240 kJ mol

your
method
reasoning
calculations: the summation of equations method and the
make
enthalpy cycle diagram method. In what situations would
reaction.
arithmetic al
you use an enthalpy cycle? In what situations would you
rather
It
than
U
the
strengths and limitations of each method? T
o what extent are
where
errors
advisable
simply
examiner
the
prefer,
in
make
sure
assessments.
when
to
c alculating
clearly
recording
the
opportunity
to
show
nal
you
Oen
show
c andidates
the
enthalpy of
your full working
answer.
This
gives
assign partial marks
n
use the summation of equations method? What are the
is
you
clearly
C
i
n
methods for representing and solving Hess’
s law
o
v
Whichever
In the previous worked example, you have seen two
Enthalpy cycle diagram
y
Communic ation skills
Figure 5
p
r
e
ATL
+
O
2
for the formation of methanol
1
(−394)
n
y
gure 5.
OH
3
the methods dierent if they represent the same concept?
applic able.
o
i
4.
C alculate
the
summation
enthalpy
of
change
equations
for
method,
the
and
following
reaction, using the
equations
1–3
below:
1
3H
u
l
a
E
v
O
f
x
t
a
r
o
d
Practice questions
2
(g)
+
2C(s)
O
+
2
(g)
→ C
2
H
5
OH(l)
2
1
1
C
2
H
5
OH(l)
+ 3O
2
(g)
→
2CO
2
(g)
+
3H
2
O(l)
ΔH
=
−1367 kJ mol
ΔH
=
−394 kJ mol
ΔH
=
−286 kJ mol
1
2
C(s)
3
H
+
O
2
(g)
→
CO
2
(g)
1
1
2
(g)
O
+
2
(g)
→
H
2
O(l)
2
5.
Determine
enthalpy
the
cycle
4NH
3
(g)
+
enthalpy
diagram
5O
2
(g)
change
method
for
the
and
following
equations
→ 4NO(g) + 6H
2
reaction, using the
1–3
below:
O(l)
1
1
N
2
N
3
2H
2
(g)
+
O
(g)
+
3H
2
(g)
→
2NO(g)
ΔH
=
+66 kJ mol
ΔH
=
−92 kJ mol
ΔH
=
−572 kJ mol
1
2
2
(g)
→
2NH
(g)
→
2H
3
(g)
1
410
2
(g)
+
O
2
2
O(l)
y
label
equations
P
Finally,
CH
2
r
2
represent
e
Then,
in
from
Reactivity
1.2
Energy
cycles
in
reactions
Using Hess’s law to nd enthalpy change
When
heated,
decomposes
and
water.
decomposition
determine
the
then
the
is
The
enthalpy
dicult
enthalpy
apply
Hess’s
decomposition
of
6.
hydrogenc arbonate
potassium
to
change
determine
change
law
c arbonate,
to
for
nd
potassium
two
the
(Note
for this
directly.
other
Repeat
the
procedure,
hydrogenc arbonate
c arbon
You will
that
reached
this
time,
bec ause
using
instead
record
the
potassium
of
potassium
the
lowest
reaction is endothermic.)
reactions,
enthalpy
change
Analysis
for
Part 1: Known enthalpy changes of reaction
hydrogenc arbonate.
Write
a
balanced
equation,
r
1.
including
Relevant skills
for
Inquiry
Record,
propagate
2:
out
and
express uncertainties
a.
potassium
c arbonate
b.
potassium
hydrogenc arbonate
acid
C arry
relevant
and
accurate data
processing
3:
Evaluate
weaknesses,
the
implic ations
limitations
and
of
methodologic al
2.
Show
3.
Process
that
enthalpy
assumptions on the
conclusions
a.
the
your
acid
is
in
excess
change
potassium
the
Anhydrous
•
The
effervesces.
prevent
loss
between
Cover
of
the
i
n
reaction
c arbonate is an irritant.
c arbonates and acids
reaction
reagents
and
v
potassium
3
•
2 mol dm
•
anhydrous
•
potassium
hydrochloric acid
hydrochloric
c alorimeter.
Add
stir
the
5.
the
the
Dispose
and
potassium
mixture,
of
of
exact
transfer
the
it
to
your
reaction
a
hydrochloric acid
and
hydrochloric
acid
thermal decomposition of potassium
Apply Hess’
s law, and your processed data from part1,
8.
Look
up
change
Record
of
the
for
the
enthalpy
theoretic al
the
determine
9.
Identify
the
two
10.
Identify
error
in
and
this
online.
percentage
assumptions
investigation
values
value
for
decomposition
hydrogenc arbonate
coffee-cup
temperature.
and
Propagate uncertainties to determine the absolute
of
reached.
mixture
hydrogenc arbonate
7
.
c arbonate to the acid solution and
the
c arbonate
potassium
decomposition of potassium hydrogencarbonate.
it
temperature of the acid.
monitoring
potassium
•
to determine the enthalpy change for the thermal
c arbonate into a
30 cm
into
•
hydrogencarbonate
6.
mass.
collect
hydrochloric
Draw an enthalpy cycle connecting the three reactions:
•
3
to
temperature
the
Show
and
Write a balanced equation, including state symbols, for
uncertainty
potassium
the
cylinder
Record
maximum
3 g
Record
acid
E
4.
5.
o
boat.
measuring
practic al.
between:
hydrochloric acid
to produce potassium carbonate and other products.
before starting.
a
a
v
O
Use
i
approximately
weighing
3.
approval
the
u
Place
for
for
l
f
x
teacher
required
and
the decomposition of potassium hydrogencarbonate
through the method and write a list of
equipment
2.
(solid)
(solid)
t
a
r
o
Read
c arbonate
hydrogenc arbonate
Method
1.
d
potassium
4.
n
U
Materials
c ases.
Part 2: Applying Hess’s law
vessel with a lid to
spillage.
reaction
hydrogenc arbonate
p
Hydrochloric acid is an irritant.
•
protection.
hydrochloric
o
•
eye
C
Wear
both
y
potassium
for
c arbonate
acid
•
in
and
experimental data to determine the
r
e
b.
S afety
hydrochloric acid
O
Inquiry
t
i
s
•
and
symbols,
y
•
C alorimetry
state
n
3:
between:
l
1:
Tool
reaction
P
Tool
•
the
y
•
c arbonate.
temperature
e
and
produce
s
s
dioxide
potassium
to
and
discuss
explain
two
of
Use
error
made
their
major
you
the
have
obtained.
enthalpy
potassium
this
of
value to
your
result.
throughout this
validity.
sources
of
systematic
investigation.
appropriately.
411
1
What
drives
chemic al
reactions?
LHA
Reactivity
Standard enthalpy changes of combustion,
⦵
ΔHc
⦵
, and formation, ΔHf
(Reactivity 1.2.3
and Reactivity 1.2.4)
must
describing
be
a
standard
enthalpy
change,
there
are a number of steps that
s
s
When
completed.
Write
3.
Include
The
standard enthalpy change of combustion ,
that
occurs
an
the
type
equation
reaction.
describe
symbols
in
the
the
reaction.
r
state
of
to
equation.
P
⦵
when
standard
The
a
substance
standard
conditions:
state
in
of
25.00 °C
its
a
ΔHc
pure
temperature,
gas
hydroc arbon
dioxide,
and
a
CO
2
,
undergoes
and
component
as
of
liqueed
follows:
13
C
H
10
(g)
O
+
2
(g)
2
standard
section
the
are
data
equation
of
4
H
10
the
(g)
O
+
2
2
(g)
the
(g)
of
with
included
exothermic
is
2
H
their
of
in
10
burned
the
form that it
(LPG).
for
on
to
produce
The
equation
for its
1
, is
−2878 kJ mol
other
substances,
. This
are
given
standard states.
c an
also
equation.
reaction,
included
oxygen
⦵
ΔHc
butane
the
with
, is a highly ammable
O(l)
butane,
combustion
value
gas
+ 5H
for
reacts
4
combustion
along
value
indic ates an
o
d
so
the
be
The
energy
is
written with
negative
released into the
product side:
1
→
4CO
2
(g)
+ 5H
2
O(l)
+
2878 kJ mol
2
i
Activity
t
a
r
o
f
x
Write equations to describe the standard enthalpy change of combustion for
the following compounds, and state the enthalpy change values by referring
u
l
a
E
v
O
412
for
13
C
booklet,
Therefore,
Reactivity 1.1.
4CO
combustion
combustion
change
surroundings.
introduced in
the
petroleum
enthalpies
n
U
reactions
of
chemic al
enthalpy
enthalpy
Exothermic and endothermic
14
of
standard
→
it
Butane, C
C
The
enthalpy
the
i
n
in
and
O.
is
is
Pa.
o
value,
v
The
2
substance
p
4
1.00 × 10
combustion,
water, H
enthalpy change
y
is
of
r
e
combustion
pressure
the
state
O
c arbon
a
t
i
s
When a
and
is
(298.15 K), which is taken as being
5
room
,
standard
n
under
oxygen.
of
y
takes
in
mole
l
completely
one
to section 14 of the data booklet. Ensure that you include state symbols and
that the equations are balanced:
a.
octane, C
b.
cyclohexanol, C
c.
methanoic acid, CH
d.
glucose, C
e.
chloroethane, C
8
H
6
18
H
12
6
O
H
12
O
2
O
2
6
H
2
Cl (hint: a strong, corrosive acid is one of the products)
5
y
Determine
2.
e
1.
Reactivity
1.2
Energy
cycles
in
reactions
LHA
Global impact of science
Much
of
the
world’
s
hydroc arbons.
will
sources
enable
the
fuel
high
are
do
and
make
with
Industrial
set
in
investigated
costs
factors
decisions
by international
as
a
possible
new
processes
the
are
alternative
Glasgow.
this
associated
measure
researching
are
with
c arbon-
technology
rarely totally
every method of
environmental impact of
most
about
important
what
fuel
for a fuel to be
sources
are
used to
produceenergy?
change
that
occurs
are
there
argon,
Ar,
and
Hg,
uorine, F
and
Two
2
(g)
change
of
is
→
of
elements
butane
4
are
are
a
solids
standard
liquid
substance
in
their
are
H
10
by
c arbon,
the
C,
(g)
formation
of
and
following
butane,
energy
for
state,
example
standard conditions:
hydrogen, H
2
.
Therefore,
equation:
⦵
ΔHf
the
standard
conditions,
under
is
from its constituent
2
described
C
under
of
formed
, is
C
5H
Br
elements
gases
1
−126 kJ mol
.
This
value,
n
U
enthalpy
butane
+
.
Most
are
,
is
i
n
4C(s)
2
bromine,
elements
of
states.
that
substance
o
constituent
formation
The
standard
elements
a
v
The
the
11
of
p
mercury,
their
mole
y
in
but
one
r
e
elements
when
O
standard enthalpy change of formation, ΔHf
t
i
s
⦵
The
n
should
we
which
targets
COP26
problem
is
greenhouse gases. This
y
Who
being
one
emissions.
x
How
sources,
sustainable?
is
and
of
l
fuel
emission
Paris
environmental
production.
dierent
in
However,
level of NO
emissions
reach
ammonia
source.
there
to
by the combustion of
community
y
energy
of
harmful
COP21
produced
P
“green”:
as
is
scientic
r
is
such
combustion
neutral
reduce
the
e
The
to
governments
agreements,
supply
s
s
energy
energy
Increasingly,
and other values for the standard enthalpy change of formation for many common
compounds,
of
these
c an
be
compounds
8
notice
of
to
that
6
O
when
combustion
necessary
correct
H
v
will
2
to
use
Cl
writing
and
balance
fractions
3
the
to
the
data
the
equation.
the
The
enthalpy
enthalpy
change
change
of
formation
benzoic acid, C
e.
c arbon
f.
methylamine, CH
equations
fractional
This
nal
is
6
H
3
of
chemic al
the
few
by
for
referring
COOH
CO
NH
standard
stoichiometric
one
5
monoxide,
for
standard states
values
d.
chemic al
formation,
balance
booklet.
section.
standard
state
a
ethanol, C
of
o
H
chloromethane, CH
changes
be
3
13
same
i
propane, C
E
You
the
and
l
c.
the
section 13 of the data booklet:
O
b.
describe
in
u
a.
to
compounds,
f
x
to
equations
following
section
t
a
the
in
given
d
Write
are
r
o
Activity
found
2
enthalpy
coecients
may
c ases when it is
equation.
413
Reactivity
1
What
drives
chemic al
reactions?
LHA
Under
standard
example,
conditions,
diamond,
the
element
graphite
and
values.
an
c arbon
exists
as
buckminsterfullerene.
several
allotropes,
Allotropes
of
for
c arbon
⦵
have
dierent
convention
The
allotropes
of
is
ΔHf
that
the
most
If
element
stable
has
allotrope
several
is
the
allotropes, the normal
standard
state.
The
enthalpy
c arbon and
1
of
structures
are
formation
is
0 kJ mol
for
all
elemental
substances
in
their
standard states.
discussed in
Graphite
is
standard
enthalpy
taken
as
the
standard
for
the
c arbon
allotropes, and so it has a
Structure2.2
1
Would
for
you
expect
allotropes
of
an
element,
such
as
diamond
⦵
averaging
the
have
dierent
∆H
values?
f
(Structure 2.2)
and
graphite,
strengths and
this
approach.
n
of
to
values.
y
limitations
enthalpy
l
Evaluate
bond
Applying Hess’s law to enthalpy changes of combustion
the
of
a
enthalpy
overall
the
of
of
the
of
i
n
the
U
is
a
direct
reactants)
applic ation
of
a
reaction,
reactants
products
⦵
= Σ(ΔHc
is
equal
so
and
the
nal
enthalpy
values of
change.
ΔHr
and
to
,
c an
be
c alculated
between the sum of the
the
the
sum
overall
of
the
enthalpy
enthalpy change
⦵
− Σ(ΔHc
Hess’
s
function,
initial
enthalpy
dierence
C
⦵
the
⦵
for
The
reaction:
ΔHr
This
of
state
between this initial and nal state of a
overall
change
combustion
combustion
a
o
of
the
data.
is
between
pathway
aect
enthalpy
v
changes
changes
ofthe
The
not
combustion
enthalpy
enthalpy
dierence
system.
does
standard
enthalpy
topic,
the
p
using
of
reaction
this
is
y
The
in
r
e
chemic al
earlier
reaction
O
discussed
change
t
i
s
As
products)
law.
n
Using
Hess’
s
i
d
formation
law
and
enthalpy
of
combustion
data,
the
enthalpy change of
−1
of
t
a
r
o
in the data
pentane
is
determined to be −177 kJ mol
,
but
the
value
given
1
booklet is
−173 kJ mol
.
Suggest
why
the
values
aredierent.
Applying Hess’s law to enthalpy changes of formation
u
f
x
The
l
a
E
v
O
414
6.
o
Practice question
standard
enthalpy
change
enthalpy
of
formation
data.
changes
of
formation
of
formation
of
the
ΔH
r
the
reactants
⦵
for
The
is
a
reaction
dierence
products
equal
to
and
the
the
∑(ΔH
f
also
sum
overall
⦵
=
c an
between
be
the
of
the
enthalpy
⦵
products)
−
∑(ΔH
f
c alculated using
sum
reactants)
of
the
enthalpy
enthalpy changes of
change
of
the
reaction:
y
rationale
.
e
the
0 kJ mol
P
on
of
Linking question
Thinking skills
Reect
formation
r
ATL
of
s
s
their
Reactivity
1.2
Energy
cycles
in
reactions
LHA
Worked example 3
C alculate the standard
enthalpy change of formation of pentane, C
5
H
12
,
using the enthalpy of combustion data from
section 14of the data booklet.
s
s
Solution
Step 1:
Write
values
Write
from
balanced
equations
chemic al
for
the
equation
combustion
for
of
the
formation
c arbon,
of
1
hydrogen
mol
of
and
pentane:
pentane
5C(s)
and
nd
+ 6H
their
2
(g)
standard
C(s)
+
O
2
(g)
→
CO
2
(g)
1
∆H
12
(l)
enthalpy change
O
+
2
(g)
→
H
2
−286 kJ mol
O(l)
equations
12
with
+
8O
either
for
the
2
the
(g)
→
5CO
2
(g)
summation
reaction.
These
of
+
6H
2
O(l)
equations
methods
∆H
method
were
formation
required
for
pentane
the
that
overall
reaction
involve
equation.
scheme.
c arbon,
Therefore
+
5O
2
your
you
(g)
c alculation,
and
should
→
5CO
2
you
pentane.
need
multiply
(g)
C arbon
the
ΔH
enthalpy
=
to
is
cycle
diagram method to
earlier in this topic.
a
use
enthalpy
reactant
of
in
values
equation
combustion
p
5C(s)
In
hydrogen
enthalpy
y
are
of
reactions
r
e
combustion
or
introduced
Summation of equations method
the
1
= −3509 kJ mol
O
change
(l)
t
i
s
enthalpy
H
n
these
the
5
y
Use
C
l
⦵
3
y
(g)
=
1
∆H
2
−394 kJ mol
2
the
H
P
1
H
=
⦵
2
Inspect
5
r
⦵
determine
C
section 14 of the data booklet.
1
Step 3:
→
e
Step 2:
a
of
corresponding to
1,
but
c arbon
by
ve moles
ve:
1
−1970 kJ mol
v
a
the
product
reactant.
As
a
pentane,
result,
you
the
should
enthalpy
the
three
equations
use
3O
+
6H
together,
(g)
→
equation
change
(g)
2
sign
2
O(l)
6H
1,
but
C
c ancelling
5
H
12
the
2
+
6H
common
the
reaction is
for
2
species
2
(l)
+
8O
2
(g)
O(l)
+
2
+
gives
the
to
be
reversed
to
make
common
pentane
1
to
both
sides
of
the
reaction, and add the
⦵
5CO
2
(g)
ΔH
c
1
=
−1970 kJ mol
=
−1716 kJ mol
⦵
6H
2
O(l)
ΔH
c
1
⦵
→
C
5
H
12
(l)
+
8O
2
(g)
ΔH
1
=
+3509 kJ mol
⦵
→
2
2
overall
equation
5C(s)
+
+
2
+ 6H
2
(g)
C
→
5
H
C
5
12
H
+
12
(l).
2
1
ΔH
Therefore,
=
the
−177 kJ mol
total
against
the
to
the
sum
E
reactants
the
diagram
of
the
kJ mol
ΔH
is
5C(s) +
is
on
the
enthalpy
right.
change
Oxygen, O
values,
2
(g),
is
following
via
the
CO
2
(g) and H
2
O(l)
the
included
arrows
intermediates.
in
6H
(g)
C
2
H
5
(I)
12
enthalpy
If
from the
you
3509
are going
5 ×
of
an
arrow,
reverse
the
sign
of
the
enthalpy
(
286)
change.
(
summarized
394)
below:
⦵
ΔH
not
6 ×
products,
direction
c alculation
a
diagrams.
C alculate
This
cycle
v
cycle
enthalpy
1
−177
l
O
enthalpy
product, not
= +3509 kJ mol
Enthalpy cycle diagram method
The
a
positive:
ΔH
→
(g)
u
f
x
the
change
2
6H
+
2
C ancelling
+
must
negative
o
+
(g)
3O
→
(g)
t
a
5C
2
+
2
1
= −1716 kJ mol
c
equation
from
species
i
2
5CO
(g)
5O
the
ΔH
⦵
→
d
r
o
6H
Total
+
⦵
O(l)
changes
values:
5C(s)
2
n
add
enthalpy
2
+
U
5CO
Now
(g)
C
For
2
i
n
6H
o
You also require six moles of hydrogen as a reactant. Equation 2 should be used, with the enthalpy change multiplied by six:
1
=
5
× (−394)
+
6
×
(−286)
+
3509
=
−177 kJ mol
5CO
(g) +
2
6H
O(I)
2
415
LHA
Reactivity
1
What
drives
chemic al
reactions?
Worked example 4
C alculate the enthalpy of combustion of pentane, C
5
H
12
(l),
using the enthalpy of formation data from
section 13 of the
data booklet.
s
s
Solution
Step 1:
Write
a
balanced
chemic al
equation
Write
enthalpy
for
change
the
formation
values
from
5
H
of
12
(l)
+
8O
pentane,
2
(g)
→ 5CO
c arbon
2
(g)
+
dioxide,
6H
and
2
O(l)
water
from their elements and nd their
r
standard
equations
section 12 of the data booklet.
+
6H
2
(g)
→
C
5
H
12
(l)
∆H
1
f
−173 kJ mol
3
H
+
O
2
(g)
→
CO
2
(g)
∆H
=
−394 kJ mol
=
−286 kJ mol
1
f
1
⦵
2
(g)
O
+
2
(g)
→
H
2
O(l)
∆H
1
f
2
these
the
equations
enthalpy
with
change
either
for
the
the
summation
of
equations
method
reaction.
Summation of equations method
c alculation,
must
be
you
and
reversed.
need
water.
As
a
to
use
Pentane
result,
the
enthalpy
is
a
product
in
an
→
equation
formation
of
2,
c arbon
5C(s)
+
5C(s)
but
+
ve
dioxide
5O
2
(g)
sign
→
6H
changes
2
(g)
moles
by
to
the
equation,
from
a
negative
2
for
⦵
∆H
the
diagram method to
that
equation
involve
1.
pentane,
Therefore,
equation
positive:
1
required
(g)
to
in
= +173 kJ mol
ve:
5CO
reactions
product
⦵
∆H
are
formation
but
overall
U
C
of
(l)
change
overall
o
is
enthalpy
12
corresponding
the
v
the
H
in
i
n
dioxide
multiply
5
values
reactant
enthalpy
C
C arbon
a
cycle
p
1
dioxide
enthalpy
y
your
c arbon
r
e
In
or
O
Use
determine
t
i
s
Step 3:
l
C(s)
y
2
n
=
⦵
equation.
Therefore,
you should
1
= −1970 kJ mol
f
6H
the
three
equations
together,
values:
6H
2
C
5
H
12
+
5C
+
(g)
+
12
+
2
3O
5O
2
→
→
(l)
5O
(g)
c ancelling
2
+
6H
the
2
O(l)
∆H
species
6H
2
+
3O
1
= −1716 kJ mol
to
both
sides
of
the
reaction, and add the
⦵
5C(s)
+
6H
2
(g)
1
∆H
=
+173 kJ mol
⦵
5CO
2
(g)
∆H
f
1
=
−1970 kJ mol
=
−1716 kJ mol
⦵
→
(g)
⦵
f
common
→
(g)
6H
2
O(l)
∆H
1
f
⦵
→
l
u
f
x
Total
H
2
t
a
5C(s)
5
3O
i
r
o
C
+
o
add
enthalpy
(g)
d
Now
2
n
You require six moles of water as a product. Equation 3 should be used, with the enthalpy change multiplied by six:
5C
2
+
6H
5
12
2
+
5CO
(l) + 8O
+ 6H
2
2
O
(g) → 5CO
2
∆H
(g) + 6H
2
1
=
−3513 kJ mol
O(l). Therefore, the total
2
1
a
O
enthalpy change for the reaction is−3513 kJ mol
ΔH
Enthalpy cycle diagram method
enthalpy
from
If
the
you
are
enthalpy
the
sum
reactants
going
of
to
the
enthalpy
the
change
products,
via
values,
the
following
C(s) and H
2
(g)
+
O
5CO
(g)
2
the
This
the
direction
c alculation
is
of
an
arrow,
summarized
reverse the sign of the
5
(
below:
(+173)
+
5
×
(−394)
+
6
×
(−286)
= −3513 kJ mol
5C(s) +
416
×
394)
1
=
(g)
2
+
6H
6H
(g)
2
O(I)
2
arrows
intermediates.
⦵
ΔH
(I)
12
6
×
(
286)
173
against
change.
H
5
diagram is on the right.
E
C alculate
cycle
v
The
C
y
5C(s)
P
⦵
1
e
C
Step 2:
for the combustion of 1 mol of pentane:
Reactivity
1.2
Energy
cycles
in
reactions
LHA
Practice questions
7
.
Calculate the enthalpy change of formation of propanone,
CH
COCH
3
9.
Consider
the
following
enthalpy
cycle
diagram:
(l), using enthalpy change of combustion data
3
Step 1
from section 14 of the data booklet.
the
summation
enthalpy
cycle
of
CH
CH
(g)
2
+
H
CH
2
3
equations method and the
diagram
method
to
support
your
7
answer.
7
+
O
(g)
2
gas
c an
be
made
using
the
2
(g)
2
2
Hydrogen
O
e
+
8.
(g)
3
s
s
Use
following
4
(g)
+
Step 2
H
2
O(g)
→
3H
of
the
enthalpies
of
formation,
+
CO(g)
booklet
lists
the
standard
2CO
⦵
ΔH
,
f
for
some
of
the
(g) +
2
species
3H
y
data
2
13
(g)
P
Section
r
Step 3
reaction:
CH
O(I)
2
l
⦵
in
the
reaction
above.
b.
Determine
no
a.
value
is
listed
for H
2
⦵
the
using
formation
value of
the
ΔH
1
,
standard
step
in
kJ mol
,
for the
2
the
using
standard
b.
Determine
enthalpy change of
enthalpy
the
standard
enthalpy
of step 1.
values.
c.
Suggest
one
reason
why
the
O utline
the
of
c alculated
enthalpy change of
ΔH
from bond enthalpies is
accurate
lessaccurate.
with
some
you
of
will
of
propan-1-ol is
the
compare
methods
this
by
Inquiry
the
C alculate
3:
Compare
accepted
3:
and
the
and
of
context
discuss
random
percentage
outcomes
scientific
Identify
systematic
interpret
errors
a
E
v
O
l
u
f
x
Inquiry
and
an
each
of
the
the
a.
Refer
b.
Conduct
c alculations to
i
•
3:
algebraic
Determine
o
and
1.
reason
why
c an
it
c an
be
,
value of
considered
be
considered
enthalpy
of
combustion
of
propan-1-ol
to
following methods:
average
real
(as
you
have
a
bond
c alorimetry
described in
Record
access
the
to
values.
experiment. This could be
Reactivity 1.1),
a
relevant
enthalpy
suitable
or
simulated, if
simulation
software.
measurement uncertainties
error
t
a
Tool
•
arithmetic
r
o
•
to
Use
problems
d
3:
solve
one
b
⦵
ΔH
described in
Relevant skills
Tool
Method
value to
n
this topic.
•
and
law in part
p
combustion
task,
, of
o
obtained
this
U
those
In
C
of
1
.
v
enthalpy
i
n
standard
−2021 kJ mol
Hess’s
approximate.
Comparing ∆H values
The
using
y
reaction
value
r
e
this
why
ΔH
change,
c alculated
⦵
c.
change,
section 14 of the data booklet.
O
reaction
C alculate
(g).
n
why
y
O utline
t
i
s
a.
and
propagate them.
investigation
c.
2.
Refer
to
Determine
standard
the
enthalpies
percentage
error
of
formation.
of
each
of
the
values
sources and impacts of
obtained in methods
3.
Discuss
and
standard
each
a,
of
your
b and
explain
enthalpy
a, b
the
of
and
c
differences
combustion
c alculated
values
of
between the
propan-1-ol and
obtained
by methods
c
417
1
What
drives
chemic al
reactions?
LHA
Reactivity
Born–Haber
Born–Haber
cycles
the
changes
an
enthalpy
ionic
are
compound.
ions
in
of
the
gaseous
and
anity,
IE,
mole
steps
is
of
with
in
the
lattice
the
state.
sometimes
cycle
For
further
include
and
enthalpy
from
metal
Hess’
s
steps
enthalpy,
standard
electrons
of
several
one
ions
ionization
These
in
ionization
enthalpy
change
mole
with
law.
involved
of
multiple
energies
occurs on the
or
positively
dened:
M(g)
→
M
(g)
+
IE
in
of
(g)
→
ionization
M
is
(g)
+
e
endothermic.
endothermic.
of
For
gaseous
a
solid
,
of
is
species,
the
an
X
2
equal
electrons
+
anity
c an
be
–
e
electron
anity
one
→
of
X
anity
found
t
a
ionic
the
atoms
EA
of
is
helium
value of
to
half
ΔH
that
c an
occurs
process is
is
at
equal to the
of
the
bond
enthalpy
enthalpy change on the addition of one
in
the
gaseous state:
< 0
typic ally
or
change
This
energy
> 0
at
standard
of
(g)
in
negative,
nitrogen.
but
Values
of
there
rst
are
and
exceptions, such
second
electron
section 9 of the data booklet.
⦵
lattice enthalpy,
occurs
is
mole
non-metals
o
the
to
–
X(g)
i
d
r
o
when
∆H
gaseous
lattice
ions
,
is
are
dened
formed
as
the
from
standard
one
mole
of
enthalpy change that
structural units of a solid
lattice:
u
l
f
x
a
E
v
O
418
ΔH
n
U
of
Electron
The
⦵
X(g)
electron anity, EA,
mole
as
→
C
The
(g)
enthalpy
formed.
> 0
is
at
ionization
⦵
the
⦵
ΔH
rst
o
i
n
1
is
of
⦵
at
(section 12 of the data booklet):
2
0
standard
element
species,
ΔH
diatomic
>
y
gaseous
IE
p
a
M(g)
v
For
→
at
r
e
M(s)
ΔH
atoms
monatomic
enthalpy of sublimation:
0
O
mole
>
values
t
i
s
enthalpy of atomization,
one
The
section 9 of the data booklet.
⦵
The
when
IE
n
process
found
M
e
l
be
:
+
+
y
The
2
2
rst,
+
MX(s)
The
process
c an
be
L attice
is
found
→ M
chemistry,
gaseous
ions
⦵
(g)
∆H
lattice
>
Experimental
0
values
of
lattice
enthalpy
at
298.15 K
section 16 of the data booklet.
are
oen
formation
the
+ X
endothermic.
in
enthalpies
exothermic
–
(g)
lattice
from
of
quoted
the
solid
enthalpy
the
solid
as
negative
lattice
always
lattice.
from
refers
values
gaseous
to
the
that
represent the
ions.
However, in DP
endothermic
formation of
y
:
the
P
1
charged
electrons,
+
IE
enthalpy of
formation.
valence
are
formation of
energy,
of
that
atoms
cycles combine
the
r
second,
one
applic ation
e
removal
The
electron
Ionization energy,
another
associated
s
s
atomization,
cycles (Reactivity 1.2.5)
Reactivity
1.2
Energy
cycles
in
reactions
⦵
have
already
substance
formed
states.
be
as
dened
the
under
enthalpy
standard
Standard
represented
the
a
change
of
single
enthalpy
that
conditions
enthalpy
by
standard
∆H
from
formation
its
of
constituent
an
ionic
elements
substance,
2
(g)
→
in
such
their
as
standard
NaCl(s),
c an
equation:
⦵
Cl
, of a
f
NaCl(s)
∆H
−
s
s
+
formation,
occurs when one mole of a substance is
1
Na(s)
of
LHA
We
1
= −411 kJ mol
f
2
correct
use
of
subject-specic
your
knowledge
terminology
and
is
essential
understanding
of
to
your ability
new concepts in
encounter
electron
the
Term
key
terms
standard
anity.
in
state,
M ake
a
this
section:
lattice
enthalpy,
glossary
that
has
an
entry
for
Denition
Equation
∆H
+
Standard
lattice
of
gaseous
add
terms
a
to
ions
solid
your
from one
lattice
chemistry
lattice
atomization,
and
enthalpy
of
enthalpy
formation
be
is
(X)
+
(g)
X(g)
f
x
at
1
X
(g)
(g)
2
2
l
a
time.
electron anity
Born–Haber
cycle. A
e
EA(X)
e
a
v
O
IE(M)
+
u
+
gure 6.
i
ΔH
in
over
energy,
construct
(g)
t
a
r
o
+
M
shown
ionization
to
X
o
cycle
combined
d
Born–Haber
of
c an
+
C
The
(g)
n
cycle
enthalpy,
glossary
U
Using the Born–Haber
M
→ M
o
to
of
i
n
continue
generalized
MX(s)
v
mole
c an
enthalpy change
that occurs on the formation
p
enthalpy
y
r
e
⦵
You
each term,
following:
Symbol
lattice
standard
enthalpy of atomization, ionization
O
and
detailing
several
formation,
t
i
s
energy
of
n
will
enthalpy
y
You
l
chemistry.
y
demonstrate
e
to
P
The
r
Self-management skills
ATL
+
M
(g)
+
X
(g)
1
+
M(g)
X
E
ΔH
(g)
2
2
ΔH
(M)
at
ΔH
f
1
+
X
2

Figure 6
(g)
2
A generalized
Born–Haber cycle
419
Reactivity
1
What
drives
chemic al
reactions?
LHA
The
be
unknown
you
know
anities
an
value
determined
ionic
values
lattice
each
the
enthalpy
of
The
the
before,
pathway
of
you
you
goes
of
nd
the
step
Much
should
like
reverse
the
in
in
a
cycle.
ionization
is
to
the
sign
of
of
the
For
the
of
sum
enthalpy
any
arrow
cycle
c an
example, if
energies,
enthalpy
equal
with
the
direction
Born–Haber
the
standard
formation
steps.
against
any
pathway
atomization,
c an
enthalpy
other
for
opposite
enthalpy
enthalpy,
of
change
the
electron
formation of
of
the
cycle
enthalpy
s
s
for
constructed
where
of
following
compound.
changes
you
the
and
by
diagrams
enthalpy changes
in
the
cycle.
e
r
Worked example 5
cycle in gure 7.
mol
Br
(g)
2
2
p
+
1
K
+419 kJ mol
1
Br
U
C
+
K(g)
(l)
2
2
(g)
1
x kJ mol
ΔH
lattice
(K)
+89 kJ mol
n
at
ΔH
f
1
o
i
d
Br
(l)
2
2
1
392 kJ mol
Born–Haber cycle to c alculate the lattice enthalpy of potassium
bromide
Solution
The
u
l
a
E
v
O
f
x
t
a
r
o
Figure 7
+
lattice
enthalpy
is
the
enthalpy
change
for
the
reaction
KBr(s)
→ K
(g)
+
−
Br
(g).
In
gure
7,
this
step
is
shown
from
the
bottom
right
of
the
cycle, to
⦵
the
middle
right,
labelled
∆H
lattice
.
To
nd
the
lattice
enthalpy,
follow the
+
alternative
Add
the
pathway,
values
of
going
the
lattice
from
enthalpy
⦵
∆H
⦵
(KBr)
= −∆H
that
arrow,
your
so
you
the
answer
bec ause
it
is
KBr(s)
clockwise
change
for
around to K
at
+ ΔH
at
−(−392) + (+89) + (+419) + (+112) + (−325)
+687 kJ mol
for
an
was
the
the
reverse
of
the
+392
enthalpy
endothermic
(g).
1
reversed:
lattice
Br
(Br) + EA(Br)
=
followed
+
⦵
(K) + IE(K)
=
sign
−
(g)
each step:
⦵
+ ΔH
f
–
Note
420
Br
1
ΔH
+

+
(g)
o
i
n
v
IE (K)
1
325 kJ mol
EA (Br)
e
(l)
e
y
+
K
r
e
1
n
+112 kJ
+
O
(Br)
at
Br(g)
t
i
s
1
ΔH
+
(g)
l
y
+
K
of
an
process.
enthalpy
instead of
ionic
of
formation
−392.
reaction
Remember
compound
as
a
to
positive
express
value,
y
P
Determine the lattice enthalpy of potassium bromide using the Born–Haber
Reactivity
F actors
lattice
10.
Interpret
and
utilize
the
values
from
the
Born–Haber
cycle
in
Energy
aecting
enthalpy
the
are
cycles
in
reactions
LHA
Practice question
1.2
value of
detailed in
gure 8 to
Structure2.1.
c alculate
oxide,
the
MgO.
enthalpy
MgO
of
has
formation
doubly
for the ionic compound magnesium
charged ions.
s
s
2+
Mg
(g)
+
O(g)
+
2e
+249 kJ mol
at
EA
1
(O)
141 kJ
mol
(1)
(g)
+
O
(g)
+
2e
2
2
(g)
+
O
+
(g)
e
y
1
IE
(Mg)
+1451 kJ mol
(2)
1
(O)
+753 kJ mol
(2)
(g)
+
O
(g)
+
e
2
2
(g)
+
O
(g)
(1)
O
(g)
2
2
(MgO)
+148 kJ mol
ΔH
(MgO)
U
f
1
+
O
2
C
(Mg)
at
i
n
1
ΔH
(g)
2
1
+3791 kJ mol
lattice
o
ΔH
v
+
p
1
Mg(g)
y
r
e
Mg
+738 kJ mol
(Mg)
2
2+
1
IE
O
t
i
s
EA
1
+
Mg
n
l
2+
Mg
y
P
1
2+
Mg
e
(O)
r
1
ΔH
1
n
x kJ mol

Figure 8
Born–Haber cycle to c alculate the enthalpy of formation
i
d
t
a
r
o
u
l
f
x
a
2
The lattice structure of crystalline magnesium oxide:
E
v
O

Figure 9
o
for magnesium oxide
grey = Mg
+
2–
,
red = O
Linking question
What
are
the
factors
that
inuence
the
strength
of
lattice
enthalpy in an ionic
compound? (Structure 2.1)
421
Reactivity
1
What
drives
chemic al
reactions?
End of topic questions
Using
the
Born–Haber
1.
Using
your
answer
knowledge
the
guiding
from the
question
as
Reactivity 1.2
fully
as
represents
cycle
the
below, determine which
lattice
enthalpy
of
s
s
c alculation
strontium
1
topic,
chloride, SrCl
2
,
in
kJ mol
possible:
2+
(g) + 2Cl(g)
698
of energy help us to predict energy changes during
r
+1064
reactions?
2+
(g) + 2Cl
Sr
l
Multiple-choice questions
Which
3
the
N–H
bond
?
3
(g)
→
N(g)
1
+
3H(g)
1
NH
B
3
(g)
Sr(g) + 2Cl(g)
N(g)
→
+
H(g)
3
D
NH
3
(g)
→
(g)
→
3
N
2
(g)
H
+
2
2
+242
Sr(g) + Cl
(g)
2
Sr(s) + Cl
enthalpies
2
(g)
c an
+
be
•H(g)
used
to
c alculate
the
enthalpy
of
the
following
Referring
of
change
of
B
(3
× 391)
−
(436
C
−[(3
D
−(6
+
equation
×
945)
+
(3
436)
Na
C
NaCl(s)
→
Na
→
Na
391)
945]
lattice
enthalpy?
+
Na
(g)
+ Cl
(g)
+
(g)
+ Cl
(g)
+
(aq)
+ Cl
+
(s)
+ Cl
(aq)
(s)
SrCl
(s)
2
A
−(−829)
B
−829
deduce
c alculation of the
u
→
×
+
a
NaCl(s)
booklet,
the
945]
l
B
E
v
O
422
+
represents
NaCl(g)
NaCl(s)
+
945]
[(3
A
D
→
436)
(g)
from bond enthalpies.
r
o
Which
391)
×
data
represents
3
o
[(3
2NH
i
−
×
the
⇌
t
a
× 391)
+
(g)
d
(6
436)
of
2
reaction
A
×
3H
12
following
f
x
LHA
4.
section
the
+
n
enthalpy
to
(g)
kJ mol
U
which
2
in
C
N
reaction,
i
n
1
change
(g)
+
C
−(−829)
D
−829
+
+
164
164
+
+
+
829
+
242
242
164
164
lattice enthalpy
+164
2
o
Bond
•NH
v
3.
3
(g)
2
p
NH
y
1
C
r
e
3
+549
O
NH
(g) + 2Cl(g)
enthalpy
t
i
s
A
represents
yplahtne
inNH
equation
n
y
+
Sr
2.
+
+
+
+
549
549
242
242
(g)
+
+
549
549
+
+
1064
1064
+
−
1064
1064
−
− (−698)
698
− 698
(−698)
y
P
Exam-style questions
e
Sr
How does application of the law of conservation
LHA
5.
Topic review
Reactivity
LHA
6.
The
Born–Haber
shown
cycle
for
potassium
oxide, K
2
O, is
1.2
Energy
cycles
Extended-response questions
7
.
+
The
+ O
following
production
2
(g)
equation
of
describes the industrial
ammonia
using
the
Haber
process.
(g)
N
2
(g)
+
3H
2
(g)
⇌
2NH
3
(g)
s
s
1
+612 kJ mol
a.
C alculate
this
+
2K
the
process
enthalpy
using
change
bond
of
reaction
enthalpy
data.
(g) + 2e
+ O(g)
and
explain
exothermic
or
whether
the
e
State
reaction is
endothermic.
1
8.
lattice
Ammonia
reacts
with
oxygen
to
steam.
enthalpy
a.
Write
a
balanced
form
chemic al
1
nitrogen and
equation
for this
showing
+
2
O
full
structural
(g)
2
bonds
1
c.
Using
data
O(s)
2
this
9.
represents
the
lattice
enthalpy
428
−(−361)
Octane, C
−361
D
−(−361)
+
428
428
+
838
+
+
+
612
838
838
428
+
−
+
612
612
838
−
612
oxygen
8
H
18
U
The
,
is
a
for
all
reactants and
nature
(l)
for
values
the
fuel,
of
from
[1]
the
covalent
[4]
section 12 of the
enthalpy
change
for
[3]
component
as
following
of
octane
liquid
is
gasoline.
When
combusted in
reaction:
25
O
+
bond
change
b.
18
the
H
Using
the
determine
burned
via
C
a.
8
is
C
C
+
+
booklet,
o
+
v
−361
i
n
B
?
enthalpy
reaction.
gasoline
1
bond
p
c alculation
inkJ