Chemistry
for
CAPE®
Unit
1
Chemistry
for
Roger
Leroy
CAPE®
Norris
Barrett
Annette
Maynard-Alleyne
Jennifer
Murray
Unit
1
3
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edition
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illustrations
moral
Oxford,
University’s
University
Original
All
the
education
Text
Street,
referenced
for
in
all
at
Murray
2012
Contents
Introduction
v
Chapter
4.
1
Module
4
Redox
Redox
reactions
reactions
48
and oxidation
1
state
Chapter
1
Atomic
1.
1
The
1.2
Isotopes
1.3
Energy
structure
structure of the
2
atom
4.2
Redox
radioactivity
levels
and
emission
6
Sub-shells
1.5
Ionisation
and
atomic orbitals
energies
8
10
Revision questions
2.
1
2
matter
and
bonding
Covalent
The
gas
laws
52
5.2
The
kinetic theory of
5.3
Using the
ideal
5.4
Changing
state
gas
gases
54
equation
56
58
6
Energetics
60
14
6.
1
Enthalpy
6.2
Bond
6.3
Enthalpy
6.4
Hess’s
6.5
More
6.6
Lattice
changes
60
energies
62
14
bonding
– dot
and
changes
by
experiment
64
cross
diagrams
law
66
16
2.3
More dot
and
2.4
Ionic
metallic
2.5
Electronegativity
and
52
and forces of
attraction
2.2
Kinetic theory
12
Structure
States of
5
5.
1
Chapter
Chapter
50
4
spectra
1.4
equations
2
Chapter
and
48
cross diagrams
bonding
enthalpy
changes
68
18
energy: the
Born–Haber
20
cycle
70
and
6.7
intermolecular forces
22
Properties of
giant
24
Properties of
simple
More
about
lattice
energy
Exam-style questions
2.6
2.7
structures
26
2.8
Shapes of
molecules
(1)
28
2.9
Shapes of
molecules
(2)
30
Revision questions
3
The
3.
1
Equations
3.2
Calculations
mole
and
Module
74
Module
Chapter
2
7
Rates of
7
.
1
Following the
7
.2
Calculating
7
.3
Collision theory
reaction
course of
a
76
reaction
76
32
concept
moles
using the
34
rate of
reaction
and
78
rate of
reaction
80
7
.4
Rate
82
7
.5
Order of
34
equations
mole
concept
reaction from
36
3.3
Empirical
3.4
Using Avogadro’s
and
molecular formulae
3.5
Solution
law
experimental data
84
Reaction
86
38
7
.6
mechanisms
40
Revision questions
concentration
titrations
42
More
44
about titrations
8.
1
Revision questions
88
and
Chapter
3.6
1
molecular
compounds
Chapter
–
72
8
Chemical
Characteristics of
equilibria
90
chemical
46
equilibria
8.2
The
equilibrium
90
constant, K
92
c
iii
Contents
8.3
Equilibrium calculations involving K
94
Chapter
96
Groups
c
8.4
The
equilibrium
constant, K
12
II,
The
IV
chemistry of
and VII
140
p
8.5
Le Chatelier’s
8.6
Solubility
principle
product,
98
12.
1
Properties of Group
12.2
Group
II
elements
140
100
K
sp
8.7
Solubility
product
II
carbonates,
nitrates
and
and the
sulphates
common
ion
effect
12.3
Chapter
9
Acid–base
equilibria
Brønsted–Lowry theory of
Group
IV
elements
and their
tetrachlorides
104
12.4
9.
1
142
102
Some
144
properties of Group
IV
acids
oxides
and
bases
12.5
9.2
Simple
pH
calculations
Relative
IV
and oxides
148
Equilibrium calculations involving
weak acids
9.4
stability of Group
106
cations
9.3
146
104
Acid
and
12.6
Group VII: the
12.7
Halogens
12.8
Reactions of
halogens
150
108
and
hydrogen
halides
152
base dissociation
constants
halide
ions
154
110
Revision questions
9.5
Changes
in
pH during
156
acid–base
titrations
112
Chapter
9.6
Acid–base
indicators
9.7
Buffer
9.8
Calculations
13
Transition
elements
13.
1
solutions
An
introduction to transition
116
elements
involving
158
buffer
13.2
solutions
Physical
properties of transition
118
elements
Chapter
10
Electrode
and
equilibrium
10.
1
Introducing
potentials
13.3
120
standard
160
Coloured
ions
and oxidation
states
13.4
162
Complex
ions
and
ligand
electrode
exchange
potential
Electrode
potentials
and
Complex
10.4
Using
and
redox
reactions
166
122
Chapter
Redox
ions
cell
potentials
10.3
164
120
13.5
10.2
158
114
reactions
and
cell diagrams
14
Qualitative tests for
124
ions
168
Ø
E
values to
chemical
10.5
predict
change
Electrode
126
potentials
electrochemical
and
cells
128
Revision questions
130
Exam-style questions
–
Module
2
14.
1
Identifying
cations
(1)
168
14.2
Identifying
cations
(2)
170
14.3
Identifying
anions
Exam-style questions
Period
11.
1
11
The
3
174
sheets
176
revision questions
178
elements of
3
134
Glossary
183
Index
186
Physical properties of the Period 3
elements
in
134
11.2
Patterns
11.3
Properties of
chlorides
iv
Module
3
Answers to
Chapter
–
132
Data
Module
172
Period
3
Period
elements
3 oxides
136
and
138
Answers to
questions
exam-style
CD
Introduction
This
Study
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and
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On
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
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est
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Chemistry
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CAPE
Chemistry.
v
1
Atomic
1.
1
The
structure
structure
Learning outcomes
Dalton’s
John
On
completion
should
be
able
of
this
describe
Dalton’s

describe
the
atom
in
relative
the
section,
atomic
structure
terms
of
charges
sub-atomic
of
theory
the
position,
and
the
atom
atomic theory
Dalton
thought
of
atoms
as
being
hard
spheres.
He
suggested
that:
you
to:

of
masses

all
atoms
of
the

atoms
cannot

atoms
of

atoms
combine
same
be
element
broken
different
down
elements
to
form
are
any
have
more
exactly
alike
further
different
complex
masses
structures
( compounds).
of
particles.
Later
changes to
Dalton’s
atomic
centuries,
Dalton’s
theory
scientists
was
have
first
atomic theory
described
modified
the
in
1807.
theory
to
fit
Over
the
the
past
evidence
two
available.
1897
Thomson
discovered
suggested
sea
of
the
the
electron
‘plum-pudding’
positive
(which
model
of
he
the
called
atom:
a
‘corpuscle’).
electrons
He
embedded
in
a
charge.
1909
Rutherford
suggested
positively-charged
planetary
model
that
nucleus
with
most
in
of
the
electrons
the
mass
middle
of
of
the
the
surrounding
atom
atom.
the
was
This
in
led
a
tiny
to
the
nucleus.
1913
Bohr
suggested
distances
that
electrons
depending
on
their
could
only
orbit
the
nucleus
at
certain
energy.
John Dalton (1766–1844)
1932
Did you know?
Scientists
now
Chadwick
think
that
matter
discovered
the
presence
of
the
neutron
in
the
nucleus.
is
1926 onwards
made
called
up
of
sub-atomic
‘leptons’
electron
is
a
and
type
of
particles
Schrödinger
,
Born
terms
probability
lepton
of
and
neutrons
are
of
the
Heisenberg
of
finding
described
it
at
any
the
one
position
point.
of
This
an
led
electron
to
the
in
idea
and
of
protons
and
‘quarks’. An
made
atomic
orbitals
(Section
1.4).
up
quarks.
electrons
electron
(–)
cloud
shells
of
electrons
electron
(–)
electron
proton
sea
of
+
charge
nucleus
Figure 1.1.1
(+)
nucleus
Three different models of the atom: a The plum-pudding model;
b The Bohr model; c The atomic orbital model
neutron
nucleus
Figure
Figure 1.1.2
The nucleus of an atom
contains protons and neutrons. The
electrons are outside the nucleus in fixed
energy levels.
2
1.1.2
shows
a
useful
model
of
an
atom
based
on
the
Bohr
model.
Chapter
Masses
and
The
actual
very
small.
charges of
masses
and
sub-atomic
charges
of
protons,
1
Atomic
structure
particles
neutrons
and
electrons
are
–27
For
example,
a
proton
has
a
mass
of
1.67
×
kg
10
and
a
–19
charge
of
+1.6
comparison.
Relative
×
C.
10
These
are
The
relative
shown
mass
in
the
masses
and
charges
give
an
easier
table.
Proton
Neutron
1
1
Electron
1
_____
1836
Relative
charge
+1
Sub-atomic
The
effect
electrons
of
is
a
particles
strong
shown
0
and
electric
field
–1
electric fields
on
beams
of
protons,
neutrons
and
below.
Exam tips
+
beam
+
of
beam
+
of
beam
of
The
electrons
protons
direction
an
electron
be found
deflection
+
towards
deflection
plate
–
towards
no
rule
from
the
deflect
by
a
the
movement
magnetic eld
of
Fleming’s
can
as
shown
direction
below.
left-hand
of
the
Remember
(1)
magnetic eld
The effect of an electric field on beams of electrons, protons and neutrons
is
Charged
in
effect
plate
The
Figure 1.1.3
of
neutrons
particles
plates
move
with
electrons
and
towards
the
same
protons,
the
plates
charge.
the
If
with
we
electrons
opposite
use
are
the
charge
same
deflected
and
voltage
to
a
away
to
N
→
points
the
S
(2)
in
that
the
the
centre nger
direction
opposite
to
electron flow.
much
first
greater
extent.
This
is
because
electrons
have
a
very
small
mass
finger
compared
with
protons.
field
left
hand
centre finger
Sub-atomic
particles
and
magnetic fields
current
thumb
Figure
1.1.4
shows
an
evacuated
glass
tube.
When
heated
by
a
low
voltage
motion
supply,
electrons
accelerated
they
are
towards
collide
with
produced
the
the
metal
from
the
cylinder
,
fluorescent
tungsten
C.
screen.
They
When
filament,
produce
the
a
north
F
.
They
glow
pole
are
when
of
a
magnet
Key points
is
brought
moves
near
the
downwards.
electron
The
beam,
direction
the
of
glow
the
on
the
fluorescent
movement
can
be
screen
predicted
from

Fleming’s
left-hand
rule.
The
direction
of
movement
is
at
right
angles
The
atomic
to t
the
direction
Remember
of
that
conventional
the
the
magnetic
direction
field
of
and
the
electron
conventional
flow
is
opposite
theory
has
developed
to
the
experimental
evidence.
current.
that
of
the

Most
is
current.
in
of
the
protons
the
mass
nucleus
and
of
the
which
atom
contains
neutrons. The
electron flow
electrons
F
are
outside
the
nucleus
C
in
shells.
conventional
magnetic field

Electrons,
protons
and
neutrons
current
have
characteristic
relative
N
masses
fluorescent
and
charges.
screen

movement
Figure 1.1.4
Beams
are
The effect of a magnetic field on a beam of electrons
of
protons
deflected
directions
Using
different
opposite
deflected
apparatus,
direction
since
to
they
the
have
beams
of
electrons.
no
protons
Beams
of
will
be
deflected
neutrons
will
in
not
the
be
by
in
and
opposite
electric
magnetic elds.
electrons
and
Neutrons
are
not
deflected.
charge.
3
1.2
Isotopes
Learning outcomes
and
radioactivity
Atomic
Atomic
On
completion
should
be
able
of
this
section,
dene
to:
‘mass
Atomic
number’
number
and
mass
number
(Z)
you
(It

number
also
number
equals
( Z)
the
is
the
number
number
of
of
protons
electrons
in
a
in
the
neutral
nucleus
atom
of
and
an
the
atom.
position
and
of
the
element
in
the
Periodic
T
able.)
‘isotopes’

dene
‘relative
‘relative
atomic
isotopic
mass’
mass’
using
and
the
Mass
Mass
number
number
(A)
( A)
is
the
number
of
protons
plus
neutrons
in
the
nucleus
12
C

scale
know
of
how
to
calculate
an
relative
So
atomic

write
and

masses
equations
some
the
number
of
neutrons
in
an
atom
is
A
−
Z
(mass
number
−
atomic
number).
involving
α-,
β-
Isotopes
γ-radiation
state
atom.
uses
of
radioisotopes.
Isotopes
are
atoms
with
the
same
atomic
number
but
different
mass
numbers.
We
show
the
mass
number
and
atomic
number
of
an
isotope
like
this:
electron
mass
number
→
atomic
number
→
23
Na
proton
Relative
hydrogen
1
atomic
and
←
element
symbol
11
relative
isotopic
mass
12
atoms
The
mass
of
a
single
atom
is
too
small
to
weigh,
so
we
use
the
C
proton
(carbon-12)
neutron
to
this
atom
standard,
Relative
as
a
standard.
which
atomic
has
mass
a
We
mass
(A
compare
of
the
exactly
12
masses
of
other
atoms
units.
)
r
Relative
atomic
mass
(A
)
is
the
weighted
average
mass
of
naturally
r
12
occurring
mass
deuterium
1
proton,
1
of
atoms
exactly
of
an
12
element
on
a
scale
where
an
atom
C
of
has
a
units.
atoms
neutron
Relative
Relative
isotopic
isotopic
mass
mass
is
the
mass
of
a
particular
isotope
of
an
element
12
on
a
scale
where
Calculating
tritium
1
proton,
Most
naturally
mass
numbers
atom
C
of
accurate
occurring
has
a
mass
relative
elements
of
exactly
atomic
exist
as
a
12
units.
masses
mixture
of
isotopes.
So
their
atoms
2
are
not
always
exact.
We
have
to
take
into
account
the
neutrons
proportion
Figure 1.2.1
an
of
each
isotope.
This
is
called
its
relative
isotopic
abundance .
The three isotopes of
The
relative
atomic
mass
for
neon
is
calculated
below
from
the
relative
hydrogen
isotopic
masses
and
relative
abundance.
Did you know?
Isotope
Isotopes
with
2,
8,
20,
28,
50,
Relative
isotopic
mass
Relative
abundance/%
82
20
Ne
and
126
neutrons
extremely
or
protons
stable. These
20
‘magic
21
Ne
numbers’
closed
are
thought
nuclear
the full
shells
shells
of
the
to
90.9
are
21
0.30
22
8.8
indicate
comparable
noble
to
22
Ne
gases.
(20.0
×
90.9)
+
(21.0
×
0.3)
+
(22
×
8.8)
______________________________________
A
=
=
r
100
4
20.2
Chapter
1
Atomic
structure
Radioactivity
Isotopes
of
some
spontaneously.
break
The
down,
table
elements
These
rays
shows
or
are
have
particles
three
nuclei
described
types
are
of
Name of
Type of
emission
emitted
Alpha
helium
as
which
break
radioactive
given
out.
These
down
(decay)
isotopes .
are
called
As
the
nuclei
emissions.
emission.
particles/rays
Stopped
by
Exam tips
(α)
nuclei
charged
(positively
thin
sheet
of
paper
A
particles)
common
α-particles
Beta
(β)
electrons
(produced
by
nuclear
6 mm
changes)
thick
are
not:
(γ)
very
thick
lead
He
be
written
for
each
of
that
atoms. They
helium
error
are
is
nuclei
to
think
that
electrons from
the
the
nucleus. The
β-particle
radiation
from
can
suggest
sheet
electrons
Equations
to
helium
are
. Another
outside
electromagnetic
is
2+
aluminium foil
high frequency
are
they
β-particles
Gamma
error
these
types
of
decay.
For
the
arise from
electron
the
nucleus
not
shells.
example:
α-decay
The
isotope
charge
of
2
produced
units
has
lower
a
mass
than
the
number
original
223
of
→
units
lower
and
a
nuclear
4
219
Ra
Rn
88
4
atom:
+
He
2
86
β-decay
The
mass
one.
This
number
stays
the
same
but
the
number
of
protons
increases
1
is
because
a
neutron
by
1
n)
(
is
changed
into
a
proton
(
0
p)
and
an
1
0
electron
e):
(
–1
1
(
For
example
carbon-14
0
1
n)
→
0
(
p)
+
(
1
decays
to
e)
–1
nitrogen-14
14
→
it
emits
a
β-particle:
0
14
C
N
6
when
+
e
–1
7
Key points
γ-decay
γ-rays
can
be
emitted
along
with
α-
or
β-particles
or
in
a
process
called

‘electron
capture’.
A
proton
is
converted
to
a
neutron,
so
the
Mass
of
number
stays
the
same
but
the
atomic
number
decreases
by
number
is
the
total
number
mass
one.
protons
+
neutrons
in
an
For
isotope.
example:
37
0
Ar
+
18

37
e
→
–1
Isotopes
are
same
number
different
Uses of
of
with
the
protons
numbers
of
but
neutrons.
radioisotopes


atoms
Cl
17
T
racers
for
searching
for
faults
in
pipelines
and
for
studying
of
certain
organs
in
the
body,
I
e.g.
atomic
are
and
isotopic
compared
using
the
12
131
working
Relative
masses
the
is
used
to
study
C
thyroid
scale.
function.


In
medicine,

Dating
for
radiotherapy
in
the
treatment
of
cancers.
The
relative
element
is
atomic
the
mass
weighted
of
an
mean
of
14
once
the
age
of
objects,
e.g.
using
C
to
date
objects
which
the
were
living.
atoms
of
the
isotopes
they
contain.
241

Smoke
detectors
often
use
Am.

When
an
unstable
isotope
decays
235

Generating
source
of
power
,
e.g.
U
is
used
in
many
nuclear
reactors
as
rays
a
or
particles
are
emitted.
energy.

α-,
β-,
and
γ-radiation
characteristic
have
properties.
5
1.3
Energy
levels
Learning outcomes
completion
of
this
section,
electrons
be
able
amount
explain
spectra
energy

explain
how
data from
give
evidence
levels for
the
lines
emission
of
discrete
electrons
of
the
Balmer
series
in
We
arranged
when
of
have
energy
a
that
energy
levels
certain
required
the
are
electrons
quantum
electron
is
energy
levels
electrons
The
absorbs
Lyman
say
in
the
them.
The
and
atoms
energy
for
fixed
a
values
change
is
of
energy.
called
a
The
smallest
quantum
of
to:
energy.

in
and
spectra
you
fixed
should
emission
Energy quanta
The
On
and
of
then
which
in
are
is
the
in
radiation
to
are
be
energy
their
it
in
or
quantised.
quantised.
lowest
then
said
discrete
levels
ground
can
an
The
move
atom
is
possible
state .
up
excited
The
to
a
state .
most
for
When
of
electron
energy
an
are
stable
each
an
higher
When
electrons
level.
excited
emission
electron
falls
to
a
lower
energy
level
a
quantum
of
radiation
is
given
out.
spectra
The

understand
how
energy
between
energy
levels
is
the
two
difference
electron
moves
to
a
lower
energy
level.
in
electrons’
related
to
the
energy
frequency
of
radiation.
electrons
ground
Figure 1.3.1
in
electrons
state
excited
in
state
An electron can jump from one energy level to another by absorbing or
emitting a fixed amount (quantum) of energy
The
difference
frequency
of
in
energy
radiation
between
by
the
two
energy
levels
is
related
to
the
relationship:
–1
energy
(J)
→
ΔE
=
hν
←
frequency
of
radiation
(s
)
↑
–34
Planck’s
In
any
atom
principal
electron
further
shells.
away
Atomic
When
an
there
They
from
electrical
frequencies.

It
is

The
several
levels
are
the
in
or
or
(6.63
×
possible
energy
principal
given
–1
J s
10
levels.
energy
numbers
n
=
)
These
levels .
1,
n
=
2,
n
thermal
radiation
An
emission
energy
is
is
emitted
spectr um
passed
only
at
differs
through
certain
3,
also
etc.
from
a
a
gaseous
normal
or
visible
light
that:
lines
up
of
called
going
sample
wavelengths
separate
converge
(get
lines.
closer
to
each
other)
as
their
increases.
lines
get
closer
together
6
=
called
are
spectra
the
made
Figure 1.3.2
are
They
nucleus.
emission
element,
spectrum
are
quantum
constant
Part of a simplified line emission spectrum of hydrogen
frequency
of
Chapter
Interpreting the
Each
line
moving
lines

in
the
from
seen
are
Lyman
hydrogen
higher
Balmer
fall
(seen
back
series
electrons
to
emission
a
lower
emission
spectrum
energy
is
level.
Atomic
structure
spectrum
a
result
Among
of
the
electrons
several
series
of
the:
series
electrons

a
hydrogen
1
fall
in
to
(seen
back
the
the
in
to
ultraviolet
n
=
the
the
1
visible
n
=
region),
energy
2
where
previously
excited
level.
region),
energy
where
previously
excited
level.
n = 5
n = 4
n = 3
n = 2
n = 1
frequency
emission
spectrum
Lyman
Figure 1.3.3
series
Balmer
series
A line emission spectrum is the result of electrons falling from higher energy
levels to lower energy levels
The
energy
associated
with
each
line
is
found
by
using
the
relationship
Did you know?
ΔE
=
hν.
Y
ou
wavelength),
can
the
see
that
more
the
energy
greater
is
the
frequency
released.
The
(the
point
smaller
where
the
the
lines
There
eventually
come
together
is
called
the
convergence
limit .
This
are
spectrum
an
electron
falling
from
the
highest
possible
energy
level.
If
the
other
more
energy
than
this,
it
becomes
free
from
the
pull
of
the
nucleus
atom.
The
The
This
Then
Bohr
energy
is
model
the
is
model of
levels
mirrored
of
atom
the
get
by
an
closer
the
hydrogen
converted
lines
an
spectral
the
infrared
called
and
Pfund
the
Paschen,
series.
ion.
n = 5
n = 4
towards
lines
shown
in
are
atom
together
atom
to
emission
of
Brackett
the
of
electron
region. These
has
sets
represents
getting
the
outside
closer
opposite,
a
of
the
together
.
quantum
of
n = 3
atom.
In
the
energy
Bohr
n = 2
moves
n = 1
the
A
hydrogen
further
The
quantum
more
When
levels,
electron
the
energy
electron
emitting
of
an
in
the
energy
electron
loses
n
=
will
of
level
excite
has,
energy,
radiation
1
it
the
(ground
the
electron
higher
will
fall
characteristic
state)
the
down
to
to
the
energy
to
the
the
n
level
lower
n
=
=
it
3
2
level.
level.
will
be
in.
energy
frequencies.
Key points
electron
shells
(principle

Electrons

When
occupy
specic
energy
levels
(quantum
levels)
in
the
quantum
Figure 1.3.4
electrons
gain
specic
quanta
of
energy
they
move from
shells,
n)
atom.
lower
The Bohr model of the atom.
to
The principal energy levels get closer
higher

When
energy
excited,
emitting
emission

The
levels. They
electrons
radiation
of
become
lose
‘excited’.
energy,
they fall
together towards the outside of the atom.
back
characteristic frequency. This
to
is
lower
the
energy
origin
of
levels
the
line
spectrum.
energy
difference
between
any
two
energy
levels
is
given
by
ΔE
=
hν
7
1.4
Sub-shells
Learning outcomes
and
Shells
The
On
completion
of
this
section,
atomic
and
principal
be
able
know
that
shell for
split

n
into
the
=
2
principal
and
shells)
know
relative
the
quantum
above
sub-shells
quantum
can
be
(subsidiary
Notice
terms

know
and
energy
energies
of
how
electronic
and

to
determine
the
term
a
hold
the
of
the
after
energy,
by
d.
level
first
the
apart
are
For
four
from
named
quantum
element
does
than
neutral
protons.
the
s,
first,
p,
d.
are
split
Figure
into
1.4.1
not
the
calcium
always
example
shells
in
the
conform
scandium
in
terms
order
to
the
of
the
4s
of
shows
the
increasing
the
sub-shells,
pattern
of
sub-shell
s
is
energy.
in
followed
at
a
by
p
lower
3d.
is
The
atom,
the
number
maximum
shown
in
the
of
number
electrons
of
is
electrons
equal
each
to
the
shell
number
and
of
sub-shell
can
table.
atoms
ions
understand
shells
These
the
sub-shells
conguration
for
that
of
followed
For
shells
quantum
(sub-levels).
to:
sub-shells

sub-shells
you
sub-shells
should
orbitals
Principal
Total
quantum
electrons
number of
shell
quantum
in
Maximum
principal
number of
electrons
in
sub-shell
‘atomic
shell
orbital’

describe
the
shapes
of
s
and
s
p
d
p
orbitals.
n
=
1
2
2
–
–
n
=
2
8
2
6
–
n
=
3
18
2
6
10
4d
5s
n = 4
4p
ygrene
3d
4s
Atomic orbitals
n = 3
3p
gnisaercni
Each
sub-shell
contains
one
or
more
atomic
orbitals .
An
orbital
is
a
3s
region
Each
of
space
orbital
where
can
hold
there
a
is
a
high
maximum
probability
of
two
of
finding
electrons.
So
an
the
electron.
number
of
2p
orbitals
n = 2
2s
each
in
other
each
in
sub-shell
shape
(see
is:
s
=
Figure
p
=
3
and
d
=
5.
Orbitals
differ
from
1.4.2).
1s
n = 1
z
prinicipal
1,
quantum
z
z
y
sub-shell
z
y
y
y
shell
x
Figure 1.4.1
x
x
x
The principal quantum
shells, apart from n = 1, are split into
sub-shells
s
orbital
p
orbital
p
x
Figure 1.4.2
Note
that:

A
2s

The
orbital
p
y
orbital
z
The shapes of s and p orbitals
orbital
three
is
larger
different
than
p
the
orbitals
1s
are
orbital
and
mutually
a
at
3p
is
right
larger
angles
than
to
the
2p.
each
other
.
Here
of
are
two
electrons
examples
present
2
in
of
how
to
write
the
type
of
orbital
and
the
number
them:
electrons
in
orbital
5
electrons
↓
↓
2
First
quantum
s
8
shell
→
1s
sub-shell
↑
5
Third
quantum
p
shell
→
3p
sub-shell
↑
Chapter
The
electronic
configuration of
1
Atomic
structure
atoms
Exam tips
The
electronic
electrons
in
a
1
sure
configuration
particular
shows
atom.
The
the
number
procedure
and
type
of
the
This
is:
diagram
remember
Make
you
know
the
order
of
the
orbitals,
1s
2s
2p
3s
etc.,
as
may
the
help
order
you
of lling
the
well
sub-shells.
as
the
number
of
electrons
in
the
atom.
1s
2
Fill
up
lowest
and
3
a
Stop
the
sub-shells
energy,
p
when
Example
e.g.
sub-shell
you
to
an
a
s
the
maximum
sub-shell
maximum
have
the
of
correct
can
amount
have
a
starting
maximum
with
of
2
those
of
electrons
number
of
configuration
of
sodium
2p
3s
3p
3d
4s
4p
4d
5s
5p
electrons.
1
Electronic
2s
6.
(Z
=
11):
6s
2
2
1s
Example
6
2s
1
2p
3s
2
Did you know?
Electronic
configuration
2
1s
2
6
2s
of
2
2p
3s
scandium
6
3p
2
4s
(Z
=
1
3d
21):
2
(or
2
1s
6
2s
2p
2
3s
6
1
3p
3d
2
4s
)
The
electronic
also
Note
that
the
4s
sub-shell
is
filled
before
the
3d
(see
Figure
be
p
,
x
For
electronic
configuration of
ions
to
the
ions
are
atom.
formed
In
one
general,
or
the
more
electrons
electrons
are
are
lost
or
p
,
removed
gained
from
from
or
the
be
p
y
example,
show
the
can
separate
.
(see
Figure
1.4.2).
z
the
conguration
can
When
to
1.4.1).
orbitals
The
conguration
written
of
electronic
sodium
(Example
written:
added
2
outer
1s
2
2s
2
2p
2
2p
x
1
2
p
y
3s
z
sub-shell.
Example
1
2
Magnesium
atom
Example
2
Nitrogen
atom
2
Note
and
that
ions
the
are
2
2
3s
2+
;
Magnesium
3
2p
electronic
less
2p
2
2s
1s
6
2s
1s
ion
3–
;
Nitride
ion
configuration
straightforward
(see
(N
of
(Mg
2
)
1s
some
Section
2
2s
2
)
2
1s
2s
6
2p
6
2p
transition
element
atoms
13.1).
Key points

The
principal
shells

Each

An
quantum
named
s,
sub-shell
orbital
is
a
p,
d,
can
shells
(apart from
the rst)
are
divided
into
sub-
etc.
hold
region
of
a
maximum
space
where
number
there
is
of
a
electrons.
good
chance
of nding
an
electron.

s
orbitals

Each

The

Electronic
to
are
orbital
spherical
can
maximum
the
hold
in
a
number
and
maximum
of
congurations
energy
shape
sub-levels
of
electrons
of
atoms
placed
in
p
have
2
in
an
hour-glass
shape.
electrons.
each
and
ions
order
of
sub-shell
is
are found
increasing
s
by
=
2,
p
=
adding
6,
d
=
10.
electrons
energy.
9
1)
1.5
Ionisation
Learning outcomes
energies
Ionisation
The
On
completion
of
this
section,
first
be
able
know
energy ,
ΔH
,
is
the
energy
needed
to
from
each
atom
in
one
mole
of
atoms
of
an
element
state
to
form
one
mole
of
gaseous
ions.
For
example:
+
explain
data
know
can
ionisation
evidence for
how
be
–1
(g)
Cl
+
e
ΔH
electron
energy
The
+1260 kJ mol
of
Size
of
charge
conguration
ionisation
value
the
ionisation
energy
depends
on:
sub-shells
nuclear
(greater
between
determined from
successive
=
i1


→
energy
how
give
one
its
influencing
Cl(g)

in
to:
the factors
ionisation
remove
i1
gaseous

ionisation
you
electron
should
energy
to
energies.
the
remove
charge:
number
nucleus
these
In
of
and
any
one
period,
protons),
outer
electrons.
So
the
electrons
and
tends
ΔH
the
higher
greater
to
the
the
the
nuclear
attractive
more
increase
force
energy
with
needed
increase
in
i1
nuclear

charge.
Distance
outer
of
the
electrons
between
them
outer
are
and
electrons
from
the
the
from
nucleus,
nucleus
and
the
the
the
nucleus:
less
The
attractive
lower
the
value
further
force
of
the
there
is
ΔH
i1

Shielding:
between
of
full
Electrons
the
in
nucleus
inner
shells,
full
and
the
inner
the
shells
outer
greater
the
reduce
electrons.
shielding
the
The
and
attractive
greater
the
force
the
lower
number
the
value
of
ΔH
i1
First
ionisation
Figure
1.5.1
shows
energies
how
the
across
values
of
a
ΔH
period
change
across
the
first
three
i1
periods.
first
Did you know?
ionisation
–1
energy/kj
He
2500
The
p
,
p
x
and
p
y
orbitals
in
mol
a
z
particular
Ne
sub-shell
are
equal
2000
in
F
energy. The
electrons
are
Ar
added
N
1500
Cl
one
by
one
to
these
separate
H
before
they
are
paired
P
Be
1000
orbitals
C
Si
S
up.
Ca
B
500
This
minimises
repulsions
between
Li
Na
K
electrons. The
electrons
in
0
the
0
separate
p
orbitals
of
nitrogen
10
5
are
shown
20
and
atomic
oxygen
15
number
here:
Figure 1.5.1
First ionisation energies for the first 20 elements in the Periodic Table
2p
We
can
energy
use
the
with
three
ideas
increasing
above
proton
to
explain
the
change
in
ionisation
number
.
2s

Across
a
period
there
is
a
general
increase
in
ΔH
.
The
increased
i1
1s
nuclear
charge
because
nitrogen
2
1s
The
2
2
1s
2p
extra
successive
of
protons)
electrons
outweighs
added
are
the
being
other
placed
two
in
factors
the
same
oxygen
3
2s
the
(number
electron
in
2
2s
4
2p
oxygen
has
outer
electron
there
is
the
shell.
same
There
number
is
of
not
much
inner
difference
electron
in
shielding
since
shells.
to

When
a
new
period
starts,
there
is
a
sharp
decrease
in
.
ΔH
The
outer
i1
go
into
an
orbital
which
is
already
electron
half-lled. The
extra
repulsion
goes
in
the
same
orbital
a
shell
further
from
is
Both
This
these
also
factors
explains
outweigh
why
10
the
effect
decreases
ΔH
i1
repulsion.
nucleus
and
there
of
greater
is
more
nuclear
called
charge.
spin-pair
the
of
shielding.
electrons
into
down
a
group.
Chapter

The
decrease
in
ΔH
from
Be
→
2p
sub-shell,
B
is
due
to
the
fact
that
the
1
Atomic
structure
ionisation
energies,
fifth
i1
electron
the
in
B
nucleus
slightly.
charge.
goes
than
These
A
into
the
two
similar
the
2s
sub-shell.
factors
reason
The
outweigh
explains
which
shielding
the
the
is
effect
dip
in
slightly
also
of
the
from
increases
increased
from
ΔH
further
Mg
→
nuclear
Al.
i1
2
Mg

The
decrease
=
in
1s
2
2s
ΔH
6
2p
2
2
3s
from
N
Al
→
O
is
=
1s
due
2
2s
to
6
2p
2
3s
1
3p
spin-pair
repulsion.
The
i1
eighth
in
it
it.
electron
The
easier
can
O
increased
to
remove
Successive
We
in
goes
into
repulsion
the
eighth
ionisation
remove
electrons
associated
with
ionisation
energies.
an
are
similarly
already
charged
atom
one
electrons
given
the
by
are
one.
called
symbols
The
the
an
electron
particles
makes
,
ΔH
energies
Exam tips
successive
ΔH
i1
For
has
electron.
an
these
They
the
which
energies
from
removing
of
orbital
,
ΔH
i2
,
etc.
i3
example:
In
+
O
2+
(g)
→
O
O
+
e
ΔH
=
+3390 kJ mol
1
3+
(g)
→
O
+
e
ΔH
=
+5320 kJ mol
can
configuration from
work
out
the
energies.
electron
Figure
ionisation
configuration
1.5.3
shows
of
how
an
this
energies
element
is
done.
from
In
successive
The
it
may
also
be
possible
to
distinguish
between
s
atoms
or
by
small
changes
in
always
successive
ionization
ions
are
always
3
The
charge
on
the
ion
formed
is
some
and
p
same
as
the
number
of
the
sub-
(successive)
shells
are
ions.
gaseous.
the
instances
ions formed
positive
2
ionisation
The
–1
(g)
i3
We
that:
–1
(g)
i2
2+
Electron
considering
remember
ionisation
energy
energies.
number
e.g. for
ionisation
is
a
4+
the fourth
energy
the
ion formed
ion.
5
lom
1–
2
electrons
difficult
to
very
remove
Jk/ygrene
Key points
4
8
electrons
easy
to
less
remove

The rst
ionisation
energy
is
noitasinoi
11+
the
1
energy
needed
to
remove
a
electron
3
easily
mole
removed
of
electrons from
gaseous
a
mole
of
atoms.
gol
01

The
ionisation
energies
required
nucleus
to
2
0
1
2
3
4
5
6
7
8
9
10
remove
electrons
time from
number
of
electrons
a
mole
of
at
a
gaseous
removed
atoms
Figure 1.5.3
one
11
are
called
the
successive
a Successive ionisation energies for a sodium atom; b How the arrangement
ionisation
energies.
Ionisation
energy
of electrons can deduced from these ionisation energies

on
the
nuclear
distance
from

Trends
be
of
the
in
the
values
charge,
outer
nucleus
and
ionisation
explained
using
depend
the
electrons
shielding.
energy
these
can
three
factors.

The
values
ionisation
atom
of
provide
electronic
successive
energies
of
an
evidence for
the
conguration.
11
Revision
1
State
and
explain
later
2
the four
main
any
the
protons,
of
Dalton’s
modifications
discoveries
Explain
tenets
concerning
direction
electrons
and
and
questions
to
the
his
atomic
theory
in
theory
light
13
of
neutrons
of
in
deflection
an
the
atoms
atom.
degree
Write
electronic
and
a
Ca
b
Zn
c
Cl
d
Mn
a
Define
b
Write
configurations
of
the following
ions:
of
electric field.
2+
3
What
and
is
the
rationale for
relative
than
their
charges
absolute
of
using
the
relative
sub-atomic
masses
and
masses
particles
rather
14
charges?
the
an
energy
4
How
many
protons,
the following
electrons
and
neutrons
are
the
equation
of
‘first
ionisation
showing
the first
energy’.
ionisation
calcium.
in
atoms?
15
27
a
term
State
the
three factors
ionisation
Al
that
influence
the value
of
the
energy.
13
39
b
K
16
19
131
c
Explain
energy
I
the
general
across
a
trend
of
an
increase
in
ionisation
the first
ionisation
period.
53
239
d
Pu
17
94
5
Calculate
the
relative
atomic
10
that
its
two
isotopes
abundances
6
a
Write
of
equations
following
of
boron
and
and
B
have
81.3%
showing
the
there
energy
between:
a
Mg
and Al
b
P
is
a
decrease
in
relative
respectively.
α-decay
of
18
the
and S
The following
ionisation
radioisotopes:
group
238
i
why
given
11
B
18.7%
mass
Explain
of
data
energies
the
show
the first
of
element X. Suggest
an
Periodic Table X
seven
belongs
successive
which
to.
U
92
222
ii
Ra
88
–1
Ionisation
b
Write
equations
following
showing
the
β-decay
of
energy
number
Enthalpy/kJ mol
the
radioisotopes:
1st
737
.7
234
i
Th
90
2nd
1450.7
14
ii
C
6
3rd
7
State
the
8
Explain
uses
how
of four
an
named
atomic
7732.7
radioisotopes.
emission
spectrum
is
4th
10 542.5
5th
13 630
6th
18 020
7th
21
produced.
9
10
How
is
the
a
Lyman
b
Balmer
in
Write
equation
the
meaning
State
held
12
series
produced
difference
11
series
in
that
all
of
two
the
maximum
an
s,
a
What
is
b
Draw
a:
12
hydrogen
between
of
the
the
i
2s
ii
2p
p
and
an
d
orbital
indicates
energy
symbols
number
orbital?
spectrum?
the
levels
energy
and
state
the
used.
of
sub-shell.
atomic
orbital
emission
electrons
that
can
be
711
Chapter
1
Atomic
structure
–
revision
questions
13
2
Structure
and
2.
1
States
matter
of
Learning outcomes
bonding
States of
Substances
On
completion
of
this
section,
be
able
understand
the
between forces
states
of
with
high
and
melting
melting
points
attraction
point
their
atoms
(or
ions).
have
strong
Substances
with
forces
low
of
attraction
melting
points
have
to:
weak

matter
of
you
between
should
and forces
forces
of
attraction
between
their
molecules.
relationship
of
attraction
and
a
weak forces
c
d
b
matter
between
strong
molecules
attractive

know
that
single
covalent
bonds
forces
are formed
electrons
by
sharing
between
a
the
pair
of
atoms
weak

understand
covalent
bonding
strong
bonding
within
molecules
in
forces
terms
of
overlap
of
atomic
Figure 2.1.1
The strength of forces and arrangement of particles in solids, liquids and gases
orbitals

understand
sigma
and
pi
Giant
structures
with
ionic
or
covalent
bonds
are
solids
bonding.
with
high
break
solid
with
the
many
are
only
within
the
molecular
solid
to
are
overcome
a
lot
between
of
energy
the
to
particles.
and
do
not
move.
melting
strong.
weak
so
points:
But
it
the
does
The
forces
forces
not
between
take
much
these.
xed
arrangement
only
low
low
are
fairly
close
together
and
do
vibrate.
melting
are
strong.
are
weak
points.
But
so
it
the
The
forces
does
not
forces
within
between
take
the
the
much
energy
to
these.
are
randomly
close
together
arranged.
They
but
are
slide
more
over
or
each
less
other
.
the
and
Gases
The
does
+
in
have
Particles
are
They
molecules
liquid
have
overcome
molecules
between
forces
arrangement
molecules
move.
Liquids
atom
xed
solids
the
Particles
not
one
in
molecules
energy
of
takes
vibrate.
Molecular
electrons
strong
It
structure
They
attractive forces
points:
a
Particles
giant
melting
have
forces
not
very
low
between
take
melting
the
much
points
molecules
energy
to
and
are
boiling
very
overcome
points.
weak
so
it
these.
+
Particles
are
far
apart
and
move
randomly.
gas
Figure 2.1.2
a The formation of a covalent bond
Covalent
A
electron
H
H
in
pair
covalent
bond
b Two ways of showing the covalent bond
in hydrogen
14
of
covalent
two
bond
is
formed
neighbouring
attractive
electron
bonds
bonding
forces
clouds
are
are
atoms
in
when
usually
by
balance
the
strong.
the
and
force
pair
with
nuclei
It
a
are
needs
a
of
of
the
a
attraction
electrons
repulsive
certain
lot
of
between
between
forces
distance
energy
to
the
between
apart.
break
nuclei
them.
The
the
Covalent
them.
Chapter
Sigma
A
bonds
covalent
bond
and
is
pi
orbital
formed
which
when
combines
atomic
orbitals
contributes
one
(Section
unpaired
1.4)
The
‘joined’
orbital
is
called
to
of
the
atomic
orbitals,
a
the
molecular
orbital .
The
greater
combining
stronger
is
the
covalent
(σ
called
its
a
line
bonds
drawn
formed
is
(σ
valency
valentia
covalent
(atoms)
an
atom
–
(from
meaning
the
powerful).
means
joining
‘the
power
together’. The
bonds)
concept
Sigma
of
bond.
of
bonds
power
the
So
Sigma
bonding
the
Latin
overlap
and
overlap.
electron
is
bond.
Structure
Did you know?
bonds
The
Each
2
bonds)
between
are
the
symmetrical
formed
two
about
by
nuclei.
a
line
the
The
overlap
of
electron
joining
the
atomic
density
two
orbitals
of
the
along
bond
nuclei.
many
of
hydrogen
together
valency
one
with
of
is
based
atoms
a
given
oxygen
oxygen
hydrogen
nucleus
valency
can
is
in
how
combine
atom,
two
combine
atoms
on
e.g.
the
because
with
two
water.
+
s
atomic
molecular
orbital
orbitals
Figure 2.1.3
When
a
p
modied
slightly
Two s-type atomic orbitals overlap to form a molecular orbital
orbital
to
altered
orbital,
one
similar
thing
a
σ
bond
combines
include
of
is
in
some
shape
the
so
‘lobes’
happens
with
s
and
that,
of
the
when
an
s
orbital,
some
on
p
combining
hour-glass
two
p
the
molecular
character
.
to
shape
orbitals
The
p
form
a
is
becomes
molecular
becomes
combine
orbital
orbital
smaller
.
‘end- on’.
In
A
each
case
formed.
Exam tips
molecular
nucleus
orbital
Although
of
+
remember
of
modified

p
a
s
π
electron
bond
has
density,
that
electron
these
density
two
you
areas
must
two
areas
represent
only
bond
orbital
one
bond.
Do
not
confuse
it
with
a
orbital
double
molecular
bond,
which
contains
one
σ
orbital
bond
and
one
π
bond.
+
Key points
modified
p
modified
orbital

p
bond
orbital

Figure 2.1.4
Solids
with
have
bonds
(π
bonds)

Liquids
bonds
(π
bonds)
are
formed
by
the
sideways
overlap
of
p
their
and
melting
Pi
The
electron
density
of
the
bond
formed
is
not
of
atoms
gases
points
points
attraction
or
have
ions.
low
because
of
the
atomic
weak forces
orbitals.
melting
strong forces
between
Pi
high
The formation of sigma bonds by ‘end-on’ overlap of atomic orbitals
between
their
symmetrical
molecules.
about
a
line
overlapping
joining
to
form
the
a
π
two
nuclei.
bond
(see
Figure
also
2.1.5
Section
shows
2
p
atomic
orbitals
2.9).

A
covalent
the force
the
of
nuclei
pair
of
bond
is formed
attraction
of
two
electrons
by
between
atoms
and
the
which forms
the
bond.
-bond

Sigma
on’

p
z
orbital
p
Pi
bonds
overlap
bonds
atomic
are formed
by
‘end-
orbitals.
by
‘sideways’
orbital
z
overlap
Figure 2.1.5
are formed
of
The formation of a pi bond by ‘sideways’ overlap of p
of
atomic
orbitals.
atomic orbitals
z
15
2.2
Covalent
Learning outcomes
bonding
Simple dot
A
On
completion
should
be
able
of
this
section,
shared
pair
–
dot
and
of
and
cross
diagrams
cross diagrams
electrons
in
a
single
bond
is
represented
know
that
is formed
a
pair
of
a
single
when
covalent
two
atoms
bond
of
share
draw
a
variety
and
of
cross
draw
compounds
bonds
dot
electron
and
is
a
bond
+
fluorine
cross
deficient
compounds
octet
atoms
fluorine
molecule
The electron pairs in a fluorine molecule
diagrams for
compounds
with
is
usually
of
only
the
outer
electrons
which
are
used
in
covalent
bonding.
an
Electron
expanded
pairs
in
the
outer
shell
which
are
not
used
in
bonding
are
called
electrons.
lone
pairs
Dot
and
shell

cross

diagrams
electrons
Use
a
cross
dot
to
Draw
pairs
in
to
the
can
covalently
represent
represent
outer
and
orbital
Example:
pair
with
only
It
and
shared
diagrams for
Figure 2.2.1

line.
electrons
dot
single
single
electrons
covalent
F

a
to:
a

by
you
the
the
the
contains
a
used
outer
outer
electrons
number
be
bonded
of
in
to
pairs.
of
the
arrangement
The
electrons
electrons
pairs
maximum
show
shell
shell
lone
to
molecules.
two
also
pairs
from
from
emphasise
It
main
of
one
another
the
reects
of
points
atom
and
a
atom.
number
the
outer
are:
fact
of
bond
that
each
electrons.
Methane

If
possible,
electrons
are
arranged
so
that
each
atom
has
four
pairs
of
H
electrons
around
conguration).
H
+
C
it
(an
octet
Hydrogen
is
of
electrons/
an
exception
the
–
it
noble
can
gas
only
electron
have
two
C
electrons

When
around
pairing
its
nucleus
electrons,
a
when
forming
covalent
bond
is
covalent
formed
bonds.
between
an
H
electron
Figure 2.2.2
from
one
atom
(dot)
and
an
electron
from
another
atom
Drawing a dot and cross
(cross).
diagram for methane
Examples of dot
and
cross diagrams
a
b
+
H
Cl
Cl
H
2
H
+
O
O
H
H
hydrogen
water, H
O
2
chloride
Did you know?
When
we
draw
dot
and
cross
(the
diagrams,
there
is
no
difference
3
in
+
H
N
N
extra
,
the
electrons
pair. The
dots
book-keeping
within
and
each
electron
crosses
exercise
to
are
just
keep
comes
from
H
a
a
metal
H
atom
when
forms
track
electr
O
H
ammonia,
NH
hydroxide
an
it
ion)
ion
3
of
the
electrons. The
electron
density
Figure 2.2.3
in
lone
pairs
of
electrons
is,
Dot and cross diagrams for: a hydrogen chloride; b water; c ammonia; d the
however,
hydroxide ion
greater
than
that
of
bonding
pairs
of
–
electrons.
For
more
information
see
Note
that
we
can
draw
dot
and
cross
structures
for
ions
such
as
OH
and
+
Section
2.8.
by
NH
the
same
rules.
The
square
brackets
4
show
16
applying
that
the
charge
is
spread
evenly
over
the
ion.
around
the
ion
Chapter
Electron deficient
Some
molecules
are
molecules
covalent
unable
bonds.
to
These
complete
molecules
the
are
octet
said
of
to
electrons
be
when
electron
example
is
boron
trichloride,
.
BCl
and
bonding
they
is
a
This
only
has
six
common
error
to
try
and
write
deficient .
dot
An
Structure
Exam tips
It
form
2
electrons
and
cross
diagrams for
ionic
around
3
compounds
the
boron
as
if
they
were
covalent
atom.
molecules.
of
+
3
Cl
B
a
metal
likely
to
ion
very
If
you
and
be
have
a
compound
non-metal
ionic
unless
then
the
it
is
metal
Cl
is
compared
To
see
small
with
how
to
and
the
highly
charged
non-metal
write
dot
and
ion.
cross
Cl
diagrams for
Section
Figure 2.2.4
ionic
compounds
see
2.4.
Dot and cross diagrams for boron trichloride, BCl
3
Molecules
Some
to
with
molecules
more
than
electrons.
An
8.
can
an
expanded octet
increase
These
example
the
molecules
is
sulphur
number
of
electrons
are
to
have
said
hexauoride,
an
.
SF
in
their
outer
expanded
This
has
shell
octet
12
of
electrons
6
around
the
sulphur
atom.
+
6
S
F
F
F
Figure 2.2.5
F
Dot and cross diagrams for sulphur hexafluoride, SF
6
Key points

A
single

When
of

bond
atoms form
is formed
covalent
when
bonds
two
each
atoms
atom
share
usually
a
has
pair
a full
of
electrons.
outer
shell
electrons.
Electron
their

covalent
deficient
outer
Atoms
atoms
in
a
compound
have fewer
than
8
electrons
in
shell.
with
an
expanded
octet
have
more
than
8
electrons
in
their
outer
shell.

Dot
and
cross
molecule
or
diagrams
show
how
the
electrons
pair
together
in
a
ion.
17
2.3
More
dot
Learning outcomes
and
Molecules
Some
On
completion
of
this
cross
section,
atoms

be
construct
for
The
describe
dot
and
cross
with
by
bonds
bond
is
sharing
two
shown
by
a
pairs
of
double
electrons.
line,
e.g.
A
for
double
oxygen,
bond
a
diagrams
double
co-ordinate
dot
and
bonding
two
bonding)
and
cross
oxygen
atoms
oxygen
molecule
diagrams
b
involving
co-ordinate
bonds.
2
+
2 O
atoms
C
atom
carbon
dioxide
c
C
H
4
H
C
+
H
C
H
H
C
ethene
4
H
Figure 2.3.1
When
The
atoms
atoms
Dot and cross diagrams for: a oxygen; b carbon dioxide; c ethene
atoms
triple
2 C
share
bond
is
three
shown
pairs
by
a
of
electrons,
triple
line,
a
e.g.
triple
for
bond
is
nitrogen
formed.
N ≡N.
+
Figure 2.3.2
A dot and cross diagram for nitrogen
Co-ordinate
A
co - ordinate
provides
bonding

one

a
The
the
both
to
atom
atom
lone
bonding
bond
the
occur
second
we
with
with
of
(dative
electrons
atom
pair
a
the
bond)
covalent
is
formed
bond.
For
when
1:
lone
pair
an
of
to
orbital.
orbital
(an
complete
electron
the
outer
decient
shell
of
atom)
both
Ammonium
ion,
NH
4
+
H
H
+
N
H
+
H
N
H
+
Figure 2.3.3
atom
electrons
unlled
unlled
electrons
H
one
co - ordinate
need:
with
the
covalent
for
+
Example
A dot and cross diagram for an ammonium ion, NH
4
18
is
O=O.
+
covalent
construct
bonds
double
bonds
(dative

form
multiple
to:
molecules
triple

able
with
you
formed.
should
diagrams
accepts
atoms.
Chapter
In
this
2
Structure
and
bonding
example:
+

The
H
ion
has

The
nitrogen

The
lone

Each
space
atom
pair
on
for
has
the
a
two
more
lone
pair
nitrogen
electrons
of
in
its
outer
shell.
electrons.
provides
both
electrons
for
the
bond.
+
H
(H
The
atom
=
2;
N
now
=
displayed
co - ordinate
has
2,
a
gas
electron
conguration.
8).
N
for mula
bond
noble
as
an
(formula
→.
The
showing
head
of
all
the
atoms
→
and
points
bonds)
away
shows
from
the
the
atom
H
Figure 2.3.4
The displayed formula for an
+
which
donates
Example
2:
the
lone
ammonium ion, NH
pair
.
Compound
4
for med
between
and
BF
NH
3
3
F
F
H
B
F
+
H
N
B
F
H
N
H
H
F
Figure 2.3.5
H
F
A dot and cross diagram for the co-ordination compound, BF
NH
3
In
this

B
example:
has
the
uorine
its
3
atoms.
outer

N
has

The

Both
a
simple
shell
lone
Example
3:
There
to
of
conguration
still
the
room
nearest
for
2
noble
2,
6
when
more
gas
bonded
electrons
to
to
be
three
added
to
conguration.
electrons.
donates
and
is
form
pair
nitrogen
boron
electronic
its
nitrogen
Aluminium
lone
pair
have
the
chloride,
to
the
unlled
simple
orbital
electronic
of
boron.
conguration
2,
8.
AlCl
3
Aluminium
atom
gas
is
2
chloride
electrons
is
an
short
electron
of
the
8
decient
electrons
molecule
required
–
the
aluminium
for
the
nearest
noble
Did you know?
conguration.
Hydrated
At
room
temperature,
aluminium
chloride
exists
as
Al
Cl
2
AlCl
6H
3
This
is
because
the
lone
pairs
of
electrons
on
two
of
the
chlorine
form
co - ordinate
bonds
with
the
aluminium
O
is
an
chloride,
ionic
compound,
2
atoms
but
can
aluminium
molecules.
6
anhydrous
aluminium
chloride
a
molecule
with
the formula Al
Cl
2
which
has
between
Cl
is
atoms.
co-ordinate
two
of
the Cl
6
bonding
and Al
Cl
Cl
atoms. The
ionic
Al
hydrated
because
the
compound
water
is
molecules
Al
3+
stabilise
the Al
co-ordinate
ions. They
bonding
do
between
this
Cl
molecules
and
the
highly
charged
3+
Al
Figure 2.3.6
A dot and cross diagram for the Al
Cl
2
ion.
molecule
6
Key points

A
double
three

In
bond
pairs
of
the
two
pairs
of
electrons
and
a
triple
bond
contains
electrons.
co-ordinate
for
contains
(dative
covalent)
bonding
one
atom
provides
both
by
water
electrons
bond.
19
2.4
Ionic
and
Learning outcomes
metallic
The formation of

On
completion
of
this
section,
When
be
able
know
that
are
compounds
are
transferred
from
metal
atoms
to
non-metal
atoms,
formed.
to:


electrons
ionic
you
ions
should
bonding
ions
are formed
The
outer
shell
of
both
ions
formed
have
the
noble
gas
electron
when
conguration:
atoms
gain
or
lose
electrons
+
K

write
dot
and
cross
e
O
+ 2e
→
O
8,
8,1
2–
[2,8,8]
2,6
[2,8]
structures


2–
+
K
+
2,
ionic
→
diagrams for
describe
the
structure
of
metals.
Positive
formed

In
an
ions
by
are
gain
ionic
formed
of
by
loss
of
electrons
and
negative
ions
are
electrons.
compound
the
number
of
positive
and
negative
charges
2+
must
balance,
e.g.
for
calcium
chloride
the
ions
present
are
Ca
and
2+
Cl
Exam tips
.
So
the
simply,
formula
for
calcium
chloride
is
[Ca
]
2[Cl
],
or
more
CaCl
2
1
Remember
that for
in Groups
to
I
III,
metal
the
atoms
charge
Dot
on
as
the
the
ion formed
group
is
the
number.
For
non-
When
show
metal
the
atoms
charge
number
in Groups V
on
–8.
the
For
ion
is
and
cross diagrams for
ionic
structures
same
writing
only
dot
the
and
outer
cross
diagrams
electron
shells
for
ionic
because
structures
these
are
we
the
usually
ones
involved
in
to VII,
electron
transfer
.
diagram
for
Figure
2.4.1
shows
how
to
construct
a
dot
and
cross
group
example,
magnesium
oxide.
Note:
S forms
2–
the
sulphide
ion,
S
(6
−
8
=
–2).

The
ions
formed
conguration
2
Remember
ions
do
not
that
+
e.g.
H
some
contain
of
have
the
a
full
nearest
outer
shell
noble
gas).
of
electrons
(electron
positive
metals,

The
charge
on

The
square
brackets
the
ion
is
placed
at
the
top
right.
+
and
NH
ions.
indicate
that
the
charge
is
spread
throughout
4
ion.
Mg
Mg
2+
2,8,2
Figure 2.4.1
2,6
2–
[2,8]
[2,8]
Constructing a dot and cross diagram for magnesium oxide
a
b
+
Cl
2–
Na
2+
+
[2,8,8]
[2,8]
O
Ca
+
Na
2–
2+
[2,8,8]
[2,8]
Cl
+
[2,8]
[2,8,8]
Figure 2.4.2
20
Dot and cross diagrams for: a calcium chloride; b sodium oxide
the
Chapter
Ionic
2
Structure
and
strong
bonding
bonding
ionic
between
The
ionic
bond
is
an
electrostatic
force
of
attraction
between
oppositely
charged
charged
ions.
formation
regularly
The
of
The
a
net
giant
arranged
electrostatic
attractive
ionic
in
a
forces
str ucture.
between
In
this
three-dimensional
attractive
forces
these
structure
lattice
between
ions
the
(see
ions
act
results
the
also
in
ions
in
ions
the
are
Section
all
bonds
oppositely
2.6).
directions
Cl
and
the
bonding
is
very
strong.
+
Na
Metallic
bonding
Figure 2.4.3
Most
metals
exist
in
a
lattice
of
ions
surrounded
a
‘sea’
of
Part of a giant ionic lattice
delocalised
of sodium chloride
electrons.
any
Delocalised
particular
atom
electrons
or
ion.
are
They
those
are
free
which
to
are
move
not
associated
between
the
with
metal
ions.
outer
electrons
of
metal
become
delocalised
‘sea’
Figure 2.4.4
of
delocalised
electrons
The structure of a typical metal. (Note: the diagram shows only one layer of
atoms for clarity. In reality there would be other layers on top of and beneath the layer
shown.)
In

this
structure:
The
number
electrons

The
positive
attraction

of
lost
This
to
strong
delocalised
by
each
charges
the
electrons
metal
are
held
delocalised
electrostatic
depends
on
the
number
of
atom.
together
by
their
strong
electrostatic
Exam tips
electrons.
attraction
acts
in
all
Remember
directions.
between

So
metallic
bonding
is
usually
that
the
metallic
big
and
difference
ionic
bonding
strong.
is:
The
strength
of
metallic
bonding
increases
with:
Ionic:

increasing

decreasing
positive
charge
on
the
the
Metallic:
size
of
the
metal
negative
charges
are
ions.
ions
the
negative
charges
are
ions
electrons.

increasing
number
of
delocalised
electrons.
Key points

Ions
are formed

Ions
generally

Ionic

bonding
when
have
is
a full
the
charged
ions.
Metallic
bonding
is
atoms
net
a
gain
outer
or
shell
lose
of
electrons
attractive force
lattice
of
positive
electrons.
between
ions
in
a
(noble
gas
positively
‘sea’
of
structure).
and
negatively
delocalised
electrons.
21
2.5
Electronegativity
Learning outcomes
and
Electronegativity
Electronegativity
On
completion
of
this
section,

be
able
the
meaning
of
terms
‘electronegativity’
‘bond
polarity’
describe
the
three
types
intermolecular forces:
dipole–dipole,
to
the
ability
attract
the
of
a
particular
bonding
pair
of
atom
involved
electrons
to
in
covalent
itself.
van
and
of

Electronegativity
increases
across

Electronegativity
decreases
down

The
a
period
from
Group
I
to
Group
VII.
the
order
of
electronegativity
is:
any
F
group.
>
O
>
N
>
Cl
>
Br…>
C
>
H.
weak
permanent
Polarity
in
molecules:
bond
polarisation
der Waals,

hydrogen
formation
to:
understand

is
you
bond
should
intermolecular forces
If
the
electronegativity
values
of
the
two
atoms
in
a
covalent
bond
are
in
a
covalent
bond
are
bonding.
the

If
same,
the
different,
a
we
say
that
the
electronegativity
we
say
that
bond
values
the
is
of
bond
non-polar
the
is
two
atoms
polar
b
We
±
H
H
H
+
–
can
also
apply
the
idea
of
polarity
to
molecules:
F

In
a
non-polar
molecule
the
centres
of
positive
and
negative
charge
coincide.
centre
of
ion
–
centre
of
centre
of

charge the
same
+
charge
–
In
a
polar
coincide.
Figure 2.5.1
molecule
the
centres
of
positive
and
negative
charge
do
not
charge
This
means
that
one
end
of
the
molecule
is
slightly
a Hydrogen is a non-polar
negatively
+
charged,
,
δ
and
the
other
end
is
slightly
positively
charged,
δ
molecule because the centres of + and
– charge coincide. b Hydrogen fluoride is
The
degree
polar because the centres of + and –
the
sign
of
→.
+
polarity
The
→
is
measured
points
to
the
by
a
dipole
partially
moment.
negatively
This
charged
is
shown
end
of
by
the
charge do not coincide.
molecule.
+
Examples:
+
δ
δ
δ
δ
→
+
H
If
we
tell
if
know
the
bonds
The
act
–
the
Cl
Cl
shape
molecule
so
that
molecule
positive
←
+
and
is
of
as
a
they
then
negative
a
–
Br
molecule
whole
oppose
is
each
non-polar
charge
is
and
polar
other
,
(see
the
the
or
polarity
non-polar
.
they
Figure
may
2.5.2)
of
If
each
the
cancel
because
bond,
we
polarities
each
the
can
of
other
centre
the
out.
of
same.

H
a
Cl
b
c

+
+


O
C
C
Exam tips
H
We
often
talk
about
bonding
+
H
+
in
Cl
+
Cl

+
Cl
general
terms
as
being
strong
Cl

Many
chemists
add
the
term
so
dipoles
add
so
dipoles
polar
molecule
polar
each
other
bonding,
especially
weak
bonding
we
are
talking
is
more
between
as forces
may
accurate
between
help
to
make
between forces
(strong
molecules,
22
refer
to
molecules.
the
within
bonding)
molecules
to
The polarity in: a water; b trichloromethane; c tetrachloromethane
about
Weak
it
non-polar
covalent
Figure 2.5.2
bonding. When
so
‘bonding’ for
molecule
stronger
cancel
however
molecule
reserve


or
dipoles
weak.
Cl

Cl

intermolecular forces
them
It
distinction
Inter molecular
in
neighbouring
forces
arise
molecules.
because
There
molecules
and forces

permanent

van

hydrogen
dipole–dipole
between
der
W
aals
forces
(weak).
bonding.
of
are
forces
the
three
attraction
types
of
between
the
dipoles
intermolecular
force:
Chapter
Intermolecular
metallic
forces
dipole
forces
bonding.
are
>
generally
van
der
are
The
in
weak
compared
comparative
the
order:
with
strength
hydrogen
covalent,
of
these
bonding
ionic
2
Structure
and
and
intermolecular
>
permanent
dipole–
W
aals.
CH
CH
3
Permanent dipole–dipole forces
3
+

C
Per manent
bonding
dipole–dipole
forces
are
the
weak
attractive
forces
+
O
CH
between

C
O
CH
3
3
+
the
of
δ
the
dipole
neighbouring
of
one
molecule
molecule
(see
Figure
and
the
δ
of
the
dipole
of
a
Figure 2.5.3
2.5.3).
Permanent dipole–dipole
forces between two propanone molecules
van der Waals forces
van
der
W
aals
molecules,

Electrons

So
the
than
forces
including
in
of
atoms
electron
attraction
noble
are
density
gas
are
atoms,
always
may
in
be
not
permanent.
have
van
der
All
atoms
W
aals
and
forces.
motion.
greater
in
one
part
of
the
molecule
another
.
Did you know?

So
an

This
instantaneous
dipole
can
dipole
induce
the
is
formed.
formation
of
a
dipole
in
a
neighbouring
molecule.

The
two
The
e.g.
neighbouring
molecules
attract
each
other
because
of
their
peculiar
lower
and
high
more
dipoles.
than
electron
here
density
and
temporarily

surface
extensive
in
many
molecules
greater
+

+

+
two
water
+
of
ice
tension,
are
hydrogen
can form
an
bonds
water
due
to
bonding
molecules. Water
two
pairs
water,
than
hydrogen
of
average
per
atoms
electrons.
of
So
two
molecule.
+
temporary
single
of
other
have
lone
hydrogen
+
properties
density
attraction
atom
+
between
Figure 2.5.4

and

van der Waals forces are only temporary because the electrons in the
molecules of atoms are in constant movement
Hydrogen
bonding
Key points
Hydrogen
bonding
is
a
special
form
of
permanent
dipole
bonding.
It
requires:


one
molecule
atom.
These
with
are
an
the
H
atom
most
covalently
electronegative
bonded
to
an
F
,
O
or
Electronegativity
an
N
atom
attract
atoms
in
a
is
the
covalent
electrons
in
ability
bond
the
of
to
bond
to
itself.

a
second
molecule
having
a
F
,
O
or
N
atom
with
a
lone
pair
of
electrons.

Electronegativity
together
a
H-band
3
shown
dashes
or
3
of
molecules
+
H
+
determine
+
polar

whether
used
a
to
molecule
is
+
or
non-polar.
H
H
+
+

H
van
der Waals forces
are
based
on
H
temporarily
molecules
structure
be
H


H
+
Figure 2.5.5
can
dots
+
lone
differences
the
b
by
+

with
induced
dipoles.
pair
Hydrogen bonding: a in pure water; b between water and ammonia

Hydrogen
between
pair
a
of
bonding
electrons
hydrogen
covalent
occurs
molecules
on
atom
bond
to
having
F, O
or
a
lone
N
and
attached
F, O
or
by
a
N.
23
2.6
Properties
of
giant
structures
Lattices
Learning outcomes
A
On
completion
should

able
describe
of

be
their
this
section,
you
giant
lattice
with
to:
The
structures
in
terms

ionic
covalent
how
the
properties
structures
of

related
to
their
and
a
regular
bonding
types
ionic,
of
three-dimensional
between
giant
e.g.
the
structures
sodium
arrangement
particles
are
called
of
particles.
giant
Lattices
str uctures .
are:
chloride,
magnesium
oxide
(see
Section
2.4
diagram)
giant
giant
covalent
silicon
(sometimes
called
giant
molecular),
e.g.
diamond,
dioxide
metals

are
three
for
structures,
is
strong
giant
lattices
understand
giant
of
metallic,
e.g.
copper
,
iron
(see
Section
2.4
for
diagram).
structure.
Properties of

Melting
the
(see

and
large
boiling
number
Section
Soluble
giant
in
of
ionic
structures
points
are
strong
bonds
high:
it
takes
between
a
lot
the
of
energy
oppositely
to
break
charged
ions
2.4).
water:
the
ions
can
form
ion–dipole
bonds
with
water
molecules.
+
+
H

H
O
+

H
O
+
+
+
+
H
H
H
H
+


O
O
H
H

O
H
+
+
Figure 2.6.1

Conduct
can
in
only
the
+
Water molecules forming ion–dipole bonds around + and – ions
electricity
move
only
when
when
molten
molten
or
or
dissolved
dissolved
in
water
.
in
water:
They
the
cannot
ions
move
solid.
solid
a
cathode
Figure 2.6.2
liquid
b
anode
cathode
anode
a The ions cannot move in the solid state. b The ions are free to move in the
liquid state.

Exam tips
Brittle:
when
ions
the
of
repulsive
A
common
suggest
error
that
in
exams
electrons
are
is
substances
sodium
chloride
such
as
conduct
force
forces.
is
applied,
charge
So
the
are
the
next
lattice
to
layers
each
breaks
of
ions
other
,
slide.
there
When
are
many
many
easily.
to
involved
a
when
a
same
b
molten
electricity.
force
First
is
ask
ionic
metal
yourself
if
(compound
with
remember
non
the
the
of
structure
metal).
‘ion’
in
applied
reactive
If
it
ionic.
is
ionic,
Figure 2.6.3
a A small force is applied. b The ions move out of position and ions with the
same charge are next to each other so the lattice breaks along this plane.
24
Chapter
Properties of

Melting
the
and
large
giant
boiling
number
of
covalent
2
Structure
and
bonding
structures
points
are
strong
covalent
high:
it
takes
bonds
a
in
lot
the
of
energy
network
silicon
to
of
break
atoms.
atom
oxygen
atom
weak forces
between
layers
Figure 2.6.4
The structures of: a diamond; b graphite; c silicon dioxide
Did you know?

Insoluble
with
in
water:
the
atoms
are
too
strongly
bonded
to
form
bonds
Each
carbon
bonded

Apart
from
electrons

some
the
layers
Apart
soft
the
ions
when
in
because
layers
Melting
that
a
and
do
are
free
they
there
are
slide
is
to
conduct
move.
These
hard:
Graphite
electrons
there
are
conducts
move
no
‘spare’
because
anywhere
it
along
weak
van
each
huge
is
three-dimensional
too
der
difcult
W
aals
other
to
forces
when
a
network
break.
between
force
is
of
Graphite
the
layers.
π
to
three
electron
carbon
the
the
directions
over
electricity:
applied.
are
different
can
not
electrons.
voltage
graphite
bonds
Properties of

they
delocalised
from
strong
So
graphite
or
has
is
atom
in
graphite
is
water
.
a
p
atom. These
delocalised
rings
in
bonding.
along
the
others. This
in
the
the
So
the
in
around
not
a
all
structure
graphite
but
a
each
electrons form
system
graphite
layers,
leaves
orbital
by
conducts
between
layers.
applied.
metals
boiling
points
are
strong
forces
high:
it
takes
a
lot
of
energy
to
break
Key points

the
large
number
the
delocalised
Insoluble
in
attraction
of
electrons
water
(see
(although
between
the
ions
of
attraction
Section
some
and
the
ions
and
2.4).
react
the
between

with
water):
delocalised
the
force
electrons
is
too
form
bonds
with
water
.
Some
metals
react
with
water
ionic
because
Giant
structures
and
Conduct
electricity
when
solid
or
molten:
the
delocalised
electrons
to
move
when
a
voltage
is
have
boiling
high
points
Malleable
(can
be
beaten
into
they
in
shape)
and
ductile
(can
be
drawn
Giant
when
a
force
is
applied,
the
layers
of
metal
ions
can
slide
other
.
The
attractive
ionic
forces
between
the
metal
ions
and
conduct
can
still
act
in
all
directions.
So
when
the
layers
slide,
or
can
easily
form.
This
leaves
the
metal
in
a
different
in
are
water. They
electricity
dissolved
in
when
water
new
because
bonds
structures
soluble
the
molten
electrons
strong
over
only
each
many
directions.
into
generally
wires):
have
many
applied.


ions.
are
bonds
free
a
of
electrons.
because

have
lattice
they
melting
lose
structures
repeating
strong

to
Giant
regularly
of
shape
the
ions
are free
to
but
move.
still
strong.

a
b
Metals
conduct
because
are free
some
to
electricity
of
move
the
electrons
throughout
the
force
structure.
applied

Giant
not
they
Figure 2.6.5
a A small force is applied to a metal. b The metal ions slide out of position,
covalent
conduct
do
not
electrons
structures
electricity
have free
or
do
because
moving
ions.
but there are still strong forces between the ions and the delocalised electrons.

Metallic
and
structures
ductile
atoms
can
are
because
slide
malleable
the
over
layers
each
of
other.
25
2.7
Properties
Learning outcomes
of
simple
Molecular
Substances
On
completion
of
this
section,
be
able

know
solids
how
the
molecular
related
to
that
with
This
a
simple
reects
a
molecular
regular
structure
packing
of
the
such
as
iodine,
molecules
in
a
I
,
can
lattice.
form
The
to:
understand
simple
lattices
2
forces

compounds
you
crystals.
should
molecular
their
compounds
of
are
structures
simple
often
properties
have
between
the
simple
molecular
brittle.
They
delocalised
do
molecules
structures
not
conduct
electrons
nor
are
the
which
weak
are
van
solids
electricity
der
at
because
W
aals
room
they
forces.
So,
temperature
have
are
neither
ions.
molecular
a
lattice
structure

understand
bonding
of

how
hydrogen
influences
the
properties
molecules
describe
the
molecular
and
solubility
of
compounds
non-polar
in
polar
solvents.
Figure 2.7.1
The lattice structure of iodine (only shown in two dimensions)
Melting
and
boiling
points of
simple
molecular
structures
Many
simple
melting
and
overcome
with
CH
CH
3
CH
3
the
weak
molecular
down
a
molecular
boiling
when
because
It
such
does
as
not
van
der
W
aals
forces
keeping
van
der
W
aals
forces
increase
are
it
intermolecular
lattices,
heated.
substances
points
forces
iodine
take
the
liquids
does
gases.
take
the
sulphur
,
are
much
energy
They
much
between
or
lattice
or
not
to
have
energy
molecules.
also
easily
overcome
low
to
Solids
broken
the
weak
together
.
with:
CH
3
3

increasing
number
of
electrons

increasing
number
of
contact
in
the
points
molecule
in
the
molecule
(contact
points
CH
3
3
3
3
are
So
b
CH
places
the
melting
CH
and
boiling
come
points
close
of
together).
down
each
group
as
the
the
noble
number
of
gases
and
electrons
the
(and
the
3
relative
H
increase
CH
2
mass)
increases.
CH
2
2
Pentane
CH
and
2,2-dimethylpropane
have
the
same
number
of
because
there
electrons,
CH
2
3
but
Figure 2.7.2
points
molecules
2
halogens
H
the
CH
2
CH
where
the
boiling
point
of
pentane
is
higher
.
This
is
are
more
2,2-dimethylpropane, a has
contact
points
for
van
der
W
aals
forces
to
act
over
.
The
total
van
der
a lower boiling point than pentane; b
because there are more contact points for
van der Waals forces to act over
W
aals
forces
Large
molecule
molecules,
temperature
number
W
aals
26
per
are
therefore
greater
.
So
the
boiling
point
is
higher
.
of
such
because
electrons).
forces
to
act
as
straight
they
have
Also
over
.
chained
very
there
are
high
polymers,
relative
many
are
solids
molecular
contact
points
at
room
masses
for
the
(large
van
der
Chapter
The
anomalous
W
ater
has
a
much
properties of
higher
boiling
point
water
Group
VI
hydrides.
The
boiling
than
expected
points
of
Structure
and
bonding
Exam tips
by
comparison
with
It
other
2
the
hydrides
from
S
H
is
not
always
safe
to
assume
to
2
that
T
e
H
increase
gradually
due
to
increased
van
der
W
aals
forces.
W
ater
compounds
with
hydrogen
has
2
bonding
a
much
higher
boiling
point
than
the
other
hydrides
because
it
always
extensively
compared
vaporise
hydrogen-bonded.
with
water
other
than
Since
hydrogen-bonding
intermolecular
the
other
forces,
Group
VI
it
takes
a
is
lot
or
dipole–dipole
bonding
is
have
a
higher
boiling
point
stronger
more
energy
to
than
those
with
van
der Waals
intermolecular forces.
hydrides.
the
boiling
point
of
For
SbH
example,
(not
3
H
400
O
H-bonded)
2
NH
is
higher
than
(H-bonded). This
is
that
of
because
we
3
)K(
H
300
Te
are
2
tniop
H
S
molecules
masses,
SbH
Se
with
being
very
more
3
2
2
than
200
10×
the
mass
of
NH
gniliob
3
100
Figure 2.7.3
0
0
2
3
4
5
Most
less
solids
dense
arranged
The
are
in
an
also
has
are
a
hydrides
order
to
is
Most
and
are
molecules
non-polar
formed
the
are
the
form
of
than
ethanol,
C
hydrogen
water
.
to
and
the
OH
and
to
if
the
in
ice
is
are
bonding.
water
.
bonding.
are
are
such
as
iodine,
and
however
,
so
simple
than
.
This
sulphur
is
in
Did you know?
are
between
soluble
because
in
they
water
.
You
can float
of
the
strong
bonding
in
the
water.
a
piece
paper
H


+
H
some
filter
needle
hydrogen
Place
paper.
on
water
the
a
Put
needle
the
surface
very
paper
still
sinks,
water. When
the
needle
the
should
+
+
H
C
remain
H
H
2
Figure 2.7.5
of
H
+
+
and
of filter
on

+
+
H
H
needle
+
H

a
because
on
+
The structure of ice
forces
3
with
Figure 2.7.4
water
those
molecules
NH
solute
dissolve
intermolecular
stronger
ammonia,
between
molecules
non-polar
will,
new
forces
solute
5
bonds
hydrogen
however
,
substances
They
Some
by
hydrogen
molecules
the
Ice,
molecules
liquid
attractive
Many
solute
themselves.
H
due
in
between
them.
hexane
solvent
2
can
in
as
as
liquids.
the
stabilised
molecular
attracted
molecules
e.g.
tension
strength
such
because
together
covalently-bonded
not
between
is
structure
greater
insoluble
solvents
solute
water
,
the
corresponding
This
close
surface
usually
themselves.
butane,
as
simple
dissolve,
solvent
their
water
.
lattice
not
high
Solubility of
and
than
(liquid)
‘
open’
molecules
W
ater
denser
than
The boiling
points of the Group VI
period
In
comparing
different
H
on
the
surface
of
the
water.
5
Hydrogen bonding between: a ammonia and water; b ethanol and water
Key points

Simple
molecular
solids
can form
lattices
solids
have
melting
with
weak forces
between
the
molecules.

Simple
are

between
Hydrogen-bonded
points

molecular
weak forces
than
total van
of
simple
a
and
low
have
many
der Waals forces
molecular
and
boiling
points
because
there
molecules.
molecules
expected
The
the
relatively
are
soluble
between
compound
at
higher
in
molecules
room
melting
and
boiling
water.
may
determine
the
state
temperature.
27
2.8
Shapes
of
Learning outcomes
molecules
VSEPR theory
The
On
completion
of
this
section,
be
able
predict
the
to
shapes
and
in
out
electron
the
pair
shapes
of
repulsion
theory
molecules.
It
uses
(VSEPR
the
theory )
following
can
be
rules:
Pairs
simple
of
electrons
in
the
outer
shells
of
the
atoms
in
a
molecule
repel
bond
each
angles
shell
work
to:


valence
you
used
should
(1)
molecules
other
and
move
as
far
apart
as
possible.
This
minimises
and
repulsive
forces
in
the
molecule.
ions.

Repulsion
than

the
Repulsion
than
the
of
various
a
below
and
lone-pairs
lone-pairs
lone-pairs
between
molecules
information
shapes
lone-pairs
between
between
repulsion
Shapes of
The
between
repulsion
and
and
bond-pairs
bond-pairs
with only
shows
how
(of
electrons)
bond-pairs
and
(of
theory
is
greater
electrons.
electrons)
bond-pairs
single
VSEPR
of
of
is
greater
electrons.
bonds
is
used
to
work
out
the
molecules.
b
H
Four
bond
pairs.
No
lone
pairs
around
C.
H
109.5°
Equal
repulsion.
So
bond
angles
all
109.5°.
C
H
Shape:
H
H
H
tetrahedral
+
ion
(NH
is
also
this
shape)
4
H
d
c
Three
bond
pairs.
No
lone
pairs
around
B.
F
120°
F
Equal
So
bond
angles
all
120°.
F
B
B
repulsion.
Shape:
trigonal
planar
F
f
e
T
wo
bond
pairs.
No
lone
So
bond
pairs
around
Be.
180°
Equal
repulsion.
angles
both
180°.
Be
Shape:
h
g
Six
F
F
bond
Equal
S
90°
S
F
linear
pairs.
No
repulsion.
Shape:
lone
So
pairs
bond
around
angles
all
S.
90°.
octahedral
F
F
F
Exam tips
j
i
Three
greater
Remember
that
‘tetra-’
bond
pairs.
One
lone
pair
around
N.
repulsion
epulsion
means four.
Greater
repulsion
between
lone
and
H
So
a
tetrahedron
has four faces.
N
bonding
pairs.
So
bond
angles
close
up
to
H
H
H
107°.
H
less
Shape:
pyramidal
107°
repulsion
+
l
k
Figure 2.8.1
A tetrahedron
Three
bond
Greater
H
means
eight.
So
pairs.
an
has
One
lone
between
pair
lone
around
and
O.
bonding
So
bond
angles
close
up
to
107°.
H
Shape:
octahedron
repulsion
H
H
‘Octa-’
pairs.
O
H
pyramidal.
(The
this
n
m
greatest
repulsion
O
also
has
3
107°
eight faces.
ion
CH
T
wo
shape.)
bond
Greatest
pairs.
T
wo
repulsion
lone
pairs
between
around
lone
pairs,
O.
less
O
H
H
repulsion
H
between
lone
pairs
and
bond
medium
and
least
repulsion
between
bonding
104.5°
repulsion
Figure 2.8.2
An octahedron
So
bond
angles
close
up
to
104.5°.
least
Shape:
repulsion
28
pairs
H
non-linear,
V-shaped
pairs.
Chapter
Shapes of
multiple
molecules
and
compound
ions
2
Structure
bonding
Did you know?
with
bonds
Experimental
The
and
information
below
shows
how
VSEPR
theory
is
used
to
work
out
the
the
data
shows
carbon–oxygen
bond
that
all
lengths
in
2–
shapes
of
another
carbon
dioxide
as
well
as
oxo
ions
(those
containing
oxygen
and
non-metal).
CO
ions
are
the
same.
Similarly
all
3
the
sulphur–oxygen
the
SO
bond
lengths
in
2–
a
ion
T
wo
180°
groups
of
two
bonding
C
No
O
lone
pairs.
Equal
to
bond
angles
both
hybridisation
same. This
is
2–
Three
orbitals.
bond
has
that
a
of
bond
a
length
double
and
single
linear
bond
d
the
180°.
between
Shape:
of
repulsion.
Each
So
2–
the
pairs.
due
c
are
4
b
groups
of
bond
pairs.
between
the
relevant
atoms.
No
O
lone
O
pairs.
Equal
repulsion.
So
120°
C
bond
C
O
angles
120°.
O
Shape:
trigonal
planar
120°
(The
nitrate
ion
is
also
trigonal
planar
.)
2–
e
2–
f
Four
O
groups
of
bond
pairs.
No
O
lone
O
S
O
pairs.
Equal
repulsion.
So
S
bond
O
angles
109.5°.
O
Shape:
tetrahedral
109.5°
O
Key points

The
shapes
theory
which
electrons

and
In VSEPR
repulsion
bond
depends
around
theory,
>
angles
a
on
in
the
particular
lone-pair
bond-pair
:
:
molecules
number
of
can
lone
be
predicted
pairs
and
using VSEPR
bonding
pairs
of
atom.
lone-pair
bond-pair
repulsion
>
lone-pair
:
bond-pair
repulsion.
29
2.9
Shapes
of
Learning outcomes
molecules
Hybridisation of orbitals
Methane
On
completion
of
this
(2)
section,
has
four
C–H
bonds
of
equal
length.
The
four
unlled
C
atomic
you
1
orbitals
should
be
able
can
be
thought
of
as
being
mixed
so
that
each
s
has
character
4
to:
3
p
and
character
.
This
process
of
mixing
atomic
orbitals
is
called
4

predict
angles
the
in
shapes
simple
and
bond
organic
allows
compounds

understand
molecules
3
hybridisation .
a
formed
the
shapes
using
hybridisation
of
the
with
of
idea
atomic
These
greater
and
equal
mixed
overlap
also
allows
repulsion
of
orbitals
atomic
each
called
orbitals
bond
between
are
to
be
when
the
orbitals.
sp
a
same.
They
use
ideas
describe
of
the
of
orbital
all
σ
is
bonds
of
orbitals
hybridisation
shape
are
them.
a

Hybridisation
molecular
b
to
benzene.
+
3
sp
orbital
Is
from C
orbital
from
C-H
H
molecular
orbital
3
Figure 2.9.1
a An sp
orbital from carbon combines with a 1s orbital from H to form a
molecular orbital making up a C–H bond; b Methane has 4 directional hybrid orbitals
Other
hydrocarbons
with
single
bonds
also
form
molecular
orbitals
from
3
hybridised
sp
orbitals.
-bonds
3
Figure 2.9.2
The structure of ethane. The molecular orbitals formed from sp
hybrids
allow each bond to be a σ-bond.
The
The
structure of
dot
and
cross
ethene
diagram
and
shape
of
a
ethene
are
shown
below.
b
H
H
117
.3°
C
C
H
Figure 2.9.3
In
ethene,
H
a Dot and cross diagram for ethene; b Shape of an ethene molecule
one
singly
occupied
2s
orbital
and
two
of
the
three
singly
2
occupied
2p
orbitals
in
each
carbon
atom
hybridise
to
make
three
3
orbitals.
form
σ
These
bonds
have
which
approximately
120°
similar
are
shapes
arranged
with
each
to
in
other
a
orbitals.
sp
plane
since
sp
2
These
making
there
is
a
sp
bond
equal
orbitals
angle
repulsion
of
of
the
electrons.
The
remaining
form
The
-bond
30
2p
orbitals
from
each
carbon
atom
overlap
sideways
to
bond.
H–C–H
expected,
Ethene has three sp
bond
because
angle
of
the
in
ethene
is
117.3°, rather
inuence
of
the
π-bond
the
of
the
carbon
than
whose
the
120°
electron
density
orbitals
in one plane and a π-bond above and
below this plane
π
-bonds
2
Figure 2.9.4
a
at
right
This
angles
bond
to
angle
that
of
allows
the
plane
maximum
overlap
and
of
the
hydrogen
orbitals.
atoms.
is
Chapter
2
Structure
and
bonding
Resonance
In
methane
particular
and
ethene
positions.
over
three
or
over
these
atoms.
Benzene,
more
H
C
6
shows
a
,
In
atoms,
These
has
six
is
structures.
electrons
are
localised,
substances,
allowing
electrons
carbon
the
some
are
atoms
of
said
i.e.
they
molecular
the
to
electrons
be
arranged
are
in
orbitals
free
extend
movement
delocalised
in
a
ring.
Figure
2.9.5(a)
6
representation
structure
the
some
a
single
The
of
benzene.
(composite)
bonds
between
The
form
the
↔
means
which
carbon
lies
that
the
between
atoms
are
actual
these
neither
two
double
nor
Exam tips
single
is
bonds.
called
a
They
are
resonance
somewhere
in-between.
The
composite
structure
hybrid
It
is
a
common
resonance
a
b
two
or
They
are
single
Figure 2.9.5
a Two possible ways of representing benzene; b A modern representation of
many
lines
the
of
think
that
mixtures
a
structures. We
by
Figure
the
of
structure.
‘in-between’
cases
(see
to
are
more forms
represent
in
error
hybrids
use
can
structure
of
dashed
2.9.5(b))
benzene
2
In
benzene,
hybrid
the
orbitals
six
carbon
(one
to
atoms
each
form
hydrogen
a
hexagon
atom
and
with
two
to
three
localised
other
carbon
sp
2
atoms).
120°.
This
These
The
The
3
leaves
orbitals
six
orbitals
sp
a
single
overlap
electrons
are
p
arranged
orbital
sideways
involved
H
can
to
on
in
each
form
move
a
a
plane.
of
the
So
six
the
carbon
delocalised
freely
around
bond
system
the
angles
are
atoms.
of
π
bonds.
ring.
H
2
Figure 2.9.6
The structure of benzene: a The p orbitals before overlap (the sp
hybrids are
shown by the straight lines); b The delocalised system of π bonds
2
In
graphite,
the
other
form
move
an
each
carbon
carbon
atoms.
extensive
along
conducts
the
atom
The
system
layers
of
when
forms
three
remaining
delocalised
a
voltage
is
p
localised
orbitals
electrons.
applied.
sp
hybrid
overlap
These
This
is
orbitals
sideways
electrons
why
on
to
to
can
graphite
electricity.
Key points

Hybridisation
orbital

The
shapes
theory

with
and
Benzene
is
of
s
and
mixed
of
simple
the
a
p
atomic
orbitals
results
in
the formation
can
predicted
of
an
character.
idea
organic
of
molecules
be
using VSEPR
hybridisation.
resonance
hybrid
with
a
planar
ring.
31
Revision
1
a
Using
the
concept
of
atomic
questions
orbitals,
explain
8
a
how:
b
the
three
types
of
intermolecular forces
of
attraction.
i
sigma
ii
pi
and
bonds
Which
your
State
of
b
two
the
attraction
are formed.
these
Identify
bonds
is
stronger?
i
Explain
answer.
type
in
of
intermolecular force
the following
of
molecules:
CO
ii
NH
iii
HF
iv
ICl
3
2
Draw
dot
and
cross
diagrams for
the following
molecules:
a
S
H
9
2
b
PCl
Magnesium
all
oxide,
solids. Using
diamond
structure
and
and
aluminium
bonding
are
explain
the
3
c
differences,
CO
if
any,
among
these
three
solids
in
terms
2
of
3
a
How
is
a
co-ordinate
the following
properties:
bond formed?
a
melting
point
b
electrical
conductivity
c
solubility
in
+
b
O
H
has
a
co-ordinate
bond.
Draw
a
dot
and
3
cross
4
Use
dot
diagram
and
cross
to
show
the
diagrams
bonding
to
in
illustrate
this
the
ion.
10
in
a
5
the following
potassium
ionic
Using
aluminium
a
An
element
has
has filled
Would
in
the
relatively
bonding
a
is
b
does
one
electron
in
its
outer
explain
Place
inner
shells.
this
c
iodine vaporises very
not
conduct
electricity
Describe
the
easily.
type
Explain
why
water
has
a
much
or
point
low?
the following
melting
of
this
Explain
element
your
be
12
answer.
the
other Group VI
Which
of
higher
melting
the following
Period
3
metals
in
order
sodium
point,
molecules
or
BCl
PCl
b
What
Using VSEPR
a
‘electronegativity’.
is
the
trend
in
O
H
3
electronegativity
as
you
go
down Group VII?
b
H
S
2
c
BCl
d
CCl
e
PH
3
c
Explain
your
answer
to
part
(b).
4
7
Identify
the following
polar:
a
BF
b
CH
c
CO
d
NH
3
Cl
3
2
3
32
would
?
have
Explain
a
your
3
theory,
state
and
explain
magnesium
+
Define
point
answer.
of
the following
a
boiling
hydrides.
point:
13
aluminium
6
higher
of
element.
melting
high
increasing
iodine:
brittle
3
c
why
shell
than
bonding
and
oxide
11
b
structure
compounds:
sulphide
b
and
water.
bonding
molecules
as
polar
or
non-
3
molecules
and
ions:
the
shapes
of
Chapter
2
Structure
and
bonding
–
revision
questions
33
3
The
3.
1
Equations
mole
Learning outcomes
On
completon
should
be
able
of
ths
concept
and
Writing
secton,
The
four
The
atom
dene
‘molar
the
chemical
steps
below
equations
show
how
counting
is
shown
by:
terms
=
,
a
H
simple
=
,
O
chemical
=
equation.
x.
1:
W
rite
down
the
formulae
of
all
reactants
and
products:
‘mole’ and
mass’
construct
balance
C
to:
CH
(g)
+
O
4

to
you
Step

moles
molecular
and
(g)
→
CO
2
(g)
+
H
2
O(g)
2
onc
Step
2:
Count
the
number
of
atoms
in
each
reactant
and
product:
equatons.
CH
+ O
4

Step
3:
Balance
one
of

the
→
CO
2
atoms
+
H
2
xx

e.g.
O
2
xx
x

hydrogen:
Exam tips
CH
+ O
4
When
balancng


Neer
alter
the formula
of
a
Step
4:
Keep
compound.

Balance
by
→

The rst
be
the
balancing,
one
type
puttng
numbers
at
+
CH
a formula.
atom
balance
n
Oxygen
s

balanced
whch
s
should
easest
combuston
Y
ou
of
to
can
the
often
instead
equation
easest
ratio
of
atom
xx
at
a
time
until
all
atoms
2O
just
if

state
you
→
CO
2
the
+
2H
2
xxxx

number
of
O
2
xx
xx

each
type
of
atom
on
either
side
prefer
.
of
reactants
of
the
and
products
reaction.
In
in
the
a
balanced
equation
equation
for
the
is
called
combustion
the
of
last.
methane
Numbers
way
O
2

to
stoichiometry

2H
xx
reactons.
The
balance

balance:
of
one
+
2
xx

4
the front
CO
2
equatons:
n front
through
formula.
the
multply
atoms
n
all
the
dioxide):
above,
2
the
stoichiometry
is
1
(methane):
2
(oxygen):
1
(carbon
(water).
the
Ionic
Ionic
equations
compounds
carbonate
as
water
,
ions
the
well
include
as
salts
acids
separate.
and
For
such
as
alkalis.
ammonium
When
ionic
sulphate
4
SO
2
(aq)
→
2NH
4
SO
2
in
2–
(aq)
+
SO
4
(aq)
→
2H
(aq)
4
+
H
sodium
dissolve
example:
+
)
(NH
and
compounds
2–
(aq)
+
SO
4
(aq)
4
+
NaOH(aq)
When
ionic
reaction.
ions.
Step
T
o
1:
compounds
The
ions
write
W
rite
an
that
ionic
the
full
react,
play
Na
only
no
(aq)
some
part
in
+
of
the
the
2:
W
rite
the
+
CuSO
charges
on
chemical
(aq)
those
2+
Mg(s)
+
Cu
(aq)
ions
reaction
→
equation,
MgSO
4
Step
OH
3:
Cancel
the
+
SO
spectator
4:
The
(aq)
→
Mg
(aq)
are
ionic:
2–
SO
(aq)
+
Cu(s)
(aq)
+
Cu(s)
2+
(aq)
equation
Mg(s)
Cu(s)
4
→
Mg
is
4
that
which
remains:
2+
34
the
ions:
Cu
ionic
+
which
4
Step
in
spectator
e.g.
2+
2–
(aq)
(aq)
substances
2+
Mg(s)
part
called
4
4
Step
take
are
equation:
balanced
Mg(s)
→
+
Cu
(aq)
2+
→
Mg
(aq)
+
Cu(s)
Chapter
3
Exam tips
For
precptaton
the
ons
s
makng
added
to
reactons
up
the
aqueous
t
s
often
precptate.
sodum
easer
For
chlorde,
to
wrte
example,
lead
an
onc
when
chlorde
s
equaton from
aqueous
lead
ntrate
precptated. The
ons
2+
whch
take
part
n
the
reacton
are
Pb
and Cl
,
so
the
equaton
s:
2+
Pb
(aq)
+
2Cl
(aq)
→
PbCl
(s)
2
State
State
the
symbols
symbols
reactants
reactant
(s)
(l)
mole
relative
grams
is
often
added
products.
to
equations
These
are
to
placed
show
after
the
the
physical
state
of
formula
of
each
solution
in
water)
product:
solid
The
The
and
are
and
liquid
and
molar
atomic
called
a
(g)
mass
mole
of
gas
(aq)
aqueous
(a
mass
or
relative
that
molecular
substance
mass
of
(abbreviation
a
substance
mol).
We
use
in
the
Exam tips
12
C
scale
One
as
mole
specific
a
is
standard
the
in
amount
particles
(atoms,
comparing
of
masses
substance
ions
or
which
molecules)
accurately,
has
as
the
there
so:
same
are
number
atoms
in
of
exactly
Accurate
relate
are
n
gen
the
atomc
masses
Perodc Table
to
12
12 g
of
C
the
isotope.
four
sgncant gures,
e.g.
A [Na]
r
=
The
molar
mass,
M,
is
the
mass
of
one
mole
of
specied
substance
a
grams.
So
the
term
‘molar
mass’
can
apply
to
ionic
compounds
as
22.99. You
wll
well
For
molecules
we
use
the
term
‘relative
molecular
mass’.
rounded-up
of
M
are
molar
g mol
mass
of
a
compound
such
as
sodium
sulphate,
Na
SO
2
ths
e.g.
alue
by
adding
the
relative
atomic
masses
together
,
taking
the
number
of
each
type
of
values:
Na
=
=
23.
In
dong
dong
mole
smple
at
ths
leel
n
chemstry
take
these
‘rounded-up’
alues
as
atom:
23,
S
=
32;
O
=16
accurate
problems
r
Secton
2Na
A [Na]
when
is
beng
A
n
into
we
account
use
4
calculatons
calculated
gen
r
Use
calculatons.
The
alue for
The
–1
units
be
as
calculatons,
molecules.
often
in
1S
wth
enough
to
gnore
sgncant gures
any
(see
3.5).
4O
–1
M
=
(2
×
23)
+
32
+
(4
×
16)
=
142 g mol
Key points

A
mole
s
the
amount
of
substance
whch
has
the
same
number
of
12
speced
partcles

Molar

Chemcal
equatons
reactants
and

In
mass
onc
s
the
as
there
mass
of
show
are
1
atoms
mole
equal
of
a
n
exactly
12 g
substance
numbers
of
n
each
of
the
C
sotope.
grams.
type
of
atom
n
the
products.
equatons
the
spectator
ons
are
not
shown.
35
3.2
Calculations
Learning outcomes
Simple
We
On
completon
should
be
able
of
ths
using
secton,
can
the
mole
nd
the
mole
concept
calculations
number
of
moles
by
using
the
relationship:
you
to:
mass
of
substance
(g)
____________________
number
of
moles
(mol)
=
–1

perform
the
calculatons
mole
based
molar
on
mass
(g mol
)
concept.
Worked
How
example
many
moles
1
of
magnesium
chloride,
MgCl
are
present
in
19.1 g
of
2
magnesium
chloride?
values:
(A
Mg
=
24.3,
Cl
=
35.5)
r
Exam tips
Step
1:
Calculate
the
molar
mass
of
MgCl
=
24.3
=
95.3 g mol
+
(2
×
35.5)
2
When
wrtng
mportant
to
about
make
moles,
clear
t
–1
s
what
type
mass
(g)
19.1
___________
Step
of
partcles
we
are
referrng
to.
2:
Use
the
relationship:
moles
=
molar
we
just
say
moles
of
chlorne,
t
clear
whether
ts Cl
atoms
=
mass
0.200 mol
95.3
s
Worked
not
_____
=
If
example
2
or Cl
2
molecules.
For
example
n
35.5 g
of
What
mass
of
calcium
hydroxide,
Ca(OH)
,
is
present
in
0.025 mol
of
2
Cl
there
are
0.5 mol
of Cl
2
but
molecules
2
1.0 mol
of Cl
atoms.
Ca(OH)
?
2
values:
(A
Ca
=
40.0,
O
=
16.0,
H
=
1.0)
r
Step
1:
Calculate
the
molar
mass
of
Ca(OH)
=
40.0
=
74.0 g mol
+
2
×
(16.0
+
1.0)
2
–1
Step
2:
Rearrange
mass
Step
3:
nd
the
mass
the
mass

the
molar

the
balanced
Worked
Fe
=
Step
of
the
O
2
O
the
a
of
×
in
terms
molar
values:
products
specic
mass
of
of
mass:
mass
mass
(g)
,
=
0.025
×
74.0
=
1.85 g
in
a
reaction
we
use:
reactant
this
reactant
equation.
3
maximum
is
formed
(stoichiometric)
example
Calculate
equation
moles
masses

oxide,
the
=
Substitute
Reacting
T
o
(g)
reduced
mass
by
of
iron
excess
formed
carbon
when
monoxide
399 g
(A
3
of
1:
Calculate
the
relevant
O
2
2:
W
rite
the
=
formula
3:
Multiply
(2
×
55.8)
stoichiometric
O
2
masses
(Fe
and
55.8,
O
Fe
each
Fe
(s)
+
3CO(g)
simple
×
16.0)
=
159.6
→
2Fe(s)
+
3CO
(g)
2
formula
mass
in
grams
by
the
relevant
number:
O
2
Use
(3
):
3
equation:
+
3CO
→
2Fe
→
2
+
3CO
3
2
159.6 g
4:
+
3
stoichiometric
proportion
to
×
55.8 g
calculate
the
(111.6 g)
amount
111.6
______
399 g
→
×
159.6
36
=
3
Fe
Step
iii)
r
Fe
Step
Fe
16.0).
2
Step
iron(
values:
399
=
279 g
Fe
of
iron
produced:
Chapter
Worked
We
can
used
be
to
do
similar
obtain
used,
but
Calculate
water
example
a
the
when
a
3
The
mole
concept
4
calculations
given
slightly
nd
of
different
minimum
methane
to
amount
mass
the
method
of
undergoes
maximum
product.
The
is
shown
methane
of
reactants
shown
above
can
here.
needed
complete
mass
method
to
produce
combustion
9.0 g
of
values:
( A
r
C
=
12.0,
Step
1:
H
=
1.0,
Calculate
O
the
=
16.0).
relevant
formula
masses
and
(CH
H
4
CH
=
12.0
+
(4
×
1.0)
=
16.0;
H
4
Step
2:
O
=
Calculate
moles
of
mass
(g)
Use
the
(g)
+ 2O
they
(g)
(g) +
2H
2
moles
18.0
the
mass
of
can
=
moles
since
also
example
Calculate
the
the
O(g)
be
×
×
molar
16
=
mass
4.0 g
calculations
applied
from
to
the
electrons
masses
mass
have
of
of
the
hardly
ions.
The
atoms
any
mass
from
of
ions
which
mass.
5
maximum
aqueous
0.25
mole
different
methane:
methane:
(g)
and
of
2
0.50 mol
method
excess
CO
the
→
signicantly
derived
calculate
0.25 mol
equations
are
→
to
2 mol
Worked
with
=
0.50 mol
→
Calculate
mole
not
16.0)
18
2
=
are
×
1 mol
mass
The
=
mass
equation
4
Ionic
(1
9.0
=
stoichiometric
CH
4:
+
____
=
molar
Step
1.0)
water:
___________
3:
×
2
moles
Step
(2
O):
2
mass
silver
of
ions
silver
formed
values:
(A
Zn
when
=
3.27 g
65.4,
Ag
of
=
zinc
reacts
108).
r
+
Step
1:
Calculate
the
relevant
formula
masses
(Zn
and
Ag
):
+
Zn
Step
2:
W
rite
the
=
65.4;
stoichiometric
Ag
Step
3:
Multiply
each
+
2Ag
atomic
stoichiometric
2+
(aq)
mass
→
Zn
65.4
Step
4:
Use
+ 2Ag
in
grams
+
by
2Ag(s)
the
relevant
2+
(aq)
g
simple
(aq)
number:
+
Zn(s)
108
equation:
+
Zn(s)
=
→
Zn
(aq) +
→
proportion
to
2Ag(s)
2
calculate
the
×
108 g
amount
(216 g)
of
silver
produced:
3.27
_____
3.27 g
×
→
216
=
10.8 g
Ag
65.4
Key points

The
mole
requred

The
concept
to form
mole
a
concept
formed from
a
can
be
used
specc
can
gen
be
used
mass
to
mass
of
calculate
of
to
the
masses
of
reactants
product(s).
calculate
the
mass
of
specc
products
reactants.
37
3.3
Empirical
Learning outcomes
and
Empirical formula
An
On
completon
of
ths
secton,
empirical
be
able
for mula
and
of
a
molecular formula
compound
shows
the
simplest
whole
number
you
ratio
should
molecular formulae
of
the
elements
present
in
a
molecule
or
formula
unit
of
the
to:
compound.

deduce
emprcal formulae
usng
A
absolute
or
molecular
for mula
in
compound

The
formula
deduce

Many
of
shows
the
total
number
of
each
type
of
atom
relate
present
masses

masses
elements
n
a
molecule.
a
molecular formulae.
simple
molecular

Many
of
an
ionic
inorganic
compound
molecules
is
always
have
the
its
empirical
same
formula.
empirical
and
formulae.
organic
molecules
have
different
empirical
and
molecular
formulae.
Some
empirical
and
molecular
formulae
are
shown
below.
Compound
Emprcal formula
Molecular formula
hydrogen
HO
H
peroxde
O
2
butane
C
H
2
benzene
C
5
2
H
4
CH
C
10
H
6
sulphur
doxde
SO
SO
2
cyclohexane
2
CH
C
2
Deducing
an
example
Calculate
the
12
1
empirical
5.95 g
H
6
empirical formula
Worked
contains
6
tin
and
formula
7.1 g
of
a
compound
chlorine
of
values:
(A
tin
Sn
=
and
chlorine
119,
Cl
=
which
35.5).
r
Step
1:
Calculate
the
number
of
moles
of
each
Sn
Cl
5.95
7.1
_____
_____
=
0.05 mol
=
119
Step
2:
Divide
each
by
the
lower
number
0.05
W
rite
the
formula
of
moles
0.2
____
=
0.1
=
0.5
3:
0.2 mol
35.5
_____
Step
element.
0.4
0.5
showing
the
simplest
ratio:
SnCl
4
Worked
In
this
example
example,
we
2
are
given
information
about
%
by
mass
rather
than
mass.
Calculate
iodine
by
the
which
mass
values:
(A
r
38
empirical
contains
C
formula
8.45%
=
of
a
compound
carbon,
12.0,
H
=
2.11%
1.0,
I
=
of
carbon,
hydrogen
127.0).
and
hydrogen
89.44%
and
iodine
Chapter
Step
1:
Divide
%
by
3
A
r
C
H
8.45
=
2:
Divide
=
by
lowest
=
2.11
0.704
______
______
1
=
0.704
the
0.704
127.0
number
=
W
rite
2.11
1.0
0.704
3:
______
0.704
______
Step
89.4
_____
12.0
Step
I
2.11
_____
3
=
0.704
formula
showing
1
0.704
the
simplest
ratio:
CH
I
3
Exam tips
1
After
Step
e.g.
P
(n
1
to
ths
2
you
2.5 O.
case
by
may
In
2).
be
ths
So
left
wth gures
case
you
must
the formula
s
P
The
emprcal formulae
farly
smple
1.46 O.
It
s
of
ratos. After
therefore
emprcal formula
of
The
molecular
example,
the
Step
2
adsable
P
you
type
of
may
round
be
left
up
the
formula
as
formula

the
empirical

the
molar

the
empirical
Worked
A
is
a
simple
formula
the
we
number
wth
1.46
a
to
norganc
rato
such
1.5. Ths
ones,
as
wll
1
are
P
to
ge
an
multiple
of
butane
of
the
H
C
empirical
formula,
need
to
empirical
has
twice
formula.
the
For
number
of
10
H
C
.
T
o
deduce
the
5
know:
formula
mass
of
the
formula
example
compound
numbers,
whole
3
2
molecular
a
molecular formula
molecular
atom
whole
get
especally
4
each
to
5
compounds,
to
not
up
O
2
Deducing the
many
are
O
2
2
that
multply
has
an
compound
mass.
3
empirical
formula
CH
and
a
molar
mass
of
84.
2
Deduce
its
empirical
formula
values:
(A
C
=
12.0,
H
=
1.0).
r
Step
1:
Find
the
Step
2:
Divide
empirical
the
molar
formula
mass
of
mass:
the
12.0
+
compound
(2
by
×
1.0)
the
=
14.0
empirical
formula
Key points
mass:

An
emprcal formula
shows
the
84
___
=
6
smplest
whole
number
rato
of
14
atoms
Step
3:
Multiply
each
atom
in
the
empirical
formula
by
the
in
Step
a
compound.
number

deduced
n
Emprcal formulae
are
deduced
2:
usng
×
CH
2
6
=
C
masses
or
relate
masses
elements
present
n
H
6
12
of
the
a
compound.

Molecular formulae
total
number
a formula

A
of
unt
show
atoms
of
a
molecular formula
can
the
formula
relate
of
f
the
the
n
compound.
deduced from
mass
the
present
be
emprcal
molecular
compound
s
known.
39
3.4
Using Avogadro’s
Learning outcomes
Avogadro’s
Avogadro’s
On
completon
of
ths
secton,
be
able
law
states
and
that
pressure
equal
have
volumes
equal
of
all
numbers
gases
of
at
the
same
molecules
to:
At

law
you
temperature
should
law
apply Aogadro’s
law
to
room
temperature
and
pressure
(r
.t.p.)
1
mole
of
any
gas
occupies
deduce
3
.
24.0 dm
the
stochometry
of
a
At
standard
temperature
and
pressure
(s.t.p.)
it
occupies
reacton
3
22.4 dm
nolng

deduce
gases
molecular formulae from
Applying
Avogadro’s
law
to
the
synthesis
of
water
from
hydrogen
and
oxygen:
combuston
data.
2H
(g)
+
O
2
2
Did you know?
2
(g)
1
volumes
number
of
atoms
n
a
1
atoms
s
ery
large:
6.02
×
We
10
–1
atoms
called
mol
. Although
the Aogadro
number
was rst
Loschmdt. That
for
s
the Aogadro
ths
alue
constant,
deduced
why
volume
2
mol
volumes
3
3
24 dm
48 dm
(at
r
.t.p.)
can
use
Avogadro’s
numbers
of
molecules
numbers
of
moles.
law
in
in
the
mole
same
calculations
volume
of
because
gas,
there
if
there
are
also
are
equal
equal
s
the
by Johann
the
constant
O(g)
2
2
mole
23
of
2H
mol
3
48 dm
The
→
2
mol
Worked
example
Calculate
the
1
symbol
s
–1
L
mass
of
ethane
(M
=
30 g mol
3
)
in
240 cm
of
ethane
gas
at
r
.t.p.
3
Step
1:
Change
cm
Step
2:
Calculate
3
to
dm
3
:
240 cm
3
÷
1000
=
0.240 dm
3
volume
(dm
)
_____________
the
number
of
moles
using:
moles
=
24
0.240
______
=
=
0.01 mol
24
Step
3:
Calculate
Worked
Calculate
ethane
is
mass:
example
the
mass
(g)
=
moles
×
M
=
0.01
×
30
=
formed
at
r
.t.p.
when
0.3 g
2
volume
completely
of
carbon
burnt
in
dioxide
excess
oxygen
values:
(A
C
=
7.50 g
of
12.0,
r
H
=
Step
1.0).
1:
Calculate
the
number
of
moles
of
ethane:
Exam tips
7.50
_____________________
=
(2
You
may nd
example
of CO
2
by
t
easer
to
12.0)
+
(6
×
0.25 mol
1.0)
do Worked
calculatng
formed from
×
the
7
.50 g
of
mass
Step
2:
W
rite
the
relevant
ethane,
stoichiometric
mole
equation
for
the
reaction
and
identify
the
ratios:
2
.e.
2C
H
2
30 g
2
×
ethane
44 g CO
produces
=
(g)
+
7O
6
(g)
→
4CO
2
(g) +
6H
2
2 mol
O(l)
2
4 mol
88 g CO
2
2
4
__
Step
So
0.25 g
ethane
produces
88
×
3:
Calculate
the
number
of
moles:
0.25 mol
ethane
×
→
0.25 mol
2
0.25
CO
2
=
22 g CO
2
CO
formed
from
0.25 mol
ethane
=
0.5 mol
2
22 g CO
=
2
22/44
=
CO
2
0.5 mol CO
2
3
Step
4:
Calculate
the
volume
of
CO
at
2
40
r
.t.p:
0.5
×
24
=
12 dm
Chapter
Deducing the
The
worked
deduce
the
Worked
stoichiometry of
example
below
stoichiometry
example
shows
of
a
how
a
we
mixture
At
to
the
end
deduce
Step
of
1:
the
the
of
hydrogen
reaction
the
reaction
can
use
Avogadro’s
there
of
unbalanced
and
is
20 cm
only
the
of
water
oxygen
present.
is
law
reacted
Use
together
.
Apply
Avogadro’s
equation:
(g)
H
law:
+
O
(g)
→
2
3:
W
rite
the
3
volumes
2H
1
1
(g)
volume
mole(cule)
+
O
2
Deducing
The
worked
deduce
the
Worked
a
O(l)
2
20 cm
mole(cule)s
equation:
H
2
40 cm
2
Step
law
reaction.
3
2:
to
Avogadro’s
2
Step
concept
3
stoichiometry
W
rite
mole
3
40 cm
of
The
reaction.
3
A
3
(g)
→
2H
2
O(l)
2
molecular formula
example
below
molecular
example
shows
formula
of
how
a
we
can
compound
use
Avogadro’s
using
law
combustion
to
data.
4
3
Propane
contains
carbon
and
hydrogen
only.
3
reacts
with
Deduce
for
the
Step
1:
exactly
the
oxygen,
125 cm
molecular
formula
W
rite
the
information
H
x
(g)
+
of
propane
Find
the
1
(g)
carbon
and
write
Deduce
→
H
xCO
4:
propane
is
formed.
balanced
equation
ratio
5
(g)
(g)
+
of
5O
of
C
O(l)
2
3
gases
and
3
(g)
→
use
Avogadro’s
3CO
(g)
+
yH
2
atoms:
law:
volumes
2
number
yH
2
volumes
y
+
equation:
75 cm
1 mol
C
O(l)
2
H
x
be
Step
of
dioxide
a
unbalanced
3
volume
x
the
125 cm
simplest
C
3:
of
2
3
Step
75 cm
below
O
y
25 cm
2:
25 cm
reaction.
C
Step
When
3
→
3 mol
CO
y
(so
x
must
2
3)
Deduce
the
number
of
H
atoms:
C
H
3
(g)
+
5O
y
(g)
→
3CO
2
(g)
+
2
O(l)
yH
2

6
of
the

so
4
oxygen

so
4
moles
which
Step
5:
W
rite
the
10
oxygen
atoms
of
come
atoms
must
water
from
balanced
are
the
react
react
with
with
formed
carbon
hydrogen
containing
8
to
form
water
hydrogen
atoms
propane.
equation:
H
C
3
(g) + 5O
8
(g)
→
3CO
2
(g) + 4H
2
O(l)
2
Key points

Aogadro’s
law
temperature
states
and
that
pressure

The
stochometry of

The
molecular formula
from
combuston
a
data
equal
reacton
of
by
olumes
contan
a
the
can
smple
of
same
all
gases
number
be deduced
molecular
by
at
of
the
applyng Aogadro’s
compound
applyng Aogadro’s
same
molecules.
can
be
law.
deduced
law.
41
3.5
Solution
Learning outcomes
concentration
The
This
On
completon
of
ths
secton,
concentration of
refers
be
able
to
the
amount
a
of
titrations
solution
solute
dissolved
in
a
solvent
to
make
a
you
solution.
should
and
In
chemistry
the
units
of
solution
concentration
are
usually
to:
–3
expressed

understand
concept
how
can
be
the
mole
used
in
moles
per
cubic
decimetre
).
(mol dm
the
is
the
molar
to
number
determne
This
concentration
concentraton
concentration
of
solution
)
(mol dm
of
moles
of
solute
(mol)
_____________________________
–3
of
=
3
volume
of
solution
(dm
)
solutons
–3
In

know
how
ttratons
to
carry
and
how
out
to
calculations
involving
process
the

to
change
mass
results

gures
in
grams
3
understand
n
the
use
chemcal
of
solution
concentration
in
mol dm
remember:
acd–base

to
change

that
moles

that
volume
to
moles
3
cm
to
dm
3
by
dividing
volume
in
cm
by
1000
sgncant
solute
=
concentration
×
volume
calculatons.
moles
of
solute
______________
of
solution
=
concentration
Simple
calculations
Worked
example
Calculate
the
involving
concentration
1
concentration
of
a
solution
of
potassium
hydroxide
( M
=
3
56.0)
containing
3.50 g
KOH
in
of
125 cm
solution.
3.50
_____
Step
1:
Convert
grams
to
moles:
=
0.0625 mol
KOH
56.0
3
Step
2:
Change
cm
Step
3:
Calculate
3
to
dm
:
3
125 cm
3
=
0.125 dm
0.0625
_______
concentration:
–3
=
KOH
0.500 mol dm
0.125
Worked
example
Calculate
the
2
3
mass
of
magnesium
chloride,
MgCl
(M
=
95.3)
in
50 cm
2
–3
of
a
0.20 mol dm
solution
3
Step
1:
Change
cm
Step
2:
Calculate
of
magnesium
3
to
dm
:
chloride.
3
50 cm
3
=
0.050 dm
Exam tips
–3
When
dealng
wth
number
sgncant gures
of
moles,
which
equals:
concentration
(mol dm
3
×
volume
(dm
)
=
0.20
×
0.050
=
0.010 mol
MgCl
2
(s.f.):
Step

zeros
e.g.
before
0.036
numbers
has
2
are
not
s.f.,
Zeros
e.g.

after
When
dong
seeral
The nal
data
42
s.f.,
s.f.
calculaton
do
wth
round
up
round
up for
answer.
answer
the
to
the
lowest
proded.
should
same
be
number
number
n
answer
signicant
not
expressed
as
3
are
steps. Only
your nal
s.f.
a
steps,
between

has
moles
to
grams:
=
0.010
×
95.3
=
0.95 g
(to
two
gures)
s.f.
a gure
0.0400
Convert
signicant
The

3:
the
of
is
to
two
gures
in
signicant
the
data
is
gures
two.
because
the
least
number
of
)
Chapter
Worked
example
3
The
mole
concept
3
–3
What
of
volume
sodium
of
a
solution
hydroxide,
of
NaOH
concentration
(M
=
0.10 mol dm
contains
2.0 g
40)?
2.0
____
Step
1:
Convert
grams
to
–2
moles:
=
5
×
10
mol
NaOH
40
moles
2:
Calculate
volume
in
dm
(mol)
_______________________
3
Step
:
=
–3
concentration
(mol dm
)
–2
5
×
10
________
=
0.10
3
=
0.50 dm
Acid–base titrations
A
titration
solution
is
of
used
concentration

to
determine
unknown
Fill
a
the
acid).
of
a
burette
the
amount
concentration.
solution
with
acid
of
of
alkali
known
The
of
substance
procedure
for
present
in
a
determining
the
is:
concentration
(after
washing
it
with
burette

Record

Put
a
the
initial
known
burette
volume
of
reading.
alkali
into
the
ask
using
a
volumetric
acid
pipette.

Add
an
acid–base

Add
the
indicator
to
the
alkali
in
the
ask.
volumetric
acid
slowly
from
the
burette
until
the
indicator
changes
pipette
colour

Record
the

(end
the
point).
nal
burette
reading
(nal
–
initial
burette
reading
is
called
titre).
Repeat
the
process
until
two
or
three
successive
titres
differ
by
no
3
more
than
0.10 cm
alkali
white

T
ake
the
average
of
these
titres.
We
usually
ignore
the
rst
titre,
and
tile
since
indicator
this
is
the
rangender
titration
(rough
titration).
For
example,
in
the
Figure 3.5.1
table
below,
the
fourth
and
fth
titres
would
be
selected
and
The apparatus used in an
averaged.
acid–alkali titration
Frst ttre
Second ttre
(rangender)
(cm
Thrd ttre
3
Fourth ttre
3
)
(cm
5th ttre
3
)
(cm
3
)
(cm
)
3
(cm
)
32.92
32.85
32.00
32.25
32.35
Key points
–3

Molar
concentraton
(mol dm
)
=
number
of
moles
of
solute
(mol)
÷
3
olume

Acd–base
colour

of
When
soluton
(dm
ttratons
rapdly
at
are
the
processng
).
carred
end
out
usng
an
ndcator
whch
changes
pont.
ttraton
results,
the
alues
selected
should
be from
two
3
or

three
The
successe
answer
gures
as
to
the
a
ttres
whose
calculaton
lowest
s
number
alues
are
expressed
of
no
to
more
the
than
same
sgncant gures
n
0.
10 cm
number
the
data
of
apart.
sgncant
proded.
43
3.6
More
about
Learning outcomes
Calculating
In
On
completon
of
ths
secton,
be
able
know
how
to
calculate
concentration
solution
concentration
by titration
from
titration
results
we
need
know:
to:


order
solution
you
to
should
titrations
to
calculate
the
volume
and
concentration
of
the
solution
in
the
burette,
e.g.
the
soluton
acid
concentraton from
ttraton

know
a
the
results
of
experments
how
ttraton
to
to
use
the
results from
calculate

the
volume

the
balanced
of
a
the
solution
equation
for
in
the
the
ask
reaction.
the
Worked
stochometry
of
example
1
reacton
3

know
some
examples
of
redox
25.0 cm
of
a
of
aqueous
solution
of
sodium
hydroxide
is
exactly
neutralised
by
3
ttratons.
–3
12.2 cm
sulphuric
acid
of
concentration
0.100 mol dm
.
–3
Calculate
the
concentration
in
of
mol dm
the
sodium
hydroxide
solution.
Step
1:
Calculate
the
moles
of
acid
(moles
=
concentration
×
volume
in
3
)
dm
12.2
_____
0.100
×
–3
=
1.22
×
10
mol
H
SO
2
4
1000
Step
2:
Use
the
stoichiometry
of
the
equation
to
calculate
moles
of
NaOH:
H
SO
2
(aq)
+
2NaOH(aq)
Na
×
SO
2
mol
2
+
2H
O(l)
2
–3
mol
10
(aq)
4
mol
–3
1.22
→
4
1
2.44
×
10
mol
3
Step
3:
Calculate
the
concentration
of
NaOH
(moles
÷
volume
in
dm
):
–3
2.44
×
10
___________
–3
=
=
0.0976 mol dm
0.0250
Using titration data to deduce
T
o
nd
the
volumes
of
Worked
stoichiometry
both
we
to
know
the
concentrations
example
2
–3
of
a
0.0250 mol dm
solution
of
a
metal
hydroxide
–3
with
1:
the
hydrochloric
hydroxide.
Calculate
was
titrated
3
0.10 mol dm
neutralise
Step
and
reactants.
3
25 cm
need
stoichiometry
the
acid.
Deduce
number
of
the
It
required
12.5 cm
stoichiometry
moles
of
each
of
of
this
acid
to
reaction.
reactant:
25
_____
moles
hydroxide
=
0.0250
×
–4
=
6.25
×
10
mol
1000
12.5
_____
moles
of
hydrochloric
acid
=
0.10
×
–3
=
1.25
×
mol
10
1000
Step
2:
Deduce
the
simplest
mole
ratio
of
hydroxide
to
hydrochloric
–4
hydroxide
6.25
OH
Step
3:
W
rite
the
M(OH)
44
10
2HCl
mol
1
→
:
hydrochloric
:
stoichiometric
+
2
×
acid:
–3
HCl
equation:
MCl
+
2
2H
O
2
acid
1.25 × 10
2
mol
Chapter
3
The
mole
concept
Redox titrations
Redox
The
titrations
end
change
point
of
one
examples
are
Potassum
involve
in
of
these
the
given
oxidation
titrations
reactants
or
reduction
is
detected
by
use
of
reactions
either
a
(see
using
special
Section
the
redox
4.1).
colour
indicator
.
Some
below:
manganate(vii) ttratons
2+
This
can
be
used
ammonium
to
iron( ii)
nd
the
concentration
sulphate.
The
of,
reaction
is
for
example,
carried
out
Fe
in
ions
the
in
+
of
acid
When
ions):
(H
Did you know?
presence
usng
solutons
manganate(vii) for
2+
MnO
(aq)
+
5Fe
+
(aq)
+
8H
2+
(aq)
→
Mn
potassum
you
3+
(aq) +
5Fe
(aq)
+
4H
4
O(l)
2
cannot
purple
very
light
very
green
pale
very
pink
keep
the
MnO
ions
in
the
potassium
manganate( vii)
are
soluton for
ery
pale
long
yellow
because
brown
manganese(iv)
The
of
ttratons,
deep
purple
deposts
oxde
of
are formed.
It
in
4
s
also
not
easy
to
see
the
menscus
2+
colour
.
When
the
potassium
manganate(
vii)
is
added
to
the
Fe
ions,
the
n
the
burette
because
potassum
2+
ions
MnO
are
converted
to
Mn
ions
which
are
almost
colourless.
After
4
manganate(vii)
2+
all
the
has
Fe
reacted,
the
purple
colour
of
the
MnO
ions
are
s
one
of
the
most
visible.
4
ntensely-coloured
3+
So
the
colour
change
at
the
end
point
is
yellowish
(because
of
the
known.
ions)
to
compounds
Fe
For
ttratons
nolng
ths
purple.
compound
burette
t
s
permssble
readngs from
the
to
top
take
of
the
Iodne–thosulphate ttratons
menscus.
These
are
used
concentration
in
of
reactions
a
solution
where
of
iodine
hydrogen
is
liberated,
peroxide
by
e.g.
to
adding
nd
the
iodide
ions:
+
2I
(aq)
+
H
O
2
The
iodine
with
a
liberated
solution
of
in
(aq)
+
2H
(aq)
→
I
2
this
sodium
reaction
as
an
thiosulphate,
(aq)
+
2Na
2
The
end
iodine
The
the
point
of
present
end
point
titration
S
2
brown
O
2
the
sodium
used
is
sharpened
ask
as
up.
the
blue-black
in
the
blue-black
to
colourless
(aq)
→
S
O
2
2NaI(aq)
O(l)
solution,
+
Na
by
end
of
S
adding
point
is
iodine
change
a
few
and
gives
a
is
The
of
starch
to
the
colourless.
solution
solution
when
end
when
brown
starch
colourless
sharper
(aq)
6
occurs
from
drops
neared.
O
4
colourless
titration
change
titrated
3
2
colour
is
:
colourless
thiosulphate
The
presence
2H
2
3
colourless
is
+
aqueous
Na
2
I
(aq)
2
not.
to
is
This
point.
Key points
Dchromate(vi) ttratons
The
reactions
acid.
For
of
potassium
dichromate
are
carried
out
in
the
presence
of

The
can
example:
the
2–
Cr
O
2
2+
(aq)
+
6Fe
+
(aq)
+
14H
3+
(aq)
→
2Cr
concentraton
be found
relatonshp
+
6Fe
(aq)
+
7H
soluton
usng
concentraton
–3
(mol dm
light
a
O(aq)
2
very
of
ttraton
3+
(aq)
7
orange
by
green
very
green
)
=
number
of
moles
pale
of
yellow
solute
(mol)
÷
olume
of
3
soluton
(dm
)
together
wth
the
2–
The
Cr
O
2
ions
in
the
potassium
dichromate( vi)
are
orange
in
colour
.
7
stochometrc
2+
When
the
potassium
dichromate
is
added
to
the
the
2–
ions,
Fe
equaton from
the
Cr
O
2
7
reacton.
3+
ions
are
converted
titration
is
sulphonate
point
is
not
+
then
to
ions
Cr
obvious
so
phosphoric
from
green
a
acid)
to
which
redox
is
are
green.
indicator
added.
The
(barium
The
colour
end
point
of
the
diphenylamine
change
at
the

The
can
end
violet-blue.
stochometry
be found
by
concentratons
are

of
a
reacton
ttraton
of
both
f
the
solutons
known.
Redox
ttratons
change
n
often
colour
of
nole
one
of
a
the
reactants.
45
Revision
1
Balance
the following
molecular
questions
equations,
9
then
Pentane
contains
carbon
and
hydrogen
3
write
the
ionic
equation for
20 cm
each:
only. When
3
of
pentane
reacts
with
160 cm
of
oxygen,
3
a
(aq)
FeCl
+
NaOH(aq)
→
Fe(OH)
3
b
(s)
+
HCl(aq)
→
MgCl
(aq)
+
H
2
Na
S
2
S(s)
NaCl(aq)
a
of
product
is
carbon
dioxide
is
produced,
the
other
O
2
(aq)
+
HCl(aq)
→
steam.
(g)
2
NaCl(aq)
+
H
3
O(l)
Deduce
+
the
molecular formula
of
pentane.
2
+ SO
(g)
2
2
100 cm
3
Mg(s)
c
+
10
Calculate
the
molar
concentrations
of
the following
solutions:
Define:
3

a
a
mole
3.33 g
KNO
in
250 cm
of
solution
3
3

b
molar
Calculate
b
mass
the
2.65 g
CO
Na
2
molar
mass
of
in
75 cm
of
solution
3
the following
11
Calculate
the
mass
of
solute
in
the following
compounds:
solutions:

)
Pb(NO
3

Ca
(PO
3
3
2
a
–3
of
25 cm
)
4
a
0.01 mol dm
solution
3
2
b
750 cm
of
NaCl
–3
of
a
0.27 mol dm
solution
of
NH
NO
4

CuSO
·5H
4
12
3
a
Calculate

7
.
1 g
the
Na
number
of
moles
in
A
student
titrated
a
standard
solution
of
3
the following:
SO
2

3
O
2
hydrochloric
acid
solution
obtained
and
against
25 cm
of
sodium
hydroxide
the following volumes
of
acid:
4
20 g CaCO
3
3
b
Calculate
the
mass

0.025 mol

2.0
of
Burette
the following:
readngs/cm
1
2
3
4
25.80
37
.90
NaHCO
3
Fnal olume
–3
×
10
mol
HCl
Intal olume
4
Calculate
burn
the
5.5 g
of
mass
of
propane
oxygen
(C
H
3
5
Write
the
following
molecular
and
required
to
0.50
Actual olume ttre
empirical formula
of
26.65
25.65
25.50
the
a
Complete
b
From
the
table
above.
compounds:
a
dinitrogen
b
disulphur
the
results
compound
in
the
table,
determine
the
tetroxide
of
acid
required
to
neutralise
the
dichloride
sodium
A
1.00
).
8
average volume
6
12.50
completely
contains
0.48 g
of
carbon
and
hydroxide.
0.08 g
–3
13
A
student
titrated
a
0.2 mol dm
solution
of
–1
of
hydrogen
and
has
a
molar
mass
of
56 g mol
.
3
potassium
Find
the
empirical
and
molecular formulae
of
manganate()
against
25.0 cm
of
the
iron()
solution. The
average volume
of
potassium
compound.
3
manganate()
7
Calculate
the
masses
of
the following volumes
at
r.t.p.
solution.
3
a
6 dm
of O
2
3
b
120 cm
of SO
2
8
When
butane
(C
H
4
steam
and
carbon
)
is
burnt
completely
in
oxygen,
10
dioxide
are
produced. Calculate
3
the volume
of
46
butane
is
of
steam
produced
completely
burnt.
at
s.t.p.
when
in
this
titration
was
27
.50 cm
of
Calculate
gases
used
56 cm
the
molar
concentration
of
the
iron()
.
Chapter
3
The
mole
concept
–
reson
questons
47
4
Redox
4.
1
Redox
reactions
reactions
Learning outcomes
What
A
On
completon
of
ths
secton,
be
able
are
simple
definition
reduction
understand
terms
of
state
reactions?
is
the
of
oxidation
loss
of
is
oxygen.
the
gain
But
if
of
we
oxygen.
look
at
redox
electron
reactons
transfer
n
are
oxide
taking
with
place
at
hydrogen,
the
same
we
can
see
that
oxidation
oxdaton
(oxdaton
know

deduce
an
the
number)
oxdaton
an
atom
Copper( ii)
oxdaton
number
oxdaton
n
a
+
H
(g)
→
Cu(s)
+
H
phrase OIL
remember
of
electron
redox
usng
Is
oxide
is
oxidised.
reduction
are
There
two
been
rules.
wll
help
reactons
n
you
losing
Loss
both
are
oxidised
and
reactions
taking
other
or
gets
are
reduced.
reactions
place
at
the
same
Is Gan
(of
gains
oxidation
oxygen
and
time.
ways
of
finding
out
whether
electron
transfer

changes
in
oxidation
or
not
a
substance
has

Oxidation
is
loss

Reduction
is
gain
of
state.
in
redox
reactions
electrons.
of
electrons.
terms
reacts
with
chlorine
electrons).
Mg
to
+
form
Cl
→
magnesium
chloride:
MgCl
2
Reducton
Hydrogen
where
reduced:

Magnesium
(of
oxygen
Redox
to
transfer:
Oxdaton
gets
Electron transfer
RIG
O(l)
2
of
Exam tips
The
reduction
rules
number
compound
number
and
state
and

definition
of
and
2
n
simple
reaction
time:
CuO(aq)
change
A
the
to:
copper( ii)

redox
oxidation
you
of
should
and
2
electrons).
Each
magnesium
atom
loses
two
electrons
from
its
outer
shell.
This
is
oxidation.
2+
Mg
Each
chlorine
complete
its
atom
outer
in
the
shell.
→
chlorine
This
is
+
Mg
2e
molecule
gains
one
electron
to
reduction.
–
Cl
+
2e
→
2Cl
2
Equations
like
separately
are
preferable,
to
these
called
write
showing
half
these
the
oxidation
equations .
half
It
is
equations
or
also
reduction
reactions
acceptable,
though
not
as
as:
2+
Mg
−
→
2e
Mg
and
Cl
→
2Cl
−
2e
2
Oxidation
We
can
extend
reactions
or
in
a
discussion
48
(oxidation
definition
covalent
of
redox
numbers)
to
compounds
include
by
oxidation
using
and
oxidation
reduction
states
numbers).
oxidation
ion
the
involving
(oxidation
An
states
state
(oxidation
compound
oxidation
to
show
number
number )
the
has
is
degree
been
a
of
number
given
oxidation.
abbreviated
as
In
to
the
each
atom
following
‘OxNo’.
Chapter
Oxidation
number
1
OxNo
2
The
OxNo
of
each
3
The
OxNo
of
an
refers
to
a
on
the
single
atom
atom
ion
ion,
in
or
an
arising
e.g.
4
The
OxNo
of
uorine
5
The
OxNo
of
an
is
6
Redox
reactons
rules
ion
in
from
a
oxygen
compound.
is
0,
single
e.g.
atom
Mg
is
=
the
0,
H
=
0.
same
as
the
3+
Mg
in
a
element
2+
charge
4
=
+2,
compounds
atom
in
a
2–
Fe
=
is
+3,
Br
always
compound
=
–1,
S
=
–2.
–1.
is
–2
(but
in
peroxides
it
–1).
The
OxNo
hydrides
it
of
is
a
hydrogen
–1),
e.g.
atom
:
CH
H
in
=
a
compound
+1,
HCl:
H
is
=
+1
+1,
(but
in
NaH:
metal
H
=
–1.
4
7
The
e.g.
sum
in
of
all
the
OxNo’s
of
atoms/ions
2
3+
The
a
compound
is
zero,
3
2–
2Al
8
in
O
Al
sum
of
=
the
2
×
(+3)
OxNos
=
in
a
+6
3O
compound
=
ion
3
×
(–2)
equals
=
the
–6
charge
on
3–
the
3
9
ion,
O
e.g.
atoms
The
most
negative
in
=
the
3
×
atoms
=
–6
electronegative
OxNo
+5
−
element
of
6
N
=
in
=
–1
a
+5
(the
and
OxNo
charge
compound
is
on
of
the
given
ion)
the
OxNo.
Applying oxidation
The
ion,
NO
(–2)
of
many
number
elements
in
Did you know?
rules
Groups
V
to
VII
have
variable
oxidation
The
numbers.
We
work
these
out
using
the
oxidation
number
rules.
of
example,
what
is
the
oxidation
number
of
S
in
SO
H
2
Roman
numbers
partcular
Applying
rules

Applying
rule
5
and
2H
6:
+
H
=
4O
+1
+
S
and
=
0
O
so
=
of
–2
2(+1)
+
4(–2)
+
S
=
that
So
OxNo
of
S
=

reactions
Increase
in
show
partcular
called
names
n
the
the
oxdaton
atom
‘Stock
(or
state
on). Ths
chlorde
nomenclature’.
shows
that
ron
+6
has
Redox
the
ons)
0
Iron(ii)

(or
?
s
7:
atoms
4
compounds

after
For
and oxidation
oxidation
number
of
an OxNo
of
+2.
Potassum
manganate(vii)
shows
that
manganese
an OxNo
the
state
an
atom
or
ion
in
a
reaction
is
Chlorne(i)
has
oxde
shows
of
+7
.
that
chlorne
oxidation.
has

Decrease
in
oxidation
number
of
an
atom
or
ion
in
a
reaction
an OxNo
of
+1.
is
reduction.
When
copper( ii)
nitrogen
change
oxide
as
reacts
with
ammonia,
each
atom
of
copper
and
shown.
Key points
N
3CuO
+
oxidised
2NH
3Cu
+
3
+2
Cu
N
+
3H
2
–3
0
O

2
0

reduced
Oxdaton
s
loss
Reducton
s
gan
Redox
nto
The
copper
in
the
copper
oxide
is
reduced
to
copper
(OxNo
change
equatons
two
to
electrons.
of
electrons.
can
be
dded
equatons,
one
from
showng
+2
half
of
oxdaton
and
the
other
0).
reducton.
The
nitrogen
in
the
ammonia
is
oxidised
to
nitrogen
gas
(OxNo
change

from
–3
to
Oxdaton
s
ncrease
n
oxdaton
0).
state.
Reducton
oxdaton

Some
atoms
other
can
oxdaton
atoms
be found
number
decrease
hae xed
numbers. The
of
s
n
state.
n
a
usng
oxdaton
number
compound
oxdaton
rules.
49
4.2
Redox
equations
Learning outcomes
Balancing
We
On
completon
should

be
able
deduce
of
ths
half
balance
equatons from

balancing

then
understand
agent
and
know
how
terms
ablty
of
the
terms
reducng
to
order
by:
the
number
the
two
of
electrons
half
lost
equations
and
gained
together
.
agent
Construct
dsplacement
,
IO
1
a
balanced
in
acidic
ionic
equation
solution
using
for
the
the
two
reaction
half
of
zinc
equations
with
shown
iodate(v)
below:
3
n
2+
reducng
reference
example
oxdsng
elements
oxdsng/
by
equations
equatons
ions,

half
putting
Worked

two
equations
you
to:
balanced
releant
secton,
half
Equation
1:
Oxidation
of
zinc
to
zinc( ii)
Equation
2:
Reduction
of
iodate( V)
ions:
Zn
→
Zn
+
2e
to
to
iodine:
reactons.
1
+
IO
+
6H
+
5e
→
I
3

Each
IO
Zn
ion
atom
gains
loses
5
2
electrons
electrons
when
when
+
3H
2
2
O
2
oxidised.
Each
I
atom
in
the
reduced.
3

T
o
balance
Multiply
the
electrons:
equation
1
by
5
(10
electrons
in
total):
2+
5Zn
Multiply
equation
2
by
2
→
(10
5Zn
+
electrons
10e
in
total):
+
2IO
+
12H
+10e
→
I
3

Put
the
two
equations
together
and
–
5Zn
+
+
6H
2
2IO
cancel
+
+
12H
the
electrons:
2+
→
5Zn
+
I
3
Balancing
Worked
+
6H
2
equations
example
O
2
using oxidation
O
2
numbers
2
Exam tips
W
rite
a
balanced
equation
for
the
reaction
of
ClO
ions
in
acidic
solution
3
2+

When
balancng
numbers,
are
per
the
oxdaton
oxdaton
atom
–
you
with
Fe
Step
1:
3+
ions
to
the
same
Cl
ions,
Fe
W
rite
down
in
the
unbalanced
oxidation
on
each
If
a
of
present,
states
identify
the
that
acd
+
Fe
+
+
H
atoms
which
3+
→
Cl
+
Fe
+
H
3
the
Step
queston
equation,
number
sde
equaton.

water
.
number
.
2+
atoms
and
numbers
ClO
of
ions
must
change
compare
form
2:
Deduce
the
O
2
changes
in
OxNo
s
oxidation
change
number
.
–6
wrte
2+
ClO
+
Fe
+
+
3+
H
Cl
+
Fe
+
H
3
O
2
+
H

When
on
the
balancng
numbers,
left.
usng
remember
+5
oxdaton
you
to
select
two
OxNo
reactants
products
whose
3:
Balance
the
is
change.
If
+1
oxidation
number
changes
so
that
a
total
oxidation
oxdaton
the
reacton
ClO
+
6Fe
+
+
H
3+
→
Cl
+
6Fe
+
3
s
the
0.
2+
numbers
change
+3
and
number
two
–1
usually
Step
need
+2
disproportionation
H
O
2
reaction
+
Step
(see
Secton
12.8)
you
wll
4:
Balance
present)
need
to
select
three
the
charges
by
balancing
and
then
other
molecules
H
such
ions
as
(or
OH
water
.
speces.
2+
ClO
+
3
50
the
only
6Fe
+
+
6H
3+
→
Cl
+
6Fe
+
3H
O
2
ions
if
Chapter
Reducing
During

A

An
The
a
redox
reducing
ability
order
of
reactions.
another
an
of
on
agent
a
its
A
electrons
gains
substance
electrode
to
oxidising
or
chemical
solution
act
as
a
reaction
For
copper( ii)
gets
is
a
better
+
reducing
Cu
gets
reduced.
agent
Section
ability
is
one
gaining
Silver
where
one
zinc
a
electrons).
not
can
out
atom
displaces
better
So
the
displace
agent
reagents
displacement
replaces
copper
from
2+
→
than
Zn
(aq)
copper
oxidising
reaction
copper
put
sulphate:
+
(Zn
is
Cu(s)
better
at
2+
is
Cu
does
oxidising
We
carrying
2+
electrons).
or
10.1).
by
example,
(aq)
agent
oxidised.
reducing
2+
Zn(s)
Zinc
reactons
agents
and
(see
reducing
reaction.
of
and
electrons
potential
displacement
a
aqueous
loses
agent
their
in
and oxidising
Redox
reaction:
oxidising
depends
in
agents
4
agent
goes
from
to
than
the
copper(
losing
2+
Zn
(Cu
is
better
at
right.
ii)
sulphate
solution
because
2+
silver
is
a
better
oxidising
agent
than
copper
and
the
ion
Cu
is
a
better
+
reducing
If
we
agent
react
salts,
we
←–––
than
different
can
build
metal
is
a
Mg
–––→
metals
up
a
with
metal
better
metal
to
ion
this
is
aqueous
reactivity
reductant
Zn
According
silver
Ag
(better
Fe
a
better
reactivity
solutions
series ,
at
different
releasing
Sn
oxidant
series,
(better
metal
electrons)
Cu
at
accepting
magnesium
will
←–––
Ag
electrons)
displace
silver
–––→
from
nitrate.
Mg(s)
+
2AgNO
(aq)
→
Mg(NO
3
But
of
e.g.
silver
will
Displacement
halogen
halide
will
ions
not
displace
reactions
displace
(see
a
Section
magnesium
can
less
)
3
also
from
involve
reactive
one
(aq)
+
aqueous
non-metals.
from
2Ag(s)
2
an
magnesium
A
more
aqueous
nitrate.
reactive
solution
of
12.6).
Key points

Redox
lost
equatons
and

Redox

A
more
reacte
In
of
ts
can
n
equatons
soluton

ganed
be
deduced
releant
can
be
f
a
reacton,
the
The
reducng
agent
t
s
by
wll
loses
by
the
balancng
dsplace
better
oxdsng
balancng
numbers
of
electrons
equatons.
balanced
element
salt
half
at
agent
a
less
losng
gans
electrons
oxdaton
reacte
changes.
element from
a
electrons.
electrons
and
number
becomes
and
becomes
reduced.
oxdsed.
51
5
Kinetic
5.
1
The
theory
gas
laws
Learning outcomes
On
completion
should
be
able
of
this
Compressing
section,
We
can
are
continuously
picture
gases
gases
as
a
collection
of
randomly
colliding
with
each
other
and
with
the
particles
walls
of
which
the
to:
container

state
Boyle’s

draw
graphical
law
and Charles’
law
representations
because
in
which
they
are
they
are
constantly
placed.
hitting
The
the
gas
particles
walls
of
force
Boyle’s
law
the
exert
a
pressure
container:
to
pressure
and Charles’
(Nm
(N)
_________
–2
illustrate
moving
you
)
=
2
area
(m
)
law
When

calculate
volumes
and
we
together
.
of
gases
under

calculate
using
different
Charles’
Boyle’s
volumes
temperatures
The
the
volume
molecules
hit
of
the
a
gas,
walls
the
of
molecules
the
container
get
squashed
more
often
closer
and
so
different
the
conditions
decrease
pressures
of
gas
pressure
increases.
law
and
gases
conditions
under
using
law.
increased
pressure
Did you know?
Figure 5.1.1
When the volume of the container is decreased, the gas molecules are
squashed closer together and hit the walls of the container more often
The
chemical
high
industry often
pressures
involving
in
gases,
industrial
e.g.
in the
uses very
processes
–2
We
measure
pressure
in
newtons
per
square
metre
(Nm
)
or
in
pascals
Haber
–2
1 Nm
process for
cases
it
is
making
convenient to
‘atmospheres’
the
is
ammonia.
unit of
pressure. This
with
Standard
is
atm)
atmospheric
larger:
1 atm
=
pascal
(Pa).
pressure
is
101 325 Pa
(often
rounded
down
to
it
numbers
Boyle’s
than
1
101 000 Pa).
as
because
smaller
=
such
use
(abbreviated
easier to deal
In
law
101 000 Pa.
Boyle’s
law
states:
temperature,
Boyle’s
law
pressure
Boyle’s
is
too
the
For
a
fixed
volume
obeyed
high.
as
of
a
long
Figure
number
gas
as
5.1.2
is
the
of
moles
inversely
temperature
shows
two
of
a
gas
at
proportional
is
not
graphical
too
a
fixed
to
its
low
pressure .
or
the
representations
of
law.
a
b
emulov
emulov
0
0
0
pressure
0
1
pressure
Figure 5.1.2
Boyle’s law shown graphically: a As the volume of gas decreases the pressure
increases; b A plot of volume against 1/pressure shows proportionality
52
Chapter
Under
different
conditions
we
can
represent
Boyle’s
law
5
Kinetic
theory
mathematically
as:
P
V
1
initial
pressure
=
P
1
×
Worked
example
A
a
V
2
initial
volume = nal
pressure
2
×
nal
volume
1
3
gas
has
volume
of
volume
this
gas
of
1.2 dm
when
the
.
Its
pressure
pressure
is
is
100 kPa.
increased
to
What
250 kPa
is
the
at
constant
temperature?
Step
1:
Substitute
the
values
into
V
P
1
=
P
1
V
2
:
100
×
1.2
=
250
100
2:
Calculate
the
new
volume:
V
V
2
×
1.2
__________
Step
×
2
3
=
=
0.48 dm
2
250
Heating
As
the
move
gases
temperature
faster
and
If
the
further
apart.
increases.
reach
a
If
point
temperature
In
pressure
So
we
chemistry
gas
the
is
to
with
the
the
gas
the
xed
of
volume
pressure),
force
the
increases
temperature
absolute
use
a
constant,
theoretical
the
often
of
(at
increased
remain
volume
called
we
increases
wall
decrease
when
is
a
the
as
a
because
molecules
the
gas,
of
the
they
the
molecules
have
must
more
get
temperature
we
gas
must
is
emulov
energy.
of
hit
eventually
zero.
This
temperature.
absolute
temperature
scale
(Kelvin
0
temperature
scale).
The
units
of
this
scale
are
kelvins
(K).
0
Kelvin
So
temperature
minus
15 °C
in
=
°C
kelvin
+
Figure 5.1.3
273.
is
–15
temperature
+
273
=
(K)
As the temperature of a gas
increases, its volume increases in direct
258 K.
proportion (Charles’ law)
Charles’
law
Exam tips
Charles’
law
pressure,
the
(kelvin)
states:
For
volume
of
temperature.
Charles’
a
a
fixed
gas
Figure
number
is
directly
5.1.3
of
moles
of
a
proportional
shows
a
gas
to
graphical
at
its
constant
absolute
representation
You
of
can
Charles’
law
into
Boyle’s
one
law
and
equation
if
law.
pressure
Under
combine
different
conditions
we
can
represent
Charles’
law
mathematically
and
temperature
both
change:
as:
P
V
1
V
P
1
2
____
2
____
=
initial
nal
volume
_____________________
volume
V
___________________
___
1
nal
T
=
initial
temperature
(K)
V
example
T
1
___
2
2
=
temperature
(K)
T
1
Worked
T
2
Key points
2
3
At
27 °C
occupy
a
at
gas
occupies
227 °C
a
volume
assuming
that
of
600 cm
the
.
pressure
What
volume
remains
does
the
gas

Standard
pressure
temperature
273 K
Step
1:
Convert
°C
to
27 °C
=
27
+
227 °C
=
227
273
+
=
and
273

Kelvin
temperature

Boyle’s
the
values
into
the
=
°C
+
273
300 K
=
500 K
V
V
Substitute
101 325 Pa
K:
___
1
2:
are
respectively.
law: At
temperature,
Step
and
constant?
600
___
2
=
equation
T
T
1
volume
of
a
gas
V
____
:
constant
the
____
2
=
300
is
inversely
proportional
to
its
500
2
pressure.
600
____
Step
3:
Calculate
the
new
volume:
=
V
3
×
2
500
=
1000 cm

Charles’
law: At
constant
300
pressure,
is
the
directly
absolute
volume
of
proportional
a
to
gas
its
temperature.
53
5.2
The
kinetic
Learning outcomes
theory
General
Most
On
completion
of
this
section,
be
able
know
the
gaseous

know
the

know
real
characteristics
the
the
and
of
gases
know
basic
either
exist
as
small
isolated
molecules,
atoms,
e.g.
e.g.
ammonia
helium
and
and
sulphur
argon.
in
a
gas:
assumptions
of

are
arranged

are
far

move
apart
randomly
from
(are
each
disordered)
other
theory
differences
ideal
the
or
particles
rapidly
and
randomly.
between
gases
The

are
gases
the
state
kinetic
properties of
to:
The

gases
you
dioxide,
should
of
ideal
gas
kinetic theory
and
ideal
gases
equation.
The
kinetic
theory
of

Gas
particles
are

Gas
particles
do

Gas
particles
have

Collisions
they
A
gas
the
which
obey
Real
at
many
moving
not
between
has
these
energy
of
Boyle’s
have
each
other
.
volume.
gas
particles
are
characteristics
the
law
made
different
gases
differences
do
particles
and
is
Charles’
are
obey
elastic,
is
a
i.e.
called
an
measure
law
no
energy
is
lost
when
ideal
of
gas.
their
In
such
a
temperature.
gas,
Ideal
exactly.

We
ignore

The
very
attraction
high
and
the
law
theory
and
and
the
very
at
is
volumes
between
of
temperatures.
noticeable
kinetic
pressure
measurements
Boyle’s
especially
The
cannot
accurate
pressures
not
temperatures.
At
randomly.
attract
no
that:
gases
Scientists
gases
states
collide.
kinetic
gases
gases
of
low
high
always
the
particles
The
Charles’
very
not
the
volumes
results
law
pressures
many
show
exactly.
obeyed
of
that
These
and
very
low
because:
particles.
is
not
zero.
temperature:
nitrogen
1.5
emulov ×

The

So
particles
the
are
volume
volume
of
the
closer
of
the
together
.
particles
container
they
is
not
are
negligible
placed
compared
with
the
in.
1.0
erusserp
the
gas
is
ideal

The
forces

Attractive

The
of
attraction
forces
between
between
the
the
particles
particles
pull
cannot
them
be
ignored.
towards
each
other
.
0.5
carbon
pressure
volume

0
0
200
400
is
lower
than
expected
for
an
ideal
gas
and
the
effective
dioxide
600
At
of
the
extremely
gas
high
is
lower
than
pressures
expected.
and
extremely
low
temperatures
the
800
particles
may
be
so
close
to
one
another
that
repulsive
forces
between
5
pressure
(kPa
×
10
)
the
Figure 5.2.1
clouds
cause
deviations
from
ideal
gas
behaviour
.
Hydrogen, nitrogen and
carbon dioxide show deviations from
Boyle’s law over a range of pressures
54
electrons
Gases
and
which
do
pressures
not
are
obey
called
Boyle’s
real
law
gases
and
Charles’
law
at
all
temperatures
Chapter
The
For
us
ideal
an
the
ideal
gas
gas
ideal
gas
5
Kinetic
theory
equation
we
can
combine
Boyle’s
law
and
Charles’
law.
This
gives
equation :
PV
=
nRT
Where:
P
is
the
pressure
V
is
the
volume
in
pascals,
n
is
the
number
R
is
the
gas
T
is
the
temperature
Pa.
3
in
of
cubic
metres
moles
of
(1 m
).
–1
mol
(8.31 J K
in
1000 dm
gas.
–1
constant
3
=
kelvin
).
(K).
mass
in
grams
(m)
_________________
Since
number
of
moles
=
we
molar
equation
mass
can
rewrite
the
ideal
gas
(M)
as:
mRT
_____
PV
=
M
If
we
PV
=
know
nRT,
four
we
of
can
the
ve
quantities
calculate
the
in
the
ideal
gas
equation,
fth.
Key points

The
kinetic
motion

An
at
ideal
gas
laws

Gases

The
theory
a
gas
states
variety
is
one
under
all
of
that
gas
particles
are
always
in
constant
random
speeds.
which
obeys
pressures
do
not
obey
the
ideal
gas
equation
gas
is
and
laws
PV
the
=
gas
laws.
Real
gases
do
not
obey
the
temperatures.
at
high
pressures
and
low
temperatures.
nRT.
55
5.3
Using
the
Learning outcomes
On
completion
should

be
able
perform
of
this
ideal
Calculations
section,
you
calculations
gas
equation
using the
Worked
example
Calculate
the
ideal
gas
equation
1
to:
using
volume
of
gas
in
a
weather
balloon
which
contains
the
0.24 kg
ideal
gas
of
helium
at
a
temperature
–1
equation
He
(A
=
4.0;
R
=
8.31 J K
of
–25 °C
and
a
pressure
of
60 kPa
–1
mol
)
r

know
how
to
determine
the
Step
relative
molecular
mass
of
1:
Change
pressure
calculate
with
low
boiling
points
or
the
inserting
relevant
data
temperature
number
of
into
the
correct
units
and
moles:
gases
60 kPa
by
and
liquids
=
60 000 Pa
–25 °C
=
–25
+
273
=
248 K
into
240
____
the
ideal
gas
equation.
0.24 kg
=
240 g
moles
of
He
=
=
60 mol
4
Step
2:
Rearrange
the
ideal
gas
equation
into
Exam tips
the
form
you
need:
nRT
____
PV
=
nRT
→
V
=
P
1
When
doing
gas
law
calculations,
60
think
of
the
units
that
you
Step
3:
Substitute
the
gures:
V
(in
)
m
×
8.31
×
248
________________
3
always
3
=
=
2.06 m
60 000
are
likely
to
have
to
change
to
as
‘molPaK’
moles
mol
pressure
Deducing
relative
We
simple
molecular
masses
Pa
temperature
can
use
a
weighing
method
to
deduce
the
relative
molecular
K
mass
of
a
gas.
3
2
Make
sure
that
you
change
cm
A
3
or
dm
3
to
m
3
(1 m
large
flask
of
known
volume
is
lled
with
gas
and
the
mass
of
the
gas
is
3
=
1000 dm
)
found.
the
If
values
method
used,
of
the
is
its
temperature
in
the
ideal
suitable
accuracy
in
is
and
gas
the
pressure
equation
school
limited
to
are
known,
nd
the
laboratory
because
of
we
value
when
problems
can
of
M.
dense
substitute
Although
gases
involving
are
the
all
this
being
buoyancy
air
.
Worked
example
A
volume
2
3
flask
flask
is
of
100 kPa
2.00 dm
and
the
contains
temperature
3.61 g
is
of
20 °C.
–1
molecular
Step
1:
mass
Change
of
the
gas
pressure,
(R
=
and
gas.
The
pressure
the
=
100 000 Pa
2.0 dm
20 °C
Step
2:
Rearrange
and
molar
the
ideal
mass,
=
gas
M)
to
PV
=
20
+
temperature
to
their
correct
2.00/1000
273
M
=
(in
the
the
3.61
the
×
×
10
3
m
form
showing
mass,
mRT
_____
→
M
=
M
Substitute
2.00
subject:
mRT
3:
=
293 K
_____
Step
units:
–3
=
equation
make
PV
values:
8.31
×
293
_____________________
M
=
=
–3
100
56
000
×
2.00
×
10
the
).
3
100 kPa
in
relative
–1
mol
8.31 J K
volume
a
Calculate
44
(to
two
signicant
gures)
m,
Chapter
Deducing the
A
similar
point
The
relative
method
can
be
molecular
used
to
mass of
deduce
the
a volatile
molar
mass
of
5
Kinetic
theory
liquid
a
low
boiling
liquid.
apparatus
is
shown
below:
thermometer
hypodermic
syringe
gas
syringe
syringe
Figure 5.3.1
The

the
bulb
Apparatus used for nding the relative molecular mass of a volatile liquid
procedure
Let
light
oven
gas
is:
syringe
Record
this

Record
volume

Draw

Inject
reach
a
particular
temperature
in
the
syringe
oven.
temperature.
some
of
some
of
the
the
of
the
air
liquid
in
the
into
liquid
gas
the
from
syringe.
hypodermic
the
syringe
hypodermic
and
syringe
weigh
into
the
it.
gas
syringe.


Allow
liquid
in
Record
the
The

the
vapour
the
mass
calculation
V
olume
of
to
gas
is
vaporise
and
record
the
nal
volume
of
air
+
syringe.
of
the
carried
vapour
=
hypodermic
out
nal
in
gas
the
syringe
same
syringe
way
and
as
volume
the
for
–
a
pressure.
gas
initial
but:
gas
syringe
volume

Mass
of
liquid
hypodermic
=
initial
mass
of
hypodermic
syringe
–
nal
mass
of
syringe
Key points


The
ideal
equation,
temperature
or
equation
known.
The
The
by
is
are
relative
volume

gas
of
relative
(at
the
(at
of
of
a
can
if
be
the
gas
mass
of
of
liquids
vapour
temperature
used
other
can
temperature
volume
known
moles
mass
known
molecular
measuring
vaporised
number
molecular
gas
PV = nRT,
to
physical
be found
and
by
pressure,
quantities
weighing
a
volume,
in
the
known
pressure).
with
low
produced
and
deduce
boiling
when
a
points
known
can
be found
mass
of
liquid
pressure).
57
5.4
Changing
Learning outcomes
state
The
The
On
completion
should
be
able
of
this
section,
liquid
molecules
describe
to:
the

liquid
state
in
are
not
motion,
proximity

explain
the
of
liquid:
distances
arranged,
due
to
although
the
effect
of
there
weak
may
be
some
order
intermolecular
over
forces
of
terms
which
are
constantly
being
broken
up
as
they
gain
kinetic
and
energy
arrangement
a
regularly
attraction
of
in
you
short

state
from
neighbouring
molecules
particles
terms
‘melting’

are

have
close
more
together
kinetic
each
other
in
and
energy
than
in
a
solid
and
so
are
able
to
slide
over
‘vaporisation’

explain
of
changes
energy
in
changes
proximity,
state
and
in
changes
arrangement
in
Most
of
fairly
random
way.
substances
that
are
liquid
at
room
temperature
are
either:
and

motion
a
terms
covalently
bonded
molecules
with
considerable
van
der
W
aals
forces
of
particles.
attraction,

e.g.
molecules
bromine
with
dipole–dipole
signicant
bonding
e.g.
hydrogen
bonding,
e.g.
water
or
propanone.
Did you know?
Water
deal
in
of
the
up
hydrogen
smaller
exist
in
to
groups
the
18
ice-like
state
structure. This
extensive
or
liquid
of
water
Changing
state
When
is
its
a
solid
molecules

The
with
energy
bonds
broken
as
is
a flickering
water.
increases
solid
the
vibration
of
the
particles.
The
increases.

The
forces

The
solid
of
attraction
between
the
particles
weaken.
more
are
melts
slide
when
over
enough
each
energy
is
transferred
to
make
the
other
.
continuously
within
The
sometimes
crystal
melting
point
is
the
temperature
when
the
solid
is
in
equilibrium
the
with
to
the
molecules.
and formed
structure. This
of
an

being
transferred
temperature
particles
Hydrogen
heated:
Larger
with
between
arranged
great
to
perhaps
molecules
structure
randomly
due
a
bonding.
structure,
water
is
has
the
liquid.
referred
structure
of
For
an
ionic
attractive
molecular
overcome
When

a
The
the
The
less
is
energy
of
energy
lattice.
energy
of
required
the
to
melting
required
intermolecular
transferred
from
is
is
So
for
forces
break
point
melting
holding
the
is
–
strong
very
just
the
ionic
high.
For
enough
molecules
a
to
together
.
heated:
The
forces
lot
the
weak
liquid
escape
a
in
solid,
particles.

solid
forces
increases
temperature
attraction
the
liquid
of
between
and
the
the
movement
liquid
the
become
particles
a
and
energy
of
the
increases.
vapour
.
weaken
This
until
process
they
is
called
vaporisation

The
liquid
boils
equilibrium
when
between
enough
the
energy
liquid
and
is
the
transferred
gas
to
allow
an
phase.
Exam tips
It
is
the
important
boiling
liquid. This
is
equilibrium
58
to
point
distinguish
some
called
with
between
particles
have
evaporation.
the
vapour,
evaporation
enough
Boiling
which
occurs
appears
and
energy
as
boiling.
to
when
bubbles
Even
escape from
the
whole
within
below
the
liquid
the
is
liquid.
in
Chapter
These
changes
temperature
because
the
molecules
of
state
stays
energy
(or
can
be
constant
ions)
absorbed
rather
shown
at
the
goes
than
on
a
heating
melting
to
break
raising
liquid
point
the
the
curve.
and
forces
at
5
Kinetic
theory
The
the
boiling
between
point
the
temperature.
vaporising
) C°(
erutarepmet
gas
heating
up
solid
liquid
heating
up
melting
0
solid
0
heating
5
up
10
15
time
Figure 5.4.1
20
(min)
The change in temperature when water is heated at a constant rate. Note
how the temperature stays constant as the solid melts and the liquid turns to vapour
When
a
vapour
attracted
It
to
is
each
cooled,
other
.
the
particles
Energy
is
lose
released
kinetic
and
the
energy
vapour
and
begin
turns
to
to
be
liquid.
condenses
When
each
a
liquid
other
is
more
sufciently
low
cooled,
and
for
Summarising
the
more
the
particles
slowly.
liquid
to
changes of
lose
kinetic
Eventually
solidify.
The
the
energy
and
slide
temperature
liquid
past
is
freezes.
state
energy
absorbed
vaporisation
melting
condensing
freezing
released
Figure 5.4.2
Energy is required to melt and vaporise a substance. Energy is released when
a substance condenses and freezes
Key points

Particles
over

in
each
Most
liquid
substances
molecular

a
Particles
are
not
ordered,
are
close
together
and
move
by
sliding
other.
which
structure
in
a
solid
are
with
are
liquid
weak
at
room
temperature
attractive forces
generally
arranged
in
a
have
between
lattice. The
the
a
simple
molecules.
particles
only
vibrate.

When
a
solid
particles

When
They
a
melts,
the
together. The
liquid
energy
particles
vaporises,
the
input
overcomes
begin
particles
to
slide
gain
the forces
over
energy
one
and
keeping
the
another.
move further
apart.
move freely.
59
6
Energetcs
6.
1
Enthalpy
Learning outcomes
changes
Enthalpy
Particles
On
completion
should
be
able
of
this
section,
have
know
that
to:

energy
enthalpy
two
main
types
change
of
energy:
you
Potential
closer

and
energy
two
is
the
attracting
energy
due
particles
are
to
to
the
position
each
other
,
of
the
the
particles.
lower
the
The
potential
changes
energy.
accompany
chemical
reactions


understand
the
meaning
of
Kinetic
energy
is
the
energy
associated
with
the
movement
of
the
particles.
term
‘enthalpy
change’
Energy

describe
the
terms
has
content
and
is
be
the
absorbed
total
to
amount
separate
of
energy
particles
from
(potential
each
and
other
.
kinetic)
in
The
heat
chemicals.
‘endothermic’
The

to
‘exothermic’
draw
energy
exothermic
prole
and
heat
content
is
also
called
the
enthalpy
(symbol
H).
diagrams for
We
endothermic
in
cannot
their
measure
normal
enthalpy
physical
by
states
itself.
at
So
it
was
101.325 kPa
decided
pressure
that
and
all
elements
298 K
have
reactions.
zero
enthalpy.
These
conditions
are
called
standard
conditions .
The
Ø
is
symbol
We
can,
used
to
however
,
show
standard
measure
conditions.
enthalpy
changes .
These
occur
when
heat
Exam tips
energy
is
exchanged
with
the
surroundings
in
a
chemical
reaction.
The
Ø
symbol
In Chemistry
it
is
useful
to
for
mean. You
letters
have
and
already
enthalpy
change
is
.
ΔH
what
between
the
enthalpy
of
the
products
sigma
and
enthalpy
change
come
pi. The Greek
and
the
is
the
reactants.
they
Ø
across
Ø
ΔH
=
Ø
H
−
H
products
two:
The
know
difference
some Greek
standard
reactants
letter
–1
The
capital
change
delta
in
enthalpy
change
Δ
is
something.
change
in
used
and
to
So
denote
ΔH
units
of
enthalpy
change
are
kJ mol
a
means
ΔT means
Exothermic
and
endothermic
reactions
temperature.
T
wo
important
–
terms

system

sur roundings
reaction
and
An
the
is
chemical
–
these
carried
anything
decrease
any
into
in
common
include
reaction
Ø
are
reactants
out,
dipping
endother mic
surroundings
which
and
the
the
the
not
reaction
absorbs
chemistry
energy
container
taking
−
from
the
in
which
in
e.g.
a
the
a
reaction
thermometer
.
surroundings.
The
temperature.
Ø
H
products
part
mixture,
Ø
H
are:
products
air
,
solvents
in
is
positive.
So
ΔH
reactants
is
positive.
reaction
Ø
For
example:
CaCO
(s)
→
CaO(s)
+
CO
3
All
e.g.
reactions
thermal
requiring
An
a
continuous
decompositions.
endothermically,
exother mic
surroundings
(g)
e.g.
Some
ammonium
reaction
increase
Ø
releases
in
of
heat
dissolve
are
in
endothermic,
water
energy
to
the
surroundings.
example:
C(s)
+
O
(g)
→
2
All
combustion
acids
60
are
reactions
exothermic.
The
Ø
is
negative.
So
ΔH
reactants
is
negative
reaction
Ø
For
572 kJ mol
temperature.
H
products
+
chloride.
Ø
–
H
=
reaction
input
salts
–1
ΔH
2
CO
(g)
2
are
–1
ΔH
=
−394 kJ mol
reaction
exothermic.
Many
reactions
of
metals
with
Chapter
Energy
Energy

the
6
Energetics
prole diagrams
prole
relative
usually
diagrams
enthalpy
includes

the
reaction

the
enthalpy
the
(enthalpy
of
change,
an
reactants
formulae
pathway
(exothermic),
the
on
a
prole
the
of
and
show:
products
reactants
and
on
the
y-axis,
this
products
x-axis
downward
upward
diagrams)
arrow,
arrow,
↑,
↓,
indicates
indicates
energy
energy
released
absorbed
(endothermic).
H
(g)
+
I
2
(g)
2
ygrene
–1
H
=
+151 kJ mol
2HI
reaction
Figure 6.1.1
pathway
Energy prole diagram for the thermal decomposition of hydrogen iodide
(endothermic)
The
diagram

The

So
for
reaction
the
the
endothermic
mixture
enthalpy
of
gains
the
reaction
energy
products
is
so
shows
the
higher
that:
surroundings
than
that
of
cool
the
down.
reactants.
Ø

So
ΔH
is
positive.
reaction
Mg(s)
+
2HCl(aq)
ygrene
–1
H
=
–456.7 kJ mol
MgCl
(aq)
+
2
H
(g)
2
Key points
reaction
pathway

Figure 6.1.2
Energy
changes
in
chemical
in
heat
Energy prole diagram for the reaction of magnesium with hydrochloric acid
reactions
result
loss
or
(exothermic)
gain:
The
diagram
for
the
exothermic
reaction
shows
that:

In
an
an
enthalpy
exothermic
transferred

The
reaction
mixture
loses
energy
so
the
surroundings
warm
So
the
enthalpy
of
the
reactants
is
higher
than
that
of
the
ΔH.
reaction,
the
heat
is
surroundings.
up.
In

to
change,
an
endothermic
reaction
products.
heat
is
absorbed from
the
Ø

So
ΔH
is
negative.
reaction
surroundings.

Exothermic
negative
reactions
ΔH
reactions
values,
have
have
endothermic
positive
ΔH
values.

Energy
prole
relative
and
diagrams
energies
products
of
and
the
the
show
the
reactants
enthalpy
change.
61
6.2
Bond
energes
Learning outcomes
Bond

On
completion
of
this
section,
making
Breaking
the
and
bond
bond
between
breaking
two
atoms
requires
energy.
It
is
you
endothermic.
should
be
able
to:


know
that
chemical
Making
new
bonds
between
two
atoms
releases
energy.
It
is
reactions
exothermic.
take
place
through
energy

changes
involved
in
If
the
energy
released
breaking
and
know
that
bond
breaking
endothermic
and
bond
making
in
new
bond
bonds,
breaking
the
is
greater
reaction
is
than
the
energy
endothermic.
If
the
energy
We
how
bond
in
absorbed
making
in
new
bond
bonds,
breaking
the
is
less
reaction
is
than
the
energy
exothermic.
is
exothermic
understand
making
is
released

in
making


absorbed
bond
can
draw
energy
prole
diagrams
to
show
these
changes.
energies
a
can
be
used
enthalpy
to
calculate
change
of
the
2H
+
2Cl
reaction
bond
making
bond

describe
energy
energy
can
and
affecting
how
affect
H
ygrene
bond
the factors
bond
chemical
+ Cl
2
2
H
2H–Cl
reactivity.
reaction
pathway
Exam tips
Do
not
make
the
mistake
in
b
thinking
4H
that
in
during
the
In
many
bonds
reaction.
in
Section
are
break
particular
7
.5 for
broken
bonds
into
the
break
sequence
details).
showing
all
+
2O
bonds
only
during
more
diagrams
the
reactions
Particular
a
all
ygrene
certain
level
reaction
compound
atoms.
form
a
bond
bond
breaking
making
2H–H
+O=O
H
2H
and
Energy
the
atoms
reaction
breaking
then
reforming
accounting’
only.
pathway
are for
Figure 6.2.1
‘atom
O
2
(see
Energy prole diagrams for: a an endothermic reaction – the thermal
decomposition of hydrogen chloride; b an exothermic reaction – the synthesis of water
from hydrogen and oxygen
Bond
Bond
energies
energy
particular
is
bond
the
in
bond
dissociation
bond
broken
is
amount
one
energy.
put
of
mole
after
of
The
this
energy
needed
gaseous
symbol
to
break
molecules.
for
bond
one
This
energy
is
is
mole
also
E.
of
The
type
to
units
the
of
bond
of
is
is
to
energies
When
the
bond
energy
required
broken.
released
type
bond
energy
V
alues
being
of
same
are
a
So
C=C
always
bonds
–1
.
kJ mol
break
are
new
E(C=C)
double
positive
are
the
amount
of
broken.
For
example:
=
the
energy
+610 kJ mol
they
amount
absorbed
2O(g)
ΔH
=
+
496 kJ mol
–1
2O(g)
62
→
O=O(g)
ΔH
=
–
496 kJ mol
refer
of
to
bonds
energy
when
–1
→
refers
bond.
because
formed,
as
O=O(g)
of
symbol.
–1
The
a
called
the
same
Chapter
Factors
For
affecting
atoms
of
the
bond
same
energy
bonds.
This
is
type,
double
because
bonds
there
is
a
have
greater
higher
bond
attractive
energies
force
than
compounds
bonding
pairs
of
electrons
and
the
nuclei
of
the
atoms
forming
compared
with
only
one
electron
if
=
+
ethene
the
pi
in
the C=C
sigma
bond
bond. You
is
weaker
can
see
than
this
pair
.
–1
E(C–C)
as
the
the
bond
such
between
bond
two
Energetics
Did you know?
In
single
6
you
compare
the
bond
energies
–1
E(C=C)
350 kJ mol
=
+
610 kJ mol
of
the
carbon–carbon
single
and
–1
E(C≡C)
The
bond
energies
of
the
=
halogens
+840 kJ mol
double
are:
exposed
–1
E(Cl–Cl)
=
+
bond
–1
244 kJ mol
E
(Br–Br)
=
+
=
+
electron
and
the
reects
density
smaller
the
of
more
the
amount
pi
of
193 kJ mol
overlap
–1
E(I–I)
bonds. This
of
the
p
atomic
orbitals
z
151 kJ mol
2
compared
The
bond
energies
decrease
down
the
group
as
the
distance
between
nuclei
increases.
This
is
because
there
is
less
force
of
the
lengthens.
bonding
The
pair
bond
of
electrons
energy
for
and
uorine,
the
nuclei
however
,
as
is
the
much
sp
bonding
in
sigma
molecular
orbitals.
(see
attraction
Section
between
the
the
the
atomic
with
2.9).
bond
lower
than
–1
expected
close
each
The
to
(E(F–F)
each
other
bond
atom
=
other
+
).
158 kJ mol
so
the
electron
This
is
clouds
because
in
the
the
F
atoms
neighbouring
are
very
atoms
repel
considerably.
energies
increases
in
of
carbon–halogen
bonds
also
decrease
as
the
halogen
size.
–1
E(C–Cl)
=
–1
E(C–Br)
+340 kJ mol
=
+280 kJ mol
–1
E(C–I)
The
bond
atoms
in
energy
a
of
a
molecule.
particular
For
=
+240 kJ mol
type
of
bond
can
be
affected
by
other
example:
–1
E(C–F)
in
CH
F
=
+
–1
452 kJ mol
E(C–F)
in
CF
3
For
this
reason
Calculating
The
‘balance
energies
is
=
+
485 kJ mol
4
we
often
use
enthalpy
sheet
shown
average
changes
method’
below
for
for
bond
energies,
using
calculating
the
bond
E
energies
enthalpy
changes
using
bond
reaction:
H
H–C–H
+
2
O=O
→
O=C=O
+
2
H–O–H
Key points
H

Bonds
broken
(endothermic
+)
Bonds formed
(exothermic
–1
−)
/kJ mol
E(C–H)
=
4
×
410
=
1640
2
×
E(C=O)
=
2
×
740
=
×
E(O=O)
=
2
×
Bond
energy
496
=
992
4
×
E(O–H)
=
4
×
460
=
to
gaseous
=
+2632 kJ
Total
=

enthalpy
change
=
+2632
–
3320
=
Bond
energy
and
is
is
the
break
energy
one
bond
in
mole
one
of
mole
a
of
molecules.
energies
calculate
–688 kJ mol
a
Bond
endothermic
making
–3320 kJ
–1
Overall
an
–1840
particular
Total
is
bond
–1480
needed
2
and
exothermic.

×
breaking
process
–1
/kJ mol
4
Bond
the
can
be
used
enthalpy
to
change
in
reaction.
reactivity

The
strength
of
the
bond
–1
The
bond
Nitrogen
break
the
energy
only
of
reacts
bonds to
decomposes
the
very
N ≡N
under
get
the
readily
triple
very
bond
harsh
reaction
because
is
high
conditions
started.
the
very
O–O
(994 kJ mol
because
Hydrogen
bond
it
is
).
difcult
between
to
peroxide
energy
is
.
ease
of
decomposition
of
hydrogen
different
halides
is
to
the
strength
of
the
hydrogen–halogen
bond
(see
is
different
A
high
value
of
bond
energy
may
also
contribute
related
in
atoms
compounds.

The
given
only
–1
150 kJ mol
slightly
two
Section
to
a
molecule
being
12.7).
relatively
unreactive.
63
6.3
Enthalpy
changes
Four types of
Learning outcomes

On
completion
of
this
section,
Standard
be
able
experment
enthalpy
enthalpy
change
change
of
reaction:
The
enthalpy
change
when
you
the
should
by
amounts
of
reactants
shown
in
the
to:
equation
react
to
give
products
Ø
under
standard
conditions.
Symbol
ΔH
r

describe
enthalpy
changes
1
of
Example:
2Fe(s)
+
1
Ø
O
(g)
→
Fe
2
2
reaction,
O
2
(s)
ΔH
3
–1
=
–824.2 kJ mol
r
combustion,

neutralisation
and
Standard
enthalpy
change
of
neutralisation:
The
enthalpy
change
solution
when
1
mole
alkali
under
of
water
is
made
in
the
reaction
between
an
acid
and
an
Ø

know
how
to
determine
these
standard
conditions.
Symbol
ΔH
n
enthalpy
changes
by
experiment
Example:
HCl(aq)
+
NaOH(aq)
→
NaCl(aq)
+
H
O(l)
2
Ø

use
experimental
calculate
data
enthalpy
–1
=
ΔH
to
–57.1 kJ mol
n
changes.

Standard
enthalpy
mole
substance
of
a
change
is
of
solution:
dissolved
in
a
The
very
enthalpy
large
change
amount
of
when
water
1
(to
Ø
innite
dilution)
under
standard
conditions.
Symbol
ΔH
sol
+
Example:
NaCl(s)
+
aq
→
Na
Ø
(aq)
+
Cl
(aq)
–1
ΔH
=
+3.9 kJ mol
sol

Standard
when
1
enthalpy
mole
of
a
change
of
substance
combustion:
is
burnt
in
The
excess
enthalpy
oxygen
change
under
standard
Ø
conditions.
Symbol
ΔH
c
Example:
–1
Ø
(g)
CH
+
2O
4
Finding
We
can
(g)
→
CO
2
enthalpy
calculate
(g)
+
2H
2
changes
many
O(l)
ΔH
2
enthalpy
=
–890.3 kJ mol
c
using
changes
a
calorimeter
from
the
results
of
experiments
thermometer
involving
can,
a
a
vacuum
procedure
plastic
calorimeter.
ask
or
A
a
calorimeter
more
can
complex
be
piece
a
of
apparatus.
cup,
The
a
metal
general
is:
lid
reaction

React
known

Measure

Calculate
the
amounts
of
reactants
temperature
the
energy
change
transferred
in
of
a
known
the
using
volume
of
solution.
solution.
the
relationship
q
=
mcΔT
mixture
Where
polystyrene
cup
m
is
q
the
is
the
mass
energy
of
transferred
solution
in
the
in
joules
(J),
calorimeter
–1
c
is
ΔT
Figure 6.3.1
polystyrene
A polystyrene drinking cup
the
is
specic
the
heat
capacity
temperature
of
change
water
(rise
or
in
grams
(g),
–1
°C
(Jg
fall)
–1
).
V
alue
of
c = 4.18 Jg
–1
°C
(°C).
Ø

Calculate
the
enthalpy
change
per
mole
of
specic
reactant
for
,
ΔH
sol
can act as a simple calorimeter
Ø
per
mole
of
water
formed
for
or
ΔH
for
the
number
of
moles
shown
n
Ø
in
the
equation
for
ΔH
r
Ø
ΔH
by
calorimetry:
worked
example
r
3
0.90 g
of
solution
was
to
zinc
in
stirred
27.9 °C.
a
was
added
to
calorimeter
.
constantly.
Calculate
75.0 cm
The
The
the
–3
of
0.25 mol dm
copper( ii)
sulphate
temperature
enthalpy
of
change
the
for
is
copper(ii)
in
excess.
solution
this
rose
reaction,
sulphate
The
from
A
Zn
mixture
18.0 °C
=
r
Step
1:
Substitute
q
Step
2:
the
=
Calculate
data
75.0
the
×
into
the
4.18
energy
per
×
equation:
(27.9
mole
of
–
q
18)
Zn:
=
=
mcΔT
3103.65 J
0.9 g
→
3103.65 J
65.4
_____
So
65.4 g
→
×
3103.65
=
225 531.9
J
0.9
–1
Standard
enthalpy
signicant
64
gures)
change
of
reaction
=
–230 kJ mol
zinc
(to
two
65.4.
,
Chapter
Enthalpy
changes
of
neutralisation
are
found
in
a
similar
way
to
the
6
Energetics
above
Exam tips
example
and
by
adding
measuring
known
the
concentrations
maximum
and
temperature
volumes
rise.
of
Enthalpy
acid
and
changes
alkali
of
In
solution
are
found
by
adding
a
known
amount
of
solute
to
a
known
calorimetry
assume
of
water
in
a
calorimeter
and
measuring
the
maximum
rise
in
experiments
we
volume
that:
temperature.
3
Standard
enthalpy
A
simple
apparatus
is
shown
in
Figure
change of
used
to
combustion
measure
the
enthalpy
by

1 cm
of
water

Aqueous
weighs
1 g.
calorimetry
change
of
if
combustion
they
solutions
are
are
treated
as
water.
6.3.2.

The
solution
specic

When
has
heat
a
the
same
capacity
solid
is
added
as
water.
to
a
lid
solution,
mineral
the
mass
of
the
solid
is
wool
ignored.
water
calorimeter
draught
screen
cap
spirit
burner
Figure 6.3.2
Measuring the enthalpy
change of combustion
The
procedure
is:
Exam tips

Weigh

Put
the
spirit
burner
and
cap
with
the
fuel.
You
a
known
amount
of
water
in
the
calorimeter
and
record
should
know
the
main
sources
its
error
in
calorimetry
experiments:
temperature.

Light
the
spirit
temperature

Remove
The
results
above.
We
the
are
need
mass
of
of
burner
,
the
spirit
in
in
the
the
and
let
calorimeter
replace
a
cap
the
similar
cap
way
rises
and
as
the
for
fuel
by
burn
about
until

the
10 °C.
the
the
water

the
temperature

the
mass

the
molar
fuel
mass
in
rise
the
of
of
heat
/
to
the
solution
air from
and from
the
the
burning fuel
worked

example

calorimeter
the
loss
water
reweigh.
know:

of
water
burner
,
processed
to
remove
loss
of
and
thermometer
heat
to
evaporation
the
calorimeter
of fuel.
water
burnt
of
the
fuel.
Ø
The
value
of
is
ΔH
calculated
per
mole
of
fuel
burnt.
is
enthalpy
c
Key points

Enthalpy
change
of
reaction,
ΔH ,
the
change
when
the
r

amounts
of
reactants
Enthalpy
change
of
shown
in
the
combustion,
equation
ΔH
,
is
the
react
to
enthalpy
give
products.
change
when
1
mole
c
of

a
substance
Enthalpy
is
change
burnt
of
in
excess
oxygen.
neutralisation,
ΔH
,
relates
to
the
enthalpy
change for
n
+
the
reaction:
H
–
+ OH
→
H
O
2

Enthalpy
change
of
solution,
ΔH
,
is
the
enthalpy
change
when
1
mole
of
a
sol
substance
dissolves
to form
a
very
dilute
solution
in
water.
Ø

The
symbol

Values for
indicates
ΔH ,
r
using
a
ΔH
,
c
the
ΔH
enthalpy
and
n
ΔH
,
change
can
often
under
standard
be found
by
conditions.
direct
experiment
sol
calorimeter.
65
of
6.4
Hess’s
law
Learning outcomes
On
completion
should
be
able
of
this
Enthalpy
section,
change of formation
Standard
enthalpy
mole
compound
change
of
for mation:
The
enthalpy
change
when
1
you
of
a
is
formed
from
its
elements
under
standard
to:
Ø
conditions.
Symbol
ΔH
f

explain
and
use
the
term
Ø
Example:
‘enthalpy

state
change
Hess’s
C(graphite)
+
2H
C(graphite)
law
(g)
→
CH
2
of formation’
is
used
here
(g)
because
–1
ΔH
4
[CH
f
graphite
is
the
(g)]
=
–74.8
kJ
mol
4
most
stable
form
of
carbon.
Ø
The
value
of
of
ΔH
an
element
in
its
normal
state
at
room
temperature
is
f

use
Hess’s
law
to
calculate
zero.
enthalpy
reaction
changes
and
of formation,
combustion.
Hess’s
Hess’s
Did you know?
law
is
an
example
of
the
to
of Thermodynamics
chemical
states
that
states
as
reactions. The
6.4.1
of
that
the
the
route
illustrates
total
by
enthalpy
which
Hess’s
law.
the
It
change
reaction
shows
for
a
chemical
reaction
is
proceeds.
that
the
enthalpy
change
is
First
the
law
law
independent
Figure
Hess’s
law
same,
whether
you
go
by
the
direct
route
or
the
indirect
route.
applied
First
law
Energy cannot be created
We
can
found
use
Hess’s
directly
by
law
to
calculate
experiment,
e.g.
enthalpy
the
changes
enthalpy
which
change
of
cannot
be
formation
of
propane.
or destroyed
Ø
Deducing
enthalpy
change of
reaction
using ΔH
values
f
A
+
B
C
direct
+
D
An
reactants
enthalpy
cycle
(Hess
cycle)
is
constructed
as
shown
in
Figure
6.4.2.
products
route
H
H
r
r
reactants
products
NH
+
HCl
NH
3
X
+ Y
H
H
f
shows
the
route
via
H
f
(NH
f
)
H
3
+H
products
Cl)
4
(HCl)
f
intermediates
1
elements
in
1
N
+
2H
2
2
+
Cl
2
2
2
and Y
standard
Figure 6.4.1
(NH
f
indirect
reactants
X
Cl
4
states
The enthalpy change is the
Figure 6.4.2
same whichever route is followed
Ø
Using
Hess’s
law
we
can
see
that
Ø
ΔH
+
f
ΔH
example
Calculate
ΔH
ΔH
f
Ø
=
–
f
products
Ø
ΔH
r
Worked
Ø
=
r
Ø
So
ΔH
reactants
ΔH
products
f
reactants
1
Ø
for
the
reaction:
SO
r
(g)
2H
S(g)
→
3S(s)
+
values:
SO
f
(g)
=
2H
2
–296.8 kJ mol
O(l)
2
–1
Ø
ΔH
+
2
–1
,
H
2
S(g)
=
–20.6 kJ mol
,
2
–1
H
O(l)
=
–285.8 kJ mol
2
The
stages

W
rite

Draw

Apply
are:
the
the
balanced
enthalpy
Hess’s
equation.
cycle
with
the
elements
at
the
law.
SO
(g)
+
2H
2
H
O(l)
2
[SO
f
2
]
2
2
×
H
f
+2
×
H
f
[H
S]
2
3S(s)
+ O
(g)
2
Figure 6.4.3
66
bottom
+
2H
(g)
2
[H
O]
2
as
shown.
Chapter
According
to
Hess’s
Energetics
law:
Ø
Ø
ΔH
6
[SO
f
] + 2ΔH
2
Ø
[H
f
S] +
ΔH
2
Ø
=
2ΔH
r
[H
f
O]
2
Ø
–296.8
+
2
×
(–20.6)
+
= 2
ΔH
×
(–285.8)
r
Ø
so
ΔH
=
2
×
(–285.8)
+
296.8
+
(2
×
20.6)
r
Ø
so
–1
=
ΔH
–233.6 kJ mol
r
Ø
Deducing
enthalpy
change of
reaction
using ΔH
values
c
The
enthalpy
cycle
is
constructed
as
shown
below
and
a
worked
example
shown.
H
r
reactants
H
product(s)
+O
c
H
+O
2
c
2
(reactants)
(products)
combustion
product(s)
Figure 6.4.4
Worked
example
Calculate
ΔH
2
Ø
for
the
reaction:
3C(graphite)
+
4H
r
(g)
→
C
2
H
3
(g)
8
–1
Ø
ΔH
values:
C(graphite)
=
–393.5 kJ mol
,
c
–1
(g)
H
=
–285.8 kJ mol
,
2
–1
H
C
3
(g)
=
–2219.2 kJ mol
8
H
r
3C(graphite)
+
4H
C
(g)
3
×
H
H
3
2
(g)
8
[C(gr)]
c
H
+
4
×
H
[C
c
(g)]
8
(g)]
[H
c
H
3
2
3CO
(g)
+
4H
2
O(l)
2
Figure 6.4.5
According
to
Hess’s
Ø
ΔH
law:
Ø
+
Ø
ΔH
r
[C
c
H
3
]
= 3ΔH
8
Ø
[C(graphite)] +
4 ΔH
c
[H
c
(g)]
2
Ø
+
ΔH
(–2219.2)
=
3
×
(–393.5)
+
4
×
(–285.8)
r
Ø
so
ΔH
so
ΔH
=
3
=
–104.5 kJ mol
×
(–
393.5)
+
4
×
(–285.8)
+
2219.2
r
Ø
–1
r
Ø
Note:
In
this
reaction,
ΔH
is
also
the
enthalpy
change
of
formation
of
Key points
r
propane.

Enthalpy
refers
Exam tips
of
a
to
of formation
the formation
compound from
under
Remember
change
standard
of
its
1
mole
elements
conditions.
that:

Hess’s
law
states
that
the total
Ø

ΔH
of
an
element
is
0.
enthalpy change in a reaction is
f
independent of the route.

In
enthalpy
correct
cycles,
make
sure
that
you follow
the
arrows
round
in
the
direction.

Hess’s
law
calculate

Look
carefully
at
the
state
symbols
Ø
e.g.
ΔH
[H
f
when
looking
–1
O(g)]
2
=
–241.8 kJ mol
up
enthalpy
ΔH
[H
f
–1
O(l)]
=
be
used
enthalpy
to
changes
which
changes,
Ø
but
can
cannot
be
determined
directly
by
–285.8 kJ mol
2
experiment,
e.g.
ΔH
from
r
ΔH
of
reactants
and
ΔH
or
f
products.
c
67
6.5
More
enthalpy
Learning outcomes
An
We
On
completion
of
this
section,
enthalpy
can
use
be
able
construct
using
enthalpy
cycle
bond
in
Figure
energies
using
bond
energies.
It
is
often
6.5.1
to
easier
,
calculate
however
,
to
an
enthalpy
use
the
‘balance
to:
sheet’

the
cycle
you
change
should
changes
Hess
cycles
using
method
(see
Section
6.2).
bond
energies
H
r
reactants

explain
of
the
term
hydration,
enthalpy
ΔH
bond
hyd

explain
the
term
lattice
products
change
of
energy,
energy
bond
reactants
of
energy
products
ΔH
latt

gaseous
use
Hess
ΔH
or
cycles
to
atoms
determine
Figure 6.5.1
Δ
H
sol
hyd
Worked
example
Calculate
ΔH
1
Ø
for
the
complete
combustion
of
ethene
to
carbon
dioxide
r
and
water
.
E(C=C)
+610,
E(C–H)
E(C=O)
H
+740,
+410,
E(O=O)
E(H–O)
+496,
+460
H
H
r
H
H
E(C=C)
+
4
×
E(C–H)
+
3
×
E(O=O)
2 C(g)
+
6O(g)
+
4
×
E(C=O)
+
4
×
E(H-O)
4H(g)
Figure 6.5.2
Ø
According
to
Hess’s
law:
ΔH
=
E(reactants)
+
E(products)
are
because
r
Note:
The
values
of
E(products)
negative
bond
formation
is
exothermic.
E(C=C)
+
610
+
4
×
E(C–H) +
3
×
E(O=O)
+
4
×
+
4
×
410
3
×
496
+
4
×
+
Ø
Δ
(–740)
+
+
4
4
×
×
E(H–O)
(–460)
–1
=
H
E(C=O)
–1062 kJ mol
r
Did you know?
We
can
calculate
theory from
the
Enthalpy
lattice
size
energy
and
change of
Standard
enthalpy
mole
specied
hydration
change
of
and
hydration:
The
lattice
enthalpy
energy
change
when
1
in
of
a
gaseous
ion
dissolves
in
enough
water
to
form
an
charge
Ø
innitely
dilute
solution
under
standard
conditions.
Symbol
Δ
H
hyd
of
the
ions
and
packed. This
the
way
they
sometimes
are
gives
slightly
2+
Examples:
Mg
2+
(g)
+
aq
→
Mg
Ø
(aq)
ΔH
–1
=
–1920 kJ mol
hyd
Ø
different
values from
the
lattice
Cl
(g)
+
aq
→
Cl
(aq)
–1
ΔH
=
–364 kJ mol
hyd
energy
obtained
by
experiment.
Lattice
‘Lattice
energy’
is
a
commonly
energy:
The
enthalpy
change
when
1
mole
of
an
ionic
compound
used
Ø
is
formed
from
its
gaseous
ions
under
standard
conditions.
Symbol
ΔH
latt
term
and
is
often
used
to
mean
2+
the
same
as
lattice
enthalpy. The
difference
between
small
we
the
two
is
Example:
often
use
the
68
to
mean
the
same
+
2Cl
(g)
→
CaCl
(s)
ΔH
–1
=
–2258 kJ mol
latt
so
energies
are
always
exothermic
two
when
terms
(g)
2
Lattice
that
Ca
Ø
thing.
new
‘bonds’
are
made.
because
energy
is
always
released
Chapter
Ø
Enthalpy
change of
of
can
calculate
hydration
the
using
enthalpy
an
change
enthalpy
of
Energetics
Ø
solution from ΔH
and
ΔH
latt
We
6
hyd
solution
or
the
enthalpy
change
cycle.
H
latt
gaseous
ions
ionic
solid
cation(s)
H
hyd
H
sol
+
anion(s)
H
hyd
ions
in
aqueous
solution
Figure 6.5.3
Here
is
a
worked
Calculate
the
example:
enthalpy
change
of
solution
of
magnesium
Ø
the
following
data:
–1
[MgCl
ΔH
latt
–1
=
–2592 kJ mol
chloride
Ø
,
using
2+
ΔH
2
[Mg
]
=
hyd
Ø
,
–1920 kJ mol
]
–1
ΔH
[Cl
]
=
–
364 kJ mol
hyd
H
latt
2+
Mg
(g)
+
2Cl
Mg Cl
(g)
(s)
2
2+
H
[Mg
]
hyd
H
+
2
[MgCl
sol
× H
[Cl
]
2
]
hyd
2+
Mg
(aq)
+
2Cl
(aq)
Figure 6.5.4
According
to
Hess’s
law:
Ø
Ø
ΔH
[MgCl
latt
]
+
ΔH
Ø
[MgCl
sol
Ø
So
ΔH
2
Ø
[MgCl
sol
]
=
[Mg
[Mg
Ø
]
+
2
×
ΔH
hyd
Ø
] + 2
×
+ 2
×
[Cl
]
hyd
Ø
ΔH
[Cl
]
−
ΔH
hyd
Ø
ΔH
2+
ΔH
hyd
–1920
So
=
2+
ΔH
2
]
2
[MgCl
latt
(–364)
−
]
2
(–2592)
–1
[MgCl
sol
]
=
–56 kJ mol
2
The
value
also
be
of
the
found
enthalpy
from
this
change
type
of
of
hydration
enthalpy

the
lattice

the
enthalpy
change
of
solution

the
enthalpy
change
of
hydration
cycle
for
if
a
we
particular
ion
can
know:
energy
of
one
of
the
ions.
Key points

ΔH
can
be found
by
using
a
Hess
cycle
involving
bond
energies
of
r
reactants

Enthalpy
and
products.
change
of
hydration,
ΔH
is
hyd
of

gaseous
Lattice
ions
energy,
dissolves
ΔH
is
in
the
water
the
enthalpy
enthalpy
change
when
1
mole
,
to form
a
change
very
dilute
when
1
solution.
mole
of
an
ionic
latt
compound

A
Hess
is formed from
cycle
can
be
drawn
its
gaseous
relating
ions.
the
enthalpy
changes
ΔH
,
sol
ΔH
ΔH
and
hyd
.
latt.
69
6.6
Lattce
energy:
Learning outcomes
On
completion
of
this
section,
Some
more
In
to
order
be
able
Born–Haber
enthalpy
calculate
lattice
cycle
changes
energy,
we
need
to
use
ionisation
energies,
you
enthalpy
should
the
change
of
formation
and
two
other
enthalpy
changes.
These
two
to:
are:

explain
the
change
of
terms
‘enthalpy
Standard
atomisation’
mole
‘electron
enthalpy
change
of
atomisation:
The
enthalpy
change
when
1
and
of
gaseous
atoms
is
formed
from
an
element
in
its
standard
state
afnity’
Ø
under
standard
conditions.
Symbol
ΔH
at

calculate
lattice
energy from
an
Ø
Examples:
appropriate
Born–Haber
Na(s)
→
Na(g)
ΔH
–1
=
+
107.3 kJ mol
at
cycle.
1
Ø
Cl
(g)
→
Cl(g)
ΔH
2
2
–1
=
+
122 kJ mol
at
Ø
V
alues
of
ΔH
are
always
positive
(endothermic)
because
energy
has
to
be
at
absorbed
to
remember
Electron
1
mole
break
that
bonds
the
afnity:
of
holding
enthalpy
The
electrons
rst
are
the
change
electron
added
to
a
atoms
of
together
.
atomisation
afnity
mole
of
is
the
gaseous
It
is
is
per
important
atom
enthalpy
atoms
formed.
change
to
to
form
when
1
mole
Ø
of
gaseous
ions
.
X
Symbol
ΔH
ea1
Ø
Examples:
Cl(g)
+
e
→
Cl
(g)
–1
ΔH
=
−348 kJ mol
=
−200 kJ mol
ea1
Ø
S(g)
+
e
→
S
(g)
–1
ΔH
ea1
As
with
anions
ionisation
with
energies,
multiple
successive
electron
afnities
are
to
form
charges:
Ø
O(g)
used
+
e
→
O
(g)
–1
ΔH
=
−141 kJ mol
=
+798 kJ mol
ea1
2–
O
(g) +
e
→
O
Ø
(g)
–1
ΔH
ea2
Exam tips
It
is
important
enthalpy
that
change
you
put
the
calculations
correct
to nd
Ø
is
generally
negative
but
ΔH
ea1
be
is
always
is
being
added
to
a
negatively
charged
and
ΔH
values
are
always
positive.
at
Ø
ΔH
are
always
negative.
latt
Ø
70
when
Make
you
sure
carry
you
because
ion. The
Ø
values
values
f
+)
positive,
overcome.
Ø
ΔH
or
out
know
these.
the
incoming
ea2
electron
ΔH
(–
energy.
Ø
ΔH
to
signs
lattice
may
be
positive
or
negative.
repulsive forces
have
Chapter
Born–Haber
We
can
lattice
of
Energetics
cycles
construct
energy
6
an
an
enthalpy
ionic
cycle
and
use
Hess’s
law
to
determine
the
solid.
ions
H
in
latt
ionic
gaseous
solid
state
H
H
f
(several
steps)
elements
in
standard
their
states
Figure 6.6.1
Ø
ΔH
consists
of
several
steps:
1
atomise
metal
→
atomise
ionise
Ø
So
for
NaBr:
ΔH
Ø
=
ΔH
1
Hess’s
law:
+
ΔH
+
=
can
show
lattice
energy
these
cycle .
of
=
ΔH
ΔH
[Na]
+
ΔH
i1
[Br]
ea1
Ø
−
ΔH
f
changes
The
sodium
+
Ø
ΔH
all
Bor n–Haber
Ø
Ø
[Br]
→
f
latt
a
atoms
ΔH
latt
Ø
We
metal
Ø
ΔH
1
So
ionise
at
Ø
ΔH
→
atoms
Ø
[Na]
at
Ø
By
non-metal
non-metal
on
1
an
energy
Born–Haber
bromide
as
cycle
shown
level
can
in
diagram.
be
used
Figure
to
This
is
called
determine
the
6.6.2.
+
Na
(g)
+
Br(g)
+
e
H
[Br]
ea1
–325
+
Na
H
(g)
+
Br
(g)
[Na]
i1
+494
H
[
at
Na(g)
+
Na(g)
+
Br(g)
]
Br
2
1
+112
H
2
Br
(g)
2
Exam tips
[Na]
at
H
1
Na(s)
+107
+
2
[Na
Br]
latt
Br
(l)
2
It
H
[Na
is
easier
to
determine
lattice
Br]
f
energy
–361
Na
the
Figure 6.6.2
by
calculating
the
sum
of
Br(s)
atomisation
energies,
ionisation
A Born–Haber cycle to determine the lattice energy of sodium bromide. All
energies
and
electron
afnities
–1
gures are in kJ mol
Ø
rst
(ΔH
Ø
). Then
apply
ΔH
1
Ø
ΔH
ΔH
1
From
the
enthalpy
change
Ø
see
that:
Ø
=
ΔH
values
ΔH
1
shown
on
Ø
[Na]
+
ΔH
at
the
Born–Haber
+
ΔH
we
[Na]
+
ΔH
i1
+
107
+
112
+
494
make
+
(–325)
kJ
mol
ΔH
Ø
=
ΔH
latt
likely
to
f
mistakes
confused
by
with
too
the
signs
many
or
enthalpy
–1
=
+388
kJ
mol
changes
1
Ø
less
[Br]
–1
Ø
=
are
ea1
be
ΔH
. You
can
Ø
Ø
[Br]
at
cycle
=
latt
Ø
–
this
way.
Ø
−
ΔH
f
1
–1
=
361
−
(+388)
=
−749 kJ mol
Key points

Enthalpy
change
of
atomisation,
ΔH
,
is
the
enthalpy
change
when
1
mole
at
of

gaseous
Electron
atoms
afnity,
is formed from
ΔH
is
the
the
element.
enthalpy
change
when
1
mole
of
electrons
is
ea
added

A
to
1
mole
Born–Haber
changes
ΔH
,
latt
of
gaseous
cycle
ΔH ,
f
is
an
ΔH
,
at
atoms.
enthalpy
ΔH
and
cycle
which
includes
the
enthalpy
ΔH
ea
71
6.7
More
about
Learning outcomes
lattce
The
energy
Born–Haber
cycle for
magnesium oxide
+
Magnesium
On
completion
of
this
section,
2
should
be
able
.
So
understand
how
Born–Haber
with

and
the
radius
to
cycles
multiple
explain
an
ion
with
a
2
when
constructing
a
charge.
Born–Haber
The
cycle
oxide
for
ion
has
magnesium
a
charge
oxide
of
we
to:
have

forms
you
involving
the
take
into
account:
of
ionic

the
rst
and
the

the
rst
and
second
second
ionisation
energies
of
magnesium
and
ions
charges
effect
on
to
construct
charge
magnitude
of
The
enthalpy
magnesium
cycle
oxide
electron
and
are
afnities
Born–Haber
shown
below.
of
cycle
The
oxygen.
to
calculate
relevant
the
lattice
enthalpy
energy
changes
(all
of
in
−1
lattice
)
kJ mol
energy.
are:
Ø
ΔH
Ø
[MgO]
=
−602,
ΔH
f
=
+148,
ΔH
=
+736,
ΔH
i1
[O]
=
+249,
at
Ø
Ø
[Mg]
ΔH
Ø
[Mg]
at
Ø
[Mg]
=
+1450,
ΔH
i2
[O]
=
−141,
ie1
Ø
[O]
ΔH
=
+798
ea2
H
2+
Mg
latt
2–
(g)
+
O
Mg O(s)
(g)
H
several
H
steps
f
1
Mg(s)
+
O
2
Figure 6.7.1
(g)
2
Enthalpy cycle to determine the lattice energy of magnesium oxide
2+
Mg
2–
(g)
+
O
H
2+
Mg
(g)
ea2
(g)
+
O(g)
+
2e
2+
H
Mg
ea1
(g)
+
O
O(g)
+
e
(g)
+
e
+
H
Mg
[Mg]
i2
(g)
+
Exam tips
H
Mg(g)
[Mg]
+
O(g)
i1

Take
care
with
the
signs for
1
H
Mg(g)
[O]
+
at
the rst
and
second
2
O
(g)
2
electron
1
H
afnities. When
added
they
positive
may
give
a
Mg(s)
[Mg]
at
together
value.
H
+
2
O
(g)
2
[MgO]
f
MgO(s)

Remember
that
an
arrow,
↓,
Figure 6.7.2
pointing
energy
downwards
is
given
out
means
and
an
arrow,
From
↑,
pointing
upwards
Born–Haber cycle to determine the lattice energy of magnesium oxide
that
means
the
enthalpy
Ø
energy
is
absorbed. Take
change
values
given
above
we
can
see
that:
that
care
to
ΔH
Ø
=
ΔH
1
Ø
[Mg] + ΔH
at
Ø
[O] + ΔH
at
[Mg]
Ø
+ ΔH
i1
[Mg]
Ø
+ ΔH
i2
Ø
[O]
+ ΔH
ea1
[O]
ea2
Ø
draw
these
correctly
if
there
is
a
ΔH
=
(+148)
+
(+249)
=
+3240 kJ mol
+
(+736)
+
(+1450)
1
−1
second
electron
afnity
involved.
Ø
ΔH
Ø
=
ΔH
=
–602
latt
Ø
−
f
ΔH
1
−1
72
−
(+3240)
=
–3842 kJ mol
+
(−141)
+
(+798)
Chapter
Born–Haber
cycles for AlCl
,
Na
3
O
and Al
2
6
Energetics
O
2
3
3+
Aluminium
chloride,
AlCl
,
has
an
Al
ion
and
3
Cl
ions.
So
in
the
3
calculation:
Ø

ΔH
Ø
[Cl]
and
ΔH
at
[Cl]
must
be
multiplied
by
3.
ea1
3+

The
energy
required
to
form
Al
Ø
(g)
from
Al(g)
is
Ø
ΔH
+
+
Sodium( i)
oxide,
Na
O,
has
2
Ø
ΔH
i1
+
ΔH
i2
i3
2−
Na
ions
and
1
O
ion.
So
in
the
+
ΔH
2
calculation:
Ø

ΔH
Ø
[Na]
and
ΔH
at
[Na]
must
be
multiplied
by
2.
i
2−

The
energy
required
to
form
O
Ø
(g)
from
O(g)
is
Ø
ΔH
ea1
3+
Aluminium
oxide,
Al
O
2
,
has
2
ea2
2−
Al
ions
and
3
O
ions.
So
in
the
3
calculation:
Ø

Ø
[Al]
ΔH
must
be
multiplied
by
2
and
ΔH
at
[O
at
]
must
be

The
energy
required
to
form
Al
multiplied
Ø
(g)
from
Al(g)
is
ΔH
2−
The
energy
required
to
form
O
+
ΔH
from
O(g)
is
Lattice
affects the value of
energy
oppositely

The
the
arises
charged
size
of
the
opposite
smaller
less
from
ions.
ions.
charge
ions.
So,
exothermic
electrostatic
value
Small
than
for
as
the
The
any
the
lattice
ions
large
given
size
of
are
ions.
+
ΔH
of
attracted
The
=
effect
is
e.g.

=
The
seen
charge
we
to
on
compare
sized
ions,
charged
the
ions
the
we
ions
of
if
the
LiCl
ions.
the
=
that
very
the
is
lattice
to
ions
higher
energy
of
on
gets
increases:
−1
−780 kJ mol
cation
=
Ions
size
is
KCl
kept
=
−711 kJ mol
constant:
−1
−848 kJ mol
with
opposite
lattice
see
is
strongly
density
−1
−1031 kJ mol
strongly
more
,
Cl
cation
−1
LiF
between
−1
NaCl
−848 kJ mol
same
attraction
charge
−1
LiCl
ea2
on:
The
anion
the
i3
energies?
force
depends
ΔH
Ø
ΔH
ea1
What
Ø
+
i2
Ø
(g)
3.
Ø
i1

by
2
3+
a
high
charge
energies
of
LiF
charge
than
and
the
lattice
energy
much
higher
than
LiBr
of
LiF
ions
=
are
−803 kJ mol
attracted
with
a
low
MgO,
which
have
MgO
which
has
which
has
singly
more
charge.
If
similar
doubly
charged
ions:
Ø
−1
[LiF]
ΔH
=
Ø
−1049 kJ mol

The
of
way
the
[MgO]
=
−3923 kJ mol
latt
the
ions
however
,
−1
ΔH
latt
ions
also
than
are
ar ranged
affects
ionic
the
in
value
charge
and
of
the
the
ionic
lattice.
lattice
energy.
The
It
arrangement
has
less
effect,
size.
Key points

When
constructing
successive
Born–Haber
values for
ΔH
and
i

The
value
lattice
of
lattice
energy
is
energy
more
cycles
ΔH
involving
must
be
used
ions
with
when
multiple
charges,
appropriate.
ea
depends
exothermic
if
on
the
the
size
ion
is
and
charge
smaller
and
of
the
has
a
ions. The
higher
charge.
73
Exam-style
Answers to
all
exam-style questions
can
questons
be found on the
–
Module
1
accompanying CD
3
10
Multiple-choice questions
of
25.0 cm
aqueous
ron(ii)
3
20.5 cm
1
What
s
the
relate
atomc
mass
(A )
of
sulphate
requred
–3
of
0.02 mol dm
potassum
manganate(vii)
bromne,
r
for
complete
reacton. The
onc
equaton for
the
79
gen
that
the
relate
abundance
of
Br
s
50.5%
reacton
s:
81
and
Br
s
49.50%?
2+
5Fe
+
–
(aq)
+
MnO
(aq)
+
8H
3+
(aq)
→
5Fe
(aq)
+
4
A
(79
×
0.495)
–
(81
×
0.505)
2+
Mn
(aq)
+
4H
O(l)
2
B
(79
×
0.495)
+
(81
×
0.505)
–1
What
C
(79
×
0.505)
–
D
(79
×
0.505)
+
(81
×
s
concentraton
(g mol
)
of
the
soluton
of
0.495)
ron(ii)
(81
the
×
sulphate?
0.495)
(A
alues:
Fe
=
56; S
=
32; O
=
16)
r
2
Whch
of
the followng
partcles
wll
be
deflected
to
25.0
1
_____
A
the
greatest
degree
by
an
0.02
__
×
×
20.5
A
Alpha
B
Protons
partcles
C
Electrons
D
Neutrons
152
×
152
5
20.5
1
_____
B
0.02
__
×
×
25.0
5
20.5
52
X
3
×
electrc eld?
s
a
radosotope. Whch
of
_____
the followng
C
26
0.02
×
represents
the
resultng
undergoes
beta
nuclde, Y,
when
×
5
×
152
×
5
×
152
25.0
52
X
26
25.0
52
D
52
Y
A
48
B
Y
27
4
_____
decay?
C
Whch
of
s
the
20.5
D
Y
24
the followng
×
56
Y
28
0.02
28
electronc
conguraton
Structured questions
for
the
copper(i)
A
[Ar]
10
1
[Ar]
9
4s
3d
9
C
on?
3d
B
[Ar]
3d
1
2
11
4s
a
4s
D
[Ar]
Descrbe
what
Whch
of
the followng
equatons
represents
onsaton
energy
of
A
A
Na(g)
→
Na
B
Na(g)
→
Na
+
+
e
+
2e
C
Na
D
Na
2+
(g)
→
Na
–
(g)
loss
of
potassum
+
has
of
an
be
used
+
2+
(g) →
Na
to nd
the
concentraton
the followng e
of
e
peroxde. The
two
half
equatons
are
–
(g) + 2e
–
+
+
8H
–
+
5e
2+
→
Mn
+
4H
H
successe
O
2
+
element
to
manganate(vii)
4
An
gan
[2]
soluton
MnO
6
or
number
–
(g)
hydrogen
+
of
sodum?
–
(g)
terms
oxdaton
agent.
standard
can
+
n
n
the
b
second
change
3d
oxdsng
5
happens
electrons AND
10
O
2
→
O
2
+
–
2H
+
2e
2
–1
onsaton
energes
measured
n
kJ mol
:
i
Usng
the
1st
736
4th
10 500
balanced
2nd
1450
5th
13 600
between
3rd
7740
group
of
the
Perodc Table
s
the
By
to
belong
V
B
Whch
hae
of
the
IV
C
the followng
largest
8
KF
B
Whch
of
lattce
III
haldes
a
and
[2]
to
the
changes
n
numbers
of
the
elements,
show
that
D
of
manganate(vii)
s
the
oxdsng
II
potassum
would
and
hydrogen
peroxde
s
the
reducng
agent.
[4]
energy?
KCl
C
the followng
KBr
descrbes
D
the
shape
KI
of
When
aqueous
soluton
of
the
soluton
was
bromne
was
potassum
added
halde,
to
KX
,
an
a
aqueous
brown
obsered.
the
2–
carbonate
wrte
reacton
element
c
A
the
to?
agent
7
aboe,
show
mangangate(vii)
reference
potassum
A
to
peroxde.
makng
oxdaton
lkely
equatons
potassum
hydrogen
ii
What
half
equaton
i
Wrte
the formula
ii
Usng
your
of
the
halde
on, X.
[1]
?
on, CO
3
A
octahedral
C
pyramdal
B
trgonal
D
tetrahedral
balanced
planar
aqueous
9
What
s
the
molar
mass
of
0.07 g
of
a
gas
answer
n
part
equaton for
bromne
and
the
i
aboe,
wrte
reacton
a
between
KX.
[2]
whch
iii
Whch
of
the
elements,
or X
Br
2
,
has
the
2
3
occupes
a olume
of
120 cm
at
r.t.p.?
greater
oxdsng
ablty?
Explan
your
answer
3
(1
mole
of
gas
occupes
a olume
of
24 dm
at
r.t.p.)
usng
A
74
14
B
16
C
28
D
32
an
approprate
half
equaton.
[4]
Module
12
a
State THREE
t
b
relates
The
knetc
deal
i
to
assumptons
theory
howeer
State
the
ii
a
does
the
knetc
deal
theory
an
all
exst
under
gases
n
whch
behaour
of
t
71.
11%
are
and
aganst
gas
P
on
the
Use
Why
b
i
the
deal
of
gas
moles
equaton
of
a
gas
to
excess
of C,
oxygen
2.22%
of
H
and
used?
[1]
‘emprcal formula’
and
‘molecular
If
the
molar
mass
ts
of
oxalc
acd
emprcal formula
s
,
90 g mol
and
molecular
formula.
the
c
The
[6]
concentraton
ttratng
the
occupes
[2]
calculate
same
calculate
whch
s
Dene
ii
[2]
number
26.67%
–1
why
2
i
contan
questions
formula’.
explan
showng
and CO
to
Exam-style
of O.
a
gases
axes.
c
s found
[4]
PV
deal
as
[3]
realty.
occur.
graph
cures for
that
not
condtons
deatons
Sketch
assumes
ths
deate from
the
of
gases.
1
t
wth
a
manganate(vii)
a
2
moles
of
of
oxalc
standard
acd
soluton. They
potassum
can
soluton
be found
of
react
n
manganate(vii)
by
potassum
a
rato
to
5
of
moles
3
olume
of
R
=
at
0.072 dm
–1
8.31 JK
–48 °C
and
3.4 atm.
of
oxalc
acd.
In
one
such
ttraton
t
was found
–1
3
mol
[5]
that
t
requred
an
aerage olume
of
21.5 cm
of
–3
ii
If
the
mass
of
the
gas
n
part

aboe
s
0.02 mol dm
0.94 g,
potassum
manganate(vii)
to
react
3
calculate
ts
relate
molecular
mass.
wth
[1]
i
13
a
State
Hess’s
law.
25.0 cm
Descrbe
The
enthalpes
and
water
of formaton
of
carbon
what
you
acd.
of
the
would
obsere
at
the
end
reacton.
[1]
doxde
ii
–1
are
oxalc
[1]
pont
b
of
Calculate
the
number
of
moles
of
oxalc
acd
–1
and
–393 kJ mol
–286 kJ mol
used
n
the
ttraton.
[3]
respectely.
–3
iii
i
Use
ths
nformaton
to
draw
an
Calculate
of
cycle
dagram for
the
standard
enthalpy
of
ethane
H
(C
2
the
oxalc
Calculate
the
standard
concentraton
)
(mol dm
[1]
).
Usng
the
molar
mass
gen
n
part

aboe,
[2]
6
calculate
ii
molar
acd.
of
iv
formaton
the
energy
enthalpy
the
mass
concentraton
of
the
oxalc
of formaton
acd.
of
c
A
ethane.
student
was
experment
of
soluton
n
of
requred
the
lab
to
to
carry
out
determne
ammonum
15
an
the
ntrate. The
enthalpy
The rst
3
are
gen
a
polystyrene
cup,
a
onsaton
energes
of
the
elements
n
Perod
as follows:
student
Symbol
was
[1]
[2]
measurng
of
Na
Mg
Al
S
P
S
Cl
Ar
496
738
578
787
1012
1000
1251
1520
cylnder
element
and
i
a
thermometer.
Why
used
was
n
a
polystyrene
the
cup
and
not
a
beaker
experment?
[1]
Frst
onsaton
energy/
ii
State
one
assumpton
that
was
made
when
–1
kJ mol
performng
the
calculatons for
ths
reacton.
iii
Gen
the
[1]
that
water
the
was
temperature
dagram,
ntal
temperature
27
.3 °C
15.
1 °C,
clearly
and
draw
showng
the nal
an
was
endothermc
or
Dene
ii
Usng
an
prole
Usng
the
nformaton
n
exothermc.
part
iii
term
‘rst
magnesum
equaton
energy
of
onsaton
(Mg)
llustratng
the
as
an
energy’.
example,
the rst
[1]
wrte
onsaton
element.
[1]
i
Usng
the
table,
state
what
the
general
trend
[4]
n
iv
the
the
b
reacton
i
was
energy
whether
a
of
aboe,
the alue
we
go
of
across
the rst
the
onsaton
perod
and
energy
account for
s
as
ths
–1
calculate
when
the
16.0 g
enthalpy
of
of
soluton
ammonum
ntrate
n
kJ mol
general
dssoles
trend.
[3]
n
ii
Study
the
table
and
account for
any
3
100 cm
of
[4]
water.
rregulartes
(A
values:
N
=
14;
H
=
1; O
=16; C
=
12.
r
The
enthalpes
onsaton
of
n
the
energes
general
across
trend
the
of
the rst
perod.
[5]
–1
formaton
of
carbon
doxde
and
water
are
–393 kJ mol
c
i
Sketch
a
graph
to
show
the rst e
–1
and
–286 kJ mol
respectely. The
standard
successe
combuston
onsaton
energes
of
magnesum
–1
of
ethane
s
–1560 kJ mol
).
aganst
ii
14
Ethanedoc
acd,
commonly
called
oxalc
How
the
do
an
organc
acd
that
occurs
naturally
n
these alues
help
anmals
and
oxygen. When
s
made
oxalc
of
acd
s
carbon,
burnt
hydrogen
n
electrons
of
remoed.
successe
[2]
to
dentfy
the
group
onsaton
to
whch
plants
magnesum
and
of
acd
energes
s
number
excess
belongs.
[3]
and
oxygen
75
7
Rates
7
.
1
Following
of
Learning outcomes
reaction
the
Rate of
A
On
completion
of
this
course
section,
study
be
able
design
reaction
reaction
of
the
rate
kinetics,
of
we
reaction
carry
is
out
called
reaction
experiments
(and
carry
out)
kinetics.
measure
In
order
the
rate
to
at
study
which
are
used
up
or
products
are
made.
suitable
change
experiments for
affect
in
concentration
studying factors
(or
which
to
to:
reactants

a
you
reaction
should
of
reaction
amount)
of
reactants
or
products
rate.
rate
of
reaction
=
time
Rate
of
reaction
−3
is
in
this
change
concentration
terms
as
−1
Following the
method
nature
expressed
for
s
mol dm
The
usually
taken
of
used
the
continuous
course of
to
follow
products
methods
and
and
the
a
reaction
progress
reactants.
sampling
of
The
a
reaction
methods
depends
fall
into
on
two
the
groups:
methods.
Continuous methods
A
particular
period

gas
syringe
of
physical
property
of
the
reaction
mixture
is
monitored
over
a
time.
Measuring volume of gas given off
The
volume
time
of
intervals
a
gas
given
using
a
gas
off
in
a
syringe
reaction
or
can
upturned
be
measured
measuring
at
various
cylinder
initially
delivery
full
of
water
.
For
example
in
the
reaction
of
calcium
carbonate
with
tube
rubber
bung
hydrochloric
various
acid,
times
as
the
the
(s)
CaCO
volume
reaction
+
of
carbon
2HCl(aq)
→
CaCl
3
marble
hydrochloric
dioxide
can
be
measured
at
proceeds.
(aq)
+
CO
2
(g)
+
H
2
O(l)
2
acid
chips

Figure 7.1.1
Colorimetry
The reaction of calcium
This
is
suitable
for
reactions
in
which
there
is
a
change
in
the
intensity
carbonate with hydrochloric acid can be
of
colour
during
the
reaction.
For
example
in
the
reaction
between
iodine
found by measuring the volume of carbon
and
dioxide given off at various times
propanone
in
aqueous
COCH
CH
3
The
‘cell’
light
iodine
is
(aq)
acidic
+
I
3
brown
solution:
(aq)
→
CH
2
in
colour
COCH
3
but
the
I(aq)
+
HI(aq)
2
other
reactants
and
products
are
sensitive
colourless.
As
the
reaction
proceeds,
the
intensity
of
the
brown
colour
cell
fades.
by
A
or
We
can
use
transmitted
lter
is
chosen
a
colorimeter
through
so
that
the
the
to
measure
reaction
correct
the
amount
of
light
absorbed
mixture.
wavelength
of
light
falls
on
the
meter
reaction
light
filter
source
Figure 7.1.2
76
mixture
reaction
mixture,
mixture
light-sensitive
A colorimeter
proceeds
the
in
more
more
cell.
the
light
As
light
is
cell.
is
the
The
more
absorbed
reaction
transmitted
intense
and
less
between
through
is
the
of
transmitted
iodine
to
colour
the
and
the
reaction
through
to
propanone
light-sensitive
cell.
the
Chapter

If
7
Rates
of
reaction
Changes in electrical conductivity
the
total
reaction,
number
we
can
conductivity .
mixture.
mixture
We
This
of
use
uses
occurs.
ions
follow
a
in
the
a
reaction
reaction
by
conductivity
alternating
Conductivity
meter
current
is
mixture
changes
measuring
so
to
monitor
that
measured
in
no
during
changes
in
the
reaction
electrolysis
siemens,
a
electrical
of
the
S.
Did you know?
dipping
conductivity
Hydrogen
electrode
ions
have
ions
very
conductivities
ions. This
measured
meter
present
hydroxide
electrical
compared
means
conductivity
reaction
and
high
changes
even
as
that
if
can
there
reactants
with
other
electrical
be
are
or
other
ions
products.
+
Hydrogen
mixture
ions
are
really
H
O
ions
3
Figure 7.1.3
(see
A conductivity electrode connected to a meter
Section
appear
An
example
of
a
reaction
that
can
be
followed
by
this
method
by
is:
to
‘chain
thought
9.
1).
move
Hydrogen
extremely
conduction’
to
be
–
ions
rapidly
they
transmitted
are
rapidly
+
(CH
)
3
CBr(l)
+
H
3
O(l)
→
(CH
2
)
3
COH(aq)
+
H
(aq)
+
Br
(aq)
3
+
between
H
O
ions
and
water
3
As
the
reaction
increases

as
proceeds,
hydrogen
the
ions
electrical
and
conductivity
bromide
ions
are
of
the
solution
molecules.
formed.
Other methods
There
may
be
a
change
in
pH
as
some
reactions
proceed.
This
may
be
+
ions
due
to
H
can
be
use
Some
to
or
OH
ions
monitor
these
reactions
can
reaction
mixture
not
accurate
very
as
be
a
being
type
followed
gas
unless
is
produced
of
by
gas
consumed.
A
pH
meter
reactions.
measuring
released.
the
or
has
This
a
the
can
be
relatively
decrease
done
high
by
in
mass
of
weighing,
molar
the
but
is
mass.
The sampling method
Small
and
samples
analysed,
after
the
are
samples
proceeding,
removed
usually
for

It
is
cooled

A
chemical
are
from
using
a
taken,
the
reaction
specic
they
mixture
chemical
must
be
at
reaction.
quenched
to
particular
times
Immediately
stop
the
reaction
example:
rapidly
so
that
the
reaction
slows
down
signicantly.
Key points
does
not
can
affect
be
added
the
which
substance
to
prevents
be
the
reaction
continuing
but
analysed.


A
catalyst
can
be
The
course
of
be followed
An
example
is
the
reaction
of
propanone
with
iodine
in
acidic
particular
solution:
or
CH
COCH
3
(aq)
+
I
3
(aq)
→
CH
2
COCH
3
sample
is
taken
at
reaction
can
I(aq)
+
a
particular
time
and
by
measuring
property
of
a
a
reactant
product.
HI(aq)
2

A
a
removed.
added
to
sodium
Suitable
methods
of following
a
carbonate
reaction
directly
changes
in
are
measuring
+
solution.
reaction.
Sodium
The
carbonate
reaction
reacts
mixture
with
can
the
then
be
H
ions
titrated
which
with
catalyse
the
to
determine
the
concentration
of
i
volume
of
gas,
ii
sodium
electrical
thiosulphate
the
conductivity,
iii
colour
iodine.
intensity

Some
by
at
or
iv
pH.
reactions
taking
a
various
reaction
can
be followed
sample for
time
analysis
intervals
as
the
proceeds.
77
7
.2
Calculating
Learning outcomes
rate
Reaction
For
On
completion
of
this
section,
many
be
able
describe
with
rate
and
reactions,
progress of
reaction
rate
a
changes
reaction
magnesium
reacts
with
dilute
as
the
reaction
hydrochloric
acid,
how
reaction
rate
varies
the
hydrochloric
acid
falls
and
the
volume
of
hydrogen
gas
given
off
time
+
2HCl(aq)
→
(aq)
MgCl
+
H
2
know
rate
When
concentration
rises:
Mg(s)

proceeds.
the
to:
of

reaction
you
excess
should
of
how
to
calculate
graphically.
(g)
2
reaction
We
use
molar
square
brackets
concentration
of
to
indicate
molar
hydrochloric
concentration.
So
[HCl]
means
acid.
−3
[HCl]
mol dm
0.8
Time/s
0.6
0
0.4
15
0.2
43
0.
1
100
180
b
a
H
2
f o emulov
[HCl]
0
0
0
Figure 7.2.1
time
0
(s)
time
(s)
The progress of the reaction beween Mg and HCl can be followed: a by
measuring the decrease in concentration of HCl or; b the increase in volume of H
2
−3
We
can
see
that
[HCl]
decreases
from
0.8
to
0.6 mol dm
in
the
rst
15
seconds.
∆
0.8
[HCl]
_______
We
can
write
this
∆
The
graph
decreasing
measure
i.e.
at
point
is
the
70 s.
on
a
curve
with
the
very
can
curve
0.6
−3
If
we
gets
small,
0.013 mol dm
−1
s
15
shallower
make
calculate
(see
=
time
which
time.
rate
We
–
_________
=
as
we
time
obtain
this
Figure
the
rate
a
by
with
time.
interval
rate
at
a
drawing
So
over
the
particular
a
rate
which
tangent
point
at
a
7.2.2).
0.6
md lom
3–

0.4
]lCH[
0.2

0
20
40
60
80
100
120
time
Figure 7.2.2
78
(s)
140
160
180
in
time
particular
0.8
0
is
we
200
Chapter
T
o
calculate

Select

Draw
(α)

a
a
on
straight
the
the
Calculate
of
reaction
the
line
line
at
at
a
particular
corresponding
this
point
(the
to
Rates
of
reaction
time:
a
particular
tangent)
so
that
time.
the
two
angles
same.
tangent
convenient

rate
point
look
Extend
the
7
line
to
meet
the
axes
of
the
graph
This
is
the
rate
(or
at
values).
the
gradient
of
the
graph.
of
reaction.
0.54
_____
In
the
example
−3
above,
−3
10
mol dm
The
negative
the
rate
of
reaction
at
70 s
is
=
−3.4
×
157
−1
s
[HCl]
sign
shows
that
the
rate
is
decreasing
with
time.
0
Initial
rate of
reaction
0
The
of
initial
the
reaction
is
calculated
by
drawing
a
tangent
at
the
start
can
nd
by
Looking
at
proceeds,
rate
rate during
how
reaction
drawing
the
the
at
a
(s)
Drawing a tangent to show
of
in
varies
at
with
several
Figure
reaction
particular
reaction
rate
tangents
gradients
rate
a
hydrochloric
Figure 7.2.3
time
initial rate of reaction
in
reactant
each
of
curve.
Changes
We
rate
7.2.4,
decreases.
time
the
concentration
different
we
In
corresponds
can
points
see
Figure
to
a
on
that
as
7.2.4(a)
particular
of
the
a
the
we
specic
graph
.
reaction
can
see
that
concentration
of
acid.
a
b
0.8
V
final
H
2
]lCH[
f o emulov
0.6
0.4
0.2
V
t
t
time
Figure 7.2.4
time
a The rate of change of [HCl] and; b rate of volume change of H
both
2
decrease with time
In
rate
experiments
particular
results
of
products
example
reactant,
experiments
to
in
show
are
interested
than
in
how
product,
measuring
what
Figure
of
V
we
rather
happens
increase
to
the
the
varies
in
concentration
with
time.
We
concentration
concentration
of
of
can
or
a
use
the
volumes
reactants.
of
For
7.2.4(b):
hydrogen
–
V
nal
of
hydrogen
is
proportional
to
[HCl]
at
time
t
t
Key points

For

To
most
reactions,
calculate
graph,
rate
tangents
the
specic
the
tangent.
of
are
rate
of
reaction
reaction
drawn
times). The
at
at
rate
decreases
specic
specic
of
the
times from
points
reaction
as
is
on
given
a
the
by
reaction
proceeds.
concentration–time
graph
the
(corresponding
gradient
(slope)
to
of
79
7
.3
Collision
Learning outcomes
On
completion
should
be
able
of
this
theory
explain
rate
of
reaction
Collision theory
section,
In
order
to
break
to
react,
the
reactant
particles
must
collide
with
enough
energy
you
specic
bonds
in
their
molecules.
They
must
also
collide
with
to:
the

and
the
effect
concentration,
of
correct
react
particle
size
on
rate
of
together
come
so
into
that
the
parts
of
each
molecule
that
are
going
to
contact.
and
The
temperature
orientation
minimum
energy
that
colliding
particles
must
have
in
order
to
react
reaction
is
called
the
activation
energy ,
symbol
.
E
At
this
minimum
energy,
the
a

explain
the
effect
of
catalysts,
molecules
including
enzymes,
on
the
rate
have
a
particular
conguration
called
the
transition
state
of
reactions
The

describe
the
role
of
enzymes
and
concentration,
pressure
and
surface
area
in
on
industrial
effect of
rate of
reaction
biological
Concentration
processes.
The
more
concentrated
(molecules
or
ions)
in
a
a
solution,
given
the
volume.
greater
is
the
number
There
are
more
increases.
So,
rate
of
chances
particles
of
collisions
a
occurring.
transition
state
collision
frequency
of
reaction
increases.
Pressure
ygrene
This
E
The
is
relevant
for
gaseous
systems.
The
higher
the
pressure,
the
greater
a
is
the
number
of
molecules
in
a
given
volume.
There
are
more
chances
of
reactants
collisions
products
reaction
pathway
reaction
transition
The
collision
frequency
increases.
So,
rate
of
increases.
Size of
The
b
occurring.
solid
greater
particles
the
total
surface
area
of
a
solid,
the
greater
is
the
number
of
state
particles
gas
that
are
molecules.
If
exposed
we
to
break
collide
a
large
with
lump
molecules
of
solid
in
into
a
solution
smaller
or
pieces
with
we
ygrene
E
a
expose
products
react
a
larger
faster
surface
than
larger
area.
So,
if
the
mass
is
the
same,
smaller
particles
particles.
reactants
Effect of temperature on
The
reaction
curve
particles
in
any
gas
or
solution
are
moving
at
different
speeds.
The
against
the
Activation energy for: a an
exothermic reaction; b an endothermic
reaction
reaction
pathway
The
Figure 7.3.1
Boltzmann distribution
rate of
energy
of
fraction
a
of
particle
depends
particles
distribution
that
on
have
a
its
speed.
particular
A
graph
energy
of
is
energy
called
a
Boltzmann
cur ve
average
energy
)E(
fo
with
of
particles
sufficient
energy to
htiw
noitcarf
ygrene
selcitrap
number
react
E
a
Figure 7.3.2
Boltzmann
distribution curve. The area under
the graph represents the total
energy
The
shaded
energy
energy.
80
to
area
react
shows
when
number of particles
(E)
the
they
proportion
collide,
i.e.
of
molecules
equal
or
that
greater
have
than
enough
the
activation
Chapter
The
effect of temperature on
Increasing
slightly.
the
An
temperature
increase
molecules.
The
much.
the
But
activation
in
Figure
the
reaction
rate
kinetic
proportion
7.3.3).
of
increases
If
there
energy,
changes
the
temperature
average
energy
activation
in
rate of
are
there
more
will
shape
with
(see
the
more
distribution
energy
does
energy
the
particles
be
of
the
however
,
molecules
markedly
Rates
of
reaction
reaction
increases
energy,
7
not
with
to
area
energy
all
curve
the
increase
equal
shaded
successful
of
or
above
under
equal
to
collisions
very
the
or
and
the
graph
above
the
increases.
T
>
T
2
1
T
1
There
are
)E(
a
E
T
at
a
T
more
greater
than
particles
energy than
at
T
2
1
2
fo
Did you know?
E
a
htiw
noitcarf
ygrene
selcitrap
with
Most
Figure 7.3.3
Boltzmann
distribution curve at a lower
1
(E)
enzymes
they
above
are
left
become
at
40 °C for
inactive
temperatures
too
long. The
and a higher
temperature, T
energy
if
weak
attractive forces
holding
the
temperature T
2
protein
are
chain
in
overcome
its
and
correct
the
position
protein
Catalysis
changes
Catalysts
path
(a
speed
up
different
activation
energy
successfully
the
rate
of
mechanism)
and
leads
to
a
therefore
reaction
with
by
lower
greater
providing
activation
proportion
of
a
different
energy.
the
reaction
particles
longer
to
Lower
colliding
reacting.
b
enzymes
are
=
uncatalysed
a
cat
=
catalysed
a
ygrene
a
fo
ygrene
E
E
cat
a
reaction
E
however,
tangled
E

pathway
energy
up
which
is
Increasing
size)
are
protein
substrate
they
catalysts.
They
have
and
bound
catalyse.
They
an
catalyses
to
the
This
catalyse
active
the
site
most
part
reaction.
enzyme).
is
on
because
(A
Enzymes
the
of
of
the
reactions
their
surface
substrate
are
very
substrate
ts
is
a

rather
in
the
in
the
like
a
key
ts
a
lock.
This
allows
the
bonds
which
are
and
formed
to
be
in
their
in
correct
reaction
the frequency
of
particle
temperature
rate
because
particles
have
than
activation
the
rate
of
increases.
Increasing
increases
more
energies
greater
energy.
active
to
The
Boltzmann
distribution
be
curve
broken
chain
others.
decrease
increases
reaction
reactant
specic
into
in
which

site
with
rapidly
protein
concentration
(or
collisions
loosely
reactions
protein
the
a A catalyst lowers the activation energy of a reaction; b A greater number
organisms.
the
about
E
Enzymes
bonds
the
many
at
Key points
a
because
living
the
to function
because
no
cat
of particles have energies greater than or equal to the activation energy
Enzymes
In
substrate,
temperatures. Above
reactant
Figure 7.3.4
able
is
conformation
reaction.
excess
site
a
htiw
noitcnarf
E
gets
active
correct
the
of
denatures
E
E
energy
the
presence
60 °C,
selcitrap
activation
in
catalyse
higher
a
shape. The
shows
the
relative
number
positions.
of
particles
having
particular
energies.
active
site
protein

Catalysts
increase
the
rate
substrate
of
Figure 7.3.5
reaction
by
providing
an
The substrate ts the active site
alternative
reaction
with
activation
pathway
of an enzyme similar to a key tting a lock
Many
the
enzymes
are
manufacture
citric
of
important
drugs
and
in
processes
in
the
such
as
manufacture
baking,
of
brewing
chemicals
and
such
as

lower
Enzymes
catalyse
are
proteins
specic
energy.
which
reactions.
acid.
81
7
.4
Rate
equations
Learning outcomes
Introducing
In
On
completion
of
this
section,
Section
be
able
understand
reaction’,
‘rate

This
saw
how
corresponded
to
to
a
calculate
particular
the
rate
of
reaction
concentration
of
at
a
particular
reactant.
If
we
to:
plot

we
equations
you
time.
should
7.2,
rate
the
‘rate
terms
‘order
equation’
of
and
a
graph
of
rate
of
reaction
of
Mg
with
acid
against
[HCl],
we
get
a
2
curve.
line
But
if
we
showing
a
plot
the
rate
proportional
of
reaction
against
we
[HCl]
get
a
straight
relationship.
constant’
interpret
concentration–time
b
md lom(
3–
reactions.
s
1–
order
md lom(
3–
second
)
graphs for
s
and
a
1–
zero, rst
and
)
concentration–rate
etar
etar
2
[HCl]
[HCl]
2
Figure 7.4.1
Rate of reaction of HCl with Mg plotted: a against [HCl]; b against [HCl]
2
W
e
can
write
this
proportionality
The
proportionality
The
overall
The
power
the
rate
expression
to
case,
hydrochloric
How do
The
which
equation
particular
order
equation.
example
constant
(in
we
is
the
called
called
the
case
that
2)
of
reaction
can
for
rate
the
is
only
the
cannot
be
of
a
conduct
By
be
found
(g)
is
order
second
reactant
of
the
order
rate
k[HCl]
is
raised
reaction.
with
In
respect
to
determined
by
at
[H
]
from
conducting
a
the
stoichiometric
series
of
experiments.
For
+
2NO(g)
→
2H
a
O(g)
+
N
varying
the
(g)
2
concentration
of
each
of
the
time.
keeping
[NO]
constant,
experiments
show
that
constant,
experiments
show
that
2
rate
=
].
k[H
2
Did you know?
By
varying
[NO]
keeping
[H
]
2
2
rate
The
rate
equations for
reactions
or
ions
may
not
include
present
in
=
k[NO]
some
compounds
the
Combining
these
two
rate
equations
we
get
the
overall
rate
equation:
chemical
2
rate
=
k[H
]
[NO]
2
equation.
order
of
In
some
reaction
instances,
can
the
be fractional.
We
say
82
that
the
reaction
is
rst
order
with
respect
to
H
,
2
with
respect
in
this
equation?
2
experiments
one
varying
=
constant
particular
the
2
reactants
reaction
reaction:
2H
We
of
equation
called
reaction
rate
rate
acid.
we deduce the
It
the
concentration
this
say
is
mathematically:
to
NO
and
third
order
overall.
second
order
Chapter
How
We
can
can
we deduce the order of
deduce
the
order
of
reaction

plotting
a
graph
of
reaction

plotting
a
graph
of
concentration
The
shape
of
these
graphs
is
rate
with
Rates
of
reaction
reaction?
respect
against
of
a
7
to
a
single
concentration
reactant
against
reactant
of
by:
reactant
time.
characteristic.
a
b
2nd
order
k[A]
rate
=
order
rate
=
k[A]
order
lom(
etar
lom(
1st
k
zero
md
order
3–
zero
)
md
3–
fo
A fo noitartnecnoc
=
s
1–
)
noitcaer
2
rate
2nd
order
(deep
curve
levelling out)
–3
concentration
Figure 7.4.2
of A
(mol
dm
)
time
(s)
a How concentration of reactant A affects reaction rate; b How
concentration of reactant A changes with time
Units of
The
For
units
the
k
of
k
are
reaction:
different
NO(g)
+
for
different
CO(g)
+
O
orders
(g)
→
of
NO
2
reaction.
(g)
+
CO
2
For
example:
(g)
2
2
the
rate
Step
1:
equation
is:
Rearrange
rate
the
=
rate
Exam tips
k[NO]
equation
in
terms
of
k:

is
easy
to
confuse
units
in
section. The following facts
rate
______
k
It
2
important:
–3
2:
Substitute
the
are
=
[NO]
Step
this
units
of
rate
and
concentration:

Units
of
rate
are
mol dm
–1
s
–1
–3
mol dm
(don’t forget
–1
the
s
).
s
___________________
=
–3
mol dm
–3
×
mol dm

Units
of
k for
zero
–3
are
Step
3:
Cancel
the
mol dm
order
reactions
–1
s
units:

–1
Units
of
k for rst
order
reactions
s
___________________
=
–1
–3
are
–3
×
s
mol dm
1
Step
4:
Rearrange
with
the
positive
index
rst:
_________

3
is the same as dm
–1
mol
–3
mol dm
3
=
dm
–1
mol
–1
s
Key points

Rate
of
reaction
is
related
to
concentration
of
reactants
by
the
rate
equation.
m

The
general form
constant,
orders

Order
[A]
with
of
of
and
[B]
respect
reaction
concentration
or
the
rate
are
to A
can
be
equation
is
concentrations
and
rate
of
=
k[A]
n
[B]
reactants
where
and
m
k
and
is
n
the
are
rate
the
B.
determined from
concentration
against
graphs
of
reaction
rate
against
time.
83
7
.5
Order
of
Learning outcomes
reaction from
Order of
For
On
completion
of
this
section,
a
zero
be
able
deduce
order
appropriate

interpret
graphs
of
reaction from
to
So
second
data
a
reaction,
given
the
a
reactant
initial
order
method
rate–concentration
from
obtained
the
when
decomposes
rate
reactions,
concentrations
concentration–time
and
order
initial
graph
of
rates
shows
a
concentration
constant
rate
against
of
time
decline
with
(see
Section
to:
7.4).

reaction from
data
you
respect
should
experimental
in
of
the
results
of
a
of
given
hydrogen
the
reaction
initial
is
rate
several
reagent.
of
an
can
by
the
be
calculated
experiments
Figure
peroxide
presence
given
at
using
7.5.1(a)
four
gradient.
by
and
tangent
different
starting
the
initial
catalyst.
rst
the
shows
different
enzyme
For
The
results
concentrations
initial
rates
are
graphs
then

perform
calculations
to nd
k
plotted
against
the
initial
concentrations
of
(Figure
7.5.1(b)).
This
shows
that
rate
=
O
k[H
2
rate
of
reaction
by

in
the
understand
rate
the
].
So
the
peroxide
reaction
is
rst
2
substituting
order
values
hydrogen
or
with
respect
to
hydrogen
peroxide.
equation
meaning
of
–3
mol
dm
s
1–
30
)
0.40
‘half-life’.
0.6
dm
–3
0.
12
mol
dm
10
3
mol
0.3
0.2
etar
0.06
0.4
fo
O fo emulov
2
noitcaer
mol
mc(
3
0.20
O
2
mc (
3
)
–3
20
0.5
dm
0.
1
0
0
0
50
100
150
time
200
0
0.2
[H
O
2
Figure 7.5.1
0.4
–3
(s)
a The initial rate of decomposition of H
O
2
O
b A plot of initial rate of decomposition of H
2
]
mol
dm
2
at different concentrations;
2
against [H
2
O
2
] shows the reaction to be
2
rst order
Deducing
We
can
a value for the
deduce
concentrations
from
the
acid
the
and
overall
initial
catalysed
rate
of
rates.
rate
constant
reaction
The
hydrolysis
from
table
of
limited
below
methyl
data
shows
data
showing
obtained
methanoate:
+
H
HCO
CH
2
+
H
3
O
–3
[HCO
CH
2
→
HCO
2
+
]/mol dm
[H
H
+
CH
2
OH
3
–3
–3
]/mol dm
Initial
rate/mol dm
–1
s
3
–3
0.50
1.00
0.54
×
1.00
1.00
1.
10
2.00
1.00
2.25
2.00
2.00
4.48
10
–3
×
10
–3
×
10
–3
×
10
+

Doubling
the
concentration
of
HCO
CH
2
doubles
So
rate
the
is
,
keeping
[H
]
constant,
3
rate.
rst
order
with
respect
to
CH
[HCO
2
].
3
+

Doubling
the
concentration
of
H
keeping
[HCO
CH
2
doubles
the
rate.
+
So
84
rate
is
rst
order
with
respect
to
[H
].
]
3
constant
also
Chapter
7
Rates
of
reaction
+
The
overall
rate
equation
is:
rate
=
k
[H
][HCO
CH
2
The
rate
constant
experimental
from
the
can
values
table
be
into
deduced
the
rate
by
substituting
equation.
].
3
any
T
aking
of
the
×
Half-life
Half-life
half
its
is
sets
line
of
of
values
above:
–3
0.54
the
rst
–3
=
10
k
(0.50)
×
and order of
the
time
original
taken
value.
(1.00)
So
k
=
1.08
×
10
3
–1
dm
mol
–1
s
reaction
for
the
Symbol:
concentration
of
a
reactant
to
fall
to
t
1/2
Figure
7.5.2
shows
decomposition
half-lives
are
of
several
successive
hydrogen
constant.
half-lives
peroxide.
This
is
Y
ou
can
characteristic
for
see
of
a
the
enzyme-catalysed
that
rst
successive
order
reaction.
0.4
0.3
H[
2
O
2
]
0.2
Did you know?
0.
1
The
half-life for
may
22s
23s
used
a rst
to nd
order
the rst
reaction
order
21s
rate
0
0
be
20
40
constant
using
the
relationship:
60
0.693
time
______
t
=
1/2
Figure 7.5.2
The successive half-lives for the decomposition of H
O
2
are constant (about
2
k
22 s). So the reaction is rst order with respect to hydrogen peroxide
For
zero
order
order
reactions
reactions
the
the
successive
successive
half-lives
half-lives
decrease.
second
]A[
]A[
t
t
time
Figure 7.5.3
For
increase.
time
a Successive half-lives for a zero order reaction decrease; b Successive
half-lives for a second order reaction increase
Key points

Order
of
reaction
experiments
the

reaction
The
rate
rate
ii
can
be found
determining
the
by
i
measuring
change
in
initial
rate
concentration
in
of
several
a
reactant
as
proceeds.
constant
can
be found
by
substituting
relevant
values
into
the
equation.

Half-life

In
a rst
is
the
order
time
taken for
reaction,
the
half-life
concentration
is
independent
of
of
a
reactant
the
to
halve.
concentration
of
reactant.
85
7
.6
Reaction
Learning outcomes
mechanisms
The
rate determining
Chemical
On
completion
of
this
section,
be
able
We
understand
the
term
‘rate
determining step’

deduce
call
this
the
in
reaction
organic
chemistry
mechanism .
Each
occur
of
in
these
a
number
steps
of
may
to:
involve

especially
you
steps.
should
reactions,
step
possible
mechanism
determining
reaction
mechanisms from
intermediates
reaction
step
which
is
are
called
controls
highly
the
the
rate
overall
reactive.
The
deter mining
rate
of
slowest
step .
reaction.
So
step
The
in
in
the
rate
the
reaction
sequence,
appropriate
Step
1
Step
2
Step
3
data.
Reactant
A
+
→
Reactant
Intermediate
X
→
slow
The
overall
reactants
The
rate
to
in
Example
Aqueous
the
of
reaction
that
rate
1: The
sodium
reaction
order
is
in
the
suggests
step.
So
hydroxide
)
on
Step
rate
1,
the
conversion
3
Chemists
First
CBr
+
OH
with
to
→
respect
OH
are
usually
(CH
)
ions.
to
So
COH
+
Br
3
2-bromo -2-methylpropane
rate
=
k[(CH
)
ion
not
is
not
react
involved
directly
in
with
the
OH
and
CBr].
3
rate
determining
ions.
3
think
step:
OH
does
those
2-bromo -2-methylpropane:
3
order
the
of
2-bromo-2-methylpropane
hydrolyses
CBr
equation
3
respect
that
)
(CH
C
fast
3
This
Product
step.
hydrolysis of
rst
with
depends
appear
3
zero
→
X.
determining
(CH
The
Y
fast
intermediate
substances
involved
Intermediate
B
that
the
ionisation
of
reaction
occurs
in
two
steps:
2-bromo -2-methylpropane
slow
+
(CH
)
3
CBr
→
(CH
3
)
3
C
+
Br
3
+
Second
step:
reaction
of
(CH
)
3
C
with
OH
3
fast
+
(CH
)
3
The
on
slow
the
step
rate
at
determines
which
(CH
rate
with
86
equation
the
because
intermediate.
+
OH
→ (CH
3
the
)
3
the
C
)
3
overall
CBr
rate
ionises.
of
COH
3
reaction.
Hydroxide
So
ions
the
do
rate
not
depends
appear
3
they
are
involved
in
the
rapid
ionic
reaction
in
Chapter
Example
2: The
Propanone
reacts
reaction of
with
iodine
propanone
in
the
with
presence
of
an
7
Rates
of
reaction
iodine
acid
catalyst.
The
+
catalyst
provides
ions
H
for
the
reaction.
+
H
CH
COCH
3
+
I
3
→
CH
2
COCH
3
I
+
HI
2
+
The
rate
equation
for
this
reaction
is:
rate
=
k[CH
COCH
3
A
proposed
mechanism
][H
]
3
is:
+
OH
O
Exam tips
fast
+
step
1:
CH
C CH
3
H
+
CH
3
3
3

intermediate A
You
need
learning
+
OH
not
any
worry
of
about
these
OH
mechanisms
at
present.
In
rate
slow
+
step
CH
2:
CH
3
3
C
CH
3
+
H
of
2
reaction
questions
you
will
be
intermediate A
given
about
OH
fast
steps
3
and
CH
4:
C
CH
3
+
the
can
see
that
this
mechanism
information
mechanism.

Remember
that
we
cannot
2
deduce
We
relevant
fast
I
2
any
is
consistent
with
the
rate
the
reaction
directly from
equation.
the
mechanism
rate
equation.
+
Intermediate
A
is
derived
from
the
CH
COCH
3
and
H
ions
which
react
3
The
best
we
can
do
is
to
see
if
the
+
together
to
form
it.
This
is
why
both
COCH
CH
3
rate
in
equation.
the
rate
Iodine
is
we
by
are
given
reference
propanone
contains
in
a
fast
H
appear
in
the
3
step
later
on,
so
does
not
appear
reaction
mechanism
with
rate
the
is
consistent
equation.
equation.
Predicting order of
If
involved
and
a
to
reaction
the
with
rate
mechanism,
determining
bromine
ions.
OH
reaction from
The
in
given
we
step.
alkaline
proposed
a
can
predict
For
the
example,
solution.
reaction
reaction
The
order
the
alkaline
mechanism
mechanism
of
reaction
reaction
of
solution
is:
slow
Step
1:
CH
COCH
3
+
OH
→
CH
3
COCH
3
+
H
2
O
2
fast
Step
2:
CH
COCH
3
The
slow
+
Br
2
step
→ CH
2
involves
one
COCH
3
Br
+
Br
2
molecule
of
CH
COCH
3
the
rate
equation
should
be:
rate
=
COCH
k[CH
3
and
one
OH
ion.
So
3
][OH
]
3
Key points

The
rate

The
rate
of
order
number

The
determining
step
in
a
reaction
with
respect
mechanism
determines
the
overall
reaction.
rate
of
of
reaction
molecules
equation
involved
provides
in
to
the
evidence
particular
rate
to
reactants
determining
support
a
shows
the
step.
particular
reaction
mechanism.
87
Revision
1
a
Benzenediazonium
chloride
questions
decomposes
above
3
5 °C
to
produce
phenol,
hydrochloric
acid
For
the
below
nitrogen
as
shown
in
the following
reaction X
+ Y
→
Z,
the
tabulated
results
and
were
obtained.
equation:
+
C
H
6
N
5
Cl
(aq)
+
H
2
O(l)
→ C
2
H
6
N
OH(aq)
+
Experiment
5
(g)
+
[X]/
used
b
The
describe
to
study
equation for
propanone
I
a
(aq)
suitable
the
rate
the
of
method
this
that
could
Initial
rate/
–3
–3
mol dm
mol dm
mol dm
1
0.030
0.015
0.
1242
2
0.010
0.015
0.0138
3
0.010
0.045
0.0414
2
Briefly
[Y]/
–3
HCl(aq).
–1
s
be
reaction.
reaction
of
iodine
with
is:
+ CH
2
COCH
3
(aq)
→ CH
3
COCH
3
I(aq)
+
2
a
What
is
the
order
with
respect
to X?
b
What
is
the
order
with
respect
to Y?
c
What
is
the
overall
d
Write
the
e
Determine
+
H
Explain
to
2
The
why
monitor
questions
shown
a
colorimetric
the
rate
of
which follow
this
are
(aq)
+
I
method
(aq).
is
suitable
order
of
the
reaction?
reaction.
based
on
the
graphs
rate
equation for
the value
of
the
the
reaction.
rate
constant for
this
reaction.
below:
I
4
The following
etar
graph
shown
questions
below,
experimental
data.
concentration
of
a
are
which
It
based
was
a
student’s
generated from
represents
reactant
on
[Y],
the
change
over
a
in
period
of
75
[reactant]
seconds.
]tnatcaer[
II
0.6
0.5
md
3
time
lom
III
0.4
0.3
]Y[
]tnatcaer[
0.2
0.
1
time
0.0
IV
20
40
etar
time
a
60
80
(seconds)
i
What
is
the
initial
ii
What
is
the
instantaneous
rate?
rate
at
[reactant]
a
Which
of
the
graphs
represent
a
zero

20
seconds,

60
seconds?
order
reaction?
b
b
How
would
represents
you
determine
a rst
or
second
whether Graph
order
reaction
i
What
is
ii
What
are
to
the
reactant
How
would Graph
IV
this
change,
if
it
were
labelled
88
a
second
axes?
term
‘half-life’?
the values
of
the rst
two
half-lives
reaction?
Giving
your
order
reaction,
with
reasoning,
state
the
order
to
reaction
represent
the
shown?
c
c
by
with
for
respect
meant
II
the
same
with
respect
to
reactant Y.
of
the
Chapter
5
The
sketch
below
distribution for
a
represents
sample
of
the
gas
Boltzmann
at
a
7
temperature
a
of
7
Consider
Step
1:
Rates
the
reaction
reaction
(g)
NO
of
+ SO
2
500 K.
Step
2:
–
revision
mechanism:
(g)
→
NO(g)
ii
Which
iii
b
a
Copy
the
diagram
and
label
the
A
What
is
+ O
(g)
is
the
is
→
2NO
the
same
axes,
sketch
the
graph
to
show
distribution
would
change
=
][F
k[NO
to
if
the
Giving
i
What
is
by
the
rate
2
your
reasoning,
is
state
consistent
the
effect
of
a
decrease
on
the
rate
of
a
Step
1:
Step
2:
+
NO
NO
F
→
FNO
2
+
F
graphs,
along
with
and
notations,
to
explain
this
Single
step
half-life
relate
to
the
concept
ii
of
+
What
F
2
mechanism:
→
2FNO
2
would
reaction
).
(t
if
2
happen
the
A
sample
tested
to
the
rate
concentration
of
of
F
the
were
2
1/2
a
slow
fast
effect.
2
questions
F
FNO
appropriate
2NO
The following
+
2
→
2
6
these
reaction?
II
symbols
of
observed
equation?
2
your
the
in
I
Use
which
with
390 K.
temperature
ii
equation,
temperature
rate
c
equation?
]
mechanisms
decreased
step?
how
i
the
fast
rate-determining
described
2
On
(g)
equation?
axes.
rate
b
the
rate
slow
2
overall
step
is
reaction
the
(g)
3
2
What
+ SO
2
2NO(g)
i
questions
of
a
after
nuclide
16
What fraction
with
a
half-life
of
4
days
is
while
the
sample
has
the
concentration
of
NO
2
remained
days.
of
doubled,
the
same?
decayed?
1
__
A
4
3
__
B
4
7
__
C
8
15
__
D
16
b
Radioactive
newly
decay
prepared
is
a rst
order
radioactive
process.
nuclide
has
–6
constant
(rate
approximate
c
A
1
hour
B
1
day
C
1
week
D
1
month
The
half-life
much
of
a
constant)
half-life
of
of
1
the
sample
10
If
a
decay
–1
s
,
what
is
the
nuclide?
of Thorium-234
12 g
×
a
would
is
24
hours.
remain
How
after
96
hours?
A
B
3 g
C
0.75 g
D
d
1.5 g
A
6 g
sample
after
its
of
a
radioactive
nuclide
is
tested
6
days
preparation.
7
__
It
is found
of
that
the
sample
has
decayed.
8
What
is
the
A
2
days
B
4
days
C
6
days
D
7
half-life
of
this
nuclide?
days
89
8
Chemical
8.
1
Characteristics
Learning outcomes
equilibria
Reversible
Many
On
completion
of
this
section,
be
able
understand
chemical
equilibria
reactions
reactions
then
continue
stops.
We
until
say
one
the
of
the
reaction
reactants
goes
to
runs
out.
completion,
The
e.g.
reacting
to:
magnesium

chemical
you
reaction
should
of
the
and
principles
of
hydrogen.
with
This
hydrochloric
reaction
is
acid
to
form
magnesium
chloride
and
irreversible.
physical–chemical
Some
reactions
are
reversible
e.g.
equilibria
forward

state
the
characteristics
of
CuSO
5H
4
system
in
dynamic
reaction
a
O
CuSO
+
2
5H
4
O
2
equilibrium.
backward
Heating
blue
copper( II)
turns
it
back
Reversible
place
react
at
to
show
hydrated
in
time
reactants
at
reactions
do
equilibrium
reaction
iodide
the
called
the
the
the
go
can
we
white
copper(
(backward
backward
the
sulphate
reaction).
reactions
reactions .
as
anhydrous
II)
The
reactants
take
products
form
completion.
use
gas
to
the
and
equilibrium
iodine
sign
vapour
to
Y.
For
produce
gas:
approach
alone.
and
time
to
it
sulphate
forward
(g)
+
I
2
products
turns
anhydrous
II)
same
hydrogen
H
We
the
equilibrium
not
reaction
of
to
copper(
which
are
sulphate
water
These
an
hydrogen
blue
same
give
example,
to
copper( II)
Adding
reactions
the
products.
T
o
hydrated
sulphate.
reaction
equilibrium
Figure
(g)
Y
2HI(g)
2
by
8.1.1(a)
starting
shows
off
how
with
the
either
reactants
concentrations
of
alone
H
or
and
2
when
a
mixture
of
hydrogen
(0.02 mol dm
)
and
iodine
tube
When
–3
)
(0.02 mol dm
reached,
and
there
products.
is
is
heated
no
The
in
further
same
a
sealed
change
in
equilibrium
at
the
is
500 K.
concentration
reached
when
equilibrium
of
we
the
start
is
reactants
with
–3
hydrogen
iodide
(0.04 mol dm
)
alone
a
(see
Figure
8.1.1(b)).
b
and
[I
]
noitartnecnoc
noitartnecnoc
equilibrium
reached
[HI]
[HI] falls
md lom(
3
2
falls
)
]
2
md lom(
3–
)
[H
equilibrium
reached
[H
]
and
2
[I
]
2
rises
rises
time
Figure 8.1.1
time
The change in concentrations of reactants and products as equilibrium is
–3
approached: a starting with 0.02 mol dm
–3
H
(g) and 0.02 mol dm
2
–3
0.04 mol dm
90
HI(g)
I
(g); b starting with
2
I
2
–3
change
Chapter
Characteristics of the

It

is
dynamic:
to
products
to
reactants.
At
molecules
and
equilibrium
equilibrium
of
molecules
the
rate
of
reactants
are
products
are
the
forward
and
Chemical
equilibria
state
continually
of
8
being
continually
backward
converted
being
converted
reactions
are
equal.

The
concentration
change.
and

They
backward
Equilibrium
where
none
mixture.
the
are
only
Many
simple
example
equilibrium
If
we
put
energy
reactions,
molecules
increases.
reached
the
we
a
where
rate
as
is
write
the
from
this
a
can
solution
liquid
equilibrium
rates
A
of
closed
escapes
take
and
in
no
from
place
gases
the
to
the
the
the
do
not
forward
in
are
system
the
is
one
reaction
open
beakers
if
involved.
of
moving
to
the
pressure
system
equilibrium
is
become
from
liquid.
its
with
the
A
by
a
the
container
.
the
most
vapour
.
molecules
condense.
is
closed
a
increasing
exerted
called
a
in
the
As
vapour
attractive
Eventually
liquid
to
more
the
forces
a
point
vapour
is
at
liquid–vapour
vapour
vapour
in
equilibrium
pressure .
For
water
,
as:
O(l)
H
Y
2
a
and
to
in
molecules
water
vapor
chemistry
vapour
They
other
vapour
The
closed
are
its
water
surface.
each
cause
physical
and
container
,
molecules
from
a
system .
products
concentration
closer
reached.
in
closed
or
at
the
same.
equilibrium
closed
molecules
liquid
can
an
the
get
products
because
however
,
in
between
a
escape,
the
same
in
place
escape
They
equilibrium
with
to
a
is
equilibrium
of
up
water
start
between
set
in
the
reactants
takes
and
This
are
occurs
the
Liquid–vapour
A
reactants
reactions
of
reaction
of
constant.
H
O(g)
2
b
c
vapour
liquid
Figure 8.1.2
liquid
a A liquid is placed in a sealed container; b Liquid molecules move from
liquid to vapour at a greater rate than they return from vapour to liquid; c At equilibrium
the molecules move from liquid to vapour at the some rate as they move from vapour to
liquid
Key points

Chemical
occur

at
equilibrium
the
Equilibrium
are
the
is
is
dynamic:
the
backward
and forward
reactions
time.
reached
when
the
rate
of forward
and
backward
reactions
same.

Equilibrium

In
a
same
closed
only
occurs
system,
a
in
a
liquid
closed
is
in
system.
equilibrium
with
its
vapour.
91
8.2
The
equilibrium
constant,
K
c
Learning outcomes
The
equilibrium
constant,
K
c
If
On
completion
of
this
section,
we
high
should
be
able
put
temperature
construct
and
both
iodine
equilibrium
with
the
product,

terms
deduce
a
sealed
and
hydrogen
tube
iodine
and
are
heat
present
it
in
at
a
the
constant
gas
phase,
of
(g)
units
of
+
I
2
concentration
the
iodide:
expressions
H
in
in
hydrogen
to:
together

hydrogen
you
When
K
we
measure
the
(g)
Y
2HI(g)
2
concentration
of
each
reactant
and
product
at
c
equilibrium,

describe
an
experiment
a
value for
nd
there
is
a
relationship
between
these
concentrations.
to
For
determine
we
this
reaction
the
relationship
is:
K
c
2
[HI]
_______
=
constant
(K
)
c
[H
][I
2

The
constant
is
called
]
2
the
equilibrium
constant ,
K
c
−3

[HI]

in
means
K
c

refers
whole
value
writing
start
equilibrium
take
care
to
solids
that
appear
of
K
is
the
is
same
called
in
the
(at
concentrations
however
the
of
in
mol dm
terms
of
concentrations.
the
equilibrium
same
expression
temperature
hydrogen,
iodine
and
and
pressure)
hydrogen
iodide
we
any
equilibrium
reaction
we
can
write
an
equilibrium
expression
the
is
based
on
the
stoichiometric
equation.
For
the
equation:
equation. Their
mA
concentration
constant
in
with.
which
stoichiometric
equilibrium
iodide
ignore
For
any
hydrogen
c
whatever
expressions,
the
relationship
Exam tips
When
to
of
c
The
The
concentration
remains
much
there
+
nB
Y
pC
+
qD
constant
is.
So
p
the
q
[D]
[C]
________
The
equilibrium
expression
is:
=
K
c
equilibrium
expression for
m
the
[A]
n
[B]
reaction:
2
]
[NH
_________
3
Example
1:
N
(g)
+
3H
2
Fe
O
2
(s) + 3CO(g) Y 2Fe(s) + 3CO
3
(g)
Y
2NH
2
(g)
K
3
=
c
3
(g)
[N
2
][H
2
]
2
2
is
[SO
]
__________
3
3
[CO]
Example
2:
2SO
(g)
+
O
2
(g)
Y
2SO
2
(g)
K
3
=
c
2
______
K
[SO
=
]
[O
2
c
]
2
3
[CO
]
2
Units of
K
c
These
vary
with
the
form
of
the
equilibrium
expression.
They
have
to
be
−3
worked
in
the
out
by
substituting
equilibrium
in
mol dm
expression.
For
each
the
‘concentration
boxes’
example:
2
−3
]
[SO
(mol dm
__________
3
=
K
of
−3
) ×
(mol dm
)
___________________________________
units
=
.
c
2
[SO
]
−3
[O
2
]
(mol dm
−3
)
×
(mol dm
)
−3
×
(mol dm
)
2
)
___________________________________
So:
units
=
=
−3
)
×
(mol dm
)
1
_________
3
=
dm
−1
mol
−3
mol dm
If
the
no
92
top
units.
and
bottom
of
the
expression
cancel
completely
then
there
are
Chapter
Determining
a value of
K
by
8
Chemical
equilibria
experiment
c
The
value
of
K
for
the
reaction
c
CH
COOH
+
C
3
can
be
The

found
using
procedure
Make
up
H
2
ethanoic
acid
a
OH
Y
CH
5
titration
COOC
3
ethanol
method.
The
reaction
of
known
concentration
and
catalyst,
e.g.
3
and

Use
a
Shake
of
ethyl
burette
the
H
O
2
water
catalysed
by
an
acid.
to
2 mol dm
HCl.
ethanoate,
water
deliver
mixture
temperature
volumes
of
the
3
water
,
2 cm
3
2 cm
ethanol,
1 cm
ethanoic
–3
of
5 cm
amounts
is
and
3
acid
+
5
ethanoate
is:
mixtures
reactants
H
2
ethyl
to
well
reach
each
and
of
(Y
ou
and
the
allow
could
to
start
with
known
acid.)
liquids
it
also
into
stand
a
for
stoppered
1
week
at
bottle.
room
equilibrium.
−3

After
1
week,
hydroxide
the

reacts
Find
the
added

of
a
Calculate
acid
in
the
number
concentrations
of
reaction
the
moles
acid
ethanoic
reaction
start
the
as
mixture
an
with
indicator
.
mixture
sodium
1 mol dm
During
decreases
as
this
the
time,
ethanoic
ethanol.
of
and
the
the
of
the
hydrochloric
acid
catalyst
you
concentrations:
hydrochloric
of
of
titration.
and
Moles
whole
concentration
similar
of
the
phenolphthalein
with
exact
by
Calculation

using
amount
acid
titrate
at
of
the
ethanoic
start
of
acid,
the
a,
ethanol,
experiment
b,
from
water
,
the
c,
volumes
used.
acid
mixture
at
by
experiment.
equilibrium
titration
Call
–
this
=
moles
moles
of
of
acid
calculated
hydrochloric
acid
at
in
the
x.
Exam tips

Moles
of
ethyl
experiment
because
ethanoate

Moles
–
for
of
is
moles
every
is
–
is
=
moles
ethanoic
of
of
acid
ethanoic
ethanoic
at
acid
at
equilibrium
acid
reacted,
a
the
(a
–
mole
start
x).
of
of
This
the
is
It
ethyl
at
for
to
equilibrium
moles
is
of
ethyl
every
=
moles
ethanoate
mole
of
of
at
ethanol
at
equilibrium
ethanol
reacted,
a
the
(b
–
mole
start
(a
of
–
important
equilibrium
formed.
because
ethanoate
of
mole
ethanol
experiment
This
ethanoate
of
the
the
equation
given. The
way
that
value
you
example,
formed.
you
of
write
you
K
have
depends
for
the
if
write
you
reaction
the
write
Moles
+
of
moles
for
every
water
of
at
ethyl
mole
of
equilibrium
ethanoate
water
at
=
moles
of
water
equilibrium
formed,
a
mole
of
(c
+
at
(a
ethyl
start
–
of
x)).
experiment
This
ethanoate
is
is
because
been
on
equation.
the
HI
as
2HI
Y
H
between
+
2
expression
I
the
For
equation
H
,
2

the
according
c
the
x)).
ethyl
that
expression
I
and
2
equilibrium
2
is:
formed.
[H
][I
2
]
2
_______
kc

The
concentrations
expression
to
nd
a
are
then
value
of
substituted
into
the
=
2
equilibrium
[HI]
K
c
Key points

For
an
equilibrium
concentration
of
reaction
reactants
there
and
is
a
relationship
products
and
the
between
the
equilibrium
constant
K
.
c
p
[C]
q
[D]
_______

The
general form
of
the
equilibrium
expression
is
K
=
c
m
[A]
[A]
and
[B]
[C]
and
[D]
p,
q,
m,
n
=
reactant
=
product
concentrations
are
number
of
stoichiometric

K
can
be
n
[B]
concentrations
moles
of
relevant
reactants/products
in
the
equation.
calculated from
the
results
of
appropriate
experiments.
c
93
8.3
Equilibrium
calculations
involving
K
c
Learning outcomes
Calculating
K
from
equilibrium
concentrations
c
On
completion
should
be
able
of
this
section,
In
some
or
moles,
know
how
to
calculations
equilibrium
present
at
calculations
equilibrium.
we
may
When
be
given
doing
the
these
amount,
in
calculations
grams
we
rst
to:
need

equilibrium
you
to
calculate
the
concentration
of
each
of
these
substances.
perform
involving
constant,
the
K
Worked
example
1
.
c
Propanone
reacts
with
CH
hydrogen
COCH
3
At
equilibrium
present
as
as
HCN
Y
to
were
form
CH
3
there
well
+
cyanide
an
addition
C(OH)(CN)CH
3
0.013 mol
0.011 mol
of
the
of
product:
3
propanone
addition
and
product,
0.013 mol
of
HCN
C(OH)(CN)
CH
3
3
CH
.
The
volume
of
the
reaction
mixture
was
500 cm
.
Calculate
K
3
c
−3
Step
1:
Calculate
CH
the
concentration
COCH
3
+
of
each
HCN
Y
CH
3
0.013
in
mol dm
:
C(OH)(CN)CH
3
1000
3
1000
_____
1000
_____
×
0.013
_____
×
0.011
500
×
500
−3
=
substance
500
−3
0.026 mol dm
−3
0.026 mol dm
0.022 mol dm
[CH
C(OH)(CN)CH
]
____________________
3
3
Step
2:
W
rite
the
equilibrium
expression:
K
=
c
[CH
COCH
3
][HCN]
3
0.022
______________
Step
3:
Substitute
the
values:
K
=
3
=
32.5 dm
−1
mol
c
0.026
Calculating
K
using
initial
×
0.026
concentrations
and
c
concentration of
Worked
0.29 mol
allowed
are
example
of
to
ethanoic
reach
present.
product
2
acid
and
equilibrium.
Calculate
0.25 mol
At
of
ethanol
equilibrium
are
mixed
0.18 mol
of
together
ethyl
and
ethanoate
K
c
Step
1:
W
rite
the
below
balanced
equation
with
CH
COOH +
C
3
Step
2:
the
necessary
information
it.
At
the
start:
At
equilibrium:
Calculate
the
(According
OH
Y
CH
5
COOC
3
H
2
0.25 mol
the
1 mol
O
2
0
of
moles
equation,
for
at
equilibrium:
every
1 mol
of
COOC
CH
3
formed,
+ H
5
0
0.18 mol
number
to
H
2
0.29 mol
of
H
O
is
H
2
5
formed.)
2
CH
COOH
=
0.29
–
0.18 mol
=
0.11 mol
(since
for
every
0.18 mol
3
of
COOC
CH
3
H
C
2
CH
OH
=
3:
formed,
0.18 mol
COOC
0.25
H
–
0.18 mol
2
W
rite
the
formed,
=
0.07 mol
equilibrium
H
][H
expression
O]
94
is
consumed)
(since
H
and
OH
for
is
every
0.18
of
consumed)
5
substitute
0.18
×
0.18
0.11
×
0.07
the
values.
=
=
4.2
(no
units)
c
that
the
COOH
___________
K
COOH][C
3
of
C
2
=
[CH
same
of
5
COOC
[CH
0.18 mol
c
Note
CH
3
_____________________
3
2
5
2
K
of
5
5
3
Step
H
2
we
do
number
of
H
2
not
have
OH]
5
to
know
concentration
equilibrium
expression.
the
terms
concentrations
in
the
since
numerator
and
there
are
the
denominator
Chapter
Worked
example
mixture
were
put
of
in
hydrogen
a
sealed
Chemical
equilibria
3
−2
A
8
(2.40
tube
×
and
10
left
−2
mol)
to
and
come
to
iodine
(1.38
equilibrium.
×
At
10
mol)
equilibrium
−3
1.20
×
mol
10
of
iodine
was
present.
Calculate
K
c
Step
1:
W
rite
the
below
balanced
equation
with
the
necessary
information
it.
H
(g)
+
I
2
(g)
the
start:
2.40
At
equilibrium:
×
Y
2HI(g)
2
−2
At
−2
mol
10
1.38
×
10
1.20
×
10
mol
0
−3
Step
2:
Calculate
Amount
the
of
number
iodine
of
moles
(1.38
so,
mol
×
at
H
−2
)
10
equilibrium:
consumed
−2
=
at
mol
−
(0.12
×
10
−2
)
=
1.26
×
10
=
1.14
×
10
equilibrium
2
−2
=
(2.40
×
−2
)
10
−
(1.26
×
10
−2
)
mol
−2
(1.26
×
mol
10
I
were
consumed
so
the
same
amount
of
H
2
consumed
because
is
2
these
have
the
same
mole
ratio
in
the
equation.)
so,
mol
HI
at
equilibrium
−2
=
2
×
1.26
×
stoichiometric
produces
Step
3:
W
rite
equation
2 mol
the
−2
mol
10
of
=
2.52
every
×
10
mole
of
expression
and
2
iodine
in
which
substitute
−2
[HI]
(2.52
_______
(since
the
reacts
HI.)
equilibrium
×10
the
values:
2
)
___________________________
=
K
mol
K
c
=
=
46.4
(no
units)
c
−2
[H
][I
2
]
(1.14
×
10
−3
)
×
(1.20
×
10
)
2
Key points

Values
of
K
can
be
deduced from
the
concentration
of
reactants
and
c
products

present
Concentrations
calculated from
at
of
equilibrium
specic
the
together
reactants
equilibrium
or
with
the
products
expression,
equilibrium
at
expression.
equilibrium
knowing
the
value
can
of
be
K
c
together
and
at
with
relevant
concentration
data
at
the
start
of
the
experiment
equilibrium.
95
8.4
The
equilibrium
constant,
K
p
Learning outcomes
Mole fraction
For
On
completion
of
this
section,
reactions
than
should
be
able
to
understand
term
measure
mixtures
of
concentrations.
gases,
When
it
we
is
easier
are
to
dealing
measure
with
pressures
mixtures
of
to:
gases

involving
you
the
‘partial
meaning
of
the
we
mixture
need
in
to
know
terms
of
the
fraction
moles.
This
is
of
a
particular
called
the
gas
mole
present
in
the
fraction
pressure’
number
of
moles
of
a
number
of
moles
of
gases
particular
gas
_________________________________________

construct
equilibrium
mole
expressions
fraction
.
=
total
in
terms
of
partial
a
mixture
of
0.2 mol
H
and
0.3 mol
O
2
deduce
units
of
the
mixture
pressures
In

in
the
mole
fraction
of
H
2
is:
2
K
p
0.2
_________

perform
calculations
involving
K
=
p
0.2
Partial
The
is
exerted
called
the
partial
pressure,
by
the
partial
partial
The
0.4
0.3
pressures
pressure
gases
+
molecules
pressure
pressure
pressures
of
all
=
the
of
total
of
a
that
particular
gas.
pressure
gases
in
the
gas
Symbol,
×
mole
mixture
in
a
mixture
of
p
fraction
add
up
to
the
total
P
T
P
=
p
T
+
p
1
+
p
2
…
3
Example:
A
mixture
of
gases
contains
1.2 mol
N
,
0.5 mol
H
2
and
0.3 mol
of
O
2
.
2
5
The
total
pressure
is
8.0
×
kPa.
10
Calculate
the
partial
pressure
of
hydrogen.
0.5
_______________
5
partial
pressure
of
=
H
(8.0
×
10
)
×
5
=
2
×
10
kPa
2
1.2
Using the
equilibrium
constant,
+
0.5
+
0.3
K
p
Equilibrium
constants
in
terms
of
partial
pressures,
K
,
may
be
calculated
p
in
a
similar
way
to
those
for
except
K
that
no
square
brackets
are
used.
c
For
example,
the
equilibrium
N
(g)
expression
+
3H
2
(g)
for
Y
the
2NH
2
reaction:
(g)
3
Exam tips
is:
2
Although
the
standard
pressure
p
is
approximately
1.01
×
10
NH
___________
3
K
=
p
5
3
Pa,
pN
×
p
H
2
industrial
chemists
often
use
2
the
Units
of
K
have
to
be
worked
out
in
a
similar
way
as
those
for
p
atmosphere
(atm)
as
a
unit
K
.
The
c
of
units
are
calculated
using
pascals,
Pa.
5
pressure,
1 atm
=
1.01
×
10
Pa.
For
the
reaction
N
(g)
2
You
should
be
prepared
to
from
calculations
using
+
3H
(g)
Y
2NH
2
(g)
,
the
units
are
worked
out
3
do
atmospheres
the
equilibrium
expression
as
follows:
as
2
the
base
unit
as
well
as
pascals.
)
(Pa)
___________
1
______________________
cancelling
_____
=
96
×
(Pa)
2
)
×
(Pa)
×
(Pa)
–2
=
3
(Pa)
(Pa)
Pa
Chapter
Worked
Sulphur
example
dioxide
reacts
with
oxygen
(g)
+
to
O
2
equilibrium,
the
Chemical
equilibria
1
2SO
At
8
mixture
form
(g)
Y
sulphur
2SO
2
trioxide:
(g)
3
contains
12.5 mol
SO
,
87.5 mol
O
2
and
2
7
100 mol
.
SO
The
total
pressure
of
the
mixture
is
1.6
×
10
Pa.
3
Calculate
K
p
Step
1:
Calculate
partial
pressures:
12.5
_____
p
7
=
×
1.6
×
6
10
=
1.0
×
10
Pa
SO
2
200
87.5
_____
7
=
p
×
1.6
×
10
×
1.6
×
10
6
=
7.0
×
10
=
8.0
×
10
Pa
O
2
200
100
____
7
=
p
6
Pa
SO
3
200
Step
2:
W
rite
the
equilibrium
expression:
2
p
SO
____________
3
K
=
p
2
p
SO
×
pO
2
Step
3:
Substitute
the
partial
2
pressures:
6
(8.0
×
2
)
10
________________________
K
=
–6
=
9.1
×
10
–1
Pa
p
6
(1.0
Worked
2 mol
of
example
nitrogen
×
10
2
)
6
×
(7.0
×
10
)
2
and
6 mol
of
hydrogen
are
mixed
and
allowed
to
reach
7
equilibrium
mixture
at
680 °C
contains
and
3 mol
2
of
Pa
×10
pressure.
ammonia.
At
Calculate
equilibrium,
the
K
p
N
(g)
+
3H
2
Step
1:
Find
the
number
(g)
Y
2NH
2
of
moles
(g)
3
of
nitrogen
and
hydrogen
at
equilibrium:
Moles
N
=
2
–
1.5
=
0.5
mol
2
1
(For
each
mol
NH
formed
mol
3
of
N
has
been
used,
2
2
1
__
i.e.
3
×
mol)
2
Moles
H
=
6
–
4.5
=
1.5
mol
2
(For
each
2
mol
of
NH
formed
3
mol
H
3
are
consumed
Key points
2
3
__
i.e.
3
×
mol)

Partial
pressure
=
total
pressure
×
2
mole fraction
Step
2:
Calculate
partial
of
=
2
×
10
=
2
×
10
0.1
=
0.2
×
10
×
0.3
=
0.6
×
10
Pa

2
7
7
=
2
×
10
the
equilibrium
of
partial
0.6
=
1.2
×
10
reactants
gures:
2
p
7
NH
=
constant
pressures
in
can
be
Pa
deduced from
(1.2
___________
3
K
The
7
×
3
Substitute
gases.
terms
NH
3:
mixture
Pa
2
Step
a
7
H
p
in
7
×
N
p
gas
pressures:
7
p
of
×10

2
In
a
and
reaction
partial
pressures
of
products.
involving
gases,
)
_______________________
=
p
3
pN
×
p
7
H
2
(0.2
×10
)
7
×
(0.6
×10
3
the
quantities
of
reactants
or
)
2
products
–13
K
=
3.3
×
10
at
equilibrium
can
be
–2
Pa
p
deduced from
expression
the
and
K
equilibrium
together
with
p
relevant
partial
pressure
data.
97
8.5
Le Chatelier’s
Learning outcomes
principle
Position of
Position
On
completion
should

be
apply
able
of
this
section,
effect
products
in
principle

concentration
If
the
of
change
pressure
temperature
If
on
an
that
the
Le
changes
concentration,
temperature
relative
amounts
of
reactants
and
of
position
products
of
increases
equilibrium
relative
moves
to
to
the
the
reactants,
we
products,
we
right.
concentration
the
of
position
reactants
of
increases
equilibrium
relative
moves
to
the
to
the
left.
Chatelier ’s
changed,
pressure
affect
principle:
the
If
position
of
one
or
more
equilibrium
factors
shifts
that
in
affect
the
an
equilibrium
direction
which
in
the
the
change
or
values
of
Le
Chatelier ’s
principle
relates
to
changes
in
concentration,
pressure
and
temperature.
K
c

the
that
opposes
and
the
equilibrium
are
K
to
mixture.
in
reaction
how
equilibrium
or
say
know
an
to

concentration,

refers
principle
you
say
the
equilibrium
Le Chatelier’s
to:
Le Chatelier’s
explain
of
equilibrium:
p
apply
the
Le Chatelier’s
Haber
Process
principle
and
to
the
Note:
of
.
K
Catalysts
They
have
only
no
speed
effect
up
the
on
the
rate
of
position
of
equilibrium
or
the
value
reaction.
c
Contact
Process.
Position of
The
effect
of
equilibrium:
changing
changing
concentrations
can
be
concentration
seen
by
referring
to
the
reaction:
COOH
CH
+
C
3

Increasing
position
More

the
of
Increasing
position
acid
concentration
ethanoate
the
of
OH
Y
CH
5
to
ethanoic
right
to
of
the
(to
are
+
H
5
acid
(or
oppose
O
2
ethanoate
water
ethanol)
the
added
shifts
the
reactant(s)).
formed.
ethyl
left.
H
2
ethyl
water
concentration
equilibrium
of
the
and
COOC
3
ethanol
equilibrium
ethyl
H
2
ethanoic
ethanoate
More
(or
ethanoic
water)
acid
shifts
and
the
ethanol
are
formed.

Removing
the
right
water
Note:
water
(to
are
from
oppose
the
the
reaction
removal
shifts
of
the
water).
position
More
of
ethyl
equilibrium
ethanoate
formed.
Changing
the
concentration
has
no
effect
on
the
value
of
or
K
K
c
Position of
Change
of
gas
(g)
in
+
I
2
Exam tips
is
a
common
error
to
a
change
in
the
(g)
on
only
position
Y
affects
each
side
of
2HI(g).
reactions
of
the
equilibrium
This
of
gas
change
on
in
the
left
pressure
and
concentration
reaction
such
because
does,
2
moles
the
the
of
however
,
the
are
different
example,
there
numbers
is
no
reaction:
stoichiometric
gas
on
change
the
equation
shows
2
the
right.
position
of
equilibrium
in
or
(g)
+
3H
2
affects
on
there
For
as:
N
pressure
where
equation.
pressure
suggest
a
that
pressure
changing the
p
2
moles
A
It
equilibrium:
molecules
change
H
in
to
and
value
of
K
(g)
Y
2NH
2
(g)
3
or
c
K
.
If
we
add
more
reactants,
more

Increase
in
pressure
shifts
the
position
of
equilibrium
to
the
right
p
where
products
are formed
until
the
K
is
restored.
Remember
it
are
fewer
gas
molecules
(2
moles
on
the
right
and
4
moles
value
on
of
there
the
left).
This
opposes
the
increase
in
the
number
of
gas
molecules
is
c
per
only
a
change
in
temperature
the
equilibrium
volume
Decrease
K
pressure
increases.
in
pressure
makes
decreases).
the
The
molecules
position
of
move
further
equilibrium
apart
shifts
(their
to
(Temperature Only
where
Changes
the
constant.
concentration
Remember TOCK
when
that

changes
unit
the
gas
molecules
are
more
concentrated.
).
c
Note:
Changing
the
pressure
has
no
effect
on
the
value
of
or
K
c
98
K
p
the
left
Chapter
Position of
A
change
in
equilibrium:
temperature
the
value
of
K
or
For
an
endothermic
value
of
K
or
K
More
products
c
.
reaction,
So
the
increase
position
of
in
p
We
temperature
equilibrium
increases
shifts
to
the
the
right.
formed.
So
for
the
Y
(g)
H
+
I
2
increasing
the
and
(g)
ΔH
2
temperature
products
an
value
exothermic
of
or
K
K
c
reactants
more
+9.6 kJ mol
effect of temperature
of
the
position
the
hydrogen
position
iodide
.
So
the
increase
position
of
in
formed.
So
for
the
increase
the
r
shifts
reaction,
of
equilibrium
in
favour
So
energy
+
O
Y
the
of
dioxide
temperature
equilibrium
2SO
and
decreases
shifts
to
the
reaction
the
the
left.
i.e.
More
in
the
(g)
+
in
3H
2

yield
can
shifts
in
=
Increasing
the
(g)
the
in
the
increase
direction
of
in
direction
energy,
energy
absorbed.
is
absorbed
in
endothermic
reactions.
position
Haber
of
equilibrium
in
favour
processes
Process:
Y
2NH
(g)
–1
ΔH
3
increased
the
goes
−197 kJ mol
r
industrial
2
be
the
–1
ΔH
Ø
N
increases
particles.
oxygen.
Maximising the yield
synthesised
(g)
3
temperature
temperature
the
reaction:
2
increasing
sulphur
(g)
in
of
opposing
Energy
The
equilibrium.
decomposes.
Ø
(g)
2
is
=
−92 kJ mol
r
by:
pressure:
more
product
(NH
)
is
formed
because
the
3
position

of
equilibrium
Decreasing
the
shifts
temperature:
in
favour
increases
of
fewer
the
molecules.
yield
of
NH
because
the
3
reaction

is
Removing
shifts

to
Using
the
a
exothermic.
ammonia
the
right
catalyst.
equilibrium
in
by
condensing
favour
This
of
speeds
it:
fewer
up
the
position
of
equilibrium
molecules.
the
reaction
rate
but
has
no
effect
on
yield.
Did you know?
We
use
compromise
although
rate
of
conditions
reaction
for
increases
the
as
synthesis
of
temperature
ammonia
increases,
because
the
yield
of
In
ammonia
decreases.
So
a
temperature
of
about
420–50 °C
is
used
to
the Contact
reasonable
yield
at
a
fast
enough
Process,
a
high
give
pressure
a
is
not
used
because:
rate.

The
yield
of
SO
produced for
3
The
key
acid
is:
reaction
in
the
Contact
Process
for
the
manufacture
of
sulphuric
SO
is fairly
high
at fairly
low
2
temperatures.
Ø
(g)
2SO
+
O
2
(g)
Y
2SO
2
(g)
–1
ΔH
3
=
−197 kJ mol
r

We
can
increase
the
yield
of
SO
by
decreasing
the
the
pressure
and
Highly
are
using
a
catalyst.
The
corrosive
theory
behind
this
is
less
as
for
the
Haber
easily
contained
oxides
in
the
the
reactor
same
sulphur
temperature,
3
increasing
vessel
at
high
pressure.
Process.
Key points

Le Chatelier’s
principle
states:
When the conditions in a chemical eq uilibrium
change, the position of eq uilibrium shifts to oppose the change.

Increase
the
in
right.
concentration
Decrease
in
of
reactants
concentration
shifts
of
the
position
reactants,
shifts
of
the
equilibrium
position
to
to
the
left.

A
change
in
temperature
affects
the
value
of
K
or
K
c
concentration

The
to
or
conditions
get
the
to
–1
=
p
are
2SO
Ammonia
principle
on
of
For
Le Chatelier’s
understand the
An

use
reaction:
Ø
the
can
p
are
2HI(g)
of
equilibria
K
c

Chemical
Did you know?
changing temperature
changes
8
best
pressure
used
in
yield
of
have
the
no
Haber
product
effect
on
Process
at
a
but
in
values.
and Contact
reasonable
changes
p
these
rate
of
Process
are
chosen
reaction.
99
8.6
Solubility
product,
K
sp
Learning outcomes
Solubility
Many
On
completion
of
this
section,
ionic
soluble
should
be
able
construct
for
equilibrium
solubility
expressions
deduce
units for
perform
calculations
solubility
to
be
dissolve
in
insoluble.
water
But
but
even
some
are
‘insoluble’
only
ionic
slightly
compounds
is
some
ions
saturated
in
when
solution
no
more
to
a
very
solute
small
will
extent.
dissolve
in
We
it.
say
that
Solubility
a
solution
is
measured
in
(or
mol dm
sometimes
in
g/100 g
water).
K
sp

appear
−3
product, K
sp

or
to:
form

compounds
you
product.

Potassium
involving
chloride
35.9 g/100 g

Calcium
hydroxide
solution

Silver
is
soluble
–
is
described
a
saturated
solution
contains
water
.
contains
chloride
is
0.113 g/100 g
regarded
as
as
sparingly
soluble
–
a
saturated
water
.
‘insoluble’.
It
is
in
fact
sparingly
soluble
−4
–
a
saturated
Solubility
When
a
solid.
they
is
The
move
contains
1.93
×
g/100 g
10
water
.
product
sparingly
equilibrium
the
solution
soluble
or
established
ions
from
move
solution
‘insoluble’
between
from
to
the
the
salt
is
ions
solid
to
added
in
to
water
,
solution
solution
at
and
the
an
the
same
ions
in
rate
as
solid.
water
molecules
–
I
not
ions
shown
+
ions
saturated
solution
+
of Ag
–
ions
and
I
ions
solid
Figure 8.6.1
At equilibrium ions move from saturated silver iodide solution to silver iodide
solid at the same rate as they move from solid to solution
We
can
write
the
equilibrium
in
AgI(s)
the
Y
diagram
above
+
−
Ag
(aq)
+
I
as:
(aq)
+
[Ag
−
(aq)][I
(aq)]
_______________
The
related
equilibrium
expression
is:
K
=
[AgI(s)]
But
the
concentration
of
solid
is
constant,
so
we
can
write
the
expression
as:
+
=
K
[Ag
−
][I
]
sp
K
is
called
the
solubility
product .
The
square
brackets
show
solubility
sp
−3
in
mol dm
Solubility
saturated
of
their
.
product
solution
relative
is
the
of
a
product
sparingly
of
the
concentrations
soluble
salt
at
298 K
1:
ion
Co(OH)
ratio
Co
2OH
Al
O
2
100
ion
the
in
a
power
2+
K
ion
3
=
[Co
=
[Al
−
][OH
]
sp
3+
2:
to
−
and
2
Example
each
concentrations.
2+
Example
of
raised
ratio
2Al
2−
and
3O
3+
K
sp
]
2
[O
2−
]
3
2
Chapter
Units of
We
solubility
calculate
the
8
Chemical
equilibria
product
units
for
K
in
the
same
way
as
for
general
equilibrium
sp
expressions
involving
K
c
Example:
2+
=
K
[Co
2
][OH
]
−3
units
are
(mol dm
−3
)
×
(mol dm
2
3
)
=
mol
−9
dm
sp
Solubility
Worked
A
product
example
saturated
calculations
1: Solubility
solution
of
lead
product from
iodide,
PbI
,
contains
solubility
0.076 g
PbI
2
solution.
Calculate
for
K
lead
iodide.
in
100 g
2
Molar
mass
of
lead
iodide
=
sp
−1
461.0 g mol
−3
Step
1:
Calculate
the
concentration
0.076
the
in
1.65
×
10
−3
mol dm
100
concentration
of
each
2+
[Pb
ion
in
−3
]
=
1.65
×
10
solution.
−3
mol dm
−3
]
[I
mol dm
−3
=
461.0
Calculate
solution
_____
×
2:
the
1000
______
Step
of
=
2
×
(1.65
×
10
−3
)
mol dm
−3
]
[I
=
3.30
×
10
−3
mol dm
2+
Step
3:
W
rite
the
equilibrium
expression:
K
=
[Pb
2
][I
]
sp
−3
Step
4:
Substitute
the
values:
K
=
(1.65
×
10
−3
) ×
(3.30
×
10
2
)
sp
−8
=
1.80
×
10
−8
Step
5:
Add
the
units:
K
=
1.80
×
10
3
−9
mol
dm
sp
Worked
example
Calculate
the
2: Solubility from
solubility
product
−3
solubility
in
mol dm
of
calcium
sulphate,
CaSO
.
4
−5
=
K
2.0
×
10
2
mol
−6
dm
sp
Step
1:
W
rite
down
the
equilibrium
expression:
2+
K
=
[Ca
2−
][SO
sp
Step
2:
Rewrite
the
equilibrium
expression
2+
In
CaSO
,
[Ca
3:
=
[SO
the
so
K
=
−5
4:
Complete
the
×
10
calculation
2+
[Ca
]
[Ca
of
one
ion
only:
2
]
sp
value:
2.0
Step
terms
2+
]
4
Substitute
in
2−
]
4
Step
]
4
2+
=
and
[Ca
add
2
]
units:
−5
=
√
2.0
×
10
−3
=
4.47
×
10
−3
mol dm
Key points

Solubility
of
ions
in
account

product
a
is
an
saturated
the
of
relative
The
value
and
concentration
K
can
The
solubility
expression
solution
number
be
of
of
a
showing
each
calculated
the
sparingly
ion
using
in
equilibrium
soluble
salt.
It
concentration
takes
into
solution.
the
relevant
equilibrium
expression
sp

of
K
and
of
a
of
ions
present
sparingly
the formula
of
at
soluble
the
equilibrium.
salt
can
be
calculated
using
the
value
salt.
sp
101
8.7
Solubility
product
and
the
common
ion
effect
Learning outcomes
Predicting
Solubility
On
completion
should

be
able
explain
the
the
of
this
section,
precipitation
product
can
be
used
to
when
two
when
we
solutions
are
mixed.
principles
common
ion
underlying
mix
sulphate?
solutions
The
possible
of
the
solubility
selective
whether
example:
soluble
reaction
will
a
precipitate
we
get
a
will
form
precipitate
salts
barium
chloride
and
sodium
is:
effect
+
Na
2
relate
For
to:
BaCl

predict
you
product
precipitation
to
of
SO
2
→
BaSO
4
+
2NaCl
4
the
ions
Sodium
chloride
is
very
soluble
in
water
but
barium
sulphate
is
sparingly
soluble.

know
how
to
determine
solubility
2+
product
by
experiment.
For
barium
sulphate
K
=
[Ba
2−
][SO
sp
2+

If
the
ionic
product
−10
]
=
1.0
×
2
10
mol
−6
dm
4
[Ba
2−
][SO
−10
]
is
greater
than
1.0
×
10
2
mol
−6
dm
4
a
precipitate
of
barium
sulphate
2+

If
the
ionic
product
][SO
[Ba
will
form.
−10
2−
]
is
less
than
1.0
×
10
2
mol
−6
dm
4
barium
Worked
Will
a
sulphate
will
example
precipitate
×
10
1.20
×
10
in
solution.
It
will
not
precipitate.
1
form
−5
6.00
remain
when
an
aqueous
solution
containing
−3
mol dm
−5
barium
chloride
is
added
to
an
equal
sodium
of
BaSO
sp
=
1.0
×
10
2
mol
−6
dm
4
2+
1:
of
sulphate?
−10
K
Step
volume
−3
mol dm
Calculate
the
concentrations:
[Ba
−5
]
=
3.00
×
10
2−
[SO
−3
mol dm
−6
]
=
6.00
×
10
−3
mol dm
4
2+
Step
2:
W
rite
the
equilibrium
expression:
K
=
[Ba
2−
][SO
sp
Step
3:
Substitute
the
=
(3.00
10
=
1.80
gures
to
give
the
−5
×
×
(6.00
Step
4:
Compare
2
mol
10
ionic
product:
−6
)
−10
×
]
4
values
of
×
10
)
−6
dm
ionic
product
and
K
:
sp
−10
1.80
so
Worked
What
silver
a
precipitate
example
from
×
10
−10
10
will
is
greater
than
1.0
×
10
form.
2
concentration
chloride
−4
1.0
×
of
a
silver
ions
solution
is
needed
of
sodium
=
1.8
in
solution
chloride
to
precipitate
containing
−3
mol dm
?
−10
K
of
AgCl
×
10
2
mol
−6
dm
sp
+
Step
1:
W
rite
the
equilibrium
expression:
K
=
[Ag
=
[Ag
][Cl
]
sp
−10
Step
2:
Substitute
the
gures:
(1.80
×
10
)
+
−4
]
×
(1.0
×
10
)
+
Step
3:
Rearrange
to
nd
[Ag
]:
−10
1.80
]
×
10
____________
+
[Ag
=
−6
= 1.8
×
10
−3
mol dm
−4
1.0
×
10
+
is
102
the
minimum
concentration
of
Ag
needed
to
cause
precipitation.
Chapter
Precipitation
We
can
use
and qualitative
solubility
product
analysis
precipitate
data
compounds
to
choose
from
suitable
their
reagents
solutions.
For
silver
nitrate
to
test
for
halide
ions
such
as
,
Cl
Br
and
I
.
stones
are
soluble
in
water
as
are
most
nitrates.
But
silver
halides
have
low
halide
solubility
ion
product.
results
in
a
So,
adding
precipitate
only
a
few
drops
of
by
precipitation
of
present
substances
silver
nitrate
in
the
urine. The
a
depends
on
many factors.
to
If
a
produced
Most
deposition
very
are
we
naturally
halides
equilibria
to
example
the
use
Chemical
Did you know?
Kidney
selectively
8
the
urine
is
too
concentrated,
e.g.
the
solubility
product
of
some
ions
+
(aq)
Ag
These
precipitates
can
be
+
Cl
(aq)
distinguished
→
by
AgCl(s)
their
characteristic
may
be
such
as
exceeded
calcium
solubility
product
by
higher
experiment
ions
The
method
is
based
on
making
a
saturated
solution
of
a
and
analysing
the
solution
to
nd
the
oxalate
than
normal
concentration
of
equilibrium
ion
which
calcium
is
present
hydroxide,
in
the
Ca(OH)
solid
as
well.
For
sp
and
of
may
up. A
calcium
also
example
in favour
shift
of
the
a
to
of
insoluble
calcium
nd
salts
for
K
build
concentration
precipitation
particular
may
particular
the
substance
precipitates
colours.
calcium
Determining
and
phosphate
such
as
calcium
carbonate
and
:
2
calcium

Add
enough
solid
is
calcium
present
as
well
reach
equilibrium.

Filter
off

Titrate
acid

of
the
samples
Calculate
the

the
titration
Find
solid
known
by
K
hydroxide
as
a
calcium
of
the
to
a
known
solution.
volume
Shake
and
of
leave
water
for
24
so
that
hours
phosphate,
conditions
become
especially
too
if
the
alkaline.
to
hydroxide.
calcium
hydroxide
solution
with
hydrochloric
concentration.
concentration
of
the
calcium
hydroxide
solution
from
results.
substituting
in
the
equilibrium
expression
sp
2+
=
K
[Ca
2
][OH
]
sp
The
The
salt
common
common
caused
common
by
with
ion
ion
effect
effect
adding
the
a
is
the
reduction
solution
dissolved
of
salt.
a
in
the
solubility
compound
The
which
solubility
of
of
has
in
BaSO
a
an
dissolved
ion
in
water
is
4
−5
1.0
×
−3
mol dm
10
−3
.
But
the
solubility
of
BaSO
in
0.10 mol dm
4
−9
sulphuric
acid,
SO
H
2
difference
by
,
is
only
1.0
×
10
−3
mol dm
.
We
can
explain
referring
to
equilibrium
2+
=
K
[Ba
expression:
2−
][SO
sp
−10
]
=
1.0
×
10
2
mol
−3
dm
4
2−
The
this
4
concentration
of
−3
SO
ions
in
the
solution
=
0.10 mol dm
4
Key points
2−
(ignoring
the
very
small
concentration
of
ions
SO
from
barium
4
sulphate).

−10
Substituting
the
concentration
2+
So
[Ba
]
−9
=
1.0
×
10
values:
1.0
×
2+
=
10
[Ba
]
×
The
common
to
the
of
a
ion
=
solubility
of
barium
sulphate
in
H
dissolved
quantitative
effect
work
salt
by
adding
a
SO
4
solution
and
practical
which
involving
preparing
with
contains
the
an
ion
dissolved
in
salt.
chemistry

In
refers
solubility
−3
mol dm
common
common
effect
in
0.1
2
The
ion
reduction
and
weighing
solids
When
the
ionic
product
in
the
(gravimetric
equilibrium
expression for
K
sp
analysis),
it
is
important
that
no
solid
is
lost.
The
common
ion
effect
can
exceeds
help
here.
For
example:
the
concentration
of
sulphate
in
aqueous
be
determined
precipitates
is
washed
will
be
with
ions
with
lost
dilute
in
the
sulphate
gravimetrically
barium
in
sulphate.
water
,
the
sulphuric
wash
some
ltrate.
acid
liquid
remains
If
a
by
barium
So
the
rather
ensures
adding
precipitate
and
that
is
water
.
the
of
chloride.
barium
sulphate
precipitate
than
barium
ions
ltered
The
will
presence
maximum
in
dissolve
then
of
amount
water
and
washed
sulphate
of
product
is
a
precipitated.
This
sulphate
and
solubility
solution
salt
can
the
barium

The
can
the
in
solubility
be found
product
by
concentration
solution
by
of
a
salt
determining
of
the
titration
ions
or
other
methods.
precipitated.
103
9
Acid–base
9.
1
Brønsted–Lowry
Learning outcomes
equilibria
completion
should

be
able
explain
the
of
this
section,
A
acid
and
acid
is
a
substance
acids
acids
which
and
neutralises
and
bases
a
base
to
form
a
salt
and
water
.
base
is
a
substance
base
which
2HCl(aq) +
neutralises
acid
to
form
a
salt
and
water:
CaO(s)
→
an
CaCl
(aq)
+
H
2
using

Brønsted–Lowry
describe
the
behaviour
acids
and
theory
differences
of
strong
and
bases
you
to:
terms
of
Simple definitions of
An
On
theory
acid
in
Another
denition
of
base
an
acid
is:
a
substance
O(l)
2
salt
water
which
ionises
in
water
to
+
weak
form
bases.
hydrogen
ions.
A
hydrogen
ion,
,
H
is
sometimes
referred
to
as
a
proton.
+
HCl(g)
Another
denition
water
form
to
of
a
+
base
hydroxide
aq
is:
a
→
H
(aq)
+
substance
Cl
that
(aq)
dissolves
or
ionises
in
soluble
in
ions.
+
KOH(s)
Many
water
metal
is
oxides
called
an
or
+
aq
hydroxides
→
K
are
(aq)
+
bases.
OH
A
(aq)
base
which
is
alkali
Did you know?
+
The
H
O
ion
is
called
Brønsted–Lowry theory of
a
3
hydroxonium
frequently
ion.
called
a
t
is
‘hydronium’
alone
we
in
talk
ion.
H
ions
aqueous
about
and
bases
rarely
An

A
acid
is
a
proton
donor
.
base
is
a
proton
acceptor
.
occur
solutions. When
hydrogen

or
+
‘oxonium’
acids
less
ions,
we
+
When
hydrogen
donated
to
the
chloride
water
.
So
gas
reacts
is
HCl
with
a
water
,
the
H
Brønsted–Lowry
ion
acid .
of
the
The
acid
is
water
(g)
+
really
mean
a
H
O
ion.
t
is
often
accepts
a
proton
so
it
is
a
Brønsted–Lowry
base.
3
convenient,
however,
especially
in
+
H
donated
+
calculations
and
talk
to
about
write
this
hydrogen
ion
as
H
+
ions.
HCl(g)
+ H
O(l)
→
H
2
acid
When
ammonia,
NH
O
(aq) + Cl
(aq)
3
base
,
reacts
with
water
,
it
accepts
a
proton
from
the
3
+
water
and
becomes
an
ion.
NH
So
ammonia
is
a
Brønsted–Lowry
base.
4
The
water
donates
the
proton,
so
it
is
a
Brønsted–Lowry
acid.
+
H
donated
+
NH
(g) + H
3
base
Y
ou
will
notice
that
Brønsted–Lowry
For
example,
following
water
NH
(aq)
+ OH
(aq)
4
do
can
not
act
is
set
as
have
methanoic
equilibrium
Y
acid
acids
when
O(l)
2
an
to
acid
acid
or
involve
reacts
a
base
aqueous
with
–
it
III)
up:
donated
+
Y
HCOOH + HClO
2
base
104
acid
+
HCOOH
2
amphoteric
solutions.
chloric(
+
H
is
ClO
2
acid,
the
Chapter
Conjugate
Every
away
acid
its
Every
equilibria
pairs
has
base
a
a
conjugate
base
–
the
ion
left
when
acid
–
the
ion
formed
an
acid
has
given
has
a
conjugate
when
an
acid
–
COOH
(aq)
+ H
3
O
(l)
Y
CH
2
acid
Strong
has
proton.
CH
they
Acid–base
proton.
accepted
Strong
9
and
acids
weak
ionise
dissociate
acids
(almost)
completely.
+
COO
(aq) +
H
3
base
and
base
(aq)
conjugate
acid
bases
completely
For
O
3
conjugate
in
solution.
We
sometimes
say
example:
+
(l)
HNO
+
H
3
O(l)
→
H
2
O
(aq)
+
NO
3
(aq)
3
+
or
more
simply
HNO
(l)
+
aq
→
H
(aq)
+
NO
3
Hydrochloric
Weak
acids
ionise
equilibrium
present
weak
lies
than
Organic
acid,
nitric
acid
and
(dissociate)
to
the
(aq)
3
left.
sulphuric
only
There
acid
partially
are
many
in
are
all
strong
solution.
more
The
molecules
acids.
position
of
acid
citric
acid
of
ions.
acids
such
as
methanoic
acid,
ethanoic
acid
and
are
acids.
Exam tips
+
CH
COOH(l)
+
H
3
O(l)
Y
CH
2
COO
(aq)
+
H
3
O
(aq)
3
t
Strong
bases
ionise
completely
in
solution,
is
important
to
distinguish
e.g.
between
the
strength
of
an
acid
and
+
NaOH(s)
+
aq
→
Na
(aq)
+
OH
(aq)
its
Weak
bases
equilibrium
ionise
(dissociate)
lies
the
to
left.
only
There
partially
are
many
in
solution.
more
The
molecules
position
of
of
base
concentration.
degree
of
Strength
to
the
or
weak. Concentration
ionisation
refers
–
strong
refers
to
3
present
than
the
ions.
number
of
moles
per
dm
. A
–3
4 mol dm
Ammonia
and
amines
are
examples
of
weak
solution
of
ethanoic
acid
bases.
is
concentrated,
but
it
is
a
weak
acid.
+
CH
NH
3
Comparing
The
table
bases.
weak
shows
Strong
+
H
acids
bases
of
same
Y
CH
and
(aq)
+
OH
(aq)
3
bases
pH
much
Strong
NH
3
typical
have
concentration.
O(l)
2
some
acids
same
the
(g)
2
values
lower
bases
for
pH
have
strong
values
much
and
than
higher
weak
weak
pH
acids
acids
values
of
and
the
than
weak
concentration.
Key points
Concentration
pH
of
acid
acid or
base
strong
pH
weak
pH
acid
strong
pH
base
weak

base
The
Brønsted–Lowry theory
states
that
donors
and
acids
are
bases
proton
are
proton
–3
0
1.0 mol dm
2.4
14
11.6
acceptors.
–3
2
0.01 mol dm
Strong
acids
have
a
3.4
higher
electrical
12
conductivity
and

10.6
react
more
Strong
calcium
carbonate
or
magnesium.
This
is
because
strong
than
hydrochloric
weak
acids
(for
acid)
have
example
a
higher
ethanoic
concentration
acid).
of
bases
are
completely
aqueous
solution. Weak
ionised
in
acids
and
rapidly
acids
are
hydrogen
only
partially
ionised
in
(for
aqueous
example
and
(almost)
bases
with
acids
solution.
ions

Strong
equal
acids
and
distinguished
their
bases
concentrations
aqueous
by
the
of
can
pH
be
values
of
solutions.
105
9.2
Simple
pH
Learning outcomes
calculations
pH
We
On
completion
should
be
able
of
this
section,
can
compare
the
relative
acidity
or
alkalinity
of
substances
using
the
you
pH
scale.
We
express
ion
concentration
this
as
the
logarithm
to
the
base
10
of
the
hydrogen
to:
−3

define
the
terms
pH
and
(in
):
mol dm
K
w
+
pH
=
−log
[H
]
10

perform
pH,
calculations
hydrogen
concentration
and
and
involving
hydroxide
K
.
ion
Most
pH
values
neutrality
we
(neither
use
in
acidic
chemistry
nor
are
alkaline)
on
a
being
scale
pH
of
7.
0
to
V
ery
14
with
concentrated
w
acids
pH
can
have
values
The
The
p
p
pH
above
values
below
0
and
very
concentrated
alkalis
can
have
14.
notation
notation
chemists
to
numbers
and
(as
in
express
pH)
is
small
is
−log
a
modied
values
shown
logarithmic
simply.
The
scale
used
relationship
by
between
the
below.
10
−12
Number
−log
The
p
1
×
(number)
notation
constants,
−7
10
1
12
can
also
be
=
pH
pH of
Strong
1
×
7
used
in
−log
[OH
]
pK
10
The
−5
10
0
10
5
relation
to
4
10
1
×
0
other
10
−4
equilibrium
e.g.
pOH
Simple
×
−log
[K
10
]
pK
a
=
−log
w
[K
10
]
w
calculations
strong
acids
=
a
are
acids
completely
ionised
in
water
.
So
the
pH
can
be
easily
calculated.
−3
1 mol dm
−1
HCl
= −log
(1
×
10
= −log
(1
×
10
−3
0.01 mol dm
) =
pH
0
pH
2
−2
HCl
−3
0.002 mol dm
) =
−3
HCl = −log
(2
×
10
) =
pH
2.7
Exam tips
+
You
must
make
sure
that
you
know
how
to
convert
pH
to
H
concentration
+
and
H
most
concentration
common
to
method
pH. This
may
vary from
calculator
to
calculator. The
is:
+
pH
e.g.
pH
3.6
to
[H
]
x
3.6
+
[H
]
106
±
→
(shift)
→
5
10
−4
→
answer
(2.51
×
10
)
−5
e.g.
1.8
×
10
1.8
f
→
you
are
not
to
→
sure
–
pH
EXP
ask
your
→
±
→
teacher.
log
→
±
→
answer
(4.74)
Chapter
Worked
example
Calculate
the
pH
of
−5
3.32
×
9
Acid–base
equilibria
1
a
solution
whose
hydrogen
ion
concentration
is
−3
mol dm
10
+
pH
=
−log
[H
]
10
−5
=
−log
(3.32
×
10
)
=
4.48
(note
that
there
are
no
units
to
pH)
10
Worked
example
Calculate
the
2
hydrogen
ion
concentration
of
a
solution
of
pH
10.5.
+
pH
=
−log
[H
]
10
+
]
[H
The
ionic
−pH
=
−10.5
10
=
product of
−10
10
=
water,
3.2
×
10
−3
mol dm
K
w
W
ater
can
act
substance
is
molecules
as
either
dissolved
can
act
as
an
in
an
acid
it
or
(see
acid
base
depending
Section
by
9.1).
donating
In
on
what
pure
protons
to
type
water
,
other
a
of
few
water
water
molecules.
+
H
O
+
H
O
2
2
acid
base
Y
H
O
+
OH
3
conjugate
acid
conjugate
base
+
We
can
simplify
this
equation
to:
H
O
Y
H
+
OH
2
+
[H
][OH
]
__________
For
which
the
equilibrium
expression
is:
K
=
[H
O]
2
Since
the
the
value
concentration
for
K.
The
of
water
expression
is
very
then
high,
we
can
incorporate
this
into
becomes:
+
K
=
[H
][OH
]
w
−14
K
is
called
the
ionic
product
of
water .
Its
value
is
1.00
×
10
2
mol
−6
dm
w
Did you know?
Using
K
to
calculate the
pH of
alkalis
w
Why
We
can
use
K
to
calculate
the
pH
of
strong
bases.
We
need
to
know
is
the
pH
of
water
7?
the
w
We
concentration
of
hydroxide
ions
in
the
solution
and
value
of
can
see
this
if
we
use
the
ionic
K
w
expression for
Worked
example
Calculate
pH
water:
+
3
[H
−14
][OH
]
=
1.00
×
]
=
[OH
]
10
+
of
a
solution
of
potassium
hydroxide
of
concentration
and
[H
−3
0.0250 mol dm
+
So
[H
2
]
−14
=
1.00
×
10
+
Step
1:
W
rite
the
ionic
product
expression
in
terms
of
[H
]
ions.
+
and
[H
−14
]
=
√
1.00
×
10
−7
=
1.00
×
10
−7
−14
K
=
1.00
×
10
2
mol
−6
−log
dm
(1.00
×
10
)
=
7
10
w
+
Step
2:
W
rite
the
equilibrium
expression
in
terms
of
[H
]
Key points
K
______
w
+
[H
]
=
[OH
]

Step
3:
Substitute
the
values
into
the
pH
is
a
measure
ion
concentration,
−14
1.00
]
×
=
the
[H
hydrogen
].
+
10
____________
+
[H
of
+
expression.
−13
=
4.00
×

pH
=
−log
[H
].
10
10
0.0250

K
is
the
ionic
product
of
water
w
−13
Step
4:
Calculate
pH:
−
log
(4.00
×
10
)
−14
=
pH
12.4
(value
10

The
=
pH
1.00
of
be found
a
=
[H
10
strong
using
+
K
×
][OH
the
2
mol
base
−6
dm
).
can
expression
−
].
w
107
9.3
Equilibrium calculations involving weak acids
Learning outcomes
The
acid dissociation
constant,
K
a
On
completion
should
be
able
of
this
section,
The
equilibrium
The
equation
law
can
be
applied
to
aqueous
solutions
of
weak
acids.
you
for
the
partial
ionisation
of
ethanoic
acid
in
water
H
O
is:
to:
+
COOH(aq)
CH

understand
the
dissociation
term
constant,
K
’
a

perform
calculations
involving
+
H
3
‘acid
Because
that
its
O(l)
Y
CH
2
water
is
present
concentration
at
is
a
COO
(aq)
+
3
very
constant
high
and
concentration,
we
(aq)
3
can
we
simplify
can
this
assume
equation
to:
+
the
use
of
K
,
pH
and
[H
].
+
a
CH
COOH(aq)
Y
CH
3
The
equilibrium
COO
(aq)
+
H
(aq)
3
expression
for
this
reaction
is:
+
[CH
COO
][H
]
_______________
3
K
=
a
[CH
COOH]
3

K
is
the
acid
dissociation
constant .
For
a
monobasic
weak
acid
such
a
−3
as
ethanoic
acid,
the
units
of
are
K
mol dm
a
−3

A
high
value
for
K
e.g.
40 mol dm
indicates
that
the
position
of
a
equilibrium
is
well
over
to
the
right.
The
acid
is
almost
completely
ionised.
−5

A
low
value
for
e.g.
K
1.3
×
10
−3
mol dm
indicates
that
the
position
a
of
equilibrium
is
well
over
to
the
left.
The
acid
is
only
very
slightly
ionised.

We
can
also
use
values
pK
to
compare
the
strengths
of
acids.
a
=
pK
−log
a
The
[K
10
general
].
a
equilibrium
expression for K
a

We
can
based
write
on
the
a
general
general
equilibrium
expression
for
all
monobasic
acid
reaction:
+
HA

HA

A
The
represents
represents
general
the
the
unionised
conjugate
equilibrium
Y
+
H
A
acid.
base
expression
of
is
the
acid.
therefore:
+
[H
][A
]
________
K
=
a
[HA]
+
Since
the
concentration
of
H
and
A
must
be
equal,
we
can
simplify
this
further:
+
[H
2
]
_____
=
K
a
[HA]
In
calculations
involving
K
we
make
two
assumptions.
a

If
the
K
value
for
the
weak
acid
is
very
small,
we
can
assume
that
the
a
concentration
as
the
of
acid
concentration
molecules
of
remaining
undissociated
acid
at
equilibrium
molecules
in
is
the
the
same
original
acid.

We
can
ignore
water
.
For
water
,
K
,
w
108
the
most
is
hydrogen
work
very
this
small.
is
ions
which
acceptable
arise
from
because
the
the
ionisation
ionic
product
of
for
Chapter
Calculations
involving
9
Acid–base
equilibria
K
a
We
can
calculate
the
value
of
K
for
a
weak
acid
if
we
Exam tips
know:
a

the
concentration

the
pH
of
acid
n
calculations
involving
K
,
a
of
the
solution.
remember
that
concentration
We
can
calculate
the
pH
of
a
solution
of
a
weak
acid
if
we
the
value
of
the
pH
and
pH
to
hydrogen
ion
concentration.
concentration
Worked
ion
converted
K
a

be
know
to

hydrogen
can
example
of
the
acid.
1
−3
Deduce
the
pH
of
a
solution
of
0.05 mol dm
−5
K
for
ethanoic
acid
=
1.74
×
ethanoic
acid.
−3
10
mol dm
a
Step
1:
W
rite
the
equilibrium
expression.
+
[H
2
+
]
[H
____________
K
2
]
_____
=
or
K
a
=
a
[CH
COOH]
[HA]
3
+
Step
2:
Rearrange
the
equation
+
[H
to
make
[H
]
=
K
×
Substitute
Step
4:
Find
the
subject:
[CH
values:
[H
COOH]
3
+
3:
the
2
a
Step
2
]
2
−5
]
=
(1.74
×
10
−7
)
×
0.05
=
8.7
×
10
+
[H
] by
taking
the
square
−7
√
8.7
×
=
10
root:
−4
9.33
×
−3
10
mol dm
+
Step
5:
Convert
[H
+
]
to
pH:
pH
=
−
log
[H
−4
]
=
−
log
10
pH
Worked
=
(9.33
×
10
)
10
3.03
example
2
−3
The
pH
of
a
0.01 mol dm
of
methanoic
acid
is
2.9.
Deduce
the
value
of
K
a
+
Stage
1:
Convert
Stage
2:
W
rite
pH
the
to
[H
+
]:
[H
equilibrium
−2.9
]
=
−3
10
=
1.26
+
[H
2
+
]
=
[H
Substitute
into
2
]
=
K
a
[HCOOH]
3:
−3
mol dm
_____
or
a
Stage
10
expression.
__________
K
×
the
[HA]
equilibrium
expression:
−3
(1.26
×
10
2
)
______________________
K
=
a
0.01
−4
K
=
1.59
×
10
−3
mol dm
a
Key points
+

For
a
weak
acid,
[H
]
and
pH
can
be
calculated
+
[H
using
the
expression:
−
][A
]
_______
K
=
a
[HA]
−
Where
[A
]
is
the
concentration
of
concentration
unionised
of
ionised
acid
and
[HA]
is
the
acid.
109
9.4
Acid
and
Learning outcomes
base
Using
dissociation
pK
constants
values
a
We
On
completion
of
this
section,
can
use
−log
a
be
able
values
to
compare
the
acidity
of
weak
acids.
a
=
pK
should
pK
you
[K
10
].
Some
examples
are
shown
in
the
table.
a
to:
−3

describe
pK
,
K
a

perform
and
pK
b
Acid
b
calculations
Equilibrium
involving
K
and
pK
b
and
solution
K
/mol dm
pK
a
+
H
SO
2
,
a
aqueous
a
sulphuric(IV)
pK
in
Y
−2
H
+
HSO
3
1.5
×
10
5.6
×
10
1.82
3
pK
b
w
+
hydrofluoric
HF
Y
methanoic
HCOOH
−4
H
+
F
+
acid
Y
3.25
−4
H
+
HCOO
1.6
×10
3.80
Exam tips
+
‘carbonic
acid’
CO
+
H
2
We
can
pH for
calculate
a
dibasic
the
acid
way
as for
a
Y
−7
H
+
HCO
4.5
such
as
H
S
in
a
The
higher
the
pK
monobasic
value,
the
weaker
the
because
the
6.35
acid.
acid.
can
use
pK
values
in
calculations
rather
than
K
a
is
10
a
We
This
×
3
approximate
2
similar
O
2
value
of
K
values.
a
is
a1
much
greater
than
K
.
a2
Worked
example
1
e.g.
−3
Deduce
the
pH
of
a
solution
of
0.001 mol dm
benzoic
acid.
pK
benzoic
a
–
H
S(aq)
Y
HS
+
(aq)
+
H
(aq)
acid
=
4.2.
2
–8
K
=
8.9
×
a1
10
–3
mol
dm
Step
1:
Convert
pK
to
K
a
–
HS
2–
(aq)
Y
S
+
H
(aq)
So
=
K
:
pK
a
+
aq)
10
−pK
a
K
=
1.2
×
a2
10
−
log
[K
10
]
a
−4.2
a
–13
=
a
K
=
−5
10
K
a
=
6.3
×
10
−3
mol dm
a
–3
mol
dm
Step
2:
W
rite
the
equilibrium
expression:
+
We
can
ignore
the
H
ions
arising
+
[H
2
]
_____
=
K
from
the
second
ionisation
their
concentration
a
because
[HA]
is
insignificant
+
Step
compared
with
those
3:
Rearrange
the
equation
to
+
the first
make
[H
2
]
the
subject:
arising form
ionisation.
[H
2
]
=
K
×
[HA]
a
+
Step
4:
Substitute
Step
5:
Find
the
values:
[H
2
]
−5
=
(6.3
×
10
−8
)
×
0.001
=
6.3
×
10
+
[H
]
by
taking
the
square
root:
−8
√
6.3
×
−4
10
=
2.51
×
10
+
Step
6:
Convert
[H
−3
mol dm
+
]
to
pH:
pH
=
−log
[H
]
−4
=
−log
10
pH
The
=
(2.51
×
10
)
10
3.6
basic dissociation
constant,
K
b
For
calculations
constant,
.
K
involving
For
a
example,
weak
in
the
alkali,
we
can
use
the
basic
dissociation
reaction:
b
+
NH
(g)
+
H
3
We
can
write
the
O(l)
Y
NH
2
equilibrium
(aq)
+
OH
(aq)
4
expression:
+
[NH
+
][OH
]
[BH
____________
4
K
][OH
]
___________
=
or
more
generally
K
b
=
b
[NH
]
[B]
3
+
Where
We
[B]
make
is
the
weak
similar
expression
(see
base
and
assumptions
Section
[BH
to
]
is
those
the
for
a
conjugate
weak
acid
acid.
in
applying
9.1).
+
Since
[BH
]
=
[OH
]
we
can
simplify
the
expression
2
[OH
]
_______
=
K
b
[B]
110
further
to:
this
Chapter
Worked
example
Calculate
the
9
Acid–base
equilibria
2
−3
pH
of
a
0.01 mol dm
solution
−4
K
(ethylamine)
=
6.5
×
of
−3
10
−14
mol dm
K
b
Step
ethylamine.
=
1.0
×
10
−3
mol dm
w
1:
W
rite
the
equilibrium
expression:
2
[OH
]
_______
K
=
b
[B]
2
Step
2:
Rearrange
the
equation
to
make
[OH
]
the
subject:
2
[OH
]
=
K
×
[B]
b
2
Step
4:
Substitute
Step
5:
Find
the
[OH
]
values:
by
taking
[OH
the
−4
]
=
(6.5
square
6.5
×
6:
Use
K
=
[H
−6
)
×
−3
=
10
2.55
×
10
calculate
[H
+
Step
10
0.01
=
6.5
×
10
root:
−6
√
×
−3
mol dm
+
][OH
]
to
]:
w
−14
K
1.0
______
w
+
[H
]
=
×10
___________
−12
=
=
3.92
×
10
−3
mol dm
−3
[OH
]
2.55
×
10
+
Step
7:
Calculate
pH:
pH
=
−log
[H
−12
]
=
−log
10
pH
=
(3.92
×
10
)
Exam tips
10
11.4
A
quick
way
question
Using
pK
and
pK
a
to
such
get
as
an
answer
‘calculate
to
the
a
pH
values
b
of
a
solution
of
a
strong
base
of
−3
A
useful
expression
which
relates
K
to
K
w
and
K
a
concentration
is
0.
1 mol dm
’
is
to
b
use
the
expression
14
–log
[OH
].
10
pK
+
pK
a
=
pK
b
w
+
This
works
log
[OH
because
−
log
[H
]
−
10
This
can
also
be
written
as:
]
=
14
10
−14
−log
K
10
Worked
−
log
a
K
=
10
b
example
3
14
(the
14
derives
from
−log
(1.0
×
10
))
10
−3
Deduce
the
value
of
K
for
a
0.01 mol dm
solution
of
aqueous
ammonia.
a
pK
for
ammonia
=
4.7
b
Step
1:
Apply
the
relationship
pK
+
pK
a
pK
+
4.7
=
14
So
pK
a
Step
2:
Convert
=
pK
b
to
calculate
pK
w
=
14
–
:
a
4.7
=
9.3
a
pK
to
K
a
:
a
−10
pK
=
−log
a
K
10
=
−log
a
(9.3)
K
10
=
5.0
×
10
−3
mol dm
a
Key points

For
a
using
weak
the
base
the
general
concentration
of
hydroxide
ions
can
be
calculated
expression:
+
−
][OH
[BH
]
__________
K
=
b
[B]
+
where

[BH
]
is
the
the
concentration
pK
=
a
−logK
and
a
concentration
of
pK
unionised
=
b
of
the
protonated
base
(salt)
and
[B]
is
base.
−logK
b
111
9.5
Changes
Learning outcomes
On
completion
should
be
able
of
this
in
pH
pH–titration
section,
you
Measuring
describe
changes
in
pH

explain
titrations
the
titrations
curves
pH
changes
changes
Figure
in
strong
or
weak
acid
strong
or
weak
base.
pH
when
reacts
with
a
a
when
9.5.1
electrode
from
all
the
the
are
is
an
graphs
acid
shows
placed
burette
time.
continuously

manually
to
by
the
is
the
in
at
The

added
base
curves
showing
how
the
pH
of
an
acid
or
base
during
changes
acid−base
acid–base
to:
Titration

during
a
pH
added
apparatus
the
acid
slow
is
using
in
and
a
data
in
the
the
an
alkali
used
the
to
or
an
follow
flask
constant
recorded
recording
alkali
to
and
rate.
alkali
these
the
The
added
to
changes.
alkali
added
solution
is
an
A
acid.
pH
gradually
kept
stirred
either:
logger
pH
attached
after
xed
to
a
computer
volumes
of
or
acid
have
been
flask.
(alkali)
Titration
The
output
to:
graphs
curves
in
Figure
weak
monoprotic
acids
in
when
it
9.5.2
acids
which
only
reacts.
The
are
one
show
the
titrated
hydrogen
exact
shape
results
with
ion
of
per
the
obtained
alkalis.
molecule
curve
when
strong
Monoprotic
of
depends
acid
on
acids
can
be
whether
and
are
donated
the
acid
meter
and
data
base
are
strong
or
weak.
logger
chart
recorder
a
pH
sensor
Figure 9.5.1
b
pH
PC
pH
14
14
12
12
acid
Monitoring the pH as a base
is added to an acid
equivalence
10
10
point
8
8
6
6
4
4
2
2
0
equivalence
point
0
0
5
10
15
20
25
30
35
40
45
50
–3
volume of
0.
10
mol
0
5
10
15
20
25
3
base
dm
added
(cm
)
c
30
35
40
45
50
–3
volume of
0.
10
mol
dm
3
base
added
(cm
)
d
pH
pH
14
14
12
12
equivalence
10
10
point
8
8
6
6
4
4
2
2
0
equivalence
point
0
0
5
10
15
20
25
30
35
40
45
–3
volume of
0.
10
Figure 9.5.2
mol
dm
50
0
5
10
15
20
25
3
base
added
(cm
)
30
35
0.
10
mol
dm
45
50
3
base
added
(cm
)
pH titration curves for: a a strong acid and a strong base; b a strong acid and
a weak base; c a weak acid and a strong base; d a weak acid and a weak base
112
40
–3
volume of
Chapter
Apart
very
from
steep
units.
The
called
the
steep
the
point
for
of
acid
the
where
at
a
the
and
weak
single
which
equivalence
portion
Strong
curve
portion
the
acid
drop
base
point .
Y
ou
and
of
has
can
weak
base
exactly
see
base,
the
changes
the
curves
pH
neutralised
that
this
by
the
show
Acid–base
equilibria
a
several
acid
corresponds
9
to
is
the
curve.
strong
base
−3
Figure
9.5.2(a)
shows
the
titration
of
0.10 mol
dm
hydrochloric
acid
−3
(strong
acid)
with
sodium
0.10 mol dm
HCl(aq)
+
NaOH(aq)
→
hydroxide
NaCl(aq)
(strong
+
H
base).
O(l)
2
We
can
see
why
the
pH
changes
so
rapidly
around
the
equivalence
point
3
by
calculation.
Just
before
the
equivalence
point
there
3
NaOH
and
is
24.9 cm
of
3
HCl
25 cm
present.
The
excess
0.1 cm
of
acid
is
present
in
3
solution.
49.9 cm
Exam tips
0.1
_____
+
So
[H
]
=
0.1
−3
×
=
0.0002 mol dm
=
pH
3.7
49.9
f
3
When
there
is
0.1 cm
+
[H
−3
of
alkali
in
excess:
−14
]
=
1.0
×
10
[OH
]
−11
/0.0002
=
5
×
10
=
0.0002 mol dm
−3
mol dm
So
pH
=
you
with
titrate
an
titration
10.3.
images
The
equivalence
point
is
at
approximately
pH
7
because
the
salt
base
(in
the flask)
(in
the
burette),
curves
will
be
of
those
that
the
are
the
pH
mirror
shown,
formed
e.g. for
is
a
acid
a
strong
acid
and
a
strong
neutral.
base:
Strong
acid
and
weak
base
pH
−3
Figure
9.5.2(b)
shows
the
titration
of
0.10 mol dm
hydrochloric
acid
14
−3
(strong
acid)
with
aqueous
0.10 mol dm
ammonia
(weak
base).
12
HCl(aq)
+
NH
(aq)
→
NH
3
The
ammonium
base
and
is
chloride
acidic
as
a
formed
result
of
is
Cl(aq)
10
4
the
salt
of
a
strong
acid
and
a
weak
8
hydrolysis:
6
+
NH
Cl(aq)
+
H
4
So
the
O(l)
→
NH
2
equivalence
point
(aq)
+
H
3
is
at
about
O
(aq)
+
Cl
(aq)
3
pH
4
5.
2
0
Strong
base
and
weak
acid
0
5
10
15
20
25
30
35
40
−3
Figure
9.5.2(c)
shows
the
titration
of
0.10 mol dm
ethanoic
acid
(weak
−3
acid)
with
sodium
0.10 mol dm
hydroxide
(strong
base).
Key points
+
CH
COOH(aq)
+
NaOH(aq)
→
CH
3
COO
Na
(aq)
+
H
3
O(l)
2

The
and
sodium
is
ethanoate
alkaline
CH
as
a
COO
formed
result
(aq)
of
+
is
the
salt
of
a
weak
acid
and
a
strong
hydrolysis:
H
3
O(l)
→
CH
2
COOH(aq)
+
OH
the
equivalence
point
is
at
shape
and
weak
pH–titration
acid
base
and
the
strength
of
curve
the
used.
(aq)
about
pH
For
a
strong
acid
and
a
strong
9.
there
sudden
base
a
on
base
Weak
of
depends
3

So
The
base
is
pH
a
very
change
large
and
corresponding
acid
to
the
end
point
of
the
titration.
−3
Figure
9.5.2(d)
shows
the
titration
of
0.10 mol dm
ethanoic
acid
(weak

For
a
strong
acid
and
a
weak
base
−3
acid)
with
aqueous
0.10 mol dm
ammonia
(weak
base).
or
a
weak
acid
and
a
strong
base,
+
CH
COOH(aq)
+
NH
3
(aq)
→
CH
3
COO
NH
3
there
(aq)
pH
The
ammonium
ethanoate
formed
is
the
salt
Both
leaving
the
the
ethanoate
equivalence
and
point
ammonium
of
at
around
ions
pH
a
weak
acid
and
a
COO
3
(aq)
+
H
O(l)
2
→
CH
large
and
sudden
can
undergo

+
to
the
point
of
the
titration.
hydrolysis
7.
COOH(aq)
corresponding
weak
For
a
there
CH
a fairly
change
end
base.
is
4
OH
weak
is
a
acid
and
gradual
a
weak
change
base
in
pH
as
(aq)
3
the
acid
is
added
to
the
base.
113
9.6
Acid–base
Learning outcomes
On
completion
should
be
able
of
this
indicators
Introduction to
section,
An
acid–base
pH
changes
know
that
indicator
is
a
indicators
substance
which
changes
colour
when
the
you
over
a
certain
range
of
pH
values.
Indicators
can
be
thought
to:
of

acid–base
specific
indicators
as
being
weak
acids
whose
conjugate
base
has
a
different
colour
.
are
+
HIn
used
to
determine
the
end
indicator
of
acid–base
titrations
Y
molecule
the
strength
of
the
In
deprotonated
indicator
depending
colour
on
+
H
point
acid
A
colour
B
and
+

In
more
acidic
(or
less
alkaline)
conditions
the
concentration
of
H
of
H
base
ions


explain
the
applied
to
explain
the
term
‘pH
range’
is
higher
.
as

So
the
position
of

So
the
indicator

In
equilibrium
moves
to
the
left.
indicators
basis for
shows
colour
A.
selecting
+
appropriate
indicators for
use
more
ions
acid–base
alkaline
(or
less
acidic)
conditions
the
concentration
in
is
lower
.
titrations.

So
the
position

So
the
indicator
Indicator
usually
indicators
to
of
be
its
a
yellow
greyish
at
change
is
That
a
is
colour
moves
to
the
right.
B.
pH
6
and
colour
the
this
the
point,
two
blue
over
colour
at
of
a
pH
the
the
colour
extremes.
pH
7.6.
range
indicator
At
For
an
of
of
one
or
should
the
two
indicator
example,
units.
change
in
may
be
bromothymol
intermediate
point
it
For
the
blue
appears
green.
colour
not
At
between
pH
The
shows
useful,
range.
intermediate
is
equilibrium
range
Indicators
middle
of
6
why
adding
acid
to
alkali

pH
7.6
↑
↑
yellow
grey-green
blue
change
good
←
↑
of
an
indicator
screened
indicator
alone
since
methyl
should
the
orange
also
colour
(a
be
distinct.
change
mixed
is
Methyl
rather
indicator)
is
orange
indistinct.
often
used
in
preference.
For
an
end
indicator
point
happen,
of
the
correspond
The
table
some
the
to
change
working
below
correctly,
titration
rapid
to
work
shows
–
the
in
the
vertical
pH
in
of
the
range
some
colour
the
part
must
of
region
the
of
change
curve.
the
end
sharply
For
this
point
at
to
must
indicator
.
colour
changes
and
the
working
range
of
indicators.
Indicator
Colour
at
pH
in the
lower
range
Colour
pH
at
in the
higher
range
pH
orange
red
at
pH
3.2
yellow
at
pH
4.4
3.7
methyl
red
red
at
pH
4.2
yellow
at
pH
6.3
5.
1
blue
phenolphthalein
yellow
at
pH
colourless
at
6.0
blue
at
pH
deep
10.
1
reddish
pH
pink
7
.6
at
at
point
methyl
bromothymol
7
.0
pH
10
9.3
8.2
alizarin
114
the
yellow
yellow
at
pH
at
pH
13.0
12.5
end
Chapter
Choosing the
pH
correct
9
Acid–base
equilibria
pH
indicator
The
selection
depends
on
of
a
correct
where
the
14
14
12
12
indicator
most
rapid
10
change
in
pH
occurs
as
the
acid
with
a
phenolph-
thalein
8
titrated
10
phenolph-
is
bromothymol
base.
Most
acid
and
indicators
strong
acid
and
strong
can
a
be
base
used
strong
for
a
methyl
base,
6
4
4
2
2
0
0
5
0
e.g.
blue
blue
6
Strong
10
15
20
25
30
35
40
45
50
–3
volume of
because
their
the
range
vertical
Phenolphthalein
0.
1
mol
added
(cm
10
15
20
25
30
35
40
45
50
–3
)
volume of
a
falls
titration
part
is
of
often
the
of
a
strong
acid
with
b
a
its
0.
10
titration
base
weak
mol
3
base
dm
added
(cm
)
colour
of
a
strong
acid
with
a
base
curve.
preferred
pH
because
5
3
base
dm
bromothymol
strong
within
0
orange,
phenolphthalein,
blue,
thalein
8
change
is
pH
more
obvious.
14
Strong
The
acid
sharpest
and
weak
change
in
base
the
14
12
12
10
10
thalein
8
curve
base
for
is
a
strong
between
Methyl
red,
acid
pH
and
3.0
methyl
a
and
orange
phenolph-
phenolph-
pH
bromothymol
weak
7.5.
blue
6
blue
are
4
methyl orange
because
the
their
vertical
part
range
of
the
falls
0
curve.
5
0
Phenolphthalein
is
10
its
colour
to
the
20
25
30
35
40
45
50
–3
change
does
vertical
part
0.
10
mol
dm
0
added
(cm
)
curve.
It
would
slowly
after
only
the
15
20
25
30
35
40
45
50
–3
0.
10
mol
3
base
dm
added
(cm
)
titration
of
a
weak
acid
with
d
a
titration
of
a
weak
acid
with
a
of
weak
base
change
Figure 9.6.1
colour
10
volume of
str
the
5
3
base
not
c
correspond
15
unsuitable
volume of
because
methyl orange
2
0
within
blue
suitable
2
indicators
bromothymol
6
or
4
bromothymol
thalein
8
Selecting a suitable indicator for different acid–base titrations: a a strong
equivalence
acid and a strong base; b a strong acid and a weak base; c a weak acid and a strong base;
point.
d a weak acid and a weak base
Weak
The
acid
sharpest
between
its
pH
colour
and
not
and
fall
colour
change
7.0
and
change
methyl
strong
within
the
11.
pH
the
are
not
vertical
before
curve
for
a
Phenolphthalein
corresponds
orange
slowly
in
base
the
to
the
suitable
part
of
weak
is
a
vertical
part
indicators
the
equivalence
curve.
acid
and
suitable
of
the
because
They
a
strong
indicator
curve.
their
would
Methyl
range
only
base
is
because
red
does
change
Key points
point.

Weak
No
acid
and
indicator
indicators
changing
is
weak
suitable
would
slowly.
show
For
base
for
a
a
colour
weak
sudden
example,
Acid–base
acid
and
change
in
a
weak
colour
bromothymol
blue
base.
None
because
starts
the
of
pH
changing
indicators
when
there
change
in
The
range
is
change
a
sharp
pH.
the
is

the
colour
pH
range
of
an
(usually
indicator
about
2
is
pH
3
when
about
of
24 cm
alkali
have
been
added
and
nishes
changing
units)
over
which
the
indicator
3
colour
when
26 cm
have
been
added.
changes

For
in
an
an
indicator
acid–base
colour
to
colour.
the
curve
change
place
where
in
to
work
must
the
pH
correctly
titration,
the
correspond
pH-titration
changes
suddenly.
115
9.7
Buffer
solutions
Learning outcomes
What
A
On
completion
of
this
section,
is
buffer
be
able
the
buffer
solution
is
solution?
a
solution
which
minimises
pH
changes
when
small
you
amounts
should
a
of
acids
or
alkalis
are
added.
to:

define
term

explain
how
control
pH
‘buffer
buffer
solution’
Acidic
buffers
solutions
An
acidic
base
(the
buffer
salt
solution
of
the
is
weak
a
mixture
acid).
An
of
a
weak
example
is
acid
an
and
its
aqueous
conjugate
solution
of
−3

describe
the
importance
of
buffer
ethanoic
acid
)
(0.1 mol dm
and
its
conjugate
base,
sodium
ethanoate
−3
solutions
and
in
biological
industrial
systems
).
(0.1 mol dm
processes.
+
CH
COOH(aq)
Y
CH
3
So
a
and
If
buffer
the
not
ratio
of
change
pH)
On
solution
conjugate
will
base.
conjugate
contains
We
much,
change
addition
acid
say
of
acid
the
very
to
relatively
that
concentration
very
not
COO
(aq)
+
H
(aq)
3
weak
of
high
are
conjugate
hydrogen
much
this
there
(see
buffer
base
ion
also
(salt)
concentrations
reserve
base
supplies
(added
salt)
concentration
Section
of
of
both
and
(and
acid
base.
acid
does
therefore
the
9.3).
solution:
+

ions
H
combine
with
CH
COO
ions.
3

So
the
position
undissociated

But
because
COO
CH
of
,
of
equilibrium
acid
is
the
the
shifts
to
the
left
and
a
little
more
formed.
relatively
ratio
of
high
base
to
concentration
acid
changes
of
very
added
base
little.
3

So
On
the
pH
addition
hardly
of
changes.
alkali
to
this
buffer
solution:
+

OH
ions
combine
with
H
ions
to
form
water
.
+

The
and
removal
a
little
of
H
more
ions
acid,
shifts
the
position
COOH
CH
of
equilibrium
dissociates
to
CH
3

But
because
COO
CH
,
of
the
the
relatively
ratio
of
to
the
right
COO
3
high
base
to
concentration
acid
changes
of
very
added
base
little.
3

A
So
the
buffer
pH
may
hardly
be
changes.
made
more
effective
by
increasing
the
concentration
of
Exam tips
both
Remember
the
buffer
and
most
base
solutions
completely
salt
that
conjugate
is
in
a
of
salt.
ionised. You
conjugate
interchangeably.
If
acidic
base
t
may
used
the
does
is
see
the
acid
ratio
not
of
the
amounts
of
position
the
conjugate
concentration
change
therefore
the
and
very
pH)
acid
of
concentration
much,
will
or
not
alkali
either
the
are
the
(salt).
conjugate
very
added,
has
acid
base
hydrogen
change
equilibrium
of
of
base
been
or
so
its
ion
much.
the
(added
and
concentration
If,
buffer
far
salt)
however
,
will
altered
conjugate
not
as
base
to
(and
very
work
an
large
because
lower
by
acid
the
unacceptable
amount.
Basic
A
buffers
basic
buffer
solution
is
a
mixture
of
a
weak
base
and
its
conjugate
acid
−3
(salt).
An
example
is
an
aqueous
solution
of
ammonia
(0.1 mol dm
−3
its
conjugate
base,
ammonium
chloride
(0.1 mol dm
+
NH
(aq)
3
weak
116
+ H
(aq)
).
+
Y
NH
(aq)
4
base
conjugate
acid
)
and
Chapter
This
and
On
buffer
solution
conjugate
addition
acid.
of
contains
There
acid
to
relatively
are
this
reserve
buffer
high
concentrations
supplies
solution,
of
the
added
the
both
Acid–base
equilibria
base
acid.
aqueous
+
removes
of
9
ammonia
+
ions.
H
A
little
more
salt,
NH
,
is
formed.
4
On
addition
of
alkali
to
this
buffer
solution,
the
aqueous
ammonia
+
removes
shifts
the
to
the
Buffers
Many
A
left
in
of
their
of
and
will
depend
number
H
a
on
with
hydrogencarbonate
more
work
The
pH
at
of
specic
the
to
the
help
buffer:
in
water
.
The
ammonia
systems
buffers
water
form
is
position
of
equilibrium
formed.
systems
only
bodies.
blood
to
little
buffer
Hydrogencarbonate
combines
ions
natural
enzymes
animals
parts
added
to
pH
keep
blood
is
maintain
Carbon
blood
form
pH
a
near
pH
constant
approximately
this
dioxide
to
values
the
(a
7.
in
So
various
pH
7.4.
pH.
product
solution
of
respiration)
of
ions.
+
(g)
CO
+
H
2
O(l)
Y
HCO
2
(aq)
+
H
(aq)
3
+
If
the
H
ion
concentration
in
the
blood
rises:
+

H

The
ions
combine
with
HCO
ions.
3
position
of
equilibrium
moves
slightly
to
the
left.
+

The

concentration
There
is
buffer
to
still
a
of
high
ions
H
enough
is
reduced
to
concentration
keep
of
the
pH
constant.
ions
HCO
to
allow
the
3
function.
+
If
the
ions
OH
to
ion
form
Phosphate
blood
concentration
water
.
The
buffers:
plasma.
The
in
the
blood
equilibrium
Several
type
equilibrium
of
rises,
moves
slightly
phosphate
involved
they
ions
2
Y
blood
gets
too
acid,
the
right.
present
in
the
+
H
4
acid
If
are
the
H
+
HPO
4
to
with
is:
2−
PO
H
combine
conjugate
position
of
base
equilibrium
shifts
slightly
to
the
left
2−
but
there
are
still
sufcient
PO
H
2
phosphates
Protein
buffers.
the
to
act
buffers:
These
amino
as
a
acids
or
HPO
4
in
solution
in
the
for
the
4
buffer
.
Some
buffers
and
proteins
can
be
and
acidic
amino
or
acids
basic
depending
blood
on
the
act
as
charges
on
Key points
proteins:

+
PH
Y
+
P
+
H
or
A
buffer
Y
P
protonated
deprotonated
protonated
deprotonated
form
form
form
form
in
A
industry
solutions
are
used
in
electroplating,
the
manufacture
of
dyes
a
addition
of
buffer
weak
or
Buffer
on
amount

solutions
minimises
pH
+ H
change
Buffer
solution
+
PH
acid
or
solution
acid
weak
and
its
base
of
a
small
base.
is
a
mixture
conjugate
and
its
of
a
base
conjugate
and
acid.
treatment
of
leather
.
They
are
also
present
in
some
detergents,
soaps
and

shampoos
where
it
is
important
not
to
damage
fabric,
skin
or
Buffer
solutions
maintain
pH
by
hair
.
keeping
acid

a fairly
and
Blood
constant
conjugate
buffer
ratio
of
base.
solutions
include
−
HCO
ions,
phosphate
ions
and
3
proteins.
117
9.8
Calculations
Learning outcomes
involving
Equilibrium
Consider
On
completion
of
this
section,
a
buffer
solutions
aspects
buffer
solution
consisting
of
a
weak
acid
and
its
conjugate
you
base:
should
be
able
to:
+
CH

know
how
to
calculate
the
pH
COOH(aq)
buffer
We
data
can
write
hydrogen
know
acid
COO
conjugate
(aq)
+
H
(aq)
base
solution from
appropriate

CH
3
weak
a
Y
3
of
how
to
calculate
the
an
equilibrium
expression
for
this
reaction
in
terms
of
ions:
acid/
+
base
ratio
required
to
make
a
COO
[CH
][H
]
_______________
3
K
=
a
buffer
solution
with
a
particular
[CH
COOH]
3
pH
[CH
[H
COOH]
[HA]
____________
3
+
So
]
=
K
_____
+
×
or
more
generally
[H
]
=
K
a

know
how
to
conduct
×
a
an
[CH
COO
]
[A
]
3
experiment
to
determine
the
pH
Since
K
is
constant,
the
ratio
of
the
concentration
of
acid
to
conjugate
a
of
a
buffer
solution.
+
base
(salt)
controls
the
ion
H
concentration
(and
thus
the
pH).
Dilution
−
of
the
buffer
has
no
Deducing the
We

can
the
calculate
value
of
effect
on
pH of
its
a
since
buffer
the
pH
of
of
the
weak
K
a
pH
buffer
and
A
HA
are
diluted
equally.
solution
solution
if
we
know:
acid
a

the
equilibrium
Worked
Deduce
concentrations
example
the
pH
of
a
of
to
weak
acid
and
conjugate
base.
1
buffer
solution
made
3
ethanoate
the
by
adding
0.20 mol
of
sodium
−3
of
500 cm
0.10 mol dm
ethanoic
acid.
K
of
ethanoic
acid
a
−5
=
1.7
Step
×
1:
−3
mol dm
10
Calculate
the
concentrations
of
the
acid
and
its
conjugate
base:
−3
[CH
COOH]
=
0.10 mol dm
=
0.2
3
1000
_____
[CH
COO
]
−3
×
=
0.40 mol dm
3
500
+
Step
2:
Rearrange
the
equilibrium
expression
terms
of
[H
]
ions:
COOH]
[CH
____________
3
+
[H
in
]
=
K
×
a
[CH
COO
]
3
Step
3:
Substitute
the
values:
0.10
+
[H
]
_____
−5
=
1.7
×
10
×
−6
=
4.25
×
10
−3
mol dm
0.40
+
Step
4:
Calculate
pH:
pH
=
−log
[H
]
−6
=
−log
10
(to
We
can
2
(4.25
×
10
)
pH
make
buffer
solution
calculations
easier
to
deal
with
if
we
expression:
[salt]
_____
pH
=
+
pK
log
a
10
[acid]
So
if
we
are
given
the
pK
or
calculate
it
from
a
calculation
much
more
K
,
we
can
do
the
a
simply.
−5
Step
1:
Convert
K
to
a
118
=
5.4
10
s.f.)
pK
:
a
−log
(1.7
10
×
10
)
pK
=
a
4.77
use
the
Chapter
Step
2:
Substitute
into
the
9
Acid–base
equilibria
equation:
Exam tips
log
(0.40)
__________
10
pH
=
4.77
+
n
buffer
solution
calculations,
do
(0.10)
not fall
pH
=
4.77
+
=
5.4
(to
2
concentration
the
of
acid–conjugate
base
ratio
same
as
conjugate
hardly
any
an
acidic
buffer
solution,
the
higher
the
acid–conjugate
base
lower
with
an
a
is
the
pH
particular
acid
or
in
that
[salt]
Worked
is
pH,
what
concentrations.
of
we
buffer
.
need
volumes
The
the
the
example
same
example
as
to
to
But
in
know
mix
order
below
to
exactly
solutions
shows
[conjugate
of
thinking
of
hydrogen
that
ions
make
how
of
the
how
to
a
buffer
much
acid
do
this.
Remember
conjugate
ionisation
of
that
base
the
added
as
a
acid.
t
salt.
solution
base
and
the
concentration
ratio,
is
the
the
base.
of
comes from
For
trap
s.f.)
is
Calculating the
the
0.6
the
pH
into
to
salt
add
of
to
known
Remember
base].
2
3
How
many
solution
of
pH
moles
of
containing
sodium
propanoate
0.10 mol
of
must
propanoic
be
acid
added
to
to
make
a
250 cm
buffer
of
solution
5.30?
−5
of
K
propanoic
acid
=
1.35
×
10
−3
mol dm
a
Step
1:
Calculate
the
concentration
of
the
propanoic
acid:
1000
_____
[acid]
=
0.10
−3
×
=
0.40 mol dm
250
+
Step
2:
Calculate
[H
Step
3:
Rearrange
+
]
the
from
pH:
[H
equilibrium
−5.30
]
=
10
−5
=
expression
5.01
with
×
[salt]
10
as
−3
mol dm
the
subject:
[acid]
_____
[salt]
=
K
×
a
+
[H
Step
4:
Substitute
the
]
values:
−5
(1.35
×
)
10
×
0.4
__________________
[salt]
=
−3
=
−
5.01
×
1.08 mol dm
5
10
250
_____
Step
5:
Calculate
number
of
moles:
1.08
×
=
0.27 mol
1000
Key points
Determining the
We
can
pH
electrode
into
the
from
that
nd
pH
of
a
connected
buffer
the
the
the
pH of
solution
meter
.
The
potential
buffer
to
a
and
a
solution
pH
the
meter
.
value
functioning
difference
buffer
of
(voltage)
experimentally
The
of
the
solution
glass
the
glass
pH
pH
found
electrode
between
the
by
using
electrode
by
is
direct
based
surface
of
a
is
glass
the
glass
The
pH
can
reading
on
the

dipped
be
of
a
equilibrium
fact
weak
and
buffer
calculated
acid
together
solution
using
the
concentrations
and
with
its
the
of
conjugate
K
the
base
value for
the
a
the
solution
varies
linearly
with
the
pH.
A
hydrogen
electrode
(see
section
weak
10.1)
can
also
be
used
to
determine
acid.
pH.

The
acid/
solution
the
base
can
values
and
the
K
of
ratio
be
of
a
buffer
calculated
pH
value
of
of
the
the
using
solution
weak
acid.
a

The
be
pH
using
to
of
a
buffer
determined
a
a
pH
glass
solution
can
experimentally
electrode
connected
meter.
119
10
Electrode
10.
1
Introducng
Learning outcomes
potentals
and
standard
Introducing
equlbrum
electrode
electrode
potental
potentials
2+
When
On
completion
of
this
section,
a
copper
equilibrium
should
be
able
rod
is
placed
in
a
solution
is
set
the
electrode
(aq)
ions,
the
following
to:
2+
define
Cu
up:
term
−
(aq)
Cu

of
you
+
2e
Y
Cu(s)
‘standard
There
potential’
are
two
opposing
reactions:
2+

describe
the
standard
hydrogen

Cu
ions
in
solution
accept
electrons
and
become
Cu
atoms.
2+
electrode
The
ions
Cu
get
reduced.
2+

describe
standard
potentials for

electrode
metal–metal
Cu
atoms
lose
electrons
and
become
Cu
ions.
The
Cu
atoms
get
oxidised.
ion
systems.
For
a
relatively
unreactive
metal
such
as
copper
,
the
position
of
this
2+
equilibrium
Cu
atoms
lies
are
to
the
right.
relatively
ions
Cu
difcult
to
are
relatively
easy
to
reduce
and
oxidise.
Did you know?
For
The
absolute
metal
and
thought
potential
its
to
ions
be
in
between
solution
caused
by
a
a
a
relatively
equilibrium
reactive
lies
to
the
metal
as
magnesium,
the
position
of
this
left.
2+
are
build
such
−
(aq)
Mg
+
2e
Y
Mg(s)
up
2+
Mg
of
electronic
charge
on
the
ions
easy
of
the
of
positive
layer
given
metals. This
attracts
ions. An
off
charge
by
a
more
metal,
difference
a
electrical
is formed. The
the
to
double
greater,
the
the
metal
When
a
the
metal
ions
in
metal
potential)
solution.
measure
system.
and
relatively
difcult
to
reduce
and
Mg
atoms
are
relatively
oxidise.
layer
electrons
between
are
surface
is
It
is
the
We
is
placed
established
not
a
this
solution
between
possible
difference
call
in
to
of
the
one
its
ions,
metals
measure
between
value
the
this
a
voltage
metal–metal
electrode
voltage
atoms
and
(absolute
the
directly.
ion
metal
But
system
ions
we
and
in
can
another
potential
solution.
electrical
metal
surface
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
–
+
The
standard
hydrogen
electrode
double
layer
If
we
we
want
need
the
a
to
standard
The
compare
standard
hydrogen
standard

hydrogen

an

a
the
at
for
of
is
electrode
101 kPa
(1
different
metals
comparison.
electrode
hydrogen
gas
ability
electrode
by
The
denition
consists
to
release
electrode
zero
at
all
electrons,
potential
of
temperatures.
of:
atmosphere)
in
equilibrium
with
−3
aqueous
platinum
the
platinum
gas
which
hydrogen
temperature
298
of
electrode
hydrogen
platinum
the
solution
and
hydrogen
coated
with
hydrogen
allows
ions
close
at
a
concentration
platinum
ions.
contact
black
(Platinum
between
in
1.00 mol dm
contact
black
the
of
is
with
nely
hydrogen
both
divided
gas
and
ions.)
K
salt
wire
bridge
hydrogen
(101
gas
kPa)
finely divided
Pt
+
H
–3
(aq),
1
mol
dm
Figure 10.1.1
black
on
Pt
The standard
electrode
hydrogen electrode
The
half
equation
for
this
electrode
is:
+
H
(aq)
1
__
−
+
e
Y
H
2
120
(g)
2
Chapter
Measuring
In
order
to
standard
measure
the
electrode
electrode
Zn
place
a
rod
of
pure
zinc
potentials
and
equilibrium
relating
to
the
half
equation:
−
(aq)
+
2e
Y
Zn(s)
−3
we
Electrode
potentials
potential
2+
10
into
a
2+
1.00 mol dm
solution
of
Zn
ions.
2+
This
is
the
half
Zn/Zn
cell.
We
then
connect
this
to
a
standard
Ø
hydrogen
electrode .
comparing
the
We
voltage
measure
of
the
standard
half
cell
with
electrode
the
potential,
standard
,
E
by
hydrogen
electrode.
V
0.76 V
salt
bridge
H
(101
kPa)
2
Exam tip
Zn
Pt

It
is
the
common
bridge
error
to
allows
think
that
conduction
2+
+
H
a
salt
Zn
(aq),
(aq)
of
electrons.
It
connects
two
–3
1
mol
dm
ionic solutions,
Ø
Figure 10.1.2
Measuring E
of
ions
the
half
between
ionic
other
cells
solution.
affect
is
porous
The
of
two
half
material
when
of
electrode
temperature
25 °C

concentration

pressure
salt
bridge.
completing
be
salt
in
a
high
solution,
a
strip
electrical
be
have
potentials.
conduction,
ionic balance.
movement
circuit
of
potassium
resistance
we
allows
electrical
is
temperature
So
electrode
This
the
bridge
saturated
potential.
comparing

a
The
soaked
should
pH
by
cells,
maintained.
ions,
standard
conditions
connected
voltmeter
Concentration
all
the
balance
inert
are
must
half cell
maintaining
two
it
2+
for a Zn/Zn
allowing
The
so
lter
so
that
paper
or
nitrate

Remember
a
cell
is
that
the
sometimes
‘potential
voltage
called
of
its
difference’.
voltmeter
.
and
to
pressure
use
These
of
gases
standard
conditions
are:
(298 K)
−3
of
of
any
ions
1.00 mol dm
gases
involved
101 kPa.
Ø
The
standard
under
other
electrode
standard
half
potential,
conditions
with
E
,
the
of
a
half
standard
cell
is
the
hydrogen
voltage
measured
electrode
as
the
cell.
2+
For
the
the
voltage
half
Zn/Zn
cell
connected
to
the
standard
hydrogen
electrode,
2+
developed
is
–
0.76 V
.
If
we
replace
the
half
Zn/Zn
cell
by
a
2+
half
Cu/Cu
depends
respect
on
to
cell,
the
whether
the
voltage
the
standard
half
is
+
cell
hydrogen
0.34 V
.
donates
The
or
electrode
sign
of
receives
(see
the
cell
voltage
electrons
Section
with
10.3).
Key points

Electrode
or

ions
The
in
potential
solution
standard
measures
to
accept
hydrogen
the
ability
of
an
element
to
in
H
release
electrons
electrons.
electrode
is
a
half
cell
which
gas
at
a
pressure
2
−3
of
101 kPa
is
in
equilibrium
with
a
solution
of
1.00 mol dm
+
H
ions.
Ø

The
standard
half
cell
electrode
under
standard
potential,
E
conditions
,
of
a
half
compared
cell
with
is
a
the
voltage
standard
the
hydrogen
electrode.
121
10.2
Electrode
Learning outcomes
potentals
A variety of
and
half
cell
potentals
cells
Ø
There
On
completion
of
this
section,
by
should
be
able
are
three
main
types
of
half
cell
whose
E
values
can
be
obtained
you
connecting
them
to
a
standard
hydrogen
electrode:
to:
2+

describe
the
standard
measurement
electrode
potentials for
non-metal–non-metal

metal–metal

non-metal–non-metal
ion

half
systems

the
measurement
electrode
involving
element
oxidation

define
cells,
the
(aq)
Zn(s)/Zn
cells
containing
ion
ions
half
half
cell
(see
Section
10.1)
of
cells
the
same
element
in
different
oxidation
of
Half
standard
same
half
states.
describe
systems
ion
of
in
ions
two
of
the
different
states
‘standard
cells
containing
non-metals
and
non-metal
ions
potentials for
cell
If
no
metal
using
a
is
present,
platinum
reaction
but
solution
of
it
its
electrical
electrode.
must
ions.
be
in
connection
The
platinum
contact
Figure
with
10.2.1
with
is
both
shows
the
the
inert.
the
solution
It
plays
element
chlorine/
is
no
made
part
and
an
chloride
in
the
aqueous
ion
half
potential’
Ø
cell

calculate
standard
cell
connected
to
the
standard
hydrogen
electrode.
The
value
E
for
this
potential,
system
is
+
1.36 V
.
Ø
E
cell
1
Ø
Cl
(g)
+
e
Y
Cl
in
terms
(aq)
E
=
+
1.36 V
2
2
Ø
Note:
It
does
equation
in
not
this
matter
way
or
as
(g)
Cl
of
+
E
values
2e
Y
2Cl
whether
we
write
the
half
(aq)
2
V
+1.36 V
salt
bridge
Cl
(g),
2
1
H
atmosphere
(g),
2
1
atmosphere
Pt
Pt
298 K
Cl
+
H
–3
1
mol
–3
(aq),
1
mol
dm
dm
Ø
Figure 10.2.1
For
a
non-metallic
bromide,
in
Measuring E
the
ions,
the
Half
When
a
states,
ions.
sulphur/
cells
in
foil
The
cell
the
equilibrium
should
same
be
with
half
applies
to
in
a
its
the
ions,
e.g.
bromine
solid
in
bromine/
liquid
equilibrium
and
half
with
its
sulphide.
containing
half
both
liquid
platinum
bromide
e.g.
for the chlorine/ chloride half cell
ions
contains
ions
ions
must
in different oxidation
of
have
the
the
same
same
element
standard
in
states
different
oxidation
concentration
of
−3
.
1.00 mol dm
A
platinum
electrode
is
used
to
make
3+
with
the
solution.
Figure
10.2.2
shows
the
connection
2+
/Fe
Fe
electrical
half
cell
connected
to
Ø
the
standard
hydrogen
electrode.
3+
Fe
Some
half
cells
MnO
−
(aq)
may
also
−
+
e
+
8H
Fe
+
for
acids
or
E
=
+
5e
Y
ions
are
included
−
The
MnO
,
4
−3
1.00 mol dm
122
is
+
0.77 V
.
alkalis
0.77 V
e.g.
Mn
Ø
(aq)
+
4H
O(l)
E
=
+
1.52 V
2
hydrogen
reaction.
system
2+
4
The
this
Ø
(aq)
−
(aq)
value
E
2+
Y
contain
+
(aq)
The
.
because
2+
Mn
they
are
essential
for
the
+
and
8H
are
all
present
at
a
concentration
of
Chapter
10
Electrode
potentials
and
equilibrium
V
+0.77 V
salt
bridge
H
(g),
2
1
3+
Fe
–3
(aq),
1
mol
298 K
dm
2+
Fe
atmosphere
–3
(aq),
1
mol
dm
+
H
Ø
Figure 10.2.2
Standard
We
can
3+
Measuring E
cell
calculate
for the Fe
–3
(aq),
1
mol
dm
2+
/Fe
half cell
potential
the
voltage
of
an
electrochemical
cell
made
up
of
two
Ø
half
cells
whose
values
E
relative
to
the
standard
hydrogen
electrode
are
known.
Ø
The
voltage
measured
is
the
difference
between
the
E
values
of
the
two
Ø
half
cells.
This
is
called
the
standard
cell
potential ,
.
E
The
standard
cell
cell
potential
two
is
standard
the
half
voltage
cells
are
developed
under
standard
conditions
when
joined.
2+
Figure
10.2.3
shows
a
2+
ion
Cu/Cu
half
cell
connected
to
a
Zn/Zn
half
cell.
temperature
298 K
+
V
–1.
10 V
salt
bridge
zinc
copper
2+
Zn
2+
298 K
(aq)
Cu
(aq)
–3
1
Figure 10.2.3
mol
–3
dm
1
mol
dm
Two half cells connected to give the standard cell potential
Exam tips
The
two
relevant
half
equations
2+
Cu
−
(aq)
+
Ø
2e
2+
Y
Cu(s)
E
−
(aq)
Zn
are:
+
=
+
0.34 V
Always
=
−
0.76 V
the
make
sure
that
you
include
Ø
2e
Y
Zn(s)
E
you
sign
are
of
electrode
doing
potential
calculations
when
involving
Ø
T
o
nd
the
value
of
E
,
we
subtract
the
less
positive
(or
more
negative)
cell
electrode
Ø
E
potentials. The
negatives
Ø
value
from
the
more
positive
(or
less
negative)
E
value.
and
positive
can
get
confusing.
Ø
So
in
this
case
E
=
+
0.34
–
(−
0.76)
=
+
1.10 V
cell
Key points

The
standard
compared

A
electrode
with
platinum
a
potential
standard
electrode
is
of
a
hydrogen
used
with
half
cell
is
the
voltage
of
the
half
cell
electrode.
half
cells
in
which
there
is
no
metal
electrode.

The
standard
conditions
cell
when
potential
two
is
the
standard
voltage
half
cells
developed
are
under
standard
joined.
Ø

Standard
cell
potential,
E
Ø
,
is
the
difference
between
the
E
values
of
the
cell
two
half
cells.
123
10.3
Redox
reactons
and
cell
dagrams
Ø
Learning outcomes
E
values

On
completion
of
this
section,
By
convention,
be
able
redox
reactions
electrode
potentials
refer
to
a
reduction
reaction,
so
you
the
should
and
electrons
appear
on
the
left
of
the
half
equations.
to:
Ø


use
standard
electrode
The
more
reduce
to
determine
the
direction
in
a
(or
less
negative)
the
value
of
,
E
the
ions
or
other
species
on
the
left
of
the
the
easier
equation.
it
is
They
to
are
of
better
electron ow
positive
potentials
at
accepting
electrons.
So
they
are
better
oxidising
agents.
cell
Ø


construct
The
more
negative
(or
less
positive)
the
value
of
,
E
the
easier
it
is
to
cell diagrams.
oxidise
the
releasing
weaker
species
on
electrons.
the
They
right
are
of
the
better
half
equation.
reducing
They
oxidising
stronger
2+
–
(aq)
Cr
oxidising
+
Y
(aq)
+
2e
Y
Zn(s)
+
e
–0.91 V
=
–0.76 V
Y
=
0.00 V
power
H
(g)
E
2
in
species
3+
2+
(aq)
Fe
+
e
Y
Fe
E
=
+0.77 V
the
+
–
(aq)
Ag
+
e
Ag(s)
E
=
+0.80 V
=
+1.36 V
(g)
+
e
Y
Cl
(aq)
E
2
2
stronger
oxidising
and
release
with
the
Zn
weaker
reducing
agent
list
are
above,
good
electrons
ions
Cl
more
readily)
agent
to
e
Ø
Cl
Cr
–
(lose
1

right
Ø
Y
more
readily)
on
Ø
(aq)
left
Referring
of
Ø
2
e
=
reducing
E
1
(aq)
increasing
E
Ø
of
H
Cr(s)
–
+
(gain
reducing
Ø
2e
2+
Zn
the
at
agent
increasing
species
better
agents.
agent
power
are
or
we
see
reducing
(the
that:
agents
reaction
is
because
further
over
they
to
are
the
more
left)
likely
to
compared
Ag.
3+

Cl
is
a
better
oxidising
agent
than
Fe
because
it
accepts
electrons
2
3+
more
readily
than
ions.
Fe
3+

Zn
releases
electrons
more
readily
than
H
and
Fe
accepts
electrons
2
+
more
readily
than
H
The direction of
When
we
connect
electron flow
two
half
in
cells
cells:
Ø

The
half
positive
cell
with
the
more
with
the
less
positive
(less
negative)
value
of
E
(more
negative)
value
of
E
is
the
is
the
pole.
Ø

The
half
negative
So
when
cell
positive
pole.
the
following
two
half
cells
+
Reaction
1:
Ag
are
–
(aq)
+
e
Ø
Y
2+
Reaction
2:
connected
(aq)
Cu
Ag(s)
E
=
–
+
2e
+
0.80 V
Ø
Y
Cu(s)
E
=
+
2+
the
Cu/Cu
0.34 V
+
half
cell
is
the
negative
pole
and
the
Ag/Ag
half
cell
is
the
Exam tips
positive
pole.
Ø
In
order
to
deduce
the
direction
of
This
is
because
according
to
values,
E
copper
is
better
at
releasing
+
electron ow,
remember
that:
the
electrons
than
silver
and
Ag
ions
are
better
at
accepting
electrons
than
2+
more
positive
electrons.
attracts the negative
Cu
goes
ions.
in
the
Therefore
backward
Reaction
1
direction.
goes
So
in
the
the
forward
direction
of
direct
electron
2+
external
124
wires
is
from
the
Cu/Cu
and
Reaction
flow
in
+
half
cell
to
the
Ag/Ag
half
cell.
the
2
Chapter
10
Electrode
potentials
and
equilibrium
e
salt
Ag
Cu
bridge
+
Ag
2+
(aq)
Cu
(aq)
+
Ag
Figure 10.3.1
2+
(aq)
+
e
(aq)
+
2e
The direction of electron ow is from the negative to the positive pole
Cell diagrams
A
convenient
way
of
For
a
write
cell
diagram
the
outside
of
reduced
the
cell
representing
diagram.
constructed
as
shown
two
below.
metal/
The
cells
metal
reduced
is
called
ion
a
cell
systems
forms
are
on
we
the
diagram.
form
oxidised
form
oxidised
2+
Zn(s)
phase
electrochemical
from

Zn
form
reduced
2+
(aq)

Cu
change
Ø
(aq)
salt
form

Cu(s)
E
=
+1.1 V
bridge
Ø
The
sign
of
E
in
these
diagrams
represents
the
polarity
(+
or
–)
of
the
2+
right
hand
electrode
in
the
diagram.
In
this
case
the
half
Cu/Cu
cell
is
2+
+
with
respect
to
the
half
Zn/Zn
cell.
If
we
write
the
cell
diagram
the
Ø
other
way
round,
the
is
E
negative.
2+
Cu(s)
The
standard
half

Cu
cell
2+
(aq)

Zn
diagram
for
Ø
(aq)
the

Zn(s)
E
hydrogen
=
–1.1 V
electrode
is:
+
Pt
[H
(g)]

2H
(aq)

2
The
hydrogen
gas
and
platinum
electrode
are
regarded
as
being
non-metals
or
ions
a
single
phase.
For
more
oxidation
aqueous
complex
states,
ions
half
the
and
a
cells
comma
2+
Pt
If
hydrogen
bracketed

or
in
Fe
hydroxide
the
is
involve
shown
separates
3+
(aq),
Fe
which
platinum
cell
as
the
a
separate
different
ions

Cu
appear
diagram,

the
Cu(s)

[Mn
half
E
=
+
4H
O(l)],
[MnO
2
reduced
e.g.
–0.43 V
equation,
they
+
(aq)
the
states
are
e.g.
2+
Pt
different
Ø
(aq)
in
in
from
oxidation
2+
(aq)
phase
(aq)
+
+
8H
(aq)]
oxidised
form

Ag
(aq)

Ag(s)
4
form
Key points


The
of
electron ow
(less
negative
electrode
Cell
ions

direction
negative
In
a
diagrams
which
cell
positive)
species
each
take
diagram
oxidised
the
on
a
cell
is from
potential
to
the
half
cell
with
the
most
the
half
cell
with
the
less
potential.
show
may
in
electrode
part
of
in
the
the
reduced
the
oxidised
and
reduced
species
and
other
reaction.
species
are
placed
on
the
outside
and
the
inside.
125
Ø
10.4
Usng
E
values to
predct
chemcal
change
Ø
Learning outcomes
Using
Figure
On
completion
of
this
section,
E
values to
10.4.1
compares
predict
some
a
reaction
oxidising
and
reducing
powers
of
some
you
Ø
elements
should
be
able
and
ions
by
comparing
values:
E
to:
Ø

use
E
values
to
predict
whether
2+
Mg
a
reaction
is
likely
to
occur
or
−
(aq)
+
Zn
species
not.
Ø
2e
2+
Y
Mg(s)
E
Y
Zn(s)
E
−
(aq)
+
on
(aq)
+
e
−0.76 V
=
0.00 V
H
(g)
E
=
+0.34 V
2
gain
right
2+
−
(aq)
Cu
+
lose
Ø
2e
Y
Cu(s)
E
electrons
electrons
3+
2+
(aq)
Fe
more
on
Ø
Y
2
left
−2.38 V
=
species
1
+
H
=
Ø
2e
+
e
Y
Fe
Ø
(aq)
E
=
+0.77 V
=
+0.80 V
=
+1.36 V
readily
more
+
−
(aq)
Ag
+
e
readily
Ø
Y
Ag(s)
E
1
Ø
Cl
(g)
+
e
Y
Cl
(aq)
E
2
2
Ø
Figure 10.4.1
As the value of E
gets more positive the species on the left are stronger
oxidising agents and the species on the right are weaker reducing agents

For
each
half
equation,
the
more
oxidised
form
is
on
the
left.
Ø

The
more
reaction
at
positive
to
move
accepting
the
in
value
the
electrons
of
than
the
,
E
forward
the
greater
direction.
species
is
This
above
the
is
it.
tendency
because
It
is
a
it
better
of
is
the
better
oxidising
agent.
Ø

The
more
reaction
losing
We
can
negative
to
move
electrons
use
these
the
in
the
than
ideas
value
reverse
the
to
of
the
greater
direction.
species
predict
,
E
below
whether
This
it.
a
It
is
is
the
is
because
a
tendency
better
reaction
will
it
is
of
reducing
take
the
better
at
agent.
place
or
Ø
not.
we
If
a
say
reaction
that
it
is
is
likely
feasible.
to
take
For
place
using
example:
Will
information
zinc
react
from
with
values,
E
copper(
II)
ions?
The
relevant
half
reactions
are:
2+
Zn
−
(aq)
+
2e
+
2e
2+
Zn(s)
E
Y
Cu(s)
E
−
(aq)
Cu
Ø
Y
=
−
0.76 V
=
+
0.34 V
Ø
2+
Cu
Zn
2+
ions
has
a
have
a
greater
greater
tendency
tendency
to
lose
to
gain
electrons
electrons
than
than
Cu.
So
Zn
the
ions
and
position
of
2+
equilibrium
of
the
Zn/
equilibrium
of
the
Cu/Cu
reaction
Zn
moves
to
the
left
and
the
position
of
2+
reaction
reaction
moves
2+
In
these
predictions,
as
occur
in
126
not
the
right.
The
overall
is:
Zn(s)
do
to
a
apply,
we
standard
we
have
+
Cu
are
2+
(aq)
→
assuming
electrochemical
to
consider
Zn
that
cell
other
(aq)
the
+
Cu(s)
standard
apply.
If
conditions
standard
information
(see
such
conditions
Section
10.5).
Chapter
The
We

anticlockwise
can
predict
writing
down
negative

the
the
the
value
direction
pattern

whether
in
reaction
half
from
from
potentials
and
equilibrium
is
feasible
reactions
by:
with
the
one
with
the
most
top
which
starting
pattern
a
the
Electrode
rule
two
at
10
the
the
the
reaction
bottom
bottom
left
is
feasible
is
given
an
anticlockwise
left.
is:
reactant
→
product
→
reactant
→
product
Ø
E
Ø
E
more
negative
(less
positive )
reactant
oxidised
Example
form
reduced
form
1
2+
Will
product
chlorine
oxidise
3+
Fe
ions
to
Fe
ions?
Ø
E
3+
+0.77
Fe
2+
(aq)
+ e−
Y
Fe
better
(aq)
reducing
agent
1
+1.36
Cl
can
agent.
see
that
Applying
the
the
+ e−
Y
Cl
(aq)
2
2
better
Y
ou
(g)
oxidising
better
agent
reducing
anticlockwise
1
agent
rule,
reacts
the
with
reaction
2+
Cl
(g)
+
Fe
the
is
better
oxidising
feasible:
3+
(aq)
→
Cl
(aq)
+
Fe
(aq)
2
2
2+
This
because
chlorine
is
better
at
accepting
electrons
than
Fe
ions
and
Exam tips
3+
Fe
is
better
at
releasing
electrons
than
Cl
ions.
If
Example
you
arrange
the
two
half
reactions
2
with
the
one
with
the
most
negative
+
Will
aqueous
iodine
oxidise
silver
to
silver
ions
Ag
?
value
be
Ø
at
the
the
top,
bottom
the
left
reactants
and
top
will
right
E
1
+0.54
species.
I
(aq)
+
e−
Y
I
(aq)
2
2
better
reducing
agent
+
+0.80
Ag
(aq)
better
Applying
the
+
e−
oxidising
anticlockwise
rule,
Y
Ag(aq)
Key points
agent
the
reaction
is
not
feasible.
The
reaction
Ø

will
be
between
silver
ions
and
iodide
ions
rather
than
between
silver
A
reaction
value for
iodide
is feasible
if
the
E
and
the
reaction
in
the
ions.
forward
direction
is
positive.
Ø
W
ar ning!
The
feasibility
of
a
reaction
based
on
values
E
only
tells
us

that
a
reaction
is
possible.
It
is
no
guarantee
that
a
reaction
will
A
happen.
reaction
Some
reactions,
although
feasible,
take
place
so
slowly
that
they
do
value
of
to
be
happening.
We
say
that
these
reactions
are
the
half
if
the
equation
not
involving
seem
is feasible
Ø
E
the
species
being
kinetically
reduced
is
more
positive
than
the
controlled.
Ø
E
the
Ø
Selecting
E
value
of
species
the
half
being
equation
of
oxidised.
values

The
direction
of
a
reaction
can
Ø
Y
ou
must
take
care
when
selecting
E
values
from
a
table
of
data.
Some
be
half
reactions
may
look
very
similar
,
especially
if
two
ions
are
involved
predicted from
e.g.
showing
2+
VO
+
+
2H
+
2H
3+
+
e
Y
V
+
e
Y
VO
Ø
+
H
+
H
O
E
=
+0.34 V
=
+1.00 V
2
+
+
VO
2
2+
diagrams
half
equations
and
their
Ø
E
values.
Ø
O
E
2
127
10.5
Electrode
potentals
and
electrochemcal
cells
Learning outcomes
Concentration
The
On
completion
of
this
section,
position
of
change
equilibrium
is
and
electrode
affected
by
potential
changes
in
concentration
and
you
Ø
temperature.
should
be
able
This
also
applies
to
redox
equilibria.
values
E
refer
to
a
to:
−3
concentration
of
ions
of
and
1.00 mol dm
temperature
of
298 K.
What
Ø

describe
how
E
values
vary
with
concentration

know
how
to
happens
when
system?
apply
the
T
ake
we
for
alter
principles
redox
equilibria
to
concentration
the
half
2+
of
+
ions
in
a
metal–metal-ion
reaction:
−
(aq)
Cr
of
the
example
Ø
2e
Y
Cr(s)
E
=
−0.91 V
energy
2+

storage
devices
If
the
concentration
of
Cr
(aq)
increases,
the
value
of
E
becomes
less
including fuel
negative.
e.g.
–0.83 V
.
cells.
2+

If
the
concentration
more
We
can
negative.
apply
reactions

If
the
the

If
the
are
the
value
of
E
becomes
principle
to
concentration
changes
in
redox
one:
of
shifts
more
the
to
ion
the
easily
on
the
right
as
accepted,
left
of
more
the
the
equation
electrons
value
of
E
are
gets
increases,
accepted.
more
If
positive
negative).
concentration
equilibrium
electrons
less
decreases,
–0.99 V
.
concentration
less
the
(or
this
(aq)
Cr
Chatelier ’s
as
equilibrium
electrons
(or
Le
such
e.g.
of
are
of
shifts
more
the
to
ion
the
easily
on
left
the
as
released,
left
more
the
of
the
equation
electrons
value
of
E
are
gets
decreases,
released.
more
If
negative
positive).
3+
For
a
system
such
as
the
concentrations
of
2+
(aq)
Fe
both
+
ions
e
Y
Fe
equally
(aq),
has
no
increasing
effect
on
(or
the
decreasing)
value
of
E.
3+
This
is
because
position
of
the
increase
equilibrium
to
in
the
the
concentration
right,
but
increase
of
in
pushes
Fe
the
the
concentration
of
2+
Fe
pushes
the
Predicting
If
reactions
difcult
to
a
position
of
reaction
are
carried
predict
if
a
equilibrium
under
out
the
left
to
non-standard
under
reaction
to
non-standard
is
feasible.
As
a
an
equal
extent.
conditions
conditions
rough
it
guide
is
we
more
can
say
that:

If
the
0.3 V
,
electrode
the
potentials
predicted
of
reaction
two
is
half
likely
reactions
to
differ
happen,
e.g.
by
the
more
than
reaction
3+
between
copper
difference
is
and
ions
Fe
is
likely
to
be
2+
+
2e
Y
Cu(s)
3+
the
less
difference
than
0.3 V
,
E
2+
(aq)
Fe
If
because
in
the
+
e
Y
electrode
Fe
potentials
feasibility
of
the
=
+0.34 V
=
+0.77 V
Ø
(aq)
E
of
two
reaction
half
may
reactions
not
be
condence,
cannot
be
e.g.
predicted
the
reaction
because
the
3+
(aq)
Fe
between
difference
2+
+
e
Y
+
e
Y
Fe
(aq)
128
(aq)
is
2+
ions
and
0.03 V
.
Ø
E
+
Ag
Ag
=
+0.77 V
=
+0.80 V
Ø
Ag(s)
E
differ
predicted
+
with
the
Ø
(aq)
Cu

feasible
0.43 V
.
Fe
ions
by
Chapter
Batteries
and
Most
batteries
They
often
and

cells
have
They
are
are

They

Some
Button
Many
a
solid
cells
The
deliver
can
we
use
do
not
operate
concentrations
One
equilibrium
of
under
standard
electrolyte.
Modern
conditions.
batteries
in
size
although
high
and
lightweight.
lightweight
voltage
for
a
ones
Car
are
considerable
batteries,
being
however
developed.
period
of
time.
recharged.
half
cells
type
cell
not
such
of
so
as
cell
there
contain
equations
a
those
relies
is
used
on
no
liquid
the
to
power
reaction
spillage
of
electrolyte
+
The
voltage
the
this
of
the
type
of
negative
pole
and
and
electrolyte.
electrolyte
is
a
high
iodine.
Most
a
It
is
button
paste.
+
e
Y
Li
E
+
e
Y
I
E
cell
cell
should
in
use,
are
button
the
=
−3.14 V
=
+0.54 V
Ø
conditions
Another
deliver
2
2
of
the
watch
lithium
Ø
I
voltage
liquid
or
1
The
a
of
are:
Li
because
and
advantages:
small
a
be
potentials
cells
state
do
many
heavy,
button
voltage.
cells
higher
often
quite
Electrode
cells
and
have
10
be
however
,
not
cell
is
silver
+0.54
−
is
(−3.14)
more
=
likely
3.58 V
.
to
be
about
3 V
standard.
the
zinc−silver
oxide
is
the
oxide
positive
cell.
The
zinc
is
the
pole.
Did you know?
Ø
Zn(s)
+
Ag
O(s)
→
ZnO(s)
+
2Ag
E
=
+
1.60 V
cell
2
The
Fuel
use
of fuel
disadvantages
In
a
cells
have
many
cells
fuel
cell,
a
fuel
(often
hydrogen)
releases
electrons
at
one
as
well
as
advantages:
electrode

The
storage
of
hydrogen
in
large
Ø
and
for
oxygen
a
fuel
gains
cell
electrons
with
an
at
acidic
the
other
electrolyte
electrode.
The
relevant
E
values
is
a
problem.
Refuelling
has
are:

+
to
be
carried
out
Ø
+
2H
quantities
2e
Y
H
E
=
0.00 V
2
more
+
often
than
with
a
petrol
Ø
+
4H
O
+
4e
Y
2H
2
O
E
= +1.23 V
2
engine.
The
overall
reaction
is:
2H
+
O
2
→
2H
2
O.
2

The
fuel
petrol
or
cells
used
diesel
in
some
cars
and
buses
have
several
advantages
over
W
ater
is
hydrogen
is
produced from
fossil fuels.
engines:


The
the
only
product.
No
carbon
dioxide
or
nitrogen
oxides
They
do
not
work
well
at
low
temperatures.
are
released.


They
produce
more

They
are
efcient.
more
direct.
very
energy
per
The
gram
of
fuel
transmission
than
of
petrol
power
to
or
the
They
are
expensive.
diesel.
engine
is
Key points
electron flow
Ø

V
The
ion
value
half
cell
negative
negative
positive
electrode
electrode
the

of
ions
Modern
as
in
a
metal/metal
becomes
the
more
concentration
of
decreases.
batteries
advantages
H
E
in
have
terms
of
small
size,
O
2
low
2

electrolyte
Fuel
by
Figure 10.5.1
cells
using
and
high
generate
the
voltage.
a
voltage
energy from
the
Simplied
diagram of a hydrogen−oxygen
water
mass
reaction
of
oxygen
with
a fuel.
fuel cell
129
Revson
1
For
each
reacton
gven,
wrte
the
questons
specfied
c
A
mxture
of
1.40 mol
and
H
1.40 mol
I
2
are
2
3
equlbrum
expresson
unts. Where
K
s
and
state
requred,
the
assume
assocated
the
placed
partal
and
n
the
a
2 dm
ask
equlbrum
pressures
a
i
are
2NO
measured
Cl(g)
Y
n
atmospheres.
2NO
2
Fe
(g)
+ Cl
2
+
I
(g);
Y
Fe
0.36 mol
K
+
I
+
equlbrum
. What
s
(g)
temperature,
Y
2HI(g)
s
2
mxture
the value
of
K
2
contans
at
ths
c
temperature?
__
(aq)
H
c
1
2+
(aq)
establshed. The
2
–
(aq)
constant
(g)
2
3+
ii
at
H
p
I
(aq);
K
2
2
c
3
d
b
i
3O
(g)
Y
2O
2
(g);
(g)
+ ClF
3
Cl
2
(g)
Y
3HF(g)
+
__
N
3
at
1
__
NH
6.00 mol SO
are
placed
n
a
3 dm
ask
2
p
1
ii
When
K
3
(g)
+
Cl
2
2
a
hgh
temperature,
40%
decomposes
to SO
2
(g);
2
2
and Cl
as
shown
by
the
equaton:
2
K
p
2+
c
i
Ba(IO
)
3
(s)
Y
Ba
+
2IO
(aq);
Al(OH)
(s)
Y
Al
3
+
3OH
(aq);
i
HF(aq)
ii
HCN(aq)
+
H
O(l)
Calculate
K
3
K
Y
H
2
O
Y SO
(g)
+ Cl
2
at
ths
temperature.
–
(aq)
+
F
(aq);
K
3
a
4
a
A
mxture
of
6.5 g
H
and
4.0 g
N
2
+
Y
H
–
(aq)
+ CN
(aq);
a
a
HS
+
H
O(l)
Y
H
S(aq)
+ OH
(aq);
K
2
5.0 dm
what
vessel.
What
do
you
are
the
the
total
pressure
Based
on
pressures
of
N
s
n
3.9 atm,
understand
by
the
term
and
H
2
For
the
equlbrum
?
2
system
‘dynamc
2NO(g)
+ Cl
mxture
at
(g)
Y
2NOCl(g),
an
equlbrum
2
equlbrum’?
b
If
partal
b
b
a
placed
–
(aq)
2
2
are
2
3
K
–
e
(g)
2
c
sp
+
d
(g)
2
sp
–
(aq)
Cl
2
K
3
+
ii
SO
–
(aq)
2
Le Chateler’s
prncple,
explan
a
constant
hgh
temperature
has
how
3
partal
pressures
of
NO
=
9.6
×
10
Pa, Cl
=
1.7
×
2
each
of
the factors
lsted
would
affect
the
4
10
4
Pa,
NOCl
=
2.8
×
10
Pa. Calculate
K
at
ths
p
equlbrum
system.
temperature.
PCl
(g)
Y
PCl
5
(g)
+ Cl
3
(g);
∆H
=
+87
.9 kJ
2
c
(Consder
yeld,
poston
of
equlbrum
and value
At
a
partcular
contans
only
temperature,
at
N
a
a vessel
partal
ntally
pressure
of
2.50 atm
2
of
equlbrum
constant.)
and O
at
a
partal
pressure
of
1.50 atm.
2
i
addng
more
PCl
The
3
ii
decreasng
iii
addng
iv
ncreasng
v
removng Cl
a
the
equlbrum
establshed
pressure
contans
What
the
?
reacton
5
NH
NO
at
t
a
s found
partal
the value
the forward
(aq)
+
H
(aq)
Y
NH
would
be
effect
(aq),
4
reacton
s
Y
of
K
that
2NO(g)
the
s
pressure
mxture
of
0.76 atm.
?
a
What
b
The
s
meant
by
the
term
solublty?
+
3
where
the
s
and
(g)
p
+
For
+ O
2
temperature
2
c
(g)
2
catalyst
the
N
exothermc,
what
of:
solubltes
gven
n
What
s
i
and
the
ii
K
n
pure
water,
of
the
substances
below
were
at
temperature for
ths
determned
at
25 °C.
each
sp
substance?
i
ncreasng
the
pressure
ii
ncreasng
the
concentraton
of
i
(aq)
NH
0.710 g
n
of AgBrO
–3
iii
ncreasng
the
400
ml
of
water.
3
3
ii
temperature?
7
.52
×
–3
mol dm
10
BaF
.
2
c
i
The
solublty
product,
for
K
MgCO
sp
3
a
For
the
system
(g)
2NO
Y
2NO(g)
+ O
2
–8
×
3
whch
has
reached
equlbrum
n
a
2 dm
10
2
mol
s
3.5
0.0222 mol
NO
,
6.34 g
NO
–6
dm
at
25 °C. What
would
be
the
ask
solublty,
contanng
(s)
3
(g),
2
at
ths
temperature
n
and
2
10.8 g O
,
what
would
be
the value
of
K
2
,

pure

a
water,
along
c
–3
wth
the
approprate
unts?
soluton
contanng
2+
Mg
0.015 mol dm
ons?
b
For
the
system A(aq)
+
B(aq)
Y
R(aq),
at
a
–2
partcular
temperature,
K
=
4.5
×
10
3
dm
–1
mol
.
ii
Explan
iii
Would
why
the
solubltes
n
i
dffer.
c
In
an
equlbrum
mxture
at
the
same
precptate
MgCO
f
a
soluton
3
–3
temperature,
the
known
concentratons
contanng
n
0.055 mol dm
magnesum
ntrate
–3
mol dm
are
[A]
=
2.
10,
[B]
=
1.45. Calculate
the
s
mxed
wth
a
soluton
–3
equlbrum
concentraton
of
R.
1.22
×
10
Explan
130
contanng
–3
mol dm
your
sodum
answer.
carbonate?
Chapters
8–10
Revision
3
6
a
The
concentratons
of
HCl
and C
H
6
COOH
are
8
A volume
of
questions
–3
30 cm
of
0.
15 mol dm
KOH
soluton
5
–3
both
0.025 mol dm
.
was
placed
n
a
concal ask
3
i
Why
s
acds,
the
pH
even
dfferent for
though
they
these
have
the
70 cm
two
and
a
total volume
of
–3
of
0.
10 mol dm
HNO
soluton
added
n
3
5.0 ml
same
alquots. The
equvalence volume
was found
3
to
concentraton?
ii
iii
What
s
What
the
s
25 °C,
pH
the
of
pH
gven
the
of
the
HCl?
the
at
K
benzoc
ths
acd
at
temperature
be
45 cm
.
a
What
s
the
ntal
b
What
s
the
pH
c
What
of
s
20 ml
of
of
×
10
What
s
HClO
the
The
pH
? What
of
s
a
the
of
a
s
soluton?
after
the
addton
the
pH
of
the
soluton
after
addng
3
soluton
0.021 mol dm
of
55 cm
of
HNO
?
3
pOH?
9
pH
KOH
3
?
4
c
the
soluton
–3
mol dm
–3
b
of
?
HNO
a
–5
6.3
pH
the
soluton
was found
to
be
a
Draw
a
dagram
of
the
apparatus
whch
would
11.4. What
be
used
to
determne
the
standard
s
electrode
4+
potental
of
the
half
cell Sn
2+
(aq)/Sn
(aq).
+
i
[H
],
b
i
Draw
a
cell
dagram for
–
ii
[OH
]
the
two
half
cells Al
2+
a
i
Wrte
base
combnaton
3+
?
and
7
the
a
sutable
equaton
phenylamne,
(C
H
6
to
show
NH
5
),
how
Zn
of
Ø
(aq)/Al(s);
E
=
–1.66 V
Ø
(aq)/Zn(s);
E
=
–0.76 V
the
dssocates
ii
Calculate
the
standard
iii
Wrte
balanced
cell
emf for
ths
cell.
n
2
water.
the
equaton for
the
cell
reacton.
ii
What
s
the
conjugate
acd
of
ths
base?
–10
iii
The
for C
K
b
H
6
NH
5
s
4.3
×
10
–3
mol dm
.
2
Calculate
–3

the
pH

the
pK
of
a
0.225 mol dm
soluton,
.
b
b
i
Explan
why
a
mxture
of
H
HC
3
acd)
and
NaC
H
3
functon
ntrc
as
acd
a
O
5
buffer
and
(sodum
O
5
(lactc
3
lactate)
would
3
whereas
sodum
a
ntrate
mxture
would
of
not.
–5
ii
The
for CH
K
a
What
by
COOH
s
1.8
×
10
–3
mol dm
.
3
s
the
mxng
pH
of
15.5 g
a
buffer
of CH
soluton formed
COOH
(ethanoc
acd)
3
wth
15.5 g of CH
COONa
(sodum
ethanoate)
3
3
n
a
250 cm
volumetrc ask?
2–
iii
The
K
for CO
b
s
the
–4
s
2.
1
×
10
–3
mol dm
. What
3
pH
of
a
buffer formed
by
mxng
–3
0.
105 mol dm
–3
Na
CO
2
NaHCO
wth
0.
120 mol dm
3
?
3
131
Exam-style
Answers to
all
exam-style questions
can
questons
be found on the
Module
2
accompanying CD
6
Multiple-choice questions
–
From
the
data
gven,
whch
of
the followng
are
true?
Ø
E
20
1
A
radoactve
source
has
1.6
×
10
atoms
of
a
4+
Sn
radoactve
sotope,
wth
a
half-lfe
of
3
days.
2+
(aq)/Sn
(aq)
+0.
15 V
How
2+
V
many
atoms
wll
decay
n
12
(aq)/V(s)
–
MnO
2–
(aq)/MnO
4
19
A
1.0
×
–1.2 V
days?
(aq)
+0.56 V
4
19
10
C
4.0
×
10
4+
–
i
MnO
s
a
stronger
oxdsng
agent
than Sn
(aq).
4
20
B
1.2
×
20
10
D
1.5
×
10
4+
2
For
the
reacton X
+
Z
→
M,
the
concentraton
ii
V
wll
iii
A
reacton
A
i,
B
Only
i
and
ii
C
Only
i
and
iii
D
Only
ii
2+
reduce Sn
to Sn
.
2+
of
2+
and V
between Sn
s feasble.
–3
reactants
tme
s
and
products
measured
n
s
measured
seconds. The
n
rate
mol dm
and
equaton
ii
and
iii
are
true.
s
are
true.
2
Rate
What
concluson
can
be
=
k[Z]
made from ths
nformaton?
3
A
The
unt for
the
rate
constant,
k,
s
dm
–1
mol
B
The
unt for
the
C
The
reacton
rate
constant
s
.
Consder
the
reacton,
.
(g)
X
+
2Y
2
rate
doubles,
f
the
concentraton
The
of
reacton
rate
reactant X
ncreases,
as
the
For
(g);
∆H
=
+58.0 kJ
2
of
the followng
the
ncreases.
reacton,
(g)
NO
Step
1:
NO
(g)
+
NO
Step
2:
+ CO(g)
→
mechansm
(g)
→
NO
2
NO
(g)
combnatons
NO(g)
+ CO
the
+ CO(g)
s
correctly
temperature?
Position of equilibrium
Yield
Value of K
shft
decreases
no
to
left
change
(g)
B
shft
to
rght
ncreases
ncreases
C
shft
to
rght
decreases
decreases
D
shft
to
left
ncreases
decreases
proposed:
+
NO(g)
→
NO
(g)
+ CO
2
rate
the
(g),
slow
(g)
fast
2
8
s
ncreasng
3
3
What
of
2
two-step
2
effect
c
2
the followng
the
concentraton
A
3
Y2XY
doubles.
predcts
D
(g)
2
of
Whch
Z
true.
–1
s
mol dm
are
true.
–1
s
7
–3
s
equaton for
ths
For
propanoc
acd,
whch
has
reacton?
–5
=
K
1.35
×
10
–3
mol dm
the
dssocaton
n
water
s
a
A
Rate
=
k[NO
][CO]
3
CH
2
B
Rate
=
k[NO
CH
3
]
COOH(aq)
+
H
2
O(l)
Y
2
2
–
CH
CH
3
C
Rate
=
k[NO
COO
+
(aq)
+
H
2
O
(aq)
3
][NO]
3
Whch
D
Rate
=
][CO
k[NO
2
of
the followng
statements
are
true?
]
2
–
[CH
CH
COO
+
(aq)][H
O
(aq)]
3
2
3
____________________________
i
4
Three
monobasc
acds X, Y
and
Z
all
have
The
K
a
=
a
[CH
CH
3
COOH(aq)][H
2
O(l)]
2
–3
concentraton
of
0.011 mol dm
. The
pK
of X
s
5.61,
a
–3
ii
The
pH
of
a
iii
Addng OH
0.010 mol dm
soluton
s
2.
–7
the
K
of Y
s
4.
13
×
10
and
the
pH
of
Z
s
4.78. What
a
–
s
the
order
of
the
acds
n
terms
of
ncreasng
would
cause
the
equlbrum
to
shft
acd
to
the
rght.
strength?
iv
A
X
< Y
<
Increasng
the
molarty
of
the
acd
would
Z
+
result
n
a
hgher
concentraton
of
O
H
ons
at
3
B
Z
< X
< Y
C
Y
< X
<
equlbrum.
D
5
Z
< Y
Gven
Z
A
i,
ii
and
B
ii,
iii
C
iii
and
D
i
iii
and
iv
are
true.
< X
the followng
standard
electrode
half
cells
and
correspondng
and
iv
iv
are
are
true.
true.
potental values,
and
ii
are
true.
Ø
half
cell
E
/V
9
3+
Mn
–
(aq)
+
Y
2+
Zn
What
s
the
solublty
of
25 °C,
f
the
solublty
product
Mn
n
Zn(OH)
(aq)
constant,
+
2e
Zn(s)
–0.76
condtons
s
3.0
Ø
the
correct
cell
dagram
and
E
?
–6
cell
2+
A
Zn(s) | Zn
B
Mn
3+
(aq) || Mn
2+
,
Mn
C
Pt(s) | Mn
D
Zn(s) | Zn
(aq),
2+
132
(aq)
6.7
B
1.7
×
–3
10
+0.73 V
mol dm
–8
×
10
–3
mol dm
2+
(aq) || Zn
2+
A
2+
(aq) | Mn
3+
(aq)
(aq) | Zn(s)
3+
Mn
–1.25 V
–6
C
5.3
×
10
D
4.2
×10
–3
mol dm
2+
(aq) || Zn
3+
(aq) || Mn
at
under
sp
Y
–16
s
K
–
(aq)
water
+1.49
these
what
pure
2
2+
e
(aq),
(s) | Zn
(aq)
+0.73 V
2+
Mn
(aq) | Pt(s)
+2.25 V
–6
–3
mol dm
×
10
3
mol
–9
dm
?
Module
10
The
solublty
of
copper(i)
bromde CuBr
–5
water
at
25 °C
followng
s
7
.28
×
conclusons
pure
12
Ths
queston
s
based
on
Exam-style
the
two
half
questions
cells
gven
–3
10
are
n
2
mol dm
correct,
. Whch
based
on
of
the
below,
ths
along
wth
ther
standard
electrode
potental
values.
Ø
data?
E
+
–5
i
The
solublty
wll
be
greater
than
7
.28
×10
,
Ag
f
(aq) + e
Y
Ag(s)
+
CuBr
s
dssolved
n
a
soluton
of
MnO
KBr.
The
numercal value
of
the
s
K
5.3
+0.80 V
2+
(aq) + 5e
Y
Mn
(aq) + 4H
4
–9
ii
(aq) + 8H
×
a
10
O(l)
+1.52 V
2
What
do
you
understand
by
the
term
‘standard
sp
2
iii
The
unt
of
the
would
K
be
electrode
–6
mol
potental’?
[2]
dm
sp
b
+
iv
The
concentraton
of
on
the Cu
at
n
pure
water,
would
be
7
.28
×
10
i
Draw
whch
–3
i,
ii
B
ii,
and
iii
and
iv
are
C
ii
D
and
and
i,
ii
iii
and
iv
are
are
iii
be
dagram
used
to
of
the
set
up
apparatus,
ths
to
determne
the
standard
cell,
cell
n
e.m.f.
[3]
true.
ii
iii
labelled
could
mol dm
order
A
a
equlbrum,
–5
On
the

Label
dagram,

Show
true.
the
anode
and
the
cathode.
[1]
electron ow.
[1]
true.
are
the
drecton
of
true.
c
Calculate
balanced
the
standard
equaton for
cell
the
e.m.f.
cell
and
wrte
a
reacton.
[2]
Structured questions
d
Wrte
e
How
the
cell
dagram.
[2]
–3
11
A
student
weak
ppetted
monobasc
25.0 ml
acd
of
0.
112 mol dm
soluton
nto
a
of
a
would
ths
was
added
an
alkalne
soluton, X, from
each
addton,
the
soluton
was
ensure
complete
mxng
and
the
pH
was
shown
on
usng
the
a
pH
graph
meter. The
the
student’s
of
n
the
respectve
half
cells
affect
the
cell
electromotve force?
the
why
KCl
would
[2]
Explan
not
be
sutable for
use
n
then
the
measured
of
swrled
f
to
concentraton
a
value
burette. After
the
concal ask.
solutons
To
changng
results
salt
brdge.
[2]
are
13
below:
a
i
State
ii
Based
the
postulates
on
the
of
collson
the
collson
theory,
theory. [2]
explan
why
14
ncreasng
the
concentraton
of
a
reactant
12
would
generally
cause
an
ncrease
n
the
rate
10
Hp
of
a
reacton.
[2]
8
b
The
results
shown
n
the
table
below
were
6
obtaned for
the
reacton
4
A(aq)
+
B(aq)
→ C(aq)
2
where
the
ntal
rate
was
determned
by varyng
0
5
10
15
20
25
30
35
40
45
50
concentratons
of A
and
B,
at
room
temperature.
3
volume
a
Gve
the
name
solution X
or formula
of
an
(cm
)
alkal
that
could
Experiment
[A]
[B]
Initial
–3
produce
b
the
s
results
i
What
ii
Determne
the
K
shown
of
the
on
the
graph.
[1]
acd?
pK
for
the
rate
–3
)
(mol dm
–3
)
(mol dm
[3]
a
the
(mol dm
acd.
–1
s
)
–2
1
0.
125
0.
125
1.45
×
10
2
0.250
0.
125
2.90
×
10
3
0.
125
0.250
5.80
×
10
[1]
a
–2
c
From
the
relevant
equvalence volume
data,
determne
the
and
any
molarty
other
of
the
base.
–2
[3]
i
d
Explan
greater
why
the
than
pH
at
the
equvalence
pont
Sketch
a
graph
rate
a
reacton
7
.
of
The
soluton
has
ts
maxmum
bufferng
reactant
where
V
s
be
how
the
ntal
determned from
the
the
change
n
concentraton
of
a
capacty
V
at
llustrate
could
[2]
montorng
e
to
s
over
tme.
[2]
equvalence volume.
2
ii
i
ii
What
What
s
the
do
you
‘bufferng
f
The
pH
ndcator
at
ths
pont?
understand
by
the
6.8–8.4. State,
red
gvng
sutable for
use
n
a
reasonng,
determne
the
order
wth
respect
to A

the
order
wth
respect
to
[2]
B.
[2]
[2]
has
your
a
pH
range
reasonng,
iii
Wrte
ths
ttraton.
the
iv
Calculate
rate
equaton for
the
reacton.
[1]
of
whether
a value for
the
rate
constant,
k,
at
t
ths
s
your

term
capacty’?
phenol
Gvng
[1]
temperature,
and
state
the
unt.
[2]
[2]
v
What
would
be
the
rate
of
reacton,
f
the
–3
startng
concentraton
of A
was
0.30 mol dm
–3
and
B
was
0.
14 mol dm
?
[2]
133
11
The
11.
1
Physical
elements
Learning outcomes
completon
of
ths
Structure
secton,

be
able
explan
3
Period
3
and
melting
elements,
their
Period
3
elements
points
physical
state
at
room
temperature
and
you
melting
should
Period
properties of the
The
On
of
points
in
kelvin
are
shown
below:
to:
the
varaton
n
some
Group
physcal
propertes
elements
and
n
terms
bondng
meltng
wth
ponts,
conductvty,
and
of
of
Perod
3
I
II
III
IV
V
VI
VII
0
sodum
magnesum
alumnum
slcon
phosphorus
sulphur
chlorne
argon
371 K
922 K
933 K
1683 K
317 K
392 K
172 K
84 K
sold
sold
sold
sold
sold
sold
gas
gas
structure
reference
to
electrcal
electronegatvty
densty.
)K(
Did you know?
exsts
n
dfferent forms
dfferent
slghtly
The
meltng
sulphur
dfferent forms
shaped
dfferent
(whch
crystals
meltng
pont
of
has
gniliob/gnitlem
(allotropes). These
have
tniop
Sulphur
and
ponts.
monoclnc
rod-lke
3000
crystals)
2500
boiling
point
melting
point
Si
2000
Al
1500
Mg
1000
Na
P
S
500
Ar
s
392 K,
but
the
meltng
pont
Figure 11.1.1
Melting and
0
of
rhombc
octahedral
sulphur
shaped
Phosphorus
also
(whch
crystals)
has
red,
boiling points of the Period
has
s
11
It
s
the
13
whte
14
atomic
386 K.
15
16
17
18
3 elements
number
and
We
black forms.
12
can
explain
the
differences
in
melting
point
in
terms
of
structure
and
whte form
bonding:
whch
melts
at
317 K.

Sodium,
The
magnesium
ions
sodium
are
to
held
and
aluminium
together
aluminium
by
there
a
is
sea
an
have
of
a
giant
delocalised
increase
in
the
metallic
structure.
electrons.
number
From
of
electrons
+
donated
form.
the
to
The
Silicon,
in
lot
of
They
b
P
increase
forces
with
is
to
large
metal
ions
number
of
between
in
melting
lattice
number
ions
order
point
similar
of
Mg
of
to
strong
forces.
the
to
the
So
the
Mg
to
Period
diamond.
covalent
Al
electrons,
and
Na
3+
and
delocalised
these
the
2+
,
Na
the
overcome
increase
highest
It
Al.
3
takes
bonds.
So
high.
chlorine
points
,
has
have
because
the
a
simple
there
of
molecular
are
molecules.
number
contact
S
and
covalent
increasing
sulphur
,
it
between
of
the
attraction
points
the
and
number
So
charge
the
very
melting
forces
molecules.
S
low
the
is
sulphur
attractive
with
has
giant
break
point
W
aals
and
c
have
to
a
when
difcult
boiling
IV
forms
Phosphorus,
ionic
more
and
energy
melting
electrons
the
Group
It
of
electrostatic
the
points
the
a
the
melting
a

is
and
elements.
sea
greater
greater
electrons

the
only
The
electrons
points
between
higher
melting
structures.
weak
van
in
der
the
van
der
W
aals
molecule
neighbouring
point
than
8
phosphorus,
P
.
Chlorine
is
a
gas
since
it
is
a
diatomic
molecule
with
4
S
a
fairly
small
number
of
electrons.
P
Figure 11.1.2
The structures of: a sulphur;
b phosphorus and; c chlorine
134

Argon
these
exists
are
only
very
as
low,
isolated
so
it
has
atoms.
a
very
The
low
van
der
melting
W
aals
point.
forces
between
Chapter
Density of
The
of
table
Cl
Period
below
and
Ar
shows
are
temperature
so
those
have
Element
3
density
their
very
The
elements
of
Perod
3
elements
the
of
11
low
of
the
liquids.
Period
They
are
3
elements.
gases
at
The
density
room
densities.
Na
Mg
Al
S
P
S
Cl
Ar
0.97
1.74
2.70
2.32
1.82
2.07
1.56
1.40
−3
Density/g cm
Density
(iii)

the
depends
way
The
and

Si
the
density
their
has
a
on:
(i)
atoms
the
are
increases
mass
lower
mass
of
packed
from
the
in
Na
atoms,
their
to
Al
(ii)
the
size
of
the
atoms,
lattice.
as
the
size
of
the
atoms
decreases
increases.
density
than
Al
because
it
has
a
more
open
lattice
structure.

The
density
Although
together
of
the
has
Electrical
P
and
atoms
an
S
(and
are
Cl
and
heavier
,
Ar
the
as
way
liquids)
the
is
relatively
molecules
are
low.
packed
effect.
conductivity of
Period
3
elements
−1
The
electrical
conductivity
Period
3
elements
hardly
conduct
as
are
(measured
shown
below.
in
siemens
Chlorine
per
and
metre,
argon
are
Na
Mg
Al
S
Conductivity/
0.218
0.224
0.382
2
P
of
the
and
S
−10
×
−17
10
1
×
−23
10
1
×
10
−1
S m
10
Na,
Mg
metallic
and
Al
are
structure
conductivity
electrons
Mg
)
gases
solids.
Element
8
S m
good
free
increases
provided
provides
all
are
2
and
by
Al
conductors.
to
from
each
move
Na
to
‘atom’
provides
3).
The
when
Al
as
a
the
increases
The
delocalised
voltage
is
number
(Na
electrical
electrons
applied.
of
in
the
The
delocalised
provides
1
electron,
conductivity
of
Si
is
poor
Key points
because
(a
it
has
substance
no
delocalised
whose
electrical
electrons.
It
conductivity
is
classed
as
increases
a
metalloid
with
increase
in

temperature).
Some
of
its
electrons,
however
,
can
move
out
of
Across
of
especially
when
‘contaminating’
atoms
are
present
in
its
Perod
lattice.
So
it
a
semi-conductor .
P
and
S
hardly
conduct
electricity
the
elements
because
covalent
molecules
with
no
delocalised
The
3
electronegativity
of
the
atoms
Na
Mg
increases
Al
a
across
Period
S
P
1.2
1.5
to
atomc.
meltng
reects
pont
1.8
2.
1
2.5
bondng
the
of
the
elements.
3.
S
and
Across
Perod
3
electrcal
Cl
alumnum
0.9
n
perod
conductvty
Electronegativity
to
elements

Element
covalent
molecular
structure
The
changes from
gant
varaton
across
Period
to
electrons.

Electronegativity of
structure
they
smple
are
the
is
metallc
called
3
position,
ncreases
then
to
decreases.
3.0

The
densty
Perod
then

3
of
the
ncreases
shows
a
to
general
Electronegatvty
a
elements
n
alumnum
decrease.
ncreases
across
perod.
135
11.2
Patterns
Learning outcomes
in
Period
Patterns
The
On
completon
of
ths
secton,
covalent
be
able
elements
atomic
radius
is
and
often
ionic
used
as
radii
a
in
measure
Period
of
the
3
size
of
an
atom.
you
This
should
in
3
is
half
the
distance
between
the
nuclei
of
two
atoms
of
the
same
to:
type.

explan
and
the
onc
varaton
radus
of
n
atomc
Perod
3
elements
r

descrbe
the
elements
reactvty
wth
of
oxygen,
Perod
3
chlorne
Figure 11.2.1
The covalent radius of an atom, r,
is half the distance between the two nuclei
and

water
predct
the
type
of
bondng
n
the
The
chlordes
and
oxdes
of
Perod
atomic
each
elements
by
dfferences
reference
n
onc
to
radus
and
there
The
the
is
type
covalent
or
radus’
depends
bond formed:
van
example, for
decrease
element
into
the
additional
in
Period
3.
As
we
move
across
the
period,
So
the
has
one
shielding
charge
size
of
more
quantum
the
of
proton
shell
outer
from
shell
Na
atoms
in
(the
to
its
nucleus.
third
electrons
Cl
pulls
decreases
But
electron
by
the
across
the
the
shell).
inner
electrons
the
added
So
shells.
closer
to
period.
Na
Mg
Al
Si
P
S
Cl
0.
156
0.
136
0.
125
0.
117
0.
110
0.
104
0.099
on
metallc,
der Waals.
across
same
nuclear
atom
‘atomc
of
no
increase
size
the
goes
nucleus.
Did you know?
term
successive
electron
electronegatvty.
The
radii
3
of
atom
For
and
nuclear
charge
sodum:
Atomic
covalent
radus
=
0.
134 nm
metallc
radus
=
0.
191 nm
(covalent)
radius
van
der Waals
radus
=
(nm)
0.230 nm
Figure 11.2.2
Covalent
when
a
rad
are
comparng
generally
atom
szes
The atomic radii of the elements from sodium to chlorine
used
across

Metal
shell.
perod.

ions
So
are
their
Non-metal
shell.
the
ions
There
atom.
which
it
formed
ions
is
are
by
losing
smaller
formed
more
This
is
are
by
repulsion
makes
the
electrons
than
gaining
negative
ionic
each
from
their
ion
outer
electron
atoms.
electrons
between
the
in
their
electrons
larger
than
outer
than
the
electron
there
atom
is
radii
successive
decrease
ion
has
from
one
4+
to
Na
more
Si
.
proton
As
in
we
its
move
across
nucleus.
But
the
period,
electronic
+
structure
pulls
A
the
is
the
same.
electrons
similar
The
closer
explanation
increase
to
the
(higher
in
nuclear
nucleus.
nuclear
So
charge
the
charge
size
and
from
of
same
the
Na
ions
4+
to
accounts
for
the
decrease
in
ionic
radius
from
3–
P
2–
S
0.20
Cl
)mn(
suidar
+
Na
0.
10
2+
cinoI
Mg
3+
Al
4+
Si
0
11
12
13
14
atomic
+
Figure 11.2.3
136
The ionic radii from Na
−
to Cl
15
number
16
17
P
to
Si
decreases.
electronic
3−
structure)
in
from
derived.
+
The
their
Cl
Chapter
The

reaction of
Sodium
reacts
Period
3
elements
with
with
steam
reacts
to
+
2H
very
form
O(l)
→
2NaOH(aq)
+
slowly
with
Perod
3
H
+
does
not
O(g)
H
react
(g)
cold
water
but
it
reacts
readily
oxide:
→
MgO(s)
+
H
2
Aluminium
of
2
magnesium
Mg(s)

elements
water
2
Magnesium
The
vigorously:
2Na(s)

11
(g)
2
with
hot
or
cold
water
but
it
reacts
with
steam:
2Al(s)
+
O(g)
3H
→
Al
2

Silicon,
phosphorus
and
O
2
sulphur
do
(s)
+
3H
3
not
(g)
2
react.
They
are
insoluble
in
water
.

Chlorine
of
dissolves
hydrochloric
slightly
in
chloric( i)
and
water
acid
and
then
reacts
to
form
a
mixture
:
+
Cl
(g)
+
H
2
The

reaction of
Sodium
O(l)
Y
2H
(aq)
+
Cl
(aq)
+
ClO
(aq)
2
reacts
Period
vigorously
3
elements
when
heated
to
with oxygen
form
sodium
oxide,
Na
O
2
(although
the
nal
stable
product
is
sodium
peroxide,
Na
O
2
2Na(s)
+
O
(g)
→
Na
2

Magnesium
and
aluminium
O
2
react
):
2
(s)
2
vigorously
with
oxygen
to
form
oxides:
2Mg(s)
+
O
(g)
→
MgO(s)
and
4Al(s)
+
3O
2

Silicon
reacts
(g)
→
2Al
2
slowly
with
oxygen:
Si(s)
+
O
(g)
→
SiO
2

Phosphorus
reacts
vigorously
with
oxygen
O
2
(s)
3
(s)
2
to
from
phosphorus(
v)
Did you know?
oxide:
The
4P(s)
+
5O
(g)
→
2P
2
O
2
bleachng
chlorne

Sulphur
burns
steadily
in
oxygen:
S(s)
+
O
(g)
→
SO
2

Chlorine
The
Apart
and
argon
reaction of
from
argon
do
not
combine
Period
(and
3
chlorine
all
of
most
with
3
chlorc(i)
due
to
the formaton
of
acd.
oxygen.
with
Period
s
(g)
2
directly
elements
itself)
acton
(s)
5
chlorine
elements
react
with
chlorine.
Sodium,
magnesium
and
MgCl
AlCl
2
and
and
aluminium
respectively,
which
react
are
vigorously
ionic
solids.
to
form
Silicon,
NaCl,
phosphorus
3
sulphur
react
more
slowly.
with
sodium:
with
silicon:
with
phosphorus:
with
sulphur:
For
example:
2Na(s)
+
Cl
(g)
→
2NaCl(s)
2
Si(s)
+
(g)
2Cl
→
SiCl
2
2P(s)
+
(l)
4
(g)
5Cl
→
2PCl
2
2S(s)
+
Cl
(s)
5
(g)
→
2
S
Cl
2
(l)
2
Key points


The
man
atomic
radius
and the dstance of the outer
Across
Perod
water

nuences on
charge
tends
Across
3,
to
Perod
the
reactvty
of
the
n
Perod
shell
3
are the
sze of the
electrons from the
elements
wth
oxygen,
nuclear
nucleus.
chlorne
and
decrease.
3,
the
chlordes
and
oxdes
change from
onc
to
covalent
compounds.
137
11.3
Properties
Learning outcomes
of
Oxidation
The
On
completon
of
ths
Period
secton,
oxides
be
able
descrbe
and

descrbe
and
n
of
the
oxdes
terms
varaton
states
oxdes
elements
is
in
more
the
Perod
3
bondng
of
of
of
n
Perod
elements
n
3
rises
chlordes
use
oxidation
across
all
leads
the
to
the
Period
3
all
exist
electronegative
an
in
than
positive
any
of
oxidation
these
states
elements.
and
in
expanded
of
each
element
This
corresponds
their
outermost
octet
for
oxides
in
to
Period
the
electron
of
P
,
S
and
3
in
their
The
oxides
ability
of
the
shells
in
bonding.
elements
to
This
Cl.
chlordes
elements
electronegatvty
radus
state
period.
electrons
Oxide
Na
O
MgO
Al
2
O
2
SO
3
P
2
O
4
SO
10
Cl
3
O
2
7
and
Max. oxidation
onc
chlorides
to:
the
oxdaton
and
states
oxygen
maximum

oxides
you
because
should
of
3
+1
+2
+3
+4
+5
+6
+7
charge
state

explan
the
behavour
trend
of
n
oxdes
acd–base
and
The
hydroxdes
across
Perod
non-metallic
oxides
can
also
form
oxides
in
lower
oxidation
states
3.
e.g.
SiO,
P
O
2
O
Cl
2
,
SO
3
and
several
lower
oxides
of
chlorine
(Cl
2
O,
ClO
2
,
2
).
6
The
chlorides
because
period.
rising
The
to
of
Period
chlorine
is
oxidation
+5
in
3
more
PCl
.
elements
also
exist
electronegative
states
The
of
the
chlorides
maximum
in
that
positive
the
other
show
oxidation
a
oxidation
elements
similar
state
for
states
in
the
pattern,
sulphur
in
its
5
chlorides,
however
,
Chloride
is
+2.
NaCl
MgCl
AlCl
2
Max. oxidation
+1
SCl
3
+2
PCl
4
+3
SCl
5
+4
2
+5
+2
state
Some
of
the
oxidation
non-metallic
states,
e.g.
elements
,
PCl
S
3
Bonding
Did you know?
The
Many
onc
compounds
have
in
structure
and
of
covalent
character
on
small
the
polarsaton. Ths
hghly-charged
electron
larger
lkely
anon.
to
cloud
Ion
occur
s
when
caton
charge
hgh
charge
when
s
i
the
a
ii
has
a
hgh
charge
and
in
2
and oxides of
of
these
of
the
atoms
Period
compounds
involved
electronegativity,
chloride
will
be
and
Mg,
or
Na,
s
are
transferred
can
be
3
related
the
relative
the
in
more
bonding.
likely
it
is
The
that
greater
the
the
oxide
ionic.
The
electronegativity
difference
between
or
oxygen
are
therefore
is
ionic.
so
the
great
that
metal
one,
atom
two
to
the
or
three
electrons
non-metal
atom.
respectively
These
The
other
Period
3
oxides
are
covalently
oxides
bonded.
has
dioxide
the
s
Al
from
most
has
a
giant
covalent
structure
(see
Section
2.6).
Although
4+
in
anon
lower
much
caton
small,
in
dstorts
of
polarsaton
and
chlorides
a
Silicon
a
form
because
difference
of
bonding
also
a
electronegativity
degree
2
chlorides
can
Cl
large.
theory,
that
the
an
ion
Si
lattice
can
energy
exist,
cannot
the
fourth
compensate
ionisation
for
it
to
energy
make
is
ionic
so
large
stable.
SiO
2
Oxides
a
of
P
,
S
and
Electronegativity
b
Period
+
3
Cl
a
differences
chlorides.
anhydrous
have
Sodium
aluminium
simple
can
also
chloride
chloride
molecular
is
a
be
and
used
structure.
to
explain
magnesium
the
structure
chloride
covalently-bonded
are
molecule,
of
ionic.
Al
Cl
2
But
.
6
3+
Al
Cl
3+
The
separate
ions
electron
ionic
up
is
very
small
and
the
Al
ion
This
high
positive
charge
density
tends
to
pull
the
chloride
ion
between
towards
the
Al
and
it
to
Cl
such
atoms.
an
extent
This
is
that
called
electron
ion
pairs
are
polarisation
The polarisation of a large
ion by a small, highly charged Al
138
highly
electrons
3+
Cl
is
‘atoms’
shared
Figure 11.3.1
aluminium
between
larger
the
of
density
charged.
builds
radius
ion
Chlorides
of
Si,
P
and
S
have
a
simple
molecular
structure.
in
the
Chapter
Reactions of
The
oxides
of
Period
sodium
and
3 oxides
magnesium
and
react
11
The
elements
of
Perod
3
hydroxides
with
water
to
form
hydroxides:
Na
O(s)
+
H
2
Magnesium
sodium
basic.
hydroxide
hydroxide.
They
react
is
less
Oxides
with
+
alkaline
and
acids
MgO(s)
O(l)
→
2NaOH(aq)
2
to
because
hydroxides
form
2HCl(aq)
a
of
salt
→
it
is
less
sodium
and
MgCl
water
.
(aq)
+
oxide
does
not
dissolve
in
water
than
magnesium
are
e.g.
H
2
Aluminium
soluble
and
O(l)
2
but
reacts
with
acids
and
alkalis:
Al
O
2
Al
O
2
(s)
+
3H
3
(s)
SO
2
+
(aq)
→
Al
4
2NaOH(aq)
(SO
2
+
3H
3
)
4
O(l)
→
+
3H
3
2
substance
which
acts
both
as
an
acid
(aq)
4
sodium
A
O
2
2NaAl(OH)
and
a
base
is
aluminate
said
to
be
amphoteric
Silicon
dioxide
is
insoluble
in
water
but
like
Al
O
2
SiO
(s)
+
2NaOH(aq)
→
Na
2
SiO
is
an
acidic
SiO
2
oxide.
It
does
not
react
reacts
(aq)
+
H
3
with
with
hot
alkali:
3
O(l)
2
acids.
2
Oxides
acidic
of
P
,
S
and
solutions
Cl
and
are
all
with
SO
acidic
alkalis
(g)
+
H
+
6H
2
O
P
4
Reactions of the
Chlorides
dissolve
of
in
sodium
water
oxides.
to
form
O(l)
→
H
→
4H
2
(g)
6
they
SO
are
water
to
form
(aq)
(aq)
3
Period
and
with
3
PO
3
magnesium
because
react
e.g.
2
O(l)
2
chlorides of
and
They
salts,
3
hydrated
aluminium
chloride
Did you know?
ionic.
Anhydrous
The
chlorides
of
silicon
and
phosphorus
are
hydrolysed
by
water
to
has
acidic
solutions.
Fumes
of
hydrogen
chloride
are
also
(l)
+
2H
+
3H
4
3
SiO
+
4H
5
(s)
+
4HCl(g)
(aq) +
3HCl(g)
→
H
PO
3
O(l)
→
2
H
PO
(aq) +
behaves
chlorde
3
3
smple
chlorde
molecular
dfferently
alumnum
2
O(l)
2
(s)
PCl
→
2
(s)
PCl
O(l)
a
structure.
released.
It
SiCl
alumnum
form
5HCl(g)
water
4
to
hydrated
chlorde. Alumnum
soluton
to form
an
s
hydrolysed
acdc
3+
The
hydrolysis
of
S
Cl
2
thionic
acid
being
is
complex,
several
sulphur
compounds
including
2
[Al(H
O)
2
]
6
by
soluton:
2+
→
[Al(H
O)
2
OH]
+
+
H
5
formed.
Key points

The
oxdaton
rest,

Chlordes
chlordes

Across

Sodum
s
states
apart from Ar,
of
Na
are
oxdes
and
Mg
3,
the
n
to form
oxdes
water
acdc
of
S,
and
P,
S
Na
to Al
are xed. The
states.
to form
onc
solutons. The
other
solutons.
change from
oxdes
oxdes
chlordes from
oxdaton
dssolve
magnesum
amphoterc. The
and
several
hydrolysed
Perod
and
of
have
basc
to
hydroxdes
and Cl
are
amphoterc
are
(Al)
to
acdc.
basc. Alumnum
oxde
acdc.
139
12
The
12.
1
Propertes
chemstry
Learning outcomes
of Groups
of Group
Physical
Elements
On
completon
of
ths
secton,
be
able
explan
the
of Group
II
structure

descrbe
araton
n
propertes
elements
n
terms
and
the
elements
and VII
elements
properties
of
in
the
periodic
orbitals,
s,
p,
d
table
or
f
in
can
be
their
put
outer
into
blocks
shell.
according
Group
II
to
elements
the
are
in
to:
the

IV
you
type
should
II
II,
s
block
outer
reactons
the
Periodic
T
able.
They
have
two
s-type
orbitals
in
their
shell.
of
bondng
wth
in
s-block
of Group
oxygen,
water
II
p-block
and
d-block
dlute

state
acds
some
uses
of
MgO, CaO,
Figure 12.1.1
The s, p, d and f
f-block
Ca(OH)
and CaCO
2
blocks n the Perodc Table
3
The
table
shows
some
physical
properties
of
these
elements.
−1
Element
Atomic
Metallic
Ionic
Density/
Melting
Ionisation
energies/kJ mol
−3
number
radius/nm
radius/nm
g cm
point/°C
First
beryllum,
Be
0.
122
0.031
1.85
12
0.
160
0.065
1.74
694
763
calcum, Ca
20
0.
197
0.099
1.55
839
590
1150
strontum,
38
0.215
0.
113
2.60
769
548
1060
56
0.224
0.
135
3.51
725
502
966
magnesum,
barum,
4
Mg
Sr
Ba
Dd you know?
Metallic
The
The
metallc
radus
s
half
between
two
lled
n
a
metallc
s
block
radus
s
gant
elements,
greater
than
electron
metallic
1450
radius
ionic
shells
radii
both
increases.
increase
The
down
ionic
the
radius
group
is
as
much
the
number
smaller
radius
because
when
forming
ions
the
electrons
in
than
the
outer
structure.
shell
For
and
1760
neghbourng
the
nucle
metallic
ionic
900
the
of
dstance
and
1278
Second
the
metallc
the
coalent
have
been
lost.
Density
radus. The
metallc
dagram
radus,
shows
the
The
r
in
density
the
Beryllium
are
r
From
and
packed
on
the
calcium
magnesium
more
efciently
mass
to
have
in
of
the
barium
higher
the
atoms
there
is
and
an
densities
lattice
with
the
way
increase
than
less
they
in
calcium
wasted
are
packed
density.
because
they
space.
r
Melting
The
As
points
melting
we
go
Ca,
the
So
Sr
and
sea
of
there
takes
follow
a
the
spoils
Ba.
are
less
high,
group
this
Going
delocalised
is
less
points
down
Magnesium
140
depends
lattice.
similar
to
there
trend
down
the
is
a
are
of
general
group
between
overcome
trend.
expected
because
electrons
attraction
energy
as
it
has
the
these
giant
metallic
decrease
a
away
melting
the
and
forces.
structure.
lattice
which
from
electrons
attractive
in
different
electrons
further
these
a
contribute
positive
the
The
points.
structure
nuclei
boiling
to
to
nuclei.
and
it
points
Chapter
First
and
go
second
As
we

further

screened
These
is
to
Some
When
group,
from
more
the
the
outweigh
the
these
Group
II
they
generally
exhibit
because
chemical
meals
act
and
aluminium
it
as
effect
to
other
Reaction
with oxygen
a
redox
they
reaction.
greater
the
as
II
of Groups
II,
IV
and VII
are:
the
The
inner
nuclear
down
two
From
the
electron
charge
shells.
and
so
it
group.
electrons
magnesium
are
outer
II
of
metals
properties
Group
go
They
the
number
increased
we
lose
agents.
lose
physical
the
electrons
of
reactions.
easier
some
than
react,
here.
is
the
electrons
considered
This
outer
by
reducing
similar
is
chemistry
nucleus
effectively
remove
The
energies
reactions of Group
So
group
the
away
factors
easier
shell.
down
ionisation
12
more
which
are
reactivity
with
So
down
Beryllium
more
its
oxygen
their
outer
barium,
reactive
electrons.
elements.
from
to
like
the
has
those
reactions
increases
they
are
of
not
down
the
group.
2Mg(s)
+
O
(g)
→
2MgO(s)
2
The
oxides
are
basic
in
character
,
CaO(s)
+
reacting
H
O(l)
→
with
water
Ca(OH)
2
Reaction
with
Magnesium
steam
to
Calcium
very
slowly
magnesium
reacts
vigorously
Ca(s)
white
hydroxides:
(s)
2
+
with
oxide
with
2H
cold
and
cold
O(l)
water
water
→
precipitate
is
but
hydrogen
to
more
(see
form
Ca(OH)
2
A
form
water
reacts
form
to
(s)
calcium
+
H
2
observed,
but
some
vigorously
Section
with
11.2).
hydroxide:
(g)
2
Ca(OH)
dissolves
to
form
a
2
weakly
alkaline
similar
manner
alkaline
down
solution.
with
the
The
other
increasing
group
as
the
elements
vigour
.
The
solubility
of
down
the
resulting
the
group
react
solutions
hydroxides
get
in
a
more
Key points
produced
increases.

Reaction
All
the
with
members
salt
of
of
the
acids
the
group
metal.
The
react
readily
reactivity
with
acids
increases
to
down
form
the
hydrogen
+
2HCl(aq)
→
MgCl
(aq)
+
II
Ba
metals
are
elements from
water
to
metal
hydroxide.
produce
H
SO
H
2
(aq)
→
BaSO
4
The
atomic
(s)
+
H
4
compounds of Group
Calcium
oxide:
making

Calcium
hydroxide:
cement
and
mortar;
making

acidic
soil;
drying
Group
II
limestone
blocks
for
metals
oxides
making
bleaching
powder;
The
the
metals
group.
burn
which
building;
down
reactivity
elements
dilute
carbonate:
of
the
in
are
air
to
more
the
group.
agent
limewater
Calcium
radii
down
II

neutralising
removing
as
SiO
of Group
with
oxygen,
hydrochloric
down
the
Many
compounds
II
water
acid
and
increases
group.
2
slag
in
the
blast
furnace
for
the
extraction
of
iron;
making
calcium

oxide
for
of Group
II
cement
metals

Magnesium
oxide:
for
lining
a
(g)
soluble

and
2
form
some
hydrogen
(g)

Uses of
to
with
2
increases
+
Mg
react
group:
2
Ba(s)
which
and

Mg(s)
Group
have
important
uses.
furnaces
141
12.2
Group
II
Learning outcomes
carbonates,
Thermal decomposition of
Group
On
completon
of
ths
secton,
be
able
II
carbonates
from
form
the
oxide
and
explan
the
araton

n
barium,
(s)
and
explan
araton
II
The
ntrates
solublty
of Group
know
the
test for
usng
barum
II
of
→
CaO(s)
+
decompose
CO
3
thermal
of Group
carbonates
the
to
on
heating
dioxide:
to:
decomposton

sulphates
carbonates
magnesium
carbon
CaCO

and
you
to
should
ntrates
temperatures
shown
at
which
(g)
2
these
carbonates
are
50%
decomposed
are
below.
the
Carbonate
sulphates
sulphate
MgCO
CaCO
3
SrCO
3
BaCO
3
3
ons
Temperature/K
813
1173
1553
1633
chlorde
Going
down
Group
decomposition.
go
down
the
The
size

The
smaller
charge
of
of
Smaller

So
II
a
carbonates
II
cations
cation
cation,
larger
are
the
the
C=O
ionic
become
have
the
more
same
resistant
charge
to
(+2)
thermal
but
as
we
increases.
better
is
at
distorting
the
electron
cloud
ion.
polarisers
with
it
of
large
smaller
ions
ions
(see
have
a
Section
greater
11.3).
degree
of
bonding.
degree
bond
the
carbonate
better
carbonates
in
greater
break
metal
the
the
Group
The
the
ions
covalence

the
Group
group:


II,
All
in
of
covalence,
the
the
carbonate
less
is
the
energy
required
to
ion.
2+
2+
Mg
Ba
2–
2–
CO
CO
3
3
2+
Figure 12.2.1
a The small Mg
on s a good polarser of the carbonate on; b The larger
2+
Ba
on s a poor polarser of the carbonate on
Thermal decomposition of
Group
form
II
the
nitrates
oxide,
from
magnesium
nitrogen
2Mg(NO
)
3
Going
down
Group
decomposition.

The
better
nitrate
142
it
is
(s)
and
barium,
decompose
the
2MgO(s)
+
4NO
(g)
+
O
2
nitrates
the
distorting
on
heating
also
become
(g)
2
more
resistant
to
thermal
cation:
the
The
greater

The
less
the
energy
ion.
degree
of
required
electron
covalence
to
break
to
oxygen:
2
smaller
at
→
to
cloud
charge
of
the
larger
ion.

nitrate
II,
The
dioxide
nitrates
a
in
the
ionic
particular
bonding.
N−O
bond
in
the
Chapter
Solubility of Group
The
solubility
Magnesium
of
Group
sulphate
II
II
is
The
chemstry
of Groups
II,
IV
and VII
sulphates
sulphates
very
12
in
soluble.
water
Barium
decreases
sulphate
down
is
the
group.
‘insoluble’.
Ø
Solubility
depends
on
the
enthalpy
change
of
solution,
∆H
.
As
a
rough
sol
Ø
guide,
the
more
endothermic
the
value
of
,
∆H
the
less
soluble
is
the
sol
salt.
The
enthalpy
change
of
solution
depends
on
the
relative
values
of
Ø
the
lattice
energy,
and
∆H
the
enthalpy
changes
of
hydration
of
the
latt
Ø
aqueous
ions,
(see
∆H
Section
6.3).
hydr
2+
Ca
2–
(g)
+
SO
(g)
4
2+
H
[Ca
]
hyd
2–
+H
[SO
hyd
]
4
H
latt
(aq)
Ca SO
4
H
sol
Ca SO
(s)
4
Ø
Figure 12.2.2
An enthalpy cycle for calculatng ∆H
sol

Lattice
energy
decreases
MgSO
down
>
the
CaSO
4
This
the

decrease
large
The
But
the
is
fairly
sulphate
enthalpy
this
size
ion
decrease
of
the
small
is
because
by
of
By
Hess’s
law:
than

So
>
order:
BaSO
it
is
4
determined
more
by
the
size
of
cation.
also
larger
the
increases
because
it
is
in
the
same
determined
order
.
more
by
anion.
Ø
=
∆H
Ø
∆H
sol
Ø
SrSO
the
4
hydration
Ø

the
relatively
cations
>
in
4
than
change
group
−
∆H
hyd
latt
Ø
∆H
−
∆H
hyd
gets
more
endothermic
down
the
group
(the
value
of
latt
Ø
gets
∆H
more
Testing for
The
relative
following
positive).
sulphate
insolubility
test
to
detect
of
So
the
solubility
decreases.
ions
barium
sulphate
sulphate
is
used
as
the
basis
for
acid
or
hydrochloric
the
ions:
Key ponts

Acidify
This
acid
the
removes
to
form
precipitate

Add

If
solution
to
aqueous
tested
contaminating
carbon
of
be
dioxide.
barium
barium
nitric
carbonate
If
these
carbonate
chloride
with
would
or
ions,
were
not
which
interfere
aqueous
react
removed
with
barium
with
acid.
the
insoluble
the

white
Group
are
test.
II
more
ions
are
present
a
white
precipitate
of
Ths
nitrate.
barium
sulphate
resstant
decomposton
s
because
degree
sulphate
carbonates
of
and
to
down
of
the
ntrates
thermal
the
group.
decreasng
polarsaton
of
the
is
2−
large CO
formed:
or
NO
3
2+
Ba
(aq)
SO
(aq)
4
→
BaSO
by
the
3
ncreasngly
2−
+
on
large
metal
caton.
(s)
4

The
decreasng
Group
II
group
can
of
relate
the
enthalpy
and

to
a
be
alues
barum
a
of
of
n
of
the
the
terms
ther
hydraton
energes.
soluton
sulphate,
down
explaned
changes
lattce
When
solublty
sulphates
chlorde
s
contanng
whte
added
a
precptate
s
formed.
143
12.3
Group
IV
Learning outcomes
elements
Some
The
On
completon
of
ths
secton,
be
able
Group
four
explan
the
propertes
elements
structure

descrbe
Group

descrbe
araton
of
n
n
physcal
the Group
terms
and
the
IV
of
in
exsts
the
n
bondng
the
of
reacton
wth
to
tn
whch
made
by
of
the
too
tn
cold
cold
may
f
their
in
the
outer
p
block
of
principal
the
Periodic
quantum
T
able.
shell.
They
They
are
all
all
at
the
room
temperature.
Some
of
their
physical
properties
are
shown
table.
Structure
Melting
Electrical
point/°C
conductivity
3550
non-conductor
the
carbon
(damond), C
non-metal
of Group
slcon,
S
metallod
1410
sem-conductor
germanum, Ge
metallod
937
sem-conductor
tn,
metal
232
conductor
metal
327
conductor
IV
has
dependng
as
a
lead,
structure
In
countres
become
are
Pb
powdery
a
clmates,
they
Sn
low
exsts
damond.
hae
in
Element
Trends

whch
are
elements
water.
seeral forms,
temperatures,
smlar
electrons
bondng
temperature. At
grey form
elements
IV
IV
ther
Dd you know?
on
properties of Group
tetrahaldes
tetrahaldes
Tn
IV
tetrachlordes
to:
solids

physical
ther
you
have
should
and
artcles
and
(see
damaged
left
Si
the
melting
Ge
have
Section
fact
outsde for
in
that
a
giant
2.6).
lot
points
of
covalent
They
all
energy
is
structures
have
very
needed
similar
high
to
to
melting
break
the
that
of
points
many
C(diamond)
due
to
covalent
the
bonds
in
structure.
long.

The
decrease
decreasing
in
melting
amount
of
point
energy
from
C(diamond)
needed
to
break
to
the
Ge
reects
−1
E(C−C)
=
the
bonds:
−1
E(Si−Si)
+350 kJ mol
=
+222 kJ mol
−1
E(Ge−Ge)

Sn
and
other
the
so
Pb
ions
have

There

Carbon
its
does
free
relatively

Si
a
in
the
move
Ge
melting
of
points
metallic
Section
melting
to
are
in
2.4).
compared
bonding
Tin
and
with
decreases
lead
have
as
most
the
fairly
size
large
of
ions,
points.
electrical
diamond
are
some
Section
of
Some
different
increase
decreases
does
used
its
conductivity
not
in
down
conduct
covalent
electrons
are
the
group.
electricity
bonding.
because
Graphite
delocalised
and
so
are
2.6).
electricity
electrons.
conductivity
metalloid
of
because
(see
with
described
trend
electrons
conduct
‘jumping’
increases
144
low
form
shell
conduct
delocalised
by
(see
general
outer
and
low
strength
+188 kJ mol
conductivity
is
to
relatively
The
increases
Electrical
all
have
metals.
=
to
of
places
in
a
very
their
in
small
the
lattice.
temperature
with
increase
semi-conductors .
extent.
electrons
in
Their
(as
can
They
move
Their
opposed
is
not
of
have
position
conductivity
to
temperature).
structure
do
out
metals,
Si
and
described
where
Ge
as
are
Chapter

Sn
and
are
is
Pb
are
electrical
delocalised.
a
greater
electrons
nuclear
second
trend
force
in
All
the
all
a
conductors
slightly
attraction
with
from
atomic
Sn
Sn.
to
This
is
This
energies
is
its
Sn.
than
in
This
a
ions
because
results
down
their
conductor
greater
than
energies
because
better
between
Pb
radius.
ionisation
The Group
are
of
ionisation
of
is
compared
charge
increase
Sn
is
group
and
the
the
effect
having
opposite
(see
shell
Pb.
In
of
of
chemstry
of Groups
II,
IV
and VII
electrons
lead
there
the
to
increased
shielding
higher
Section
The
delocalised
effect
the
lead
outer
than
12
the
rst
and
and
general
1.5).
IV tetrachlorides
elements
in
Group
IV
form
tetrachlorides,
XCl
.
The
tetrachlorides
4
simple
covalent
molecules
with
tetrahedral
structure.
Cl
Ge
Cl
Cl
Cl
Figure 12.3.1
All
the
This
der
The tetrahedral structure of germanum(iv) chlorde
tetrachlorides
is
because
W
aals
dipoles
they
forces
cancel.
are
are
volatile
simple
between
their
Although
liquids
covalent
and
CCl
SiCl
,
SnCl
4
,
and
PbCl
4
,
they
They
have
are
boiling
only
non-polar
lower
boiling
points.
weak
van
because
points
the
than
4
there
is
no
=
76 °C
SiCl
4
simple
pattern
in
=
57 °C
GeCl
4
=
SnCl
the
boiling
points:
PbCl
=
87 °C
105 °C
4
Reactivity of the Group
tetrahalides,
=
4
114 °C
4
the
low
with
4
CCl
All
have
molecules
molecules.
4
GeCl
−
with
IV tetrahalides
the
exception
of
CCl
with
,
are
water
readily
hydrolysed
by
4
water
to
chloride
the
are
oxide
also
in
the
+4
produced,
(l)
SiCl
oxidation
hydrolysis
of
the
+
2H
(l)
+
ease
of
hydrolysis
→
SiO
fumes
of
hydrogen
(s)
+
4HCl(g)
2
tetrahalides
2H
4
The
O(l)
2
heavier
GeCl
Acidic
e.g.
4
The
state.
O(l)
Y
is
GeO
2
reversible.
(s)
+
For
example:
4HCl(g)
2
increases
down
the
group
from
SiCl
to
PbCl
4
the
metallic
chloride
lead( ii)
is
nature
of
the
hydrolysed,
chloride
also
a
Group
little
IV
atom
increases.
decomposition
of
the
When
lead(
iv)
as
4
lead(
iv)
chloride
to
occurs.
Key ponts

C,
S

Down Group
and
and Ge
a
hae
IV,
general

Group
IV

The Group
gant
the
coalent
elements
ncrease
n
elements form
excepton
IV
of
show
metallc
are
a
Sn
general
character
coalently
tetrahaldes
carbon
structures;
bonded
hydrolysed
and
Pb
are
decrease
and
n
metals.
meltng
electrcal
pont
conductty.
tetrachlordes.
rapdly
by
water
wth
the
tetrachlorde.
145
12.4
Some
propertes
Learning outcomes
The Group
The
On
completon
of
ths
secton,
oxides

be
able
descrbe
and
of
IV oxides:
the
Group
IV
IV
oxdes
structure
elements
and
exist
in
stability
the
+2
and
+4
oxidation
you
states.
should
of Group
The
table
shows
the
type
of
structure
of
these
oxides.
to:
the
thermal
trends
n
stablty
bondng
of
+2 oxidation
state
+4 oxidation
state
the
CO
smple
molecular
CO
molecular
2
Group

IV
descrbe
oxdes
the
trends
character
of Group
oxdaton
states
n
IV
SO
acd–base
oxdes
smple
molecular
SO
gant
coalent
gant
coalent
gant
coalent
wth
onc
character
gant
coalent
wth
onc
character
2
n
GeO
onc
wth
some
GeO
2
+2
and
+4.
coalent
SnO
gant
character
onc
SnO
2
PbO
gant
onc
PbO
2
Down
each
character
series
of
the
of
oxides,
Group
IV
the
structures
elements
get
more
increases.
ionic
and
SnO
as
the
PbO
2
degree
from
of
covalent
CO
and
character
,
CO
are
in
solids
their
with
ionic
giant
structure.
structures
metallic
have
some
2
All
the
and
oxides,
high
apart
melting
2
points.
CO
and
are
CO
gases
at
room
temperature,
because
they
have
a
2
simple
molecular
molecules

in
Carbon
the
structure
only
solid
and
liquid
monoxide,
CO
has
decompose
on
a
weak
forces
between
their
states.
strong
triple
bond
and
does
not
heating.
Germanium( ii)

with
oxide
disproportionates
on
2GeO(s)
heating:
→
GeO
(s)
+ Ge(s)
2
Oxidation
Tin( ii)

oxide,
heating
oxides

in
in
Carbon
SnO,
the
the
numbers
and
lead( ii)
absence
presence
dioxide,
of
of
air
.
+4
oxide,
They
PbO,
are
do
0
not
readily
decompose
oxidised
to
on
higher
oxygen.
has
CO
+2
strong
double
bonds
and
does
not
2

decompose
on
The
in
oxides
down
the
heating.
the
group.
+4
Only
oxidation
lead( iv)
state
tend
oxide,
PbO
to
,
decrease
however
,
in
stability
undergoes
2
signicant
thermal
decomposition:
PbO
(s)
→
PbO(s)
+
2
lead(iv)
Acid−base
The
+2
+
table
and
CO
summarises
oxidation
2 oxidation
ery
lead(ii)
(g)
oxide
properties of the oxides
below
+4
O
2
oxide
state
weakly
the
acid–base
properties
of
the
oxides
states.
+
acdc
4 oxidation
CO
state
acdc
2
SO
−
SO
ery
weakly
2
GeO
amphoterc
GeO
amphoterc
2
SnO
amphoterc
SnO
amphoterc
2
PbO
amphoterc
PbO
amphoterc
2
146
acdc
in
the
Chapter

The
the
oxides
in
the
+2
corresponding
CO
only
reacts
oxidation
oxides
with
hot
in
state
the
+4
are
less
acidic
oxidation
concentrated
(more
state.
sodium
For
basic)
12
The
chemstry
of Groups
II,
IV
and VII
than
example:
hydroxide,
but
CO
2
forms
an
acidic
solution
in
water:
+
(g)
CO
+
H
2
CO
also
reacts
O(l)
Y
HCO
2
with
(aq)
+
H
(aq)
3
dilute
alkalis:
2
(g)
CO
+
2NaOH(aq)
→
Na
2

The
oxides
group
as
in
the
CO
2
both
oxidation
metallic
states
character
of
(aq)
H
O(l)
2
become
the
+
3
more
Group
IV
basic
atom
down
the
increases.
For
example:
Carbon
silicon
dioxide
dioxide
reacts
only
(s)
SiO
+
with
reacts
aqueous
with
2NaOH(aq)
hot
→
alkalis
(hydroxide
concentrated
Na
2
SiO
2
Reactions of the
(aq)
+
H
3
sodium
ions)
but
alkali:
O(l)
2
silicate
amphoteric oxides of Group
IV
Oxidation state +2

GeO,
SnO
character

They
For
all
and
in
PbO
the
react
are
order
with
all
amphoteric,
GeO
acids
to
<
SnO
form
<
a
but
have
increasingly
salt
with
oxidation
+
2HCl(aq)
→
(aq)
SnCl
+
H
2
They
For
state
+2.
example:
SnO(s)

basic
PbO.
all
react
with
alkalis
to
form
oxo
O(l)
2
ions
with
oxidation
state
+2.
example:
2−
GeO(s) +
2OH
(aq)
→
GeO
(aq)
+
H
2
O(l)
2
germanate(ii)
ion
2−
SnO(s)
+
2OH
(aq)
→
SnO
(aq)
+
H
2
O(l)
2
stannate(ii)
ion
2−
PbO(s)
+
2OH
(aq)
→
PbO
(aq)
+
H
2
O(l)
2
plumbate(ii)
ion
Oxidation state +4

GeO
,
SnO
2

They
and
PbO
2
all
are
all
amphoteric.
2
react
with
acids
to
form
a
salt
with
oxidation
state
+4.
For
example:
(s)
SnO
+
4HCl(aq)
→
SnCl
2
SnO
reacts
with
dilute
acid
but
PbO
2
They
For
+
2H
O(l)
2
will
only
react
with
2
concentrated

(aq)
4
all
acid,
react
oxidising
with
alkalis
it
to
to
chlorine
form
oxo
(see
ions
Section
with
12.5).
oxidation
state
+4.
Key ponts
example:

2−
(s) +
GeO
2OH
(aq)
→
GeO
2
(aq)
+ H
3
The
relate
stablty
of Group
O(l)
2
germanate(iv)
IV
ion
oxdes
n
oxdaton
state
+4
2−
SnO
(s)
+
2OH
(aq)
→
SnO
2
(aq)
+ H
3
+
2OH
(aq)
→
PbO
the
group.
ion

2−
(s)
2
down
2
stannate(iv)
PbO
decreases
O(l)
(aq)
3
+ H
The
acdc
character
of
the Group
O(l)
2
IV
plumbate(iv)
oxdes
n
oxdaton
state
+2
ion

and
+4
and
the
Group
state
decreases
basc
IV
+2
character
oxdes
are
n
more
correspondng
state
down
the
group
ncreases.
oxdaton
basc
oxdes
n
than
the
oxdaton
+4.
147
12.5
Relate
stablty
of Group
IV
catons
and
oxdes
Learning outcomes
Stability of
Going
On
completon
of
ths
secton,
be
able
and
descrbe
Group
IV
,
+4 oxidation
the
oxides
the
oxides
of
the
the
relate
stabltes
oxdes
+4
the
+2
oxidation
oxidation
state
state
become
less
become
stable
more
with
to
the
+2
state.
of
oxidation
the
in
states
to:
respect

and
you
stable
should
down
+2
and
aqueous
state
+2
oxidation
state
+4
catons
CO
decreasing
CO
decreasing
2
of
the Group
IV
elements
n
stability
and
SiO
stability
SiO
and
2
oxdaton
states
+2
and
+4
wth
Ge
better
GeO
better
2
Ø
reference
to
E
alues
reducing
SnO
SnO
agent
PbO
PbO
oxidising
2

descrbe
based
some
on
uses
slcon(iv)
of
ceramcs
oxde.
agent
2

Carbon
monoxide,
readily,
e.g.
oxidised
to
at
a
CO,
high
carbon
is
a
good
dioxide
which
O
Fe
2
Oxidation

Carbon
numbers:
dioxide
is
a
reducing
temperature
it
is
(s)
more
It
iron(
loses
iii)
electrons
oxide
to
iron.
It
is
stable.
+ 3CO(g)
→
2Fe(l)
+
3CO
3
(g)
2
+3
poor
agent.
reduces
+2
reducing
agent.
0
It
loses
+4
electrons
with
great
difculty.

Lead( iv)
oxide,
PbO
,
is
a
good
oxidising
agent.
It
accepts
electrons
2
readily,
e.g.
it
oxidises
hydrogen
chloride
to
chlorine:
(s) + 4HCl(aq) → PbCl
PbO
2
Oxidation

Lead( ii)
oxide
is
a
–1
poorer
(aq) + Cl
2
numbers: +4
oxidising
agent
in
(g) + 2H
2
+2
O(l)
2
0
comparison
with
lead(
iv)
oxide.
Reactions of
some Group(IV)
cations
Ø
In
Section
10.4,
feasibility
of
Group
of
a
IV
we
showed
reaction.
metals
in
We
how
can
E
values
apply
oxidation
can
these
states
+2
be
ideas
and
used
to
the
+4.
See
to
predict
aqueous
Figure
the
cations
12.5.1.
E
+
increasing
–0.30
Ge O
+
4H
2+
+
2e
Ge
+
2H
2
O
2
oxidising
increasing
4+
power
of
species
+0.
15
Sn
2+
+
2e
Sn
reducing
on
the
power
left
of
(better
ions
on
the
at
right
4+
accepting
+1.69
Pb
(better
at
2+
+
2e
releasing
Pb
electrons)
electrons)
Figure 12.5.1
Worked
The reducng and oxdsng powers of some Group IV ons
examples
1
to
3
together
with
Figure
12.5.2
show
how
we
can
Dd you know?
4+
compare
the
ability
of
Pb
4+
ions
and
Sn
ions
to
oxidise
other
species.
Ø
Tables
of
E
alues
often
hae
tn
E
and
lead
ons
wrtten
n
the form
2+
4+
Pb
–0.76
4+
and
Sn
.
Many
tn
and
Zn
(aq)
+
2e
Zn(s)
(aq)
+
2e
Sn(s)
+
2e
Sn
lead
2+
–0.
14
compounds
n
the
+4
Sn
oxdaton
4+
+0.
15
state
are
nsoluble
n
water.
So
Sn
2+
(aq)
3+
hae
to
be
dssoled
n
concentrated
+0.77
Fe
+1.69
Pb
2+
(aq)
+
e
4+
acds
or
alkals
soluton.
to
get
these
ons
nto
Figure 12.5.2
(aq)
Fe
(aq)
2+
+
2e
Pb
(aq)
An electrode potental dagram for determnng the feasblty of some
reactons nolng Group IV catons
148
(aq)
they
Chapter
Worked
example
12
The
chemstry
of Groups
II,
IV
and VII
1
Exam tps
4+
Will
2+
Pb
ions
oxidise
an
aqueous
solution
containing
Fe
ions
to
3+
aqueous
An
ions?
Fe
easy
way
reacton
Ø

According
to
the
to
determne
occurs
or
not
s
f
a
shown
4+
E
values,
Pb
ions
are
better
at
accepting
electrons
4+
3+
than
ions.
Fe
below.
4+
Pb
ions
are
better
oxidising
For
example:
wll
Pb
(aq)
agents.
2+
ons
Ø

According
to
the
than
The
values,
ions.
Pb
Fe
ions
Fe
are
reaction
with
the
more
2+
direction,
ions
better
are
better
reducing
at
releasing

i.e.
(aq)
Fe
negative
→
Fe
Wrte
down
equatons
E
value
goes
in
the
ons?
overall
reaction
is:
the
two
noled
half
makng
that
you
reerse
the
sgn
Pb
2+
(aq)
+
2Fe
2+
(aq)
→
Pb
showng
electron
the
loss
3+
(aq)
+
2Fe
(aq)
ths
case,
equaton
ii).
4+
i
example
of
(aq)
(n
Worked
sure
reverse
3+
4+
The
(aq)
electrons
agents.
equaton

Fe
2+
Ø

wth
2+
E
2+
react
2+
(aq)
Pb
+
2e
Y
Pb
(aq)
2
Ø
E
4+
Will
Sn
=
+1.69V
2+
ions
oxidise
an
aqueous
solution
containing
Fe
ions
2+
to
ii
Fe
3+
(aq)
Y
Fe
(aq)
+
e
3+
aqueous
ions?
Fe
Ø
E
Ø

According
to
the
values,
Fe
ions
are
better
at
accepting
–0.77V
electrons

4+
than
=
3+
E
Add
the
two
Fe
ions
are
better
oxidising
agents.
+1.69
Ø

According
to
the
values,
Sn
ions
are
better
at
releasing
Sn
ions
are
better
reducing

agents.
Ø

The
reaction
with
the
more
negative
2+
direction,

So
no
i.e.
=
+
0.92 V
E
value
goes
in
the
If
the
alue
poste,
reverse
of
the
the
oltage
reacton
wll
s
occur.
4+
(aq)
Sn
reaction
Worked
0.77
electrons
2+
ions.
Fe
–
2+
E
2+
than
oltages
3+
ions.
Sn
→
Sn
4+
(aq)
So
Pb
(aq)
2+
reacts
wth
Fe
(aq).
occurs.
example
3
4+
Will
Zn
reduce
Sn
ions?
Ø

According
to
the
E
values,
Zn
are
better
at
releasing
electrons
than
4+
ions.
Sn
Zn
is
a
better
reducing
agent.
Ø

The
reaction
with
the
more
negative
E
value
goes
in
the
reverse
2+
direction,
i.e.
Zn(s)
→
(aq)
Zn
4+

The
initial
reaction
is:
Zn(s)
+
Sn
2+
(aq)
→
Zn
2+
(aq)
+
Sn
Ø

But
according
to
E
(aq)
2+
values
and
the
anticlockwise
rule,
Sn
can
be
Ø
reduced
further
by
Zn
to
tin
–
(E
0.14 V).
4+

The
overall
Uses of

reaction
ceramics
Fur nace
linings:
is:
2Zn(s)
+
based on
SiO
is
a
good
Sn
2+
(aq)
→
2Zn
(aq)
+
Sn(s)
silicon dioxide
thermal
insulator
and
has
a
very
high
2
melting

point
Abrasives:
due
is
SiO
to
the
hard
many
and
strong
has
a
covalent
high
melting
bonds.
point
(heat
is
given
out
2
on

grinding).
Manufacture
of
glass
and
porcelain:
is
SiO
relatively
unreactive
and
2
is
easily
moulded.
The
high
melting
point
is
useful
for
ovenware.
Key ponts

The
relate
stabltes
of
the
oxdes
and
aqueous
catons
of
the Group
IV
Ø
elements

Down Group
agent

can
wth
Ceramcs
be
IV
explaned
oxdes,
respect
based on
to
the
the
wth
+4
+2
reference
oxdaton
oxdaton
slcon(iv) oxde
are
to
E
state
alues.
becomes
a
better
oxdsng
state.
used for furnace
lnngs
and
abrases.
149
12.6
Group VII:
Learning outcomes
the
Physical
The
On
completon
of
ths
halogens
secton,
elements
be
able
explan
the
physcal
and

aratons
propertes
halogens
n
T
able.
terms
of
of
n
the
They
(molecules
(the
seven
halogens)
electrons
in
are
in
their
the
the
relate
halogens
descrbe
p
outer
block
of
principal
the
quantum
are
all
non-metals
having
two
which
exist
as
diatomic
the
as
reactty
oxdsng
use
of
molecules
atoms).
the
structure
−3
Halogen
State
at
20 °C
Density/g cm
Atomic
Melting
radius/nm
point/°C
0.072
–220
of
fluorne,

VII
have
bondng
explan
the
Group
All
to:
shell.

of
you
Periodic
should
properties
F
gas
1.51
(at
gas
1.53
lqud
3.
12
sold
4.93
85 K)
2
agents
bromne
chlorne, Cl
(at
238 K)
0.099
–101
2
water.
bromne,
Br
(at
293 K)
0.
114
–7
2
odne,
I
(at
293 K)
0.
133
+114
2
Colour
The
and
colour
F
:
solubility of the
gets
yellow;
darker
Cl
2
Cl
reacts
and
Br
2
is
organic
Br
:
reddish-brown;
I
water
.
called
Chlorine
‘chlorine
reacts
water ’
:
grey-black.
2
slightly
and
with
‘bromine
water
.
water ’
Solutions
respectively.
iodide.
in
water
.
This
such
Iodine
solution
as
is
solution
brown.
cyclohexane,
it
is
iodine
When
appears
dissolved
iodine
is
purple.
in
aqueous
dissolved
Iodine
in
vapour
is
purple.
and
atomic
shells
atomic
radius
increases
electrons
Melting
the
van
with
are
solvents
Density
As
yellowish-green;
2
insoluble
potassium
The
group:
2
Iodine
also
the
2
Fluorine
of
:
down
halogens
the
there
inner
points
W
aals
increases
and
halogens
der
points
by
get
intermolecular
is
down
the
between
down
the
forces.
The
boiling
larger
,
group
as
the
shielding
density
of
number
the
increases
outer
down
of
electron
shell
the
group.
points
increasing
the
group
The
the
increased
shells.
and
forces
increase
radius
number
molecules
as
boiling
it
takes
points
of
electrons
stronger
.
more
show
So
energy
a
the
to
similar
makes
the
melting
break
trend,
these
and
F
Cl
2
being
solid.
150
gases
at
room
temperature,
bromine
a
volatile
liquid
and
iodine
2
a
Chapter
Chemical
The
reactivity of the
halogens
halogen
can
solution.
are
less
displace
These
Halogen
are
reactive
a
less
redox
Halide
chemstry
of Groups
II,
IV
and VII
down
the
halogen
group.
from
an
A
more
reactive
aqueous
halide
reactions.
ion
(aq)
chloride, Cl
bromide,
–
Cl
The
halogens
going
reactive
12
turns
Br
(aq)
iodide,
orange
turns
I
(aq)
brown
2
no
Br
reacton
–
turns from
orange
2
to
no
I
reacton
no
reacton
brown
–
2
The
reactions
The
cyclohexane
can
be
can
be
conrmed
layer
by
dissolves
adding
the
cyclohexane
halogen
only,
so
to
any
solutions.
changes
veried.
Aqueous
chlorine
potassium
displaces
bromine
from
an
aqueous
solution
(aq) + 2KBr(aq)
→
2KCl(aq)
+ Br
2
equation:
Iodine
will
iodine
is
numbers:
not
2Br
(aq)
→
2Cl
(aq)
+ Br
not
as
good
–1
bromine
an
(aq)
2
0
displace
(aq)
2
(aq) +
Cl
2
Oxidation
of
bromide:
Cl
Ionic
the
colour
–1
from
oxidising
0
potassium
agent
as
bromide
because
the
bromine.
Ø
We
can
use
oxidising
values
E
agents,
e.g.
to
explain
does
the
chlorine
relative
oxidise
reactivity
iodide
of
ions
halogens
to
as
E
iodine?
/V
I
+0.54
(aq)
+
2e
2I
(aq)
2
Ø

As
E
values
get
more
positive
(less
negative),
the
halogens
on
the
left
+1.07
Br
(aq)
+
2e
2Br
(aq)
+
2e
2Cl
(aq)
2
become
better
oxidising
agents.
They
accept
electrons
more
readily.
+1.36
Ø

As
values
E
become

better
accepts
Cl
get
less
positive
reducing
electrons
(more
agents.
more
negative),
They
readily
release
than
I
halides
electrons
and
2
the
I
ions
on
more
the
(aq)
2
+2.87
readily.
release
Cl
right
(F
+
2e
2F
electrons
Ø
2
Figure 12.6.1
more
readily
than
Cl
)
2
Usng E
alues to predct
ions.
the feasblty of reacton between
Ø

The
reaction
direction,
with
the
more
negative
E
value
goes
in
the
reverse
halogens and haldes
i.e.
(aq)
I
→
I
(aq)
2

So
according
to
the
Cl
anticlockwise
(aq)
+
2I
(aq)
rule,
→
the
2Cl
reaction
(aq)
+
I
2
is
feasible:
Key ponts
(aq)
2

The
use of
Unsaturated
Bromine
When
an
bromine
compounds
water
is
used
water to test for C=C
contain
to
test
for
unsaturated
compound
bromine)
the
changes
bromine)
to
solution
one
or
double
is
more
shaken
colour
C=C
bonds
from
in
with
bonds
double
organic
bromine
orange
(the
bonds.
Halogens

Down
the
compounds.
become
water
darker
colour
of
exst
as
datomc
molecules.
(aqueous
group,
less
n
the
olatle
halogens
and
are
colour.
the

The
halogens
are
good
oxdsng
colourless:
agents,
CH
=CH
2
+
2
Br
→
2
CH
Br – CH
2
Br
decreasng
2
ethene
bromine
1,2-dibromoethane
(colourless)
(orange)
(colourless)
the

Bromne
for
oxdsng
down
water
s
unsaturated
carbon
the
ablty
group.
used
(C=C)
to
test
bonds
n
compounds.
151
12.7
Halogens
Learning outcomes
and
hydrogen
Reaction of
Sodium
On
completon
of
ths
secton,
thiosulphate,
be
able
descrbe
of
iodine
with
O
2
by
,
sodium thiosulphate
can
be
used
to
determine
(see
Section
the
redox
reacton
of
of
bleach.
wth
sodum
Figure
12.7.1
shows
that
descrbe
the
halogens
3.6)
this
explan
thosulphate
reacton
wth
the
stabltes
to
of
reaction
O
S
ions
are
better
reducing
agents
than
analyse
of
I
ions
feasible
and
I
3
2
are
better
oxidising
agents
than
S
O
ions.
6
the
hydrogen
relate
is
2−
molecules
E
/V
2–

and
2−
because
4

the
3
titration
to:
2
odne
S
2
samples

Na
you
concentration
should
halogens
haldes
S
+0.09
thermal
2–
O
(aq)
6
4
+
2e
2S
O
4
(aq)
3
the Group VII
+1.54
I
(aq)
+
2e
2I
2
hydrdes.
Figure 12.7.1
Dd you know?
So
Starch
s
a
complex
contanng
chans
the
reaction
2Na
carbohydrate
of
whch
of
may
glucose
be
helcal form.
Iodne forms
O
2
(aq)
+
I
3
(aq)
→
Na
2
S
2
O
4
a
and
chlorine
oxidation
of
are
stronger
sodium
oxidising
thiosulphate
to
complex
s
a
these
dark
blue-black
O
2
helces. Ths
can
complex
stll
sodum
odne
react
s
so
weak
wth
speces
thosulphate.
has
colourless
been
used
starch
that
So
up,
agents
than
sulphate
iodine
and
cause
ions:
2−
(aq)
+
4Cl
3
(aq)
+
5H
2
O(l)
→
2SO
2
−
(aq)
+
8Cl
+
(aq)
+
10H
(aq)
4
colour.
The
The
2NaI(aq)
weak
2−
wth
+
n
S
complex
(aq)
6
unts,
arranged
further
a
S
2
Bromine
some
is:
estimation of
chlorine
in
bleaches
odne
such
when
only
as
all
the
the
Commercial
commonly
bleaches
called
determined
by
usually
‘sodium
titration.
contain
sodium
hypochlorite’.
The
procedure
The
chlorate(
chlorine
i),
in
NaClO.
bleach
This
can
is
be
is:
remans.

Dilute

Add
the
bleach
excess
by
acidied
a
known
potassium
−
NaClO(aq)
+
with
iodide
distilled
solution
to
water
.
liberate
iodine:
+
(aq)
2I
amount
+
2H
(aq)
→
I
(aq)
+
NaCl(aq)
+
H
2

Titrate
the
liberated
concentration

The
end
point
indicator
halogens.

is
turns
Hydrogen
Hydrogen
using
with
indicator
.
when
the
blue-black
halides
are
formed
reactions
uoride,
HF:
when
are
colour
hydrogen
slower
The
chloride,
down
reaction
(g)
+
F
2
Hydrogen
thiosulphate
of
the
of
a
known
starch−iodine
colourless.
H

sodium
halides
These
Hydrogen
iodine
starch
O(l)
2
(g)
is
→
combines
the
halogen
explosive
even
directly
with
the
group.
in
cool
conditions.
2HF(g)
2
HCl:
The
reaction
is
explosive
in
the
presence
of
sunlight.
(g)
H
+
Cl
2

Hydrogen
bromide,
(g)
→
2HCl(g)
2
HBr:
H
gas
and
Br
2
vapour
react
slowly
on
2
heating.

Hydrogen
iodide,
HI:
gas
H
and
2
closed
container
to
form
H
(g)
2
152
I
vapour
react
slowly
2
an
equilibrium
+
I
(g)
2
Y
mixture:
2HI(g)
on
heating
in
a
Chapter
Hydrogen
uoride
laboratory
halides
HF
are
All
gases
compared
hydrogen
at
with
at
19.5 °C
may
room
the
be
and
either
so
temperature.
other
its
liquid
The
hydrogen
state
or
gas.
under
The
much
halides
is
to
chemstry
of Groups
II,
IV
and VII
normal
other
higher
due
The
hydrogen
boiling
its
point
of
extensive
bonding.
hydrogen
water
boils
conditions
12
halides
are
acidic
–
they
form
acids
when
they
dissolve
in
e.g.
+
HCl(g)+
O(l) →
H
H
2
O
(aq) +
Cl
(aq)
Cl
(aq)
3
+
or
more
simply:
The thermal
The
thermal
atom
HF
and

HBr
aq
stability of
stability
increases

HCl(g)+
in
HCl
of
the
→
H
hydrogen
hydrogen
(aq)
+
halides
halides
decreases
as
the
halogen
size.
are
not
decomposes
decomposed
slightly
at
at
about
temperatures
450 °C
to
form
of
1500 °C
hydrogen
and
bromine.

HI
decomposes
rapidly
at
about
2HI(g)
Y
450 °C
(g)
H
to
+
2

When
a
purple
hot
grey-black
from
a
platinum
vapour
gas
of
solid
or
wire
iodine
on
vice
the
is
is
side
versa
of
in
seen,
the
omitting
hydrogen
and
iodine:
(g)
2
placed
rst
I
form
a
tube
which
tube.
the
of
hydrogen
then
The
liquid
turns
direct
phase
iodide
directly
formation
is
a
to
of
a
a
solid
called
sublimation
The
the
ease
bond
of
thermal
energies
of
decomposition
the
of
the
hydrogen
hydrogen−halogen
halides
is
related
to
bond:
−1
Bond
The

bond
The
energy
larger
hydrogen

So
down
nuclei

So
the
the
and
going
halogen
H–F
562
431
H–Br
366
H–I
299
halogen
halogen
group
the
down
energy/kJ mol
H–Cl
decreases
and
Bond
this
atom,
way
the
because:
greater
is
the
distance
between
the
nuclei.
there
bonding
the
in
is
a
smaller
attractive
force
between
the
electrons.
group,
it
takes
less
energy
to
break
the
carbon–
bond.
Key ponts

The
reacton
of
odne
wth
sodum
thosulphate
can
be
explaned
by
Ø
reference
to
E

The
halogens

The
relate
group
as
alues.
react
wth
thermal
the
hydrogen
stabltes
of
hydrogen–halogen
to form
the
bond
hydrogen
hydrogen
strength
haldes.
haldes
decrease
down
the
decreases.
153
12.8
Reactons
Learning outcomes
of
halde
The
reaction of
When
On
completon
of
ths
secton,
silver
nitrate
be
able
descrbe
ons
the
wth
reactons
aqueous
of
sler
halde
soluble

descrbe
by
aqueous
the
ions
in
wth
coloured
cannot
aqueous
to
an
aqueous
silver
nitrate
precipitates
of
solution
silver
of
Cl
halides
,
are
Br
or
I
ions,
formed.
be
tested
solution
for
and
in
no
this
way
because
precipitate
is
silver
uoride
is
formed.
procedure
is:
ammona
reacton
of
halde

Add
dilute
soluble
ons
added
with
ntrate
The
followed
is
ions
to:
Fluoride

halide
you
characteristically
should
ons
concentrated
nitric
acid
to
contaminating
the
solution
carbonates
under
or
test.
This
hydroxides
removes
which
may
any
form
sulphurc
insoluble
silver
compounds.
acd

descrbe
wth
the
cold
solutons
reacton
and
of
hot
of
Add
a
few

Record

Add

If
drops
of
aqueous
silver
nitrate
until
a
precipitate
is
seen.
chlorne
aqueous
sodum

hydroxde.
the
colour
dilute
the
of
aqueous
precipitate
the
precipitate.
ammonia
does
not
to
see
if
redissolve,
the
add
precipitate
redissolves.
concentrated
aqueous
ammonia.
Halide
ion
Colour
of
precptate
Conrmatory
test
Cl
Br
I
whte
cream
pale
dssoles
dssoles
n
concentrated
concentrated
NH
NH
dlute
(aq)
NH
3
If
left
exposed
to
The
silver
chloride

The
silver
bromide

The
silver
iodide
each
case,
nsoluble
(aq)
3
n
(aq)
3
sunlight:

In
n
yellow
the
precipitate
precipitate
precipitate
equations
AgNO
for
(aq)
+
turns
turns
does
the
grey
grey
not
rapidly.
very
turn
grey.
precipitation
KBr(aq)
→
slowly.
reaction
AgBr(s) +
are
KNO
3
similar
e.g.
(aq)
3
+
Ionic
The
ion
equation:
precipitate
is
formed
(aq)
Ag
of
AgCl
(See
+
Br
dissolves
Section
(aq)
in
→
AgBr(s)
excess
ammonia
because
a
complex
14.3).
Exam tps
Reaction
Remember
between
nole
the
haldes
the
aqueous
that
sold
redox
and
sulphurc
haldes
solutons).
with
concentrated
(not
acd
the
The
strength
>
Br
ion,
is
Cl
the
less
of
with
easier
force
of
halide
iodide
it
is
to
ions
H
lose
H
SO
2
.
ions
An
are
not
acid−base
reducing
the
an
best
and
the
solid
agents
reducing
electron
between
SO
2
Chloride
as
being
attraction
Concentrated
acid
from
follows
agent.
their
nucleus
outer
and
the
The
the
pattern
larger
shell
outer
I
the
>
halide
because
there
electrons.
NaCl
4
strong
enough
reaction
reducing
agents
to
reduce
occurs:
4
NaCl(s)
+
H
SO
2
154
sulphuric
reactons
(l)
4
→
NaHSO
(s)
4
+
HCl(g)
the
S
in
Chapter
Concentrated
H
SO
2
When
bromide
and
solid
ions
react
NaBr(s)
with
H
+
H
SO
2
agents
reaction
to
occurs
reduce
state
S
in
chemstry
of Groups
II,
IV
and VII
NaBr
SO
the
is
rst
(l)
→
NaHSO
4
because
from
there
the
an
acid–base
sulphur
reaction:
4
(s)
+
HBr(g)
4
bromide
+6
ions
are
oxidation
strong
state
in
enough
H
SO
2
oxidation
The
4
2
Further
12
dioxide,
to
reducing
the
+4
4
SO
2
2HBr(g)
+
H
SO
2
OxNo:
−1
Concentrated
H
SO
2
Hydrogen
ones
iodide
above.
They
are
strong
oxidation
is
Iodide
S.
A
state
rst
ions
(g)
+
Br
2
solid
formed
in
better
(l)
+
2H
2
O(l)
2
0
NaI
to
the
an
acid–base
reducing
reducing
SO
H
of
SO
+4
and
are
enough
in
mixture
→
4
2
H
(l)
4
+6
agents
−2
reaction
agents
to
reduce
oxidation
than
S
state
similar
from
in
to
bromide
the
the
hydrogen
the
ions.
+6
sulphide,
4
products
may
be
formed:
2
2HI(g)
+
H
SO
2
6HI(g)
+
(l)
2
OxNo:
−1
+
H
The
cold
dilute
+
SO
I
(l)
+
2H
2
+
3I
(l)
→
H
4
+
4H
O(l)
2
0
S(g)
+
4I
2
(s)
+
4H
2
−2
Sodium
O(l)
2
(s)
2
chlorine
alkali:
(g)
S(s)
0
+6
reaction of
With
→
4
2
−1
SO
2
(l)
+6
8HI(g)
OxNo:
→
4
SO
H
with
O(l)
2
0
alkalis
chloride
and
sodium
chlorate(
i)
are
formed:
Cl
(aq)
+
2NaOH(aq)
→
NaCl(aq)
+
NaClO(aq)
+
H
2
OxNo:
With
are
hot
−1
concentrated
alkali:
Sodium
+1
chloride
and
sodium
chlorate(v)
formed:
(aq)
3Cl
+
6NaOH(aq)
→
5NaCl(aq)
+
NaClO
2
OxNo:
Both
a
O(l)
2
0
−1
these
single
reactions
substance
reduced
to
a
reduced
to
Cl
Oxidation
(aq)
+
3H
3
0
are
disproportionation
becomes
lower
oxidised
oxidation
ions
number
and
has
state).
oxidised
been
to
In
to
a
reactions
higher
these
either
abbreviated
O(l)
2
+5
examples
chlorate( i)
to
(reactions
oxidation
OxNo
in
state
chlorine
or
in
has
chlorate( v)
the
above
which
and
been
ions.
equations.
Key ponts

Haldes
react
precptates
wth
whch
aqueous
dssole
sler
to
ntrate
dfferent
to form
extents
characterstcally
n
aqueous
coloured
ammona.
−

Halde
as

ons
are
exempled
Chlorne
and
ncreasngly
by
reacts
chlorate(v)
Chlorde
ons
ther
wth
ons
good
ablty
cold
and
to
reducng
reduce
hot
are formed
n
both
these
n
the
concentrated
aqueous
respectely. These
agents
solutons
are
order Cl
sulphurc
to form
−
<
Br
−
<
I
acd.
chlorate( i)
dsproportonaton
reactons.
reactons.
155
Reson
1
Whch
of
of
the followng
decreasng
best
solublty
of
questons
descrbes
the Group
the
II
order
5
metal
Anhydrous
bonded
hydroxdes?
Whch
A
Ba > Sr > Ca > Mg > Be
B
Ba > Sr > Ca > Be > Mg
ths
alumnum
chlorde
s
a
coalently
compound.
of
the followng
statements
correctly
explans
statement?
3+
A
The
radus
of
B
The
charge
on
the Al
s
large.
3+
C
Sr > Ca > Ba > Mg > Be
densty
of
the Al
on
dstorts
the
–
electron
D
charge
of
the Cl
on.
Be > Mg > Ca > Sr > Ba
3+
C
2
Whch
of
the followng
statements
relatng
to
The
the
of
the
+2
and
+4
oxdaton
states
of
dfference
between Al
ons
s ery
hgh.
the
–
D
Group
IV
elements
A
Tn(ii)
B
Lead(iv)
C
The
s
are
readly
oxdsed
to
are
easly oxdsed to
the
+2
oxdaton
state
queston
a
Explan
Germanum
the
As
the
exhbts
elements
on.
after
readng
pages
162–7
terms
to
‘monodentate’
lgands,
and
ge
an
and
‘bdentate’
example
of
each
character.
sem-conductor
n
Perod
Classficaton
s
the followng
statements
3
of
the
traersed from
A
Electronegatty
B
Meltng
C
Atomc
D
Screenng
ponts
s
Perodc
left
to
b
rght,
whch
i
Ge
of
use
can
of
an
arrow,
the
ste from
whch
of
ii
steadly.
4
occur.
the
metal
true?
decreases.
rse
by
propertes.
bondng
3
6
appled
showng,
D
the Al
ncreases
as
metallc
polarses
lead(ii).
6
wth
effectely
tn(iv).
Answer
compounds
of
3+
on
The Cl
correct?
compounds
stablty
and
–
Cl
stablty
electronegatty
name
atom
and
6
of
a
complex
exhbts
a
on
n
whch
co-ordnaton
the
number
respectely.
Sketch
the
shape
of
the
spatal
arrangement
of
ons
n
i
showng
the
bonds.
–
rad
c
decrease.
of alence
electrons
The
the
ncreases.
chlorde
blue
green
4
The
relate
HX,
s
acd
strength
represented
by
HF
Whch
of
<
of
the
hydrogen
the followng
HCl
the followng
<
HBr
<
ths
The thermal
s
hexaaquacopper(ii)
i
Wrte
an
dsplace
on
to
water
produce
n
the
on.
equaton for
ii
State
the
atom
n
ths
dsplacement.
co-ordnaton
number
of
the
copper
HI
statements
prodes
the
each
on.
best
Suggest
a
reason for
the
dfference
n
order?
stablty of
HX
ncreases
as the
group
number
n
the
tetrachloro-
complex.
descended.
Hydrogen
odde
has
the
largest
standard
a
Suggest
dssocaton
bolng
rse
n
the
and
ge
maxmum
examples
oxdaton
of,
the
states
enthalpy.
across
The
reasons for,
bond
steady
C
to
haldes,
7
B
able
tetrachlorocuprate(ii)
co-ordnaton
A
s
order:
iii
explanaton for
,
on, Cl
pont
of
HX
ncreases
as
the
group
the
Perod
3
elements.
s
b
The
pH
of
aqueous
solutons
of
the
chlordes
ascended.
of
D
Hydrogen
bondng
decreases
as
the
group
the first
three
elements
of
Perod
3
s
gen
s
below:
descended.
Element
Na
Mg
Al
pH
7
7
5
Use
a
knowledge
explan
c
the
Descrbe
the
elements
water,
156
of
aboe
ease
sodum,
gng
structure
and
bondng
to
nformaton.
of
reacton
alumnum
releant
between
and
equatons.
the
chlorne
wth
Chapters
8
a
The
table
elements
below
shows
n Group
Element
Na
3,
the
denstes
sodum
Mg
Al
to
of
the
10
a
chlorne:
S
P
The
table
below
temperatures
S
Cl
of
Carbonate
11–
12
ndcates
Reson
the
decomposton
the four Group
MgCO
1.0
1.7
2.7
2.3
1.8
2.
1
<
0.
1
Temperature/K
(ii)
CaCO
3
Density/
questons
carbonates:
SrCO
3
813
BaCO
3
1173
3
1553
1633
–3
g cm
Carefully
Draw
a
graph
to
ndcate
the araton
of
b
across
Perod
3,
elements
sodum
to
explan
the
trend
shown.
densty
i
State
the
trend
of
solublty
of
the Group
II
chlorne.
sulphates.
b
Refer
to
ther
structure
and
bondng
to
explan
ii
Use
nformaton
proded
by
the
releant
the:
enthalpy
i
steady
rse
of
densty
between
sodum
changes
to
explan
ths
trend.
and
c
Lst
one
commercal
use for
each
of
the followng:
alumnum
ii
large
decrease
n
densty
between
sulphur
i
calcum
oxde
ii
magnesum
iii
calcum
and
oxde
chlorne.
9
State
reasons
wheneer
a
the
and
ge
an
possble, for
tetrahaldes,
example,
the
MCl
,
usng
obseratons
of
the Group
an
n
IV
equaton
d
(a)–(c):
are olatle,
ii
ncrease
group
b
Carbon
than
s
n
non-polar
ther
elements:
i
wth
of
water
as
the
descended.
monoxde, CO,
carbon
s
a
better
reducng
carbonate
knowledge
the
wth
ii
the
solublty
the
ease
and
ts
of
II
chemstry
propertes
water,
resultng
iii
agent
of Group
the followng
reacton
compounds
reacton
your
predct
4
i
Use
and
of
the
to
radum:
subsequent
pH
soluton
of
the
chlorde
decomposton
and
of
carbonate
the
ntrate(v)
products.
doxde, CO
2
11
c
Slcon(iv)
oxde
has
a ery
hgh
meltng
pont
a
Explan
wth
reference
the araton
does
not
conduct
electrcty. Use
structure
to
structure
and
bondng
but
of
meltng
pont
n
the Group VII
and
elements.
bondng
to
ge
an
explanaton for
ths
statement
b
about
the
i
State
be
d
Copy
and
what
complete
the followng
table
wth
i
the
acd
added
ons formed
and
ii
see
f
bromne
were
to
to
when
the
solutons
oxdes
chlorde
of
potassum
odde
and
respectely.
react
ii
Name
iii
How
the
type
of
reacton
noled.
alkal.
would
respect
Oxide
would
by
potassum
nsertng
you
oxde.
reactions
i
wth
acd
ii
wth
to
you
your
explan
answer
your
to
obseratons
part
n
ii?
alkal
c
A
black
heated
powder,
wth
P,
conc.
contanng
HCl
manganese
eoled
a
greensh
when
gas
,Q,
Germanum(ii)
whch
dssoled
n
water. A
soluton
of
ths
gas, S,
oxde
turned
Tn(ii)
oxde
S
when
oer
Lead(iv)
ltmus
paper
allowed
water
n
a
to
red
before
stand
beaker
n
an
beng
bleached.
nerted
produced
test
bubbles
of
tube
a
gas,
oxde
T,
on
exposure
i
Suggest
ii
Name
iii
Wrte
a
to
sunlght.
name
the
and formula for
P.
gas Q.
an
equaton
to
explan
an
equaton for
the
producton
of Q.
iv
Wrte
the
reacton
producng
soluton S.
v
Deduce
the
the
oxdaton
anons found
vii
Name
viii
Wrte
the
an
type
states
of
the
halogen
n
n S.
of
reacton
equaton for
the
whch
produced S.
producton
of
the
gas T.
157
13
Transition
13.
1
An
elements
introduction
Learning outcomes
Electronic
A
On
completon
of
ths
secton,
transition
be
able
descrbe
element
is
a
d
block
element
which
ions
with
an
incomplete
d
electron
of
the rst
elements
and
of
the
rst
row
transition
elements
Element
Electronc
ttanum, T
1s
the
oxdaton
states
anadum, V
1s
chromum, Cr
1s
manganese,
1s
2
2s
descrbe
row
the
transton
more
are
shown
below:
transton
configuraton
6
2p
2
3s
6
3p
2
3d
2
4s
of
2

or
electron
row
2
the rst
one
The
ther
ons
descrbe
forms
sub-shell.
electronc
conguraton

elements
to:
the
transton
elements
configuration of transition
congurations

transition
you
stable
should
to
2
2s
6
2p
2
3s
6
3p
3
3d
2
4s
elements
characterstcs
2
of
2
2s
6
2p
2
3s
6
3p
5
3d
1
4s
elements.
2
Mn
2
2s
2
ron,
Fe
1s
2
2s
2
cobalt, Co
1s
nckel,
1s
1s
6
6
2
2s
6
2p
6
7
2
4s
8
3d
6
3p
2
4s
3d
6
2
3s
6
6
3p
2
4s
3d
3p
2
3s
5
3d
3p
2
3s
2p
6
3p
2
3s
2p
2
2
copper, Cu
6
2
2s
2
3s
2p
2s
2
N
6
2p
2
4s
10
3d
1
4s
Did you know?
Scandum,
Sc,
and
znc,
Zn,
are
not
transton
elements
although
they
are
d
3+
block
elements.
Sc forms
only
one
on,
Sc
,
wth
no
electrons
n
ts
d
sub-shell.
2+
Zn forms
The
only
one
electronic
pattern
of
on,
Zn
,
and
congurations
lling
the
d
has
of
sub-shell.
a
Cr
complete
and
For
Cu
Cr
,
with
one
having
electron
one
of
the
in
d
each
orbital,
orbitals
sub-shell.
not
follow
the
arrangement
of
,
gives
lled.
For
a
greater
Cu,
the
d
sub-shell
greater
stability
electron
When
with
in
than
which
lost
atom:
1s
2
2s
2
V
atom:
T
ransition
one
1s
2s
stability
in
of
each
the
d
orbital,
orbitals
1
4s
[Ar]3d
with
than
only
a
,
gives
a
single
2+
:
can
2
1s
2s
2
2p
6
3p
2
3s
form
3s
2
3d
6
3p
6
electrons
to
form
ions,
it
is
the
4s
e.g.
2
3s
6
2p
lose
rst,
6
2p
2
elements
Fe
158
having
elements
are
2
Ti
electrons
sub-shell
it.
transition
electrons
paired
d
arrangement
10
a
expected
a
1
4s
[Ar]3d
completely
do
the
5
3d
2
4s
3
3d
4s
more
2
3p
6
3d
2+
Ti
2
2
ion:
1s
3+
V
than
ion:
one
6
:
1s
type
3+
Fe
2
2s
2
2
1s
2s
of
6
2p
2
2s
2p
ion,
2
2p
6
3s
2
3s
6
6
3p
2
3s
3p
e.g.
2
3p
6
3d
5
2
3d
6
3d
2
of
Chapter
The
range of oxidation
13
Transton
elements
Ti
states
V
T
ransition
elements
may
have
several
different
oxidation
states.
Cr

The
most
common

The
maximum
oxidation
state
of
transition
elements
is
+2.
Mn
the
oxidation
involves
all
4s
From
onwards,
and
state
3d
of
the
transition
elements
up
to
Mn
Fe
electrons.
Co

Fe
electrons
become
the
+2
oxidation
increasingly
state
harder
to
dominates
remove
as
because
the
the
nuclear
3d
Ni
charge
Cu
increases.
Figure 13.1.1

Higher
oxidation
states
of
transition
elements
are
found
in
complex
The range of oxidation
states in transition element compounds in
2−
ions
(see
Section
13.3)
or
compound
ions
such
as
and
MnO
CrO
4
Period 4. All oxidation states are positive.
4
The commonest oxidation states are
ringed.
Characteristics of transition
T
ransition
properties
elements
which
have
typical
them
apart
They
T
ransition
element
ions
form
coloured

T
ransition
element
ions
form
complex

T
ransition
elements
often
related
with
properties
other

is
compounds
metallic
from

(this
form
set
elements
and
to
different
their
the
oxidation
in
are
elements
have
very
high
density.

T
ransition
elements
have
very
high
melting

They
Most
when
they
compounds
placed
do
not
in
a
magnetic
and
ions
of
magnetic
retain
their
the
eld,
transition
they
magnetism
Co
after
and
the
align
when
S
points
good
catalysts
states).
and
boiling
points.
elements
are
themselves
the
paramagnetic :
with
magnetic
eld
the
is
eld.
But
removed.
S
N
transition
Fe,
often
properties:
N
Figure 13.1.2
states.
oxidation
T
ransition
typical
some
ions.

have
have
compounds.
compounds
differences
but
metals:
element
A small rod of a transition element aligns itself in a magnetic eld
Ni
are
magnetic
fer romagnetic .
eld
has
been
They
retain
permanent
magnetism
withdrawn.
Key points

Transton
of

When
the

elements form
one
or
more
ons
wth
an
ncomplete
d
sub-shell
electrons.
4s
a
transton
sub-shell
Transton
and
elements
compounds
magnetc
element forms
then from
exst
(contanng
propertes
n
the
on,
3d
arous
complex
and
an
electrons
are rst
lost from
sub-shell.
oxdaton
ons),
catalytc
the
and
states, form
often
hae
coloured
characterstc
actty.
159
13.2
Physical
Learning outcomes
On
completon
should
be
able
of
ths
properties
Ionisation
secton,
In
Periods
in
the
explan
the
changes
radus
of
the
n
energy,
and
values
3
of
atomic
the
Periodic
and
T
able
elements
ionic
there
is
radii
of
the
rst
ionisation
energy
and
a
considerable
the
atomic
and
variation
ionic
radii.
to:
relately
atomc
and rst
small
and
values
from
elements,
titanium
to
however
,
show
only
small
changes
in
these
copper
.
onc
onsaton
transton
transition
energy
elements
Element
T
V
Cr
Mn
Fe
Co
N
Cu
661
648
653
716
762
757
736
745
0.
132
0.
122
0.
117
0.
117
0.
116
0.
116
0.
115
0.
117
0.090
0.090
0.085
0.080
0.076
0.078
0.078
0.069
across
Ø
a
transition
you
The

2
of
–1
/kJ mol
∆H
perod
1

descrbe
qualtately
physcal
propertes
the
Atomc
of
transton
radus/nm
elements
compared
wth
calcum.
Ionc
2+
radus
of
ons/nm
Ø
Across
this
series
of
transition
elements,
the
rst
ionisation
energy,
∆H
,
i1
generally
increases,
successive
increase
the
case
of
d
going
the
sub-shell.
one
ionic
compared
increased
the
element
forces
and
to
the
of
the
caused
the
d
is
the
block
elements
between
of
block
Period
them.
transition
s
elements
by
the
4
The
are
elements,
in
metals,
iron
and
s
on
the
p
II
are
some
is
an
of
which
in
block
by
the
increased
going
sub-shell,
ionisation
the
in
atomic
electrons
The
from
electrons.
is
placed
and
marked
properties
of
the
just
before
transition
differences
calcium
with
nickel.
ron
nckel
839
1540
1450
Densty/g cm
1.55
7
.86
8.90
Atomc
0.
197
0.
116
0.
115
0.099
0.076
0.078
590
762
736
pont/°C
–3
Ionc
radus/nm
radus/nm
–1
Frst
160
onsaton
energy/kJ mol
one
increased
calcum
Meltng
the
energy
slightly
elements.
the
3d
should
added
calcium
calcium
some
d
only
the
with
a
each
electron
Similarly,
outer
cancelled
Group
to
but
and
repulsion
there
there
of
This
since
additional
series,
Although
compares
nucleus.
difference
elements
4.
But
smaller
.
the
almost
extra
Period
table
the
Atoms
markedly
electrons
much
in
their
the
the
nucleus
is
element
in
make
along
the
by
adding
next
next
Comparing transition
Calcium
to
in
sub-shell.
differences
effect
to
is
amount.
electrons
caused
of
increase
large
attractive
transition
repulsive
with
effect
small
proton
outer
same
elements,
tend
very
more
the
the
a
electrons
The
element
radii
by
one
of
into
outer
transition
between
and
on
only
have
attraction
are
repulsion
into
element
the
electrons
but
two
Chapter
The
transition
calcium.
This
elements
s
compared
electrons
elements
is
a
to
with
The
form
can
greater
charged)
force
atomic
from
the
greater
which
of
and
the
in
ionic
lling
the
d
radii
for
of
of
attraction
to
pull
the
ionisation
because
and
the
this
element
The
electronegativity
with
slightly
less
Electrical
table
can
4s
of
but
3d
(and
points
elements
than
transition
release
and
small
sea
in
only
its
two
outer
transition
sub-levels.
often
electrons
of
There
highly
in
comparison
are
all
compares
the
8
the
in
outer
especially
better
at
to
shielding
So
the
is
and
outer
is
the
the
related
to
the
elements.
the
there
a
electrons
relatively
outer
electrons
nucleus.
elements
greater
than
This
transition
nucleus
transition
have
the
than
force
are
of
calcium.
smaller
transition
Ca
=
across
higher
attraction
than
that
between
1.0,
the
radii
This
than
elements
Fe
and
series
Ni
from
is
=
Ti
is
due
to
the
fact
calcium.
signicantly
1.8.
to
higher
The
Cu
as
the
elements
character
.
electrical
conductivity
elements
of
calcium
with
some
of
good
conductors
electrical
Ca
T
Mn
Fe
N
Cu
0.30
0.024
0.007
0.
10
0.
15
0.63
–1
S m
transition
single
the
closer
larger
elements.
Conductty/10
of
the
electrons.
denser
e.g.
Element
a
in
good
p
much
elements.
electrons.
increases
metallic
are
conductivity of transition
transition
Most
of
ions
calcium,
electronegativity
less
or
relatively
outer
elements
compared
s
between
energies
of
transition
good
melting
electrons’
the
sub-shell
electrons
transition
are
and
higher
bonding
the
the
transition
are
than
that
with
of
both
calcium
d
sub-shell
charge
force
rst
of
the
The
the
‘sea
from
ions
have
Calcium
between
typical
tends
nucleus
The
and
metallic
delocalised
electrons
element
radii
nuclear
calcium
get
calcium.
attraction
transition
Electrons
the
harder
stronger
Transton
calcium.
presence
of
the
of
are
the
with
release
corresponding
The
elements
reects
13
s
elements
electron
conductors.
than
are
and
a
So
it
is
too
and
conductors
lled
copper
manganese
conductor
,
good
half
or
and
to
electricity.
chromium
nickel.
reactive
of
completely
Although
be
used
in
full
are
d
Those
shell
relatively
calcium
is
electrical
a
wiring.
Key points

Across
the rst
onc
radus
each
electron
row
of
and rst
added
transton
onsaton
goes
nto
elements,
energy
a
d
are
the
changes
relately
sub-shell
and
so
n
atomc
small. Ths
the
s
sheldng
and
because
effect
s
mnmsed.

The
s
physcal
block
propertes
element
of
because
transton
of
the
elements
presence
of
dffer from
the
d
those
of
a
typcal
sub-shell.
161
13.3
Coloured
Learning outcomes
ions
and
The various oxidation
T
ransition
On
completon
of
ths
secton,
be
able
descrbe
can
states of vanadium
is
one
such
form
ions
element,
in
the
arous
oxdaton
solid
exhibit
characteristic
ammonium
colours.
vanadate,
of
than
ions
in
one
their
V
anadium
VO
NH
4
states
more
whose
oxidation
various
state.
oxidation
to:
states

elements
states
you
V
anadium
should
oxidation
.
When
( v)
is
usually
ammonium
supplied
vanadate
as
is
3
+
anadum
acidied
the
vanadium
becomes
part
of
a
positive
,
ion,VO
in
which
2

explan
the formaton
ons
transton
by
of
coloured
vanadium
has
an
oxidation
state
of
+5.
elements.
+
+5
↑
yellow
VO
2
2+

+4
VO
+3
V
+2
V
0
V
blue
3+
oxidation
state
green
2+


The
redox
potential
chemistry
of
vanadium
can
purple/mauve
be
explained
using
an
electrode
chart.
Ø
E
/v
2+
–0.76
–
Zn
+
2e
Zn
(aq)
(s)
3+
2
V
–0.26
+
e
V
(aq)
(aq)
2+
–0.
14
–
Sn
+
2e
Sn
(aq)
(s)
2+
+
VO
+0.34
+
–
2H
+
(aq)
3
e
V
+
e
+
(aq)
(l)
Fe
(aq)
+
VO
+
2
e
(aq)
+
Figure 13.3.1
O
2
2
Fe
+1.00
H
(aq)
3+
+0.77
+
(aq)
–
2H
(aq)
+
2
e
VO
+
(aq)
H
(aq)
O
2
(l)
An electrode potential chart showing the oxidation states of vanadium and
other compounds
Ø
By
reference
to
vanadium( v)
E
values,
ions
to
a
we
lower
can
2+

Addition
of
Fe
select
oxidation
suitable
compounds
+
ions
to
VO
will
reduce
the
VO
2+
ions
to
VO
2
2+
ions
Fe
reduce
+
ions
2
2+
ions.
to
state:
are
better
reducing
agents
than
VO
+
ions
and
VO
2
3+
ions
are
better
oxidants
than
2+
+
ions.
Fe
So
the
reaction
VO
→
VO
2
Ø
goes
in
the
forward
direction
and
the
reaction
2+
value
will
goes
be
in
from
the
reverse
yellow
to
direction
with
Fe
).
Addition
of
Sn
to
The
+
ions
VO
will
reduce
the
ions
to
V
ions.
2
+
the
E
change
3+
VO
2
Using
higher
colour
blue.
+

the
3+
→
(Fe
anticlockwise
rule,
the
reaction
2+
→
VO
VO
3+
→
V
goes
2
Ø
in
the
forward
direction
and
the
reaction
direction
(Sn
→
with
the
higher
value
E
2+
goes
in
from
the
reverse
yellow
to
blue
to
+
+
(aq)
2VO
The
colour
change
will
be
green.
+
Sn(s)
).
Sn
+ 4H
2+
(aq)
→
2VO
2+
(aq)
+ Sn
(aq) + 2H
2
then:
Sn(s)
+
2VO
O(l)
2
2+
+
(aq)
+ 4H
3+
(aq)
→
2V
2+
(aq)
+ Sn
(aq) + 2H
O(l)
2
+

Addition
of
Zn
to
VO
+
ions
will
reduce
the
VO
2
2+
ions
Using
the
anticlockwise
to
V
ions.
2
2+
+
rule,
the
reaction
→
VO
VO
3+
→
V
2+
→
V
2
Ø
goes
in
the
forward
direction
and
the
reaction
with
the
higher
E
2+
value
will
162
goes
be
in
from
the
reverse
yellow
to
direction
blue
to
(Zn
green
to
→
Zn
purple.
).
The
colour
change
Chapter
The formation of
Compounds
appear
coloured
coloured
when
13
Transton
elements
ions
they
absorb
energy
that
corresponds
3+
to
certain
appear
wavelengths
purple
spectrum.
and
red
because
Most
of
regions.
of
they
the
So
light
the
absorb
light
the
in
light
which
solution
visible
mostly
passes
appears
a
spectrum.
in
the
through
is
Aqueous
green
from
region
the
ions
Ti
of
violet,
the
blue
purple.
b
noitprosba
violet
blue
red
green
blue
yellow
violet
appears
purple
red
absorbs
400
500
600
700
green
wavelength
nm
3+
Figure 13.3.2
a The absorption spectrum of aqueous Ti
ions; b Only violet, blue and red
light are transmitted through the solution.
What
causes the
colour?
3+
T
ransition
element
number
water
(dative
of
covalent)
molecules
are
ions,
such
molecules.
bond
called
with
as
Ti
Each
the
ligands
,
in
water
solution
molecule
transition
and
the
are
element
resulting
ion
bonded
forms
ion.
is
a
to
a
denite
co - ordinate
These
called
a
water
complex
3+
ion.
The
complex
ion
in
the
−water
Ti
complex
has
the
formula
3+
[Ti(H
O)
2

]
6
The
3
d
orbitals
degenerate

The
presence
orbitals
pushed
of
to
orbitals

When
of
an
of
the
an
the
isolated
They
into
in
a
energy
the
complex
ion.
element
same
affects
Orbitals
levels
than
ion
average
the
described
electrons
close
those
are
as
energy.
to
the
further
in
the
ligands
d
are
away.
The
to
orbital
groups.
moves
light
transition
have
element
higher
two
electron
energy,
all
ligands
transition
slightly
split
higher
in
orbitals .
is
from
a
d
absorbed
orbital
in
the
of
lower
visible
energy
region
of
a
d
the
Key points
spectrum.

The
frequency
of
the
light
absorbed
depends
on
the
energy
difference

between
the
split
d
levels.
Different
ligands
split
the
d
energy
levels
An
acded
soluton
ammonum
different
amounts.
So
different
ligands
may
cause
different
colours
anadate(v)
can
to
react
be
of
by
wth
reducng
agents
absorbed.
to form
anadum
dfferent
a
ons
oxdaton
wth
states
b
Ø
dependng
alues for

A
lgand
on
the
s
a
the
relate
half
E
reactons.
molecule
or
on
Energy difference
E
=
wth
hn
of
a
one
or
electrons
2+
a Degenerate orbitals in a Cu
lone
whch
co-ordnate
transton
Figure 13.3.3
more
bond
metal
par
can form
wth
a
on.
2+
ion; b The ligand (water) in a [Cu(H
O)
2
]
6

Transton
element
compounds
ions causes d orbital splitting
are
often
d-orbtal
coloured
splttng
because
caused
of
by
lgands.
163
13.4
Complex
Learning outcomes
ions
More
A
On
completon
of
ths
secton,
and
about
ligand
is
be
able
explan
the
prncple
of
lgand
know
d
that
depends
lgand
on
donate
the
n
well
exchange
water
,
bonds
from
attracts
stablty
an
are
electrons
and
constants
to
covalent)
exchange

molecule
or
and
ion
complex
with
one
or
ions
to
a
transition
more
element
lone
ion.
pairs
of
Examples
of
electrons
simple
to:
ligands

ligands
exchange
you
available
should
a
ligand
a
in
the
lone
ammonia
are
formed
transition
nuclear
pairs
complex
of
ion
and
the
elements
charge.
So,
electrons
is
chloride
with
formed.
do
the
not
Co - ordinate
element
shield
relatively
strongly
Some
ions.
transition
outer
poorly
enough
examples
the
to
are
(dative
ions
because
electrons
shielded
form
nucleus
co - ordinate
shown
in
the
very
the
bonds
table
equlbrum
below.
reacton
where
compete for
the
lgands
bondng
wth
the
Transton
transton

descrbe
element
some
element
on
Lgand
Formula of
complex
on
examples
of
lgand
2+
2+
water,
Fe
H
O
[Fe(H
2
O)
2
]
6
exchange.
3+
3+
ammona,
Co
NH
[Co(NH
3
2+
3+
NH
]
6
2–
Cu
3
)
3
chlorde
on, Cl
[CuCl
]
4
H
3+
N
3–
Cr
3
hydroxde
on, OH
[Cr(OH)
]
6
+
H
+
ammona,
Ag
NH
Co
N
NH
[Ag(NH
3
3
)
3
]
2
3
2+
2+
damnoethane,
N
[N(NH
CH
2
CH
2
NH
2
)
2
]
3
NH
3
NH
CH
2
CH
2
NH
2
2
NH
3
3+
[Co(NH
)
3
Figure 13.4.1
Co-ordnaton
]
number
6
A complex ion formed from
The
co - ordination
number
of
a
complex
ion
is
the
number
of
co -
a cobalt 3+ ion and six ammonia molecules
ordinate
bonds
a
ligand

Monodentate

Bidentate
forms
ligands
with
form
one
the
central
bond
per
transition
ligand,
e.g.
metal
water
,
ion.
ammonia.
Exam tips
These
Remember
that
the
charge
ligands
ligands
form
have
two
two
bonds
lone
pairs
per
ligand,
available
to
e.g.
diaminoethane.
form
co - ordinate
on
bonds.
a
complex
charges
and
all
on
the
on
s
the
the
sum
transton
of
the
metal
on

Hexadentate
ligands
form
6
bonds
per
ligand,
e.g.
EDT
A.
lgands.
a
b
H
H
C
C
N
N
H
O
H
H
O
H
H
H
Figure 13.4.2
164
Two bidentate ligands: a 1,2-diaminoethane; b benzene-1,2-diol
Chapter
Ligand
If
there
Transton
elements
exchange
is
more
transition
ligands:
13
than
metal
O
H
one
cation.
and
NH
2
transition
.
ligand
For
in
a
solution,
example,
The
better
they
aqueous
the
ligand
can
compete
ammonia
is
at
for
contains
competing
for
a
two
the
3
element
ion,
the
more
stable
is
the
complex
formed.
When
we
2+
add
a
few
drops
of
concentrated
HCl
to
an
aqueous
solution
of
ions
Cu
2+
(which
contains
the
complex
ions
O)
[Cu(H
2
equilibrium
is
set
V
],
following
up:
2+
O)
[Cu(H
the
6
2
]
2–
(aq)
+
4Cl
(aq)
Y
[CuCl
6
]
(aq)
+
6H
4
blue
O(l)
2
yellow-green
2+
Addition
of
HCl
to
the
[Cu(H
O)
2
equilibrium
to
the
right
and
]
complex
shifts
the
position
of
6
so
the
colour
changes
from
blue
to
green
as
2–
the
complex
]
[CuCl
(aq)
is
formed.
Addition
of
water
shifts
the
position
4
of
equilibrium
The
to
equilibrium
constant,
the
left.
constant
for
this
reaction
is
called
the
stability
K
stab
2–
[[CuCl
]
(aq)]
_________________________
4
K
=
stab
2+
[[Cu(H
O)
2
The
larger
complex
Ammonia
ammonia
blue
the
and
value
the
has
a
will
complex
of
the
more
higher
shift
ion
the
is
stability
likely
the
]
4–
]
×
[Cl
(aq)]
6
constant,
complex
stability
constant
position
of
the
will
more
stable
than
Cl
equilibrium
ions.
to
the
+
4NH
+ 2H
3
O(l)
Y
[Cu(NH
2
)
3
(H
4
occur
as
exchanges
complexes
are
similar
with
to
water
,
those
of
a
of
deep
2
]
ions
and
copper
+
4Cl
(aq)
Y
[CoCl
+ 4Cl
(aq)
ammonia.
complexes:
]
(aq)
+
6H
4
O(l)
2
blue
2+
O)
2
(aq)
2–
(aq)
6
pink
[Co(H
]
2
blue
Cl
the
2+
O)
[Co(H
O)
2
deep
ions
ligand
addition
and
2+
(aq)
4
The
the
formed:
yellow-green
Cobalt( ii)
So
right
2–
]
[CuCl
is
form.
]
Key points
2+
(aq)
+ 6NH
6
(aq)
Y
[Co(NH
3
)
3
pink
]
(aq)
+
6H
6
O(l)
2
yellow

Transton
elements form
complexes
Lgand
exchange
n
one
The
red
blood
pigment
haem
is
found
as
a
group
attached
to
the
in
red
blood
cells.
The
molecule
has
an
Fe

ion
or
more
with
number
Lgands
of
6.
Four
of
the
co - ordination
positions
are
are
is
with
a
position
atoms
in
a
nitrogen
is
with
complex
atom
an
in
oxygen
ring
the
system.
protein
molecule.
A
fth
molecule.
The
co - ordination
The
oxygen
sixth
molecule
and
carried
to
the
cells
for
respiration.
Carbon
the
element
co-ordnate
on
by
one
or
bonds.

Lgand
exchange
depends
on
weakly
the
bound
to
position
coordination
is
bonded
to
more
nitrogen
wth
a
transton
co - ordination
combnng
lgands.
protein
2+
haemoglobin
by
haem
monoxide
will
stablty
constants
n
an
also
equlbrum
reacton
where
the
2+
bind
to
the
ion.
Fe
It
has
a
stability
constant
about
200
times
higher
lgands
than
O
.
In
the
presence
of
carbon
monoxide,
hardly
any
oxygen
will
the
respiration
is
inhibited.
The
result
is
often
death,
even
if
of
carbon
monoxide
is
fairly
transton
element
wth
on.
the

concentration
bondng
bind
2
and
compete for
A
colour
change
s
often
low.
obsered

when
lgands
are
lgands
n
Examples
a
one
or
other
complex.
of
lgand
exchange
2+
nclude Cu
more
exchanged for
2+
(aq)
or Co
(aq)
wth
2+
NH
and CO/O
3
complex
wth
the
Fe
2
n
haem.
165
13.5
Complex
Learning outcomes
ions
The
The
On
completon
should

be
able
deduce
ths
secton,
shapes
planar,
of
complexes
tetrahedral
redox
shapes of
shape
of
a
reactions
complex
complex
ion
ions
depends
on:
you
to:
the
(square
of
and

its
co - ordination

the
type
of
number
ligand
which
bonds
to
the
transition
element
ion.
or
The
shape
of
the
complex
cannot
be
predicted
using
the
VSEPR
theory
octahedral)
because
their
3+

descrbe
the
use
of
Fe
MnO
/Mn
and Cr
as
O
2
redox
inuence
on
in
d
orbitals
differ
from
those
in
s
and
p
orbitals
in
structure.
,
3+
2–
4
electrons
2+
/Fe
2+
the
/Cr
The
most
common
co - ordination
numbers
are
2,
4
and
6.
7
systems.
Co-ordnaton
Complexes
number
involving
gold,
2
silver
or
copper
in
oxidation
state
+1
usually
+
have
a
linear
structure,
e.g.
)
[Ag(NH
3
]
,
CuCl
2
2
+
[H
N
→
Ag
←
NH
3
Co-ordnaton
These
shape
may
is
be
more
number
either
4
square
common.
It
]
3
planar
is
or
found
tetrahedral.
in
many
The
square
ii),
nickel(
planar
copper( ii)
and
2–
platinum( ii)
complexes.
The
tetrahedral
form
is
found
in
the
[CoCl
]
4
2–
ion
and
the
molecule.
Ni(CO)
The
reason
why
the
[CoCl
4
coordination
chloride
ion
transition
is
of
4
relatively
element
rather
large.
than
So
6
fewer
(see
Figure
ligands
13.4.1)
ion
can
t
is
has
round
that
a
the
the
central
ion.
2
a
]
4
number
2
b
N
Cl
N
C
C
Cl
Ni
Co
Cl
C
C
N
CO
O
2
Cl
N
C
2
Cu
2–
Figure 13.5.1
a [Ni(CN)
]
2–
has a square planar shape; b [CoCl
4
CO
O
2
A square planar complex of
copper with 2-hydroxybenzoate ions
Square
Each
planar
ligand
complexes
forms
Co-ordnaton
This
Did you know?
is
the
Examples
sometmes
use
two
can
also
coordinate
number
commonest
be
formed
bonds
to
excreted
n
EDTA
transton
posonng. The
the
complex
urne.
ligands.
element
ion.
type
of
coordination.
It
is
always
octahedral.
O)
]
,
3–
[Fe(CN)
6
]
2+
and
[Ni(NH
6
)
3
]
6
as
metal
ons
are
An
octahedral
bond
with
the
Octahedral
Each
166
bidentate
transition
6
3+
antdote
the
the
2
lgand
using
with
are:
[Co(H
hexadentate
an
has a tetrahedral shape
2
Figure 13.5.2
Doctors
]
4
C
ligand
shape
is
central
complexes
forms
usually
formed
transition
two
can
also
if
element
be
coordinate
the
ion
formed
bonds
ligand
is
using
with
atom
relatively
three
the
forming
the
small.
bidentate
transition
ligands.
element
ion.
Chapter
13
Transton
elements

CH
3
a
H
2
b
CH
O
NH
2
2
2
H
H
O
N
NH
2
2
2
CH
2
Co
H
OH
Co
O
CH
2
2
2
H
N
NH
2
OH
2
CH
2
NH
2
2
CH
OH
2
2
Figure 13.5.3
Octahedral complexes of cobalt: a with a monodentate ligand, water;
b with a bidentate ligand, diaminoethane
Transition
When
compounds
reagents,
the
Different
useful
ions
in
a
of
state
states
demonstrating
as
way.
a
A
solution
few
manganate( vii)
of
of
the
iron( ii)
of
of
are
iron( ii)
agent
ions
to
+
MnO
(aq)
+
8H
may
sulphate
such
as
iron( iii)
→
5Fe
change.
are
often
containing
can
be
acidied
iron(
used
ii)
in
potassium
ions.
3+
(aq)
suitable
complexes
solution
+
(aq)
with
element
element
A
redox
treated
ammonium
oxidising
oxidise
and
transition
reactions.
2+
5Fe
elements
transition
redox
drops
will
compounds
transition
oxidation
oxidation
such
such
element
2+
(aq)
+
Mn
(aq)
+
4H
4
O(l)
2
2+
The
colour
of
the
solution
changes
from
the
light
green
of
the
Fe
ions
to
3+
the
reddish-brown
products
conrms
precipitate
agents
colour
is
such
the
formed.
as
zinc
of
ions.
Fe
presence
The
react
of
iron(
opposite
iii)
to
of
when
change
ions
3+
2Fe
addition
ions
colour
iron( iii)
with
The
+
Zn(s)
→
2Fe
a
to
the
red-brown
occurs
form
when
iron( ii)
2+
(aq)
alkali
reducing
ions
e.g.
Did you know?
2+
(aq)
+
Zn
(aq)
Potassum
The
to
iron( ii)
the
ions
solution
formed
when
a
in
this
reaction
grey-green
can
be
precipitate
tested
is
for
by
adding
alkali
formed.
usually
manganate(vii)
because
Potassum
managnate(vii)
hgh
MnO
ions
present
in
a
solution
of
acidied
t
can
degree
prmary
The
dchromate
preferred
to
n
be
of
redox
ttratons
obtaned
purty.
standard
s
potassum
It
s
to
a
used
because
t
as
a
s
also
to
ge
potassium
4
stable
manganate( vii)
dioxide
and
can
be
hydrogen
used
as
an
sulphide
oxidising
as
well
as
agent
being
to
test
useful
for
in
redox
titrations
2+
(see
to
page
45).
indicate
It
the
permanent
is
reduced
end
pink
point
to
of
a
colouration
Mn
ions.
redox
titration.
is
seen
due
The
to
colour
At
the
the
change
end
presence
can
point
of
n
ar,
readly
soluble
sulphur
be
used
a
stable
mass
soluton,
and
ttrates
has
a
hgh
molar
reproducbly.
a
excess
MnO
4
Key points
ions.
acid),
An
example
C
H
2
is
in
estimating
the
concentration
of
ethanedioic
(oxalic
O
2
4

+
2–
5C
O
2
(aq) + 2MnO
4
(aq) + 16H
The
shape
→
10CO
4
(g) + 2Mn
(aq) + 8H
2
O(l)
complexes
2
number
Potassum dchromate(vi)
Potasium
dichromate( vi)
is
a
bonds
good
oxidizing
agent.
2–
the
orange
O
Cr
2
of
transton
element
2+
(aq)
In
acidic
conditions,

The
depends
and
to
type
the
shapes
of
metal
of
on
the
lgand
whch
caton.
complexes
are
3+
ions
are
reduced
to
green
Cr
ions.
It
is
used
as
in
lnear,
7
tetrahedral,
square
planar
2+
volumetric
analysis
2–
Cr
O
2
to
determine
2+
(aq)
+
6Fe
the
concentration
+
(aq)
+
14H
of
3+
(aq)
→
2Cr
ions.
Fe
and
octahedral.
3+
(aq)
+
6Fe
–
(aq)
+
7H
7
O(l)
2

MnO
used
A
redox
indicator
such
as
barium
diphenylamine
sulphonate
is
rst
drop
of
excess
dichromate
converts
this
to
a
deep
as
O
2
can
be
7
oxdsng
agents
and
added.
to
The
2–
and Cr
4
test for
reducng
agents
n
bluish
the
laboratory.
Durng
these
solution.
reactons
changes
characterstc
colour
occur.
167
14
Qualitative
tests for
14.
1
dentifying
cations
Learning outcomes
Qualitative
There
On
completon
should
be
able
of
ths
secton,
are
descrbe
to:
used
to
how

a ame
dentfy
test
partcular
can
descrbe
catons
how
usng
soluton
of
to
dentfy
an
analysis
stages
in
analysing
the
elements
present
in
a
compound:
Preliminary
the
tests:
action
of
these
heat
include
on
the
the
appearance
of
the
solid,
flame
test
compound.
be
metal
catons

(1)
you
and

four
ions
specc

Making
a
solution
of
the

T
esting
for
the
cations

T
esting
for
the
anions
substance
under
test.
present.
present.
(See
Section
14.3)
aqueous
carbonate
ons,
know
The flame test
the
prncples
based
these

and
on
wrte
whch
ths
s
equatons for
descrbe
a
test for
A
s
reactons
ammonum
flame
block

test
of
Clean
ons.
can
the
a
used
platinum
hydrochloric
a
be
Periodic

Place

Hold
the

Note
any

Observe
identify
or
The
some
cations,
procedure
Nichrome
wire
by
especially
those
in
the
is:
dipping
it
in
concentrated
acid.
sample
wire
of
on
colour
the
to
T
able.
the
compound
the
in
edge
the
coloured
of
a
on
the
end
of
non-luminous
the
wire.
Bunsen
flame.
flame.
flame
through
a
diffraction
grating
or
spectroscope.
non-luminous
flame
Pt
wire
sample
Figure 14.1.1
The
flame
colours
I
sodum
for
selected
Group
–
orange-
–
llac
visible
to
the
yellow
168
cations
–
you
look
see
the
region
most
lines
at
the
(see
brck
the
lines
of
1.3).
in
sodium
–
red
copper
greensh-
red
apple-green
the
The
the
–
element
blue
crmson
through
lines
Section
obvious
in
flame
coloured
–
are:
Transition
strontum
barum
will
metal
(orange-red)
potassum
you
loop
II
calcum
yellow
When
on
Carrying out a flame test
typical
Group
with
a
diffraction
line
colours
line
spectrum
seen
emission
are
grating
emission
in
or
spectroscope,
spectrum
the
spectrum.
particularly
in
flame
For
distinct.
the
test
are
due
example,
the
Chapter
Making
a

Attempt

If
it
does

If
it
is
Many
So
it
to
dissolve
not
for
the
dissolve,
insoluble
tests
is
solution of the
in
water
,
cations
important
to
solid
warm
and
know
Qualtatve
tests for
ons
solid
in
a
few
cubic
centimetres
of
water
.
gently.
try
to
anions
the
14
dissolve
depend
solubility
it
in
on
nitric
acid.
solubility
rules
for
and
precipitation.
common
inorganic
compounds:
+

Common
salts
of
Group
I
cations
and
are
NH
soluble.
4

All

Sulphates
nitrates

Chlorides,
are
soluble.
2+
are
soluble
except
those
of
2+
,
Ca
Ba
2+
,
Sr
2+
and
Pb
+
bromides
and
iodides
are
soluble
except
those
of
Ag
and
2+
Pb

Carbonates
and
cations
NH
sulphites
are
insoluble
except
those
of
Group
I
+
and
4
+

All
hydroxides
are
insoluble
except
those
of
Group
I
cations,
NH
,
4
2+
2+
and
Sr
Using
Ba
sodium
Aqueous
sodium
carbonate to test for
carbonate
is
sometimes
used
cations
to
conrm
the
presence
or
+
absence
of
a
Group
I
cation
or
.
NH
Most
carbonates
are
insoluble
in
4
water
.
+

An
aqueous
solution
of
Group
I
cations
or
will
NH
not
form
a
4
precipitate

on
are
soluble.
An
aqueous
precipitate
addition
solution
because
of
sodium
containing
their
carbonate
other
carbonates
because
metal
are
cations
insoluble.
these
will
carbonates
give
Group
II
Did you know?
a
carbonates
When
give
a
white
precipitate
and
transition
element
carbonates
may
n
coloured
precipitates,
e.g.
is
MnCO
ammonum
salts
are
heated
give
a
test
tube
they
almost
always
pink.
3
decompose.
But
when
the
vapours
2+

Mg
ions
can
be
distinguished
from
other
Group
II
ions
by
adding
are
ammonium
carbonate
in
the
presence
of
ammonium
chloride.
cooled further
sold
these
conditions,
no
precipitate
is
up
the
salt
s
reformed,
+
HCl
→
NH
3
ammonium
These
ions
off,
ammonium
salts
decompose
on
warming.
Ammonia
gas
is
by
Cl(s)
4
compounds
pured
Many
the
e.g.
formed.
NH
A test for
tube,
Under
can
therefore
be
sublmaton.
given
e.g.
Key points
Cl(s)
NH
→
NH
4
(NH
)
4
(g)
+ HCl(g)
3
SO
2
(s)
→
2NH
4
(g) +
SO
3
(g)
+
H
3
O(l)
2

Ammonium
oxide
All
rather
nitrate
than
ammonium
is
an
exception.
It
decomposes
to
form
nitrogen(
i)
a
ammonia.
salts
give
off
ammonia
when
warmed
with
dilute
When
colours
dependng
the
put
n
are
man
seen,
colours
n
aqueous

Cl(aq)
+
NaOH(aq)
→
NH
4
lne
emsson
(g)
+
NaCl(aq)
+
H
3
spectrum.
Specc
metal
ammonia
can
be
detected
by
the
fact
catons
can
be
O(l)
2
dented
that
it
turns
damp
red
usng
an
aqueous
litmus
soluton
paper
on
are
Bunsen ame,
hydroxide:
NH
The
elements
characterstc
the
sodium
some
non-lumnous
of
carbonate
ons.
blue.

When
a
compound
ammonum
aqueous
ons
sodum
ammona
s
s
contanng
warmed
wth
hydroxde,
released.
169
14.2
dentifying
Learning outcomes
cations
Tests
Many
On
completon
of
ths
secton,
using
metal
be
able
descrbe
catons
how
to
usng
dentfy
sodum
metal
aqueous
explan
(if
aqueous
may
any)
be
formed
solution
redissolve
identied
by
the
by
observing
addition
of
the
dilute
colour
sodium
of
the
hydroxide
in
of
the
excess
substance
sodium
under
test.
Some
of
the
to
precipitates
hydroxide.
If
no
precipitate
is
formed:
A
cation
from
Group
I
or
ammonium
ions
ammona
may

can
hydroxide
hydroxde

or
cations
sodium
to:
an

aqueous
you
precipitate
should
(2)
the
prncples
on
be
present.
whch
(The
Group
I
cations
can
be
distinguished
by
flame
+
tests
and
ions
NH
by
warming
the
alkaline
solution.)
4
the
above
test
s
based


deduce
onc
symbols for
equatons
the
wth
reacton
of
state
If
a
white
precipitate
metal
2+
wth
()
sodum
aqueous
is
insoluble
in
excess
sodium
Ca
2+
,
2+
Sr
or
Ba
may
be
present
hydroxde
(these
()
which
2+
,
Mg
catons
forms
hydroxide:
can
be
distinguished
by
a
flame
test).
ammona.

If
a
white
precipitate
forms
which
is
soluble
in
excess
sodium
hydroxide:
3+
2+
,
Al
2+
Pb
or
Zn
ions
may
be
present.
Exam tips

If
a
coloured
the
A
lot
of
memory
work
s
precipitate
is
formed
the
colour
may
be
used
to
identify
cation:
requred
3+
to
learn
all
these
qualtatve
Cr
tests.
→
grey-green
precipitate
→
pale
→
green
→
rusty-brown
→
cream
→
green
2+
One
way
of
dong
ths
s
to
wrte
Cu
the
blue
precipitate
2+
on
and
test
on
a
sheet
of
paper
Fe
n
precipitate
(turning
brown)
3+
separate
columns,
then
cover
Fe
one
(reddish-brown)
precipitate
2+
of
the
columns
up
and
see
how
Mn
well
precipitate
(turning
dark
brown)
2+
Ni
you
can
remember
the
precipitate
tests.
The
precipitates
insoluble
in
formed
water
are
apart
of
from
metal
those
hydroxides.
of
Group
I
or
Metal
hydroxides
ammonium
are
ions.
Did you know?
T
ypical
equations
are:
2+
The
reacton
of
some
transton
(aq)
Fe
+ 2OH
(aq)
→
Fe(OH)
+ 2OH
(aq)
→
Cu(OH)
(s)
2
2+
metal
catons
wth
sodum
(aq)
Cu
hydroxde
(s)
2
3+
s
qute
complcated
because
complex formaton. The
(aq)
Cr
of
+ 3OH
(aq)
→
Cr(OH)
(s)
3
actual
The
precipitates
of
aluminium,
zinc
and
lead
hydroxides
dissolve
in
2+
reacton
of Cu
ons
wth OH
ons
excess
may
be
more
accurately
sodium
hydroxide
because
soluble
aluminates,
zincates
represented
plumbates
are
formed:
as:
(s)
Al(OH)
+
NaOH(aq)
→
NaAl(OH)
3
(aq)
4
2+
[Cu(H
O)
2
]
+2OH
→
sodium
6
aluminate
2+
[Cu(H
O)
2
(OH)
4
]
+
2H
2
O(l)
2
Zn(OH)
(s) + 2NaOH(aq)
→
2
Na
Zn(OH)
2
Pb(OH)
(s)
2
Exam tips
+ 2NaOH(aq)
→
Na
and
of
plumbates
alumnates,
are
–
wrtten AlO
,
2–
ZnO
2
respectvely. You
to
use
170
zncates
sometmes
2–
and
PbO
2
2
should
these formulae
as
be
prepared
well.
zincate
Pb(OH)
2
sodium
The formulae
(aq)
4
sodium
(aq)
4
plumbate(ii)
and
Chapter
Tests
The
using
effect
of
aqueous
conrmatory
formed
by
tests
the
substance
the
precipitate
the
under
test
allows
reactions
ammonia
some
are
of
contains
and
the
metal
The
aqueous
as
ions
of
ammonia
effect
for
due
cations
colour
of
distinction
same
OH
on
ions.
some
the
Qualtatve
tests for
ons
ammonia
ammonia
for
addition
the
of
aqueous
14
to
to
an
excess
between
sodium
the
in
the
solution
provides
precipitate
aqueous
aqueous
cations
hydroxide
(if
any)
solution
ammonia
to
be
made.
because
of
on
Most
aqueous
reaction:
+
NH
(aq)
+
H
3

The
concentration
O(l)
Y
NH
2
of
(aq)
+
OH
4
hydroxide
ions
in
dilute
aqueous
ammonia
is
+
low
in
comparison
with
NaOH,
so
test
solutions
containing
Ca
2
+
ions,
+
and
Sr
Ba
2
even
no
ions
may
form
only
slight
white
precipitate
or
precipitate.
3+

a
2
2+
can
Al
be
distinguished
from
Zn
by
the
use
of
aqueous
ammonia.
3+
ions
Al
form
a
white
precipitate
which
is
insoluble
in
excess
2+
aqueous
ammonia.
redissolves
which
is
in
excess
soluble
ions
Zn
in
aqueous
water
is
form
a
white
ammonia.
A
precipitate
colourless
which
complex
ion
formed:
2+
(aq)
Zn
+
2OH
+
4NH
(aq)
→
(aq)
→
Zn(OH)
(s)
2
2+
in
excess:
(s)
Zn(OH)
2

Copper
which
hydroxide
are
soluble
[Zn(NH
3
dissolves
in
water
.
in
excess
The
)
3
ammonia
complex
ion
is
]
–
(aq)
+
2OH
(aq)
4
to
a
form
deep
complex
blue
ions
colour
.
2+
(aq)
Cu
+
2OH
(aq)
→
Cu(OH)
(s)
2
in
excess:
2+
Cu(OH)
(s) + 4NH
2
often
(aq) + 2H
3
simplied
O(l)
→
[Cu(NH
2
)
3
(H
4
O)
2
]
(aq) + 2OH
(aq)
2
to:
2+
(s)
Cu(OH)
+
4NH
2

Other
(aq)
→
[Cu(NH
3
transition
element
)
3
hydroxides
]
(aq)
+
2OH
(aq)
4
such
as
and
Co(OH)
Ni(OH)
2
will
but
also
redissolve
in
does
Mn(OH)
excess
not
aqueous
ammonia
to
2
form
complexes,
redissolve.
2
Key points

Some
metal
precptates
catons
on
n
aqueous
addton
of
soluton form
NaOH(aq)
or
NH
characterstcally
coloured
(aq).
3

Dependng
n
excess
on
the
metal
NaOH(aq)
or
caton,
NH
these
precptates
may
or
may
not
dssolve
(aq).
3

Alumnum
and
complex

and
copper(ii)
Ionc
ons
znc
hydroxde
hydroxde
redssolve
redssolves
n
n
excess
excess
sodum
ammona
hydroxde
because
soluble
are formed.
equatons
can
be
wrtten
to
show
precptaton
and
complex
on
formaton.
171
14.3
dentifying
Learning outcomes
anions
Testing for
When
On
completon
should
be
able
of
ths
secton,
an
acid
carbonates
is
added
to
a
carbonate,
carbon
dioxide
is
released:
you
to:
CaCO
(s)
+
2HCl(aq)
→
CaCl
3
(aq)
+
CO
2
(g)
+
H
2
O(l)
2
2–

descrbe
tests
to
dentfy CO
,
3
2–
NO
,
SO
3
Carbon
dioxide
turns
limewater
(a
dilute
solution
of
calcium
hydroxide)
2–
,
SO
3
, Cl
,
Br
,
I
and
4
milky.
The
cloudiness
in
the
solution
is
a
ne
precipitate
of
calcium
2–
CrO
ons
carbonate:
4

explan
the
prncples
on
whch
Ca(OH)
(aq)
+
CO
(g)
2
these

tests
deduce
are
onc
equatons
wth
CaCO
(s)
+
H
3
O(l)
2
state
Testing for
symbols for
→
2
based
these
nitrates
reactons.
There
are
several
tests
for
nitrates:
3

W
arm
a
little
of
the
solid
with
a
2 cm
of
concentrated
sulphuric
acid.
Exam tip
Then
,
NO
You
wll
not
be
expected
add
are
a
small
piece
formed,
a
of
copper
.
nitrate
is
If
brown
probably
as
the
you
for
to
complex
ones for
should
smple
work
()
the
equatons
ntrate
know
Cu(s)
such
()
test.
be
But
+
2NO
(s)
+
4H

A
conrmatory
suspected
(or
able
→
present:
Cu
sutable
Secton
+
2NO
(g)
+
2H
2
test
nitrate
Devarda’s
(aq)
NO
is
to
and
alloy).
add
then
On
aqueous
either
sodium
zinc
warming,
hydroxide
powder
ammonia
or
gas
to
the
aluminium
is
O(l)
2
powder
released:
+
4Zn(s)
+
7OH
(aq)
+
6H
O(l)
→
NH
2
(g)
+
4Zn(OH)
3
(aq)
4
nformaton
The
(see
(aq)
2–
equatons from rst
gven
dioxide,
2+
(aq)
3
prncples
nitrogen
3
the formulae
compounds
out
of
2
to
+
remember
fumes
ammonia
can
be
identied
by
using
damp
red
litmus
paper
.
If
4.2).
ammonia
is
Testing for
The
solution
present,
the
to
be
tested
is
carbonates
barium
are
of
barium
turns
blue.
sulphates
contaminating
nitrate
litmus
then
sulphate
is
acidied
present.
added.
formed
If
a
(see
nitric
Section
acid
barium
sulphate
2+
is
present,
remove
a
any
or
white
aqueous
precipitate
12.2).
+
SO
(aq)
→
BaSO
4
Testing for
to
chloride
2–
(aq)
Ba
with
Aqueous
(s)
4
sulphites
2–
The
sulphite
ion
has
the
formula
SO
.
Sulphites
are
more
unstable
to
3
heat
than
sulphite.
If
the
sulphates
Sulphur
sulphite
solution
gas
are
and
is
indicates
dioxide
in
the
so
simple
gas
solution,
solution
the
a
is
presence
is
hydrochloric
a
to
heated.
acid
The
dioxide
conrmed

by
Bubbling
solution

Soaking
The
If
through
turns
a
+
2H
choking
a
(aq)
acidic
solution
from
piece
of
‘dichromate
sulphur
is
rst
release
→
SO
(g)
+
2
a
sample
of
the
solid
added
of
to
the
sulphur
dioxide
smell.
H
O(l)
2
Its
identication
can
be
either:
purple
lter
paper ’
dioxide
green-blue.
172
has
a
+
(aq)
3
Sulphur
heat
sulphite:
2–
SO
is
released.
then
of
test
is
of
to
paper
is
in
then
present,
potassium
manganate(
Vii).
The
colourless.
potassium
placed
the
paper
over
dichromate
the
turns
mouth
from
solution.
of
the
orange
to
test
tube.
Chapter
Testing for
The
test
The
with
to

see
either
silver
procedure
suspected
is
tests for
ons
aqueous
in
full
Aqueous
the
give
in
nitrate
white
Section
silver
precipitate
a
silver
or
aqueous
lead
nitrate.
nitrate
given
halide.
whether
Chlorides
Qualtatve
halides
involves
Testing
14
12.8.
nitrate
dissolves
precipitate
is
in
of
Nitric
then
added
aqueous
silver
acid
is
added
followed
by
to
the
testing
ammonia.
chloride:
+
(aq)
Ag
This
precipitate
formation
of
a
dissolves
complex
+
Cl
(aq)
readily
in
→
AgCl(s)
aqueous
ammonia
due
to
the
ion:
+
AgCl(s)
+
(aq)
2NH
→
[Ag(NH
3

Bromides
give
a
cream
)
3
precipitate
of
]
(aq) +
Cl
(aq)
2
silver
bromide:
+
Ag
The
silver
bromide
(aq)
+
precipitate
Br
(aq)
→
dissolves
AgBr(s)
only
in
excess
aqueous
ammonia.

Iodides
in
give
aqueous
a
pale
yellow
precipitate
of
silver
iodide
which
is
insoluble
ammonia:
+
Ag
Testing
Nitric
then

with
acid
is
lead
added
(aq)
+
I
(aq)
→
AgI(s)
nitrate
to
the
suspected
nitrate.
A
solution
of
lead
nitrate
is
added.
Chlorides
give
a
white
precipitate
of
lead(
ii)
chloride:
2+
Pb
(aq)
+
2Cl
(aq)
→
PbCl
(s)
2

Bromides

Iodides
The
give
give
addition
a
of
a
pale
deep
iodide
yellow
yellow
ions
precipitate
precipitate
to
a
of
of
solution
lead(
lead(
of
lead
ii)
ii)
bromide.
iodide.
ions
can
also
act
as
a
2+
conrmatory
test
Testing for
for
Pb
ions.
chromates
2–
The
chromate
ion
has
the
formula,
CrO
.
We
add
a
few
cubic
4
centimetres
chromate.
chromate
If
of
dilute
the
ions
sulphuric
chromate
turns
ion
orange
is
to
the
present,
due
2–
to
the
solution
the
+
2H
of
yellow
formation
+
(aq)
2CrO
acid
of
the
suspected
colour
of
the
dichromate
ions:
2–
(aq)
→
Cr
4
O
2
chromate
(aq)
+ H
7
O(l)
Key points
2
dichromate
Ions
When
a
drop
of
dilute
a
colouration
hydrogen
peroxide
is
added
to
the
can
be
followng
blue
is
seen
which
quickly
turns
dented
usng
the
dichromate,
aqueous
solutons:
green.
2–

CO
usng
hydrochlorc
acd
and
3
then
lmewater
–

NO
usng Cu
and
H
3
SO
2
or Al
4
–
and OH
2–

SO
usng
Ba(NO
4
)
3
2
2–

SO
usng
H
3

SO
2
Haldes
4
usng AgNO
or
Pb(NO
3
)
3
2–

CrO
usng
4
acd
and
H
O
2
2
173
2
Exam-style
Answers to
all
exam-style questions
can
questions
be found on the
Multiple-choice questions
1
Which
of
the following
are
chemically
ii
very
high
iii
high
tensile
Structured questions
properties
of
ceramic
6
a
i
Define
ii
State four
inert
the
term
characteristics
melting
point
iii
Write
the
strength
electronic
i,
ii
B
i,
ii,
D
i,
transition
configuration for
Mn
of
ions,
showing
electrons
in
the
only
iii,
Use
the
information
provided
in
part
3+
ions
Fe
are
easily
oxidised
to
Fe
represents
the
correct
order
of
–
3–
,
P
3+
, Al
resist
oxidation
to
Mn
[2]
ions.
[2]
increasing
When
iron(III)
chloride
B
P
C
Al
D
Na
–
, Cl
is
dissolved
in
water
a
red-
Na
3+
Na
+
–
, Cl
3+
, Al
chloride
gives
a
in
above
P
ii
PbCl
B
SnCl
above
Write
an
[2]
equation
to
represent
the
reaction.
–
acid
white
[1]
solution:
or
grey
7
precipitate
with
a
aqueous
chloride
aqueous
the following
potassium
chlorides
manganate(VII).
will
undergo
b
the
Define
the
terms
i
co-ordination
ii
ligand
complex
concentrated
added
to
is
[2]
[2]
When
sulphate
reactions?
A
the
, Cl
decolourises
of
explanation for
3–
,
P
mercury(II)
Which
an
observation.
3–
,
Give
3+
, Al
+
,
is formed.
Na
+
,
precipitate
+
,
i
3–
ii
ions
radius?
Cl
i
ions.
3+
Mn
brown
A
explain
iv
c
3
to
iv
ii
A
iii
observations:
2+
ionic
the
[2]
2+
ii,
Which
of
iii
i
2
the
orbitals
shell.
the following
ii,
a
compression
b
C
of
[2]
3+
and
Fe
valence
A
metal’.
[4]
the
occupation
resists
‘transition
metal.
3+
iv
3
accompanying CD
material?
i
Module
a
pale
until
blue
hydrochloric
solution
present
in
of
acid
is
slowly
copper(II)
excess,
a
yellow
solution
produced.
4
i
Write
the formulae
of
the
species
responsible
2
for
C
PbCl
D
SnCl
the
pale
blue
and
yellow
solutions
2
respectively.
[2]
4
4
Which
of
the
A
Carbon
B
Lead(IV)
statements
monoxide
below
turns
is
blue
ii
Write
the
equation for
iii
Using
the
concept
litmus
is
more
stable
to
heat
The
chemistry
affected
D
The
is
5
An
by
not very
i
The
ii
t
of
the
screening
element
germanium
inert-pair
effect
of
is
i
significantly
t
chloride
does
Which
A
Be
B
Mg
C
Sr
D
Ba
174
the
above
colour
change.
what
would
the
inner
is
lone
easily
not
electrons
in
properties:
hydrolysed.
pairs forming
react
element
at
all
best fits
to
lead
ii
Write
co-ordination
readily
the
with
above
an
is
be
observed
slowly
aqueous
the formula for
formed.
the following
ammonia
[3]
when
added
solution
of
until
copper(II)
sulphate.
compounds.
iii
explain
Describe
excess
effect.
effective.
exhibits
accepts
[2]
constant,
chloride.
aqueous
C
reaction.
stability
than
c
silicon(IV)
of
pink
carefully
chloride
the
correct?
water.
properties?
[3]
the final
copper
species
[1]
Module
8
a
Explain
the
difference
in
the
reaction
(if
any)
of
and SiCl
4
with
water
and
write
Exam-style
questons
10
Element/symbol
CCl
3
b.pt.
density
equations for
4
any
reactions
involved.
The
standard
electrode
[5]
F
b
4+
Ge
potential for
the
2+
(aq)/Ge
(aq)
values for Sn
and
system
Pb
to
is
–1.6 . Use
deduce
the
Cl
similar
relative
Br
stability
Group
of

the
+2
and
+4
oxidation
states
of
the
elements.
[3]
I
c
Chlorine
into
a
disproportionates
hot
concentrated
when
solution
bubbled
of
potassium
a
i
hydroxide.
i
Write
ii
an
the
ionic
and
chlorine
Write
two
equation
state
species
ionic
process
the
and
to
show
what
oxidation
physical
number
of
involved.
half
equations
thereby
a
Give
i
explanations for
is
BaSO
less
group
[2]
to
explain
the
soluble
than
term
Explain
and
b
A
properties
indicate
trends
of
the
of
by
the
the
use
above
elements
as
the
descended.
your
[2]
answers
and
to
bonding
part
with
i
in
terms
regards
to
of
b.pt.
[4]
crystalline
solid A
sulphuric
white fuming
gas
when
acid
treated
readily
with
evolves
a
MgSO
.
B
as
well
as
a
reddish vapour.
[2]
4
Magnesium
carbonate
strontium
temperature
Explain
and
the
density.
white
decomposes
at
when
the
is
trends
carbonate’s
dissolved
in
water forms
a
solution C
350 °C
which
b
table
arrows
concentrated
B
while
above
statements:
4
ii
is
structure
[5]
the following
ii
describe
‘disproportionation’.
9
the
of vertical
happens
the
Copy
produces
a
colourless
gas,
D,
with
sodium
decomposition
1340 °C.
carbonate,
which
when
solution
calcium
bubbled
through
a
[4]
of
hydroxide
produces
a
white
in
precipitate.
i
atomic
ii
ionic
radius
[2]
When
radius
cream
of
aqueous
A
the Group
compound
heating
oxide,
of

is
a

brown
containing
element
gas
and
10.46%
a
(M)
on
added
to C
dissolves
a
in
white
to
produce
oxygen
a
by
ammonia.
i
State
the
ii
State
the formulae
names
of
the
two
gases
B
and
D.
[2]
metallic
mass.
colourless
aqueous
strong
of
the
species
responsible
N
for
dissolves
which
elements.
a Group
evolves
N,
readily
nitrate
precipitate forms
concentrated
c
silver
[1]
the
production
of
the
gases.
aqueous
dentifying
the
observations
underlying
the
solution, S.
above
S
when
reacted
with
an
appropriate
dilute
M
on
crystallising from
the
[2]
acid
iii
produces
answers.
Write
two
ionic
equations
to
explain
the
resulting
observations
associated
with
the
reaction
solution.
of C
i
State
the formula
ii
Deduce
of
the
brown
gas.
c
the
atomic
mass
of
the
with
element
M.
State
iv
Write
the
name
and formula
of
Use
relevant
M.
equation
formation
of
which
M from S.
represents
[2]
to
information
explain
the
involving
relative
electrode
oxidising
[1]
power
the
nitrate.
[2]
potentials
iii
silver
[1]
of
the
halogens.
[3]
the
[2]
175
Data
Selected
sheets
bond
energies
Selected
electrode
potentials
Ø
Diatomic
molecules
Polyatomic
molecules
–1
Electrode
–1
Bond
436
C–C
350
Mg
N≡N
994
C=C
610
Al
O=O
496
C–H
410
V
F–F
158
C–Cl
340
Zn
Cl–Cl
244
C–Br
280
Fe
Br–Br
193
C–I
240
V
I–I
151
C–N
305
N
H–F
562
C–O
360
Sn
H–Cl
431
C=O
740
Pb
H–Br
366
N–H
390
2H
Bond
H–H
energy/kJ mol
energy/kJ mol
E
/V
+
Bond
Bond
reaction
+
K
e
Y
K
–2.92
2+
+
2e
Y
Mg
–2.38
3+
+
3e
Y Al
–1.66
2+
+
2e
Y V
–1.2
2+
+
2e
Y
Zn
–0.76
2+
+
2e
Y
3+
Fe
–0.44
2+
+
e
Y V
–0.26
2+
+
2e
Y
N
–0.25
+
2e
Y
Sn
–0.
14
+
2e
Y
Pb
–0.
13
2+
2+
+
+
2e
Y
H
0.00
2
2–
H–I
299
N–N
160
S
O
4
2–
+
2e
Y
2S
6
O
2
+0.09
3
2+
O–H
460
Cu
+
O–O
150
VO
2e
Y Cu
2+
+0.34
+
+
2H
3+
+
e
Y V
+
H
O
+0.34
2
+
I
2e
Y
2I
+0.54
2
3+
2+
+
Fe
e
Y
Fe
+0.77
+
+
Ag
e
Y Ag
+
+0.80
+
+
VO
2H
2+
+
e
Y VO
+
H
2
+
Br
O
+1.00
2
2e
Y
2Br
+1.07
2
2–
O
Cr
2
+
Cl
+
+
14H
3+
+ 6e
Y
2Cr
+
7H
7
O
+1.33
2
2e
Y
2Cl
+1.36
2
–
4
176
+
+
MnO
8H
2+
+
5e
Y
Mn
+
4H
O
2
+1.52
sheets
Data
0.571
301
rL
muicnerwal
]262[
0.371
201
oN
muilebon
]952[
9.861
101
dM
muivelednem
]852[
3.761
001
mF
muimref
]752[
9.461
99
sE
muinietsnie
]252[
5.261
89
fC
muinrofilac
]152[
9.851
79
kB
muilekreb
]742[
3.751
69
mC
muiruc
]742[
0.251
59
mA
muicnema
]342[
4.051
49
uP
muinotulp
]442[
]541[
39
pN
muinutpen
]732[
2.441
29
U
muinaru
0.832
9.041
19
aP
muinitcatorp
]132[
eC
munec
1.041
09
hT
munoht
0.232
901
tM
muirentiem
]86 2 [
muimso
2.091
801
sH
muissah
]962[
muinehr
2.681
701
hB
muirhob
]462[
netsgnut
8.381
601
gS
muigrobaes
]662[
mulatnat
9.081
501
bD
muinbud
]262[
fH
muinfah
5.871
401
fR
muidrofrehtur
]162[
75
aL
munahtnal
9.831
98
cA
muinitca
]722[
65
aB
muirab
3.731
88
aR
muidar
]622[
74.58
55
sC
muiseac
9.231
78
rF
muicnarf
]322[
85
rP
muimydoesarp
muidiri
2.291
aT
27
95
dN
muimydoen
1.591
W
37
06
mP
muihtemorp
munitalp
eR
47
16
mS
muiramas
dlog
0.791
sO
57
26
uE
muiporue
yrucrem
6.002
rI
67
36
dG
muinilodag
muillaht
4.402
tP
77
46
bT
muibret
dael
2.702
uA
87
56
yD
muisorpsyd
htumsib
0.902
gH
97
66
oH
muimloh
]902[
lT
08
76
rE
muinolop
bP
18
86
muibre
]012[
iB
28
96
mT
muiluht
enitatsa
oP
38
07
bY
muibretty
nodar
]222[
tA
48
17
uL
muitetul
nR
58
73
bR
68
rS
01.93
muidibur
83
26.78
Y
80.04
muitnorts
93
muirtty
rZ
69.44
19.88
04
22.19
bN
78.74
muinocriz
14
muiboin
oM
49.05
19.29
24
49.59
cT
00.25
munedbylom
34
]89[
uR
49.45
muitenhcet
44
1.101
hR
58.55
muinehtur
54
9.201
dP
39.85
K
muissatop
muidohr
64
aC
muiclac
4.601
gA
96.85
cS
muidnacs
muidallap
74
iT
muinatit
revlis
dC
55.36
V
muidanav
9.701
84
rC
muimorhc
4.211
nI
93.56
nM
esenagnam
muimdac
94
nori
muidni
nS
27.96
eF
8.411
05
oC
tlaboc
nit
bS
16.27
iN
lekcin
7.811
15
uC
reppoc
8.121
eT
29.47
cniz
ynomitna
25
nZ
6.721
I
69.87
aG
muillag
muirullet
35
eG
muinamreg
enidoi
eX
09.97
sA
cinesra
9.621
45
eS
muineles
nonex
0 8 . 38
rB
enimorb
3.131
rK
notpyrk
13.42
02
210.9
muihtil
149.6
3
iL

eB

4
muillyreb
11
91
99.22
12
92

22
03
 
cimota
21
32
13

42
23

52
33

62
43
89.62
72
53
90.82
82
63
79.03
evitaler

70.23
aN
muidos
yeK
)notorp(
cimota
eman
cimota
rebmun
lobmys
ssam

54.53
gM
muisengam
800.1

59.93
lA
muinimula
31
iS
nocilis
41
P
surohpsohp
norob
18.01
51
S
nobrac
10.21
61
ruflus
10.41
71
lC
enirolhc
negortin
81
rA
nogra
negyxo
B
00.61
C
eniroufl
N
5
00.91
O
6
noen
F
7
81.02
eN
8

9

  
negordyh

01
muileh
1
H

300.4
2
eH

177
Answers to
revision questions
z
b
Module
1
Chapter
1
1
All
z
y
(page
atoms
of
the
x
12)
same
element
are
exactly
cannot
be
atoms
of
atoms
combne
broken
dfferent
down
elements
hae
more
orbital
x
dfferent
complex
orbital
p
any further,
masses,
2
13
to form
x
alke,
s
atoms
y
a
1s
b
1s
structures
2
2s
2
6
2p
2
2s
2
6
3s
6
2p
2
3p
2
3s
2
4s
6
3p
10
3d
c
1s
d
1s
2
2
2s
2
4s
6
2p
2
2s
2
3s
6
2p
6
3p
2
3s
6
3p
5
3d
(compounds).
14
a
The rst
onsaton
energy,
ΔH
,
s
the
energy
needed
1
Atoms
are
not
ndestructble.
Atoms
are
splt
n
nuclear
to
reactons,
and
they
are
made
up
of
een
smaller
remoe
all
are
of
the
same
dentcal),
element
that
s,
can
they
hae
dfferent
possess
masses
Electrons
b
are deflected to the
poste
plate
because they
because they
not deflected
are
postely
charged
and
of
at
of
all
because they
deflecton
of
an
because
an
electron
hae
no
electron
s
gaseous
lghter
nuclear
The
absolute
Ths
s
due
to
the
gaseous
state
mole
to form
of
one
masses
and
charges
of
small,
charges
so
t
when
s
much
makng
easer
more
than
n
ons.
+
e
use
comparsons
charge,
dstance
of
the
outer
electrons from
the
ncrease
n
nuclear
charge
as
one
goes
that
a
The
electron
n Al
s
remoed from
the
hgher
energy
of
orbtal
compared
to
Mg
where
the
electron
to
be
mass.
sub-atomc
to
(g)
charge.
s
comes from
relate
and
the
lower
energy
3s
orbtal.
It
partcles
requres
less
energy
to
remoe
the
electron
masses
n Al,
and
ts
perod.
therefore
are ery
n
sheldng
remoed
3
element
are
3p
proton
one
negate
neutrons
17
a
an
→ Ca
nucleus,
across
degree
of
of
Ca(g)
Sze
the
16
The
n
sotopes.
are negately charged, protons are deflected to the
plate
atom
+
(not
15
2
each
partcles).
mole
Atoms
electron from
partcles
atoms
(subatomc
one
therefore
t
has
a
lower rst
onsaton
energy.
dong
b
The
electron
to
be
remoed from S
comes from
the
3p
x
calculatons.
orbtal
4
a
13
protons,
13
electrons,
14
b
19
protons,
19
electrons,
20
c
53
protons,
53
d
94
protons,
94
(10
×
electrons,
neutrons
neutrons
78
electrons,
neutrons
145
+
(11
×
t
The
repulson
ths
electron
whch
s
also
s
spn-pared
between
s
n
easer
the
these
to
3p
wth
two
remoe
orbtal
another
electrons
than
but
s
the
not
electron.
means
electron
that
n
P,
spn-pared.
neutrons
18
18.7)
where
Group
II
81.3)
______________________
5
=
10.8
Chapter
2
(page
32)
100
234
238
6
a
i
U
→
b
i
4
Th
+
Th
→
Ra
Pa
→
+
e
ii
+
He
C
→
N
+
a
i
Sgma
Iodne-131
s
used
to
study
s
used for
datng
objects
whch
were
once
s
used
n
Uranum-235
used to
bonds)
are formed
by
the
end-on
orbtals.
P
bonds
(π
smoke
of
bonds)
p
are formed
atomc
by
the
sdeways
orbtals.
lng
b
Amercum-241
(σ
atomc
thyrod functon
oerlap
Carbon-14
of
–1
ii
7
bonds
oerlap
e
7
6
1
2
0
14
14
–1
4
Rn
86
88
0
91
90
ii
2
234
218
222
He
90
92
234
Sgma bonds are stronger because the electron densty s
detectors
symmetrcal about a lne jonng the two nucle, whereas
s
generate
energy
n
nuclear
reactors
n a p bond the electron densty s not symmetrcal.
8
Electrons
are
arranged
n
energy
leels
whch
are
2
a
lone
pair
quantsed.
of
When
an
electron
absorbs
partcular frequency
moes
The
to
a
electron
moes
As
up
t
seen
back
does
as
a
or
hgher
so,
lne
the
of
the
emsson
a
Preously
to
a
t
of
energy
becomes
of
a
excted
and
leel.
loses
lower
energy
that
quantum
energy
eentually
down
a
waelength
electrons
that
ths
quantum
energy
t
of
energy
and
bond
leel.
loses
s
gen
partcular frequency
or
out
and
s
waelength
of
pair
electrons
n
spectrum.
P
9
energy
excted
electrons fall
back
to
the n =
1
electrons fall
back
to
the n =
2
O
leel.
Cl
b
Preously
energy
10
ΔE
=
ΔE
s
excted
leel.
hν
3
the
energy
dfference
h
s
Planck’s
ν
s
the frequency
11
2,
12
a
6
and
A
between
the
two
energy
constant.
of
the
a
leels.
A
co-ordnate
both
the
+
b
radaton.
10
O
regon
of
space
where
there
s
a
hgh
probablty
of
H
ndng
178
an
electron
bond
s formed
electrons for
the
when
coalent
one
atom
bond.
prodes
Answers to
3+
+
10
2–
2–
Al
K
Iodne
a
s
a
smple
Weak an
molecular
revision questions
sold:
der Waals forces
exst
between
the
O
molecules,
S
+
allowng
the
sold
to
break
easly.
3+
b
There
c
Because
are
no free
ons
or
delocalsed
electrons.
Al
K
2–
of
attracton
between
the
3+
2–
+
2 [2,8]
[2,8,8]
2 [2,8,8]
the forces
molecules
O
to
turn
are
the
so
sold
weak
nto
that
a
not
lqud
much
and
the
energy
lqud
s
needed
nto
a
apour.
2–
11
Water
has
a
much
hgher
bolng
pont
due
to
the fact
that
O
t
s
extensely
whch
only
hydrogen
possess an
bonded
unlke
der Waals forces
the
other
between
hydrdes
ther
2–
3 [2,8]
5
a
Because
the
element
has
one
electron
n
ts
outer
molecules.
shell
12
PCl
would
hae
a
hgher
meltng
pont
than
BCl
3
t
s
a
metal,
therefore
the
bondng
s
of
In
ths
type
of
bondng
a
lattce
of
metal
ons
the
greater
by
a
‘sea’
of
delocalsed
The
meltng
pont
would
be
of
electrons
t
possesses,
produce
more
stronger an
energy
to
der Waals forces
break. Also
t
can
of
PCl
s
greater
than
that
of
BCl
3
strong
electrostatc
attracton
between
the
and
the
delocalsed
electrons
acts
n
all
results
n
the
bond
beng
strong
and
hard
to
As
a
result
a
the
meltng
magnesum,
pont
wll
be
n
on
a
Electronegatty
noled
n
bondng
b
s
the
coalent
par
of
As
you
go
ablty
the
bondng
par
from
the
sze
of
of
a
partcular
towards
to
ts an
an
der Waals forces.
–
the
there
are
oxygen
three
bond
pars
and
one
lone
atom.
on
–
the
there
are
sulphur
two
bond
pars
and
two
lone
atom.
atom
attract
c
Trgonal
d
Tetrahedral
planar
e
Pyramdal
–
there
are
three
bond
pars.
the
–
there
are four
bond
pars.
tself.
of
the
electrons
attracton
of
electronegatty
atoms
ncrease
beng
the
on
–
the
there
are
three
phosphorus
bond
pars
and
one
lone
atom.
decreases
resultng
n
the
Chapter
3
(page
46)
ncreasngly further
1
nucleus.
a
FeCl
(aq)
+
3NaOH(aq) →
Fe(OH)
3
a
non-polar
c
non-polar
b
polar
d
polar
(s)
+
3NaCl(aq)
3
3+
Fe
7
of
decreases.
down Group VII,
because
strength
Non-lnear
par
c
the
causng
hgh.
bond formaton
electrons
Electronegatty
agan
alumnum
pars
6
the
Pyramdal
b
sodum,
,
3
break.
par
c
whch
that
drectons
13
and
sad
metal
ncrease
ons
be
hgh.
mass
Ths
whch
electrons.
requre
b
number
s
therefore
surrounded
because
3
metallc.
–
(aq)
+
3OH
(aq)
→
Fe(OH)
(s)
3
b
Mg(s)
+
2HCl(aq)
→
MgCl
(aq)
+
H
2
+
Mg(s)
+
2H
(g)
2
2+
(aq)
→
Mg
(aq)
+
H
(g)
2
8
a
Permanent
dpole–dpole forces, an
der Waals forces,
c
Na
S
2
hydrogen
O
2
(aq)
+
2HCl(aq)
→
2NaCl(aq)
+
H
3
O(l)
+
2
bondng
S(s)
+ SO
(g)
2
b
i
an
der Waals forces
2–
S
O
2
ii
hydrogen
bondng
iii
hydrogen
bondng
2
a
i
+
(aq)
A
mole
permanent
Magnesum
a
gant
→
H
O(l)
+ S(s)
+ SO
s
the
amount
of
(g)
2
of
substance
speced
partcles
whch
as
has
there
the
are
dpole–dpole forces
chlorde
has
a
gant
onc
structure,
coalent
structure
and
alumnum
n
exactly
12 g
of
the
carbon-12
sotope.
damond
ii
has
(aq)
number
atoms
9
2H
2
same
iv
+
3
s
Molar
mass
s
the
mass
of
1
mole
of
a
substance
n
metallc.
grams.
a
All
three
solds
hae
hgh
meltng
ponts.
In
magnesum
–1
b
chlorde,
forces
of
strong
onc
attracton
bonds
due
between
to
the
the
must
bonds
be
broken.
between
the
In
oppostely
damond,
carbon
strong
atoms
331 g mol
250 g mol
must
–1
ii
310 g mol
ii
20/100
–1
charged
3
ons
i
iii
electrostatc
a
i
7
.
1/142
b
i
0.025
ii
2.0
=
0.05 mol
=
0.2 mol
coalent
be
broken
×
84
=
2.
1 g
and,
–3
n
of
alumnum,
attracton
delocalsed
b
make
or
up
the fact
free
to
between
chlorde
dssoled
the
Damond
metallc
electrons
Magnesum
molten
strong
t
metal
must
wll
n
be
are
conduct
possesses
ons
to
and
because
4
the
to
electrcty
ons
or
Balanced
C
H
3
the
ons
any
10
×
36.5
(g)
equaton for
+
5O
8
=
0.073 g
(g)
→
reacton:
3CO
2
Number
when
of
From
whch
(g)
+
of C
balanced
of O
=
5.5/44
equaton,
1
mole C
electrcty
state
electrons
n
both
0.
125 mol
H
reacts
wth
5
moles
8
.
due
that
So
to
number
of
moles
of O
=
0.
125
the
×
5
=
0.625 mol
2
Mass
are
of O
=
0.625
×
32
=
20 g
2
a
Molecular formula
s
N
O
2
conduct
=
8
2
5
wll
O(g)
2
H
3
moe.
Alumnum
4H
2
moles
3
moe.
n
×
the forces
electrcty
now free
no
due
broken.
conduct
water
structure
cannot
that
the
bonds
sold
Emprcal formula
s
.
4
NO
.
2
and
molten
free
to
state
because
the
delocalsed
electrons
are
b
Molecular formula
s S
Cl
2
c
moe.
Magnesum
Emprcal formula
chlorde
s
soluble
n
water
due
to
the fact
6
that
the
ons
present
can form
on–dpole
.
2
s SCl.
bonds
C
H
wth
0.48
0.08
_____
Number
water
of
moles
_____
=
0.04
12
Damond
s
not
soluble
n
water
because
the
atoms
0.08
=
2
1
are
0.08
0.04
_____
held
=
molecules.
together
to form
bonds
wth
Mole
water.
rato
_____
=
1
0.04
Alumnum
s
not
soluble
n
water
because
the force
0.04
of
Therefore
emprcal formula
s CH
.
2
attracton
between
the
metal
ons
and
the
delocalsed
Emprcal formula
electrons
wth
s
too
strong
to
allow
the
ons
to form
mass
s
12
+
2
=
14
bonds
water.
179
Answers to
revision questions
Therefore
56
___
Dde molar mass by emprcal formula mass =
=
0.0055 mol
potassum
manganate(vii)
reacts
4
2+
wth
14
Multply
each
Therefore
atom
n
emprcal formula
molecular formula
s C
7
a
Number
of
moles
=
6/24
=
by
×
0.0055
=
0.0275 mol
Fe
3
4.
Ths
s
the
number
of
moles
3
H
4
5
Conertng
8
Therefore
0.25 mol
molar
n
25.0 cm
of
ron(ii).
3
25.0 cm
to
dm
3
:
25.0/1000
concentraton:
=
0.025 dm
0.0275/0.025
=
–3
Mass
=
0.25
×
32
=
1.
10 mol dm
8 g
3
b
Conertng
3
120 cm
to
dm
3
120/1000
Number
Mass
=
=
of
Module
0.
12 dm
moles
0.005
×
at
64
r.t.p.
=
=
0.
12/24
Volume
of
Chapter
3
56 cm
of
moles
7
(page
88)
3
of
butane
n
dm
=
of
butane
at
s.t.p.
56/1000
=
0.056 dm
1
Number
2
0.005 mol
0.32 g
3
8
=
=
0.056/22.4
a
Set
up
apparatus
smlar
to
Fgure
7
.
1.
1,
substtute
the
=
dfferent
reactants
and
measure
the olume
of
N
at
2
0.0025 mol
specc
Balanced
equaton for
tme
nterals.
reacton:
b
The
a
I
I
(aq)
s
b
Determne
brown
n
colour.
2
H
2C
4
(g)
+
13O
10
(g)
→
8CO
2
(g)
+
10H
2
O(g)
2
2
From
balanced
equaton,
2
H
moles C
4
moles
of
H
react
to
produce
and
III
10
10
at
least
the rst
two
half-les.
For
a rst
constant.
For
a
O.
2
10
___
Therefore
0.0025
moles C
H
4
produce
×
0.0025
mole
of
reacton,
the
half-les
are
order
reacton,
the
half-les
become
second
=
10
2
0.0125
order
longer.
O
H
2
3
Volume
of
steam
at
s.t.p.
=
0.0125
×
22.4
=
3
9
C
H
x
(g)
+
O
y
(g)
→
xCO
2
(g)
yH
The
a
Second
graph
b
Frst
c
Thrd
d
Rate
would
be
an
upward
cure.
order
O(g)
2
3
3
160 cm
1 ol
+
2
3
20 cm
c
0.28 dm
order
100 cm
8 ol
order
5 ol
2
C
H
5
(g)
+
8O
y
(g)
→ 5CO
2
(g)
+
yH
2
=
k[X]
[Y]
O(g)
2
Rate
10
moles
of O
reacts
wth C
so
the
other
6
moles
_______
must
e
k
6
=
=
–2
mol
9200 dm
–1
s
2
react
so
6
wth
[X]
H,
moles
H
O
are formed
whch
contan
12
atoms
of
H.
2
Therefore
pentane
must
be C
4
a
H
5
Draw
may
12
a
[Y]
tangent
to
be aratons
tangent
s
the
n
cure
the
at
the
answer
tme
a
Number
of
moles
of
KNO
–2
=
=
0.033 mol
[There
on
how
drawn.]
3.33
_____
10
gen.
dependng
i
For tme = 0 s, Intal rate = 1.
1 × 10
ii
For
tme
=
20 s,
Rate
=
7
.6
For
tme
=
60 s,
Rate
=
3.4
–3
mol dm
–1
s
3
–3
101
×
10
250
3
Conertng
dm
3
=
=
1000
b
i
0.033
concentraton
=
ii
of
moles
of
Na
CO
2
=
=
5
_____
3
to
75 cm
dm
3
:
=
a
Change
Frst
a
Knetc
b
Refer
c
=
=
Number
Mass:
of
to
dm
moles:
0.00025
×
Change
6
750 cm
25/1000
0.01
3
b
×
=
0.025
=
=
0.025 dm
0.00025 mol
0.015 g
3
to
t
=
35 s, Second
order,
Mass:
of
0.20
moles:
×
80
=
s
energy
a
Burette
dm
0.27
i
Rate
ii
Refer
a
D
b
C
c
C
d
=
the alue.
35 s
constant.
partcles
(y-axs)
7
.3.
would
decrease.
to Secton
7
.3.
A
3
:
750/1000
=
0.75 dm
7
a
×
0.75
=
0.20 mol
16.0 g
readings/
t
(x-axs), fracton of
to Secton
i
NO(g)
ii
Step
1
Rate
=
+ SO
(g)
+ O
iii
1
2
3
4
k[SO
Final volume
27
.
15
25.80
37
.90
26.50
Initial volume
0.50
0.
15
12.50
1.00
Actual volume
26.65
25.40
25.50
(g)
→ SO
2
][NO
2
12
ntal
½
half-lfe
2
Number
the
half
3
:
58.5
to
0.33 mol dm
3
25 cm
taken for
decrease
–3
0.075
3
11
Frst
c
0.025
______
concentraton
tme
to
0.075 dm
1000
Molar
the
3
106
Conertng
–1
s
0.025 mol
75
3
s
½
2.65
_____
Number
–3
mol dm
0.
13 mol dm
0.25
b
Half-lfe
concentraton
–3
=
10
0.25 dm
______
Molar
×
–1
s
–3
_____
3
to
250 cm
–3
mol dm
],
(g)
+
NO
3
the
NO
2
(g)
2
functons
as
a
2
catalyst.
3
b
cm
i
Mechansm
ii
The
Chapters
25.65
(25.40
+
8–10
Aerage olume
of
acd
I
would
(page
25.50)
double.
130)
2
_______________
b
rate
[NO
(g)]
[Cl
(g)]
2
2
________________
=
1
a
i
K
–3
=
Unts:
mol dm
Unts:
mol
Unts:
atm
Unts:
atm
Unts:
mol
Unts:
mol
c
2
2
[NO
2
3
=
Cl(g)]
25.45 cm
2+
[Fe
½
(aq)][I
(aq)]
2
________________
13
Number
of
moles
of
potassum
manganate(vii)
ii
used:
K
3+
[Fe
1½
dm
–
(aq)][I
27
.50
(aq)]
2
______
0.2
½
=
c
×
=
(p
0.0055 mol
)
O
3
______
b
1000
i
K
–1
=
p
3
(p
)
O
From
balanced
2
equaton:
3
2+
(aq)
MnO
+
5Fe
(p
+
(aq)
+
8H
½
)
(
p
HF
(aq)
)
½
(p
N
)
Cl
2
2
__________________
→
4
ii
K
2
=
p
2+
Mn
3+
(aq)
+
5Fe
(aq)
+
4H
(p
O(l)
)
(p
NH
2
–
1 mol of
potassum
manganate(vii)
reacts wth
5 mol
ron(ii)
c
i
K
=
[IO
sp
K
sp
180
3
2
(aq)]
2+
[Ba
3
(aq)]
–9
dm
3
3+
ii
)
ClF
3
=
[Al
–
(aq)][OH
3
(aq)]
4
–12
dm
the
Answers to
+
O
[H
2–
(aq)][F
(aq)]
iii
3
_________________
d
i
K
K
Unts:
–4
of CO
b
–3
=
revision questions
=
2.
1
×
–3
10
mol dm
3
mol dm
a
pH
[HF(aq)]
+
(aq)][CN
[H
8
(aq)]
________________
ii
K
=
10.3
a
pH
=
13.2
b
pH
=
12.7
c
pH
=
1.9
a
Dagram
–3
=
Unts:
mol dm
a
[HCN(aq)]
S(aq)][OH
[H
(aq)]
2
__________________
e
K
–3
=
Unts:
mol dm
4+
9
b
[HS
smlar to
Fgure
10.2.2, wth the Sn
(aq)/
(aq)]
2+
Sn
2
a
Dynamc
equlbrum
s
the
state
that
exsts for
system,
where
the
rate
of
the forward
connected to the
a
3+
b
closed
(aq)
i
Al(s)
reacton
 Al
standard
hydrogen
electrode.
2+
(aq)

Zn
(aq)

Zn(s)
Ø
ii
E
=
+0.90 V
iii
2Al(s)
cell
or
process
s
equal
process. The
at
equlbrum
represented
K
s
only
an
s
N
by
(g)
Yield
i
decrease
of
of
+
a
and
the
system
3H
(g)
reerse
and
the
system
can
Y
temperature
n
or
2+
2NH
+
3Zn
3+
(aq)
→
2Al
(aq)
+
3Zn(s)
be
of
Module
of
3
the
dynamc
2
Chapters
11–12
(page
156)
(g).
3
Position of
Value of
equilibrium
equilibrium
shft
no
to
reacton
reactants
constant K. The alue
changng
of
the
products
constant
2
b
rate
equlbrum
example
equlbrum
the
reman
by
affected
system. An
to
concentraton
left
constant
1
A
6
a
effect
2
C
3
Monodentate
central
C
lgand
atom,
e.g.,
4
can
C
only form
ammona
5
one
B
bond
molecule,
H
to
the
N:→
3
ii
ncrease
shft
iii
no
no
effect
to
rght
effect
no
effect
no
effect
bdentate
atom;
lgand
e.g.,
can form
two
bonds
ethane-1,2-damne,
H
to
iv
c
ncrease
v
ncrease
i
No
of
ii
effect
shft
to
rght
ncrease
shft
to
rght
no
on
poston
equlbrum
Yeld
of
shft
n
effect
equlbrum,
b
yeld
or alue
i
tetrachlorocuprate(ii)
ii
to
NH
2
2
rght,
no
and
↓
hexaaquaron(iii)
2
Cl
poston
central
CH
2
↓
constant
ncreases,
on alue
of
the
NCH
2
3
OH
2
Cl
H
effect
OH
O
2
2
constant
Cu(II)
iii
Yeld
decreases,
shft
to
the
left, alue
of
Fe(III)
equlbrum
constant
decreases
–3
3
a
K
b
[R]
c
K
=
Cl
Cl
15.3 mol dm
c
–3
=
0.
14
mol dm
H
O
OH
2
=
2
33.4
c
OH
–3
d
K
a
p
=
0.53 mol dm
2
c
2+
4
=
3.7 atm
,
p
H
=
c
K
[Cu(H
O)
2
]
–
(aq) + 4Cl
2–
(aq) → CuCl
6
(aq) + 6H
4
O(l)
2
2
–4
K
i
N
2
b
c
0.2 atm
=
5.0
×
=
0.24
–1
10
Pa
ii
Co-ordnaton
iii
The
no.
6;
co-ordnaton
no.
4
p
chlorne
on
s
much
larger
than
the
water
p
molecule
5
a
The
maxmum
amount
of
a
solute
that
dssoles
n
and
the
copper
gen olume
of
i
×10
a
solent
–5
K
=
5.67
=
1.70
at
2
a
partcular
therefore
of
these
only
ons.
temperature
–6
mol
can
a
accommodate four
b
atom
7
dm
a
In
Perod
3,
oxdaton
states
rse
steadly
as
the
sp
–6
ii
K
×
i
For
pure water,
For
a
3
10
–9
mol
dm
elements
ether
lose
or
gan
In
electroposte
all
ther alence
electrons.
sp
–4
c
solublty
s
1.9
×
10
–3
soluton
of
ii
The
iii
MgCO
s
2.4
common
2+
0.015 mol dm
–6
solublty
Mg
mol dm
effect
2+
would
precptate,
[Mg
2–
][CO
3
]
>
case
of
oxdaton
ons,
–3
×10
on
–3
mol dm
electrons
from
to
+4
a
i
HCl
s
a
strong
acd
and
dssocates
are
+1
to
shared
Na–Al,
+3; from
n
poste
slcon
coalent
bonds
to
rsng
+7
.
Na
O
Al
2
sp
oxdaton
6
rse from
chlorne
e.g.
K
3
states
elements,
number
O
2
+1
P
3
O
2
+3
Cl
5
O
2
+5
7
+7
completely,
b
For
NaCl
and
MgCl
:
structure
onc
crystal
lattce.
2
whereas C
H
6
COOH
s
a
weak
acd
and
dssocates
5
Bondng
–
strong
onc
(electroalent)
bonds,
ons
held
partally.
together
ii
pH
=
1.6
iii
pH
=
2.9
On
pH
c
i
=
[H
1.7
,
ii
[OH
pOH
=
–12
=
3.98
×
=
2.51
wth
–3
10
of
a
water
neutral
the
crystal
soluton
lattce
s
contanng
destroyed
hydrated
a
pH
of
ons
7
.
mol dm
–3
]
n
12.3
+
]
electrostatc forces.
addton
resultng
b
by
For
–3
×10
alumnum
chlorde:
sold
lattce
structure.
mol dm
Bondng
–
onc
wth
apprecable
coalent
character.
+
7
a
i
C
H
6
NH
5
(aq)
+
H
2
O(l)
Y C
2
H
6
NH
5
(aq) + OH
(aq)
3
On
addton
of
water
the
small
on
(large
charge
and
hghly
charged
+
ii
C
H
6
iii
NH
5
pH
=
3
alumnum
9.0,
pK
=
i
The
weak
hexaaquaalumnum(iii)
acd
HC
H
3
added
and
ts
O
5
would
neutralse
would
neutralse
any
bond
base. C
H
any
acd
O
5
from
strong
acd
and
the
NO
added.
, from
the
be
Ntrc
salt
3
functon
as
a
whch
weakens
the
the O–H
n
the
water
molecules
thus
enablng
protons
to
the
3
acd
extracted
does
. Ths
results
n
acd
behaour
as
s
exhbted
a
on,
base
3
conjugate
3
salt
whch forms
9.4
b
b
densty)
by
a
pH
of
5.
not
One
of
the
possble
equlbra
s
represented
by
the
base.
equaton:
–5
ii
K
of CH
a
COOH
=
1.8
×
10
–3
mol dm
3
3+
[Al(H
–3
[CH
COOH]
=
1.03 mol dm
O)
2
(aq)
+
H
O(l)
–3
COO
]
=
→
2
2+
[Al(H
[CH
]
6
,
3
0.756 mol dm
O)
2
OH]
5
(aq)
+
+
H
O
(aq)
3
,
3
∴pH
=
4.61
181
Answers to
c
Sodum
revision questions
reacts
explosely
wth
water
producng
10
a
hydrogen:
As
the
group
s
temperatures
2Na(s)
+
2H
O(l)
→
2NaOH
(aq)
+
H
2
Carbonates,
(g)
descended,
of
the
MCO
,
the
decomposton
carbonates
ncrease.
decompose
as
ths:
3
2
MCO
(s)
→
MO(s)
+ CO
3
Alumnum
s
protected
by
an
oxde lm
and
does
The followng
react
wth
water,
howeer
at
eleated
(g)
2
not
ponts
are
useful
n
explanng
the
trend:
temperatures
2–
The
steam
wll
conert
the
metal
to
ts
smaller
2Al(s)
+
3H
O(g)
→ Al
2
has
a
yellow-green
contans
a
low
soluton,
mxture
Cl
(g)
+
H
2
8
a
The
O
2
solublty
of
n
called
(s)
→
+
the
oxde
on, O
,
allows
t
to
(aq)
+
Hence
water,
and
a
attached
to
the
metal
be
caton.
makng
the
crystal
structure
more
thermally
stable.
whch
chlorc(i)
HClO
strongly
(g)
2
and forms
chlorne
HCl
3H
3
water
hydrochlorc
O(l)
of
oxde:
more
Chlorne
sze
The
acds:
resultant
further
(aq)
larger
strengthens
The
tendency for
wll
decrease
charge
the
these
densty
bondng
of
to
the
the
carbonates
oxde
metal
to form
on
on.
the
oxde
2
correct
graph
should
be
drawn from
the
down
the
group
and
therefore
these
gen
carbonates
become
more
resstant
to
decomposton
coordnates.
as
b
i
Rse
n
densty from
sodum
to
the
group
s
descended.
alumnum:
Or
Structure:
gant
metallc
structures
Large
Bonding:
metallc
bondng
–
atomc
whle
charge
ncreases
charge
resultng
the
large
metallc
bonds
leadng
to
close
packng
II
metals
seen
as
a
steady
ncrease
n
Decrease
n
densty
between
sulphur
and
sulphur
possesses
structure. Chlorne
exsts
as
a
sold
polarsaton
the
decrease
as
the
group
s
catons
of
descended,
to
ther
ablty
produce
to
carbon
polarse
small,
doxde
the
and
larger
the
carbonate
oxde
results
n
the
hgher
on
thermal
stablty
of
coalent
smple
the
carbonates
i
Solublty
lower
down
the
group.
coalent
b
gaseous
of
chlorne:
decreases. Ths
Structure:
allow for
densty
densty.
on
ii
catons
charge
of
therefore
atoms
of
anons. The
n
Group
stronger
denstes
radus
of
decreases
of
the Group
II
sulphates
decrease
as
the
molecules.
group
Bonding:
sulphur’s
sold
structure
coalent
bondng
whle
descended.
ndcates
ii
stronger
s
only
The
solublty
depends
on
the
enthalpy
change
of
weak
soluton,
ΔH
.
soln
ntermolecular an
der Waals forces
hold
the
By
applyng
Hess’s
law
;
ΔH
.
=
ΔH
soln
gaseous
chlorne
molecules
together. Ths
results
Both
ΔH
and
ΔH
hyd
the
sgncant
reducton
n
–
ΔH
hyd
latt
n
decrease
as
the
group
s
latt
densty.
descended,
howeer
the
decrease
by
ΔH
s
more
hyd
9
a
i
All
tetrahaldes,
MCl
,
exst
as
coalently
bonded,
4
sgncant
than
that
of
ΔH
(ΔH
latt
smple
tetrahedral
molecules
wth
weak an
ntermolecular
bonds,
hence
on
the
der
sze
Waals
depends
hyd
of
the
caton). Therefore
as
the
group
s
ther olatlty.
descended
ΔH
becomes
more
poste
resultng
soln
The
symmetry
of
the
tetrahedral
structure
leads
to
n
a
net
dpole
moment
of
zero,
thus
ther
decrease
solublty
of
the
sulphates.
non-polar
c
i
calcum
oxde:
nclude
reducng
acdty
of
sols,
nature.
paper
ii
The
reactty
of
the
tetrahaldes
ncrease
as
s
descended
–
the
elements
to
metallc,
and
hence
become
more
magnesum
oxde:
unstable,
ndgeston),
thus
of
hydrogen
chlorde
calcum
gas
along
wth
the
oxde
of
oxdaton
state
4
(CCl
carbonate:
+
2H
4
+
d
In
→ SO
+
O
→
i
the
+
+
4HCl
+2,
4
thus CO
readly
state
s
oxdsed
more
to CO
stable
.
the
better
reducng
glass,
producton
reducng
acdty
n
of
sols
should
react ery gorously
wth
water,
13.
Radum
chlorde
–
soluble,
radum
carbonate
nsoluble
Radum
ntrate(v)
would
be
most
dfcult
to
s
decompose.
agent.
11
Slcon(iv)
the
than
Hence CO
2
c
(blast furnace),
Radum
–
oxdaton
s
n
(g)
2
iii
the
(concrete),
housng),
(g)
ii
PbO
2
carbon
4HCl
2
2H
4
b
O
2
PbCl
cement,
(road,
unreacte).
pH
SCl
(relee
cables
s
4
beng
medcaton
are formed
steel
exceptonal,
laxate,
nsulaton for
wth
constructon
water, fumes
nsectcdes
these
iii
compounds
hdes,
change from
acd
non-metallc
de-harng
the
ii
group
manufacture,
a
Meltng
ponts
of Group VII
ncrease
as
group
s
oxde:
descended.
Structure:
macromolecule
wth
each
slcon
atom
Structure: Changes from
surrounded
by four
others
coalent
Bonding:
a
Very
strong
tetrahedral
aboe
molecules
coalent
bonds
wth
slcon
atom
Intermolecular forces
become
stronger
as
enronment.
group
The
coalent
solds.
Bonding:
n
smple
(damond-lke).
propertes
lead
to ery
hgh
meltng
s
descended,
–
sze
of
molecules
ncrease.
pont;
b
i
Br
(l)
added
to
KI
(aq)
–
brown
soluton formed.
2
the
absence
of
any
unpared
electrons
account for
ts
Br
(l)
added to
KCl
(aq)
–
reddsh
brown
colour.
2
nsulatng
d
propertes.
Oxides
Reactions
With
acid
With
ii
Redox
reacton
iii
Iodne
dsplaced
odde
ons.
by
bromne,
bromne
–
2I
–
(aq)
+
Br
(l)
→
I
2
2+
Germanum(ii)
oxde
(aq)
Ge
oxdses
alkali
(aq)
+
2Br
(aq)
2
2–
GeO
(aq)
c
2
i
P
=
ii
Q
MnO
2
2+
Tn(ii)
oxde
Sn
2–
(aq)
SnO
= Cl
2
(aq)
2
iii
MnO
iv
Cl
+
4HCl
→
MnCl
v
Ox.
vi
Dspropotonaton
vii
2HClO(aq)
2
2+
Lead(iv)
oxde
Pb
(aq)
+ Cl
2
+
3
(aq)
(g)
+
H
2
O(l)
→
HClO(aq)
O
2
+
HCl(aq)
2
state
+1
→
2HCl(aq)
–1
+ O
(g)
2
182
2H
2
2–
PbO
to
Glossary
Born–Haber
cycle
An
enthalpy
Dipole
cycle
A
separaton
of
charge
n
a
A
used
Acid
A
proton
Acid–base
(hydrogen
indicator
changes
colour
acd–base
A
when
ttraton
Acid dissociation
on)
donor.
substance
the
pH
changes
constant,
Boyle’s
whch
n
an
The
equlbrum
partcles
to
allow
Alkali
A
constant for
energy
must
a
The
hae
reacton
soluble
a
weak
mnmum
when
to
they
law
nersely
lattce
The olume
of
proportonal
proton
donor. A
base
a
to
gas
ts
An
s
molecule. One
energy.
a
has
s
pressure.
acd
s
acd.
energy
collde
Buffer
solution
alkals
A
soluton
changes
are
n
added
to
when
the
acds
example,
of
the
are
allotropes
of
The
wth
an
acd
as
wthout
change
well
whch
A
postely
charged
as
The olume
negately
charged
The
same
of
atoms
chemcal
A
redox
type
oxdsed
The
nto
of
and
reacton
atom
n
n
a
reduced
breakng
ons,
up
of
especally
dssocatng
a
gas
a
referrng
nto
H
and A
wth
splitting
proportonal
to
energy
system
A
system
n
splttng
hgher
and
of
lower
leel,
lgands
caused
energy
s
not
around
a
by
the
central
presence
transton
whch
element
or
a
kelns.
on.
number
nto
the
an
n
The
s
d-orbtals
matter
(Z)
a
on.
of
of
number
s
acds
d-orbital
law
can
Closed
A
n
ons.
alkal.
Atomic
reacton
group
+
temperature
Anion
A
the
to
atoms decay.
oxde
and
charge.
the
to
drectly
react
ncreases
nucle
Charles’
An
reacton
charge
tself.
He
radoacte
the
molecule
Cation
emtted when
Amphoteric oxide
that
carbon.
2+
Alpha radiation (α)
substance
and
of
or
another
molecule
negate
reaction
atom
Dissociation
A
poste
the
smultaneously.
same
damond
of
partal
Disproportionation
speces
rate
graphte
one
whch
soluton.
occur.
Catalyst
For
a
end
reacton.
that
pH
base.
Dfferent forms
element.
end
where
C
Allotrope
partal
replaces
mnmses
or
a
other
Displacement
a
proton
acceptor.
a
Activation
calculate
Brønsted–Lowry theory
rapdly.
K
to
ganed
on.
or
of
Dot
and
cross diagram
A
dagram
lost.
protons
n
the
nucleus
of
the
atom
of
space
showng
Common
Atomic orbital
A
regon
the
nucleus
where
there
s
probablty
of ndng
one
or
solublty
s,
p,
types
d
of
electron. Orbtals
salt
Half
orgn
on
n
the
the
same
nucle
type
n
a
of
of
the
compound
outer
showng
of
the
electrons
by
dots
or
common.
Dynamic
ion
An
on
contanng
a
(equilibrium)
element
bonded
to
an
products
to
reactants
as
are
well
as
bonds.
two
reactants
A
reacton,
lgands
conerted
co-ordnate
In
central
the
by
between
of
a
has
equlbrum
radius
n
soluble
whch
can
Compound
atoms
sparngly
or f.
radius/covalent
dstance
a
another
electrons
crosses.
transton
Atomic
of
addng
arrangement
n
the
by
Complex
be
reducton
two
an
specc
The
a
salt
good
effect
shell
the
outsde
ion
the
substance
to
products.
contanng
coalent
atoms from
two
or
more
dfferent
bond.
E
elements
Avogadro
constant
The
number
bonded
of
Condensation
atoms,
ons
or
molecules
atoms,
ons
or
molecules.
n
a
mole
Conjugate
law
Equal olumes
of
the
same
number
change
n
state
when
changes
pair
An
to
a
acd
lqud.
Electrical
conductivity
and
substance
base
on
metals
dffer
alence
to
The
and
conduct
ablty
graphte
electrcty.
of
a
In
delocalzed
gases
each
contan
The
of
a apour
Avogadro’s
together.
sde
of
an
equaton
whch
electrons
are
responsble for
of
+
from
each
other
by
Co-ordinate
bond
A
a
H
on.
electrcal
conducton.
In
molten
and
molecules.
where
both
by
same
the
coalent
electrons
bond
are
aqueous
proded
B
Base
A
proton
(hydrogen
on)
the
radiation
(β)
Electrons
gen
the
nucleus
when
a
Bidentate
the
number
The
number
bonds formed
central
Lgands
whch
can form
a
of
transton
by
lgands
element
on
complex.
electrode
co-ordnate
bonds
wth
the
Electron
afnity
afnty
element
bond
the
Boltzmann distribution
the
the fracton
of
one
(single)
sharng
curve
A
mole
A
bond formed
partcular
partcles
energes
of
electrons
mole
of
gaseous
two
of
a
par
of
electrons
atoms.
aganst
radius
See
atomc
radus.
mole
of
gaseous
partcular
off
the
humped
of
energy
a
Electronic
Dative
covalent
bond
mole
of
needed
partcular
gaseous
n
molecules
(under
Degenerate orbitals
the
same
energy
Delocalised
condtons).
polarisation
the
A
bondng
coalent
Atomc
orbtals
equally
due
bond
electrons
to
of
a
bond
to
tself.
of
The
electrons
e.g.
1s
are
not
dfferences
2
2s
n
shells
and
5
2p
are free
to
seen
n
a
more
adjacent
n
Diatomic
A
seres
of
spectroscope
brght
due
to
leel.
electrons
moe
spectrum
at
Electrons
between
electrons fallng from
hgher
to
whch
three
energy
leels.
or
The formula
showng
atoms.
the
shared
to
conguration
Empirical formula
where
ablty
coalent
bond.
to
bond
lower
Bond
a
2
excted
standard
ons.
See Co-ordnate
lnes
one
one
one
energy.
gradually.
The
mole
to
to form
cure
Emission
a
n
electrons
sub-shells
energy
1+
The
atom
arrangement
break
when
added
wth
D
characterstc
‘tals’
s
atoms
Electronegativity
attract
Bond
electron
change
graph
Covalent
whch
The rst
enthalpy
on.
between
a
s
central
by
transton
has
See Standard
potental.
two
Covalent
It
conducton.
potential
decays.
n
of
are
radoacte
to
sotope
responsble for
Electrode
ons
off
co-ordnate
from
moble
acceptor.
Co-ordination
Beta
atom.
salts,
Elements
or
smplest
rato
of
atoms
n
a
compounds
molecule.
electronegatty. Shown
by
an
arrow
whch
hae
molecules
contanng
only
Endothermic
+
and
the
sgns
δ
and
δ
two
atoms.
For
example,
energy from
carbon
A
reacton
whch
absorbs
chlorne,
the
surroundngs.
monoxde.
183
Glossary
Energy
level
The
regons
Hybridisation
at arous
The
combnaton
of
M
dstances from
electrons
the
hae
nucleus
dened
where
amounts
the
mxed
of
prole diagram
Shows
enthalpy
change from
products
along
Enthalpy
n
a
chemcal
a
the
change
reacton
The
pathway.
energy transferred
Shows
chemcal
a
to
H
A
atom
,
The
compound
calculated from
A
K
c
present
gas
A
K
the
gas
whch
obeys
the
gas
law
under
equation
and
n xed
of
pressure
and
number
products
and
back
Mole
acd–base
ndcator
reacton
whch
to
the
Half
of
dstance
two
atoms
whose
ncreases
n
n
a
wth
electrcal
an
temperature.
The
n
mass
of
a
mole
of
a
grams.
The
amount
of
number
substance
of
there
are
atoms
carbon-12
speced
hang
bond
The
electrostatc
exactly
partcles
12 g
of
oppostely
charged
each
type
of
atom
Molecular orbital
An
equaton
those
n
ons
and
molecules
are
The
the
by
atoms.
In
a
mxture of
molecules
equlbrum
(or
constant for
orbtal formed
orbtal’s from
shown.
surroundngs.
water
An
atomc
Mole fraction
product of
number
molecule.
takng
dfferent
reacton
the
a
whch
out
the
n
ons.
reactons
equation
Shows
attracton
of
n
n
sotope.
Molecular formula
Ionic
the
nucle
molecules.
are
part
the
electrons.
substance
mass
same
the
ges
metal
Weak forces
n
only
energy
A
combnng
A
a
to
ndcator.
equal.
Exothermic
n
ons
and
the
Ionic
metal
nRT.
See
as
reacton
of
of
of
concentratons. The
the forward
the
number
nucleus.
condtons.
relatonshp
between
are
radius
a
bond formed
delocalsed
substance
Ionic
rate
all
concentratons
A
A
total
n
Relates olume,
Molar
PV =
The
attracton
ncrease
Ideal
redox
reaction
bond
the
of
(A)
protons
law
equlbrum.
reactants
present
Boyle’s
reacton.
to
by
Metallic
water.
equlbrum
A
Metallic
sea
a
or
between
whch
wth
wth
atom.
+
metal.
Intermolecular forces
Equilibrium
of
N
Metalloid
p
at
reacton
or
number
neutrons
conductty
constant,
expression
or
molecules
F, O
proceed.
moles.
K
a
I
Indicator
lnkng
to
breakdown
by
temperature,
a
ntermolecuar
between
routes
p
expresson for
weak
catalyst.
constant
Equilibrium
wth
ΔH
c
K
orbtal
Mass
A
bonded
and Charles’
Equilibrium
an
between
alternate
reacton to
proten
bond
Hydrolysis
Ideal
Enzyme
to form
character.
force formed
the
reactants
reacton,
Enthalpy cycle
for
orbtals
Hydrogen
energy.
Energy
atomc
onsaton
of
atoms), the
number of
moles of
a
water,
partcular
F
molecule dded
by the total
K
w
number of
Feasibility
The
lkelhood
or
not
of
Ionisation
a
energy
The rst
reacton
occurrng
as
predcted
by
E
moles of
Monodentate
energy
s
the
energy
needed
to
1
alues.
mole
of
electrons from
1
mole
placed n a magnetc eld, lnes tself up
gaseous
atoms
(under
molecules.
whch form
bond
wth
the
one
central
of
transton
A substance whch, when
Lgands
remoe
co-ordnate
Ferromagnetic
all the
onsaton
o
element
on.
standard
condtons).
N
wth the eld and retans ts magnetsm
Ion
polarisation
The
dstorton
of
the
Non-polar
electron
when the magnetc eld s remoed.
cloud
of
an
anon
by
a
A
Isotopes
hang
radiation
charged)
(γ)
Radaton
Atoms
of
dfferent
the
of
wth
no
charge.
caton.
Nucleon
G
Gamma
hghly
molecule
(small
separaton
and/or
same
numbers
number
See
Mass
number.
element
of
protons.
gen
O
off
an
sotope
decays
and
a
proton
s
K
conerted
an
Gas
to
a
neutron
by
capturng
Kinetic theory
electron.
constant
gas
Giant
The
equaton,
ionic
Order
constant
n
the
deal
R
structure
A
gant
gases
are
those
n
n
Partcles
constant
n
lquds
and
moement
and
solds brate.
structure
(with
respect to
substance)
the
power
In
to
a
rate
whch
rased. The
oerall
s
the
these
of
particular
equaton
the
s
sum
a
ths
s
concentraton
order
of
reacton
powers.
L
made
Giant
up
of
poste
structure
dmensonal
metallc
or
and
These
hae
network
onc
negate
of
a
Oxidation
ons’.
Lattice
three-
of
coalent,
A
regular
atoms,
ons
H
The
half
energy
when
1
The
mole
of
an
an
element
or
on
oxdaton
state
another
oxdaton
and
an
on
An
the
an
ts
onc
gaseous
ons
s
(under
a
oxdaton
redox
principle
affectng
When
the
equaton
or
are
reducton
equlbrum
balanced
changed,
shfts
to
the
concentraton
the
decrease
to
taken for
the
An
on
of
a
lmtng
lone
or
molecule
half
ts
wth
The
co-ordnate
one
s
enthalpy
change for
or
whch
ndependent
the
reacton
can
bonds
element
wth
a
central
on.
of
the
takes
exchange
lgand
The
replacement
of
by
another
n
a
transton
complex.
A
par
of
electrons
n
the
outer
place.
shell
of
an
atom
whch
n
s
not
bonded.
a
A
wth
ts
magnetsm
eld
of
s.
ncreases
the
another
s
substance
the eld
whch,
but
when
does
the
lnes
not
when
tself
retan
magnetc
remoed.
pressure,
nddual
gas
p
n
The
a
pressure
mxture
of
of
an
gases.
Permanent dipole–dipole
Intermolecular force
molecules
whch
dpoles.
+
pH
–log
[H
10
184
t
that
magnetc eld,
up
forces
route
pair
Paramagnetic
Partial
a
Lone
by
whch
elements.
element
reacton
electrons
ntal alue.
The Group VII
law
of
reactant
one
Hess’s
par
amount
Ligand
Halogens
(or
whch
reduced
P
placed
transton
to
number)
poston
mnmse
part
form
or
or
reactant
electrons
number
of
wth
more
tme
oxdsed
A
A
atom
showng
Ligand
The
how
agent
remoes
ncrease
reactant.
the
poston
electrons.
Half-life
Oxidising
partcular
oxdaton
change.
of
a
descrbes
change
compound
state/number
to
or
state.
equaton
of
ether
Oxidation
electrons
n
equlbrum
equation
of
enthalpy
condtons)
condtons
Half
three
of
state.
n
Le Chatelier’s
one
n
Loss
oxdaton
electrochemcal
standard
contanng
molecules
n
gen
formed from
cell
cell,
or
arrangement
dmensons.
‘bonds’.
Lattice
Half
repeatng
]
hae
between
permanent
Glossary
pH
range
about
(of
2
indicator)
pH
ndcators
unts
The
oer
change
range
whch
of
partcular
of
the
carbon-12
exactly
Relative
colour.
12
sotope
has
a
mass
of
unts.
molecular
mass
(M
)
The
mass
Standard
enthalpy
change of
hydration
The
1
speced
mole
of
a
enthalpy
change
gaseous
when
on
s
r
pi
bond
A
coalent
sdeways
pK
–log
a
oerlap
–log
w
s
A
a
of
molecule
whch
an
sotope
12
measured
atom
has
a
of
the
mass
on
a
scale
on
ons
carbon-12
of
conerted
(under
Standard
exactly
to
an
aqueous
standard
enthalpy
soluton
of
condtons).
unts.
change of
neutralisation
The
when
water
enthalpy
change
w
or
molecule
separaton
of
Principal quantum
outsde
whch
orbtals.
a
a
bond
space
atomc
of
K
10
Polar
of
nolng
K
10
pK
bond
may
the
up
whch
there
A
regon
nucleus
to
a
a
Resonance
of
charge.
level
contan
electrons
n
of
an
certan
a
hybrid
compound
between
of
atom
number
maxmum
two
Reversible
The
s
or
products
the
reactants.
structure
somewhere
the
n
can
A
be
1
reacton
n
changed
whch
back
mole
addton
(under
more forms.
reaction
the
actual
Standard
the
of
an
standard
The
speced
standard
S
Q
Standard
Salt
Quantum
Energy
of xed alues
bridge
An
nert
support
soaked
acd
to
an
alkal
change of
enthalpy
molar
stochometrc
number.
s formed from
condtons).
enthalpy
reaction
to
of
change
amounts
equaton
when
n
react
the
(under
condtons).
enthalpy
change of
n
solution
The
1
a
enthalpy
change
when
only,
KNO
used
to
make
electrcal
contact
3
whch
can
be
absorbed
or
emtted
mole
between
two
half
substance
s
dssoled
n
cells.
excess
an
of
by
water
(under
standard
atom.
Semi-conductor
Substances
whch
hae
condtons).
low
electrcal
conductty
at
room
Standard
R
temperature
Radioactive decay
an
atomc
The
nucleus,
breakng
emttng
up
of
partcles
conductty
Shielding
but
as
ncrease
temperature
(screening)
The
hydrogen
cell
under
has
a
standard
process.
nner
shells
of
electrons
half
to
whch
Pt
electrode
dppng
nto
of
–3
the
A
condtons
ncreases.
ablty
+
1 mol dm
n
electrode
n
reduce
H
ons
n
equlbrum
wth
the
H
gas.
2
Radioactive
isotopes
Isotope
of
effecte
nuclear
charge
on
other
Stoichiometry
elements
whch
hae
nucle
that
break
electrons,
especally
those
n
the
reactant
down
spontaneously
and
constant
The
proportonalty
Sigma
bonds
Bonds formed
by
n
the
rate
equaton.
end-on
oerlap
of
atomc
of
n
a
step
The
slowest
step
Solubility
product
(K
)
equaton.
acid/base
Acds
or
bases
whch
orbtals.
are
Rate determining
rato
shown
the
Strong
constant
mole
products
shell.
balanced
Rate
The
outer
An
completely
onsed
n
soluton.
equlbrum
sp
Sublimation
n
a
reacton
mechansm.
expresson
showng
the
The
sold from
Rate
equation
An
equaton
showng
the
of
each
on
n
a
sparngly
soluble
between
the
soluton
rased
to
the
power
of
state
of
reactants
whch
relate
the
rate
of
reacton
and
the
rate
Spectator
ions
Ions
whch
play
no
constant.
n
The
amount
or
a
Regons
or
change
of
a
specc
constant
The
shell
or
reactant
per
expresson
second.
of
less
showng
reagents
the
noled
mechanism
A
seres
of
In
showng
the
stages
n
prncpal
hae
a
chemcal
n
the
reacton
contaner,
the
concentraton
and
the
ar
surroundng
the
lgand
mxture
as
well
as
exchange.
nstruments
steps
the
electrons
energy.
nclude
reacton
Reaction
wthn
where
equlbrum
solent
product
a
reacton.
Stability
these
concentraton
a
beng formed.
Surroundings
reaction
of
wthout
part
more
Rate of
or ce ersa
concentratons.
quantum
affect
gas
ther
Sub-shell
concentratons
a
salt
lqud
relatonshp
drect formaton
concentraton
a
Standard
cell
potential
The
dppng
nto
the
reacton
dfference
mxture.
reacton.
n
the
standard
electrode
potental
of
System
Real
gas
A
gas
whch
does
not
obey
the
two
half
The
reactants
and
products
n
a
cells.
chemcal
reacton.
5
gas
laws
and
low
especally
at
hgh
pressure
temperature.
Standard
and
conditions
temperature
Pressure
of
10
Pa
298 K.
T
Redox
reaction
A
reacton
where
Standard
electrode
potential
The
Transition
reducton
and
oxdaton
occur
at
the
electrode
potental
of
a
half
elements
whch forms
same
tme.
compared
wth
a
standard
agent
donates
A
reactant
electrons
oxdaton
(or
number)
of
that
stable
d
block
ons
wth
element
an
hydrogen
ncomplete
Reducing
A
cell
d
sub-shell.
electrode.
decreases
the
another
Standard
change
enthalpy
whch
change
takes
An
place
at
enthalpy
V
pressure
5
reactant.
of
10
Pa
and
temperature
298 K.
Valence
Reduction
Gan
of
electrons
or
decrease
Standard
enthalpy
shell
electron
(VSEPR) theory
n
oxdaton
state.
atomisation
The
enthalpy
atomic
mass
(A
)
The
weghted
when
1
mole
of
gaseous
atoms
mass
of
the
par
atoms
of
an
formed from
ts
repulsion
element
order
repulson
s
of
lone
par-lone
s
r
aerage
The
change
decreasng
Relative
pair
change of
>
lone-par
bond-par
>
bond-par
(under
bond-par.
element
measured
on
a
scale
on
whch
standard
condtons).
van der Waals forces
an
atom
of
the
carbon-12
sotope
has
Standard
enthalpy
attracton
a
mass
of
exactly
12
unts.
combustion
The
enthalpy
isotopic
proporton
or
abundance
percentage
of
The
when
a
1
mole
completely
of
a
substance
combusted
between
molecules
of
caused
change
by
Relative
Weak forces
change of
temporary
dpole-nduced
dpoles.
s
(under
W
partcular
sotope
n
a
mxture
of
standard
sotopes.
Relative
Standard
isotopic
mass
partcular
sotope
measured
on
a
The
mass
of
of
an
element
scale
on
whch
an
a
The
condtons).
enthalpy
enthalpy
substance
atom
(under
change of formation
change
when
s formed from
standard
1
ts
mole
of
a
Weak
are
acid/base
partally
Acds
onsed
or
n
bases
whch
soluton.
elements
condtons).
185
Index
Key
n
terms
the
are
n
bold
and
are
also
lsted
glossary.
beta
radiation
bidentate
blood
A
absolute
acid
104,
bond
equlbra
Brønsted–Lowry
dssocaton
pH
pH–ttraton
acid–base
acd–base
theory
indicators
ttratons
acd–conjugate
bond
bond
energies
2,
coalent
perod
114–15
bond
112–13
rato
119
)
energy
curve
62–3,
covalent
3
elements
108–11
Boyle’s
law
cycle
and
coalent
7
bonding
18–19
number
bonding
covalent
18–19
cross
conductty
99
lmt
co-ordination
68
electrcal
105
Process
co-ordinate
62
16–17
polarisation
see
pairs
conergence
80
energes
bondng
14,
Born–Haber
(k
Contact
7
dot
co-ordnate
constant
conjugate
117
bondng
108–11
112–13
43,
base
acid dissociation
104–5
106–7
cures
model
dssocaton
see also
104–19
constants
calculatons
solutons
atomc
59
conductty
Boltzmann distribution
53
141
acd–base
buffer
Bohr’s
temperature
condense
5
164
164,
dagrams
radius
166
14
see
structures,
16–19
atomic
gant
radius
25
137
D
22
70–3
Dalton’s
52
dative
atomc
theory
covalent
bond
2
18–19
a
acdc
buffers
activation
acte
ste
alzarn
alkali
buffer
99,
165
ons
carbon
104
rule
127
aqueous
sodum
171
hydroxde
emsson
spectra
number
(Z )
atomic orbitals
radius
3
17
4
160
2–3
142,
monoxde
cell
potentals
dot
125
onc
129
Charles’
law
53
chemcal
equatons
chemcal
equlbra
23,
energes
10–11
calculatons
94–5,
4–5
equilibrium
constant
(K
equilibrium
constant
(K
solubility
atomc
8–9
theory
atomsaton
change
chlordes
product
2
see
of
standard
cobalt
law
40
)
92–3
)
96–7
(K
)
100–1,
base
seres
change
common
104
basc
buffers
basc
dssocaton
complex
116–17
constant
(K
)
110
redox
structures
double
bond
ductle
25
18,
20
19
of
80
102–3
electrical
group
conductivity
IV
perod
3
elements
elements
redox
electron
standard
enthalpy
effect
ions
163,
reactons
103
164
166–7
2,
electron
cells
benzene
186
129
31
compounds
2
concentraton
80,
98,
128
120–9
chemcal
3,
126–7
6
decent
electron ow
128–9
change
124–5
70
molecules
electronegativity
electron
128–9
cells
22,
135
124–5
conguration
b
batteres
161
3
afnity
electronic
135
elements
reactons
electron
77
144–5
potentials
predctng
combuston
ion
90–91
E
electrode
162–3
see
16
16–19
(equilibrium)
electrochemcal
76
ons
155
163
bondng
electrc elds
91
theory
51
110–11
electrochemcal
combuston
7
155
165
coloured
Balmer
137
,
45
19
splitting
transton
173
colormetry
B
139
system
collson
40–1
134,
chromates
closed
14
constant
Avogadro’s
enthalpy
atomsaton
attracte forces
Avogadro
chlorne
138,
reaction
108–9
sp
orbtals
ttratons
90–103
p
sotopes
(VI)
cross diagrams
dynamic
35
c
onsaton
and
coalent
122–3
135
22
d-orbital
168–9
dagrams
149
elements
150
dissociation
cell
31
140
disproportionation
148
80
ceramcs
3
dsplayed formula
148
146,
21,
150
displacement
172
146,
electrons
elements
dchromate
dipole
equlbrum
6–7
perod
64
carbon
cells
136
elements
leel
6–7
2
structure
energy
II
diatomic
doxde
cations
8–9
elements
transton
atomc
group
carbon
catalysts
134
perod
129
144
carbonates
172–3
bonding
163
densty
160–1
calormetry
169
ammona
atomic
delocalised
C
calcum
aqueous
atomic
104–5
116–19
5
134
antclockwse
atomc
cells
co-ordinate
degenerate orbitals
halogens
amphoteric oxide
argon
solutions
button
ammonum
anions
see also
114
134
radiation
ammona
151
blue
Brønsted–Lowry theory
114
104
alumnum
water
bromothymol
80
81
yellow
allotropes
alpha
bromne
116
energy
transfer
electroplatng
48
117
9
17
Index
emission
spectra
endothermic
energetcs
bond
half
6–7
empirical formula
reactions
60
halde
energies
62–3,
changes
63,
64–5,
Hess’s
Hess’s
law
63,
law
lattice
hydraton
66–7
energy
68,
6–7
energy
prole diagrams
enthalpy
cycle
enzymes
81
equilibrium
equlbrum
66
94–5,
equilibrium
constant
(K
equilibrium
constant
(K
equilibrium
expression
equilibrium
reactions
108–9
)
92–3
)
96–7
152–3
30
standard
enthalpy
change
bonding
23
buffers
hydrogen
electrode
120
hydrogen
haldes
84,
117
152–3
86,
on
113,
145
104
ethene
30,
excted
exothermic
expanded
92
90–1
113
gas
54–57
ideal
gas
equation
nsoluble
60
octet
55–7
processes
ame
test
meltng
pont
group
II
group
IV
odne
ion
127
159
168
(A)
14,
effect
(common)
ionic
bonding
ionic
equations
ionic
product of
45
58
144
150
bonding
radius
25,
140
elements
21
140
135
136
30
methyl
orange
methyl
red
114
114
mass
35
35
concept
34–45
law
calculatons
40–1
36–7
empirical formula
103
38–9
molecular formulae
20–1
23,
4
elements
Avogadro’s
ttratons
159
metallic
mole
22–3
26
72
3,
metallic
mole
intermolecular forces
17
F
ferromagnetic
59
molar
99
100
odne–thosulphate
feasibility
number
metals
114
ndustral
6
theory
25
meltng
methane
ideal
indicator
63
state
malleable
metalloids
I
p
pont
oxde
halogens
c
equalence
134
magnesum
mass
30
hydrogencarbonate
hydroxonum
magnesum
magnetc elds
hydraton
hydrolysis
90
calculatons
Brønsted–Lowry
M
150–1,
see
hydrogen
61
60
see
7
173
66–7
hydrocarbons
levels
enthalpy
of
70–3
energy
seres
154
hybridisation
68–9
theory
Lyman
152–3,
halogens
68
60–1,
Lowry
121
85
ons
haldes
60–73
enthalpy
equation
half-life
38
34,
water
soluton
37
(K
)
107
39
concentratons
molecular formula
38,
42–3
41
w
forces
of
attracton
formaton
see
14–15
standard
onc
enthalpy
of formaton
fuel
cells
change
3
129
ionic
G
gas
radiation
constant
gases
55
knetc
laws
isotopes
theory
ionic
giant
structures
ground
structure
14,
21,
24
24–5,
134
25
analyss
state
103
II
IV
group
IV
elements
group
IV
oxdes
group
IV
tetrachlordes
elements
catons
of
of
non-polar
matter
standard
nuclear
55–7
172
molecules
28
136
22,
charge
nucleon
58–9
enthalpy
neutralsaton
3
142,
non-metals
equaton
see
non-lnear, V-shaped
52–9
52–3
gas
states
23,
27
10
number
see
L
lattice
144–5
146–7
150–1
Process
99
lattices
lead
144–5
energy
21,
ntrate
mass
number
lnear
122–3
orbtals
atomic
degenerate
principle
exchange
163,
98–9,
165
14,
pairs
128
164–5
see also
58
order
16,
29
91
163
hybrdsaton
order
equlbrum
8–9
molecular
28
lquds
lone
O
70–3
24
Le Chatelier’s
ligand
68,
173
lqud-apour
165
cells
ntrates
54–5
laws
ligands
half
neutralsaton
change
140–1
148–9
H
haem
N
35
6
group
group VII
gases
deal
18
168–73
23,
4–5
kinetic theory
gas
164
bonds
neutron
group
Haber
tests
ions
K
giant
grametrc
multple
28–31
96
monodentate
160
15
14
shapes
mole fraction
54–5
25
solds
molecules,
24
138
52–3
structures
26
162–3
spectator
52–3
53
coalent
graphte
73,
qualtate
heatng
gant
ons
21,
10
elements
polarisation
molecular
160
giant
energy
lattces
molecular orbital
136
elements
structure,
transton
ion
5
14
compressng
gas
elements
transton
onsaton
gamma
molecular
rad
perod
30
15
d-orbital
splitting
82
of
reacton
82–3,
oxidation
48
oxidation
numbers
84–5
148
187
Index
oxidation
states
coloured
perod
ons
3
oxdaton
48,
122,
146-148
elements
states
redox
ttratons
reducing
162–3
reduction
137
(oxdaton
numbers)
state
45
agents
51,
symbols
Stock
154
35
nomenclature
stoichiometry
48
mass
(A
strong
relative
atomic
relative
isotopic
abundance
relative
isotopic
mass
relative
molecular
)
4
41,
49
44
acids/bases
105,
113,
115
r
48–9
oxdes
138,
139,
146–7
(M
)
structure
4
and
covalent
4
bondng
bonds
14–31
14,
16–17
r
oxidising
oxygen
agents
137
,
51
141
resonance
reversible
P
paramagnetic
partial
perod
( p)
elements
phenolphthalen
salt
pH
range
pH
scale
polar
117
cures
112–13
22
equlbrum
98–9
potassum
dchromate
(vi)
potassum
manganate
(vii)
dfference
precptaton
pressure
52,
102,
80,
167
45,
167
121
134,
doxde
matter
of
sulphtes
154,
sulphur
173
compounds
sodum
carbonate
sodum
hydroxde
sodum
thosulphate
3
153
8
172
acd
area
system
99,
154
80
surroundings
169
17
28
partcles
134
surface
134
21
14–15
172
sulphurc
26–7
134
22–3
81
sulphates
149
sodum
60
60
152
T
14
group
98
states
substrate
molecular
24–5,
20–1
bonding
sub-shells
10
144
ntrate
solds
bonding
sub-atomc
144
15
slcon
solublty
170
(screening)
bonds
smple
6
135,
14,
metallic
VSEPR theory
121
slcon
sler
15
potental
ionic
2
sublimation
shielding
114
of
theory
8
sigma
constant
poston
bridge
semi-conductor
shells
106–7
bonds
Planck’s
114
134
pH-ttraton
pi
23
structures
22
intermolecular forces
90
S
134–9
buffers
phosphorus
atomc
96
permanent dipole–dipole forces
phosphate
giant
31
reactions
Rutherford’s
electronegativity
56–7
159
pressure
3
hybrid
mass
27
,
II
solubility
100
temperature
sulphates
product
143
(K
)
80,
tetrachlordes
100–1,
102–3
99
145
tetrahaldes
145
tetrahedron
28
sp
principal quantum
proten
proton
buffers
level
6
soluton see
117
soluton
shape,
molecules
28
concentratons
solutons
Q
quantum
analyss
103,
stability
168
radioactive decay
82,
equations
rates of
order
reacton
of
of
theory
see
enthalpy
redox
reactions
188
ons
standard
50–1
48–51,
166–7
124–5
starch
states
redox
60
potential
change
120–1
60–1,
43,
63,
112–13
ons
158–61,
167
163
164,
exchange
reactons
anadum
trgonal
44–5,
elements
ons
142
153
166
164–5
167
162
planar
28
V
change of
valence
change of
anadum
change of
change of
162
23,
26
58
apour
pressure
VSEPR
see valence
91
shell
electron
pair
V-shaped
molecules
water
27
,
28
change of
W
64
enthalpy
64,
repulsion
28
repulsion
64
enthalpy
pair
(VSEPR theory)
aporsaton
68
enthalpy
electron
van der Waals forces
66–7
enthalpy
shell
theory
change of
change of
69
hydrogen
weak
electrode
152
of
lgand
64
enthalpy
solution
54
stablty
complex
123
70
enthalpy
reaction
standard
51
equatons
complex
enthalpy
neutralisation
standard
151
seres
gases
redox
86–7
63
halogens
reactty
standard
standard
76
mechanisms
reactty
enthalpy
hydration
knetcs
reaction
change
thermal
68–9
formation
84–5
reacton
reacton
real
standard
standard
combustion
82–3,
decomposton
coloured
165
conditions
standard
80
thermal
transition
potential
electrode
standard
76–87
reacton
cell
atomisation
86
44
163
standard
standard
step
82–3
reaction
collson
5
42–3,
35
standard
64–5,
84
rate determining
rate
5
isotopes
constant
23,
constant
standard
R
rate
of
ttratons
ions
splitting, d-orbital
6
radioactive
change
169
spectator
qualtate
enthalpy
neutralsaton
3
pyramdal
standard
matter
14–15,
58–9
120–121
23,
107
,
137
,
acids/bases
141,
105,
145
108–9,
113,
115
Chemistry
for
Unit
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