Oxford Resources for IB DP Chemistry: Course Book 9781382016469

Table of contents :
Cover
Oxford Resources for IB DP Chemistry: Course Book
Copyright
Contents
Introduction
How to use this book
Structure 1. Models of the particulate nature of matter
Structure 1.1 Introduction to the particulate nature of matter
Structure 1.2 The nuclear atom
Structure 1.3 Electron configurations
Structure 1.4 Counting particles by mass: The mole
Structure 1.5 Ideal gases
Structure 2. Models of bonding and structure
Structure 2.1 The ionic model
Structure 2.2 The covalent model
Structure 2.3 The metallic model
Structure 2.4 Fram models to materials
Structure 3. Classification of matter
Structure 3.1 The periodic table: Classification of elements
Structure 3.2 Functional groups
Tools for chemistry
Tool 1: Experimental techniques
Tool 2: Technology
Tool 3: Mathematics
Reactivity 1. What drives chemicaI reactions?
Reactivity 1.1 Measuring enthalpy changes
Reactivity 1.2 Energy cycles in reactions
Reactivity 1.3 Energy from fuels
Reactivity 1.4 Entropy and spontaneity (AHL)
Reactivity 2. How much, how fast and how far?
Reactivity 2.1 How much? The amount of chemical change
Reactivity 2.2 How fast? The rate of chemical change
Reactivity 2.3 How far? The extent of chemical change
Reactivity 3. What are the mechanisms of chemical change?
Reactivity 3.1 Proton transfer reactions
Reactivity 3.2 Electron transfer reactions
Reactivity 3.3 Electron sharing reactions
Reactivity 3.4 Electron-pair sharing reactions
Cross-topic exam-style questions
The inquiry process
The internal assessment (IA)
Index
Periodic TabIe
Cover back

Citation preview

Oxford

Resources

for

IB

Diploma Programme

2 0 2 3

E D I T I O N

C H E M I S T RY

CO U R S E

CO M PA N I O N

Sergey Bylikin

Gary Horner

Elisa Jimenez Grant

D avid Tarcy

Oxford

Resources

for

IB

Diploma Programme

2 0 2 3

E D I T I O N

C H E M I ST RY

CO U R S E

CO M PA N I O N

Sergey Bylikin

Gary Horner

Elisa Jimenez Grant

D avid Tarcy

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Acknowledgements

Chem.

Educ.

2016,

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I would

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Contents

Structure 1. Models of the particulate nature of matter

2

Structure 1.1

Structure 1.2

Structure 1.3

Structure 1.4

Structure 1.5

Structure 2. Models of bonding and structure

94

Structure 2.1

Structure 2.2

Structure 2.3

Structure 2.4

Structure 3. Classic ation of matter

228

Structure 3.1

Structure 3.2

Tools for chemistry

308

Tool 1:

Tool 2:

Tool 3:

Reactivity 1. What drives chemic al reactions?

386

Reactivity 1.1

Reactivity 1.2

Reactivity 1.3

Reactivity 1.4

Reactivity 2. How much, how fast and how far?

460

Reactivity 2.1

Reactivity 2.2

Reactivity 2.3

Reactivity 3. What are the mechanisms of chemic al change?

536

Reactivity 3.1

Reactivity 3.2

Reactivity 3.3

Reactivity 3.4

Cross-topic exam-style questions

652

The inquiry process

655

(authored by Maria Muñiz Valcárcel)

The internal assessment (IA)

(authored by Maria Muñiz Valcárcel)

668

Index

686

Periodic Table

708

Answers:

www.oxfordsecondary.com/ib-science-support

iii

Introduction

The diploma programme (DP) chemistry course is aimed at students in the 16

to 19 age group.

The curriculum

of the nature of science,

chemistry and

contexts.

seeks to develop

a conceptual understanding

working knowledge of fundamental principles of

practic al skills that

c an be applied in familiar and unfamiliar

As with all the components of

the DP,

this course fosters the IB learner

prole attributes (see page viii) in the members of

the school community.

Nature of science

Nature of science (NOS)

purposes and

is concerned with methods,

outcomes that

•

F alsication

Hypotheses c an be proved false using other

are specic to science.

NOS is a central theme that is present

across the

evidence,

entire course.

NOS features

denitely true.

throughout

You will nd

the book and

suggested

but

they c annot be proved to be

This has led

science throughout

are encouraged to come up

to paradigm shis in

history.

with further examples of your own as you work through •

Models

the programme. Scientists construct

NOS c an be organized

explanations of

into the following eleven

contain assumptions or unrealistic simplic ations,

aspects:

but •

models as simplied

their observations. Models oen

the aim of science is to increase the complexity

Observations and experiments

of

the model,

and

to reduce its limitations.

Sometimes the observations in experiments are

unexpected

and lead

to serendipitous results.

•

Theories

A theory is a broad •

explanation that

takes observed

Measurements

patterns and hypotheses and uses them to generate Measurements c an be qualitative or quantitative,

predictions. but

These predictions may conrm a

all data are prone to error. It is important to

theory (within observable limitations) or may falsify know the limitations of your data.

it.

•

Evidence

•

Science as a shared activity

Scientists learn to be sceptic al about their

Scientic activities are oen c arried out in observations and they require their knowledge to

collaboration,

such as peer review of work before

be fully supported by evidence.

public ation or agreement on a convention for clear

•

Patterns and trends

Recognition of a pattern or trend

communic ation.

forms an

• important part of

Global impact of science

the scientist’s work whatever the

Scientists are responsible to society for the science.

consequences of

•

Hypotheses

Patterns lead

environmental,

to a possible explanation. The

hypothesis is this provisional view and

further veric ation.

iv

it

requires

knowledge must

and

fairly.

their work,

whether ethic al,

economic or social. Scientic

be shared

with the public clearly

Syllabus structure

Topics are organized

below.

into two main concepts:

structure and reactivity.

This is shown in the syllabus roadmap

The skills in the study of chemistry are overarching experimental,

skills that

are integrated

experimental work,

into the course.

inquiries and

Chemistry is a practic al subject,

technologic al,

mathematic al and inquiry

so these skills will be developed

through

investigations.

Skills in the study of chemistry

Structure

Reactivity

Structure refers to the nature of matter

Reactivity refers to how and why

from

simple to more complex forms

chemic al reactions occur

Structure determines reactivity, which

in turn transforms structure

Structure 1.

Structure 1.1 — Introduction to

Reactivity 1.

Reactivity 1.1 — Measuring

Models of the

the particulate nature of matter

What

enthalpy changes

particulate nature

drives

chemic al Structure 1.2 — The nuclear atom

of matter

Reactivity 1.2 — Energy cycles in

reactions? reactions

Structure 1.3 — Electron

Reactivity 1.3 — Energy from fuels

congurations

Structure 1.4 — Counting

Reactivity 1.4 — Entropy and

particles by mass: The mole

spontaneity (Additional higher

level) Structure 1.5 — Ideal gases

Structure 2.

Structure 2.1 — The ionic model

Models of

bonding and

Reactivity 2.

Reactivity 2.1 — How much? The

How much,

amount

of chemic al change

how fast and Structure 2.2 — The covalent

Reactivity 2.2 — How fast? The

structure

how far? model

rate of chemic al change

Structure 2.3 — The metallic

Reactivity 2.3 — How far? The

model

extent

of chemic al change

Structure 2.4 — From models to

materials

Structure 3.

Structure 3.1 — The periodic

Reactivity 3.

Reactivity 3.1 — Proton transfer

Classic ation of

table:

What

reactions

Classic ation of elements

matter

are the

mechanisms Reactivity 3.2 — Electron transfer

of chemic al reactions

change? Structure 3.2 — Functional

Reactivity 3.3 — Electron sharing

groups:

reactions

Classic ation of organic

compounds Reactivity 3.4 — Electron-pair

sharing reactions

Chemistry concepts are thoroughly interlinked.

roadmap

above,

You are therefore encouraged

new and

help

For example,

“Structure determines reactivity,

as shown in the

which in turn transforms structure”.

to continuously reect on the connections between

prior knowledge as you progress through the course.

you explore those connections.

identify and

In assessment tasks,

apply the links between dierent

examples of DP-style exam

topics.

Linking questions will

you will be expected to

On page 652,

questions that link several dierent

there are three

topics in the course.

v

How to use this book

The

aim

of

this

development

book

and

understanding

Feature

aims,

boxes

by

to

through

and

for

develop

conceptual

opportunities

to

understanding, aid in skills

cement

knowledge and

practice.

sections

signposting

opportunities

is

provide

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relating

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designed to support these

ideas and concepts, as well as

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features:

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boxes

in

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Guiding questions margin

to Each topic begins with a guiding question to get you thinking.

other

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discussed

there and

other parts of

the connections between

the course.

Nature of science

These illustrate NOS using issues from both modern science and science

history, and show how the ways of doing science have evolved over the

centuries. There is a detailed description of what is meant by NOS and the

dierent aspects of NOS on the previous page. The headings of NOS feature

boxes show which of the eleven aspects they highlight.

Theory of knowledge

This

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world.

these

LHA

issues.

Parts of the book have a coloured

question.

This indic ates that

Chemistry Higher Level.

vi

bar on the edge of the page or next to a

the material is for students studying at DP

AHL means “additional higher level”.

need,

dierent

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Linking questions within each topic highlight

content

will

or

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and

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to

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as

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of

and

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guidance

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on

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information.

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for

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experimental or inquiry skills,

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the

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this

margin

your mathematic al,

especially through

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strategies.

book

to

These contain ways to develop

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Practice questions

Worked examples

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examples of how to answer

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using your chemistry knowledge and should

review these examples c arefully,

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Use these questions at the end

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each topic to draw together concepts from

that topic and

to practise

answering exam-style questions.

vii

Course book denition

The IB Learner Prole

The IB Diploma Programme course books are resource

The aim

of

materials designed

minded

people who work to create a better and

to support

students throughout

all IB programmes to develop internationally

their two-year Diploma Programme course of study

more peaceful world.

in a particular subject. They will help students gain an

develop

understanding of what is expected

described below.

an IB Diploma Programme subject

from the study of

of

the programme is to

while presenting

Inquirers: content

The aim

this person through ten learner attributes, as

They develop

their natural curiosity. They

in a way that illustrates the purpose and aims

acquire the skills necessary to conduct inquiry and of the IB.

They reect

the philosophy and

approach of

research and the IB and

encourage a deep

understanding of

snow independence in learning. They

each

actively enjoy learning and subject

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learning will be

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their lives.

providing opportunities for critic al thinking.

Knowledgeable:

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The books mirror the IB philosophy of viewing the

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doing,

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means that

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the several

This usually involves

you use into dierent

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resources, and works of art)

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physic al and

your work c an nd

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Essay.

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What constitutes malpractice? Reective:

They give thoughtful consideration to their M alpractice is behaviour that

own learning and

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you or any student

and

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personal development. plagiarism and collusion.

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or work of another person as your own.

the ideas

The following

A note on ac ademic are some of the ways to avoid plagiarism:

integrity ●

It

appropriately credit

the owners of

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ideas of another person to support one’s

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that

words and

arguments must be acknowledged

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in your work.

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are quoted

verbatim must

owners be enclosed within quotation marks and

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on your individual and

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be based

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others?

theatre arts or visual arts

creative use of a part

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original artist

you have used the

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ideas of other people is through the use of footnotes Collusion is dened and

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or closely summarize the information provided in

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You do not

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That

is,

denitions do not

malpractice include any action that

gives

need to be you an unfair advantage or aects the results of another

footnoted

as they are part of

the assumed knowledge. student.

Examples include,

taking unauthorized

material into an examination room, misconduct during Bibliographies should

include a formal list of the

an examination and resources that

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ix

Experience

the

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with

oÏer

for

DP

future

of

education

Oxford’s

digital

Science

You’re already using our print resources,

but

have you tried

our digital course on

Kerboodle?

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in cooperation with the IB and

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Learn

enabling success in DP and

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anywhere

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access

with

onscreen

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mobile-

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x

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xi

Structure

1

Models

the

nature

of

of

particulate

m a tt e r

Introduction to the

Structure 1.1

particulate nature

of matter

How c an we model the particulate nature of matter?

The universally accepted idea that all matter is composed of

atoms came from experimental evidence that could only be

explained if matter were made of particles.

Early classical theory suggested that all matter was composed

of earth, air , re, and water . However , this theory lacked

predictive power and could not account for the great variety

of chemical compounds, so it was eventually abandoned.

The systematic study of chemical changes led to the

discovery of many chemical elements that could not be

broken down into simpler substances. The fact that these

elements could only combine with one another in xed

proportions suggested the existence of atoms. It was this



way of processing knowledge through observation and

Figure

most

experimentation which led to the modern atomic theory

1

In 2021,

detailed

scientists at

Cornell University c aptured the

picture of atoms to date.

What

do models show us that

microscope images c annot?

Understandings

Structure 1.1.1 —

of

matter,

which

Elements

c annot

be

Structure 1.1.2 —

are the primary constituents

chemic ally

broken

to

down into

and

simpler substances.

Compounds

chemic ally

consist

bonded

explain

of

atoms

together

of

in

dierent elements

a

xed

ratio.

The

physic al

gases)

and

kinetic

properties

changes

of

Structure 1.1.3

—

average

energy (E

kinetic

molecular

of

matter

theory is a model

(solids, liquids,

state.

Temperature

(in

K)

is

a

measure of

) of particles. k

Mixtures

in

no

so

contain

xed

c an

be

ratio,

more

than

which

separated

are

by

one

not

element

or

chemic ally

physic al

compound

bonded

and

methods.

The composition of matter (Structure 1.1.1)

M atter

and energy

C h e mi s t r y

We

and

a re

touch

th o u g h

to

th e

of

matter

In

contrast,

are

considered

study

of

matt e r

see

of

i t.

we

in

energy

is

energy

are

a

and

The

its

c o mp o s i t i o n .

c o n s u me

ma tt e r.

u n de rs t a n di n g

shown

as

of

ma tte r,

fo r m s

c annot

our

and

produce

up

m a ny

we

ex pa n d

M atter

is

made

Air

is

a

u n i ve rs e

of

i t,

of

made

and

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s u r ro u n ds

fo r m

is

ma tte r

it

its

ma tte r

of

us,

th a t

m a tt e r

we

and

pro pe r t i e s .

is

e ve r y w he re.

and

we

c an

kn ow

is

c he mi s tr y

The

see

t he re,

seeks

characteristics

gure 2.

anything

closely

property

of

that

exists

but

associated

matter,

such

does

with

as

not

each

the

have

other,

ability

to

these

and

properties.

energy is oen

perform work or

heat.

3

Structure

1

Models

of

the

particulate

nature

of

matter

Although Chemic al

reactions

are

nuclear in

mass

and

energy

c an

be

converted

into

one

another

(for

example, in

introduced reactors

or

inside

stars),

chemistry

studies

only

those

transformations of

Reactivity 1.1. matter

where

both

products

have

the

from

form

to

one

mass

same

and

energy

mass

another

as

rather

are

conserved. In

starting

than

materials,

created

or

chemic al reactions, the

and

the

energy

is

transformed

destroyed.

made up of

particles –

atoms,

molecules,

or ions

u

Figure 2

The characteristics of matter

particles are

occupies a

in constant

volume in

MATTER

motion

space

has a

mass

Thinking skills

ATL

2

The

(E)

famous

are

Einstein

chemic al

reactions

8

(3.00 × 10

changes

This

the

is

eect

without

=

relatively

mc

the

small

,

shows

energy

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that

mass (m)

released

the

or

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).

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a

result,

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loss

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speed of light (c)

in

mass

c aused

by

is

very

large

chemic al

negligible.

of

a

demonstrates

certain

factor

compromising

other

examples

inchemistry?

4

is

E

However,

–1

m s

example

What

equation,

interconvertible.

the

of

the

is

importance

minor,

nal

it

c an

of

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oen

be

ignored

in

c alculations

result.

negligible

eects

have

you

encountered

Structure

1.1

Introduction

to

the

particulate

nature

of

matter

The atomic theory

The

law

always

of

of

conservation

combine

elements.

but

could

form

not

water,

consumed

1.0 g

of

It

be

of

mass

denite

was

and

broken

the

monoxide,

the

chemic ally.

of

that

water

with

with

observation that certain substances

led

to

elements

showed

mass

react

and

that

down

experiments

would

and

proportions

theorized

equalled

c arbon

c arbon

in

of

of

idea

Hydrogen

the

mass

formed.

1.33 g

2.66 g

the

that

combined

Other

oxygen

oxygen

of

to

to

and

was

oxygen

hydrogen

c an

and

experiments

through

form

matter

composed

form other substances

react to

oxygen

showed that

combustion

c arbon

to

form

dioxide.

It was proposed that elements, such as hydrogen, oxygen or carbon, are the The

internal

structure and

primary constituents of matter, and they cannot be chemically broken down into characteristics of atoms will be simpler substances. The idea of denite proportions suggested that particles of one discussed in

Structure 1.2

element, called atoms, would combine with atoms of another element in a xed,

simple ratio, and that atoms of one element have a dierent mass than atoms of a

dierent element. This, and other experimental evidence, led to the atomic theory.

The

atomic

c annot

be

reactions.

and

theory

states

created

or

Physic al

that

all

matter

destroyed,

and

but

chemic al

is

composed of atoms. These atoms

they

are

properties

rearranged

of

during

chemic al

matter depend on the bonding

arrangement of these atoms.

Evidence

Ancie nt

atomists,

Uddāl ak a

Ā runi

Democ ritus

w as

made

postu late d

8th

small

objects

be

of

Similarly,

into

BCE,

5th

that

increasingly

splittable”,

that

next

2000

from

mass

atoms

atomic

in

the

is

wor ld

until

units,

be

on

broken

into

must

What

to

is

is

snap

as

are

due

to

a

atomic

to

to

have

seashell

powder

further.

D alton.

dierent

“kana”.

said

“atomos”, “not

any

of

John

experiments

their

particles

producing

known

development

evidence

theories?

the

Democritus

parts

classied

based

What

natu ra l

matte r

They

parti cles.

c alled

credited

knowledge

evidence.

that

pa rticl es.

successively

not

conservation

“elements”,

s age

proposed that “particles too

He

could

could

be

In dian

re a sone d

the

BCE,

smaller

later,

could

Scientic

Āruni

indivisible

stage

years

the

philos ophers

mass together into the substances and

one

of

in

these

century

composed

The

chan ges

seen

them

Gre ek

indi visible

experience”.

in

observed

tiny,

betwe en

century

to

the

Leuci ppus,

of

that

interactions

In

and

up

among

and

theory,

D alton

over

drew

propose that

types

known as

masses.

be

supported

was

used

evidence?

to

Is

by

veriable

develop these

evidence

shaped



by

our

Figure 3

Top:

Āruni lived

in what

is now modern day

perspective? Northern India,

depicted

by the Ganges river.

Bottom:

Democritus is

in a Renaissance-era painting

5

Structure

1

Models

of

the

particulate

nature

of

matter

Chemic al symbols

In

modern

which

example,

the

Symbol

chemistry,

consist

the

chemic al

of

one

atoms

or

chemic al

symbol

and

two

symbol

for

elements

letters

iron

is

for

Fe

and

are

are

hydrogen

(the

rst

represented

derived

is

two

H

from

(the

letters

rst

of

by

the

the

letter

the

same

element

of

L atin

symbols,

names.

For

hydrogen), and

ferrum

“iron”).

Name Common

chemic al

elements

and

their

symbols

are

listed in table1; the full list is

given in the data booklet and in the periodic table at the end of this book. H

hydrogen

C

c arbon

O

oxygen

Na

sodium

Atoms

are

the

properties.

and

form

smallest

While

magnesium

S

sulfur

Cl

chlorine

Fe

iron

bound

is

Table 1

Common chemic al elements



Figure 4

M agnesium

by

chemic al

substance,

another

magnesium



matter

exist

as

elementary

sulde

bound

formula

magnesium

it

atomic

is

forces.

contains

a

still

possess

certain

Elementary substances

substance

(MgS)

chemic ally

of

that

chemic al

individually, they tend to combine together

chemic al compounds

together

elementary

(S)

of

c an

chemic al substances.

element, while

Mg

units

atoms

contain

For

one

type

composed

Mg

of

contain atoms of a

two

or

single

more elements

example, magnesium metal is an

only

chemic al

species,

atoms

of

of

compound,

and

S

atom,

sulfur

as

Mg.

atoms

it

Similarly, sulfur

only.

consists

of

(gure4). MgS is the

In

contrast,

two

dierent,

chemic al

sulde.

(le), sulfur (middle) and magnesium sulde (right)

Pure substances and mixtures

M atter

of

c an

particle

be

classied

arrangement

matter

as

a

pure

substance

or

a

mixture, depending on the type

(gure 5).

– any substance that

occupies space and has mass

pure substance

– has a

uniform chemical

element

–

composed of one

kind

of

atoms,

e.g.,

magnesium (Mg),

sulfur (S)



6

Figure 5

definite and

mixture

composition

compound

of

two

or

atoms

–

composed

more kinds of

in

a

fixed

ratio,

e.g., magnesium sulfide

(MgS),

How matter is classied

– a

combination of two or more pure

substances that retain their indiidual properties

water (H

)



homogeneous

– has

uniform composition

and

properties

throughout,

water,

according to the arrangement of particles

e.g.,

metal

sea

alloy

heterogeneous

– has

nonuniform composition

and

e.g.,

arying

paint,

properties,

salad

dressing

Structure

1.1

Introduction

to

the

particulate

nature

of

matter

Pure substances cannot be separated into individual constituents without a chemical

reaction, which alters their physical properties. In contrast, mixtures can be separated

into individual components that retain their respective physical properties.

Data-based questions

A

student

had

quantitative



two

pure

substances,

observations

were

A

made

and

and

B.

They

were

heated

in



Substance A



M ass M ass

Observations

of

and

heating

substance

of

Red

Green



Table 2

1.

C alculate

2.

State

3.

Melting

the

4.

A

colour

Results from

a

the

and

colour

ice

is

change

B

a

to

were

qualitative and

Change

in

Observations

aer

aer mass / g

heating

/ g

26.12

±

0.02

26.62

±

0.02

Black colour

27.05

±

0.02

25.76

±

0.02

Black colour

heating substances A and B

qualitative

changes

some

crucible

contents

/ g heating

B

and

Substance B

crucible and

A

crucibles

Appearance aer heating each of the two substances

Substance before

separate

recorded in table 2.

in

mass

for substances A and B.

observation

physic al

change

substances

both

pure

from

A

the

while

and

B

experiment

rusting

iron

represented

substances,

not

a

mixtures.

is

performed on A and B.

a

chemic al

physic al

Discuss

change.

change

or

whether

a

the

Explain,

chemic al

using

the

observations, whether

change.

experiment

shows

that

A

and

B

are

elements.

5.

Both

A

same

and

B

turned

black

on

heating.

C an

it

be

concluded

that

the

heating

of

these

two

substances

produced the

substance?

7

Structure

1

Models

of

the

particulate

nature

of

matter

Melting point determination

Melting

purity

point

of

melting

a

points,

t e m p e ra t u r e

v a l u e.

l ow e rs

over

a

data

The

its

c an

s u b s t a n c e.

which

that

presence

melting

used

of

to

they

matches

impurities

point

a ss e ss

substances

me ans

closely

t e m p e ra t u r e

Relevant

be

Pure

and

melt

the

in

c auses

a

Method

the

h av e

at

a

sharp

(Your teacher will provide specic instructions, depending

on

specific

the oretic al

substance

melting

to

occur

2.

Prepare

3.

•

Inquiry

observations

and

the

solids

of

being

two

analysed.)

organic

solids

(A

and

B)

for

analysis.

skills

and

of

samples

samples

of

each

solid

in

two

separate

c apillary tubes.

Tool 1: Melting point determination

Identify

identity

Obtain

ra n g e.

•

2:

the

1.

record

sucient

amounts

relevant

relevant

Following

4.

qualitative

Prepare,

your

of

in

teacher ’s instructions, mix small

the

a

two

third

solids

together.

c apillary

tube,

a

small

sample of the

mixture of the two solids.

quantitative data

5.

Determine

the

melting

point

of

your

three

samples

Materials (A,

•

Melting

•

C apillary tubes

•

S amples

point

B

and

the

mixture).

apparatus Questions

1. of

two

known

organic

solids,

for

Record

relevant

appropriate aspirin

and

salol

(phenyl

2.

Comment

points

Wear

eye

•

Note

that

•

You

protection.

the

teacher

melting

will

give

3.

point

you

apparatus

further

gets

safety

and

(for

example,

environmentally

salol

and

aspirin

very hot.

this

on

the

pure

the

results, comparing the melting

substances

structural

information

to

with

impure substances.

formulas of A and B and use

explain

the

dierence in their

4.

To

what

analyse

extent

the

could

success

melting

of

an

point

organic

data

be

used to

synthesis?

hazardous).

contain

more

than

one

element

or

compound

in

no

xed

ratio, which

for determining the are

melting

quantitative data in an

melting points.

are irritants

Mixtures Methods

of

Research

prec autions,

depending on the identity of the solids being

analysed

and

format.

2-hydroxybenzoate)

S afety

•

qualitative

example,

point

of

a

substance

not

chemic ally

Mixtures discussed in the

bonded

and

so

c an

be

separated

by

physic al methods.

are c an be

homogeneous,

in

which

the

particles

are

evenly

distributed.

Tools for chemistry Air

is

a

mixture

of

nitrogen,

oxygen, and small amounts of other gases. Air is a

chapter. homogeneous

oxygen

If

the

the

to

is

particles

mixture

the

E ach

mixture,

consistent

top,

is

are

and

not

evenly

referred to as

which

component

reveals

of

its

composition

regardless

a

of

where

air

distributed,

of

is

such

heterogeneous.

that

milk

mixture

is

a

roughly

as

in

a

its

nitrogen and 20%

mixture of two solids, then

Natural

milk

heterogeneous

maintains

80%

sampled.

physic al

will

have

the

cream rise

mixture.

and

chemic al

properties.

For

The most common homogeneous example,

hydrogen, H

,

is

explosive,

and

oxygen, O

2

, supports combustion. 2

mixtures, aqueous solutions, will When

these

substances

are

present

in

a

mixture,

their

properties

stay

the

same.

be discussed in Reactivity 3.1, and In

contrast,

water, H

O,

is

not

a

mixture

of

hydrogen

and

oxygen

but

a

chemic al

2

the properties of metal alloys in compound

formed

by

bonding

two

hydrogen

atoms

with

one

oxygen atom.

Structure 2.4. The

new

gas,

is

with

its

water

8

substance

not

has

explosive,

own

and

properties

without

a

none

it

and

chemic al

of

the

does

the

properties

not

support

hydrogen

reaction,

which

of

hydrogen

combustion.

and

oxygen

creates

or

It

oxygen. It is not a

is

c annot

a

pure substance

be

separated

new substances.

from

Structure

1.1

Introduction

to

the

particulate

nature

of

matter

Separating mixtures

Mixtures

mixture

c an

has

separated

in

of

the

Two

the

is

a

c an

be

solid

as

is

be

a

a

sand

them.

The

sulfurous

if

we

s u ga r

It

sulfur

is

powders

not.

iron(II)

maintains

pure

This

sulde,

none

of

c an be

dierence

FeS, is not

the

properties

substance.

understand

sugar

and

placed

each component of the

sulfur

compound

sugar

is

and

while

bec ause

between

bec ause

iron

smell.

individual

sugar

and

of

magnetic

separated

from

means

mixture

is

new,

attractions

of

A

Iron

have

it

usually

m i x tu re

physic al

separate

not

separated

intermolecular

Th e

to

does

components

c an

by

properties.

magnet.

used

and

solids

S and

separated

using

property

magnetic

be

unique

their

will

intermolecular

dissolve

in

forces.

water, due to

water.

in

water

and

t he

s u ga r

Intermolecular di ss o l ve s .

T he

solution

c an

th e n

be

po u re d

t h ro u g h

fi l t e r

paper

placed

funnel,

not

in

p a ss

th e

w a te r

by

a

p ro c e ss

th ro u gh

w a te r

will

and

p a ss

e v a p o ra te s

e v a p o ra t i n g

th ro u gh

(fi gu re

t he

c alled

re m a i n

on

t h ro u gh

l e av i n g

th e

f i l te r

fil t ra t i on

w a te r

pa p e r.

th e

t he

behind

f ro m

S u ga r

(fi gu re

fi l t e r

f i l te r

th e

the

6 ).

p a p e r,

p a p e r.

pu re

fil tra t e

c r ys ta l s

T he

—

the

fo r m

sand

w h e re a s

Th e

sand.

will

l a rg e

wet

Th e

s u ga r

solution

in

the

sand

t hi s

p a r ti c l e s

sugar

is

which

be

are

discussed

Structure 2.2

will

d i ss o l ve d

dried,

c an

forces

i n s i de

in a

and

th e

o bt a i n e d

pa ss e d

c r yst al li za t i on

pro c e ss

7 ).

filter paper

filter funnel

residue

(We define

a

residue

as a substance

that

remains

aer

evaporation,

distillation,

filtration

similar

or

any

process)

filtrate



evaporating

Figure 6

Filtration apparatus

sugar solution

basin

solution from

evaporating basin

cold tile

leave for a

few days t

Figure 7

The crystallization

heat for

sugar to crystallize process

9

Structure

1

Models

of

the

particulate

nature

of

matter

Distillation can be used to separate miscible liquids with dierent boiling points,

such as ethanol and water. Ethanol has a lower boiling point and will evaporate rst.

Once the vapours rise up a cooling column, they can be condensed to a liquid. As

shown in gure 8, cold water surrounds the condenser and allows the vapours to

condense to liquid ethanol. The water remains mostly in the distillation ask.

thermometer

distillation u

Figure 8

water out

Distillation apparatus

flask condenser

ethanol

and

water in

water

distillate

heat

(mostly ethanol)

Paper chromatography can be used to separate substances such as components

Paper

chromatography will in inks. A piece of chromatography paper is spotted with the mixture. The bottom

be

discussed

in

more detail in of the paper, below the spot, is placed in a suitable solvent as in gure 9(a).

Structure 2.2

The

substances

in

the

mixture

phase) and the paper (the

Figure 9

dierent

anities

intermolecular

forces of attraction

between

and

or

9(c)

ve

u

have

the

solvent

the

for

the

solvent (the

mobile

stationary phase). The anity depends on the

paper.

Figure

the

shows

a

pure

substances

mixture

that

was

in

the

mixture

composed of

pure substances.

The stages in 2D paper

(b)

(a)

(c)

chromatography

paper

some

some

hours

hours

later

later

solvent

drop of

mixture

turn

and

paper

use

a

90°

clockwise

different

solvent

Data-based questions

Look

10

at

gure

9.

1.

Which colour dot had the strongest anity for both solvent 1 and solvent 2?

2.

Which

colour

3.

Which

had

a

dots

had

stronger

a

stronger

anity

for

anity

solvent

for

2

solvent

than

1

than

solvent 1?

solvent 2?

Structure

Table

3

shows

a

summary

of

the

separation

techniques

1.1

Introduction

to

the

particulate

nature

of

matter

discussed.

Components

Technique

Description

removed

mixture

ltration

is

le

poured

through a paper lter or

liquid(s)

solid(s)

soluble

insoluble

substance(s)

substance(s)

solvent, the solution

more soluble

less soluble

cools

substance(s)

substance(s)

other

porous material

dissolution

mixture

(solvation)

or

an

is

added

organic

mixture

is

to

water

solvent

dissolved in

Activity hot

water

or

an

organic

Suggest

a

suitable

method

for

crystallization down, and the

crystals

formed

isolated

mixture

by

is

separating

are

each

of

the

following

mixtures:

ltration

heated up

solid(s) and/

evaporation or

a.

salt and pepper

b.

several

c.

sugar

water-soluble

dyes

volatile until

one

or

more of its

or

distillation

non-volatile

and

water

liquid(s) components

mixture

is

vaporize(s)

liquid(s)

d.

For

placed on

iron and copper lings

each

mixture,

describe

the

less soluble a

paper

side of the paper is

chromatography

submerged

a

separation

piece of paper; one

in

more soluble

component(s)

component(s)

move(s)

move(s) faster

or

how

each

technique

and

component

is

outline

isolated.

slower

water or stay(s) in

solvent; components place

move along the paper



Ta b l e

3

Summary of separation techniques



Figure 10 An advanced

ltration technique c alled

water for millions of people.

provided

However,

by fossil fuels. Why might

it

reverse osmosis extracts salt

this process requires vast

be important

from

amounts of energy,

seawater,

most

providing fresh

of which is currently

to consider alternative energy sources?

11

Structure

1

Models

of

the

particulate

nature

of

matter

Planning experiments and risk assessments

Relevant skills

•

Tool

1:

Separation

•

Tool

1:

Addressing

of

mixtures

safety

of

self,

others

and

the

environment

Instructions

1.

Using

to

the

c alcium

Once

a

risk

in

you

have

Identify the

Assess

•

Determine

•

Identify

you

your

doing

of

each

decided

•

the

chapter,

devise

containing

In

properties

•

If

this

mixture

assessment

and

3.

a

c arbonate.

chemic al

2.

ideas

separate

on

protocol

so,

of

a

in

a

sand,

you

must

these

iron

that

would

lings

consider

and

the

allow

you

powdered

physic al and

four substances.

method,

which

method

salt,

identify

the

hazards and complete

you:

hazards

level of

risk

relevant

suitable

control measures

disposal

methods

aligned

with

your

school’ s

health

safety policies.

have

time,

try

methodology

it

out!

and

Remember

risk

that

assessment

your

teacher

should

validate

beforehand.

Extension

You

of

could

each

aer

the

them

evaluate

the

component

separation.

together.

components

the

masses

are

before

salt,

Measure

Then

the

eectiveness

(sand,

mix

all

and

the

them

dry,

iron

to

your

mass

of

together,

and

aer

of

lings,

by

c alcium

comparing

c arbonate)

the

mass

before and

each component prior to mixing

c arry

measure

c alculate

method

and

the

the

out

your

mass

of

separation,

each

percentage

again.

recovery

make

sure

Compare

of

each

component.

Linking questions

What

factors

considered

of

How

c an

products

How

do

Why

are

contain

the

a

mixture?

of

intermolecular

between

12

are

components

two

a

alloys

generally

choosing

reaction

forces

substances?

metallic

in

be

method

to

separate the

purified?

influence

the

type

(Tool 1)

of

mixture

that

forms

(Structure 2.2)

considered

bonding?

a

(Tool 1)

(Structure

to

be

2.3

mixtures,

and

even though they often

Structure 2.4)

Structure

1.1

Introduction

to

the

particulate

nature

of

matter

States of matter (Structure 1.1.2)

Solids, liquids and gases

Matter is composed of particles. The types of interactions between these particles

determine the state of matter of a substance: solid, liquid or gas. All substances

can exist in these three states, depending on the temperature and pressure.

The

states

of

formula: (s)

matter

for

of

solid,

substances

(l)

•

Water

is

a

solid

•

Water

is

a

liquid

•

Water

is

a

gas

for

liquid

below

are

and

0 °C: H

shown

(g)

for

by

gas.

letters

For

in

brackets aer the

example:

O(s) 2

between

0

and

O(l)

100 °C: H 2

above

100 °C: H

O(g). 2

A special symbol, (aq), is used for molecules or other species in aqueous solutions.

For example, the expression “NaCl(aq)” tells us that sodium chloride is dissolved

in water while “NaCl(s)” refers to the pure compound (solid sodium chloride). The

properties of the three states of matter are summarized in gure 11.

solid

liquid

•

fixed

•

fixed

•

fixed shape

•

no

c annot be

•

•

volume

compressed

•

attractive

are

forces

•



not

move

Figure 11

attractive

are

vibrate in

no

fixed

•

no

fixed shape

•

•

around

Steam,

forces

liquid

c an be

•

attractive

forces

between particles

weaker than

particles

volume

compressed

are

those in solids

fixed positions but

do

c annot be

•

between particles

strong

particles

volume

fixed shape

compressed

between particles

•

gas

•

vibrate,

negligible

particles

rotate,

vibrate,

and

move

rotate, and

around faster than

move

in a

water and

around

liquid

ice are the three states of water

Changes of state

Substances

ice

will

positions,

but

is

reached.

A

further

reverses

At

c arbon

this

the

certain

melting.

more

This

as

states

it

the

of

until

ice

vaporizes

of

conditions,

change

CO

of

(s),

matter

heated.

a

as

The

they

(changes

accelerates

and

absorb

particles

temperature

melts

temperature

changes

dioxide,

is

violently,

point,

in

water

these

their

energy

increase

eventually

Under

change

absorb

becomes

a

release

to

energy. Solid

vibrate

in

xed

known as the melting point

its

the

or

continue

state

from solid to liquid).

movement of particles, and

gas.

The

decrease

in

temperature

state.

solid

state,

substances

known as

gure12),

c an

turn

into

sublimation,

which

is

is

commonly

gases

directly, without

typic al

used

for

for

dry

ice

(solid

refrigerating ice

2

cream

and

biologic al

samples.



Figure 12

Sublimation of dry ice

13

Structure

1

Models

of

the

particulate

nature

of

matter

The

process

water

opposite

vapour

in

the

to

air

sublimation

solidies

and

is

c alled

forms

deposition.

snowakes

of

At

low

temperatures,

various shapes and sizes

(gure13).

When

state,

when

are

a

substance

energy

a

solid

state,

a

becomes

releasing

A snowake,

the

a

a

from

by

the

liquid

changes

lose

energy

or

from

energy

intermolecular

liquid

a

more

condensed

particles

a

gas,

from

and

the

when

state

to

a

less

condensed

surroundings. This happens

a

liquid

becomes a gas. These

processes.

particles

substance,

Figure 13

becomes

substance

the

changes

absorbed

endothermic

When



is

or

to

a

solid,

the

a

to

less

the

forces

and

condensed

surroundings

become

when

a

to

and,

for

a

more

a

condensed

molecular

stronger. This happens when a gas

liquid

surroundings is an

state

becomes

exothermic

a

solid.

The

process of

process.

the product of

The

changes

of

state

occurring

in

these

transformations

are

shown

in

gure 14.

deposition of water

Non-Newtonian uids

Some

do

substances,

not

behave

Newtonian

to

them.

known

uids

You

as

like

will

maize

known as

non-Newtonian uids,

typic al liquids. The

varies

make

starch

depending

a

on

viscosity

the

2.

of non-

force

or

“oobleck”,

and

add

Continue

applied

a

non-Newtonian uid commonly

slime

Slowly

thick

starch

explore its

3.

properties.

to

or

more

some

smoothly

if

It

as

starch and mix.

the

by

mixture

adding

achieves

more maize

needed.

exploring

should

stirred

maize

until

Adjust

water,

time

mixture.

the

water

consistency.

Spend

your

water

adding

the

harden

if

properties of

tapped,

and

ow

slowly.

Relevant skills

•

Questions

Inquiry 1: Identify dependent and independent

1.

variables

•

Inquiry

2:

Identify

and

record

relevant

qualitative

observations

Describe

the

matter

each

of

properties and identify the state of

•

powdered

•

water

•

the

of

the

following:

maize

starch

S afety

Wear

eye

maize

starch–water

mixture.

protection. 2.

Suppose

question

you

were

relating

asked

to

a

to

maize

develop

a

research

starch–water

mixture.

Materials

Consider

•

Spoon

•

250

•

Powdered

•

Water

or

variables.

3

cm

possible independent and dependent

large spatula

beaker

3. maize

Research

non-Newtonian uids and identify other

starch

examples of these substances.

4.

How

has

this

experience

changed

the

way

you think

Method about

1.

Ad d

t h re e

or

fo u r

h e a pe d

spoons

of

to

t he

b e a ke r.

N o te

i ts

of

matter

and

their

properties?

Reect on

ma i z e this,

s ta rc h

states

a pp e a ra n c e

completing

the

following

sentence

starters:

and

•

I

•

Now, I think...

used to think...

c o n s i s t e n c y.

Linking questions

Why

are

some

conditions?

Why

are

some

(Structure

14

substances

solid

while

others

are

fluid

under

standard

(Structure 2.4)

2,

changes

of

state

Reactivity 1.2)

endothermic

and

some

exothermic?

Structure

1.1

Introduction

to

the

particulate

nature

of

matter

gas (g)

freezing

melting

solid (s)

liquid (l)

 

Figure 14

Endothermic and

Figure 15

Orange growers spray their fruit

with water on cold nights.

exothermic Freezing of

water is an exothermic process that

releases energy (in the form of

changes of state heat) to the fruit,

protecting it against cold

Kelvin temperature sc ale (Structure 1.1.3)

As

temperature

measure

of

the

particles

of

a

move

faster,

When

a

water

solid

solid

while

is

the

in

a

energy

to

is

between

a

energies

kinetic

vibrate

in

gas

heated,

changes

added

forces

rises,

average

used

the

they

there

liquid

to

is

and

of

particles

energy

lattice

move

no

increase.

particles.

more,

As

particles

Temperature

substances

in

a

liquid

is a

absorb

vibrate

energy,

more and

faster.

temperature change during the periods when

when

disrupt

of

the

a

liquid

solid

changes

lattice

and

to

a

gas

(gure 16). The

overcome

the

intermolecular

molecules in the liquid.

vaporization steam

100

kg

+

m

c

/

water

d

C°

condensation

steam

erutarepmet

water

+

s

om

ice

water

melting

K A 0

freezing 

ice

Figure 17

are kilogram

The seven base SI units

(kg) for mass,

length, second

energy input

(s) for time,

electric current, 

Figure 16

meter (m) for

ampere (A) for

kelvin (K) for temperature,

Graph of the heating curve for water mole (mol) for amount

of substance, and

c andela (cd) for luminous intensity. All units

There

were

many

attempts

to

measure

relative

temperature,

but

the

rst widely of

accepted

temperature

sc ale

was

introduced

by

the

Polish-born

Dutch

measurement

c an be derived

from these

physicist seven base units

D aniel

Gabriel

F ahrenheit.

You The

kelvin

is

the

base

unit

of

temperature

of

Units

measurements

(SI).

c an

There

be

are

will

learn

more about the mole

measurement in the International

in System

Structure 1.4.

seven base units, and all other units of

derived

from

these

(gure 17).

15

Structure

1

Models

of

the

particulate

nature

of

matter

Measurement

3

M aking,

recording,

and

communic ating

volume (m

measurements

2

greatly

benets

International

from

the

Bureau

French

mesures),

from

agreed

of

Weights

in

the

organisation

continuously

sc ales. The

and

Measures

=

(BIPM,

rene

late

19th

which

The

century, is an

–3

),

density

(kg m

),

energy

( joule,

J,

where

1 J

–2

s

base

)

and

units

including

seeks to set up and

measurement

1 kg m

so

on,

are

derived

from

the

seven base

units.

Bureau international des poids et

established

international

upon

are

several

dened

that

Boltzmann constant,

standards.

constant,

N

;

and

you

k;

the

according

will

to

seven constants,

recognize, such as the

speed of light,

Plank constant,

c;

the

Avogadro

h

A

The

International

System

of

Units

(SI,

from

the

French

The use of universal and precisely dened units is very

Système international d’unités) is the most commonly

used

system

seven

time

base

of

units:

(second,

(kelvin,

K),

intensity

measurement.

s),

length

electric

amount

of

Its

(metre,

current

substance

building

m),

mass

blocks

important, as it allows scientists from dierent countries

are the

to understand one another and share the results of their

(kilogram, kg),

(ampere,

A),

studies. What other advantages are there to internationally

temperature

shared and continuously updated measurement systems in

(mole, mol) and luminous

the natural sciences? You might want to look up the Mars

(c andela, cd). All other units, such as those of

Climate Orbiter.

Thinking skills

ATL

Throughout

history,

developed,

each

several

with

universal

dierent

temperature

reference

points.

sc ales

Some

have

of

been

these

are

summarized in table 4.

Sc ale

D ate

Reference

Newton

1700s

H

O

points

freezing point = 0°

2

Human

F ahrenheit

1700s

H

O

body

temperature = 12°

freezing point = 32°

2

H

O boiling point = 212° 2

Delisle

1700s

H

O

freezing point = 150°

2

H

O boiling point = 0° 2

Celsius

1700s

O

H

freezing point = 0°

2

H

O boiling point = 100° 2

Kelvin

1800s

Absolute

CGPM

1950s

Triple

BIPM

2018

Kelvin

Table 4

Examples

Temperature

in

the

units



Figure 18

A platinum–iridium cylinder

in the US was used

mass.

to dene a kilogram of

This standard

and

m

energy,

and

and

practic al

s.

It

v arious

to

joules

has

practic al

reasons

( J),

been

could

energy

which

are

decided

reasons”.

273.16 K

k.

t e m p e ra t u r e

thermal

=

to

What

sc ales

and

in

as

turn

keep

do

you

such

it

could

be

expressed

dened in terms of the base

kelvin

think

as

an

some

SI

of

base

these

unit

“for

historic al

c arefully

at

table

4

be?

above.

Identify

one

thing

you

see, one thing it makes

when the kilogram and all other SI

units were redened

think

about,

as exact quantities

your

16

for

of

related

water

bec ame obsolete in

you

based

kg,

historic al

Look

2019,

unit

is

of

dened in terms of the Boltzmann

constant,



zero = 0

point

on physic al constants

class.

and

one

thing

it

makes

you

wonder.

Share

your

ideas with

Structure

Kelvin

is

temperature

considered

Absolute

any

to

zero

kinetic

hence

an

an

Under

water

(0 K)

energy

c annot

increase

normal

get

in

is

proportional

absolute

any

that

at

collisions.

colder.

temperature

pressure,

373.15 K.

zero

this

average

kinetic

temperature

M atter

An

of

water

Absolute

the

Introduction

to

the

particulate

nature

of

matter

energy of particles and

sc ale.

implies

on

to

1.1

at

increase

1

degree

boils

on

at

the

absolute

in

the

particles

zero

c annot

temperature

Celsius.

0 °C

is

of

1

c annot

lose

kelvin

equal

to

transfer

heat and

is

equivalent

273.15 K.

100 °C, so that makes the boiling point of

Celsius

sc ale

is

–273.15 °C.

t

Figure 19

The Celsius and

Kelvin

sc ales for temperature (all values are

rounded

to whole numbers)

400 K water 100 °C

373 K boils

350 K

50 °C

300 K water 0 °C

273 K freezes

40 °C

250 K

50 °C

You

will

learn

more about the

200 K

dry ice

kinetic 78 °C

energy of particles in

195 K solid CO 2

Reactivity 2.2.

100 °C

150 K

150 °C

100 K liquid 191 °C

82 K air

200 °C

50 K

250 °C

absolute 273 °C

0 K zero

Kelvin

Celsius

Linking questions

What

is

sample

the

graphic al

at

fixed

a

distribution

temperature?

of

kinetic

energy

values of particles in a

(Reactivity2.2)

What must happen to particles for a chemical reaction to occur? (Reactivity 2.2)

17

Structure

1

Models

of

the

particulate

nature

of

matter

End-of-topic questions

5.

Which

changes

of

state

are

opposite

to

each other?

Topic review A.

1.

Using

your

answer

the

knowledge

guiding

from the

question

as

Structure 1.1

fully

as

melting

and

condensation

topic, B.

vaporization and deposition

C.

deposition and sublimation

D.

sublimation

possible:

How can we model the particulate nature of matter?

6.

Which

of

the

and

freezing

following

statements

is

incorrect?

Exam-style questions A.

solids

and

liquids

are

almost

incompressible

Multiple-choice questions

2.

Which

of

the

following

are

examples

of

B.

particles

C.

liquids

D.

particles

in

both

solids

and

liquids

are mobile

homogeneous and

gases

have

no

xed shape

mixtures?

I.

in

solids,

liquids

and

gases

c an

vibrate

Air

7. II.

Which

elements

c an

be

separated

from

each

other

by

Steel physic al

III.

Aqueous

KMnO

methods?

potassium manganate(VII), A.

oxygen

B.

hydrogen

and

nitrogen in air

C.

c arbon

D.

magnesium and sulfur in magnesium sulde

(aq). 4

3.

A.

II only

B.

III only

C.

I and II only

D.

I, II and III

What

correctly

(c arbon

8.

Which

and

change

equivalent

to

decrease

B.

increase

C.

decrease

D.

increase

in

water

temperature

by

increase

in

on

the

Celsius

temperature

by

sc ale is

20 K?

by

20 °C

20 °C

dioxide)?

endothermic?

process

A

exothermic

CO

B

exothermic

CO

C

endothermic

CO

(s)

→

CO

(s)

→

C(g) + O

(s)

→

CO

Extended-response questions

(g)

(g)

9.

Explain

why

the

proportional

Celsius

of

the

obtain

solid

following

(s)

→

C(g) + O

methods

chloride

could

be

temperature

temperature

in

temperature

kinetic

is

directly

energy but the

is

not,

increment

is

even

the

though

same

in

a

1-degree

each

sc ale?

[2]

used to Ionic

salts

c an

be

broken

down

in

electrolysis. The

from a solution of sodium unbalanced

chloride

Kelvin

average

(g) 2

10.

sodium

to

(g) 2

2

Which

293.15 °C

293.15 °C

2

2

CO

by

2

2

endothermic

by

Equation describing the

2

4.

in

the

A.

oxygen

oxygen in dry ice

describes the sublimation of dry ice

Exothermic or

D

and

ionic

equation

for

the

electrolysis of molten

water? lead(II)

I.

evaporation

II.

ltration

III.

distillation

bromide is:

2+

Pb

a.

+

Br

One

→

of

the

formula

A.

Pb + X

of

I and II only

C.

I and III only

b.

Balance

c.

The

the

lead,

Pb. State the

[1]

I, II and III

equation.

electrolysis

c arriedout

point

and

matter

18

is

X.

I only

B.

D.

products

product

of

at

of

molten

380 °C.

boiling

each

of

Write

equation

gave

data,

(b).

bromide is

reference to melting

species

state

in

lead(II)

With

point

the

temperature.

you

[1]

deduce the state of

in

the

symbols

equation at this

in

the

balanced

[2]

Structure

11.

The

kinetic

mass

×

energy

the

square

of

of

particles

the

is

c.

equal to half of their

Once

to

2

mv

=

.

Determine

how

much

the

the

obtain

Introduction

excess

removed,

velocity of the particles:

1 E

1.1

the

the

copper(II)

student

pure

to

particulate

oxide

needed

crystals

of

had

to

nature

matter

been

gure

copper(II)

of

out

sulfate

how

from

speed of

k

2

the

molecules

in

a

pure

gaseous

substance

will

solution.

couldfollow

when

the

Kelvin

temperature

is

Describe a method the student

increase

doubled.

to

obtain

pure, dry copper(II)

[2] sulfatecrystals.

12.

Pure

c aeine

is

a

white

[3]

powder with melting point 14.

Study

the

gure

below.

235 °C.

vaporization State

A

the

melting

chemist

is

point

of

c aeine

investigating

the

in

kelvin.

ec acy

of

[1]

C°

a.

b.

three

100

condensation

/

yield

in

all

method

yield

extraction

three

once

and

methods.

c ases

and

melting

is

0.960 g.

collects

point

The

of

the

the

theoretic al

She

uses

following

each

data

for the

product:

erutarepmet

c aeine

water

+

steam

water

ice

+

water

melting

0

Method 1

Method 2

Method 3

0.229

0.094

0.380

freezing ice Mass of c aeine

energy input

obtained / g

Melting point of 190–220

229–233

188–201

caeine product / °C a.

Explain

why,

input,the i.

C alculate

ii.

C alculate

the

mean

and

range

of

the

constant of

c aeine

obtained.

the

your

of

answer

signic ant

to

an

gave

c.

Suggest

error

in

the

one

this

gures.

giving

purest

way

solution

to

a

A

student

a

at

0 °C

for

5.00 g

(at

of

dilute

a

period

sample

energy

remains

of

time.

[2]

c aeine

product.

minimize

acid

chloride

atmospheric

in

100.0 g of

temperature and

[2] has

the

following

properties:

the



melting



boiling

point:

–3 °C

[1]

point:

101 °C

random

b.

[1]

copper(II)

sulfuric

sodium

standard

Sketch

sulfate

solution

a

show

graph

the

similar

heating

with

excess

to

the

curve

one

for

a

in

gure 16

sample of this

by sodiumchloride

reacting

increasing

the

reason, which method

experiment.

prepares

of

water

to

13.

the

of

appropriate number

pressure)

Determine,

of

[2]

pure

iii.

spite

percentage yield of Method 1. A

Give

in

temperature

mass

copper(II)

solution.

[2]

oxide. 15.

Elemental

iodine

exists

as

diatomic

molecules, I

.

At

2

Copper(II)

The

word

oxide

is

insoluble

equation

for

this

in

water.

reaction

is

as

follows:

room

temperature

black

solid

gently.

sulfuric acid + copper(II) oxide → copper(II) sulfate + water

cold

When

surfaces

pressure, a.

Write

a

balanced

chemic al

that

and

readily

cooled,

without

solid

iodine

pressure,

forms

it

is

violet

a

lustrous purple-

fumes

when

heated

gaseous iodine deposits on

condensing.

melts

at

Under

114 °C

to

increased

form

a

deep-

equation, including violet liquid.

state

symbols,

for

this

reaction.

[2]

a. b.

The

acid

was

heated,

then

copper(II)

Formulate

state powder

could

was

be

added

observed

until

it

was

in

use

to

a

the

that

represent all changes of

above.

[3]

excess and

reason, a method the student could

remove

mentioned

b.

State

the

melting

c.

Suggest

how

gaseous

iodine.

point

of

iodine

in

kelvin.

[1]

suspended in the solution,

quickly sinking to the bottom of the beaker. Suggest,

giving

equations

oxide

excess

copper(II)

oxide.

liquid

iodine

c an

be

obtained

from

[1]

[2]

19

The nuclear atom

Structure 1.2

Understandings

How do nuclei of atoms dier?

Structure 1.2.1

The

answer

to

this

question

was

obtained

by

nucleus

100

years

of

brilliant

In

the

late

of

rearranged

in

the

atoms

idea

that

electrons

that

Structure 1.2.2

matter

dierent

was

chemic al

theory)

positively

neutrons

charged, dense

(nucleons).

Negatively

was

occupy the space outside the nucleus

—

numbers

Isotopes

of

are

atoms

of

the

same element with

neutrons.

were indivisible and

reactions

gaining

—

M ass

spectra

are

used to determine

(known as the

atomic

a

and

more fascinating than

Structure 1.2.3

the

contain

protons

LHA

composed

is

what is known

1800s,

of

research. Sometimes, the

how we know

the question of

Atoms

composed

charged

question of

—

over

relative

atomic

masses

of

elements

from their isotopic

popularity. The composition.

discovery

scientists

of

to

electricity

study

the

and

radioactivity

structure

of

the

allowed

atom

itself.

The structure of the atom (Structure 1.2.1)

An

atom

and

contains

neutrons

which

occupy

electrons

are

The

key

1.

It

is

very

2.

It

is

a

3.

It

has

In

an

beam of

of

vast

the

small

in

charged

known

region

as

nucleus,

nucleons).

outside

nucleus

of

the

which

Atoms

nucleus.

itself

also

The

contains

contain

protons,

experiment

structure

neutrons and

particles

made

the

atom

containing

itself.

virtually

all

the

mass of the atom.

charge.

designed

alpha

to

are

by

Ernest

were

given

in

red

Rutherford

toward

a

in

1911,

sheet

of

positively

gold

charged

foil. The main

gure 1.

movable Rutherford’ s explanation detector

+

alpha

protons

electrons,

are:

comparison

dense

positive

observations

alpha particles

the

highly

radioactive

positively

known as subatomic particles.

factors

a

a

(collectively

+

source

Most alpha

particles

are

undeflected

atom +

gold

foil

Some alpha

vacuum +

particles

are

deflected

slightly

A

few alpha

undeflected

+ large

slight

deflection

deflection

particles

bounce +

off nucleus



20

Figure 1

Rutherford’s gold

foil experiment

Structure

1.2

The

nuclear atom

F alsic ation

experiment

preceded it, namely the “plum-pudding model”.

vulnerable

The

plum-pudding

claim

amorphous

present

atoms

model

positively

throughout.

particles

red

at

the

undeected.

existing

falsied

model,

If

suggested

charged

this

gold

were

foil

the

way

that

blob

the

would

Rutherford’ s

paving

the atomic model

claims

gold

that

an

foil

Scientic

The

atom

was

that

never

electrons

be

stands

knowledge

have

degree

the

gone

through its

of

new model of the atom.

is

falsiable.

up

to

true

means

directions.

C an

a

means

testing

absolute

always

The

that

This

that

they

are

contradicts them. A scientic

severe

therefore

new

single

that

with

uncertainty.

knowledge

contradicted the

development of a

are

evidence

proven

c ase, all alpha

results

for

the

with

to

is

accompanied

provisional

further

counterexample

strong

c an

a

by a

nature of scientic

evidence

falsify

but

certainty. Scientic

c an

steer it in

claim?

Activity

The

lists

below

properties

which

of

show

the

the

observations

nucleus.

Determine

in

the

which

gold

foil

experiment and the

observation

is

explained

by

property.

Observation

Property

Nearly all the alpha particles went

The

nucleus

has

a

positive

charge.

Occ asionally, some of the alpha

The

nucleus

is

particles

comparison to the size of the atom.

straight through the gold foil.

The

bounced

alpha

when

straight back.

particles

closely

are

repelled

The

approaching the

nucleus

virtually

all

is

the

very small in

very

dense, containing

mass of the atom.

nucleus.

In

1911,

Rutherford

summarized

planetary model of the atom,

In

this

model,

nucleus

of

the

the

in

the

solar

entire

nucleus

negatively

same

system’s

atom.

by

way

charged

as

mass,

However,

electrostatic

the

also

of

electrons

planets

the

results

orbit

atomic

instead

his

experiments

known as the

of

the

orbit

Sun.

nucleus

by

the

Just

positively

as

contains

gravity,

by

proposing the

Rutherford model

the

the

over

Sun

contains

99.9%

electrons

(gure2).

charged atomic

are

of

held

the

99.8%

mass of

around the

attraction.

–

+

–

electron

–

proton

–

neutron

+

+

– nucleus



Figure 2

The Rutherford model of the atom

21

Structure

1

Models

of

the

particulate

nature

of

matter

Models

Scientists

use

models

to

represent

natural phenomena. All

Atoms

themselves

are

extremely small. The diameter of

–10

models

have

understood.

2.

The

useful

size

limitations,

Consider

of

model

the

of

which

the

nucleus

the

should

depiction

is

be

of

identied and

the

exaggerated

atom

but

it

in

gure

serves as a

most

unit

atoms

used

to

is

in

the

range

1 × 10

–10

to

5 × 10

m. The

describe the dimensions of atoms is the

picometre, pm:

–12

nuclear atom.

1 pm = 10

The vast space in the atom compared to the tiny size of

the nucleus is hard to fully appreciate. Rutherford’ s native

In

X-ray

m

crystallography

dimensions

New Zealand is a great rugby-playing nation. Imagine

is

the

a

commonly

angstrom,

used

unit

for atomic

symbol Å:

–10

1 Å = 10

m

being at Eden Park stadium (gure 3) and looking down at For

example,

the

atomic

radius of the uorine atom is

the centre of the pitch from the top row of seats. If a golf –12

60 × 10

m

(60 pm).

To

convert

this

toÅ

we

c an use

ball were placed at the centre of the eld, the distance dimensional analysis,

using

the

conversion

factors

between you and the golf ball would represent the given

above:

distance between the electron and the nucleus. –12

10

m

1 Å –1

The

relative

volume

of

open

space

in

the

atom

is

vast, and

60 pm ×

×

=

0.60 Å

=

6.0 × 10

Å

–10

1 pm our

simple

representation

of

gure

2

is

obviously

spite

tiny

of

its

limitations,

Rutherford’ s

work

has

formed the

unrealistic. The nucleus occupies

basis a

m

Rutherford’s atomic model

In in

10

of

much

of

our

thinking

on

the

structure of the atom.

volume of the atom and the diameter of an atom is

Rutherford approximately

is

rumoured

to

have

said to his students:

100 000 times the diameter of the nucleus.

All science is either physics or stamp collecting!



Figure 3

of the eld

22

Eden Park,

Auckland,

New Zealand.

If the atom were the size of the stadium,

the nucleus would

look like a golf ball in the centre

Structure

1.2

The

nuclear atom

TOK

All

the

could

two

models

be

we

argued

physicists,

Switzerland

This

Nobel

Gerd

that

gave

Prize

in

discussed

objects

Binnig

invented

microscope

level.

have

that

the

scientists

in

and

assume

only

that

“real”

Heinrich

sc anning

generates

Physics

are

atoms

when

Rohrer,

tunnelling

are

they

real.

c an

working

microscope

However, it

be

at

seen. In 1981

IBM in Zurich,

(STM),

an

electron

three-dimensional images of surfaces at the atomic

the

ability

1986

was

to

observe

awarded

individual

to

Binnig

atoms

and

directly. The

Rohrer

for their

groundbreaking work.

You

c an

nd

an

atomic

sc ale

lm

created

by

IBM

c alled

A Boy and his Atom

on

the internet.



Figure 4

Has

A still from A Boy and his Atom

technology

extended

human’ s

c apacity

to

make

observations of the

natural world?

How

important

are

material

tools

in

the

production or acquisition of

knowledge?

Other

experiments

particle,

masses

the

and

have

neutron,

charges

shown

with

of

that

nearly

the

the

the

nucleus

same

subatomic

mass

particles

also

as

are

contains

the

a

proton.

neutral subatomic

The

relative

shown in table 1.

t

Particle

Relative

mass

Relative

charge

Table 1

the proton,

proton

1

+1

neutron

1

0

negligible

–1

Relative masses and

charges for

Loc ation

neutron and

electron

nucleus

electron

The

electric

charge

c arried

by

a

single

electron

is

outside nucleus

known as the

elementary The

–19

charge

(e)

and

it

has

a

value

of

approximately

1.602 × 10

C.

The

actual

particles

are

commonly

expressed

in

elementary

charge

units.

of

a

the

charge

proton as +e.

electrons

and

of

The

protons

an

electron

symbol

have

e

is

c an

oen

charges

of

be

represented as –e,

omitted,

–1

and

so

+1,

it

is

and

customary

the

to

particles

and

c an

be

charges of

found in the

For data

example,

masses

charges of these

subatomic

booklet.

charge

say that

respectively.

23

Structure

1

Models

of

the

particulate

nature

of

matter

How small is small?

Relevant

skills

•

Tool

3:

Apply

•

Tool

3:

Use

and

and

use

SI

prexes and units

interpret scientic notation

Instructions

1.

A

variety

lengths,

objects

of

small

but

in

lengths

rather

order

of

are

based

size,

shown in table 2. Without looking at their

on

what

from

you

smallest

Item

proton,

sheet

onion

charge

of

paper,

cell,

printed

c arbon

full

bond,

atom,

0.10 mm

250 µm

length

267 pm

diameter

0.30 mm

diameter

fullerene,

C

each item, list these

0.84 fm

thickness

stop,

about

largest.

Length

radius

diameter

iodine-iodine

know

to

150 pm

diameter

0.71 nm

60



2.

Table 2

Lengths of

various small items

Convert the length values into metres and state them in standard form to

two signicant gures. Refer to the following conversion factors:

–3

•

milli, m: 10

•

micro, µ: 10

•

nano, n: 10

•

pico, p: 10

•

femto, f: 10

–6

–9

–12

–15

3.

List

the

gave

4.

for

a

web

than

Provide

values

question

Conduct

smaller

5.

length

the

in

table

2

in

order

of

increasing

size.

Was

the

list

you

correct?

search

the

full

1

values

to

nd

given

reference

for

three

in

more

table

your

2,

values to add to the list: one

one

larger,

information

and

one

intermediate.

sources in question 4,

ATL following

your

school’s

Atomic number

citing

and

referencing

and the nuclear

system.

symbol

As of 2023, there are 118 known elements, given atomic numbers 1 to 118. The atomic

number of an element is also the number of protons in the nucleus of that atom. Gold,

atomic number 79, has 79 protons, while carbon, atomic number 6, has 6 protons. As

all the relative mass is in the nucleus, the dierence between the atomic number and

mass number is the number of neutrons in the element. Gold has atomic number 79 and

mass number 197 . Therefore, it has 197 – 79 = 118 neutrons. Each element is neutral,

with no charge, so the number of electrons in a neutral atom must equal the number

ofprotons.

24

Structure

1.2

The

nuclear atom

Activity

Determine

Atomic

the

missing

symbol

values

Atomic

from

the

table.

number

M ass

number

Protons

Neutrons

O

Electrons

8

13

27

85

37

80

35

27

32

120

Pb

80

207

69

100

A

Chemists

frequently

use

nuclear

symbol notation,

X, to denote the number of Z

neutrons,

isotope,

for

protons

Z

is the

example,

and

electrons in an atom.

atomic

with

mass

number,

number

and

197

X

is

and

A

the

represents

chemic al

atomic

the

mass number of the

symbol

number

79,

(gure 5). Gold,

would

have

a

nuclear

197

symbol notation of

Au. 79

mass number

A

N

=

=

+

Z

number

of

N

chemic al

where

symbol

for the element

neutrons

X Z atomic number

number



of

Figure 5

Atoms

atoms

form

=

protons

The nuclear symbol notation

compounds

sometimes

are

protons.

For

the

compound

ionic

example,

a magnesium

in

the

no

ion

nucleus

is

sharing

magnesium

a

2+

two

or

transferring

neutral,

magnesium

with

(12)

by

longer

atoms

oxide.

charge,

greater

having

as

than

react

electrons.

more

with

the

number

number

loses

of

of

As

fewer

oxygen

M agnesium

the

or

a

result, these

electrons than

atoms

two

to

produce

electrons

positively

negatively

to

charged

charged

form

protons

electrons

remaining (10).

The

resulting

charge

mass

is

also

number:

displayed

in

the

nuclear

symbol

24

24 (12

protons

+

12

notation

2

below:

+

charge: 2+

electrons)

(12

protons

–

10

electrons)

Mg 12

atomic number: 12

(12

The

oxygen

atom

chemic al element: Mg

protons)

gains

the

(magnesium)

two

electrons

lost

by

magnesium

to

produce

an

16

ion

with

a

2–

negative

charge.

The

nuclear

symbol

for

the

oxide ion is

oxide

2–

O

.

8

25

Structure

1

Models

of

the

particulate

nature

of

matter

The overall chemical equation for the reaction between magnesium and oxygen is

1

2+

O

Mg + Ionic

in

bonding

is

→

Mg

2–

+ O

2

discussed further

2

Structure 2.1

2+

2–

Mg

+ O

two

ions

Ionic

is

more

result

bonds

in

a

hold

commonly

force

the

of

ions

written

attraction

together

as

MgO,

between

to

form

as

the

them

solid

opposite

charges on the

known as an

magnesium

ionic bond.

oxide.

Activity

Linking questions

Deduce

the

notation

for

protons,

21

nuclear

an

ion

symbol

with

electrons,

What determines the different chemical properties of atoms? (Structure 1.3)

24

and

28 How

does

the

atomic

number

relate to the position of an element in the

neutrons. periodic

table?

(Structure

3.1)

Isotopes (Structure 1.2.2)

Isotopes are dierent atoms of the same element with a dierent number of neutrons.

As a result, they have dierent mass numbers, A, but the same atomic number , Z.

35

Chlorine, for example, has two isotopes: one with mass number 35,

Cl, and one 17

37

Cl. They have similar chemical properties, as they are both

with mass number 37 , 17

chlorine atoms with the same number of electrons, but dierent physical properties,

such as density, because atoms of one isotope are heavier than atoms of the other .

Naturally

occurring

(protium)

and

(gure6),

is

hydrogen

hydrogen-2

radioactive,

consists

of

(deuterium).

so

it

does

not

two

The

stable

third

occur

in

isotopes,

isotope

nature

in

of

hydrogen-1

hydrogen, tritium

signic ant quantities.

Activity

Copy

the

table

below

and

complete

it

by

deducing

the

nuclear

symbols and/

or composition of these isotopes.

Isotope

Nuclear

Z

symbol

N

A

1

hydrogen-1

(protium) H 1

hydrogen-2

(deuterium)

hydrogen-3

(tritium)

Atomic

numbers

of

1

3

isotopes

are

oen

omitted

in

nuclear

symbol

notation.

For

37

example,

‘Cl’

17 ,

tells

so

you

isotope

the

including

written

A ,

the

with

listed

a

for

of

chlorine

isotope

the

chlorine

atomic

hyphen,

each

is

number

such

element

with

as

on

mass

and

is

therefore

not

37

must

necessary.

chlorine-37,

the

number

periodic

or

c an

is

not

written as

Cl.

have an atomic number of

These

Cl-37. The

table

be

a

isotopes

c an also be

relative atomic mass,

whole

number

bec ause it

r

is

the

weighted

average of all isotopes of that element.

Natural abundance (NA) 

Figure 6

A portable tritium

The radioactive dec ay of tritium

light

of

an

isotope

is

the

percentage of its atoms among

source.

produces

high-energy electrons (beta particles).

all

atoms

of

the

abundances

given

for

all

element

isotopes

of

found

an

on

our

element,

planet.

we

c an

If

we

know

c alculate

the

the

natural

average

A

of that r

These electrons hit

a uorescent material

element.

The

opposite

task

(c alculation

of

natural

abundances

from

A ) r

and

26

make it

glow in the dark

only

if

the

element

is

composed

of

two

known isotopes.

is

possible

Structure

1.2

The

nuclear atom

Worked example 1

C alculate

the

A

r

for

iron

Isotope

using

N atural

the

values

in

the

following

table.

abundance (NA)/ %

54

Fe

5.845

Fe

91.754

56

57

Fe

2.119

58

Fe

0.282

Solution

We

know

The

A

natural

r

=

average

abundance

of

the

natural

values

add

abundance

up

to

57

×

100%

of

so

each

we

isotope

divide

by

multiplied

100

to

by

obtain

their

the

mass

numbers.

average.

Therefore:

54 A

r

× 5.845

+

56

×

91.754

+

2.119

+

58

×

0.282

=

= 55.91

100

Worked example 2

There

are

two

abundance

stable

(NA)

of

isotopes

each

of

chlorine:

isotope

given

Cl-35

that

A

and

for

Cl-37.

chlorine

C alculate

is

the

natural

35.45.

r

Solution

(A A

of isotope 1

×

NA

of isotope 1)

+

(A

of isotope 2

×

NA

of isotope 2)

= r

100

Therefore:

(35 × NA

of

Cl-35) + (37 × NA

of

Cl-37) =

35.45

100

Let

x

=

NA

of

Substituting

Cl-35, then 100

in

35x + 37(100

the

above

x

=

equation

NA

of

Cl-37.

gives:

x) =

35.45

100

Expanding

3700

the

brackets

and

resolving the

x

terms

gives:

2x =

35.45

100

Then

rearrange in terms of

3700 x

x:

3545

= 2

x =

77 .5 and 100

x =

22.5.

Therefore,

35

The

actual

natural abundances of

the

natural

abundance

of

Cl-35

is

77.5%

and

Cl-37 22.5%.

37

Cl and

Cl

are

75.8

and

24.2%,

respectively. Average

A

values

for all elements

r

The results of our calculations are slightly dierent because we used mass numbers, are 35

which

are

rounded

values

for

the

actual

masses of the

given in the data booklet and in

37

Cl and

Cl

atoms. the periodic table at the end of this

book.

27

Structure

1

Models

of

the

particulate

nature

of

matter

Density

at

Melting

Boiling

point / °C

point / °C

Compound –3

4 °C / g cm

1

H

O

1.000

0.00

100.0

O

1.106

3.82

101.4

2

2

H 2



Table 3

Physic al properties of normal and

heavy water

235

Naturally

occurring

dierences 

Figure 7

A pellet

of enriched

in

properties

of

these

are

used

for the

U. The

enrichment

uranium

235

used

isotopes

238

U and

uranium consists of two main isotopes,

physic al

as fuel in nuclear reactors

(increase

in

the

proportion of

238

U

over

U)

of

nuclear

fuel

(gure7), as most

235

nuclear

reactors

contains

only

Enriching

to

track

0.72%

one

the

require

type

of

of

uranium

this

at

least 3% of

U,

while

natural

uranium

isotope.

isotope

mechanisms

with

and

in

a

particular

progress

of

substance

reactions.

c an

This

is

also

make

oen

it

possible

referred to as

isotope labelling.

Global impact of science

Developments

in

environmental,

Nuclear

ssion,

colossal

amounts

development

Element

woman

109,

in

(gure8).

Figure 8

Her

nuclear

bomb

“I

will

to

work

with

splitting

one

(Mt),

a

technology

is

as

such

well

Frisch

In

later

being

up

the

have

nuclei

as

the

aer

to

years,

the

in

large

It

from

atoms

releasing

led tothe

bomb.

the

discovery

the

has

Meitner,

Meitner

developed

ethic al,

consequences.

of

atomic

Lise

doctorate

led

may

economic

development.

named

physics

Otto

1939.

and

was

US.

the

second

University of Vienna

of

nuclear

ssion,

invited to work on

She

declined, famously

have nothing to do with a bomb!”

think

of

other

ethic alimplic ations?

28

is

applic ations

cultural

energy,

receive

Austrian-Swedish physicist

Lise Meitner in 1906

their

involves

energy,

Nature in

stating

you

of

of

and

social,

meitnerium

atomic

C an



which

history

published in

science

politic al,

scientic

developments

that

have had important

Structure

1.2

The

nuclear atom

LHA

Practice questions Linking question

1.

State

the

nuclear

symbols

numbers

of

2.

Naturally

occurring

protons

and

for

potassium-39

neutrons

in

the

and

copper-65.

nucleus

of

each

Deduce the

How

isotope.

c an

provide sulfur

has

32

abundances:

C alculate

isotopes

33

S(95.02%),

the

four

average

A

with

the

following

34

S(0.75%),

value

for

mechanism?

36

S(4.21%)

and

isotope

evidence

tracers

for

a

reaction

natural

(Reactivity 3.4)

S(0.02%).

sulfur.

r

3.

The actual

A

value

of

sulfur

is

32.07.

Suggest

why

your

answer to the

r

previous

question

diers

from

this

value.

M ass spectrometry (Structure 1.2.3)

The

mass spectrometer

abundance

of

isotopes

(gure

in

a

9)

is

an

instrument

used

to

detect

the

relative

sample.

detector

lightest particles

positive

ions

(stage 5)

are (deflected most)

accelerated

field

heating

filament

to

in

the

electric

(stage 3)

vaporize magnet

sample

inlet

to

(stage 4)

(stage 1)

inject

heaviest particles

sample

(deflected

least)

N

electron

ionize

beam to

sample

(stage 2)

S



Figure 9

The

sample

within

a

the

result,

is

injected

sample

the

known as

Schematic diagram

are

atoms

c ations.

into

then

lose

For

of

the

a mass spectrometer

instrument

bombarded

some

of

example,

their

and

with

vaporized

high-energy

electrons

copper

atoms

to

form

c an

be

(stage 1). The atoms

electrons

positively

ionized

as

(stage

2).

As

charged ions,

follows:

+

Cu(g) + e

→

Cu

(g) + 2e

The resulting ions are then accelerated by an electric eld (stage 3) and deected by

a magnetic eld (stage 4). The degree of deection depends on the mass to charge

ratio (m/z ratio). Particles with no charge are not aected by the magnetic eld and

therefore never reach the detector . The species with the lowest m and highest z will

be deected the most. When ions hit the detector (stage 5), their m/z values are

determined and passed to a computer . The computer generates the mass spectrum

of the sample, in which relative abundances of all detected ions are plotted against

their m/z ratios(gure 10).

29

Structure

LHA

u

1

Models

Figure 10

of

the

particulate

nature

of

matter

M ass spectrum of a

sample of copper

100

80 ytisnetni

60

evitaler

40

20

0

0

60

62

64

66

m/z

The

operational

examination

details

of

the

mass

spectrometer

will

not

be

assessed in

papers.

Worked example 3

Figure

A

,

of

11

shows

boron

a

from

mass

this

spectrum

mass

from

a

sample

of

boron.

C alculate

the

relative

atomic

mass,

spectrum.

r

100

80.1

ytisnetni evitaler

50

19.9

0

0

2

4

6

8

10

12

m/z



Figure 11

M ass spectrum of boron

Solution

First,

we

number

mass

We

need

of

10,

number

c an

then

to

derive

which

of

11,

the

has

a

which

c alculate

A

information

relative

has

by

a

from

graph.

abundance

relative

finding

the

of

abundance

sum

of

The

19.9%.

the

of

peak at

The

m/z

peak at

=

10

m/z

represents

=

11

an

isotope

with

a

mass

represents an isotope with a

80.1%.

relative

abundance

of

each

isotope

multiplied

by

its

mass

r

number. The relative abundance values add up to 100%, so we divide the result by 100 to obtain the average.

11

×

80.1

+

10

×

19.9 =

100

30

10.8

68

Structure

1.2

The

nuclear atom

LHA

Data-based questions

1.

Estimate

atomic

the

relative

mass,

A

,

for

abundance

this

of

element

each

and

isotope

identify

the

from

gure

12.

Use

your

estimates

to

c alculate

the

relative

element.

r

t

Figure 12

M ass spectrum

of unknown element

6

5

ytisnetni

4

evitaler

3

2

1

0

204

203

205

206

207

208

209

m/z

2.

M ass

spectrometry

including

those

of

is

used

cosmic

for

discovering

origin.

For

the

presence

example,

cobalt

of

and

specic

nickel

elements

are

in

common

geologic al

samples,

components

of

iron

meteorites

(gure14).

Cobalt

and

nickel

compositions

of

have

these

similar

two

properties

metals

are

and

very

nearly

dierent,

identic al

so

they

relative

c an

atomic

easily

be

masses.

However,

distinguished

by

the

mass

isotopic

spectrometry

(gure13).

100

100

80

ytisnetni evitaler

ytisnetni evitaler

80

nickel

60

40

cobalt

60

40

20

20

0

0

0

58

60

0

62

58

60



Figure 13

M ass spectra of cobalt (le) and nickel (right)

Estimate

relative

the

relative

atomic

abundance

mass,

A

and

of

hence

each

isotope

deduce

for

whether

nickel.

cobalt

Use

or

your

nickel

estimates

has

the

The

actual

A

value

for

nickel

to

c alculate

larger

r

3.

62

m/z

m/z

its

A r

is

58.69.

Suggest

why

your

result

in

question

2

is

dierent.

r

31

1

Models

of

the

particulate

nature

of

matter

LHA

Structure



Figure 14

Tamentit

iron meteorite,

found

in 1864 in the S ahara Desert

Mass spectra

M ass

a

spectra

chance

to

c an

be

practice

found

in

various

c alculating

databases

average

atomic

on

the

mass

internet,

values

giving

you

from authentic

data.

Relevant skills

•

Tool

2:

Identify

•

Tool

3:

Percentages

and

extract

data

from databases

Instructions

1.

Using

a

database

of

your

choice,

search

for

the

mass

spectra

of

three

dierent elements.

2.

From

the

mass

spectra,

c alculate

the

relative

atomic

mass

of

each

element.

3.

Compare

booklet.

your

c alculated

Comment

on

relative

any

atomic

dierences

mass

you

to

that

stated in the data

observe.

Linking question

How does the fragmentation pattern of a compound in the mass spectrometer

help in the determination of its structure? (Structure 3.2)

32

Structure

1.2

The

nuclear atom

End-of-topic questions

5.

Which

of

the

following

statements

are

correct?

Topic review

I. 1.

Using

your

knowledge

from the

Structure 1.2

Nearly

all

mass

of

the

atom

is

contained within

topic, its nucleus.

answer

the

guiding

question

as

fully

as

possible:

II.

The

How do nuclei of atoms dier?

mass

number

shows the number of

protons in an atomic nucleus

III.

Isotopes

of

the

same

element

have

equal

Exam-style questions numbers

Multiple-choice questions

63

2.

What

is

correct

for

2+

Cu

A.

I and II only

B.

I and III only

C.

II and III only

D.

I, II and III

of

protons.

?

29

Protons

Neutrons

Electrons

A

29

34

27

B

29

34

31

6.

Which

of

C

34

63

31

D

34

29

27

1

3.

Which

values

are

the

same

for both

2

H

and

H

2

I.

the

following

species

contain

equal

numbers

A.

cobalt-58 and nickel-58

B.

cobalt-58 and nickel-59

C.

cobalt-59 and nickel-58

D.

cobalt-58 and cobalt-59

? 2

boiling point

II.

∆H

III.

number

Extended-response questions

of combustion

of

protons 7.

IV.

of

neutrons in their nuclei?

density

The

at

gold

gold

page

A.

I and III only

B.

I and IV only

C.

II and III only

D.

I, II and III

a.

foil

foil.

experiment

This

involved ring alpha particles

experiment

is

depicted

in

gure 1 on

20.

An alpha particle is a helium nucleus. State the

nuclear

b.

symbol

Suggest

that

the

would

for

an

results

have

of

been

alpha

the

particle.

gold

foil

observed

in

[1]

experiment

each of the

6

4.

The

naturally

occurring

isotopes

of

lithium

are

following

Li and

alternative scenarios:

7

Li.

Which

shows

the

correct

approximate

percentage i.

abundances

for

Atoms

are

instead

hard,

dense, solid balls

lithium? of

Percentage

Percentage

6

abundance of

ii.

positive

Atomic

charge.

nuclei

are

[1]

instead

negatively

7

Li

abundance of

75

25

B

50

50

Li

charged.

[1]

39

8.

There

are

two

stable

isotopes

of

potassium:

K and

41

K. The

A

of

potassium

is

39.10.

Use

this

information

LHA

A

r

C

35

65 to

D

10

determine

isotopes

90

and

the

relative abundances of the two

sketch

the

mass

spectrum

of

potassium

metal.

9.

“Dutch

14%

[3]

metal”

zinc.

oen

used

Dutch

mass

is

This

for

metal

an

alloy

alloy

making

c an

be

spectrometry.

composed of 86% copper and

closely

resembles gold, so it is

costume

jewellery.

distinguished

Explain

how

from gold using

[2]

33

Electron congurations

Structure 1.3

How c an we model the energy states of electrons in atoms?

This question is complex with many layers. What are electrons? How do we know they exist in energy states? What various

models about these energy states are there?

According

behaviour

sizes

of

to

modern

has

these

no

views,

analogues

clouds

electrons

in

depend

our

on

are

quantum

everyday

the

life,

energies

of

objects

we

c an

that

behave

visualize

electrons,

which

as

both

electrons

c an

in

have

particles

atoms

only

as

and

fuzzy

certain,

waves.

clouds.

predened

Although

The

such

shapes

and

v alues.

Understandings

Structure 1.3.1

atoms

emitting

return

to

lower

—

Emission

photons

energy

spectra

when

are

produced

electrons

in

by

excited states

Structure 1.3.5

—

state

electron

—

The

line

emission

provides

evidence

given

for

the

and

Sublevels

E ach

c an

orbital

hold

contain

has

a

dened

conguration

two

a

xed

a

high

and

energy

chemic al

electrons of opposite

number

of

orbitals,

regions

spectrum of

of hydrogen

a

environment,

levels.

spin. Structure 1.3.2

for

space

where

there

is

probability of nding an

existence of

electron. electrons

in

discrete

energy

levels,

which

converge at

energies.

Structure 1.3.3

—

The

main

energy

level

is

given an

Structure 1.3.6

—

In

of

at

higher

—

Successive

convergence

an

emission

spectrum, the limit

frequency

corresponds to

2

integer

number,

n,

and

c an hold a maximum of 2n

ionization.

electrons. Structure 1.3.7

ionization

energy data

Structure 1.3.4 — A more detailed model of the atom for

an

element

give

information

about

its

electron

describes the division of the main energy level into s, p, d conguration.

and f sublevels of successively higher energies.

Emission spectra (Structure 1.3.1)

Much

of

studies

that

our

understanding

involving

sunlight

prism.

This

c an

be

which

each

example

of

gaseous

pure

glow —

as an

it

to

the

within

spectrum

a

prism

and

1c)

the

is

subjected

series

a

continuous

(gure

into

and

spectrum

words,

a

into

congurations

In

the

1600s,

dierent

next,

the

as

and

atoms

Isaac

has

come

Newton

from

showed

coloured components using a

(gure

appears

in

Sir

a

no

rainbow.

1a).

This

type

continuous

gaps

The

are

of

series

visible.

spectrum

of

The

colours,

classic

wavelength of visible light

700 nm.

element

other

produces

down

merges

emission spectrum

between

lines

400 nm

in

light.

continuous spectrum

continuous

A

prism,

34

a

from

electron

with

wavelengths,

colour

ranges

will

broken

generates a

contains light of all

in

of

interaction

it

of

will

lines

(gure

source

to

emit

a

high

light.

against

1b)

of

spectrum

In

dark

contrast,

visible

will

a

voltage

When

light

appear.

under

this

light

reduced

passes

background.

when

of

all

This

a

cold

This

gas

pressure

through a

is

is

known

placed

wavelengths, a series of dark

is

known as an

absorption

LHA

higher

Structure

1.3

Electron

congurations

a continuous

emission

spectrum

spectrum

b hot gas

c cold gas



Figure 1

The spectra generated

absorption

from

(a) visible light

of

all wavelengths (b) a heated

spectrum

gas (c) visible light

of all wavelengths

passing through a cold gas



Figure 2

The aurora borealis (Northern Lights) in Lapland,

drawn by the E arth’s magnetic eld

to the polar regions,

Sweden.

Charged

high-energy particles from

where they excite atoms and

the Sun are

molecules of atmospheric gases,

c ausing them to emit light

35

Structure

1

Models

of

the

particulate

nature

of

matter

Emission spectra

Emission

spectra

handheld

Discharge

ionized

c an

be

spectroscope

lamps

when

a

observed

by

contain

voltage

through a simple

holding

it

up

low-pressure

is

to

a

gases

light

Method

source.

which

1.

Observe natural light through the spectroscope. Note

2.

O bs e r ve

are

down the details of the spectrum you observe.

applied.

L E D. Relevant

a r ti fi c i a l

No te

d ow n

light

the

f ro m

a

details

c o m pu te r

of

th e

s c re e n

s pe c tr u m

or

yo u

skills o bs e r ve.

•

Tool

•

Inquiry

3:

Construct

graphs

and

draw lines of best t 3.

2:

Identify

and

record

relevant

Observe

down observations

and

sucient

relevant

light

the

Inquiry

2:

Identify

and

describe

patterns,

Inquiry

2:

various

of

the

discharge lamps. Note

emission

colours,

lines

you

observe,

wavelengths and number of lines.

trends and

Q uestions

relationships

•

details

quantitative data. including

•

from

qualitative

Assess

accuracy

1.

Sketch

2.

Describe

the

spectra

each

as

you

a

observed.

continuous,

emission or

S afety absorption

•

Wear

•

The

eye

protection. 3.

discharge

lamps

will

get

very

hot.

Look

up

the

discharge

lamps

spectra of the elements in

you

observed.

Compare the

c are. theoretic al

•

emission

Handle them the

with

spectrum.

Further

safety

prec autions

will

be

given

by

depending

on

the

exact

observed

emission lines, commenting

your on

teacher,

and

the

number,

colours and positions of the

nature of the emissionlines.

discharge lamps. 4.

Next,

you

will

wavelengths

compare

of

the

the

theoretic al

emission

lines.

and

observed

Construct

a

graph

M aterials

of

•

Discharge lamps

•

Handheld

theoretic al

Draw

a

line

of

wavelength

best

t

vs

observed

through

wavelength.

your data.

spectroscope

5.

Comment on the relationship shown in your graph.

6.

Comment

on

the

accuracy

of

the

observed

wavelength data.

E ach

to

element

identify

orange

same

the

light

emission

36

Figure 3

Sodium streetlights (le) and

its

own

element.

with

Like

example,

colour

barcodes

spectra

characteristic

For

wavelengths

yellow-orange

substance.



has

c an

be

in

of

589.0

appe ars

a

used

shop

to

line

in

and

a

that

sodium

test

be

of

used

chemic al

which

atoms

589.6 nm

ame

c an

identify

the line emission spectrum of sodium (right)

spectrum,

excited

c an

emit

(gure3,

any

to

be

used

yellow-

right).

The

sodium-containing

identify

elements.

products,

line

Structure

1.3

Electron

congurations

Observations

Chemists

oen

properties

directly

of

through

instruments.

boundaries

the

of

our

light

in

the

helium

emission

human

features

through

from observing the

senses

in

a

lamps

detail.

As

is

also

of

of

is

in

the

dierence

between

observing

a

natural

directly and with the aid of an instrument?

expand the

to

vapour lamps

gure 3, observing

reveals

the

orange

helium

is

phenomenon

(oen sight), or with

Sodium

seen

region

What

revealing otherwise

spectroscope

yellow

spectrum

or

light.

c an be made

technology

observations,

orange-yellow

emission

from

the

data

Observations

Advancements

imperceptible

emit

generate

matter.

a

strong

spectrum. The light

the

more

naked

complex

eye but the



Figure 4

Helium emission spectrum

(gure 4).

Flame tests

Flame testing is an analytical technique that can be used to

Materials

identify the presence of some metals. The principle behind

•

Flame

ame tests is atomic emission. Electrons are promoted

•

Small

to a higher energy level by the heat of the ame. When

•

Bunsen

they fall back to a lower energy level, photons of certain

•

Small

wavelengths are emitted. Some of these photons are in the

KCl,

test

wire

portion

burner

samples

C aCl

(platinum

of

dilute

and

of

, SrCl 2

or

nichrome)

hydrochloric acid

heatproof mat

various

, CuCl 2

metal

salts

(e.g.

LiCl,

NaCl,

) 2

visible region of the spectrum. Method

1.

Clean the end of the ame test wire by dipping it into

the HCl solution and placing it in a non-luminous Bunsen

burner ame. Repeat until no ame colour is observed.

2.

Dip

the

end

samples,

Bunsen

metal



Figure 5

Flame test

of

and

burner

in

the

the

ame

place

it

in

ame,

salt

and

test

the

noting

the

wire

into

one

of

the

salt

edge of the non-luminous

down the identity of the

colour(s)

you

observe.

colours for dierent elements

Relevant skills

•

Inquiry

2:

Identify

and

record

relevant

qualitative

observations

S afety

•

Wear

•

Take

eye

protection.

suitable

•

Dilute

•

A

prec autions

3.

Clean the wire again and repeat with other salt samples.

4.

Clear

around open ames. up

as

instructed

by

your

teacher.

hydrochloric acid is an irritant.

variety

of

dierent

chloride

salts

will

be

used, some Q uestions:

of

which

are

irritants —

avoid contact with the skin. 1.

•

Dispose

•

Further

of

all

substances

Look up the emission spectra of the metals you tested.

appropriately. Compare these to the colours you observed. Comment

safety

prec autions

will

be

given

by

your on any similarities and dierences.

teacher,

depending

on

the

identity

of

the

salts being 2.

Explain

why

the

dierent

metals

show

dierent

analysed. amecolours.

37

Structure

1

Models

of

the

particulate

nature

of

matter

TOK

One

of

the

ways

knowledge

Inductive reasoning

up”:

they

take

is

involves

specic

developed

drawing

observations

is

through

conclusions

and

build

reasoning.

from

general

inductive

Reasoning

experimental

principles

reasoning

c an

be

deductive

observations.

or

Inductive

(“bottom-up”

3.

1.

For

example,

you

might

the

following

Lithium

chloride

Lithium

sulfate

gives

a

red ame test.

Lithium

iodide

gives

a

red ame test.

From

these

observations,

Deductive arguments

when

a

asked

to

apply

are

approach):

c an

“top

make

down”:

scientic

the

conclusion

they

infer

knowledge

hypothesis

about

lithium

salts:

in

that

specic

a

new

all

lithium

salts

conclusions

give

from

red ame tests.

general

reasoning

(“top-down”

approach):

pattern

4.

example,

Lithium

From

What

C an

this,

are

suppose

bromide

salts

you

the

give

is

your

a

scientic

lithium

knowledge

includes

the

following

premises:

salt.

propose

advantages

always

be

and

that

lithium

bromide

disadvantages

neatly

classied

of

into

gives

each

these

a

type

two

red ame test.

of

reasoning?

types?

On

what

grounds

might

we

doubt

a

claim

reached

through

inductive

On

what

grounds

might

we

doubt

a

claim

reached

through

deductive

Visible light is one type of

light,

are

microwaves,

all

The

part

of

energy

1 E

∝

λ

38

existing

observation

red ame tests.

could

reasoning

You do this all the time

hypothesis

3.

Lithium

premises.

context.

theory

2.

For

theory

pattern

observations

deductive

1.

“bottom

red ame test.

you

your

are

observation

make

gives

arguments

from them.

4.

2.

inductive.

the

of

reasoning?

reasoning?

electromagnetic (EM) radiation. In addition to visible

infrared

radiation

electromagnetic

the

radiation

is

(IR),

ultraviolet

(UV),

X-rays

and

gamma

spectrum.

inversely

proportional

to

the

wavelength,

λ:

rays

Structure

Electromagnetic

waves

all

travel at the

speed of light,

8

of

light

is

approximately

frequency

of

the

equal

radiation,

f,

to

by

3.00 × 10

the

c,

in

a

vacuum.

The

1.3

Electron

congurations

speed

–1

m s

following

.

Wavelength

is

related to the

equation:

c = f × λ

High

energy

EM

waves,

such

as

gamma

rays,

have

short

wavelengths and high

frequencies while low energy waves, such as microwaves, have long wavelengths

and

low

frequencies.

f

λm

1

10 4

10

gamma

rays

14

10 

10

(γ

rays)

λnm

1

10 0

10 400

10 1

10

X-rays

10



ultraviolet

10

00

1

elbisiv

10 ygrene

(UV)



10 14

10 infrared

00 (IR)

4

10 1

10

 10

Activity

10

microwaves

10

00

Compare

0

the

colours

red and

10 

10

green



in

gure 6. Determine which

colour has:

10 

10

radio

waves

a.

the

highest

wavelength

b.

the

highest

frequency

c.

the

highest

energy

4

10 4

10



Figure 6

The wavelength (λ) of electromagnetic radiation is inversely

proportional to both frequency and

energy of that

radiation

Data-based questions

Look

at

the

spectra

below.

Explain

how

we

know

that

stars

are

partly

composed

of

hydrogen.

3900

4000



Figure 7

7600

4500

5000

The hydrogen emission spectrum

5500

(top) and

6000

6500

the absorption spectrum generated

from

7000

7500

the Sun (bottom)

39

Structure

1

Models

of

the

particulate

nature

of

matter

The line emission spectrum of hydrogen

(Structure 1.3.2 and 1.3.3)

E ach

line

which

the

A

in

idea

that

photon

radiation

E

=

the

emission

corresponds

is

as

to

a

spectrum

specic

electromagnetic

a

quantum

of

of

an

element

amount

radiation

energy,

of

comes

which

is

has

energy.

in

a

specic

This

is

wavelength,

c alled

quantization:

discrete packets, or quanta.

proportional

to

the

frequency of the

follows:

h × f

Where

E =

the

specic

energy

possessed

by

the

photon,

expressed in joules, J

–34

h =

f

=

Planck’ s

constant,

frequency

of

the

6.63 × 10

radiation,

J s

expressed

in

hertz,

Hz,

or

inverse

–1

seconds, s

In

1913,

Niels

emission

1.

The

electron

These

2.

orbits

When

3.

of

the

at

that

an

This

c an

are

main

exist

the

for

in

a

a

the

his

only

in certain

the

orbit

stationary

orbits

the

lowest

moves

to

to

the

a

lower

energy

orbiting

making

any

electrons

Since

electrons

attempt

not

electrons

dened

to

in

spectra.

of

the

specic

By

energies

For

a

of

of

model

transitions

the

energy

higher

energy

overcome

based on its

around the nucleus.

level absorbs a photon

energy

level

and

remains

atoms

of

level, it emits a photon of light.

between

the

main

Classic al

the

two

levels.

problem of the

electrodynamics

predicted

energy and quickly fall into the nucleus,

when

wavelengths,

measuring

the

energy

Bohr

their

radiate

existence

radiate

energies,

photons

would

prolonged

did

a

dierence

Rutherford model of the atom (Structure 1.2).

that

atom

were:

time.

returns

rst

it

hydrogen

theory

discrete energy levels

with

energy,

represents

was

of

of

the

of

short

electron

photon

model

postulates

associated with

amount

level

theory

proposed

The

electron

right

When

Bohr ’s

Bohr

spectra.

impossible.

Bohr

postulated that

staying in stationary orbits.

the

atom

could

have only certain, well-

between stationary orbits could absorb or emit

producing

wavelengths

of

characteristic lines in the atomic

these

lines,

it

was

possible

to

c alculate

electrons in stationary orbits.

hydrogen

atom,

the

electron

energy (E

)

in

joules

could

be

related to the

n

energy

level number (n)

by

a

simple

equation:

1 E

= –R n

H

2

n

–18

where

R

≈

2.18 × 10

J

is

the

Rydberg

constant.

This

equation

clearly

represents

H

the

quantum

only

discrete,

of

the

quantized

half-integer

parameters,

number (n)

c an

numbers

40

nature

mean

take

values.

where

These

known as

only

higher

atom,

positive

energy.

the

energy

values

are

of

an

electron

characterized

quantum numbers. The

integer

values

(1,

2,

3,

c an

by

have

integer or

principal quantum

…),

where

greater

Structure

The

most

electron

stable

has

ground state

c alled

return

state

the

of

lowest

of

the

atom.

excited states.

to

the

the

ground

hydrogen atom is the state at

possible

In

contrast,

Atoms

state

energy.

by

in

the

excited

emitting

This

energy

energy

states

of

=

is

unstable

specic

1,

Electron

congurations

where the

known as the

levels with

are

photons

n

level

1.3

n

=

and

2,

3,

…

are

spontaneously

wavelengths

(gure8).

+energy e

+

+

p

+

p

e

p

excitation

dec ay

hf



Figure 8

Energy

rungs.

stand

levels

An

in

atoms

Electrons

between

specic,

the

Electrons returning to lower energy levels emit a photon of light, hf

amount

electron

energy

the

discrete

same

c an

level.

spectrum

of

resemble

c annot

exist

rungs

amount

of

be

of

of

ladders

between

a

ladder.

energy,

with

varying

energy

Jumping

and

levels,

up

distances

between the

much

how

each

jumping

rung

down

a

like

or

rung

level

or

you

c annot

requires a

level

releases

energy.

excited

Electrons

hydrogen

to

any

energy

returning to

n

=

2

level,

will

n,

and

return

to

any

lower

produce distinct lines in the visible

(gure 9).

Note that the red line has a longer wavelength and lower frequency than the violet

line. The energy of the photon released is lower when an electron falls from n = 3 to

n = 2, than from n = 6 to n = 2. In both cases, it represents the dierence between

two of the allowable energy states of the electron in the hydrogen atom.

colour

wavelength / nm

transition

from

violet

blue

cyan

red

410

434

486

656

n

= 6

n

= 5

n

= 4

n

n

= 6

n

= 5

n

= 4

n

= 3

n

= 2

n

=

= 3

1

◂

Figure 9

The visible lines in the

emission spectrum

of hydrogen

show electrons returning from higher

energy levels to energy level n = 2

41

Structure

1

Models

of

the

particulate

nature

of

n = 7

matter

Electron

transitions

to

the

ground

state,

n =

1,

release

higher

energy, shorter

n = 6 wavelength

ultraviolet

light,

while

electrons

returning to

n =

3

produce lines in

n = 5 the

infrared

region

It

important

of

the

electromagnetic

spectrum

(gure 10).

n = 4

n = 3

is

required IR

to

note

that

electrons

between

will

allowable

absorb

energy

or

rele ase

states.

Any

only

the

excess

exact

will

not

energy

be

radiation absorbed,

n

to

move

not

= 2

and

if

an

insucient

amount

of

energy

is

supplied

the

electrons

will

move.

visible light

Energy

levels

closer

to

the

nucleus

hold

fewer

electrons. The maximum number

2

of

electrons

holds

n n

=

3

to

has

a

any

two

energy

level,

electrons, at

maximum

of

18

n

n, is 2n

=

2

. For

there

example,

could

electrons, and

n

=

be

4

a

has

the

energy

maximum

a

of

maximum

level with

eight

of

32

n

=

electrons.

radiation

Figure 10

Electron transitions for the

Communic ation skills

ATL hydrogen atom.

Notice how the allowable

energy levels get closer together when

When

explaining

concepts,

we

sometimes

use

diagrams,

graphs or images to

the electron moves further away from the

help nucleus.

us

convey

our

ideas

more

clearly.

The energy dierence between

Prepare n = 3 and

a

written

explanation

of

atomic

emission that does not include

n = 2 is much smaller than that

any between n = 2 and

diagrams.

Exchange

it

with

a

partner.

Give

each

other

feedback,

n = 1

concentrating on:

•

Use

•

Order

•

Whether

When

of

you

feedback

or

scientic

in

any

have

to

diagram

whether

which

or

are

important

shared

make

to

voc abulary

ideas

each

it

other ’ s

improvements

accompany

not

given

concepts

adds

to

your

the

to

are

missing

from

the

explanation.

feedback, spend some time using the

your

work.

explanation.

Finally,

Discuss

choose

why

a

graph, image

you chose it and

explanation.

Linking questions

What

qualitative

such

as

gas

from

gaseous

How

does

period

42

quantitative

do

an

tubes

elements

different

How

and

discharge

emission

and

element’s

in

the

data

c an

prisms

in

be

the

collected

study

of

from instruments

emission

spectra

from light? (Inquiry 2)

spectra

elements?

number

and

provide

evidence

for

the

existence of

(Structure 1.2)

highest

periodic

occupied

table?

main

(Structure

energy

3.1)

level

1

electrons,

1

UV



=

up

in

relate to its

Structure

1.3

Electron

congurations

The quantum mechanic al model

of the atom (Structure 1.3.4)

The

Bohr model

atoms.

energy

of

It

was

levels.

narrow

several

1.

It

3.

It

bec ause

model

more

attempt

than

and

could

one

assumed

the

to

explain

quantization:

allowable

problems

The

2.

in

an

on

According

lines

dierences

was

based

to

the

Bohr,

the

incorrect

electron

It

energy

that

emission

levels.

of

states

electrons

spectra

these

However,

lines

this

of

of

electrons in

existed

in

hydrogen

discrete

consisted

corresponded to the

model

was

limited

by

assumptions:

predict

electron.

the

idea

wavelengths

energy

not

the

the

was

was

a

emission

only

spectra of elements containing

successful

subatomic

with

particle

in

the

a

hydrogen atom.

xed orbit about the

nucleus.

could

not

account

for

the

eect

of

electric and magnetic elds on the

spectral lines of atoms and ions.

4.

It

5.

Heisenberg’s

could

not

explain

molecular

bonding

and

geometry.

The uncertainty

principle

states

that

it

is

impossible

to

principles

bonding know

the

loc ation

and

momentum

of

an

electron

simultaneously.

and

stated

that

electrons

exhibited

xed

momentum

in

molecular

geometry

are

Bohr ’ s Structure 2.2

explained in model

behind

precisely

specic

circularorbits.

Bec ause

the

of

these

modern

limitations,

quantum

the

Bohr

theory

has

been

eventually

superseded

by

mechanic al model of the atom.

TOK

The

modern

quantization

quantum

with

the

mechanics

combines

Heisenberg’s uncertainty principle

impossible

and

the

that

the

the

less

more

we

it

the

of

of

we

a

particle

know

is

possible

not

of

its

an

probability

the

position

momentum,

to

pinpoint

electron

of

nding

in

the

an

an

of

and

This

an

means

electron,

vice

versa.

electron

we

in

the

the

boundaries

of

waveforms),

are

the

limits

of

human

Wave–particle duality

subatomic

species

to

is

the

The

of

around

obstacles),

all

(combination

obstacles) and

of

which

are

Einstein

(1879–1955)

described

by the

in

equation

1926

of

the

by

the

electron

is

equation,

Austrian

quantitatively

which

was

physicist Erwin

(1887–1961). Solutions to the Schrödinger

give

functions,

duality

Schrödinger

a

series

known as

states

and

of

three-dimensional

wave functions,

energies

of

mathematic al

which

describe the

electrons in atoms.

knowledge?

ability

behave

characteristics

released

even whole atoms and

interference

(bending

through

or

nature.

waves.

wave–particle

as

of

both

The

concept

that

objects

these

of

wave–particle

duality

illustrates the fact

electrons and other

particles

and

species,

such

as

of

study

do

not

always

fall

neatly into the

waves.

discrete Certain

and

of

absorbed

phenomena of light, but together they do.

possible What

to

be

particulate

separately neither of them fully explains the

Schrödinger

knowledge?

c apable

diraction

to

their

electrons,

are

(passing

formulated

(1908–1974)

tendency

suggest

We have two contradictory pictures of reality;

Albert

implic ations of this uncertainty principle on

the

photons,

characteristic

space.

Bronowski

and

entities

molecules,

tunnelling

each

has been to prove that this aim is unattainable.

are

of

c an

exact picture of the material world. One achievement …

Jacob

small

loc ation or

atom,

discrete

However,

One aim of the physical sciences has been to give an

What

momentum

as

states that it is

simultaneously.

about

about

trajectory

idea of

accurately both the momentum

know

c alculate the

region

determine

position

Although

predict

to

the

following key principles.

c ategories

we

have

developed.

What

is

the

role

mass,

of

c ategorisation

in

the

construction

of

knowledge?

43

Structure

1

Models

of

the

particulate

nature

of

matter

Schrödinger ’s

probability

wave

density,

electrons

are

path,

theory

this

region

of

region

in

There

of

four

space

gives

at

a

several

types

orbitals

in

of

is

a

order

of

that

that

a

an

electrons

electron

probability

orbitals,

has

electrons in atoms in terms of their

idea that the momentum and position of

follow

will

from the nucleus. An

high

atomic

orbital

are

the

saying

distance

there

E ach

of

probability

certain

orbitals,

Subsequent

the

where

electrons.

atomic

describe

Heisenberg’s

uncertain. Instead

space

are

two

functions

using

and

of

increasing

theoretic al,

and

energy

these

orbital

shape

are

are

a

dened

found

in

a

travel

specic

atomic orbital

nding

each

characteristic

be

an

is a

electron.

c an hold a maximum

and

energy.

labelled

labelled

The

rst

s, p, d, and f.

alphabetic ally

(g, h, i, k and so on).

The

principal

main 

Figure 11

energy

quantum

levels.

number,

These

n,

energy

introduced

levels

are

by

split

the

into

Bohr

model

sublevels

represents the

comprised of

An s orbital is spheric al. The

atomic

orbitals.

For

example,

for

n

=

1,

2

and

3,

the

s

atomic

orbitals

are 1s, 2s

sphere represents the boundary space

and

3s.

As

n

increases,

the

s

orbitals

are

further

distanced

from the nucleus.

where there is a 99% probability of nding

an electron.

The s orbital c an hold two

Figure

12

shows

that,

for

1s,

there

is

a

high

probability

of

nding

electrons close

electrons

to

the

away

nucleus

from

although

nucleus.

The

the

this

is

a

There

is

zero

is

probability

nucleus.

there

same

from

the

and

true

for

nucleus

For

small

the

with

two

the

of

that

of

zero

probability

an

nding

highest

regions

reaches

highest

probability

probability

3s,

and

2s,

never

electron

the

is

at

we

move further

somewhat

could

electron

probability

zero

when

an

be

further

away,

found closer to the

between

even

the

two

peaks.

greater distance

probability.

1s

0

50

pm

2s

0

50

100

pm

average

radius

3s

0

50

100

pm



44

Figure 12

The plots of

the wavefunctions for the rst

three s orbitals

150

Structure

Imagine

that

8.00am.

they

At

could

you

are

8.15am,

be.

a

Some

possibly

students

no

from

•

may

•

might perhaps be at the airport

•

might

be

Although

at

their

even

the

of

nding

to

dene

might

be

certain

orbital

the

a

exact

school

region

A

of

the

of

the

the

the

town

region

of

or

suggest

or

so

that

laboratory,

in

the

lesson

you

the

or

school

to

congurations

begin at

wonder

where

teacher:

the

library

c ar park

Pole!

teacher

the

is

is

where

could

a

town

99%

with

be

a

the

it

drawn

your

possible

is

a

nding

is

of

draw

this

cluster

them.

an

a

probability

This

loc ated,

Similarly,

probability

to

high

around

of

school

airport.

high

is

there

chance

where

includes

space

unknown,

areas

surface

there

that

chemistry

teacher,

Electron

centre

showing

where

perimeter,

class

oce

North

dots

DP

your

chemistry

town

boundary

space

around

represents

electron

of

the

to

loc ation

cluster

the

principal’ s

in

gone

teacher.

region

the

house

have

three-dimensional

room,

your

of

your

could

school

for

sign

is

the

sta

waiting

still

•

in

the

is

•

be

in

student

there

1.3

or

a

atomic

nding

an

(gure13).

t

Figure 13

Representation of a 1s atomic orbital as

y

y

a cluster of dots (le) and

a sphere that encloses 99%

of the dots (right)

x

x

z

z

A

p orbital

is

orientations

dumbbell

shaped

parallel to the

x,

y

There

and

z

are

axes

three

(gure

p

orbitals,

14).

These

each

are

described with

labelled p

, p x

p

.

These

shapes

all

describe

boundaries

with

the

highest

and y

probability of nding

z

electrons in these orbitals.

t

Figure 14

The three p atomic orbitals

are dumbbell shaped,

aligned along the

z z

z

x, y and

x

x

x

z axes.

There is zero probability

of

nding the electron at

of

the axes between the two lobes of the

dumbbell.

E ach of

the p

the intersection

orbitals c an hold

two electrons

y

y

y

p

x

orbital

p

y

orbital

p

z

orbital

45

Structure

1

Models

of

the

particulate

nature

of

matter

Theories and models

Current

before.

the

atomic

natural

theories

are

theory

Theories

are

world.

are

evolved

Contrary

substantiated

amassed,

from

previous

comprehensive

to

by

documented

the

vast

and

systems

use

of

the

amounts

models,

of

ideas

word

of

each

that

“theory”

observations

communic ated

by

a

superseding

model

in

and

everyday

and

the

one

explain

tested

an

that

c ame

aspect of

language, scientic

hypotheses, which

large number of scientists.

+ + +

+

+ +

+

+

+ +

+

+ +

800–400 BCE

Āruņi’ s

kana

Democritus’ atomos

1897

1913

Thomson’ s “plum

ohr model

pudding” model

“billiard ball”

mechanic al

model

1803

D alton’s

1930

uantum

1912

1926

Rutherford’ s model

Heisenberg’s uncertaint

model

and

regions

of

probabilit

model



Figure 15

The atomic theory has seen the idea of

model where electrons have specic energies and

What

other

examples

of

theories

c an

you

atoms evolve from

are found

think

indestructible spheres to the quantum

mechanic al

in regions of high probability

of ?

Linking question

What

is

the

periodic

46

relationship

table?

between

(Structure

3.1)

energy

sublevels

and

the

block

nature of the

Structure

1.3

Electron

congurations

Electron congurations (Structure 1.3.5)

E ach

atomic

spheric al

oriented

higher

orbital

and

it

has

type

the

dierently.

in

energy

has

There

than

s

a

characteristic

lowest

or

are

possible

ve

d

shape

energy.

orbitals

and

There

and

energy. The s orbital is

are

seven

f

three

p

orbitals,

orbitals,

and

each

these

are

p.

z

y

x s

z

z

z

y

y

x

x

p

z

x

x



E ach

of

z

types

there

of

level

(table

orbital:

are

x

four

dened

1).

s

For

and

types:

n

p.

s,

by

=

1,

For

p,

d,

the

d

=

the

3,

and

s

x

are

three

For

types:

n

s,

=

n,

2,

p,

p

and

d.

n

of

of of

3

to be known

types

are two

For

n = 4,

number

(n)

orbitals

M aximum

number

of

orbitals per

per

energy

electrons

within

type 2

level

s

1

s

1

p

3

2

3

x

f 2

f.

sublevel

1

x

f

orbitals need

c an hold

there

y

Number Type

quantum

number

number,

exists.

Total Principal

x

1

Only the shapes of s and

quantum

z

y

f 0

orbital

there

z

y

f

f orbitals.

principal

only

n

and

2

z

–1

p,

x d

y

f

The shapes of the s,

x

1

z

x

y

d

y

–2

energy

orbitals

x

0

y

f

Figure 16

y

d –1

x

z

y

d

z

–3

z

y

–2

f

1

z

d

y

p 0

y

z

x

p –1

z

y

s

1

p

3

d

5

(n

2

)

energy

level

1

2

4

8

9

18

(2n

)

t

Table 1

c an hold 2n

sublevels,

s

1

p

3

d

5

Each energy level,

dened

by

2

n,

electrons. The number of

or atomic orbital types, is equal

to n. For n = 4 there are four types of orbitals

(s,

4

16

p,

d,

and

f ) with 16 atomic orbitals in total

32 2

occupied

by a maximum of 2(4)

= 32 total

electrons

f

7

47

Structure

1

Models

of

the

particulate

nature

of

matter

Activity

State

the

following

for

the

energy

level with

a.

the

sublevel types

b.

the

number

c.

the total number of atomic orbitals

d.

the

of

maximum

atomic

orbitals

number

of

in

each

electrons

at

n

= 5:

sublevel

that

energy

level.

Orbital diagrams

For

to

convention,

represent

arrangement

u

Figure 17

In orbital diagrams,

represents an orbital.

each box

This diagram

s

sublevel

an

how

of

(one

“arrow

in

electrons

electrons

box

box”

are

in

notation

arranged

orbitals

is

in

c alled an

atomic

orbital diagram

orbitals

is

used

(gure 17). The

c alled electron conguration

representing an s orbital)

shows

the number of orbitals for each sublevel.

Arrows are drawn in the boxes to represent

electrons.

A maximum of two electrons

c an occupy each orbital,

so each box has a

maximum of two “arrows”

p

sublevel

(three

d

sublevel

(five

f

sublevel

Atomic

two

orbitals

electrons

of

are

solves

Hence

downwards

not

this

with

of

space

charged

be

problem

each

orbital

half-arrow,

three p orbitals p

five

the

d

where

to

by

using

spins

box

is

(gure

a

f

±

, and p

y

)

z

orbitals)

there

and

occupy

, p

x

orbitals)

seven

negatively,

able

opposite

the

the

representing

regions

should

electrons

directions.

one

are

representing

representing

boxes

Electrons

mechanics

pair

boxes

(seven

electrons.

boxes

the

is

a

like

high

same

spin

probability of nding

charges

region

notation

for

repel

of

each

space.

each

other, so

Quantum

electron. A

behave like magnets facing in opposite

shown

18).

This

with

is

an

upwards

known as the

half-arrow,

, and

Pauli exclusion

principle:

Only two electrons c an occupy the same atomic orbital and those electrons

must have opposite spins.

48

Structure

N

t

S

Figure 18

1.3

Electron

congurations

Electron spin is represented

by an arrow pointing up

(positive spin) or

down (negative spin)

S

N

N

S

S

N

magnet analogy

half-arrows

representing 3d

electrons

of opposite spin

in an orbital degenerate

3p

Electron

own

spin

axis.

is

oen

However,

interpreted

this

as

the

interpretation

rotation

has

no

of

the

physic al

electron

basis:

ygrene

TOK

around its

degenerate 3s

electrons in

2p atoms

the

as

behave

spin

they

nor

have

like

the

no

waves,

wave-like

and

a

wave

behaviour

analogues

in

our

c annot

of

rotate.

electrons

everyday

life

Unfortunately, neither

c an

and

be

c an

visualized

be

in

any

way,

2s

expressed only in

degenerate

mathematic al

theory

the

but

power

form. This lack of visualization does not undermine the quantum

rather

of

shows

the

mathematics

limits

as

the

of

human

language

perception

of

and,

at

the

same

1s

time,

science. 1

To

what

extent

does

mathematics

support

knowledge

2

3

development in the n

natural

sciences?



Figure 19

The three 2p orbitals are

degenerate as they have the same energy.

E ach

of

the

Orbitals

atomic

with

the

orbitals

same

of

the

energy

same

are

type

in

one

referred to as

sublevel

are

degenerate

of

equal

orbitals

energy.

(gure 19).

These three degenerate atomic orbitals

have lower energy than the three 3p orbitals

49

Structure

1

Models

of

the

particulate

nature

of

matter

An atom of boron (B) has ve electrons, and its orbital diagram is drawn as follows:

1

2p

2

2s boron

(B)

2

1s

The

single

equal

the

electron

energies.

together

in

2p

to

The

show

lemost

Hund’ s rule

states

boron

c an

degenerate

their

box,

in

energy

although

that

every

2p

occupy

any

orbitals

are

equivalence.

it

is

a

matter

of

personal

become

electron

doubly

with

the

occupied

orbitals,

by

the

as

boxes

they

have

joined

half-arrow

is

drawn

preference.

degenerate orbital in a sublevel is singly

occupied orbitals have the same spin. This

one

three

Traditionally,

of

occupied before any orbital is doubly occupied

have

the

represented

same

with

spin

an

in

each

electron

and that

means

of

of

that

them

all electrons in singly

the

three p orbitals must

before

opposite

spin

any

orbital

c an

(gure 20).

Practice questions

1.

Look

Why

2.

at

gure

do

State

you

which

of

conguration

State

the

20.

think

the

The

the

and

and

diagrams

based

reason

1s

1s

for

on

the

2s

2s

below

Hund’s

four

orbitals

orbitals

are

are

fully

represents

rule

and

incorrect

occupied

lled

the

a

before

correct

Pauli

diagrams

by

the

orbitals?

electron

exclusion

being

electrons.

2p

principle.

wrong.

A.

1s

2s

2p

1s

2s

2p

1s

2s

2p

1s

2s

2p

1s

2s

2p

B.

C. 1s

2s

2p

D.

1s



distributed

2p

2s

Figure 20

2p

The electrons are evenly

across the three degenerate

E.

orbitals in nitrogen before an orbital is

doubly occupied

The Aufbau principle states that as electrons are added to atoms, the lowest

available energy orbitals ll before higher energy orbitals do. The third and

fourth energy levels contain d and f orbitals (gure 21). These orbitals are typically

lled aer the s orbitals of the following levels because they are higher in energy.

As shown in gure 21, the 3d sublevel has a higher energy than 4s but lower than

4p, so 4s is lled with electrons rst, followed by 3d and nally 4p. For the same

reason, 4d orbitals are lled aer 5s, and 4f orbitals are lled only aer 6s.

50

Structure

t

Figure 21

energy and

1.3

Electron

congurations

The 4s sublevel has a lower

will ll before the 3d

sublevel

4f 6s

5p

4d 5s

ygrene

4p

3d 4s

3p

3s

2p

2s

1s

This

is

consistent

c alcium,

Ca

with

have

experimental

electrons

in

the

data

4s

that

show

that

potassium, K, and

sublevel, not in 3d.

t

Figure 22

Potassium

orbital lling diagram showing the outermost

4p electron in the 4s orbital bec ause 3d

orbitals are higher in energy

3d

4s

3p

3s

ygrene

2p

2s

1s

Generally,

the

following

order

is

observed:

Activity

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p …

Copy

the

orbital

diagram

from

In the IB Diploma Programme, only the electron congurations of atoms and ions gure21

and

complete

it

for the

up to Z=36 will be assessed. Electrons in these species can ll sublevels up to 4p. following

elements

in

their

ground

states: Electron

sharing

reactions,

ion.

There

1.

Full

so

it

are

is

and

important

three

electron

transfer

ways

are

to

to

fundamental

know

show

the

the

to

electron

electron

understanding

chemic al

a.

aluminium, Al

b.

chlorine, Cl

c.

iron,

conguration of an atom or an

conguration:

Fe

conguration Refer to the periodic table at the

back 2.

Condensed

electron

of

this

number

3.

Orbital

The

orbital

lling

lling

diagram

diagram

book

to

deduce the

conguration

(“arrows

for

in

of

electrons

in

each atom.

boxes” notation)

potassium

is

given

in

gure 22.

51

Structure

1

Models

of

the

particulate

nature

of

matter

Full electron congurations

To

write

the

a

full

electron

electrons

Rule

and

the

in

conguration,

successive

Pauli

orbitals

exclusion

we

use

the

according

periodic

to

the

table,

Aufbau

and

“build

principle,

up”

Hund’s

principle.

Worked example 1

Determine

the

full

electron

conguration

for

the

c alcium

atom.

Solution

The Aufbau principle states that as electrons are added to atoms, the lowest

available energy orbitals ll before higher energy orbitals do. From the Pauli

exclusion principle, we know that each orbital will have a maximum of two

electrons.

The atomic number of calcium is 20. Let’ s split the 20 electrons evenly across

each orbital, starting with the lowest energy rst. When writing electron

congurations, the number of electrons within each sublevel is given in

superscript, next to the sublevel:

2

•

The

1s

•

The

2s

orbital

•

The

three

•

The

3s

•

The

three

has

two

electrons: 1s

2

orbital

also

has

two

electrons: 2s

6

2p

orbitals

have

two

electrons

each, six in total: 2p

2

orbital

has

two

electrons: 3s

6

3p

orbitals

have

three

electrons

each, six in total: 3p

Practice question

This 3.

Determine

the

full

brings

us

up

to

18

electrons,

with

two

le

over to go into the orbital with

electron 2

conguration

the

next

So,

for

lowest

energy, 4s: 4s

for the phosphorus 2

c alcium,

the

full

electron

conguration is 1s

2

2s

6

2p

2

3s

6

3p

2

4s

atom.

Condensed electron congurations

As

the

gets

ions

atomic

longer

is

mostly

electrons,

electron

inner

it

c an

of

18

an

be

electrons

as

is

element

by their

inner

to

to

the

the

full

electron

conguration

write. The chemistry of atoms and

valence electrons, that is, the outermost

core electrons.

highlight

having

(known as the

increases,

time-consuming

determined

rather than the

congurations

core

group

number

and

the

same

A

valence

electron

more

convenient

electrons

and

conguration

way of writing

represent the

as

the

previous

noble gases) element in the periodic table:

Condensed electron conguration = [previous noble gas] + valence electrons

Table 2 shows some more examples of full and condensed electron congurations.

u

Table 2

Examples of

full and

condensed

Condensed electron congurations for selected

Atomic Element

Full

electron

conguration

electron

elements

number conguration

2

O

8

1s

Ne

10

1s

Mn

25

1s

2

2s

2

2

52

35

1s

2

4

2p

[Ne]

6

2p

2

2s

2

[He] 2s

6

2p

2s

2

Br

4

2p

2

2s

2

3s

6

2p

6

3p

2

3s

2

4s

6

3p

5

4s

2

3d

2

[Ar] 4s

10

3d

5

4p

5

3d

2

[Ar] 4s

10

3d

5

4p

Structure

the

condensed

electron

conguration

for

the

c alcium

Determine

electron

Solution

worked

congurations

atom. 4.

In

Electron

Practice question

Worked example 2

Determine

1.3

the

condensed

conguration

for the

phosphorus atom.

example

1,

2

c alcium to be 1s

we

2

2s

determined

6

2p

2

3s

6

the

full

electron

conguration of

2

3p

4s

.

The previous noble gas in the periodic table is argon, which has an atomic

2

number of 18. Argon has an electron conguration of 1s

2

2s

6

2p

2

3s

6

3p

, and we

2

can therefore write the condensed electron conguration of calcium as [Ar] 4s

Orbital

diagrams

c an

also

sometimes

be

shortened

by

using

a

.

condensed The

electron

conguration.

The

condensed

orbital

diagrams

for

periodic

according manganese

are

shown

table

is

structured

oxygen and to

the

type

of

sublevel

below. that

valence

appear

in.

electrons of elements

This

is

discussed further

oxygen: [He] in

2s

Structure 3.1.

2p

manganese: [Ar]

4s

3d

Self-management skills

ATL

The

ideas

in

mechanic al

Write

a

Write

three

List

the

Write

this

chapter

key

ve

chapter.

topic

model

of

span

the

summary,

key

M ake

an

answer

to

of

the

should

that

test

key,

concepts

how

longer

from

you

questions

range

no

takeaways

voc abulary

brief

a

atom,

to

than

a

and

skills:

electron

sheet

of

A4

from the quantum

congurations.

paper.

chapter.

know

your

then

write

try

from

this

chapter.

understanding

them

out

on

of

one

the

of

ideas in this

your

peers.

Exceptions to the Aufbau principle

The

Aufbau

most

in

principle

elements.

the

correctly

However,

predicts

when

atoms

the

lose

order

of

to

of

atomic

form

ions,

sublevel with the highest principal quantum number ( n)

2

So,

lling

electrons

for

Mn,

with

electron

conguration

[Ar] 4s

are

orbitals

the

lost

for

electrons

rst.

5

3d

,

the

4s

electrons will be lost

2+

rst.

This

gives the manganese ion, Mn

2

not [Ar] 4s

All

you

These

look

at

sc andium

are

electron

conguration

[Ar] 3d

,

.

with

are

the

(Sc)

3d

valence

electrons

known as the 3d

periodic

to

oxidation states

There

with

3

3d

elements

ions.

5

,

some

table

copper

(Cu).

at

tend

to

lose

two

4s

transition elements, or

the

back

These

of

this

transition

book,

metals

electrons

to

form 2+

transition metals. If

these

c an

elements

also

have

are

from

variable

in compounds.

exceptions.

With

only

one

electron

in

its

3d

orbital,

sc andium

3+

readily

forms only Sc

ions,

by

losing

this

3d

electron

and

the

two

4s

electrons.

53

Structure

1

Models

of

the

particulate

nature

of

matter

The Ionization

and

oxidation

ground

those covered in

Structure 2.1

state

congurations

of

copper

and

chromium

are

also

dierent

from

are predicted

by

Aufbau

principle.

and

2

Structure 3.1

The predicted electron conguration of copper is [Ar]4s

9

3d

, as the Aufbau principle

suggests that the lower-energy 4s orbital should be lled rst. However , the observed

1

ground-state electron conguration for copper is [Ar]4s

2

chromium, the predicted conguration is [Ar]4s

10

3d

(gure 23). For

4

1

3d

and the observed is [Ar]4s

5

3d

Activity (gure 23). In each case, promoting a 4s electron to a 3d level leads to a more stable

electron conguration. In the case of copper , this gives a full d sublevel, and in the Deduce

the

electron

conguration case of chromium, there are no paired electrons but rather six half-occupied orbitals,

2+

of the Cu

c ation. each containing an electron with the same spin.

u

Figure 23

and

The expected

observed

Cu (Z = 29)

Cr ( Z = 24)

electron

expected [Ar]

congurations of copper and

[Ar]

configuration chromium

9

2

3d

4

2

3d

4s

4s

observed [Ar]

[Ar]

configuration

10

1

The

electron

two

exceptions

LHA

sublevels

congurations

are

that

lled

5

4s

3d

you

with

of

need

chromium

to

1

3d

(Cr)

and

copper

4s

(Cu)

are the only

know. In all other elements up to

electrons

according

to

the

general

Z=36, the

order.

Ionization energy (Structure 1.3.6 and 1.3.7)

The

quantum

mechanic al

discontinuities

is

the

in

minimum

molecule

in

the

rst

energy

itsground

model

of

the

atom

helps

to

ionization energies (IE) of

required

to

eject

an

electron

explain

the

elements.

out

of

a

trends and

Ionization

energy

neutral atom or

state.

+

X(g)

Ionization

energy

→

X

(g) + e

energy and periodic The

table

+

trends

are

columns

in

the

periodic

table

are

known

as

groups,

and

the

rows

are

known

discussed further in as

periods.

Going

across

the

periodic

table,

the

groups

are

numbered

from 1 to

Structure 3.1 18.

The

periodic

sublevels

s,

electrons

for

First

p,

table

d,

and

each

ionization

f

c an

be

(gure

element

energy (IE

)

shown

24).

are

as

The

also

four

blocks

sublevels

corresponding

holding

the

to

the

outermost

four

valence

shown.

generally

decreases

down

the

groups of the periodic

1

table

and

Going

down

electrons

energy

the

increases

are

the

group,

shielded

sublevels

shielding

outermost

Going

a

and

number

from

(so-c alled

therefore

ac ross

a

period,

the

less

of

pull

“inner

the

ele ctrons

charge.

At

constantbe c ause

54

the

sublevels

of

the

electrons”).

energy

is

increases. The outermost

nucleus

by

The

required

the

more

to

electrons

sublevels,

remove

in

the

the

electrons

lower

greater

from the

sublevel.

outermost

nucle ar

across the periods.

the

the

number

are

held

s ame

time,

number

of

of

protons

closer

the

inner

to

the

in

the

nucleus

nucleus

shielding

ele ctrons

by

ee ct

does

the

inc re ases,

inc re ase d

remains

not

ne arly

change.

so

Structure

1.3

Electron

congurations

LHA

s-block

1

18

transition elements 1s

1s 13

2

2s

14

15

16

17

2p

d-block

3s

3p 3

4

5

6

7

8

9

10

11

12

4s

3d

4p

5s

4d

6s

5d

6p

7s

6d

7p

4f

f-block

5f



Figure 24

Therefore,

ionization

The

more

a

energy

energy

general

across

The blocks of the periodic table correspond

trend

period

is

is

re quire d

inc re ases

of

ac ross

decreasing

shown

Period

in

to

the

remove

to the sublevels s,

outermost

p, d and f

ele ctrons,

so

period.

ionization

energy

down

a

group

and

increasing

gure 25:

2

Period

3

Period

4

Period

5

2500

He

1

Ne lo

2000

  gree

Ar 1500

N Kr

 o i a  i  o i

Xe

O H

1000

Be

sriF

B

500

Al Li

Na Rb

K

0

10

2

18

36

54

Aoi ber



Figure 25

Plot

of

rst

ionization energy against

atomic number for the elements from

hydrogen to xenon

55

1

Models

of

the

particulate

nature

of

LHA

Structure

matter

There

are

two

clear

discontinuities

1.

Between the group 2

The

valence

The

paired 2s

across the period:

and group 3 elements

2

electron

conguration

of

2

beryllium is 2s

while

for

boron it is 2s

1

2p

.

2

making

the

electrons

electron

shield

slightly

the

easier

single

to

2p

electron

in

boron

from the nucleus,

remove.

+

Be

Be 0

0

2p

2p

2

1

2s

2s

+

B

B 1

0

2p

2p

2

Scientists

trends

The

—

in

look

the

presence

results

overall

that

of

to

for

patterns

they

and

collect.

discrepancies

do

pattern

conclusions

out

data

—

not

t

the

allows

be

2

2s

Patterns and trends

The

s ame

trend

c an

2s

be

observe d

in

comparing

group

2

to

group

2

in

any

period.

aluminium,

For

so

example,

the

rst

the

3s

ionization

3

elements

1

ele ctrons

energy

of

shield

the

aluminium

is

lone

3p

lower

ele ctron

than

that

in

of

magnesium.

further

Suppose you have a two-story building and you need to remove one oor to meet

drawn.

new height regulations. Which oor would you remove? Obviously, it will be the What can be inferred from the

top oor, as the building would collapse otherwise! The same reasoning can be patterns in successive ionization

applied to the ionization of atoms — electrons are removed rst from the highest energies?

occupied energy level, and from the highest energy sublevel within that level.

2.

Between the group 15 and 16 elements

From

group

15

to

16

there

is

also

2

conguration

Nitrogen

lled

from

u

Figure 26

A half-lled

p

has

of

a

more

sublevel,

nitrogen

stable

and

(gure

the

same

region

of

the

three

electrons

space

in

2s

the

drop

This

and

2p

more

is

in

ionization

energy.

The

3

2p

electron

therefore

26).

a

2

nitrogen is 1s

2

while

energy

is

the

increased

orbitals

oxygen it is 1s

conguration

bec ause

have

for

do

not

than

required

paired

to

remove

electrons

into

in

close

N 3

2

2p

2p

2

2

2s

electrons

2s

+

O

O 4

3

2p

2p

2

2s

56

2

2s

electron

in

occupy

nitrogen,

proximity.

+

N

an

oxygen

However,

p sublevel is

more stable than p sublevels with 2 or 4

4

2p

oxygen as it has a half-

repulsion.

come

electron

2

2s

Structure

1.3

Electron

congurations

6

most

stable

p

orbital

conguration is p

,

a

completely

lled

p

LHA

The

sublevel,

3

followed

by p

,

a

half-lled

10

example, d

why

sublevel.

This

is

generally

true

for

other

sublevels.

For

5

and d

chromium

and

electron

copper

congurations

do

not

obey

the

are

also

Aufbau

stable,

which

principle

partly

explains

(gure 23).

C alculating ionization energy from spectral data

As

the

principal

between

lines

the

quantum

levels

converging

level

in

number

converges

the

to

hydrogen

of

a

energy

levels

continuum.

emission

increases, the distance

This

c an

spectrum,

be

observed

shown

in

by

gure

spectral

27.

t

∞

level 5

Figure 27

Ultraviolet and visible light

transitions in hydrogen and

the resulting

level 4 emission spectrum

level 3

level 2

level 1

high energy

low energy

ultraviolet

light

visible light

–8

The spectral lines in the hydrogen emission spectrum converge at 9.12 × 10

m, or

912 Å (gure 28). This represents the wavelength of light at which the hydrogen

atom is ionized.

This

wavelength

c an

be

used

to

c alculate

the

rst

ionization

energy

of

hydrogen.

t

912 Å

Figure 28

Hydrogen is ionized at

the wavelength where the spectral lines

converge in the emission spectrum

900

950

1000

1050

1100

1150

1200

1250

Wavelength / Å

57

1

Models

of

the

particulate

nature

of

matter

LHA

Structure

Worked example 3

8

Spectral

lines

converge

at

9.12 × 10

m

in

the

emission

spectrum

of

the

1

hydrogen

atom.

C alculate

the

rst

ionization

energy

of

hydrogen

in

kJ

mol

Solution

First,

c alculate

the

frequency

of

radiation using

to

3.00

8

of

light,

approximately

8

3.00

equal

1

× 10

m s

=

Then,

3.29 × 10

c alculate

f

the

=

6.63 × 10

=

2.18 × 10

=

f

λ, where

×

c

is

the

speed

×

9.12 × 10

m

1

Hz

(s

)

energy

using

Planck’s

34

E

c

1

m s

8

=

15

f

× 10

15

J s

×

3.29

constant

and

the

equation

E

=

h

×

f

1

× 10

s

18

Alternatively,

The

values

of

the

J

these

two

steps

c an

be

merged

into

one

by using the

speed of light, h × c equation

Planck’ s

constant

constant

are

booklet.

The

constant

are

and

=

Avogadro’ s

λ

This

given in the data

mole

E

and

represents

absorbed

Avogadro’s

level,

discussed further in

or

in

the

energy of a single photon of light which would be

exciting

removing

the

one

electron

electron

in

a

hydrogen

atom

to

the

convergence

from the atom.

1

Ionization

Structure 1.4.

energies

are

usually

given

in

kJ mol

.

You

c an

convert the ionization

1

energy

value

to

kJ mol

using

Avogadro’s constant (N

, the number of atoms A

in

1mol)

and

the

following

equation:

1

The

ionization

(energy

energy

in

needed

kJ mol

to

remove

one

electron

from an atom)

× N A

= 1000 –18

2.18 × 10

23

J × 6.02 × 10

–1

mol

= 1000 3

=

1

kJ mol

1.31 × 10

Worked example 4

1

The

rst

ionization

booklet.

C alculate

spectrum

in

energy

the

of

Na

is

496

wavelength

of

kJ

mol

as

given

convergence

for

by

the

the

IB

data

sodium

atom

Å.

Practice questions Solution

5.

In

the

the

emission

helium

spectrum of

atom,

the

First,

nd

from

kJ

the

energy

of

ionization

for

one

atom

by

converting

the

given

value

spectral to

J

and

dividing

it

by

Avogadro’ s constant.

–8

lines

converge

at

5.05 × 10

m.

1

496 000 J mol

C alculate

the

23

/

6.02 × 10

1

mol

19

=

8.24 × 10

J

rst ionization h × c Then

c alculate

the

wavelength of light using

E

=

–1

energy,

in

kJ mol

, of helium.

λ

–34

6.63

× 10

8

J s × 3.00

× 10

–1

m s

19

6.

The

rst

ionization

8.24 × 10

energy of

J

=

λ –1

c alcium

is

590 kJ mol

.

C alculate 7

λ = 2.41 × 10 the

wavelength

of

= in

Å,

for

the

m

convergence,

2410 Å.

c alcium atomic

This

corresponds

to

the

UV

region

in

the

electromagnetic

spectrum.

spectrum.

Successive ionization energies

It

requires

atom

while

58

more

bec ause

the

energy

the

to

remove

number

of

electron–electron

the

protons

repulsion

second and

exceeds

the

decreases.

successive

number

of

electrons

remaining

from an

electrons

Structure

a

the

so

result,

electron

increased

that

only

remove

the

clouds

electrostatic

the

stable

next

are

noble

electron

pulled

attraction.

gas

closer

Once

to

all

conguration

increases

sharply,

as

the

the

nucleus

valence

remains,

shown

the

in

and

held

electrons

energy

gure

tighter

are

Electron

congurations

LHA

As

1.3

by

removed

required to

29.

t

Figure 29

ygrene

from

Removing 10 electrons

magnesium

gives the noble-gas

2

conguration 1s

or [He].

There is a

) 1–

n o it a z i n o I

lom Jk(

considerable increase in energy required to

remove the 11th electron

0

1

2

3

4

5

6

7

Numer

o

8

9

10

eletron

11

12

remoe

Worked example 5

The

rst

ve

successive

ionization

energies

for

an

unknown

element

X

have

1

the

following

the

group

values:

403,

2633,

3860,

5080

and

6850 kJ mol

.

Deduce

Practice question of

the

periodic

table

in

which

element

X

is

likely

to

be

found.

7 .

The

rst

ve

energies

of

successive ionization

an

unknown element

Solution have

the

following

values:

1

The

largest

increase

in

energy

occurs

from

the

rst

ionization

(403 kJ mol

) to 801,

2427,

3660,

25 026 and

1

the

second

(2633 kJ mol

).

This

means

that

the

second

electron is likely to be –1

32 827 kJ mol removed

from

a

stable

noble

gas

conguration

of

the

atom.

.

Deduce the

Therefore, the group of the periodic table in

outermost

energy

level

of

the

element

contains

one

electron, so the element which this element is likely to

belongs

to

group

1

of

the

periodic

table. befound.

Data-based question

Using

gure

successive

30

and

the

ionization

periodic

energies

table,

for

explain

the

two

large

jumps

in

the

sodium.

6.0

5.5

5.0

EI

4.5

gol

4.0

3.5

3.0

2.5

1

2

3

4

number



Figure 30

5

of

6

7

electrons

8

9

10

11

removed

Successive ionization energies for sodium

59

LHA

Structure

1

Models

of

the

particulate

nature

of

matter

Ionization energy data

Relevant

Part 3: Graphing the logarithm of the ionization

skills

•

Tool

2:

Extract

•

Tool

2:

Use

represent

•

data

energies

from databases

6.

spreadsheets to manipulate data and

data

in

graphic al

Title

the

third

(ionization

form

column

energy)”

in

as

your

spreadsheet “log

shown

below:

T ool 3: Construct and interpret graphs

A

Instructions

Part 1: Data collection

1.

B

C

element name:

1

Identify a database that contains successive 2

ionization energy data for the elements (for example,

ionization

log

energy/

(ionization

WebElements).

2.

Choose

one

of

the

following

elements:

sulfur,

–1

chlorine,

3.

Collect

argon,

potassium

successive

spreadsheet,

or

ionization

labelling

the

energy data in a

columns

as

4

1

follows: 5

2

6

A

energy)

kJ mol

ionization

3

c alcium.

etc

B

7 .

Compute

the

logarithm

of

each

ionization

energy

element name:

1

using

2

8.

ionization

the

spreadsheet

Construct

a

ionization

energies

vs

graph

ionization

LOG

showing

by

(or

LOG10) function.

the

plotting

logs

log

of

successive

(ionization

energy)

number.

energy/

9.

–1

Answer

the

following questions:

kJ mol

ionization

3

a.

4

Identify

the

large

increases

in

ionization

energy

1 that

5

2

6

b.

indic ate

a

change

in

main

energy

level.

Why is it useful to plot the logs of the ionization

etc

energies?

Part 4: Evidence for the existence of sublevels

10.

Part 2: Graphing successive ionization energies

4.

Plot

a

M ake

axis

5.

line

sure

that

labels,

Answer

a.

graph

IE

and

present

sc ales

your

and

a

graph

the

suitably, with

descriptive

values

that

outermost

ionization

11.

graph

C an

you

existence

with

graph

that

electrons

and

see

ionization

electrons.

increases

a

the

“zoom

will

in

in”

allow

energy

to

you to closely

level

inspect

12.

each

any

is

later

the

of

role

of

large

Explain

data

is

oen

transformed

into

graphic al

of

representations

how

does

the

properties

trend in

of

metals

IE

values

and

across

non-metals?

a

period

and

(Structure

LHA

60

How

the

do

patterns

variable

of

successive

oxidation

states

of

ionization

these

Enlarge

down

a

group

explain

the

scientic

knowledge?

employed

in

other

trends

] and ionization energies? (Tool 3, Reactivity 3.1)

energies

of

transition

(Structure

3.1)

elements

help

to

forms.

What

representations in the

3.1)

elements?

relate to the

organized into tables

graphic al

+

Why are log scales useful when discussing [H

2.

increases

increases in

they

Linking questions

in

=

sublevels.

advancement

existence of main energy levels in the atom.

unusually

energy?

Experimental

and

electron.

Explain how the graph provides evidence for the

How

n

the

closely.

title.

correspond to the

energy

Construct

examine

vs ionization.

explain

Are

subject

graphic al

areas?

KOT

why

successive

c.

you

energy

following questions:

Identify the

Explain

ionization

suitable

the

innermost

b.

of

Structure

1.3

Electron

congurations

End-of-topic questions

5.

What

is

the

maximum

possible

number

of

electrons in

Topic review the

1.

Using

your

answer

knowledge

the

guiding

from the

question

as

Structure 1.3

fully

as

third

energy

level?

topic, A.

3

B.

6

C.

9

D.

18

possible:

How can we model the energy states of electrons

inatoms?

6.

Exam-style questions

What

in

the

is

the

electron

ground

conguration

of

chromium (Z = 24)

state?

Multiple-choice questions 7

A.

2.

Which

row

is

correct

for

the

following

[Ar] 3d

regions of the 2

electromagnetic

B.

[Ar] 4s

C.

[Ar] 4s

D.

[Ar] 4s

4

3d

spectrum? 1

Ultraviolet

(UV)

Infrared

short

low

energy

wavelength

energy

1

(IR)

5

4p

low 7.

A.

Which

of

the

low

energy

frequency

low

energy

A.

IE

>

correct?

IE

B.

Molar

C.

The

4

long

B.

high

is

frequency

3

high

following

LHA

high

5

3d

ionization

energies

are

measured in kJ.

wavelength

short

high

third

ionization

energy

of

the

atom

X

represents

long

C.

the frequency

wavelength

energy

wavelength

high

long

low

low

frequency

wavelength

frequency

process:

2+

X

3+

(g)

→

X

–

(g) + e

energy

D. D.

Ionization

energies

decrease

across a period going

from le to right.

3.

Which

of

the

following

sources

of

light

will

produce a 8.

line

spectrum

when

placed

behind

a

Which

statement

nitrogen

I.

a

II.

an

gas

and

oxygen

rst

is

ionization

energies of

correct?

IE

(N)




IE

1

(O)

bec ause

oxygen

has

1

ame electrons

A.

I and II only

B.

I and III only

C.

IE

in

(N)

its




IE

1

4.

An

electron

n = 2

in

an

transition

between

atom

energy

levels

n = 4 and

produces a line in the visible

9.

The

bec ause

rst

sublevel

oxygen

(O)

than

loses

an

electron

bec ause

an

electron

nitrogen

oxygen

loses

1

from

isolated

sublevel

1

from

D.

(O)

lled

a

higher

ve

unknown

sublevel

successive

element

are

than

nitrogen

ionization

578,

1817,

energies

2745,

for an

11 577 and

–1

spectrum.

likely

to

Which

produce

electron

a

line

A.

from

n = 4 to

n = 1

B.

from

n = 4 to

n = 3

C.

from

n = 3 to

n = 2

D.

from

n = 5 to

n = 3

in

transition

the

UV

in

the

same atom is

spectrum?

14 842 kJ mol

this

element

A.

1

B.

2

C.

13

D.

14

is

.

In

which

likely

to

group of the periodic table is

be

found?

61

Structure

1

Models

of

the

particulate

nature

of

matter

16.

Extended-response questions

Sketch

an

number

10.

Explain,

in

your

own

words,

why

orbital

of

lling

unpaired

diagram

for

Al

and

deduce the

electrons.

[2]

gaseous atoms

3+

produce

line

spectra

instead

of

continuous

spectra.

[3]

17 .

A

transition element ion, X

,

has

the

electron

5

conguration [Ar] 3d

11.

State

the

full

and

condensed

electron

congurations

for element

the

following

species

in

their

selenium

c.

silicon

d.

Ti

[1]

atom

Sketch

the

condensed

orbital

lling

diagram

for

[1]

germanium atom

and

deduce the total number of p orbitals

[1] containing

atom

one

or

more

electrons.

[2]

[1]

19.

3+

c ation

Describe,

in

your

own

words,

how

the

rst ionization

[1]

energy

of

an

atom

c an

be

determined

from

its

emission

2–

e.

S

anion

[1] spectrum.

12.

Determine

which

of

the

congurations

20. impossible.

Explain

why

it

c annot

exist.

Using

the

2

2

2s

7

2p

2

3s

13.

2

2s

Deduce

6

2p

and

2

1s

1s

14.

Sketch

15.

The

2s

2

3s

6

2p

the

these

which

represents

6

2p

2

period

explore

3

the

elements,

rst ionization

from sodium to

6

3p

2

3s

of

a

of

10

3d

2

4s

an

s

the

following

ground

2

4s

6

3p

shape

Explain

the

general

trend and discontinuities in

6

3p

explain,

2

2s

2

booklet,

the

5

2

3s

congurations

of

3p argon.

2

1s

data

[2] energies

1s

[2]

below is

state.

5

4p

10

4p

21.

[2]

The

rst

four

unknown

1

[3]

successive

element

X

are

ionization

given

in

energies

table3.

for an

Deduce the

group of the periodic table in which element X is likely

5s

6

3d

electron

energies.

to

1

be

found.

[1]

5s

orbital.

[1] –1

n

IE

/ kJ mol n

diagram

electron

below

energy

(not

levels

in

to

sc ale)

the

represents some of the

n

= 7

n

= 6

n

an

energy

arrow

spectrum

62

on

transition

of

the

in

diagram

the

hydrogen.

738

1451

visible

to

represent

region

of

the

the

3

7733

4

10543

= 5 

Draw

1

2

hydrogen atom.

n

= 4

n

= 3

n

= 2

n

=

1

lowest

emission

[1]

Table 3

Successive ionization energies for element X

LHA

titanium

b.

X.

ground states:

18. a.

. Determine the atomic number of

Counting particles

Structure 1.4

by mass: the mole

How do we quantify matter on the atomic sc ale?

Atoms

are

contains

more

in

of

all

a

extremely

huge

atoms

of

the

in

a

glass

oce ans

substance,

small,

number

the

of

of

so

any

these

water

than

combined.

mole,

physic al

particles.

The

enables

glasses

unit

of

chemists

object

comfortably

There

the

of

the

to

are

water

same

relative

amount

for

with

time,

large

the

molecular

expressing

numbers

concepts

masses

masses

of

of

of

very

molar,

allow

the

atomic

small

particles.

relative

use

of

atomic

small

At

and

numbers

species.

de al

Understandings

Structure 1.4.1

of

substance.

elementary

— The mole (mol) is the SI unit of amount

One

mole

entities

contains

given

by

the

Structure 1.4.4

exactly the number of

gives

Avogadro constant.

in

the

that

—

M asses

of

atoms

are

The

ratio

compound.

number Structure 1.4.2

—

simplest

of

atoms

The

of

empiric al

of

atoms

formula of a compound

of

molecular

each

each

element

formula

element

present

present

gives the actual

in

a

molecule.

compared on a

12

sc ale

relative to

mass (A )

and

C

and

relative

are

expressed

formula

as

relative atomic

Structure 1.4.5

mass (M ).

r

by

the

amount

—

of

The

molar

solute

and

concentration

the

is

determined

volume of solution.

r

–1

Structure 1.4.3 — Molar mass (M) has the unitsg mol

Structure 1.4.6

volumes

of

of

all

—

Avogadro’s

gases

temperature

and

law

measured

pressure

states

under

contain

the

that

equal

same conditions

equal

numbers

ofmolecules.

The mole (Structure 1.4.1)

Atoms

and

Even

million

a

molecules

atoms

are

of

so

small

lead,

Pb,

that

the

their

masses

heaviest

c annot

stable

be

element,

measured

would

directly.

have

a

mass

–16

of

only

3.4 × 10

analytic al

g.

balance.

This

At

is

the

too

small

same

to

time,

be

the

weighed

number

of

even

Pb

on

the

atoms

in

most

1 g

of

sensitive

lead is

21

huge,

about

chemists

masses

19th

2.9 × 10

need

and

a

unit

very

century

,

that

large

and

which

is

allows

hard

to

them

imagine,

to

work

let

alone

comfortably

numbers of atoms. This unit, the

quickly

bec ame

one

of

the

count.

most

with

mole,

useful

Therefore,

both

was

concepts

very small

devised in the

in

chemistry.

The mole (with the unit “mol”) is the SI unit of amount of substance that contains

23

6.02214076 × 10

c an

be

an

will

use

atom,

a

elementary

molecule,

entities

an

of

electron

that

or

substance. An elementary entity

any

other

species. In this book, we

23

the

rounded

value

of

the

mole:

1mol

=

6.02 × 10 

Figure 1

One mole quantities

of dierent substances (le to right):

aluminium,

water,

copper, sucrose and

sodium chloride

63

Structure

1

Models

of

the

particulate

nature

of

matter

Avogadro’s constant

Prex

Symbol

(N

)

is

the

conversion factor linking the number of particles

A

F actor

–1

and

amount of substance

in moles. It has the unit of mol

:

–12

pico

p

10

23 –9

nano

n

10

micro

µ

10

milli

m

10

centi

c

10

N

=

–1

6.02 × 10

mol

A

–6

In

–3

–2

chemic al

other

we

c alculations,

conversion

need

to

factor

multiply

substance (n)

into

the

the

Avogadro’s

(table1).

mass

in

number

For

kg

of

constant

is

example,

by

1,000.

atoms

or

used

to

in

Similarly,

any

the

convert

other

same

way

kilograms

to

as

into

any

grams,

convert the amount of

structural units (N),

we

need

–1

deci

d

10 to

multiply

that

amount

by

N

: A

3

kilo

k

10

mega

M

10

N =

n×N

6

9

giga

G

10

A

In

chemistry

texts,

the

term

“amount

of

substance”

is

oen

abbreviated to just

“amount”.



Table 1

Decimal prexes

Worked example 1

C alculate

the

amount

of

lead

(Pb),

in

mol

and

mmol,

in

a

sample

containing

21

2.9

×

10

atoms

of

this

element.

Solution

To nd

n,

we

c an

rearrange

the

equation

N

=

n

×

N

as

follows:

A

N n

= N A

21

2.9

×

10 3

Therefore,

n(Pb)

=

≈

4.8 × 10

mol

23

6.02

×

10

3

According

The

use

gures

of

is

chemistry

correct

signic ant

discussed in the

Tools for

to

table1,

1 mmol

=

3

mol, so 4.8

10

×

10

mol

=

4.8 mmol.

3

In this example, both answers (4.8 × 10

mol and 4.8 mmol) have been

rounded to two signicant gures, the same as in the least precise value used in

21

chapter.

the division (2.9 × 10

Research skills

ATL

The

are

).

mole

so

is

a

small.

convey

just

huge

number,

Measuring

how

and

it

amounts

is

of

useful

for

everyday

counting

objects

particles

in

moles

bec ause they

c an help use to

large this number is.

Activity Choose

one

of

the

following

and

C alculate:

approximate

a.

the number of atoms in

•

How

many

moles

of

grains

2.5 mol of copper metal

•

How

many

moles

of

water

conduct

the

necessary

research

to

reach an

answer.

of

sand

are

molecules

in

are

a

desert

in

a

of

large

your

sea

or

choice?

ocean

of

choice? b.

the

number

64

molecules in

•

One

•

What

the number of atoms in

•

How

tall

0.25 mol

•

How

many

0.25 mol

c.

of

of

of

mole

of

human

cells

represents

roughly

how

many

water

water

is

the

is

age

a

of

the

universe,

stack

of

one

moles

of

air

mole

are

in

in

of

moles

of

seconds?

sheets of paper?

your school building?

people?

your

Structure

Relative molecular

1.4

Counting

particles

by

mass: the mole

mass and molar mass

(Structure 1.4.2 and 1.4.3)

In

Structure 1.2,

we

introduced the concept of

relative atomic mass,

A , which r

is

the

ratio

atom.

of

the

Similarly,

mass

of

a

certain

atom

relative molecular

to

mass,

one-twelh

M ,

is

the

of

ratio

the

of

mass

the

of

mass

a

c arbon-12

of

a

molecule

r

or

other

A

and

multiatomic

M

r

are

ratios,

species

so

they

to

one-twelh

of

the

mass

of

a

c arbon-12 atom. Both

have no units.

r

To nd the

M

of

a

molecule,

we

need to add together the

A

r

in

that

values

for all atoms

r

molecule.

Worked example 2

C alculate

the

M

for

a

molecule

of

water.

r

Solution

Water, H

O,

is

composed

of

two

hydrogen atoms (A

2

atom (A

You

=

1.01)

and

one

oxygen

r

=

16.00).

Therefore

M (H

r

r

should

always

use

the

O)

=

2

×

1.01

+

16.00

=

18.02.

2

actual

(not

rounded)

values of

A ,

which

are

given

r

in

the

keep

data

all

booklet

signic ant

and

the

gures

periodic

in

table

at

c alculated M

the

end

values

of

and

this

book.

never

Similarly,

round them to the

r

nearest

If

a

integer

substance

is

number.

composed

of

ions

instead

of

molecules, the

M

for that substance r

is

c alculated using the smallest

formula unit.

For

example,

c alcium chloride

2+

(C aCl

)

is

an

ionic

compound

that

consists

of

many

c alcium

c ations

(C a

) and

2

2+

twice

as

many

and two Cl

bec ause

chloride

ions.

the

The

masses

anions

ions

of

(Cl

have

).

Its

smallest

approximately

electrons

are

formula

the

unit

same

contains

masses

as

one

Ca

neutral atoms

negligible.

The

in

Therefore,

M (C aCl r

M any

form

ionic

is

A (C a) + 2×A (Cl) r

compounds

coordination

hydrates

) = 2

copper(II)

40.08

+

(2×35.45)

=

and

compounds

will

structure of

be

discussed

Structure 2.1.

110.98.

r

form

bonds

=

composition

ionic

hydrates:

(Structure

sulfate

compounds

2.2)

with

the

in

ions.

pentahydrate, CuSO

which

•5H 4

water

molecules

One of the most common

O.

Copper(II) sulfate

2

Activity pentahydrate

coecient

forms

“5”

large,

clear,

before “H

O”

deep-blue

means

that

crystals

one

(gure2). The

stoichiometric

formula unit of copper(II) sulfate is C alculate the

2

M

values

for the

r

bound

with

ve

molecules

of

water.

Therefore, the

M

value

for

this

be

c alculated

as

hydrate

c an following

r

species:

follows: a.

ammonia, NH

b.

sulfuric acid, H

3

M (CuSO r

•5H 4

O) = 2

A (Cu) + r

A (S) +

4×A (O) + 5×M (H

r

r

r

O) 2

SO 2

=

63.55

+

=

249.72

32.07

+

(4

×

16.00)

+

(5×18.02)

c.

sodium

sulfate

Na

•10H

SO 2

4

4

dec ahydrate,

O 2

65

Structure

1

Models

of

the

particulate

nature

of

matter



Figure 2

Crystals of

copper(II) sulfate pentahydrate, CuSO

•5H

4

Molar

Molar

mass,

mass

M,

is

of

a

chemic al

numeric ally

substance

equal

to

is

relative

the

mass

of

molecular

O 2

1 mol

mass

of

that

substance.

(for substances with

molecular and ionic structures) or relative atomic mass (for substances with atomic

–1

structure).

For

example,

M(Na)

=

22.99 g mol

–1

and

M(H

O)

=

18.02 g mol

2

Science as a shared endeavour

A

shared

understanding of common terminology helps scientists to

communic ate

Hi s to r i c a l l y,

c o n ta i n e d

or

o t he r

eectively.

th e

as

mo l e

ma ny

pa r t i c l e s)

This

was

terminology

defined

e l e me n ta r y

as

t h e re

as

entities

w e re

is

th e

constantly

amount

(a to ms ,

a to ms

in

of

being

mo l e c u l e s ,

0.0 12 k g

updated.

substance

(o r

t ha t

ions,

12 g )

of

e l e c tro n s

c a r b o n -1 2 .

23

How e ve r,

to

be

s c i e n ti s t s

On

t he

re v i s e d

16

to

n u me r i c a l

v alue

f re qu e n tl y,

m e a s u re

November

as

m a ss

2018,

of

th e

th e

mo l e

( a p prox i m a t e l y

i m prove me n ts

with

gre a te r

scientists

from

here

physic al

that

all

constants

SI

base

instead

units,

of

)

had

a l l ow e d

pre c i s i o n .

including

physic al

6 .0 2  ×  10

i n s t r u me n t a t i o n

more than 60 countries met at the

General Conference on Weights and Measures

agreed

in

the

objects.

in

Versailles,

mole,

were

France.

It

was

dened in terms of

Following these changes, one

23

mole

of

entities

The

no

a

substance

of

2018

that

match

equals

two

the

now

dened

exactly

as

6.02214076 × 10

12 g

exact

SI

numeric al

of

the

mole

exactly.

As

quantities,

v alues

of

me ans

a

that

result,

the

their

the

kilogram

the

mass

experimentally

dierences

determined

and

respective

between

these

mass

of

numeric al

a

A

the

or

are

1 mol

v alues

mole)

M

no

of

of

c arbon-12

M

(dened

longer

(dened

through

r

c arbon-12

v alues

of

numeric al

r

the

elementary

substance.

redenition

longer

through

is

so

atom).

small

However,

the

(approximately

–8

4 × 10

are

How

do

existing

66

%)

Why

that

they

constants

scientists

c an

and

be

achieve

denitions?

ignored

values

a

for

all

continuously

shared

practic al

being

purposes.

revised

and

updated?

understanding of changes made to

Structure

The amount (n),

mass (m)

and

molar

mass (M)

of

any

substance

are

1.4

Counting

particles

by

mass: the mole

related as

follows:

m n

= M

This

all

is

probably

the

stoichiometric

the

masses

of

most

common

c alculations.

chemic al

expression

Although

substances

are

the

in

chemistry,

base

SI

traditionally

unit

of

as

it

is

mass

expressed

in

is

used in almost

the

kilogram,

grams, and

–1

molar

masses

in

g mol

Worked example 3

Table

of

sugar

sucrose.

C

H 12

O 22

.

is

oen

Sucrose

sold

is

an

in

the

form

organic

of

cubes

compound

that

with

are

made

the

almost

molecular

entirely

formula

C alculate:

11

a.

the molar mass of sucrose

b.

the amount

c.

the number of

of

sucrose in one cube (2.80 g) of sugar

oxygen atoms in one cube of sugar

Solution

a.

M (C r

H 12

O 22

)

=

12

×

12.01

+

22

×

1.01

+

11

×

16.00

=

342.34

11

1

M(C

H 12

O 22

)

=

342.34 g mol

11

Activity

m b.

n

= M

C alculate:

2.80 g H

n(C 12

O 22

)

=

≈

11

0.00818 mol a.

1

the

molar

mass

of

sulfuric

acid,

342.34 g mol

H

SO 2

c.

One

mole

of

sucrose

contains

11 mol

of

oxygen atoms, so

b.

n(O)

=

11

×

n(C

H 12

=

11

×

O 22

11

n(O)

0.00818 mol

×

N

=

amount

1.00 g

≈

of

of

substance

sulfuric

in

acid

0.0900 mol c.

=

the

)

23

N(O)

4

0.0900mol

×

6.02 × 10

–1

mol

the

number

of

hydrogen

22

atoms

≈ 5.42 × 10

in

1.00 g

of

sulfuric

acid

A

19



Figure 3

There are more oxygen atoms in one sugar cube than the estimated

total insect

population on Earth (10

) and

total grains of

21

sand

on Earth’s beaches (10

)

67

Structure

1

Models

of

the

particulate

nature

of

matter

Empiric al formula, molecular

formula and

chemic al analysis (Structure 1.4.4)

The

composition

represented

of

each

element

formula

present

shows

in

the

substance

empiric al

ratio

of

by a

of

in

the

chemic al

the

c an

be

in

the

the

of

ratio

The

identic al

is

substance

formula,

molecule

simplest

substance.

formula

ions

a

molecular

dierent

as

compound

the

a

molecular

of

the

and

In

contrast, the

dierent

empiric al

(table2).

formula

structure

c an be

shows the actual number of atoms

substance.

atoms

molecular

or

same

that

of

with

which

empiric al

elements

formulas

of

that

the

are

same

For ionic compounds, the

unit,

which

represents the simplest

(gure4).

Substance

Molecular

oxygen

O

ozone

O

water

H

formula

Empiric al

formula

O 2

O 3

O

H

2

hydrogen

peroxide

O

H 2

butane

4

glucose

C

sucrose

C

HO 2

H

C

C 10

H 6

Table 2



Figure 4

is used

The

Molecular and

12

H

6

number

of

atoms

(N

=

of

a

C 11

).

H 12

supplement

certain

n×N

O 2

O 22

element

Therefore,

11

with the empiric al formula NaF. It

to prevent

is

the

O 22

selected substances

Sodium uoride is an ionic compound

element in mol

5

CH

empiric al formulas of

in some countries as a food

H 2

O

12



O 2

tooth dec ay

proportional to the amount of that

empiric al

formula

also

shows the

A

mole ratio

water, H

of

O,

elements

contains

in

two

a

chemic al

atoms

of

compound.

hydrogen

and

For

example,

one

atom

of

one

molecule of

oxygen, so the

2

atomic

68

ratio

of

hydrogen

to

oxygen

in

water

is

2:1.

Similarly,

one

mole

of

water

Structure

contains

mole

The

two

ratio

of

moles

elemental

mass,

mole

which

ratio

of

hydrogen

hydrogen

to

composition

is

c an

atoms

oxygen

of

a

in

and

compound

referred to as the

be

c alculate

to

mole

of

Counting

particles

by

mass: the mole

oxygen atoms, so the

water is also 2:1.

commonly

used

one

1.4

the

is

oen

expressed

in

percent

percentage composition,

by

ω. The

percentage composition of a compound.

Worked example 4

C alculate

the

percentage

composition

of

water.

Solution

Let

n(H

O)

=

1 mol, then

n(H)

=

2 mol and

n(O)

=

1 mol. Using

m

=

n

×

M:

2

1

m(H)

=

2 mol

=

1 mol

×

1.01 g mol

=

2.02 g

1

m(O)

×

16.00 g mol

=

16.00 g

1

m(H

O)

=

1 mol

×

18.02 g mol

×

100%

=

18.02 g

2

2.02 g

ω (H) =

≈

11.2%

18.02 g

ω (O) = 100%

In

practice,

11.2%

chemists

=

88.8%

more

oen

face

the

opposite

problem

of

deducing

Practice question the

empiric al

other

formula

experimental

destruction

for

a

data.

analysis,

in

compound

The

from

percentage

which

the

its

percentage composition or

composition

compound

is

c an

be

combusted

or

determined

by

C alculate

decomposed, and

the

percentage

composition of sulfuric acid, H

SO 2

the

masses

The

mass

a

the

combustion

percentages

analytic al

In

of

techniques,

typic al

products

into

mass

decomposition

elements

such

experiment,

combustion

converted

of

or

the

are

as

fully

sample

trapped

percentages

in

a

sample

automated

is

burned

and

of

products

c an

in

excess

weights

elements

in

by

elemental

oxygen,

These

the

4

measured.

determined

combustion

weighed.

chemic al

be

are

and

various

analysis.

the

volatile

are then

original

sample.

Worked example 5

Iron

and

formula

oxygen

of

an

form

oxide

several

that

compounds

contains

72.36%

(iron

of

oxides).

Deduce

the

empiric al

iron.

Solution

If

ω (Fe) = 72.36%, then

Let

m(Fe

O x

)

=

ω (O) = 100%

100 g, then

m(Fe)

=

72.36%

72.36 g and

=

27.64%.

m(O)

=

27.64 g

y

m Use

n

=

to

determine

the

amount

of

each element:

M

72.36 g n(Fe)

≈

=

1.296 mol

–1

55.85 g mol

27.64 g n(O)

=

≈

1.728 mol

–1

16.00 g mol

The

mole

ratio

Therefore,

the

x : y

=

1.296 : 1.728

empiric al

formula

of

≈

1 : 1.333

the

oxide

≈

is

3 : 4

Fe

O 3

. 4



Figure 5

Fe

O 3

is the main component

of the mineral

4

magnetite, a common iron ore

69

Structure

1

Models

of

the

particulate

nature

of

matter

Worked example 6

Hydroc arbons

unknown

to

are

organic

hydroc arbon

produce

26.41 g

of

compounds

has

undergone

c arbon

dioxide,

of

c arbon

and

combustion

CO

,

and

in

hydrogen.

excess

13.52 g

of

An

oxygen

water,

H

2

Deduce

the

empiric al

formula

of

the

O. 2

hydroc arbon.

Solution

1

M(CO

)

=

12.01

+

2×16.00

=

44.01 g mol

2

26.41 g n(CO

)

=

≈

0.6001 mol

2 –1

44.01 g mol

n(C)

=

n(CO

)

=

0.6001 mol

2

1

M(H

O)

=

2×1.01

+

16.00

=

18.02 g mol

2

13.52 g n(H

O)

=

≈

0.7503 mol

2 –1

18.02 g mol

n(H)

×

n(H

c arbon

=

2

and

O)

=

2×0.7503 mol≈1.501 mol

2

All

the

hydrogen

hydroc arbon, C

H x

The

mole

ratio

Therefore,

the

x:y

atoms

in

the

combustion

products

=

0.6001 : 1.501

empiric al

formula

of

≈

the

1 : 2.5

=

express

known as

comprised

number

ratio.

a

of

ratio,

This

factor

two

you

gives

by

a

this

number

formulas

In

worked

as

whole

example

number

5,

the

ratios.

ratio

we

H

values:

1.296 and 1.728.

divide

term

the

you

ratio

ratio

each

of

1

:

1.333.

should

by

of

3,

in

Then,

multiply

and

then

the

ratio

you

Whole

To

numbers

are also

c alculated

was

convert it to a whole

by the smallest number in the

c an

ratio

5

initially

non-integer

ratio

which

Multiplying

whole

empiric al

integers.

from

2 : 5

hydroc arbon is C 2

We

originate

, so: y

to

use

trial

obtain

subsequently

and

the

rounding

error to determine

whole

the

number

result,

ratio.

gives a

3 : 4.

The molecular formula of a compound can be deduced from the empirical formula

if we know the molar mass of the compound. For example, you might determine

experimentally that the molar mass of the hydrocarbon in worked example 6 is

–1

58.12 g mol

. The molar mass of the empirical formula can be calculated:

–1

(12.01

The

×

2)

value

have

+

of

twice

(1.01

×

5)

29.07 is

the

=

29.07 g mol

roughly

number

of

half

atoms

of

as

58.12,

the

therefore

empiric al

the

molecular

formula: C

Determining

the

molar

substances

is

10

masses of Table2

gaseous

formula must

H 4

suggests

that

this

hydroc arbon

could

be

butane, C

H 4

discussed in c annot

be

sure

about

it

without

further

analysis,

as

there

is

.

However, we

10

another

hydroc arbon,

Structure 1.5. methylpropane,

c an

be

with

distinguished

comparing

their

the

by

same

molecular

formula.

Butane

and

methylpropane

measuring their boiling points ( Structure 1.1) or

infrared

spectra (Structure 3.2).

Practice questions

1.

Deduce

a.

an

b.

a

the

empiric al

oxide

of

formulas

manganese

hydroc arbon

that

of

that

the

following compounds:

contains

produces

36.81%

5.501 g

of

of

c arbon

oxygen

dioxide

and

2.253 g of

water upon complete combustion

2.

Deduce

the

molecular

–1

is

70

42.09 g mol

formula

of

the

hydroc arbon

from

1b

if

its

molar

mass

Structure

1.4

Counting

particles

by

mass: the mole

Experimental determination of empiric al formula

Relevant

skills

•

Tool

1:

•

Tool

3:

Instructions

Measure

C arry

mass

out

c alculations

involving

decimals

1.

Weigh

2.

Obtain

andratios

and

•

Tool

3:

Use

•

Tool

3:

Construct

and

•

Inquiry

3:

realistic

an

clean,

a

piece

1.0 g)

from

dry

of

crucible.

magnesium

your

teacher.

ribbon

(between

Measure

its

exact

0.3 g

mass.

approximation and estimation

interpret

3.

Twist the magnesium into a loose coil and place it

4.

Heat

graphs inside

to

a

Explain

and

relevant

the

crucible.

improvements the

crucible,

with

its

lid

on,

over

a

roaring

investigation Bunsen

air

to

ame.

enter

Periodic ally

the

li

the

crucible

lid

to

allow

crucible.

S afety

•

Wear

•

Take

•

The

eye

5.

protection.

suitable

prec autions

equipment

prec autions

will

get

around open ames.

very

hot.

around it and do not touch it while it

•

M agnesium

burns

with

a

heating until the magnesium no longer

lights

Then,

up.

crucible

Take suitable

ishot.

Continue

6.

When

7 .

Heat

to

the

the

directly at it.

Repeat

remove

for

a

the

heat

source

and

allow the

few minutes.

crucible is cool, weigh it.

crucible

additional

very bright light. Do not look

cool

and

minute.

this

its

contents

Allow

to

cool

strongly

and

for an

re-weigh.

heating-cooling-weighing

cycle until the

mass is constant. M aterials

•

crucible and lid

Q uestions

•

balance

1.

•

pipeclay triangle

•

tripod

•

heat-proof mat

•

tongs

•

magnesium ribbon

(±0.01 g)

Process

of

2.

Compare

actual

3.

data

a

your

to

determine

the

empiric al

formula

oxide.

experimental

empiric al

formula to the

one.

Obtain

Plot

•

the

magnesium

mass

graph

data

of

from

mass

of

other

members

magnesium

of

oxide

your

vs

class.

mass of

Bunsen burner magnesium.

lid

4.

Identify

t

5.

crucible

any

on

Explain

of

6.

line

anomalies

the

what

the

magnesium

Explain

why

(if

applic able)

and

draw a best

graph.

graph

shows about the composition

oxide.

you

repeatedly

heated

and

weighed the

coiled magnesium

crucible

until

a

constant

mass

was

achieved.

ribbon

7 .

Identify

and

explain

two

major

sources

of

error in this

procedure.

8.

Suggest

that

realistic

could

improvements to the methodology

minimize

the

sources

of

error

you

have

Bunsen burner

identied.

9.

Reect

on

empiric al



Figure 6

The experimental set-up

round

C an

to

the

role

formula

the

of

approximation

c alculations.

nearest

whole

and

rounding in

When is it suitable to

number?

When is it not?

you come up with a rule of thumb of when to

round

and

when

not

to

round?

71

Structure

1

Models

of

the

particulate

nature

of

matter

Measurement

Atoms,

molecules

impossible.

particles

As

with

to

all

Consider

The

and

ions

concept

mass,

which

mass

of

a

so

the

c an

measurements,

the

are

of

be

is

easily

mass

sample

small

mole

has

of

that

counting

powerful

them

bec ause

directly is virtually

it

relates number of

measured.

an

uncertainty

c alcium

associated with it.

c arbonate,

C aCO

,

is

found to be

3

3.500

up

to

g

±

0.001

0.001

moles

does

c alculation

terms

Is

a

of

g.

in

it

represent?

and

either

nd

measurement

doing

means

g

particles,

questions

This

as

you

it

direction.

out.

is

How

You

quite

that

mass

This

many

will

is

measurement

clearly

particles

see

that

in

a

c an

minuscule

does

moles

it

be

inaccurate

mass.

How

by

many

represent? Do a quick

the

uncertainty

is

tiny, but in

large.

uncertainty

proceed

experiments

the

ever

negligible?

through

involve

the

DP

making

If

so,

when? Think about these

chemistry

course, particularly when

measurements.

Solutions and concentration

(Structure 1.4.5)

M any

chemic al

handle

aect

and

the

are

consists of a

component

the

and

are

of

of

the

solutes.

For

gases.

dissolved

and

the

or

so

a

in

solutions.

Sometimes

substances

more

the

solvent.

example,

out

mixtures

one

solution,

of

c arried

or

of

or

two

properties

The

other

solution

of

Solutions

solvent

more

The

of

the

in

are

used

in

easier to

bec ause

chemic al

components.

it

c an

reactions.

E ach solution

solvent is usually the major

whole

components

sugar

is

participate

or

solutes.

a

water

of

is

solution

the

are similar

solution

more

like

are

water

(clear

heterogeneous

colourless mixtures

are

solids

homogeneous

solvent

properties

c alled Homogeneous

than

properties

Solutions

to

reactions

mix

liquid)

than

sugar

(white

crystalline

powder),

so

water

is

the

solvent

discussed in

while

sugar

(from

the

is

the

solute. In this topic, we will consider only

aqueous solutions

Structure 1.1.

L atin

aqua

meaning

“water ”),

in

which

the

solvent

is

water.

solute



In

some

the

and

water

present

it

is

is

not

water ”

72

c ases,

ethanol

the

identity

water,

in

major

rather

Figure 7

each

the

of

of

the

these

mixture,

component.

than

How a solution is formed

“4%

it

solvent

liquids

is

For

solution

of

is

unclear:

c an

be

traditionally

example,

for

c alled

example, if we mix

a

solvent.

regarded

we

as

the

However, if

solvent,

even if

say “96% solution of ethanol in

water in ethanol”.

Structure

Solutions

solute

solute,

small

the

are

and

oen

and

so

has

proportion

term

solute

much

classied

solvent. A

a

of

high

ratio

solute,

“concentrated”

per

100 g

less

than

of

the

10 g

according

to

the

concentrated solution

of

of

and

refers

solvent,

the

solute

so

to

has

the

per

or

mole

a

ratio

with

term

100 g

of

much

of

the

to

more

Counting

particles

by

mass: the mole

between the

proportion of

dilute solution

solute

“dilute”

ratio

large

solvent, while a

low

solutions

and

solute

to

a

mass

contains

1.4

solvent.

than

has a

Generally,

10 g of the

refers to solutions with

solvent.

TOK

Some

words

do

interpretation

are

not

is

not

precisely

chemists

would

have

context

dened

c all

a

precise denitions and their choice and

dependent.

and

The

should

solution

of

be

terms

used

“concentrated”

with

c are.

5 g of sulfuric acid (H

For

SO 2

“dilute”,

used

in

as

much

higher

laboratories.

At

proportions

the

permanganate (KMnO

)

in

same

100 g

of

time,

of

sulfuric

a

acid

solution

water

would

of

to

5 g

be

)

and

“dilute”

example, most

in

100 g

of

water

4

water

of

are commonly

potassium

considered

very

4

concentrated

by

permanganate

The

To

any

in

antiseptic

concentrations

what

extent

Quantitatively,

amount

of

a

of

the

concentration.

in

does

communic ation

medic al

the

worker,

solutions

examples

expressing

typic al

are

less

above

concentrations

than

could

quantity

0.1 g

be

per

of

100 g

expressed

potassium

of

water.

numeric ally.

numeric ally help or hinder the

knowledge?

composition

Molar

solute

a

as

to

of

solutions

concentration,

the

c,

is

also

expressed in terms of

known as

molarity,

is

the

ratio of the

volume of the solution:

n solute

c

= solute

V solution

–3

The

most

common

units

for

molar

concentration

are

mol dm

–1

as

mol L

also

be

(which

is

the

–3

).

For

very

dilute

solutions,

smaller

units

same

–3

(mmol dm

or

µmol dm

)

c an

used:

–3

1 mmoldm

–3

=

1 × 10

–3

–3

mol dm

–6

1 µmoldm

=

1 × 10

–3

mol dm

–3

The

units

of

molar

concentrations

are

sometimes

abbreviated

as

M

(for

mol dm

)

–3

or

mM

(for

mmol dm

).

For

example,

the

expression

“2.5 MNaOH”

means that

3

each dm

of

Note

the

that

whole

the

solution

term

solution.

contains

“molar

For

2.5 mol

concentration”

example,

it

is

of

sodium

refers

incorrect

to

to

say

a

hydroxide.

specic

that

“the

substance, not the

concentration of a

–3

sodium

about

chloride

the

solution

concentration

is

of

1.0 mol dm

sodium

”,

as

chloride

it

is

or

not

clear

water.

The

whether

we

are talking

correct statement

–3

would

Molar

be

“the

concentration

concentration

is

oen

of

sodium

chloride

represented

by

in

a

square

solution

brackets

is

1.0 mol dm

”.

around the solute

–3

formula.

For

example,

the

expression [NH

]

=

0.5 M

refers

to

a

0.5 mol dm

3

–

solution

of

ammonia.

Similarly,

the

expression

[Cl

]

refers to the molar

concentration of chloride ions in a solution.

73

Structure

1

Models

of

the

particulate

nature

of

matter

Worked example 7

–3

C alculate

a

the

solution

molar

concentration

prepared

by

dissolving

of

sodium

3.60 g

of

chloride,

N aCl(s)

in

in

mol dm

water

to

,

in

make

3

25.0 cm

of

the

nal

solution.

Solution

First,

c alculate

the

molar

mass of sodium chloride:

1

M(NaCl)

=

22.99

+

35.45

=

58.44g mol

m Then use

n

=

to

c alculate the amount of solute:

M

3.60 g n(NaCl)

≈

=

0.0616 mol

–1

58.44 g mol

3

Convert

the

volume to dm

by

dividing

Activity

by

1,000:

3

V(solution)

=

25.0 cm

3

=0.0250 dm

n C alculate

the

mass of sulfuric acid,

Use

c

=

to

c alculate

the

concentration:

V 3

H

SO 2

,

in

50.0 cm

of a

solution 0.0616 mol

4

3

c(NaCl)

–3

where [H

SO 2

]

=

≈

=

1.50 mol dm

2.46 mol dm

3

0.0250 dm

4

The

composition

concentration,

of

a

ρ

solution

,

of

the

is

sometimes

solute,

which

expressed as the

is

the

ratio

of

the

mass

mass of the solute to

solute

the

volume of the solution:

m solute

ρ

= solute

V solution

Worked example 8

C alculate

worked

the

mass

example

concentration

of

sodium

chloride

in

the

solution

from

7.

Solution

If

Activity

we

know

c alculate

the

the

mass

mass

of

the

solute

concentration

and

as

the

volume

of

the

solution,

we

c an

follows:

3.60 g 3

ρ(NaCl) C alculate

the

molar

=

=

concentration

144 g dm

3

0.0250 dm –3

of

sulfuric

acid,

in

in a solution with

mol dm

ρ(H

SO 2

Alternatively,

) =

the

concentration

mass

and

concentration

molar

mass,

using

of

NaCl

the

c an

be

found

relationship

4

ρ

from its molar

= solute

c

×

M

solute

: solute

–3

0.150 g cm

3

ρ(NaCl) = c(NaCl)× M(NaCl) = 2.46 mol dm

1

×58.44 g mol

–3

The

most

common

concentration

mass,

as

follows:

ρ solute

c

= solute

M solute

74

and

units

molar

for

mass

concentration

concentration

of

the

are

same

g dm

solute

3

≈ 144 g dm

–3

and

are

g cm

related

.

M ass

by molar

Structure

1.4

Counting

particles

by

mass: the mole

Worked example 9

A

standard

solution

was

prepared

by

dissolving

6.624g

of

sodium

c arbonate,

Na

CO 2

a

,

in

deionized

water

3

3

using

3

250 cm

volumetric

ask.

An

analytic al

pipette

was

used

to

transfer

10.0 cm

sample

of

this

solution

to

a

3

100cm

volumetric

ask,

and

the

ask

was

topped

up

to

the

graduation

mark

with

deionized

water.

C alculate

the

–3

concentration,

in

moldm

,

of

sodium

c arbonate

in

the

new

solution.

Solution

First,

we

need

to

nd

the

concentration

of

sodium

c arbonate

in

the

standard solution:

1

M(Na

CO 2

)

=

2×22.99

+

12.01

+

3×16.00

=

105.99 g mol

3

6.624 g n(Na

CO 2

)

=

≈

3

0.06250 mol

–1

105.99 g mol

3

V

=

3

250 cm

=

0.250 dm

standard

Note

that

the

accuracy

of

a

typic al

volumetric

ask

is

three

signic ant

gures.

0.06250 mol 3

c

(Na standard

CO 2

)

=

=

3

0.250 mol dm

3

0.250 dm

Then

we

need

to

c alculate

the

concentration

of

sodium

c arbonate

in

the

new solution.

3

First,

c alculate

the

amount

of

Na

CO 2

in

3

V

=

10.0 cm

the

sample.

Remember

to

convert

all

volumes to dm

3

3

=

0.0100 dm

sample

3

c

(Na standard

CO

)

2

3

CO

)

=

c

(Na sample

CO 2

)

=

0.250 mol dm

3

3

n

(Na sample

When

the

2

sample

=

0.250 mol dm

3

×0.0100 dm

=

0.00250 mol

3

is

diluted

with

deionized

water

to

produce

the

new

solution,

the

amount

of

solute

does

not

change.

Therefore

n

(Na sample

Now

you

c an

CO 2

)

=

n

3

work

out

(Na new

the

CO 2

)

=

0.00250 mol

3

concentration

of

Na

CO 2

volume

of

the

3

V

in

the

new

solution

by

dividing

the

amount

of

Na

3

CO 2

by the 3

new solution:

=

3

100 cm

=

0.100 dm

new

0.00250 mol 3

c

(Na new

CO 2

)

=

=

3

0.0250 mol dm

3

0.100 dm

It

is

a

common

practice

to

store

chemic als

in

the

form

of

concentrated solutions

Practice question (so-c alled

needed.

stock solutions)

Stock

solutions

and

with

a

dilute

them

known

to

the

required

concentration

of

concentration when

the

solute

are

c alled

3.

standard solutions.

A

standard

by

copper(II) To

determine

the

concentration

of

the

standard

solution

in

worked

example

did

the

following

two

was

prepared

2.497 g of

sulfate

pentahydrate,

9, CuSO

we

solution

dissolving

• 5H

O,

4

2

using

a

in

deionized

c alculations:

3

water

100 cm

volumetric

3

ask. 1.

n

=

c

sample

× sample

A

5.00 cm

sample of this

V sample 3

solution

was

diluted

to

250.0 cm

.

n new

2.

c

C alculate

= new

–3

mol dm

nal know that

n

= sample

gives

the

following

c

n

,

so

you

c an

substitute

equation

1

into

, of copper(II) sulfate in

the

solution.

equation 2. This

new

expression:

The

× V sample

c

concentration, in

V new

You

the

process

for

preparing

standard

sample

=

solutions

new

is

discussed in the

Tools

V new

for chemistry

Therefore,

need

to

to

c alculate

know

the

solution,

and

the

c

c

V

× 1

V

= 1

× 2

the

original

volume

concentration

concentration

of

the

of

of

a

solute

the

in

solute,

a

new

the

solution,

chapter.

you just

volume of the original

new solution. In summary:

2

75

Structure

1

Models

of

the

particulate

nature

of

matter

C ase study: spectrophotometry and c alibration curves

Spectrophotometry

the

is

intensity

of

commonly

is

visible,

used

for

an

analytic al

ultraviolet

technique

and

determining

based

near-infrared

concentrations

on

the

radiation.

of

measurement of

This

technique

coloured substances in

solutions.

A

spectrophotometer

through

intensity

is

a

a

of

value

standard

curve

is

and

the

in

solute.

and

of

their

used

the

general

absorbance,

the

the

light

and

studied

for

of

of

a

certain

solution.

absorbed

substance

are

determining

wavelength,

The

are

by

These

c alibration curve

the

unknown

passes

measures the

absorbance. Absorbance

the

prepared

measured.

which

photodetector

converts it into the

light

producing a

sample.

Initially,

several

by serial dilution

absorbances

(gure8).

concentration

of

are

The

the

plotted

c alibration

coloured

studied solution.

or

a

c alibration

electric al

unknown

the

light

studied

absorbances

c ase,

pH

The

plotting

the

amount

the

concentrations,

then

of

transmitted

solutions

substance

In

the

produces

sample

describing

(gure9),

against

small

result

of

curve

relates

conductivity)

concentration

the

of

c an

measurement

a

measurable

the

be

on

solution

found

the

by

to

property

the

measuring

c alibration

(such as

concentration of

that

property

curve.

0.40

ecnabrosba

0.30

0.20

0.10

0

0

0.10

0.20

0.30

0.40

0.50

3

concentration/mmol dm



Figure 8

A typic al c alibration curve

Data-based question 

Figure 9

A series of standard

solutions of

potassium permanganate

The calibration curve in gure8 Ideally, the calibration curve should be linear, pass through the origin and have a was obtained using a series of tilt of approximately 45°. If the curve does not meet any of these requirements, it standard solutions of potassium should be constructed again using a slightly dierent wavelength of light and/or permanganate, KMnO

. A solution

4

dierent set of standard solutions. Sometimes linearity can only be achieved within with unknown concentration of a narrow range of concentrations. In this case, the studied solution can be diluted, KMnO

has an absorbance of

4

so the concentration of the studied substance falls within the range of calibration 0.285. Determine the concentration curve. In the last case, some additional calculations will be required to relate the of KMnO

in that solution.

4

concentrations of the studied substance in the diluted and original solutions.

76

Structure

Another

technique,

spectrophotometry

colorimetry,

but

“spectrophotometry”

correct

but

is

based

on

the

1.4

Counting

particles

by

mass: the mole

same principles as

limited to visible light. The terms “colorimetry” and

are

oen

used

interchangeably,

which

is

not

entirely

very common.

Concentration uncertainty of a standard solution

A

standard

solution

In

this

activity,

of

copper(II)

you

is

a

will

sulfate,

solution

prepare

each

by

of

known

two

using

concentration.

standard solutions

dierent

equipment.

M aterials

•

Wash

•

Weighing boats (2)

•

100 cm

•

Stirring

•

Funnels (2)

•

Pipettes

•

Spatula

•

Reagent

•

Blank labels

•

Colorimeter

•

Cuvettes

•

C alibration

bottle

containing

distilled

water

3

By

propagating

assess

will

the

determine

using

a

the

measurement

precision

the

of

Relevant

the

This

concentration

will

then

values.

of

allow

you will

You

your solutions

you

to

assess the

beakers (2)

rods (2)

bottles (2)

skills

Tool

1:

Measuring

•

Tool

1:

Standard

Tool

uncertainties,

concentration

concentrations.

•

•

the

actual

colorimeter.

accuracy

of

3:

volume

solution

C alculate

and

and

mass

preparation

interpret

percentage

error and curve

relating

concentration of copper(II)

percentage uncertainty sulfate and absorbance

•

Tool

3:

Express quantities and uncertainties to an

•

Copper(II)

sulfate

pentahydrate, CuSO

•5H 4

appropriate

number

of

signic ant

O 2

gures

•

Tool

3:

Record

•

Tool

3:

Propagate uncertainties

•

100 cm

•

Inquiry

2:

•

Milligram

measurement uncertainties

Additional equipment for solution 1:

3

Assess

accuracy

and

precision

S afety

volumetric ask

balance

(three

decimal

places)

Additional equipment for solution 2:

3

•

Wear

•

Solid

the

eye

protection.

copper(II)

sulfate

is

an

irritant

and

toxic to

•

100

•

Centigram

cm

measuring

balance

cylinder

(two

decimal

places)

environment

Instructions

•

Dispose

of

all

solutions

appropriately.

1.

Use

the

equipment

copper(II)

sulfate

provided

to

prepare two

standard solutions, both with

–3

concentration

solution

1,

milligram

0.020 mol dm

you

should

balance.

use

For

.

the

When

preparing

volumetric ask and

solution

2,

use

the

measuring

meniscus of the solution cylinder

2.

etched

line

indic ating

Record

and

the

centigram

balance.

measurements

you

make

along

the

way,

including their uncertainties.

3

volume,

e.g.

250 cm

3.

Following

the

your

teacher ’s

colorimeter,

instructions

measure

the

on

how to use

absorbance

of

your

solutions.

4.

a

fixed

volume of solution

when the meniscus is on

the

etched

Refer

to

the

c alibration

concentration

volumetric flask contains

of

curve to determine the actual

your solutions.

Q uestions

1.

line,

Determine

the

uncertainty

of

the

concentrations of

solutions 1 and 2.

3

e.g.

250 cm

2.

C alculate

the

percentage

error

of

the

concentrations

of solutions 1 and 2.

3.

Assess

the

precision

and

accuracy of the

concentrations of solutions 1 and 2.

77

Structure

1

Models

of

the

particulate

nature

of

matter

5. 4.

Consider

the

way

you

have

presented

The

construction

of

c alibration

curves

involves

your

ATL preparing c alculations

for

the

questions

above.

Do

samples

concentrations. think

they

convey

your

thinking?

Do

reader

would

be

able

to

easily

solutions

that

cover

Instead

of

measuring

a

and

range of

dissolving

you think

a a

of

you

follow

certain

mass

of

solute

the

way

you

have done

your

here, chemists oen start with a stock solution and thought

process?

How

could

you

improve

perform a the

presentation

of

your

c alculations?

want

to

look

serial dilution.

the

advantages

disadvantages of using a serial dilution in the

through some of the

preparation worked

Discuss

You

and may

examples

in

this

textbook

for

of

samples

for

a

c alibration

curve.

ideas.

Avogadro’s law (Structure 1.4.6)

In

1811,

at

the

This

Amedeo

same

Avogadro

temperature

hypothesis

has

suggested

and

been

pressure

conrmed

that

equal

contain

in

many

volumes

equal

of

any two gases

numbers

experiments

of

and

is

molecules.

now

known as

Avogadro’s law

Since

to

the

each

volumes

are

amount

other,

of

two

a

substance

amount

reacting

proportional

n

of

the

to

the

of

a

and

gas

gaseous

amounts

is

the

number

species

of

of

proportional

these

particles

to

measured

its

are

volume.

under

the

proportional

Therefore, the

same conditions

species:

V 1

1

= n

V 2

In

turn,

the

2

amounts

stoichiometric

know

of

the

other

of

reactants

coecients in

volume

gaseous

of

any

gas

a

and

products

balanced

consumed

substances

c an

be

are

chemic al

or

proportional to their

equation.

produced

found

without

in

the

As

a

result, if we

reaction,

the

volumes

c alculating their amounts.

Worked example 10

The

combustion

of

hydrogen

sulde,

H

S,

proceeds

as

follows:

2

2H

S(g)

+

3O

2

(g)

→

2H

2

C alculate

the

O(l)

+

2SO

2

volumes

of

(g) 2

oxygen,

O

(g),

consumed

and

sulfur

dioxide,

2

SO

(g),

produced

if

the

volume

of

hydrogen

sulde

combusted

was

2

3

0.908 dm

.

All

volumes

are

measured

under

the

same

conditions.

Solution

Practice question The

ratio of the stoichiometric coecients of H

you

c an

S and O 2

4.

is

2 : 3.

Therefore,

2

3

Incomplete combustion of multiply

the

volume

of

combusted H

S

by

to

nd

the

volume of

2

2

hydrogen

sulde

elemental

sulfur

produces combusted O

: 2

instead of sulfur 3

V(O

)

=

3

V(H

2

S)

=

2

The

3

×

0.908 dm

3

≈

1.36 dm

2

dioxide:

2

ratio of the stoichiometric coecients of H

S and SO 2

2H

S(g)

+

O

2

(g)

→

2

2H

O(l)

+

is

1 : 1.

Therefore,

2

2S(s)

2

the

volume

of

combusted H

S

is

the

same

as

the

volume

of

produced SO

2

C alculate

the

3

)

V(SO

=

V(H

2

combusted

of

S)

=

0.908 dm

2

hydrogen sulde if the Note

volume

oxygen

that

the

volume

of

liquid

water

c annot

be

found

in

the

same

manner, as

consumed in Avogadro’s

law

applies

to

gases

only.

3

this

reaction

was

1.25 dm

Linking question

Avogadro’s

behaviour

78

: 2

volume of

law

of

a

applies

real

gas

to

ideal gases. Under what conditions might the

deviate

most

from

an

ideal

gas?

(Structure 1.5)

Structure

1.4

Counting

particles

by

mass: the mole

End-of-topic questions

C alculate:

Topic review a.

1.

Using

your

knowledge

from the

Structure 1.4

The

molar

KAl(SO

) 4

answer

the

guiding

question

as

fully

as

mass

of

potassium alum,

topic, •12H

2

O.

[1]

2

possible: b.

The

c.

The

d.

The

How do we quantify matter on the atomic scale?

amount

potassium

total

of

substance,

in

mol,

in

1.00 g of

alum.

[1]

number

of

atoms

in

1.00 g of

Exam-style questions potassium

alum.

[1]

Multiple-choice questions

2.

What

is

the

copper(II)

number

sulfate

of

oxygen

atoms

pentahydrate,

in

CuSO

amount

of

0.400 mol of

by

•5H

potassiumalum.

4

O?

complete

water,

in

mol,

that

decomposition

of

c an

be

produced

1.00 g of

[1]

2

24

A.

3.60

C.

2.16

×

B.

9

D.

5.40

e.

10

The

potassium

24

×

A

sample

containing

0.70 g

of

c alcium

composition,

by

mass, of

alum.

[2]

10

8. 3.

percentage

To

visualize

the

mole,

a

chemistry

student

decided to

nitrate, 23

pile

3

C a(NO

) 3

,

is

dissolved

in

water

to

a

volume

of

200 cm

is

6.02 × 10

needed

–

What

up

grains

of

sand. Estimate the time

.

2

the

concentration of NO

ions

in

this

to

complete

this

project

if

an

average

grain

solution?

3

ofsand –3

A.

3.5 g dm

B.

7.0 g dm

0.021 mol dm

D.

0.043 mol dm

50 kg –3

For

which

the

molecular

5 mg,

and

the

student

c an

shovel

of

sand

per

minute.

[3]

–3

9.

4.

weighs

–3

C.

molecule

is

the

empiric al

formula

the

Deduce

the

empiric al

formulas

for

the

following

same as compounds:

A.

CH

CH 3

B.

CH

formula?

CH 2

OH

C.

CH

2

COOCH 3

CH 3

CH 2

D.

CH

3

CH 2

a.

an

b.

an oxygen-containing organic compound, 5.00 g of

sulfur

that

contains

59.95%

of

oxygen

[1]

COOH which produces 9.55 g of carbon dioxide and 5.87 g

3

Which volume of a 5.0 mol dm

of

3

–3

5.

oxide

CH 2

sulfuric acid (H

SO

2

of water upon complete combustion.

)

[2]

4

3

stock solution is required to prepare 0.50 dm

of a 10.

A

standard

solution

of

potassium

sulfate, K

SO 2

,

was

4

solution whose concentration of hydrogen ions is

3

prepared

from

8.714 g

of

the

solid

salt

using

a

250 cm

–3

0.10 mol dm

? volumetric

3

A.

0.010

3

cm

C.

5.0

B.

6.

A

0.0050

student

an

D.

obtained

experimental

g dm

3

cm

the

of

the

empiric al

the

mass

concentration, in

–3

,

and

potassium

10 cm

11.

following data during an

determination

C alculate

–3

cm

3

ask.

The

ve

formula of

molar

sulfate

c alibration

standard

potassium

oxide of tin:

the

in

concentration,

the

curve

in

solutions,

nal

mol dm

, of

solution.

gure7

in

in

was

which

the

permanganate, KMnO

,

[2]

constructed using

concentration of

varied

from

0.100

4

–3

to



M ass

of

tin



M ass

of

oxide

before

heating

=

0.500 mmol dm

these

mass=

of

tin

aer

.

Describe

how

you

would

prepare

1.78 g

solutions

using

serial

dilution.

[3]

heating to a constant

12.

2.26 g

C arbon

monoxide,

produces

c arbon

CO,

is

dioxide,

a

toxic gas. Its combustion

CO

(g). 2

According

to

these

data,

what

is

the

correct

formula of

a. the

oxide

of

Deduce

balanced

equation

SnO

C.

c arbon

monoxide.

[1]

SnO 3

3

b. B.

for the combustion

tin?

of A.

the

SnO

D.

C alculate

the

volumes, in dm

,

of

consumed

c arbon

SnO

2

5

monoxide

and

oxygen

if

the

combustion

produced

3

2.00 dm

Extended-response questions

7 .

Alums

are

XAl(SO

) 4

salt

hydrates

•12H 2

O,

of

the

where

X

measured

general

is

an

of

c arbon

under

dioxide.

the

same

All

volumes

conditions.

are

[2]

formula

alkali metal or other

2

singly-charged

c ation.

decompose

follows:

XAl(SO

as

) 4

•12H 2

When

O(s) 2

→

heated, most alums

XAl(SO

) 4

(s) 2

+

12H

O(l) 2

79

Structure 1.5

Ideal gases

How does the model of ideal gas behaviour help us to predict the behaviour of real gases?

As

is

with

a

any

theoretic al

simplic ation

many

c ases,

precision

it

that

predicts

sucient

model,

has

for

its

the

the

concept

properties

most

of

an

ideal gas

at

advantages and limitations. In

practic al

of

real gases with a

purposes.

low

real

temperatures

gases

ideal

gas

deviates

model

and

high

pressures

signic antly

c annot

be

from

the

the

behaviour of

prediction, so the

used under these conditions.

However,

Understandings

Structure 1.5.1

collisions

—

An

between

ideal

gas

particles

consists

are

gases

of

moving

particles

with

Structure 1.5.2

—

Real

Structure 1.5.3

—

The

molar

Structure 1.5.4

—

The

relationship

deviate

volume

from

of

an

the

ideal

ideal

between

gas

gas

the

is

a

model,

pressure,

gas

equation

pV

=

nRT

and

the

combined

gas

at

volume,

V 1

ideal

volume

particularly

constant

p the

negligible

and

no

intermolecular

forces. All

considered elastic.

a

low

temperature

temperature

temperature

p 1

law

at

specic

and

and

amount

and

high

pressure.

pressure.

of

an

ideal

gas

is

shown in

V 2

2

= T

T 1

2

Assumptions of the ideal gas

model (Structure 1.5.1)

The

ideal

gas

model

states

that

an

ideal

gas

conforms

to

the

following

ve

assumptions:

1.

Molecules of a gas are in constant random motion

This

until

2.

means

they

gas

molecules

with

another

or

inelastic

sound.

perfectly

collisions

However,

elastic

and

of

larger

the

no

(0

°C)

and

energy

stationary.

or

They

the

move

side

of

a

in

straight lines

container.

is

energy

between

lost

from

c an

be

transferred

molecules

the

in

an

ideal

as

heat

gas

are

system.

100 kPa

gas

space

in

occupies

phase

but

is

the

which

the

about

In

both

same

volume

the

1600

times

the

volume

of

liquid

water

at

273.15K

pressure (standard temperature and pressure, STP).

conditions.

same

they occupy

occupies

the

changed,

4.

water

Nitrogen

gaseous

are

objects,

collisions

volume of the container

forces

not

molecule

The volume occupied by gas molecules is negligible compared to the

Vaporized

Intermolecular

are

gas

Collisions between molecules are perfectly elastic

In

3.

that

collide

gas

650

and

of

times

c ases,

the

the

There are no intermolecular

size

gas

molecules

the

is

are

the

volume

number

of

to

of

liquid

nitrogen under

molecules in liquid and

individual

>99.9%

free

of

molecules has not

empty

space. This is the

move.

forces between gas particles

studied in For an ideal gas, the intermolecular forces are negligible compared to the kinetic

Structure 2.2. energy of the molecules. As such, an ideal gas will not condense into a liquid.

80

Structure

5.

1.5

Ideal gases

The kinetic energy of the molecules is directly proportional to Kelvin

temperature

This

relationship

is

studied in

Reactivity 2.2

Pressure–volume relationships

Robert

Boyle

pressure

of

(1627–1691)

a

given

relationship,

now

established

amount

of

known as

a

gas

is

that,

at

constant

inversely

Boyle’s law,

c an

temperature, the

proportional

be

expressed

to

as

its

volume. This

follows: 

Figure 1

An ideal gas consists of

1 p

or

∝

pV = k (a

constant)

or

p

V 1

=p 1

particles that

V 2

collide elastic ally,

have no

2

V intermolecular forces and

In

gure

the

1,

walls

pressure.

of

space,

walls,

so

the

of

If

molecules

the

the

so

the

of

container.

volume

every

is

is

gas

are constantly striking and bouncing o

force

halved,

second

pressure

a

The

there

these

there

are

doubled

of

are

twice

impacts

twice

as

as

produces

many

a

measurable

molecules

in

volume when compared

each unit

many impacts with the container

(gure 2).

pressure

Figure 2

to the volume of

the gas (the container)

volume



occupy negligible

halved

doubled

Halving the volume of a container doubles the pressure

p

p

,erusserp

,erusserp

TOK

Models are simplied

representations of natural

phenomena. The ideal gas model is volume, V

reciproc al

volume, 1/V built on certain assumptions related



Figure 3

Graphs showing the inverse relationship between pressure and

to the behaviour of ideal gases.

volume of an ideal gas

What

in

–2

The

SI

other

unit

units

of

atmosphere

inch

(psi).

found

in

pressure is the

of

pressure

(atm),

databases

and

pasc al (Pa),

commonly

millimetres

Standard

temperature

are

of

temperature

for

mercury

and

comparative

100.0 kPa

where

used

in

1 Pa

=

1 N m

–3

=

1 J m

.

M any

dierent countries, including the

(mm Hg),

pressure

purposes.

bar,

and

conditions

STP

for

pounds

(STP)

gases

is

are

0 °C

per

square

frequently

or

the

is

the

role

of

assumptions

development of scientic

models?

What

not

are

the

implic ations of

acknowledging

a

model’s

limitations?

273.15 K

pressure.

81

Structure

1

Models

of

the

particulate

nature

of

matter

Worked example 1

3

A

weather

is

released

balloon

at

sea

lled

level.

with

The

32.0 dm

balloon

of

helium

reaches

an

at

a

pressure

altitude

of

of

100.0 kPa

4500 m,

where

3

the

atmospheric

balloon

helium

at

in

that

the

pressure

altitude.

balloon

is

57.7 kPa.

Assume

remain

that

C alculate

the

the

volume,

temperature

and

in

the

dm

,

of

amount

the

of

constant.

Solution

V

× p 1

From

Boyle’s

law,

it

follows that

V

1

=

, so:

2

p 2

3

32.0 dm

× 100 kPa 3

=

V

≈

55.5 dm

2

57.7 kPa

Practice question

1.

At

a

certain

altitude,

a

weather

balloon

has

a

temperature

of

–35.0 °C and

3

a

volume

contains

of

0.250 m

16.0 g

of

.

C alculate

helium,

the

pressure,

in

kPa, inside the balloon if

He(g).

Real gases vs ideal gases (Structure 1.5.2)

When

begin

move,

of

the

to

volume

occupy

of

a

relationship

of

between

doubling

the

gas

the

against

no

of

become

pressure.

pressure

pressure

decreases

proportion

forces

reducing

pressure

real

large

intermolecular

collisions,

graph

a

and

volume

longer

signic antly,

the

container.

signic ant.

This

means

volume

for

a

real

halves

no

and

molecules

so

a

real

applies.

an

ideal

gas,

gas.

For

volume.

ideal gas

erusserp

0.5p

V

2V

0 volume 0

 Figure

an

82

ideal

4

gas

Doubling

but

not

for

the

a

pressure

real

gas

halves

the

the

Figure

real gas

p

may

little space to

decreases the number

for

longer

gas

the

This

that,

the

With

volume

for

4

inverse

shows a

the

real gas,

Structure

For

a

gas

to

deviate

intermolecular

the

molecules

from

forces

ideal

and/or

themselves.

gas

a

behaviour,

signic ant

This

there

volume

commonly

must

of

occurs

the

at

a

be

Ideal gases

detectable

gas

low

1.5

must

be

occupied

by

temperature and high

pressure.

Low temperature:

is

reduced.

form

and

As

As

volume

of

molecules

At

the

low

may

high

relationship

not

considered

be

pressure,

an

c annot

keep

for

temperature.

in

the

At

low

container,

negligible.

At

so

high

intermolecular

space

of

be

ideal

are

of

more

molecules

signic ant

volume

is

the

gas

forces

molecules

of

attraction

elastic ally.

compressed,

and

energy

intermolecular

rebound

a

gas

there

far

are

the

attraction

to

apart

part

only

no

and

behaviour

occupied

temperature,

forces

there

molecules

an

pressure,

the

kinetic

of

in

the

the

a

reduced

space

longer

space.

volume of the gas.

between them,

inverse, so the gas is

ideal gas.

them.

conditions

the

another,

becomes

pressure

Ideal gas conditions

The

one

necessarily

molecules

between

to

with

not

themselves

the

temperature,

collide

molecules

High pressure:

The

At

they

very

by

few

the

prevent

are

low

molecules

molecules

molecules

are

interaction

between

pressure and high

per

unit

of

volume

themselves is

moving

too

fast

to

allow

for

form.

Activity

1.

2.

Outline

Discuss

the

deviations

3.

main

what

Consider

assumptions

conditions

from

how

ideal

each

of

behind

pressure

the

and

ideal gasmodel.

temperature

are

likely

to

lead to

behaviour.

of

the

following

might

affect

the

validity

of

the

ideal

gasmodel:

a.

b.

4.

For

Strong

L arge

each

ideal

a.

intermolecular

molecular

of

the

gas

at

low

volume

following

behaviour

and

forces

give

pairs,

a

predict

which

is

more

likely

to

exhibit

reason:

pressure

or

gas

b.

gas

at

at

high

low

pressure

temperature

or

gas

c.

at

high

hydrogen

temperature

fluoride,

HF(g)

or

hydrogen

d.

bromide,

methane, CH

HBr(g)

(g) 4

or

dec ane, C

H 10

e.

(g) 22

propanone, CH

COCH 3

(g) 3

or

butane, C

H 4

(g) 10

83

Structure

1

Models

of

the

particulate

nature

of

matter

Real gases

Gases

that

deviate

from

the

ideal

gas

model

are

known as

n

real gases.

(

2

V – nb

) (

p + a V

)

nRT

=

Relevant skills

measured •

Tool

2:

•

Inquiry

Use

spreadsheets to manipulate data. pressure

1:

Select

sufficient

and

relevant

correction

sources of correction

information.

forces

for

for

of

between

volume

molecules

molecules •

Inquiry

1:

Demonstrate

creativity in the designing, measured

implementation

or

presentation

of

the

investigation. volume

Instructions

Parameter

The

relationship

temperature

Waals

of

between

real

gases

pressure,

is

volume, amount and

modelled

by

the

parameter

van der

and

b

for

a corrects

b

corrects

various

for

for

gases

intermolecular

molecular

are

force

volume.

shown in table 1.

equation:

–1

Substance

a

/

× 10

ammonia, NH

6

Pa m

–2

mol

–3

b

/

× 10

3

m

4.225

0.0371

1.355

0.0320

13.89

0.1164

3

argon,

Ar

H

butane, C 4

10

H

butan-1-ol, C 4

OH

20.94

0.1326

7.566

0.0648

5.580

0.0651

9

Cl

chloromethane, CH 3

H

ethane, C 2

6

H

ethanol, C 2

helium,

OH

12.56

0.0871

0.0346

0.0238

4.500

0.0442

3.700

0.0406

9.565

0.0739

5.193

0.0106

5

He

hydrogen

bromide,

hydrogen

chloride,

hydrogen

uoride,

krypton,

HBr

HCl

HF

Kr

methane, CH

2.303

0.0431

9.476

0.0659

0.208

0.0167

4

OH

methanol, CH 3

neon,

Ne

H

pentane, C 5

19.09

0.1449

OH

25.88

0.1568

9.39

0.0905

OH

16.26

0.1079

5.537

0.0305

4.192

0.0516

12

H

pentan-1-ol, C 5

11

H

propane, C 3

8

H

propan-1-ol, C 3

7

O

water, H 2

xenon,

Xe

 Table 1

84

strength and

Values of

Van der Waals parameters, a and

b,

for a selection of gases

–1

mol

a

Structure

1.

Use

of

a

the

selection

factors

instance,

of

the

data

affecting

the

in

table

1

values of

to

a

explore some

and

You

b. For

will

how

you could look at:

to

choose

and/or •

intermolecular force strength and the value of a

•

molar

•

the

mass

and

the

need

value of

to

analyse

to

decide

it.

how

much

Depending

explore,

you

may

on

data

which

need

to

to

1.5

Ideal gases

select, and

option

perform

you

c alculations

look up additional data.

b

2.

Consider

how

you

could

present

your data

ATL graphic ally. effect

of

volume

on

the

deviation

from

your gas

Prepare a one-page summary of

ideal

exploration

to

share

with

your

class.

behaviour.

Linking question

Under

comparable

behaviour

than

conditions,

others?

The molar

why

do

some

gases

deviate

more

from

ideal

(Structure 2.2)

volume of an ideal gas

Avogadro’s

law

is

covered in

Structure 1.4.

(Structure 1.5.3)

Avogadro’ s law states that equal volumes of any two gases at the same

temperature and pressure contain equal numbers of particles. The molar volume

of an ideal gas is a constant at specied temperature and pressure. For example,

3

at STP , the molar volume of an ideal gas, V

, is equal to 22.7dm

–1

mol

.

m

28.3 cm

H

CH

He

2

O

4

Cl

2

2

3

V = 22.7 dm

 Figure 5 1

1

 Figure 6

1

4.00 g mol

2.02 g mol

Molar volume of

1

16.05 g mol

any gas is identic al at

Molar volume of an ideal gas

1

32.00 g mol

70.90 g mol

a given temperature and

compared with a soccer ball

pressure

Worked example 2

3

A

2.00 dm

sample

of

an

unknown

gas

at

STP

has

a

mass

of

2.47 g.

1

Determine

the

molar

mass,

in

g mol

,

of

the

gas.

Solution

3

V n

=

2.00 dm =

= 3

V m

m M

=

2.47 g =

n

0.0881 mol

1

mol

22.7dm

1

=

28.0 g mol

0.0881 mol

Practice question

–1

2.

Determine

the

molar

mass,

in

g mol

,

of

an

elemental

gaseous substance

–3

that has a density of 3.12 g dm

at STP. Identify the substance if its molecules

are diatomic.

85

Structure

1

Models

of

the

particulate

nature

of

matter

Hypotheses

Amedeo

contain

and

Avogadro

equal

pressure.

postulated

numbers

This

of

that

particles

bec ame

equal

under

known

as

volumes

the

of

same

Avogadro’ s

dierent gases would

conditions

of

temperature

hypothesis.

A hypothesis is a tentative and falsiable explanation or description of a

phenomenon,

be

used

What

to

test

from

the

predictions

which

predictions

c an

be

deduced.

Predictions

natural

c an then

hypothesis.

might

be

derived

from

Avogadro’s

hypothesis?

Experimental determination of the molar mass of a gas

The

ideal

molar

gas

mass

of

equation

a

gas

by

c an

be

used to determine the

collecting

a

known

Instructions

volume of it 1.

under

known

conditions

of

temperature

and

Measure

ambient

Alternatively, this

practic al

you

will

of

butane

you

c an

with

search

a

barometer.

loc al

weather

data

for

experimentally determine the molar atmospheric

mass

pressure

pressure. In

found

in

disposable

plastic

pressure

in

your

geographic

loc ation on

lighters. the

day

you

do

the

experiment.

Relevant skills 2.

•

Tool

3:

range

Record

to

an

uncertainties

appropriate

in

measurements as a

in

precision

and

the

Tool

•

Inquiry

3:

C alculate

Fill

the

measuring

Assess

and

interpret

accuracy

and

Inquiry

3:

Identify

systematic

and

percentage

Inquiry

3:

and

discuss

random

Evaluate

Measure the

the

sources and impacts of

implic ations

limitations

it

in

the

cylinder

to

the

brim

with

water

trough so that its mouth is under

If

done

clamp

be

full

correctly,

of

the

measuring

cylinder

water. Hold it in this position with a

(figure 7).

error.

of

Submerge

and

the

lighter

in

water, then take it out again

methodologic al and

weaknesses,

water.

error.

4. •

with

water.

precision. should

•

trough

the

propagate

water.

2:

of

processed data. andinvert

•

plastic

temperature

3. uncertainties

Half-fill

dry

it

thoroughly

with

a

paper

towel.

Weigh the

assumptions on lighter.

conclusions.

5. •

Inquiry

3:

Explain

realistic

and

relevant

Hold

an

the

lighter

under

water

and

press

the

button

improvements

on to

the

lighter

to

release the gas so that it bubbles up

investigation.

inside the measuring cylinder (figure 7). Continue until

S afety

3

you

eye

have

exact

•

Wear

protection.

•

Butane gas is flammable. Keep away from open flames

collected

around

100 cm

of

gas.

Record the

volume.

6.

Release

7.

Dry

8.

If

the

gas

in

a

well-ventilated

area.

and sparks.

the

lighter

as

thoroughly

as

possible and

Materials reweighit.

•

Disposable plastic lighter

•

L arge

container,

for

•

100 cm

•

Balance (±0.01 g)

•

Clamp and stand

•

Thermometer

example

a

large

plastic

you

have

time,

repeat

to

get

three

sets

of

results.

trough

3

measuring

cylinder

3

100 cm

cylinder

•

Barometer

(if

available)

water

 Figure 7

86

Experiment

apparatus

measuring

Structure

1.5

Ideal gases

Questions

1.

Design

a

suitable

results

table

for

8.

your data.

Suggest

realistic

determining 2.

Process

the

your

molar

data

mass

of

to

obtain

an

experimental

value

9.

The

the Propagate the uncertainties.

4.

Compare

pressure

sum

partial

value

by

5.

Assess

6.

Discuss

your

experimental

c alculating

the

the

accuracy

value

to

percentage

and

precision

the

theoretic al

adjust

error.

of

the

relative

your

impacts

of

of

Comment

on

the

vapour

pressure

your

of

data

additional

need

to

this

method

for

the

measuring

pressure

butane.

processing

data

and

to

of

cylinder is in fact

water and the

How

could

account

information

for

do

you

this?

you

research?

systematic

and

Consider

alternative

methods

for determining the

random

mass of a gas that could be done in a school

results.

laboratory.

7.

inside

the

What

molar on

to

mass of a gas.

your data.

10.

errors

improvements

molar

for

butane.

3.

the

at

least

two

major

If

you

have

time,

show

your

ideas

to

your

sources of teacher and try them out.

experimental

error.

Linking question

Graphs

data

c an

be

points.

presented

What

representation?

are

the

(Tools

2

as

sketches

advantages

and

3,

or

as

and

accurately

limitations

plotted

of

each

Reactivity 2.2)

Pressure, volume, temperature and amount

of an ideal gas (Structure 1.5.4)

There

are

four

variables

of

ideal

1.

The

pressure

2.

The

volume the gas occupies,

3.

The

absolute

4.

The amount of the gas,

The

eect

keeping

up

with

of

the

any

changed.

of

two

He

of

two

law:

a

gas

by the gas,

that

he

gas

these

V

variables

This

performed

observed

each other:

T

n

constant.

were

aect

p

temperature of the gas,

other

Boyle’s

temperature

exerted

an

kept

that

the

an

is

on

each

what

other

Robert

experiment

constant,

pressure

but

and

the

c an

Boyle

be

investigated

did

when

he

by

c ame

where the amount and the

volume

volume

of

of

the

the

gas

container

were

was

inversely

proportional.

87

Structure

1

Models

of

the

particulate

nature

of

matter

Graphing the gas laws

Online

simulations

relationships

temperature,

amount

ideal

of

gas

and

gas.

In

direct

this

data

and

you

to

easily

pressure

pressure

simulation,

spreadsheet

about

allow

between

task,

and

you

which

analysis

inverse

and

temperature

will

will

collect

allow

skills,

as

3.

explore the

volume,

you

well

data

to

as

for

a

for

fixed

a

five

from an

the

simulation,

pressure

certain

amount

different

vary

and

the

record

of

gas.

temperatures

temperature at a

the

resulting

Collect

in

a

data

suitable

volume

for

at

table

least

in

your

spreadsheet.

practice

reinforce

Using

constant

volume and

ideas 4.

Compute the temperature values in both °C and K.

5.

Construct

proportionality.

two

graphs of

V

vs

T; one with

T

in °C and

Relevant skills the other with

•

Tool

2:

Generate

•

Tool

2:

Use

data

in K.

from simulations. 6.

Using

the

simulation,

vary

the

volume at a constant

spreadsheets to manipulate data. temperature

3:

T

•

Tool

Understand

direct

and

inverse

•

Inquiry 1: Identify dependent, independent and

a

proportionality.

certain

five

and

amount

different

record

the

of

Collect

gas.

volumes

in

a

resulting

data

suitable

pressure

for

table

at

in

for

least

your

spreadsheet. controlled

variables.

7.

Construct

8.

Use

a

graph of

p

vs

V

Materials

1 •

Simulation

that

allows

you

to

change

your

spreadsheet

to

compute

values

for

.

pressure, V

volume

and

temperature

for

an

ideal

gas.

It

must

1

have 9.

an

option

to

hold

one

variable

constant

and

Construct

a

graph of

p

vs

V

vary the

Questions other

•

two.

Spreadsheet

1.

software

What

were

variables

your dependent and independent

in

each

c ase?

Which

variables

were

Instructions controlled?

1.

Using

the

simulation,

vary

the

temperature at a 2.

constant

volume

and

record

the

resulting

Describe

a

certain

amount

of

gas.

Collect

data

the

relationship

shown

in

each

graph

pressure as

for

for

at

direct

proportionality,

inverse

proportionality,

least orother.

five

different

temperatures

in

a

suitable

table

in

your

3.

spreadsheet.

When

studying

temperature 2.

Construct

a

graph of

p

vs

gases,

values

it

into

is

SI

important

units

to

convert all

(kelvin).

Discuss

why

T. this

is

the

volume

c ase

units

for

c an

temperature,

vary

whereas

depending

on

the

pressure and

source.

The combined gas law

We

have

seen

proportional

that

to

pressure

absolute

is

inversely

proportional

to

volume

temperature.

1 p

∝

;

p

∝

T

V

Combining

the

two

relationships

gives:

pV

pV ∝ T

= k (a

or

constant) or

T

p

V

p 1

V 2

1

2

=

T

T 1

This

88

equation

is

known as the

2

combined gas law

and

directly

Structure

1.5

Ideal gases

Experiments

The

gas

laws

variables

arose

were

manipulated.

from

experiments in which certain

controlled,

Inspect

the

while

others

apparatus

were

shown

c arefully

in

gure8.

pressure gauge What

might

is

independent

the

be

explored

with

variable?

this

set-up?

What

thermometer

What

variables must 250

be

controlled?

What

is

the

purpose

of

mL

round-bottomed flask

each of the containing air

itemsdepicted?

water bath

u Figure 8

experiment

Apparatus for conducting an

into the behaviour of a gas

Worked example 3

3

A

of

weather

balloon

100.0 kPa

is

lled

released

altitude

of

35,000 m,

–50 °C.

C alculate

with

at

32.0 dm

sea

where

level.

the

of

The

helium

balloon

pressure

is

at

25 °C

and

eventually

475 Pa

and

the

a

pressure

reaches

an

temperature

is

3

the

volume,

in

m

,

of

the

gas

in

the

balloon

under

these

conditions.

Solution

List

the

conditions

of

the

gas

in

the

p

weather balloon.

=

100.0 kPa

1

3

V

=

32.0 dm

1

Remember

to

convert

temperature to kelvin:

T

=

25

+

273.15

=

298.15 K

1

Then,

list

the

Remember

conditions

to

make

of

sure

the

the

gas

units

in

the

weather

balloon

at

35,000 m.

are consistent with the initial conditions of

the balloon.

p

=

0.475kPa

2

V

=

unknown

2

T

= –50

+

273.15

=

223.15 K

2

Substitute

the

numbers

into

the

combined

p

p

V 1

gas

law:

V 2

1

2

=

T

T

3

100.0 kPa

×

Practice question

2

1

0.475kPa

32.0 dm

×

V 2

3.

=

A

the

of

an

ideal gas has

3

a

Rearranging

sample

223.15 K

298.15 K

expression in terms of

V

volume

of

1.00 dm

at

STP.

3

gives:

C alculate

the

volume, in dm

,

2

3

V

≈

5.04

×

10

3

dm

3

=

of

that

sample

at

50.0 °C and

5.04 m

2

50.0 kPa.

89

Structure

1

Models

of

the

particulate

nature

of

matter

TOK

Throughout

were

of

this

chapter

developed

alpha

particles

you

through

with

have

explored

observations

gold

atoms,

to

of

the

models

the

related

natural

world

manipulation

of

to

or

the

particulate

obtained

gases

in

the

nature

through

gas

laws,

of

matter.

M any of these concepts

experimentation:

to

from

the

interaction

explorations of subatomic particles

atCERN.

 Figure 9

The set-up used

The ATLAS detector at

by Joseph Louis Gay-Lussac to investigate the thermal expansion of gases (le)

CERN used

to investigate elementary particles (right)

How do scientists investigate the behaviour of particles that are too small to be observed directly? How have advances in

technology inuenced scientic research into what matter is made up of ?

Ideal gas equation

The

the

combined

three

gas

law

parameters,

suggests

p,

V

or

T,

that

for

aects

any

the

given gas, the change in one of

other

two

in

such

a

way that the

pV

expression

remains

constant.

The

exact

value of that constant must be

T

proportional to the amount of the gas,

pV

pV ∝ n

or

=

T

nR

T

where

is

n:

R is the

universal gas constant, or simply

known as the

pV

=

ideal gas equation,

which

is

gas constant.

traditionally

The

written

as

last

expression

follows:

nRT

The value and units of

R depend on the units of

p,

V, T and

n. If all four parameters

3

are

expressed

in

–1

R

≈

The

8.31 J K

same

standard SI units (p

in

Pa,

V

in m

,

T

in K

and

n in mol), then

–1

mol

.

value and units of

R

c an

be

used

if

pressure

3

volume in dm

is

expressed

in

kPa and

3

,

as

the

two

conversion

factors (10

–3

for

kPa

to

Pa and 10

3

for dm

3

m

)

c ancel

each other out.

Linking question

How

c an

the

experimental

90

ideal

gas

data?

law

be

used

to

(Tool 1, Inquiry 2)

c alculate

the

molar

mass

of

a

gas

from

to

Structure

1.5

Ideal gases

Worked example 4

A 3.30 g sample of

an unknown organic compound

was vaporized at

T

=150 °C and

p =101.3 kPa to produce

3

1.91 dm

3

of a gas.

The gas was combusted

in excess oxygen to produce 3.96 g of water,

2.49 dm

of

c arbon dioxide

3

and

1.25 dm

of

nitrogen at STP.

Determine the following for the compound:

a.

molar mass

b.

empiric al formula

c.

molecular formula

Solution

a.

To

determine

the

molar

mass,

we

need

to

nd

out

the

amount

of

the

compound

using

the

ideal

gas

equation:

pV

n

=

T

=

n

=

RT

150

+

273.15

=

423.15 K

3

101.3 kPa

×

1.91 dm ≈

–1

0.0550 mol

–1

8.31 J K

mol

× 423.15 K

3.30 g –1

Therefore,

M

=

=

60.0 g mol

0.0550 mol

b.

All

c arbon,

amounts

hydrogen

of

these

O)

nitrogen

in

atoms

c arbon

in

the

dioxide,

combustion

water

and

products

nitrogen

are

originate

the

same

from

as

the

those

organic compound, so the

in

the

original

sample:

3.96 g

m n(H

and

elements

=

=

≈

0.220 mol

2 –1

M

n(H)

=

2

×

18.02 g mol

O)

n(H

=

2

×

0.220 mol

=

0.440 mol

2

3

2.49 dm

V n(CO

)

=

=

≈

2

3

V

22.7 dm

0.110 mol

–1

mol

M

n(C)

=

)

n(CO

=

0.110 mol

2

3

V n(N

)

1.25 dm

=

=

≈

0.0551 mol

2 3

V

22.7 dm

M

n(N)

= 2

×

n(N

)

=

2

×

–1

mol

0.0551

=

0.110 mol

2

The

of

original

the

three

compound

elements

could

also

(hydrogen,

contain

c arbon

oxygen.

and

To

check

nitrogen)

with

this,

the

we

mass

need

of

to

the

compare

original

the

total

mass

sample:

–1

m(H)

=

0.440 mol

×

1.01 g mol

≈

0.444 g

–1

m(C)

=

0.110 mol

×

12.01 g mol

≈

1.32 g

–1

m(N)

=

m(total)

0.110 mol

=

×

0.444 g

Therefore,

the

14.01 g mol

+

1.32 g

organic

+

≈

1.54 g

1.54 g

compound

≈

did

3.30 g

not

contain

oxygen,

so

its

formula

c an

be

represented as C

H x

x : y : z

The

=

0.110 : 0.440 : 0.110

empiric al

=

N y

. z

1 : 4 : 1

formula of the compound is CH

N. 4

–1

c.

M(CH

N)

=

12.01

+

4

+

1.01

+

14.01

=

30.06 g mol

–1

.

This

value

is

half

the

experimental

value

(60.0 g mol

),

4

so

the

molecular

formula

of

the

compound

will

have

twice

the

number

of

atoms

of

each element: C

H 2

N 8

. 2

91

Structure

1

Models

of

the

particulate

nature

of

matter

End-of-topic questions

0.58

Topic review

×

8.31

×

373

B. 3

100 ×

1.

Using

your

knowledge

from the

Structure 1.5

6

×

10

250

×

10

topic, 3

answer

the

guiding

question

as

fully

as

0.58 ×

100

0.58 ×

100

×

10

6

×

250

×

10

×

10

C.

possible:

8.31

×

100

3

How does the model of ideal gas behaviour help us to

×

10

6

×

250

D. predict the behaviour of real gases? 8.31

Exam-style questions

6

Which

the

Multiple-choice questions

2.

Which

of

the

following

are

graph

pressure

×

373

correctly

and

shows

volume

of

the

an

relationship

between

ideal gas, at constant

temperature?

assumptions

of

the

ideal gas

A.

B.

model?

P

P I.

The

volume

occupied

by the gas particles is

negligible

II.

There

are

no

intermolecular

forces

V

V

between

gas particles C.

III.

Zero

particle

D.

movement

P A.

I and II only

B.

I and III only

C.

II and III only

D.

I, II and III

The

temperature

7. 3.

of

an

ideal

gas

is

the

pressure

27 °C.

P

of

the

gas

aer

Which

is

of

volume

is

A.

162 °C

B.

450 °C

C.

1527 °C

D.

1800 °C

the

of

following

hydrogen

balloons

atoms,

at

contains

constant

the

largest

temperature

doubled and and

the

1

V

What is the number

temperature

1

V

pressure?

tripled?

3

4.

A

gas

syringe

contains

40

cm

of

an

ideal

gas

at

27 °C. H

What

will

the

volume

of

the

gas

be

aer

it

is

NH

(g)

2

(g)

at

constant

1.9

3.6

C.

44

cm

D.

84

cm

HF(g)

3

2 dm

3

1 dm

4 dm

pressure?

8.

3

A.

B.

(g)

4

3

3

2 dm 57 °C

CH

3

warmed to

What

are

the

conditions

for

the

ideal

gas

behaviour of

cm

real

gases?

3

cm

3

3

5.

A

0.58

g

sample

of

an

ideal

gas

at

100

kPa

and

a

volume

expressions

0.58

of

is

×

250

equal

8.31

×

cm

to

.

Which

the

100

A. 3

100 ×

92

10

6

×

250

×

10

molar

of

the

mass

following

of

the

Low

temperature

and

low

Low

temperature

and

high

pressure.

C.

High

temperature

and

low

D.

High

temperature

and

high

pressure.

pressure.

100 °C

3

has

A.

B.

gas?

pressure.

Structure

9.

Which

are

of

the

following

statements

about

an

14.

ideal gas

At

forms

constant

temperature,

the

several

gaseous

empiric al

compounds

p I.

C arbon

Deduce

correct?

using

and

the

compounds

molecular

data

from

the

1.5

with

formulas

table

Ideal gases

uorine.

for these

below.

[3]

= constant V 3

C arbon /

M ass

of

1.00 dm

p

Compound II.

At

constant

volume,

= constant

mass

%

at

STP / g

T

X

13.65

Y

24.02

3.88

4.41

Z

17 .40

6.08

V III.

At

constant

pressure,

= constant T

A.

I and II only

B.

I and III only

C.

II and III only

D.

I, II and III

15.

An

organic compound

9.1%

of

hydrogen

vaporized

and

sample of

A

contains

36.4%

A

with

a

of

54.5%

oxygen

mass

of

of

by

c arbon,

mass. A

0.230g occupies

3

a

volume

of

at

0.0785 dm

T=95 °C and

p=102 kPa.

Extended-response questions

10.

Explain,

from

in

ideal

your

own

words,

behaviour

at

low

why

real gases deviate

a.

Determine

the

empiric al

b.

Determine

the

relative

formula of

c.

Using your answers to parts a and b, determine the

molecular

A.

[2]

mass of

A.

[1]

temperatures and high

pressures.

[2] molecular formula of A.

11.

A

c ar

tyre

inated

to

2.50 bar

(250 kPa)

at

[1]

10 °C contains 16.

A

closed

steel

cylinder

contains

0.32 mol

of

hydrogen

3

12.0 dm

of

compressed

air.

Aer

a

long

journey, the gas

tyre

temperature

increases

to

25 °C

and

the

and

0.16 mol

of

oxygen

gas.

The

volume of the

pressure 3

cylinder to

261 kPa.

Determine

the

tyre

mixture conditions.

Assume

that

is

25 dm

there

was

no

air

is

[2]

a.

C alculate

mixture Ammonium

c arbonate, (NH

) 4

CO 2

when

(s),

the

in

) 4

initial

the

temperature of the gas

pressure,

in

kPa, of the gas

cylinder.

[1]

decomposes

When

the

gas

mixture

is

ignited,

both

reactants

are

heated:

consumed

(NH

initial

3

b. readily

the

25 °C.

loss during the

journey.

12.

and

volume under these

CO 2

(s)

→

2NH

3

(g)

+

CO

3

(g) + H 2

the

O(l)

completely

cylinder

rises

to

and

800

the

°C.

temperature inside

C alculate

the

pressure

2

inside

the

cylinder

at

that

moment.

[2]

3

Determine

the

volumes, in dm

at

STP, of the individual

–3

gases

produced

ammonium

on

decomposition

of

17.

2.25 g of

c arbonate.

[2]

An

unknown gas

4.00 g

sample of

X

has

X

was

a

density

of

2.82 g dm

combusted

in

excess

at

STP. A

oxygen to

3

produce

13.

The

gases

produced

in

question12

were

2.50 g

of

hydrogen

uoride

and

2.84 dm

of

transferred to a c arbon

dioxide

at

STP.

3

sealed

vessel

with

a

volume

of

1.50 dm

,

and

the

vessel

Determine was

heated

vessel

react

at

up

that

with

to

200 °C.

C alculate

temperature.

each

other.

the

the

following

for

X:

pressure in the

Assume that the gases do not

a.

molar

b.

empiric al

mass

c.

molecular

[1]

[2] formula

formula

[2]

[1]

93

Structure

2

Models

bonding

and

of

structure

Structure 2.1

The ionic model

What determines the ionic nature and properties of a compound?

Ionic compounds are characterized by the presence

the strong electrostatic attractions between oppositely

of positive and negative ions, which attract each other

charged ions. Once liquid, however, ionic compounds are

electrostatically. In solid ionic compounds, these ions

electrical conductors due to the presence of mobile ions.

are arranged in rigid crystalline lattices. Melting these

Due to their charge, ions interact strongly with polar water

solids requires a large amount of thermal energy due to

molecules, so ionic compounds are oen water-soluble.

Understandings

Structure 2.1.1 —

When

metal

form

c alled

c ations.

gain

positive

ions

electrons,

they

Structure 2.1.2 —

form

The

atoms

negative

ionic

lose

electrons, they

Structure 2.1.3 —

When non-metal atoms

bond

ions

is

dimensional

c alled anions.

formed

Ionic

lattice

compounds

structures,

exist

as

represented

three-

by

empiric al

formulas.

by

electrostatic attractions between oppositely charged ions.

Binary

ionic

followed

by

compounds

the

anion.

are

The

named

anion

with

the

c ation

adopts

the

sux

rst,

“ide”.

Introduction to bonds and structure

Atoms

ways.

rarely

exist

Atoms

c an

elements.

The

them

rise

us

is

give

in

be

isolation.

bonded

varying

to

.

are

atoms

arrangements

certain

nitrogen, N

They

to

dierent

However,

in

connected

of

of

the

same

atoms

properties.

agriculture,

and

For

together

type,

in

several

or

to

atoms

features

of

the

example,

nitrogen

78%

fertilizers

dierent

bonds

of

are

dierent

of

the

between

air

around

added to soils

2

to

help

are

crops

dierent

grow.

to

that

This

of

is

the

bec ause

the

nitrogenous

structure

and

compounds

bonding

found

in

of

nitrogen in air

fertilizers.

Atoms are held together by chemical bonds. This chapter discusses three dierent

bonding models: ionic, covalent and metallic. These lead to four types of structure:

ionic, molecular covalent, covalent network and metallic. Y ou may be wondering

why there are four types of structure, given that there are only three types of bonds.

This is because covalent substances can be found in two arrangements: a continuous

3D network, or discrete groups of atoms known as molecules.

types of ionic

covalent

metallic

bond

types of

molecular

covalent

covalent

network

ionic structure

metallic

metal ion

deloc alised

electron

t Figure 1

bonds and

There are three types of

four types of

structure

95

Structure

2

Models

of

bonding

and

structure

Chemic al bonds Models

Chemic al

Structure

models

2 .1 ,

of

Scientific

the

so

All

d i s c u ss

s t r u c t u r e.

and

in

a

bonds

substance.

positively

depends

charged

on

help

we

This

bonding

models

does

forces

species

which

and

species

of

bonds

are

attraction that hold atoms or

occur

due

negatively

to

electrostatic

charged

ions

together

attractions

between

species. The type of bonding

involved (table 1).

us

to

The electrostatic attraction between…

c annot

is

one

of

models

Type of bonding

Positively charged

are

Negatively charged

species

ionic

c ations

species

models

covalent

atomic

metallic

c ations

to

of

a

model.

As

some

limitations

of

these

of

the

the

electrons

electrons

All bonding types involve a positively charged

are electrostatic ally attracted

species and

a negatively charged

to each other

strengths and

various bonding

Ions (Structure 2.1.1)

Sodium

are

chloride

crystalline

Ionic

these

For

and

and

compounds

electric al

ionic

sodium

elemental

chlorine is a

Sodium

are

sulfate

which

also

poor

when

compounds

instance,

 Figure 3

copper(II)

brittle,

conductors

However,

C alifornia,

of

sections,

models.

 Figure 2

pair

you species that

through

shared

deloc alized

understand the  Table 1

identify

nuclei

necessarily make

inadequate, but it

important

work

anions

have limitations.

not

weaknesses

96

strong

chemic al

useful.

the

and

are

All

phenomena.

that

d i r e c t l y.

re asons

This

is

and

models

things

2.3

simplify

c o m p l ex

Sometimes

observe

and

bonding

models

represent

visualize

2.2

are

are

electric al

molten

very

chloride,

sodium

is

are

or

ionic compounds. They

characteristic of ionic compounds.

conductors when solid, but good

dissolved.

The

reactions

and

properties of

dierent to those of their constituent elements.

the

a

examples of

properties

main

so

ingredient

metal

that

in

reacts

table

salt,

violently

is

water-soluble.

with

water, and

poisonous gas.

Sodium

chloride and

chloride crystals on a tree branch and

copper(II) sulfate crystals.

copper(II) sulfate are ionic compounds

Oshore oil platform in

USA.

What examples of structure

bonding are present

in the photo?

Before

rst

discussing

look

into

ionic

what

ions

bonds

are.

and

the

characteristics

of

ionic

structures, we will

Structure

a

C ations and anions

2.1 The ionic model

b

+

Sodium chloride contains sodium ions, not sodium atoms. Sodium atoms and

sodium ions have dierent numbers of electrons, and therefore behave dierently. Na

Na

+

You

will

notice

1.

number

2.

electron

3.

charge.

of

three

dierences

between

Na

and

Na

:

electrons

 Figure 4

Sodium

(a) sodium

atom

(b) sodium ion

arrangement

atoms

are

neutral.

Sodium

ions

have a 1+

charge,

indic ated

by a

+

superscript

+

sign

next

to

the

symbol:

Na

.

Worked example 1

Determine the number of subatomic particles to show that

a.

sodium atoms are neutral

b.

sodium

ions

have

a

1+

charge.

Solution In a.

In

a

11

protons

11

In

atom

11

sodium

protons

10

ion

11

=

=

=

(charge

Structure 1.2,

you

learned

are: that

protons

have

and

electrons

You

c an

a

1+

charge

11+)

there

(charge

electrons

O verall

=

(charge

charge is 11

a

there

(charge

electrons

O verall

b.

sodium

11

have

a

1–

charge.

)

0

charge

ignore

neutrons in ionic

c alculations

as

these

are

uncharged.

are:

11+)

=

charge is 11

10

10

=

)

1+

Worked example 2

Deduce

the

electron

conguration

of

a

sodium

atom

and

a

sodium

ion.

a

b 2–

Solution

2

Na: 1s

2

6

1

2s

2p

3s

2

2

6

S +

Na

: 1s

C ations

2s

are

S

2p

ions

with

more

protons

than

electrons.

This

means

that

c ations

are

 Figure 5

positively

the

10

charged,

combined

electrons,

as

the

negative

the

combined

charge

overall

of

positive

electrons.

charge

As

of

sodium

protons

ions

is

have

(a) sulfur atom

(b) sulde ion

greater than

11

protons and

charge is 1+

Activity Anions

than

are

negatively

protons.

Figure

charged

5

ions.

They

contain

a

greater

number

of

electrons

shows a sulfur atom and a sulde ion. The sulde ion has a

Show

that

the

sulfur

atom

is

neutral

2

2

a

charge,

slightly

atom.

denoted

dierent

This

is

by

the

name:

followed

superscript

the

rst

part

by the sux

in

the

symbol S

corresponds

to

. Note that anions adopt

the

name

of

their

parent

ide.

and

of

the

2–

by

sulde

counting

particles.

now,

we

will

only

consider

monatomic

ions.

You

will

look

at

charged

atoms

(c alled

polyatomic ions)

in

a

later

their

of

the

a

charge

subatomic

the

electron

sulfur

atom

groups and

of

has

Determine

conguration For

ion

sulde

ion.

section.

97

Structure

2

Models

of

bonding

and

structure

Predicting the charge of an ion

The

18.

main

The

group

elements

electron

corresponding

are

in

congurations

c ations

are

periodic

for

shown

table

some

main

groups 1, 2, 13, 14, 15, 16, 17 and

group element atoms and their

below:

1e

+

Na

2

Na

2

1s

6

2s

1

2p

2

3s

2

1s

6

2s

2p

+

Na

has

the

same

electron

conguration

as

neon,

Ne.

Two

dierent

species with

+

same

electron

conguration

are

c alled

isoelectronic.

Therefore,

Na

and

Ne

are

isoelectronic.

1e

2e

+

Li

2

1s

2+

Li

1

Ca

2

2s

2

1s

1s

2

2s

Ca

6

2

2p

3s

6

3p

2

1s

2

2s

6

2

2p

3s

6

3p

2+

+

Li

2

4s

is

isoelectronic

with

helium,

Ca

He.

The

resulting

“closed”)

this

all

outermost

negatively

formation

Anions

where

is

are

the

have

When

noble gas electron

their

are

c ations

sublevels.

example

formed

parent

noble

main

isoelectronic

gas

group

with

congurations.

elements

form

electrons.

the

resulting

atoms

gain

gain

they

have

c ations

of oxidation

when

atoms

As

are

bec ause

it

electrons.

electrons

in

lost

Noble

involves

to

at

gases

have all

positively

the

full

(or

achieve

done so by losing

and

electrons

loss

of

noble

C ation

electrons.

examples

a

have

oen

charged.

the

achieve

Ar.

they

electrons,

Look

order

argon,

ions,

conguration. The atoms above

valence

charged,

an

is

below

gas

electron

conguration:

+2e

+1e

2

Cl

2

1s

2

2s

6

2p

2

3s

O

Cl

5

2

3p

1s

2

6

2s

2p

2

3s

2

6

1s

3p

O

2

2s

4

2

2p

1s

–

Cl

2

2s

6

2p

2–

is

isoelectronic

with

argon,

Ar.

Atoms

To obtain a noble gas conguration,

the

O

that

gain

formation

electrons

of

anions

is

become

a

is

anions.

reduction

isoelectronic

As

reduction

is

with

the

neon,

gain

of

Ne.

electrons,

process.

a chlorine atom gains an electron.

The

formation

of

an

ionic

compound

from its elements is a

redox

reaction.

Chlorine would also have a

Consider

the

formation

of

sodium

chloride

from its elements:

noble gas conguration if it lost

the seven outermost electrons. 2Na(s)

+ Cl

(g)

2NaCl(s)

2

However, the removal of so many

+

electrons from the attractive pull

Sodium

of the positively charged nucleus

anions, Cl

requires a large amount of energy

other

while the addition of a single

redoxreaction.

is

chloride,

a

.

The

NaCl,

half

reduction

is

made

equations

and

up

are

therefore

of

sodium

shown

the

c ations,

below.

formation

The

of

Na

rst

NaCl

is

, and chloride

an

oxidation and the

from its elements is a

electron releases energy. This is +

2Na

+

2Na

2e

Electron

loss

=

oxidation

why chlorine instead will gain an

+

Cl electron to become a chloride ion.

2e

2Cl

Electron gain

=

reduction

2

The energetics of these processes, Once

you

have

learned about

oxidation states

(Structure 3.1),

you should also be

called ionization energy and able

to

see

that

the

sodium

is

undergoing

oxidation

bec ause

its

oxidation state

bec ause

its

oxidation state

electron anity, are discussed in increases

(from 0 to

+1)

and

Structure 3.1 and relevant in the decreases

construction of Born–Haber cycles

(Reactivity 1.2).

98

(from 0 to

1).

the

chlorine

is

reduced

Structure

Atoms

tend

to

achieve

a

noble

gas

electron

conguration

2.1 The ionic model

through gaining,

TOK losing,

or,

to as the

as

we

will

octet rule.

see in

It

is

Structure 2.2,

c alled

the

octet

sharing

rule

electrons.

bec ause

This

most

is

noble

oen

gases

referred

have

General eight

rules

in

chemistry

(such

electrons in their outer shell.

as

There

exists

element

a

and

relationship

its

periodic

between

table

the

group.

charge

In

of

the

ion

formed

by

a

main

group

general:

the

octet

rule)

exceptions.

have

to

oen

How

exist

for

a

have

many

rule

exceptions

to

cease to

be useful? •

Elements

in

groups

1,

2

•

Elements

in

groups

15,

•

Elements

in

group

and

16

13

and

form 1+, 2+

17

form

3–,

and 3+

2–

and

ions,

1–

respectively

ions,

respectively

The

18

(noble

gases)

do

not

electron

conguration of

2

form ions

c arbon, 1s

2

2s

2

2p

, suggests that

c arbon atoms could lose or gain The

relationship

between

periodic

table

group

and

ionic

charge

is

illustrated in four

electrons

in

order

to

achieve

gure 6. a

noble

would 1

gas

conguration. This

result

in

the

formation

18 4+

of C

4–

or C

Although

ions,

this

is

respectively.

possible,

c arbon

1 2

13

14

15

16

more

17

commonly

through

a

forms compounds

process

c alled

covalent

+ 3

Li

2

2

N

O

bonding,

F

ion

+

3

Na

4

K

5

Rb

6

Cs

2+

+

either

discussed in

2

P

formation.

S

does

not

involve

Covalent bonding is

Structure 2.2

Cl

2+ 2

Se

Br

2+ 2

Sr

Te

+

Hydrogen

3

Al

Ca

+

 Figure 6

3+

Mg

which

I

2+

Ba

The charges of some common ions

atoms

losing

that

have

only

electron

one

or

electron

gaining

in

one.

the

1s

sublevel.

Electron

loss

They

leads

to

form

the

ions

by

formation of

+

H

,

which

is

simply

a

hydrogen

nucleus:

a

proton

with

no

electrons

surrounding

+

it.

The

charge density of a H

combine

with

other

species.

ion

is

One

therefore

such

very

example

high,

is

the

so

these

c ations

readily

formation of acidic

+

hydronium ions, H

O

,

formed

when

hydrogen

c ations

bond

with

water.

3

Hydrogen

atoms

c an

also

gain

an

electron

to

achieve

a

noble

forming

hydride

anions, H

gas

conguration,

Hydride

–

thus

anions

are

very

strong

.

bases.

You

bases in

will

learn

more about

Reactivity 3.1

hydrogen

–

+ +

e –1

+

ion, H

+ 1 e

hydride hydrogen

+ ion, H

atom

+

 Figure 7

The formation of H

and H

99

Structure

2

Models

of

bonding

and

structure

Practice questions

1.

Determine

the

charge

a.

lithium,

b.

magnesium, Mg

c.

aluminium, Al

d.

uorine, F

e.

nitrogen, N

f.

selenium, Se

g.

barium,

the

of

the

ion

formed

by

each

of

the

following elements.

Li

Ba

2.

State

name

3.

Complete the table:

of

ions

d,

e

and

f

above.

Number

Number Electron

Name

Symbol

of

of

protons

electrons

Charge conguration

beryllium

0

+

K

8

2

15

18

+

H

4.

For

each

electron

2

a.

1s

b.

1s

2

2s

6

2p

conguration

2

3s

given,

identify

three

isoelectronic

species:

6

3p

2

5.

A

Explain

why

noble

transition element

to

main

group

gases

is

an

elements,

a

do

not

element

form ions.

with

transition

a

partially

element

c an

lled

d

sublevel.

2+

dierent

charges.

For

example,

iron

commonly

In

contrast

form multiple ions with

forms

Fe

3+

and

Fe

ions

(gure 8).

+ 2 –

–

Fe

: [Ar]

e 2

3d

4s

Fe: [Ar] iron(II) ion

3d

4s

– 3 e – + 3

iron atom Fe

: [Ar]

3d

4s

iron(III) ion

 Figure 8

100

Iron atoms c an form

ions with a 2+ charge and

ions with a 3+ charge

Structure

Consider

the

electron

congurations

of

the

rst-row

transition elements. As

2+

Most

of

them

areformed.

contain

This

two

helps

to

4s

electrons,

explain

why

which

most

of

are lost when the M

these

2.1 The ionic model

elements

seen in chapter

the

ions

commonly

form

3d

4s

sublevel

lls

Structure 1.3,

up

before the

sublevel.

2+c ations.

t Table 2

Electron congurations of the

Electron conguration

Symbol

rst-row transition elements

Element

Atom

2

Sc

sc andium

[Ar] 4s

Ti

titanium

[Ar] 4s

2

vanadium

[Ar] 4s

Cr

chromium

[Ar] 4s

Mn

manganese

[Ar] 4s

Fe

iron

[Ar] 4s

Co

cobalt

[Ar] 4s

Ni

nickel

[Ar] 4s

2

5

6

7

elements

7

[Ar] 3d

8

8

3d

[Ar] 4s

transition

6

[Ar] 3d

3d

1

row

5

[Ar] 3d

3d

2

rst

4

[Ar] 3d

5

3d

2

the

[Ar] 3d

3d

2

When

3

3d

2

copper

2

[Ar] 3d

3

1

Cu

1

[Ar] 3d

3d

2

V

2+ ion

1

3d

[Ar] 3d

10

9

3d

are

[Ar] 3d

ionized,

the

4s

electrons

are

lost

before

Practice questions the

3d

electrons.

Further

successive

ionizations

occur

in

many of these elements

6. bec ause

the

3d

sublevel

is

similar

in

energy

to

the

4s

Deduce

the

abbreviated

conguration

Transition

elements

characteristic

Let’ s

focus

c an

on

the

have

be

variable

explored

transition

by

oxidation states (Structure 3.1). This

examining

elements

in

successive

period

4.

electron

sublevel.

They

ionization

have

each of the

following:

energy data.

variable

of

2+

a.

Mn

b.

V

c.

Cu

d.

Cu

oxidation 3+

states

bec ause

the

4s

and

3d

sublevels

are

close

together

in

energy,

as

shown +

by

successive

ionization

energy

data

(gure

9).

It

is

important

to

realize that 2+

ionization

rarely

investment

is

formation.

If

compared

to

happens

usually

a

in

oset

certain

isolation.

by

other

ionization

Ionization

processes

only

requires

absorbs

that

a

energy,

release

small

but

energy,

amount

of

this

such

energy

as

additional

lattice

7.

energy

Zinc

a.

only

forms 2+ ions.

Deduce

the

full

electron

2+

leads

to

a

the

previous

subsequent

ionization,

exothermic

then

it

could

be

energetic ally

favourable if it

conguration of Zn

process.

b.

Explain

.

why zinc is not a

transition element.

8.

The

ion

of

a

transition metal

3

lom

40

has

mass

number

55,

electron

5

conguration

[Ar] 3d

and a

Jk 3

charge of 2+.

30 01 / ygrene

a.

b.

symbol using

Identify a 1+ ion that has the

n o it a z i n o i

same

electron

conguration

10 as

the

above.

0

2

4

number

3d

its

nuclear notation.

0

 Figure 9

Write

20

D ata for the rst

of

6

8

electrons

10

12 ionization energies of iron.

electrons are very close together in energy.

12

removed

As you c an see, the 4s and

The large jump between the 8th and 9th

electrons occurs bec ause the 9th electron is removed

from

the 3p

energy level, which is

closer to the nucleus

101

Structure

2

Models

of

bonding

and

structure

Communic ation skills

ATL

You may have noticed that we c an refer to charge using dierent formats

depending on context. When using chemical symbols, charges appear as

3+

a superscript number followed by + or –, for example, Fe

writing, we say “the ion has a

. In speech or

3+ charge”. Charge is related to oxidation state

(Structure 3.1), where the + or – sign is given rst followed by the magnitude.

For example, “oxygen has an oxidation state of –2”.

Roman

numerals

are

also

used

compounds (Structure 3.1

to

and

indic ate

oxidation states in the names of

Reactivity 3.2).

For

example,

the

symbol

for a

2+

copper(II) ion is Cu

Write

your

referring

own

to

,

its

charge

example

ionic

to

is

help

2+,

you

and

its

oxidation state is +2.

remember

these

distinct

ways of

charge.

Linking questions

How

does

charge

How

of

the

its

does

explain

position

ion(s)?

the

their

of

an

element

(Structure

trend

in

variable

in

the

periodic

table

relate to the

3.1)

successive

oxidation

ionization

states?

energies

of

transition elements

(Structure 1.3)

Ionic bonds (Structure 2.1.2)

C ations

their

and

anions

opposite

Therefore,

bond

if

a

are

electrostatic ally attracted

charges.

given

ionic ally

to

This

attraction

element

forms

results

c ations,

in

to

the

and

each

other

bec ause of

formation of ionic bonds.

another

forms

anions,

they

c an

form an ionic compound.

ecnereffid

ionic 3.0 bonding

greater

Electronegativity (χ)

ionic

ytivitagenortcele

One

way

look

at

to

estimate

whether

a

bond

between

two

given elements is ionic is to

character the

dierence

in

electronegativity

between

the

two.

Electronegativity

2.0 ( χ)

is

a

measure

electrons.

periods

of

Within

and

up

the

the

the

ability

of

periodic

groups.

an

atom

table,

This

to

attract

a

pair

electronegativity

means

that

uorine

is

of

covalently

increases

the

most

bonded

across the

electronegative

1.0 element,

One

0

of

Values

has  Figure 10

an

so

the

in

it

has

a

high

tendency

electronegativity

the

Pauling

sc ale

electronegativity

to

sc ales

are

value

attract

used

by

pairs

4.0.

The

covalently

chemists

dimensionless

of

of

and

is

bonded

c alled the

range

from

electronegativity

0.8

of

electrons.

Pauling sc ale.

to

4.0. Fluorine

c aesium, one of the

If two elements have an

electronegativity dierence greater than 1.8,

the bonding between them

will have a high

least

electronegative

electronegativity

elements,

is

0.8.

Noble

gases

are

generally

not

assigned

values.

ionic character

The

larger

the

compound,

Electronegativity and other

periodic

trends

greater detail in

are

discussed in

Structure 3.1

bonding

than

the

of

102

1.8

is

dierence

the

assumed

(gure

main

type

bonding

greater

10).

of

to

In

in

ionic

occur

reality,

bonding

present.

electronegativity

the

in

character

when

the

bonding

the

between two elements in a

of

the

bond

dierence

occurs

compound

is

in

across

ionic,

between them. Ionic

electronegativity

a

continuum,

but

there

so

is

greater

above 1.8

may be other types

Structure

2.1 The ionic model

Data-based question

Predict

which

of

the

compounds

Compound

sodium

table

Dierence

χ

uoride,

(Na)

∆

NaF

sodium

in

χ

chloride,

χ

aluminium

chloride, AlCl

=

χ

χ

have

ionic

structure.

electronegativity

0.9 and

∆

χ

(F)

=

(∆

χ

)

4.0

=

χ

0.9 and

(Cl) = 3.2

= 2.3

(Al) =

∆

will

3.1

(Na)

∆

NaCl

χ

=

in

3

1.6 and

χ

(Cl) = 3.2

= 1.6

3

 Table 3

Electronegativity dierences for selected metal chlorides

Periodic table position

You

at

c an

the

large

qualitatively

positions

of

dierences

distance

from

its

in

approximate

constituent

how

electronegativity

each

ionic

elements

are

in

a

compound

the

periodic

generally

found

will

be

by looking

table. Elements with

at

a

greater horizontal

other.

Worked example 3

Compare the ionic character of bonding in the following pairs of compounds:

a.

c aesium uoride,

b.

magnesium

C sF,

oxide,

and

MgO,

c aesium iodide,

and

C sI

c arbon monoxide,

CO

Solution

a.

b.

Qualitative comparison:

Cs

and

than

the

Cs

F

Cs

are

and

a

I

are

dierence

and

F,

gre ater

in

in

distance

the

from

periodic

table.

electronegativity

me aning

the

bond

e ach

is

Therefore,

larger

between

between

them

is

Qualitative comparison:

In

other

more

the

from

table,

other

than

electronegativity

must

be

bond

ionic.

periodic

e ach

larger

and

and

dierence

than

between

Mg

C

Mg

that

and

O

Q uantitative comparison:

χ(Mg)

∆χ(C sF)

χ(C s)

=

∆χ(C sI)

Both

in

=

higher

χ(F)

=

4.0

∆χ

3.2

0.8

∆χ

both

=

and

and

χ(I)

=

χ(C)

2.7

=

are

gre ater

compounds

percentage

are

than

ionic.

ionic

1.8,

so

the

However,

character

than

bonds

C sF

C sI.

has

a

Mg

1.3

2.6

∆χ(CO)

1.9

v alues

=

(MgO)

=

and

than

1.8

ionic ally

O

and

=

χ(O)

=

further

C

be

away

Therefore,

between

must

χ(C s)

0.8

are

are.

between

Q uantitative comparison:

=

O

O

Mg

and

more

and

O,

the

O

and

the

ionic.

3.4

2.1

and

χ(O)

=

3.4

0.8

bond

for

this

ionic ally

bec ause

compound.

bec ause

∆χ

is

C

lower

and

than

∆χ

O

is

do

gre ater

not

bond

1.8.

103

Structure

2

Models

of

bonding

and

structure

Activity

Determine

using

the

i.

whether

the

following

pairs

of

elements

are

likely

to

bond

ionic ally

following two methods:

look at their positions in the periodic table

ii.

refer

to

their

electronegativity

values in the data booklet.

a.

Li and F

d.

As and S

b.

Rb

e.

P and Cl

c.

C a and I

f.

Ag

oen

and

Ga

It

is

a

non-metallic

incorrectly

said

element

chloride, AlCl

,

that

that

bond

do

not

only

ionic

together.

t

this

bonds

There

and

Br

form when a metallic element and

are substances, such as aluminium

description. Aluminium is a metal and chlorine

3

is

a

non-metal,

has  Figure 11

Polarized

light

properties

you

that

would

are

expect

them

characteristic

of

to

bond

covalent

ionic ally.

But the compound

compounds,

such

as

a

low

micrograph

melting of ammonium

so

point

and

high

volatility.

The

electronegativity

dierence

between these

nitrate crystals. Ammonium

+

nitrate contains two polyatomic ions: NH

two

elements

(1.6)

suggests

they

do

not

bond

ionic ally.

4

and NO

.

Its uses include fertilizers and

3

Polyatomic ions rocket

propellants

Some

ionic

compounds

contain

more

than

two

elements.

For

instance, NH

Cl, 4

+

+

which is made up of NH Name

Formula

c ations and Cl

anions. Ammonium ions, NH

4

polyatomic ions.

As

,

are

4

their

name

suggests,

polyatomic

ions

are ions that contain

+

ammonium

NH 4

several atoms.

hydroxide

OH

nitrate

NO

You

are

expected

to

know

the

names

and

formulas

of

the

polyatomic

ions

shown

3

in table 4.

hydrogenc arbonate

HCO

c arbonate

CO

3

2–

3

2–

sulfate

Self-management skills

ATL

SO 4

3–

phosphate

PO 4

You

will

need

polyatomic  Table 4

to

ions

spend

in

some

table

4.

time

Some

memorizing

students

like

the

to

names

use

and

formulas of the

ashc ards,

others make

Common polyatomic ions

up

mnemonics.

actively

engage

What

with

strategies

will

you

use?

How

will

you

make

sure

you

them?

Naming ionic compounds Name

potassium

uoride

Formula

Consider

KF

patterns magnesium

uoride

the

in

list

their

of

ionic

compounds

shown

in

table

5.

C an

you

notice

any

names?

MgF 2

You should notice that, in the names of ionic compounds: c alcium

c arbonate

C aCO 3

barium

hydroxide

Ba(OH)

•

the

c ation

name

•

c ations

•

monatomic

is

given

rst

and

is

followed

by the anion

2

iron(III)

oxide

Fe

O 2

silver(I)

sulde

Ag

adopt

the

name

of

the

parent

atom

and

the

name

remains

unchanged

3

S

anions

adopt

the

rst

part

of

the

name

of

the

parent atom,

2

followed

 Table 5

Names and

ionic compounds

104

by the sux -ide.

If

the

anion

is

polyatomic,

refer to table 4

formulas of some

•

the

name

of

the

compound

does

not

reect

the

number

of

ions

in

the

formula.

Structure

Practice questions

Anions

are

common 9.

State

the

name

of

each

of

the

2.1 The ionic model

conjugate bases

acids.

The

of

strength of

following compounds: acids and stability of their anions

a.

RbF

d.

Sr(OH) 2

b.

Al

S 2

c.

e.

c an

using

3

AlN

be

compared

quantitatively

BaCO

3

f.

NH

HCO 4

K 3

,

their

which

dissociation constants,

will

be

introduced in

a

Reactivity 3.1

The formulas of ionic compounds

The

name

ratio

is

of

of

the

an

ionic

ions

in

remembering

charges

anion

total

and

and

that

compound

The

the

negative

the

charge

it.

c ation,

of

basis

net

of

must

work

you what elements it contains, but not the

working

charge

charges

then

tells

for

the

out

c ancel

out

how

the

formula of an ionic compound

compound

out.

many

First,

of

is

zero,

so

determine

each

ion

you

the

the

positive

charge of the

need

to

reach a

zero.

Worked example 4

Deduce the formulas of the following ionic compounds:

a.

c alcium

oxide

b.

c alcium nitride

c.

sodium c arbonate

d.

aluminium

nitrate

Solution

a.

To

deduce

the

the

formula

of

c alcium

oxide,

work

through

following steps.

The

the

second

method

charges

subscript,

and

is

turn

ignoring

the

criss-cross

them

the

into

the

rule.

other

Swap

ion’s

sign:

Step 1: Determine the charges of the c ation and

the anion

Ca C alcium

2

1s

has

2

2s

shell

an

6

2p

electron

2

3s

6

3p

electrons,

conguration

of

2

4s

so

.

C alcium

they

form

atoms

ions

with

have

a

2+

two

2

Oxygen

has

Oxygen

atoms

form

ions

an

with

electron

have

a

2

six

conguration

outer

shell

of

outer

charge.

1s

2

2s

electrons,

Ca

4

2p

so

.

they

Then,

charge.

2+

Therefore, calcium ions = Ca

simplify

the

ratio:

2

and oxide ions = O

O

Ca 1

1

Step 2: Determine how many of each ion are needed Step 3: Check that the net charge is zero

in order to achieve a net charge of zero You

check

your

working

by

adding

up

the

charges

of

There are two methods you can use for this step. The rst e ach

individual

ion.

If

you

did

Step

2

correctly,

the

is the bar diagram method. Write out the ions as blocks charges

will

add

to

zero:

equal to the number of charges on each individual ion:

2+

Ca

+ 2

Ca Total

positive

charge

=

2+

2

2–

O

O

Total

The

bar

diagram

contains

one

c alcium

ion

and

Net oxide

ion,

so

compound

is

the

ratio

of

c alcium

1:1.

to

oxide

in

negative

charge

=

2

one

charge

=

2

2

=

0

the

Step 4: Write the formula

Ca

This

O 1

is

a

straightforward

example

where

the

1

magnitude

anion,

and

of

charge

hence

the

is

equal

formula

for

is

the

c ation

and

C aO.

105

Structure

b.

2

Models

C alcium

ions

as

of

bonding

nitride

have

is

a

dierent

and

structure

more

complex

charges.

Work

example as the

c.

through the steps

Sodium

must

before.

c arbonate

not

split

polyatomic

and

draw

up

contains

or

cluster.

brackets

change

a

polyatomic

the

ion.

You

ratio of atoms in the

Treat it like an indivisible entity

around

it

if

the

formula contains

Step 1: Determine the charges of the cation and more than one such ion. the anion

2

C a:

2

1s

2s

6

2

2p

3s

6

3p

Step 1: Determine the charges of the cation and

2

4s

the anion

C alcium

atoms

have

two

outer

shell

electrons,

so 2

N a: they

form

ions

with

a

2+

1s

2

2s

Sodium 2

N:

1s

2

2s

atoms

have

1

3s

ve

atoms

have

one

outer

shell

electron,

so

3

2p they

Nitrogen

6

2p

charge.

outer

shell

electrons,

form

ions

with

a

so

1+

charge.

2

C arbonate

ions,

CO

,

have

a

2

charge.

3

they

form

ions

with

a

3

charge. +

Sodium

ions

=

2

Na

and

c arbonate

ions

=

CO 3

2+

C alcium

ions

=

Ca

3

and

nitride

ions

=

N Step 2: Determine how many of each ion are

needed in order to achieve a net charge of zero

Step 2: Determine how many of each ion are

needed in order to achieve a net charge of zero

Bar

diagram

method

Bar diagram method

+

+

Na

+ 2

+ 2

Ca

Na

+ 2

Ca

Ca 2–

CO

3

3

N

N The

bar

diagram

c arbonate

The

bar

two

nitride

3

diagram

ions,

contains

so

the

three

ratio

of

c alcium

c alcium

ions

to

in

and

the

ion,

contains

so

compound

the

is

two

ratio

of

the

compound

is

Ca

CO 3

rule

N 3

2

CO

Na

rule

3

Remember,

Ca

N

poly atomic

does

the There

is

is

no

alre ady

need

in

its

to

simplify

simplest

one

3:2 Criss-cross

Criss-cross

and

c arbonate

2:1

2

in

ions

to

nitride Na

ions

sodium

sodium

the

ratio

here

bec ause

form.

not

ratio

you

ion

c an

to

change.

draw

remind

Again,

1

brackets

yourself

there

is

around

that

no

its

the

formula

need

to

simplify

here.

it

Step 3: Check that the net charge is zero

Step 3: Check that the net charge is zero

+

Na

2

CO

+

Na

3

2+

Ca

2+

3

Ca

T otal positive charge = 2+

3

Ca

N

T otal positive charge = 6+

T otal negative charge = 6

Net

charge

=

2

2

=

The

fo r mu l a

is

Na

CO 2

charge

=

6

6

=

the re

a re

no

.

Note

th a t

in

th e

fi n a l

3

bra c ke ts

a ro u n d

th e

0 po l y a t o m i c

one

Step 4: Write the formula

Ca

N 3

106

0

Step 4: Write the formula

a n sw e r

Net

T otal negative charge = 2

N

2+

2

ion

c arbonate

be c a u s e

ion.

t he

fo r mu l a

c o n ta i n s

only

Structure

d.

The

nal

example,

polyatomic

ion.

aluminium

Follow

the

nitrate, also contains a

same

steps

as

2.1 The ionic model

Criss-cross rule

before.

1

Al Step 1: Determine the charges of the cation and

the anion

2

Al:

1s

2

2s

6

2p

2

3s

1

3p

1 Aluminium

so

they

atoms

form

ions

have

with

three

a

3+

outer

shell

Again,

Nitrate

ions,

,

NO

have

a

3

electrons,

charge.

1

there

is

no

need

to

simplify

the

ratio

here.

charge.

3

Step 3: Check that the net charge is zero

3+

Al

ions

=

Al

and

nitrate

ions

=

NO 3

NO 3

NO If

you

do

not

remember

the

3

charge on a

NO

3+

polyatomic

learn

these

ion,

revise

formulas

table

and

4.

M ake

charges

o

sure

by

that

Al

you

heart.

3

T otal positive charge = 3+

Net

charge

=

3

3

=

T otal negative charge = 3

0

Step 2: Determine how many of each ion are

needed in order to achieve a net charge of zero

Step 4: Write the formula

Bar

The

diagram

method

formula

is

)

Al(NO 3

to

indic ate

the

.

Note

that

brackets

are

used

3

presence

of

more

than

one

nitrate

+ 3

Al

ion.

The

formula

should

not

be

written

as

AlN

O 3

NO

The

bar

three

in

diagram

nitrate

the

NO

3

contains

ions,

so

compound

is

the

NO 3

one

ratio

3

aluminium

of

ion

aluminium

and

to

nitrate

1:3.

Al

(NO 1

) 3

3

Practice questions

Names of ionic compounds that

contain 10.

Deduce

the

formula

9

of

each

of

the

transition

elements

have

following compounds: the

a.

magnesium

e.

lithium

b.

strontium chloride

oxide

f.

barium

c.

sodium sulde

g.

ammonium

d.

lithium nitride

oxidation number of the

nitrate transition

hydrogenc arbonate

is

metal

covered in

ion

in

brackets. This

Structure 3.1

phosphate.

Linking questions

Why

is

the

reaction?

formation

of

an

ionic

compound

from

its

elements

a

redox

(Reactivity 3.2)

LHA

How

is

formal

sulfate?

LHA

Polyatomic

the

charge

used

to

predict

the

preferred

structure of

(Structure 2.2)

anions

relationship

are

conjugate

between

dissociation constant,

K

their

?

bases

stability

(Reactivity

of

common

and

the

acids.

conjugate

What is

acid’ s

3.1)

a

107

Structure

2

Models

of

bonding

and

structure

Ionic lattices and properties of ionic

compounds (Structure 2.1.3)

L attices

Within

ionic crystals,

continuous,

negative

charge

u Figure 12

ions arranged

the

ions

are

three-dimensional

ions.

The

exact

arranged in a

networks

arrangement

of

of

lattice structure.

repeating

ions

in

a

units

of

L attices

are

positive and

lattice depends on the size and

ratio of the ions.

Ionic compounds are made of

in a lattice structure

Th e

fo r mu l a

ra ti o

of

N aCl,

of

e ach

c an

ionic

ty pe

e asily

of

c o mp o u n d s

ion

in

contain

a

t he

is

an

e mp i r i c a l

s tr u c tu re.

qu a dr i l l i o n

A

ions

fo r m u l a :

single

g ra i n

a r ra n ge d

in

it

of

a

i n d i c a te s

sodium

th e

c h l o r i d e,

c o n ti n u o u s

l a tti c e.

+

Th e

1:1

fo r mu l a

i n d i c a te s

th a t

t he

Na

and

Cl

ions

a re

p re s e n t

in

t he

l a tt i c e

in

a

ra ti o.

Thinking skills

ATL

Research

and

oen

involves

nding

information

but

also

evaluating

its

usefulness

reliability.

Consider the statement “each grain of NaCl can easily contain a quadrillion ions”.

–

+

–

+

+

–

–

+

•

Does

this

•

What

information

•

How

•

Come

could

+

–

+

–

+

–

+

–

+

–

Ionic

charged

108

Ionic bonding is non-

Each ion attracts all oppositely

ions around it

would

to

you

you?

need

up

you

with

reliably

your

nd

own

this

to

fact-check the statement?

information?

estimate

of

the

number

of

ions

in

a

grain

of

salt

– compare

dierent?

directional.

reasonable

+

and

 Figure 13

sound

bonds

Is

are

the

species

Bec ause

of

in

all

an

the

ionic

forces

in

this

to

one.

dierence

surrounding

How

it,

lattice

ionic

are

anions,

very

lattice,

This

with

they

the

compare?

two

means

the

quality,

and

strong.

but

do

between

non-directional

surrounding

an

this

non-directional.

charged

the

it

Figure

remember,

that

an

attraction

each

vice

values

13

This

shows

actual

signic ant?

ion

will

being

c ation

versa.

Why might they be

in

the

means

a

2D

lattices

attract

equal

ionic

the

in

all

all

lattice

forces

oppositely

directions.

attracts

of

representation

are

3D.

attraction

of

the

Structure

2.1 The ionic model

L attice enthalpy

L attice enthalpy

values

tell

us

how

strong

the

ionic

bonds

are in particular ionic



lattice.

L attice

enthalpy,

∆

,

is

the

standard

enthalpy

change

that

occurs on

lattice

the

formation

the

strength

all

the

need

of

of

gaseous

an

ionic

electrostatic

to

shown

be

ions

bond

forces

overcome.

A

of

from

one

bec ause,

attraction

general

mole

in

of

order

the

for

between

equation

for

solid

the

ions

c ations

the

lattice.

to

and

lattice

It

is

a

measure of

become

anions

in

dissociation

gaseous,

the

lattice

process is

below:

+

MX(s)

M



(g)

+

X

(g)

∆H

> 0 lattice

The

process is

some

quoted

from

this

a

as

book

L attice

forces

we

shall

is

enthalpy

ionic radius

values

that

increases

the

the

to

ionic charge.

shown

The

enthalpy

formation

in

gure

formation

at

enthalpies

of

of

14.

298 K

for

are oen

the

lattice

However, in

gaseous

ions

from

given in the data booklet.

required

increases.

lattice

L attice

exothermic

that

denition

energy

of

booklet.

endothermic

the

ions

values

data

process

the

with

as

the

represent

only

between

oppositely

in

opposite

consider

and

Experimental

found

consistent

attraction

between

be

ions — the

which

of

c an

negative

gaseous

lattice,

are

endothermic.

compounds

Two

strength

to

overcome

factors

of

the

the

aecting

electrostatic

lattice

electrostatic

enthalpy

attraction

charged ions:

 Figure 14

•

increases

with

•

decreases

increasing

with

ionic

increasing

charge

ionic

Lattice enthalpy is the energy

required

to overcome the electrostatic

forces of

attraction holding ions together in

radius. the lattice

Look

at

the

variations

in

lattice

enthalpy of the compounds in table 6.

–12

Ionic radius/10

Ionic 

m

Ionic charge

–1

∆H

/kJ mol lattice

compound

C ation

Anion

C ation

Anion

KF

829

138

133

+1

–1

NaF

930

102

133

+1

–1

C aF

2651

100

133

+2

–1

2

 Table 6

NaF

has

a

Lattice enthalpies of selected compounds

greater

lattice

enthalpy

than

KF

bec ause

the

c ations

in

NaF

are

smaller,

+

and

therefore

The

lattice

the

electrostatic

enthalpy

of

C aF

is

attraction

between

considerably

larger

Na

ions and F

than

that

of

ions

is

greater.

KF. This is in part

2

2+

due

to

the

smaller

ionic

radius

of

Ca

+

compared to K

.

However, it is mainly due

2+

to

the

greater

attraction

charge

between

on

the

the

Ca

c ations

c ation,

and

which

anions

in

results

in

greater

electrostatic

C aF 2

109

Structure

2

Models

of

bonding

and

structure

F actors aecting the lattice enthalpy of the group 1 chlorides

Charge

density

volume.

of

the

In

this

group

enthalpies

1

of

is

a

term

task,

you

c ations

the

used

will

and

to

describe

explore

relate

the

this

to

Part 3: Analysis

charge per unit

charge density

the

trend

in

5.

C alculate

the

volume

6.

C alculate

the

charge

7.

Relevant skills

Plot

two

graphs:

between

•

Tool 2: Use

•

Tool 3: General mathematics

•

Inquiry 1:

•

Inquiry 2:

8.

State

and

Identify,

and

each

ionic

one

radius

density

c ation.

graph

and

of

each

c ation.

showing

lattice

the

relationship

enthalpy, and the

spreadsheets to manipulate data other,

trends

of

lattice

group 1 chlorides.

explain

describe

Describe

and

charge

explain

density

the

and

trends

lattice

enthalpy.

shown in the

graphs.

predictions

and

between

explain

9.

patterns,

Discuss

the

differences

between the two

graphs.

relationships

10.

Instructions

Evaluate

of

the

your

graphs

prediction, including a comparison

you

obtained

in

7

and

the

sketched

Part 1: Prediction graphs

1.

For

the

group

1

c ations,

predict

the

you

obtained in 2.

relationship

11.

Consider possible extensions to this investigation:

between:

what other aspects of ionic radius, charge density and a.

ionic radius and lattice enthalpy of their chlorides

b.

charge

lattice enthalpy could you explore? density

and

lattice

enthalpy of their

chlorides.

2.

Sketch

the

graphs

relationships

you

expect

to

obtain

for the

above.

Part 2: Data collection

3.

Collect the following data for the group 1 chlorides:

ionic charge, ionic radius, lattice enthalpy. Possible

sources of information include the data booklet and

online databases. Cite each source appropriately.

4.

Input

a

your

suitable

data

into

a

spreadsheet

and

organize it into

table.

Practice questions

1.

Write

equations,

enthalpies

of

including

KBr,

state

symbols,

that

represent

the

O

a

lattice

C aO and MgCl 2

2.

State

and

explain

whether

you

expect KF or K

to

have

lower

lattice

2

enthalpy

3.

State

have

value.

and

the

explain

which

of

greatest

lattice

enthalpy

the

following

value:

ionic

compounds

NaCl, MgCl

,

Na

2

4.

you

O

or

expect to

MgO.

2

Describe and explain the trend in lattice enthalpy of the group 1 chlorides

down the group from LiCl to CsCl.

Properties of ionic compounds

The

properties

contain

in

110

a

c ations

lattice.

of

ionic

and

compounds

anions

held

are

due

together

by

to

their

strong

structural

features: they

electrostatic

attractive

forces

Structure

2.1 The ionic model

Volatility

Global impact Volatility

(from

the

Latin

volare,

to

y)

refers to the tendency of a substance to

of science vaporize

(turn

electrostatic

into

a

forces

gas).

of

For

an

attraction

ionic

compound

holding

the

ions

to

turn

into

together

a

gas,

must

be

the

strong

overcome. Some

The

volatility

of

ionic

compounds

is

therefore

very

low:

they

are

ionic

compounds

uncharacteristic ally volatile”.

This

also

means

they

have

said to be “nonlow melting

have high boiling points. points

and

Ionic compounds typically have high melting points too. The melting point of

solvents.

sodium chloride is approximately 1 075 K. Magnesium oxide, frequently used in

be

furnaces due to its ability to withstand high temperatures, melts at around 3 098 K.

bec ause

c an

Such

described

means

they

they

be

used as

ionic liquids

as

“green

are

c an

c an

solvents”

non-volatile. This

be

more

easily

Electrical conductivity contained

In order to conduct electricity, substances must contain charged particles that are

able to move. Ionic compounds contain charged particles, cations and anions. In

a solid ionic lattice, cations and anions can vibrate around a xed point, but they

cannot change position. Solid ionic compounds do not conduct electricity because

ions in a solid lattice are not mobile. When molten or aqueous, both cations and

This,

and,

however,

mean

they

are

does

recycled.

not

necessarily

harmless. Their

manufacturing,

transportation

oen,

disposal and

c an

have

signic ant

environmental impacts.

anions are free to move past one another , allowing them to conduct electricity

when a potential dierence is applied.

A solid ionic

compound

+ cannot conduct

+ –

– +

electricity

+ –

because the

–

– +

ions cannot

+ –

– +

move. They

+ are fixed in the

+

 Figure 15

These are waste separation

–

– +

+

bins in Jakarta, Indonesia. The lemost bin

regular

– (red) is for batteries.

Electrolytes in batteries

–

–

arrangement.

+

+ –

conduct

electricity bec ause they contain

– mobile ions.

+

separated

heat

Used

from

batteries are frequently

other types of waste to

dissolve in prevent

them

from ending up in landll.

water This is bec ause batteries contain valuable

metals and

+

other substances,

be recycled.

+

which c an

Do you separate your used

+ batteries from other household

+

+

+

waste?

+ + + +

+

+ +

+ + +

+

water

+

+

+

molecule + +

+

+ + + +

+

+ When an ionic compound is heated

When an ionic compound is

strongly and melts, the ions can

dissolved in water, it can conduct

move around and the molten

electricity because its ions can

compound conducts electricity.

move among the water molecules.

t Figure 16

Molten and

aqueous ionic

compounds are electric al conductors

111

Structure

2

Models

of

bonding

and

structure

Solubility

Ionic

compounds

insoluble in

Water

Polarity

is

discussed

in

a

typic ally soluble in

polar

solvents

solvent.

The

such

as

dierence

polar

solvents

such

as

water, and

hexane.

in

electronegativity

between

the

oxygen

greater and

detail in

is

are

non-polar

hydrogen

atoms,

combined

with

the

bent

geometry

of

the

water

molecule,

Structure 2.2. result

and

in

the

partial

Imagine

water

molecule

positive

an

ionic

charges

compound

themselves

so

their

positive

are

partial

pulled

c ase

of

a

out

the

that

of

their

and

lattice

solvent

a

partial

point

added

to

towards

is

no

molecules,

water.

charges

become

there

negative

charge

on

the

oxygen atom

hydrogen atoms.

negative

and

solvent,

the

the

being

partial

charges

the

non-polar

compound

having

on

the

The

point

anions.

surrounded

attraction

so

the

water

As

by

a

and

the

result,

water

between

c ations

molecules

towards

the

position

c ations,

molecules.

ions

anions

and

individual

of

the

remain

In

ions

the

ionic

within

lattice.

Activity

Draw

a

labelled

diagram δ+

explaining

why ionic compounds

δ‒ H

δ+

H

O conduct

O

δ‒

electricity when molten or

dissolved, but not when solid.

H

H

δ+

δ+

Cl

δ+

δ+

H

H

H

O

δ+

H

O δ‒

δ‒ δ+

H

δ+ δ+

O

H

δ‒

δ‒

O

H +

Cl

Na

+

δ+

H

Cl

δ+

Na

δ+

Cl

Cl

+

H

δ+

H

Na δ+

+

+

Na

δ‒

H

+

Cl

δ‒

O

Na

O

δ+ H

+

Na

Cl

Na Cl

+

Na +

Cl

+

Na

Na Cl

δ‒

Cl

δ‒

Cl

+

Na

δ+

δ+

+

+

Na

O

H

O

H

Na +

+

Cl

Na

Na

Cl H

H

δ+

δ+

Cl

 Figure 17

The dissolution of ionic compounds in water involves interactions between ions and

Not

all

ionic

competing

•

ionic

•

the

Ionic

and

and

of

are

water

in

the

insoluble

the

lattice

molecules.

c alcium

in

water.

This

is

bec ause

there

are two

present:

c ations

between

anions

include

dissolve

attraction

between

compounds

water

112

bonds

association

c ations

ions

compounds

forces

water molecules

and

ions

anions

and

when

are

the

the

of

and

the

lattice

partial

charges

electrostatic

stronger

Examples

c arbonate

in

ionic

silver

than

the

water

attractions

association

compounds

chloride.

of

that

molecules

between the

between the

are insoluble in

Structure

ATL

Research skills

H e av y

Some

metal

•

ions,

such

w a s t ew a t e r

re m o v i n g

•

2.1 The ionic model

h e av y

Use

the

and

their

as

le ad

t re a t m e n t

metals

internet

to

out

and

nickel,

p ro c e ss e s

of

industrial

re s e a rc h

other

often

take

form

insoluble

adv antage

effluents

ex a m p l e s

of

t h ro u g h

of

this

salts.

p ro p e r t y,

p re c i p i t a t i o n .

p re c i p i t a t i o n

re a c t i o n s

uses.

Describe

and

p re c i p i t a t e

is

ex p l a i n

the

changes

that

a re

o b s e r ve d

when

a

forme d.

Linking questions

What

experimental

compounds?

data

demonstrate

the

physic al

properties of ionic

(Tool 1, Inquiry 2)

How can lattice enthalpies and the bonding continuum explain the trend in

melting points of metal chlorides across period 3? (Structure 3.1)

 Figure 18

Close-up

photograph of the

formation of a lead(II) chromate precipitate

in the reaction between aqueous solutions

of

 Figure 19

mineral form

ions,

S alt

ats at

of sodium

S alar de Uyuni in Bolivia,

chloride.

particularly lithium.

Lithium-ion batteries c an be used

potassium

chromate

which are mainly made of halite, the

The brine below the rock

Global demand

lead(II) nitrate and

for lithium

salt

crust

is rich in dissolved metal

is increasing due to its use in batteries.

to power mobile phones,

laptops and

electric vehicles

113

Structure

2

Models

of

bonding

and

structure

Solubility of ionic salts

The

patterns

ionic

in

aqueous

compounds

are

solubility

often

of

several common

referred to as

Instructions

solubility Part 1: Solubility rules

rules.

us

We

c an

deduce

use

the

these

differences in solubility to help

identity

of

will

different solutions of ionic

an

unknown ionic compound. General

In

this

task

you

mix

table

compounds

(known as a

and

observe

precipitate)

whether

is

an

insoluble

7.

solubility

For

rules

c an

be

inferred

from the data in

example:

product

•

All

•

Sulfates

nitrates

are

soluble.

produced.

are

generally

insoluble,

except

group 1

Relevant skills sulfates

•

Tool 1: Recognize

and

address

relevant

issues

in

an

ammonium

sulfate.

safety and Study

environmental

and

table

7

and

infer

at

least

three

more

general

investigation solubility rules.

•

Inquiry

2:

Interpret

qualitative data Part 2: Identification of ionic compounds

S afety You

•

Wear

•

Dilute

eye

protection.

A,

will

B,

C,

sodium c alcium

nitrate

and

silver

nitrate

solutions

be

provided

and

D.

with

These

chloride,

samples

solutions

silver

nitrate

of

are

and

solutions

potassium

c alcium

labelled

c arbonate,

nitrate,

but

you

are

do

not

know

which

is

which.

Your

job

is

to

identify

each

irritants.

solution,

•

You

should

take

You

bec ause

you

do

not

know

may

materials

from

listed and the solubility rules.

your

knowledge

of

other

areas

are all potential irritants.

Collect

and

retain

any

precipitates

you

do

not

have to mix the solutions inside test

formed. tubes.

by

•

the

draw

chemistry.

Note

•

also

exactly which is which. of

They

using

c are when handling all solutions

You

mixing

c an

a

prepare

drop

of

small-sc ale

each

on

a

mixtures of solutions

plastic

sheet.

If

a

precipitate

Dispose of waste solutions and precipitates according

is to

your

formed,

the

solution

will

become

opaque. This will be

school’s guidelines.

easily

observable,

particularly

if

you

lay

the

sheet on a

Materials black

•

Clear

•

Pipettes

•

Small

•

Dilute acid solution

•

S amples

plastic

sheet

on

a

black

background.

background Devise a method, present it clearly, and show it to your

teacher . If they approve it and if you have time, try it out!

piece

of

of

copper

solutions

wire

(~0.5

cm long)

labelled A, B, C and D

C ations

group 1 c ations

ammonium,

barium,

lead, 2+

c alcium, C a +

(Li

+

, Na

+

, K

)

+

NH

+

silver, Ag

2+

2+

Ba

Pb

4

–

nitrate, NO

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

soluble

insoluble

insoluble

soluble

soluble

soluble

slightly

soluble

insoluble

insoluble

soluble

soluble

insoluble

slightly

soluble

soluble

soluble

insoluble

insoluble

3

ethanoate,

–

CH

COO 3

–

Anions

chloride, Cl

–

hydroxide, OH

2–

sulfate, SO

slightly

soluble

insoluble

4

2–

c arbonate, CO 3

 Table 7

114

Aqueous solubility of common ionic compounds

insoluble

insoluble

Structure

2.1 The ionic model

End-of-topic questions

7.

What

is

the

formula

of

sodium

nitrate?

Topic review

A.

NaNO

B.

NaNO

C.

Na

D.

S

2

1.

Using

your

answer

the

knowledge

guiding

from the

question

as

Structure 2.1 topic,

fully

as

possible:

3

What determines the ionic nature and properties of a

N 3

compound? N 3

2.

Explain

why

ionic

substances

are

2

always compounds. 8.

Which

compound

has

the

largest

value

of

lattice

enthalpy?

Exam-style questions A.

C aS

B.

C aO

C.

K

Multiple-choice questions

3.

The

elements

which

in

group

17

generally

form ions with

charge?

D. A.

K

7+

1+

C.

1–

O 2

9. B.

S 2

Which

A.

statement

L attice

D.

B.

when

the

radii of the

increase.

Lattice

enthalpy

protons.

Give

the

full

electron

2

2s

6

2p

2

3s

an

electron

2

1s

C.

1s

D.

1s

2

2s

6

2p

2

2s

2

3s

2

6

2p

2

2s

6

3p

2

3s

6

2p

energy

needed to

L attice

enthalpy

decreases

when

the

charge of the

2

3s

statement

L attice

2

of

3d

6

3p

increases.

2

6

3p

ions

4s

2

4s

2

10.

3d

about

ionic

compounds

is

the

Which

of Which

the

c ation to an anion.

6

D.

2

a

3p component

B.

from

conguration of this ion.

C. 2

1s

represents

charge and contains 20 transfer

5.

increases

ions

7–

The ion of element X has a 2 +

A.

correct?

enthalpy

component

4.

is

enthalpy

increases

component

equation

potassium

ions

correctly

when

the

charge density

increases.

represents

the

lattice

enthalpy

oxide?

correct?

+

A.

K

2–

O(s)

→

2K

(g)

O(s)

→

2K(s)

O(s)

→ K

+

O

(g)

2

A.

Ionic

bonding

attraction

results

between

from

the

electrostatic

c ations and anions.

B.

K

C.

K

+

½O

2

B.

C alcium

uoride

is

made

up

of

C aF

molecules.

(g) 2

2+

2

2–

(g)

+

O

(g)

2

2

C.

Ionic

structures

molten

or

contain

deloc alized

D.

electrons when

Ions

are

held

electrons

6.

What

is

the

to

together

bec ause

anions

transfer

List

the

→

2K(g)

+

½O

(g) 2

lithium

halides

in

order of

increasing

lattice

enthalpy.

c ations.

name

of

C aSO

A.

LiF,

LiCl,

LiBr,

LiBr,

LiCl,

LiI

B.

LiF,

C.

LiI,

LiBr,

LiCl,

LiF

D.

LiI,

LiCl,

LiBr,

LiF

? 4

A.

O(s) 2

11.

D.

K

dissolved, but not when solid.

LiI

c arbon sulte

B.

c alcium sulte

C.

c arbon sulfate

D.

c alcium sulfate

115

Structure

12.

2

Models

Which

of

bonding

substance

has

Melting

and

an

structure

ionic

structure?

Electric al

Electric al

conductivity

conductivity

Solubility

point / °C

in water when molten

when solid

A

36

high

none

none

B

186

low

none

none

C

1083

high

good

none

D

1710

low

good

good

Extended-response questions

13.

C alcium

uoride

a.

Deduce

the

b.

Describe

has

applic ations in the eld of optics.

formula

for

c alcium

uoride.

[1]

c.

Deduce

d.

Potassium

the

Chromium the

charge

of

the

dichromate(VI)

is

a

dichromate(VI)

contains

transition element that commonly

2+

uoride.

Explain

why

form Cr

solid

conductor,

c an

conduct

The

lattice

but

molten

among

others.

Write

a

the

abbreviated

chromium

electron

conguration

electricity.

enthalpy

of

atom.

[1]

c alcium uoride

[2]

ii.

Copy

the

arrows d.

ions

c alcium uoride is a poor of

electric al

3+

ions and Cr

[3]

i. c.

[1]

structure and bonding in solid forms

c alcium

ion.

chromium.

in

diagram

the

below

boxes

to

and

draw

represent the

c alcium uoride is electronconguration in the 3d and 4s

1

2 651 kJ mol

. 2+

orbitals of a Cr

i.

Write

that

an

shows

lattice

ii.

equation,

the

process

enthalpy

Explain

why

including

of

the

state

uoride.

[2]

4s

Write

[1]

enthalpy

of

c alcium

oxide,

the

full

electron

conguration of a

3+

Cr

lattice

3d

process in part (i) is iii.

The

[1]

associated with the

c alcium

endothermic.

iii.

ion.

symbols,

ion.

[1]

C aO, 15.

The

equation

below

shows

the

formation of lithium

1

is3 401 kJ mol

.

Explain

why

the

lattice uoride

enthalpy

that

14.

Certain

ionic

of

types

of

of

c alcium

c alcium

is

potassium

[2]

contain

the

bright

dichromate(VI), K

Write

the

full

potassium

electron

orange

Cr 2

a.

from

Li(s)

+

F

(g)

→

Explain

the

atom

7

b.

Identify

the

the

charge

equation.

c.

Identify

the

oxidized

116

of

the

[1]

lithium

ion.

[1]

conguration of a species

and

the

reduced

[1]

dierence

in

mass

and

a

in

this

reaction.

[1]

between a Sketch

a

diagram

showing

the

structure of

potassium ion is lithiumuoride.

negligible.

standard conditions.

2

Balance

d.

potassium

under

O 2

ion.

why

elements

LiF(s)

a.

species

b.

its

greater than

uoride.

breathalyser

compound

oxide

[1]

[2]

Structure 2.2

The covalent model

What determines the covalent nature and properties of a substance?

Covalent bonds lead to a vast range of dierent substances.

Substances

From water to diamond to nitrogen gas, from oils to

characterized

plastics to polyatomic ions, these species contain atoms

low

held together by strong covalent bonds. Covalent bonds

molecular

lead to the formation of two dierent types of structure:

low melting points and boiling points. Their solubility and

covalent network structures (also known as giant covalent

volatility

structures) and molecular covalent structures

forces.

In

general,

covalent

substances

are

poor

electric al

conductors.

In

with

volatility

and

and

will

we

poor

structures,

vary

greatly

Structure 2.2,

how

covalent

you

represent

explain

also

their

learn

network

structures

are also

by high melting points and boiling points,

solubility

on

the

depending

will

learn

molecules

shapes

about

in

other

and

covalent

water.

hand,

on

what

as

their

a

well

Substances with

generally

have

intermolecular

covalent bond is,

as

how

we

intermolecular

network

describe

forces.

You

structures.

Understandings

Structure 2.2.1 —

electrostatic

and

the

The

octet

A

covalent

attraction

positively

bond

between

a

is

formed

shared

pair

by the

of

electrons

London

(dispersion)

forces




In

1

will

proceed

contrast, when

forward

The

reaction

value of

reactants

be

the

are

K

and

be

from

0

(endothermic)

⦵

reactants

522

+

Q

⇌

products

ΔH

r

product, and

Reactivity

An

increase

in

so

and

equilibrium

the

the

temperature

system,

Conversely,

has

the

a

equilibrium

position

decrease

opposite

will

in

eects

increase

position

of

the

of

the

“concentration”

exothermic

endothermic

temperature

on

the

the

above

of

2.3

How

far?

The

extent

of

chemic al change

heat in the

reaction will shi to the le,

reaction will shi to the right.

decreases

the

“concentration”

of

heat and

equilibria.

The eect of c atalysts on equilibrium

A

c atalyst

provides

activation

In

a

reversible

pathway

same

the

energy.

in

an

extent.

is

increases

process,

opposite

system

alternative

This

the

forward

directions,

Therefore,

achieved

in

pathway

the

the

faster,

so

rate

and

the

the

a

the

of

a

reaction

chemic al

reverse

rates

presence

but

for

of

of

reactions

both

of

this

therefore

follow

reactions

c atalyst,

position

and

lower its

reaction (Reactivity2.2).

the

the

same

increase to the

equilibrium state of

equilibrium and the

K

value

remainunchanged.

The

eect

analogy.

products

of

the

of

a

c atalyst

Imagine

(gure

forward

two

8).

and

on

the

equilibrium

communic ating

The

levels

reverse

reactants

no

of

position

vessels

liquid

in

the

c an

that

vessels

by a simple

reactants and

represent

reactants

c atalyst

no

products

the

relative

rates

products

c atalyst

reactants

products

c atalyst

c atalyst

p Figure 8

illustrated

reactions.

products

reactants

be

represent

Communic ating vessels as a model of chemic al

equilibrium

The

ow

of

connecting

will

tube.

become

Similarly,

faster

The

liquid

a

but

same

respect

If

equal

aect

c an

but

a

will

equilibrium

be

will

some

be

of

this

the

If

nal

for

we

the

limited

state

of

system

of

that

the

liquid

will

levels

the

levels

of

system

reach

liquid

will

the

not

state

of

in

the

be

vessels

aected.

equilibrium

equilibrium.

to

ow

by the diameter of the

the

to

illustrating

add

liquid

until

is

diameter,

position

used

favoured”)

be

vessels

chemic al

the

concentrations.

concentration”),

the

increase

allows

not

analogy

we

faster,

c atalyst

does

to

reaction

between

Le

Châtelier ’s principle with

the

to

le

the

become

vessel

right

(“increase

vessel

equal

(“the

again

(“a

a

reactant

forward

new

chemic al

established”).

523

Reactivity

2

How

much,

how

fast

and

how far?

The

eects

of

reaction

conditions

on

the

equilibrium position and

K

value

are

summarized in table 2.

Change in condition

concentration

of

reactant

of

product

of

reactant

of

product

Shi

increases

(towards

concentration

K

of equilibrium

to the right

no change

products)

decreases

concentration

to the le

decreases

(towards

concentration

reactants)

increases

pressure

increases

to the side with a smaller

number volume

pressure

decreases

to

the

gas

side

number volume

of

molecules

decreases

with

of

gas

a

greater

molecules

increases

⦵

temperature

increases

ΔH

r

⦵




0: to the right

ΔH

=

0: no change

ΔH




0: to the le

ΔH

=

0: no change

ΔH

⦵

ΔH

r

⦵

ΔH

decreases

ΔH

r

r

r

c atalyst

is

p Table 2

added

no

The eects of

Châtelier ’s

product

by

ammonia

between

principle

altering

on

an

achieved.

This

increases

=

0: no change

r

r

r




0:

decreases

=

0: no change

⦵

change

r

no change

process

allows

the

chemists

reaction

industrial

pressure,

decreases

0:

reaction conditions on the equilibrium position and K value

C ase study: the Haber

Le

r

0:

>

⦵

⦵

ΔH




1

ΔG

K




certain

be

K.

reversible

Which

reaction

changes

approaching

will

in

occur

a

closed

equilibrium

at

when

system,

the

constant

system will

temperature?

possible: A

Q

will

decrease

B

K

will

increase

C

Concentrations

of

reactants

will

decrease

D

Concentrations

of

products

will

increase

How can the extent of a reversible reaction

be inuenced?

Exam-style questions

⦵

Multiple-choice questions

7 .

At

1 000 K,

the

forward

reaction

has

a

negative

ΔG

value: 2.

Which

of

the

following

statements

is

correct?

2NO A

The

changes

microscopic

at

2

(g)

⇌

2NO(g)

+

O

2

(g)

equilibrium stop only at the

level

but

not

at

the

macroscopic

Which

level.

expression

is

correct

for

the

equilibrium at this

temperature? B

The

changes

at

equilibrium stop at both the 2

macroscopic

C

The

D

The

rates

of

level

both

equilibrium

are

from

the

forward

equal

equilibrium

deduced

and

to

and

reverse

level.

reactions at

zero.

constant

the

microscopic

expression

reaction

c an be

A

[NO

B

[NO

C

[NO]

D

K

>

2

2

2

]

>

]




[OH

]

+

neutral

7

]

[H

=

[OH

]

+

basic

p Table 2

> 7

Ac i d i c ,

n e u t ra l

and

basic

[H

aqueous

solutions

at

]




[OH

].

−

An

addition

of

a

base

will

increase

the

concentration of OH

(aq) ions in the

+

solution

and

decrease

the

concentration of H

+

solution, [H

(aq)

ions.

Therefore, in a basic

−

]

will

be

lower

than

[OH

].

Properties of acids and bases (Reactivity 3.1.6)

Properties of acids

All

Brønsted–Lowry

bound)

hydrogen

Exchangeable

atoms,

such

as

exchangeable

546

acids

atom

hydrogen

oxygen,

must

that

contain

c an

atoms

usually

halogens

hydrogen

atoms

at

detach

or

are

least one

from

form

sulfur.

exchangeable

rest

bonds

In

bonded

the

the

with

almost

to

of

all

acid

highly

(weakly

molecule.

electronegative

organic acids,

oxygen.

Reactivity

For

example,

following

hydrogen

structural

chloride,

sulfuric

acid

and

ethanoic

(acetic)

acid

3.1

Proton

transfer

reactions

have the

formulas:

H H

O

Cl

H

O

O

H

S

O

H

C

C

O

O

H The

nomenclature,

structure and

H

properties hydrogen chloride

sulfuric acid

of

discussed in

The

exchangeable

hydrogen

atoms

are

shown

in

red.

+

hydrogen

atoms

remaining

Here

are

part

three

dissociate

of

the

acid

examples

and

form H

produces

of

acid

an

organic

acids

are

ethanoic acid

In

Structure 3.2

aqueous solutions, these

+

(aq)

(or H

anion,

3

O

also

(aq))

c ations, while the

known as the

acid residue.

dissociation:

+

HCl(aq)

→

H

(aq)

+ Cl

(aq)

+

H

2

SO

4

(aq)

→

2H

2

(aq)

+

SO

(aq)

4

+

CH

Notice

is

that

ethanoic

exchangeable.

hydrogen (χ

To

=2.2),

3

COOH(aq)

acid

contains

explain

this,

c arbon (χ

⇌

H

four

you

=2.6)

(aq)

+ CH

3

COO

(aq)

hydrogen atoms, but only one of them

need

and

to

consider

oxygen (χ

the

electronegativities of

=3.4).

Hydrogen

and

carbon

have similar electronegativities (Δχ = 2.6 − 2.2 = 0.4), so the C–H bond has low

Electronegativity and bond polarity

polarity and does not break easily. In contrast, the dierence in electronegativity

are

between hydrogen and oxygen is signicant (3.4

The

− 2.2 = 1.2), so the O–H

discussed in

Structure 2.2.

electronegativity

bond is highly polar. Since the bonding electron pair is shied towards the more

all

elements

electronegative O atom, the less electronegative H atom develops a partial

booklet.

are

values

for

given in the data

+

positive charge. As a result, it dissociates readily to form an H

In

inorganic

bonded

to

acids

containing

oxygen,

and

so

oxygen (oxoacids),

are

all

(aq) ion.

hydrogen

atoms

are usually

exchangeable.

Activity

O

H

Cl

O

H

Cl

O Draw

hypochlorous acid

the

the

following

(HClO Depending

on

the

number

of

structural

exchangeable

hydrogen

atoms,

acids

are

3

),

(H

monoprotic

•

diprotic

(two

•

triprotic

(three

In

contrast

and

exchangeable

exchangeable

inorganic

acids

hydrogen

are

acids,

acids

atoms.

even

atom),

atoms),

hydrogen

organic

hydrogen

monoprotic,

atoms,

hydrogen

hydrogen

exchangeable

nonexchangeable

ethanoic

four

to

(one

for

atoms),

oen

For

for

for

2

SO

example, H

contain

PO

4

2

CO

3

4

),

) and phosphoric

).

example, HCl

example, H

for

3

oxoacids: chloric

perchloric (HClO

c arbonic (H classied as:

•

formulas

chlorous acid

both

3

4

PO

4

exchangeable

example, both methanoic and

though

their

molecules contain two and

respectively.

547

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

The

formulas

Along

are

with

shown

and

the

in

names

of

systematic

common

names,

acids

many

and

their

organic

anions

acids

given in table4.

brackets.

Acid

Formula

are

have trivial names, which

Anion

Name

Strength

Formula

Name

−

HF

hydrogen

uoride

uoride

weak

F

chloride

strong

Cl

bromide

strong

Br

strong

I

−

HCl

hydrogen

chloride

−

HBr

hydrogen

bromide

−

HI

hydrogen

iodide

iodide

2

H

2

S

hydrogen

sulde

weak

cyanide

weak

−

S

sulde

−

HCN

hydrogen

CN

cyanide

−

HNO

nitric

strong

NO

nitrous

weak

NO

sulfuric

strong

SO

sulfurous

weak

SO

3

nitrate

3

−

HNO

2

nitrite

2

2

H

2

SO

4

sulfate

2

H

2

SO

3

3

PO

4

phosphoric

weak

PO

sulte

3

PO

3

phosphorous

weak

PO

perchloric

strong

−

phosphate

4

3

H

−

3

3

H

−

4

−

phosphite

3

−

HClO

4

ClO

perchlorate

4

–

HClO

chloric

strong

ClO

chlorous

weak

ClO

hypochlorous

weak

c arbonic

weak

3

chlorate

3

–

HClO

2

chlorite

2

–

HClO

ClO

hypochlorite

2–

H

2

CO

3

CO

c arbonate

3

–

HCOOH

methanoic

(formic)

weak

HCOO

methanoate

(formate)

−

CH

3

COOH

ethanoic

(acetic)

weak

CH

weak

C

3

COO

2

H

2

C

2

O

p Table 4

ethanedioic

4

Common

acids

and

(oxalic)

their

anions.

Although the names

HCl,

“hydrogen

chloride”,

normal

Therefore,

acid”

it

already

they

have

we

incorrect

slightly

and

dierent

O

−

ethanedioate

4

are

(acetate)

s h ow n

in

re d

hydrochloric acid

meanings

in

(oxalate)

refer

chemistry.

to

the

When

same

we

say,

mean an individual compound, HCl, which is a gas

conditions,

is

protons

hydrogen chloride

substance,

under

E xc h a n g e a b l e

2

ethanoate

while

to

say,

“hydrochloric

“a

solution

of

acid”

is

a

solution

hydrochloric

acid”,

of

HCl

as

in

water.

“hydrochloric

refers to a solution.

Similar problems may arise when we talk about sulfuric acid, which is oen used as

an aqueous solution but can also exist in pure form (so-called “100% sulfuric acid”).

When this dierence is important, we should always say, “aqueous sulfuric acid” when

we refer to a solution, or “anhydrous sulfuric acid” when we refer to pure H

SO

2

An important characteristic of any acid is its strength.

.

4

Strong acids, such as

hydrogen chloride, dissociate completely in aqueous solutions. If we dissolve

one mole of HCl in water, the resulting solution will contain one mole of hydrogen

cations and one mole of chloride anions but no HCl molecules. In other words,

the dissociation of HCl is irreversible, which is represented by the single arrow:

+

HCl(aq)

In

addition

are

548

→

to

advised

H

−

(aq)

+ Cl

hydrogen

to

(aq)

chloride,

memorize

their

six

other

strong

acids

formulas and names.

are

listed

in

table4.

You

Reactivity

Weak acids,

dissolved

in

such

as

water.

ethanoic

For

acid,

example,

acid) contains both CH

3

COOH

dissociate

table

vinegar

molecules

only

(an

and

to

a

small

3.1

Proton

transfer

reactions

extent when

aqueous solution of ethanoic

the

products

of

their

dissociation,

+

H

(aq) and CH

represented

3

by

COO

the

(aq)

ions.

The

reversible

nature

of

this

process is

equilibrium sign:

+

CH

Almost

in

3

COOH(aq)

all

organic

table4,

are

not

it

is

H

and

safe

discussed

⇌

to

in

(aq)

many

CH

3

COO

inorganic

assume

DP

+

that

it

is

(aq)

acids

weak.

are

weak,

There

are

so

a

if

an

few

acid

is

not

listed

Activity

exceptions, but they

chemistry. Formulate

The

of

strengths

their

of

solutions,

“concentrated”

and

acids

“weak

so

and

the

and

bases

terms

“dilute”

solution”

will

not

have

no

“strong”

(table5).

be

direct

and

The

accepted

relationship

“weak”

should

colloquial

in

the

IB

to

the

not

phrases

be

the

concentrations

dissociation

confused with

hydrogen

“strong solution”

acid.

Do

arrow

assessments.

HCl

0.1 mol dm

3

The

Examples

strength

of

of

CH

solutions

oxoacids

of

3

COOH

acids

generally

0.1 mol dm

atom.

In

turn,

a

higher

with

different

increases

with

strength

the

and

3

contains

nitrous acid, HNO

+3.

An

+3 to

H

2

SO

addition

+5

4

,

and

has

a

more

2

of

oxidation

,

is

oxygen

bound

another

produces

higher

to

atoms.

two

oxygen

state

usually

For

c o n c e n t ra t i o n

oxidation state of the

states

oxygen

atoms

increases

the

state

of

sulfur

3

and

the

and

nitrogen

has

oxidation

.

in

2

SO

of

nitrogen

acids

trends:

For

oxygen

atoms

than

Activity

weak

hydrogen

uoride

are

in

in

is

a

consist

of

increases

only

along

two

the

elements)

period

third period, phosphine (PH

aqueous

chloride

(HF)

that

the

down

the

formulas of

3

strength

example,

properties

HI)

(acids

their

were

from

strong sulfuric acid,

all

Binary

elements

Structure 3.1

weak

oxidation state of

state

Similarly,

more

an

atom

Write sulfurous acid, H

of

means that the acid

example,

strong nitric acid, HNO

oxidation

weak acids.

COOH

introduced in molecule

for

HCl

CH

Oxidation central

sign

3

10 mol dm

p Table 5

forget to use a single

3

10 mol dm

Weak

bromide,

Dilute

3

Strong

for the

strong acids and an

equilibrium

Concentrated

hydrogen

cyanide and methanoic

not

for

equations

of

solutions,

(HCl)

weak

is

acid

a

3

)

while

acid.

the

Similarly,

other

three

down

does

hydrogen sulde (H

strong

demonstrate

and

2

S)

the

not

is

down

show

a

clear periodic

group

any acidic

weak acid, and

group17,

hydrogen

(gure3).

halides

oxoacids of chlorine and

phosphorus

the

from

table

3.

Deduce

oxidation states of these

elements

in

how

aect

they

each acid and outline

the

acid

strength.

hydrogen

(HCl,

HBr and

strong acids.

groups

2

CH

16

15

14

4

NH

3

H

H

H

4

O

2

HF

S

HCl

Se

HBr

2

2

sesaercni htgnerts

soirep

PH

3

3

17

HI

5

strength increases

p Figure3

Pe r i o d i c

trends

in

the

strength

of

binary

acids

549

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Diprotic

or

triprotic

weak

acids

dissociate

stepwise,

for

example:

+

H

2

SO

3

(aq)

⇌

H

(aq)

+

HSO

+

HSO

The

(aq)

3

second

⇌

H

proton

(aq)

3

2

(aq)

+

SO

(aq)

3

dissociates

to

a

much

smaller

extent

than

the

rst,

so

nearly

+

all H

Activity

The

ions

produced

reason

for

that

by

a

polyprotic

becomes

clear

if

acid

we

are

formed

consider

on

the

its

rst

charges

dissociation

on

the

ions

step.

involved

+

Formulate

the

the

stepwise

equations

for

dissociation of

in

the

ions

above

exert

equations.

electrostatic

The

rst

step

attraction

on

produces H

each

other,

(aq)

so

and HSO

pulling

(aq) ions. These

3

them

apart

requires

+

phosphoric acid.

some

energy.

On

the

second

step,

the

electrostatic

attraction

2

and SO

a

result,

likely

to

(aq)

2

(aq)

3

between H

the

ions

is

second

much

step

greater, as the anion SO

requires

more

energy,

(aq)

3

which

is

doubly

makes

this

charged.

process

As

less

occur.

Properties of bases

Despite

acids

in

the

dierence

terms

of

their

in

chemic al

behaviour

properties,

in

aqueous

bases

show

solutions.

many similarities to

Where

an

acid

produces

+

an H

(aq)

ion,

(aq)

ion

a

base

either

produces an OH

(aq)

ion

(Arrhenius

base) or accepts

+

an H

these

and

(Brønsted–Lowry

processes

remain

base).

same

However,

and

thus

c an

the

be

general principles behind

explained

by similar concepts

equations.

Most

inorganic

one

or

and

the

Na,

K,

more

OH

Rb

bases

OH

group

and

C s)

n

metal

ionic

depends

and

most

so

)

they

and

hydroxides

soluble

metal

The

hydroxides, which contain a metal atom and

nature

on

the

group2

form

of

the

metal

chemic al

metals

ionic

bond

between the metal

electronegativity.

(Mg,

C a,

hydroxides.

Sr

Alkali

and

Such

Ba)

metals

have

(Li,

very

low

hydroxides consist of a

+

c ation (M

readily

are

groups.

electronegativities,

All

the

in

one

are

water

or

more

strong

and

hydroxide

bases.

fully

Except

dissociate

anions

for

into

(OH

Mg(OH)

ions,

for

2

).

and

C a(OH)

2

,

they

are

example:

+

NaOH(aq)

→

Na

(aq)

2

Ba(OH)

C alcium

is

almost

2

(aq)

→

hydroxide

insoluble.

heterogeneous

Ba

is

If

+

OH

(aq)

+

(aq)

only

an

+

2OH

slightly

excess

equilibrium

of

(aq)

soluble

such

between

in

water,

hydroxide

the

solid

and

is

base

magnesium

added

and

to

hydroxide

water, a

aqueous ions is

established:

2

Mg(OH)

2

(s)

⇌

Mg

2

C a(OH)

These

⇌

Ca

hydroxides

are

molecules

is

550

c aused

2

of

by

(s)

Mg(OH)

low

+

(aq)

+

2OH

(aq)

+

(aq)

+

strong

2

or

2OH

bases,

C a(OH)

solubility

of

(aq)

2

.

these

so

their

The

solutions

reversible

bases

in

water,

contain

no

nature

of

not

their

by

the

undissociated

above

low

processes

strength.

Reactivity

Less

active

covalent

are

linked

Fe(OH)

In

3

metals,

together

are

addition,

their

basic

Ammonia

As

such

hydroxides,

by

covalent

these

nature

(NH

discussed

accepting

in

a

3

)

as

beryllium,

which

a

polar

is

are

one

of

the

few

aqueous

from

an

3

(aq)

+

H

(aq)

+

H

bond.

Almost

all

virtually

and

and

For

all

the

transition

oxygen

example,

covalent

insoluble

of

both

water,

so

OH

Fe(OH)

hydroxides

in

elements,

the

are

they

transfer

reactions

form

group

2

and

weak bases.

only

show

inorganic bases that does not contain a metal.

ammonia

acid

+

NH

atom

Proton

reactions with acids.

earlier,

proton

aluminium

metal

covalent

hydroxides.

hydroxides

in

the

3.1

or

acts

as

a

weak

Brønsted–Lowry

base

by

water:

+

(aq)

⇌

NH

4

(aq)

+

NH

In

3

chemic al

The

3

(aq)

organic

three

O(l)

⇌

equations,

hydroxide, NH

NH

2

+

4

2

O(l)

⇌

NH

4

OH

unstable

H

H

oen

of

methyl

properties

at

the

N

CH

as

amines

3

),

ethyl (–CH

are

similar

2

H

3

amines

CH

to

3

(aq)

+

H

one, two or

CH

3

N

CH

H

3

dimethylamine

),

may

contain

phenyl (–C

those

+

2

contain

nitrogen atom:

CH

groups (–CH

such

of

NH

represented as ammonium

exists only in solutions:

OH(aq)

methylamine

substituents,

3

is

and

H

N

CH

(aq)

ammonia

is

substituents

ammonia

Instead

which

+

derivatives of ammonia, amines (Structure 3.2),

hydroc arbon

H

H

(aq)

4

aqueous

OH(aq),

H

NH

of

6

H

5

ammonia,

other

and

for

N

CH

3

trimethylamine

any

)

C

3

3

so

hydroc arbon

on.

Acid–base

example:

+

(aq)

⇌

CH

3

NH

3

(aq)

+

CH

3

NH

2

(aq)

+

H

2

O(l)

⇌

CH

3

NH

3

(aq)

+

OH

(aq)

Practice questions

8.

Amines

a.

are

organic

Formulate

molecular

hydrochloric

b.

derivatives of ammonia.

acid

and

with

i.

dimethylamine

ii.

trimethylamine.

ionic

the

equations

for

the

reaction of

following amines:

Identify conjugate acid–base pairs in each ionic equation and state the

role (Brønsted–Lowry acid or Brønsted–Lowry base) of each species.

551

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

The

anions

ethanoate

of

weak

ion

is

acids

c an

produced

also

by

the

act

as

Brønsted–Lowry

bases.

For

example, the

dissociation of ethanoic acid:

+

CH

If

we

3

COOH(aq)

reverse

this

⇌

H

(aq)

equation,

+

the

CH

3

COO

basic

(aq)

nature

of

the

ethanoate

ion

will

become

obvious:

+

CH

3

COO

(aq)

+

H

(aq)

⇌

CH

3

COOH(aq)

The driving force of acid–base reactions is the formation of weak conjugates. For

example, ethanoic acid, CH

COOH, is a weak acid. Therefore, the dissociation

3

of ethanoic acid is an unfavourable process, so only a small proportion of the acid

exists as ions while most CH

COOH molecules remain undissociated. In contrast,

3

the reaction of a CH

COO

(aq) ion (base) with a proton produces CH

3

COOH

3

(weak conjugate acid), so the equilibrium of this process is shied almost

completely to the right.

In

aqueous

CH

3

solutions,

COO

(aq)

+

ethanoate

H

2

O(l)

⇌

ions

CH

3

react

with

COOH(aq)

water,

+

OH

producing OH

(aq) ions:

(aq)

Practice questions This

CH 9.

Formulate the equations,

3

reaction

involves

two

COOH (acid) and H

ethanoic

acid,

so

the

2

conjugate

acid–base

O (acid)/OH

equilibrium

of

(base).

this

pairs: CH

Water

reaction

is

is

a

3

COO

(base)/

weaker acid than

shied to thele.

showing the states of all species,

in which the following ions act Anions

of

polyprotic

acids

behave

as

polyprotic

bases,

for

example:

as Brønsted–Lowry bases:

+

2

CO a.

(aq)

3

+

H

(aq)

⇌

HCO

(aq)

3

cyanide ion, CN

+ 3−

b.

phosphate

ion,

HCO

PO

(aq)

3

+

H

(aq)

⇌

H

2

CO

3

(aq)

4

c.

hydrogenphosphate ion,

These

processes

are

similar

to

the

stepwise

dissociation

of

weak

polyprotic acids,

2

HPO

4

except

that

all

reactions

are

now

reversed.

Patterns and trends

Chemists

classify

substances

acid–base

classic ation

Lowry and

Lewis.

based

systems

on

have

patterns

evolved

they

over

observe.

time:

Three main

Arrhenius,

Brønsted-

+

in

aqueous

According

solution,

to

whereas

Arrhenius

bases

theory,

release OH

acids

ions.

release H

ions

Brønsted–Lowry

+

theory

ions).

is

based

Both

aqueous

denes

pairs

these

media.

acids

and

related,

is

ability

dierent,

For

you

other

of

are

in

species

relevant

theory

bases

(which

terms

of

independent

classic ation

instance,

Brønsted–Lowry

scientists

552

Lewis

therefore

but

equilibria

What

the

theories

and

disadvantages.

whereas

on

Lewis

theory

to

in

the

you

their

of

donate

study

will

to

theory

each

covers

underpins

of

acid–base

accept

solvent.

systems,

accept

learn about in

ability

the

or

many

These

with

a

or

its

the

systems in

Reactivity 3.4)

donate

theories

broad

of

protons (H

own

advantages and

range

pH

electron

represent

of

reactions,

c alculations and

are familiar with.

classic ation

oen

classify

systems

their

have

you

encountered

objects of study?

in

chemistry?

Why do

Reactivity

3.1

Proton

transfer

reactions

Reactions of acids and bases (Reactivity 3.1.7)

You

have

addition

of

weak

already

to

acids,

acids with

seen

these

how

such

as

+

react

most

acids

c arbonates

active metals

Mg(s)

acids

reactions,

2HCl(aq)

produce

→

MgCl

2

and

with

+

with

in

neutralization

metals,

metal

hydrogenc arbonates.

salts

(aq)

bases

react

and

H

2

hydrogen

gas

reactions. In

oxides

The

and

salts

reactions of

(gure4),

for

example:

(g)

+

For

strong

c an

the

be

acids,

shown

solution

the

by

are

actual

reacting

ionic equations. In the

+

Ions

not

that

do

strong

c ancelled

hydrogen ions, H

total ionic equation,

all

(aq), which

ions

present in

+ 2

(aq) + 2Cl

a

are

shown:

Mg(s) + 2H

of

species

participate

acid

out

(aq) → Mg

are

to

in

the

spectator

give the

(aq) + 2Cl

reaction

ions,

and

(aq) + H

(g)

total ionic equation

2

are

c alled

hence

the

spectator ions. The anions

chloride

anions

c an be

p Figure4

net ionic equation:

acid +

Mg(s)

In

+

contrast,

molecular

Mg(s)

+

2H

2

(aq)

ionic

form,

2CH

3

→

they

are

COOH(aq)

(right) of equal concentrations

+

Mg

equations

as

(aq)

+

H

involving

less

→

Mg(CH

3

2

(g)

net

weak

likely

to

ionic

acids

equation

must

show the acids in the

dissociate:

COO)

2

(aq)

+

H

2

(g)

molecular

equation The

2

Mg(s)

+

2CH

In

last

3

COOH(aq)

→

+

Mg

to (aq)

+

2CH

3

COO

(aq)

+

H

2

(g)

ionic

example,

equations

are

there

are

no

term

spectator ions, so the total and net ionic

in

and

identic al.

activity

Copper,

weak

acids

c an

be

distinguished

by

comparing

the

with

an

active

metal

(gure4).

However,

such

comparison

will

be

if

the

concentrations

reactant

of

both

acids

are

equal,

as

the

of

MgO(s)

2HCl(aq)

acids

with

→

metal

MgCl

2

oxides

(aq)

+

+

MgO(s) + 2H

+

given

silver and other metals

not

reaction

react with most acids and

produce

hydrogen gas in

rate depends on reactions.

2H

produce

H

2

salts

and

water.

For

O(l)

example:

molecular

equation

+ 2

(aq) + 2Cl

(aq) → Mg

+

MgO(s)

is

concentrations (Reactivity 2.2).

Reactions

+

which

valid

such the

series,

hydrogen in the activity series

never only

refers

hydrogen in

rates of their do

reactions

metals”

above

section 19 of the data booklet.

aer Strong

“active

elements

equation

the

the

Reaction of magnesium metal

with hydrochloric acid (le) and ethanoic

2

(aq)

→

(aq) + 2Cl

(aq) + H

O(l)

total ionic equation

2

+

Mg

(aq)

+

H

2

O(l)

net

ionic

equation

+

The

last

as

Brønsted–Lowry

a

equation

classied

as

shows

that

base.

neutralization

magnesium

Therefore,

oxide accepts two H

reactions

of

acids

with

ions and thus acts

metal

oxides

c an be

reactions.

Practice questions

10.

11.

Formulate

the

molecular

and

ionic

equations

a.

lithium metal with ethanoic acid

b.

aluminium metal with dilute sulfuric acid.

Formulate

iron(III)

the

oxide,

molecular

Fe

2

O

3

and

ionic

equations

for

the

reactions of

for

the

reaction of

(s), with:

a.

hydrochloric acid

b.

dilute sulfuric acid.

553

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Metal carbonates also react with acids, producing unstable carbonic acid, H

Na

2

CO

C arbonic

bubbles

H

2

These

Reaction of baking soda with

acid

3

two

Again,

2

(aq)

out

CO

Na

p Figure5

3

+ 2HCl(aq)

quickly

of

the

(aq)

→

reactions

CO

the

3

(aq)

net

+

2

(g)

are

into

+

H

2

water

CO

and

3

:

3

(aq)

c arbon

dioxide, which

(gure5):

+

H

oen

2HCl(aq)

ionic

2NaCl(aq)

decomposes

solution

CO

→

CO

2

2

O(l)

combined together:

→

equation

2NaCl(aq)

reveals

the

+

CO

nature

2

(g)

of

+

this

H

2

O(l)

process:

an acid

+

2Na

+

2

(aq)

+

CO

3

(aq)

+

2H

+

(aq)

+

2Cl

(aq)

→

2Na

(aq)

+

2Cl

+

CO

2

(g)

+

H

2

O(l)

total

ionic

equation

net

ionic

equation

+

2

CO

(aq)

(aq)

3

+

2H

(aq)

→

CO

2

(g)

+

H

2

O(l)

2

As

in

the

previous

Brønsted–Lowry

dioxide

Metal

the

NaHCO

+

Na

3

(aq)

+

+

HCO

3

(aq)

+

H

by

the

HCl(aq)

→

anion

accepting

of

two

weak

c arbonic

protons

before

acid,

CO

, acts as a

3

decomposing

to

c arbon

water.

hydrogenc arbonates,

same

+

(aq)

and

example,

base

way

as

such

as

baking

soda,

NaHCO

3

,

react with acids in

c arbonates:

NaCl(aq)

+

CO

2

(g)

+

H

2

O(l)

molecular

equation

+

(aq)

+

Cl

(aq)

→

Na

(aq)

+

Cl

(aq)

+

CO

2

(g)

+

H

2

O(l)

total

ionic

equation

net

ionic

equation

+

HCO

In

(aq)

3

this

by

+

H

c ase,

(aq)

the

accepting

a

→

CO

2

(g)

+

H

2

O(l)

hydrogenc arbonate

proton

before

ion,

HCO

decomposing

,

3

to

acts

as

c arbon

a

Brønsted–Lowry base

dioxide

and

water.

Global impact of science

Acid

deposition,

including

(the

on

primary

climate

dierent

of

rain,

acid

a

secondary

snow,

pollutants)

patterns,

continents.

rain

may

deforestation

may

may

be

be

There

occur

and

pollutant,

c an

take

many

dierent

forms

fog and dry dust. The components of acid deposition

are

away

pollution

generated in one country and, depending

deposited

no

from

of

in

neighbouring

boundaries

the

lakes

actual

and

for

acid

source

river

countries

or

deposition.

leading

systems.

to

even

The

eects

widespread

National

and

regional

Practice questions

12.

Formulate

ionic

the

molecular and

equations

for

the

reaction

of dilute sulfuric acid with

environmental

protection

eort

understand

to

better

Protection

Asia

(EANET)

secondary a.

How

c an

websites

pollutants

our

problems

554

and

the

and

control

Acid

provide

and

their

throughout

acid

Deposition

data

that

politic al

the

world

deposition.

Monitoring

c an

be

used

in

collaborate in an

The

US

Environmental

Network

the

in

E ast

discussion of

implic ations.

potassium

hydrogenc arbonate

b.

Agency

agencies

c alcium

c arbonate.

understanding

such

as

acid

of

chemistry

deposition?

help

to

address

environmental

Reactivity

3.1

Proton

transfer

reactions

Antacids

Heartburn

and

acid

stomach.

in

the

as

antacids

as

metal

any

c arbon

symptoms

These

(gure6).

oxides,

compounds

Like

other

The

the

pharmaceutic al

dioxide

active

hydroxides,

neutralize

of

indigestion

symptoms

c an

be

ingredients

c arbonates

are

in

and

c aused

alleviated

by

by

antacids

excess

hydrochloric

medicines

are

known

weak bases, such

hydrogenc arbonates. All these

excess acid.

drugs,

produced

in

antacids

the

body

have

from

various

the

side

eects.

In

particular,

reaction of stomach acid with

p Figure6

c arbonates

and

hydrogenc arbonates

of

metal

ions

disturbs

the

“Milk

of

magnesia”,

a

c auses bloating and belching, while the

suspension

intake

balance

of

electrolytes

in

the

of

magnesium

hy d r ox i d e

in

body. w a t e r,

is

a

common

antacid

TOK

Pharmaceutic al

tests

prior

includes

two

half

are

active

Half

are

drug

or

that

drug

placebo

in

(for

by

the

example,

the

the

any

who

should

received

health

trials,

participants

inactive

not

results

the

the

from

rigorous eciency

process oen

are

placed into

given the drug and the other

whether

the

two

reactions

drug,

to

This

participants

Participants

desired

active

subject

where

know

physiologic al

have

are

authorities.

are

placebo.

do

The

observed

it

clinic al

therefore

placebo.

antacids)

relevant

study

an

and

works,

participants

do

are

not

on

know which

have

groups

due

therapeutic

but

not

they

are

to

received the

compared to

the

treatment.

eect on the

the

members of the

group.

Sometimes,

placebo.

a

This

approaches

eect

of

administered

they

ascertain

the

drugs

approval

placebo-controlled

groups.

group

If

to

on

therapeutic

is

c an

the

eect

known as the

be

used

to

is

observed

in

placebo eect.

assess

and

people

who

Statistic al

are

and

given the

methodologic al

control the impact of the placebo

results.

How could a participant’ s (or a doctor ’ s) awareness of the existence and

administration of placebos aect the results of the trial? To what extent is bias

inevitable in the production of knowledge? The mechanism of the placebo

eect is not fully understood. Are some things unknowable?

When

balancing

acid–base

1.

Balance

all

nonmetals

2.

Balance

all

metals.

deduced

3.

Balance

earlier,

If

equations,

except

you

you

should

hydrogen

need

to

and

change

use

the

following steps:

oxygen.

any stoichiometric coecients

return to step1.

hydrogen.

Again,

if

you

need

to

change

any

coecients,

return to

step1.

4.

At

this

point,

oxygen

the

atoms

in

equation

the

should

reactants

be

and

balanced

products.

If

already.

their

To

verify it, count the

numbers do not match,

return to step1.

In

most

This

c ases,

strategy

change

their

this

strategy

works

well

oxidation

for

produces

a

acid–base

state.

It

c an

balanced

equation

with

the

fewest trials.

reactions, in which none of the elements

also

be

used

for

most

reactions of acids with

metals.

555

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Worked example 3

Deduce the balanced

equation for the following chemic al

H

3

PO

+

4

C aCO

→

3

Ca

3

(PO

4

)

+

2

CO

+ H

2

2

O

reaction:

Solution

Besides

P

and

H

C,

and

O,

which

the

equation

should

be

2H

involves two nonmetals,

balanced

rst.

There

Steps atoms

on

the

right

but

only

one

on

the

le,

so

3

PO

we

“2”

in

front of H

3

PO

4

3

PO

+

4

C aCO

3

→

1

Ca

3

(PO

4

)

2

+

CO

2

+

H

2

seems

to

be

and

right,

balanced

already

(one

atom

on

so

we

c an

3

is

right),

we

Ca

so

is

3

PO

4

(one

write

+

are

complete,

we

need

+

4

this

to

3C aCO

atom

“3”

in

on

the

front

le,

of

oxygen.

so

write

3

→

so

we

3C aCO

3

→

Ca

3

the

“3”

the

le,

changes

one

3

2

2

+

3CO

+

2

H

2

O

next element is

2

O:

Ca

3

(PO

4

)

2

+

3CO

+

2

3H

2

O

need

to

check

the

last

remaining

4

)

2

the

balance

are 2×4=8 O atoms in 2H

×

3

=

+

9

O

atoms

of

in

3C aCO

3

on

the

right),

so

we

,

so

we

3

PO

4

have a total of

:

CO

+

2

H

2

O

4×2=8

O

O

atoms

atoms

on

in

c arbon

the

Ca

2

3

le.

(PO

4

)

2

On

the

right,

, 3×2=6

there

O

are

atoms

in

3CO

2

O, so the total number of O atoms

(three atoms Therefore,

oxygen is

need to write “3” balanced

beforeCO

)

before H

on the right is also 8+6+3=17.

on

4

three atoms on the

C aCO

(PO

There

and 3 O atoms in 3H

However,

(PO

are six H atoms on the le but only two on

complete,

8+9=17

2H

3

now look at metals. The only metal in the

and 3 so

2

There

PO

element, equation

Ca

each

Step3 side),

→

O

2H

C arbon

3

:

the

2H

3C aCO

need to

hydrogen. write

+

4

are two P

and

so

is

the

equation.

:

Identic ation of parent acids and bases The

balancing

(those

of

redox

equations

involving changes in the

oxidation state of participating

S alts

are

parent

oen

acid

produced

and

base

for

by

a

neutralization

particular

salt

is

reactions.

to

One

formally

split

way to identify the

the

salt

into

c ation(s)

+

elements)

will

Reactivity 3.2.

be

discussed in

and

anion(s).

For

example,

sodium

sulfate,

Na

2

SO

4

,

consists

of

two

Na

c ations

2

and one SO

anion:

4

+

Na

Now

their

2

SO

we

4

→

c an

2Na

add

2

+

SO

4

hydroxide

ions

to

c ations

and

protons

to

anions

according to

charges:

+

Na

+ OH

→

+

2H

NaOH

2

+

Therefore,

SO

the

4

→

H

parent

2

SO

4

base

and

acid

for

Na

2

SO

are

4

NaOH and H

2

SO

4

,

respectively.

The

same

result

The

word

“sodium”

the

word

For

ammonium

“sulfate”

be either NH

be

556

could

accepted

3

in

be

obtained

“sodium

salts, such as NH

IB

analysing

refers

refers to sulfuric acid, H

(ammonia) or NH

in

by

sulfate”

assessments.

4

4

Cl

2

the

to

SO

4

name

hydroxide,

of

the

salt.

NaOH, while

(table4).

(ammonium

OH

systematic

sodium

chloride),

(ammonium

the

hydroxide).

parent base could

Both

answers will

Reactivity

pH curves (Reactivity 3.1.8)

unknown

concentration

of

an

acid

or

base

Proton

Standard solution

with The

3.1

in

a

solution

c an

be

determined

a

known

is a

(Reactivity 2.1)

using

a

standard

solution

of

a

base

or

acid,

reactions

solution

concentration of the

by

solute (Structure 1.4). titration

transfer

Analyte

is the

respectively.

analysed substance or the solution The

reaction

progress

c an

be

monitored

using

a

digital

pH

meter

(gure7) and

of a

data

logger,

which

automatic ally

records

the

pH

of

the

reaction

this

substance

concentration. standard

solution

is

added

to

the

pH

added

The

of

data

overall

the

collected

volume

of

the

shape

reactants

of

and

during

a

standard

the

on

pH

the

unknown

Titrant

is the

analysed solution.

reactant

The

with

mixture as the

titration

experiment

solution,

curve

depends

addition

c an

producing a

on

the

be

plotted against the

pH curve

strengths

(gures8 and 9).

and

added to the analyte or a

standard

solution

of

that

reactant

(Reactivity 2.1).

concentrations

order.

When a strong acid, such as HCl, is titrated with a strong base, such as NaOH,

the curve intercepts the y-axis at a low pH value (gure8), as the analysed solution

is strongly acidic. Typical concentrations of acids and bases used in titration

3

experiments are from 0.01 to 1 mol dm

, so the initial pH may vary from 2 to 0.

14

NaCl

+

NaOH

equivalence point

(NaCl

only)

Hp

7

pH jump

intercept

(HCl

only) p Figure7

HCl

+

NaCl

pH

Ac i d – b a s e

t i t ra t i o n

with

a

meter

0 V(NaOH)

p Figure8

Ty p i c a l

pH

curve

for

the

t i t ra t i o n

of

a

Activity

strong

acid

(HCl)

with

a

strong

base

The

pH

curve

shown

in

gure8

(N aOH)

represents

At

the

beginning

as

the

solution

of

the

titration,

the

pH

of

the

mixture

increases

very

the

slowly,

3

0.1 mol dm

still

contains

large

excess

of

acid.

For

titration of

example, if the initial

HCl(aq) with

3

0.1 mol dm

NaOH(aq).

Copy the

3

concentration

of

the

acid

were

0.1 mol dm

and

half

of

the

acid

were

neutralized, axes

the

pH

of

the

mixture

would

increase

from

1.0

to

and

curve

from

gure9 and

approximately 1.5. sketch

the

second

pH

curve

for the

3

titration of 0.01 mol dm As

the

acid

concentration

decreases,

the

pH

curve

becomes

HNO

3

(aq)

progressively 3

with steeper.

At

the

equivalence

point,

the

pH

raises

sharply

to

0.01 mol dm

KOH(aq).

7.0, as the acid is

Explain whether the changes in the neutralized

completely,

and

the

reaction

mixture

contains

only

NaCl(aq):

nature

HCl(aq)

+

NaOH(aq)

→

NaCl(aq)

+

H

2

O(l)

and

reactants

features The

pH

excess

continues

of

becomes

the

to

NaOH(aq)

very

standard

large,

rise

sharply

makes

the

NaOH(aq)

the

curve

immediately

solution

attens

solution,

aer

basic.

out

typic ally

and

the

When

concentrations of the

will

of

aect

the

the

following

curve:

equivalence point, as an

the

gradually

excess

of

the

•

y-axis

intercept

•

pH

at

equivalence

•

pH

at

which

titrant

approaches the pH of the

curve

attens

between 12 and 14. out.

To

construct

the

an

acid–base

the

indic ator

acid

or

base

equivalence

in

the

equivalence

titrations

complete

by

indic ator

changes

is

as

curve,

excess

must

be

colour.

c alculated

point,

is

pH

adding

In

using

stopped

both

the

explained in

discussed

in

the

we

AHL

need

titrant.

In

at

or

c ases,

volume

to

continue

contrast,

near

the

of

a

the

the

Reactivity 2.1.

beyond

experiment with

equivalence point, when

concentration

the

titration

titration

of

the

analysed

standard solution at the

The

use

of

acid–base

indic ators

section of this topic.

557

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

When The

construction

and

a

strong

(gure9). of

pH

curves

involving

bases

will

not

be

SL.

The

pH

shapes

and

curves

are

titrated

with

solution

a

is

strong

basic,

acid,

the

the

curve

pH

curve

is

inverted

intercepts the

value

and

declines

gradually

as

the

titrant

is

added.

The

y-axis at a

equivalence is

at

salt

the

(in

same

our

pH

value

example,

(7 .0), as the solution at that point contains only a

NaCl).

At

the

end

of

the

titration,

the

curve

attens at

discussed in the a

AHL

is

initial

features of neutral

these

the

assessed achieved

at

Since

weak acids high

and

base

interpretation

low

pH,

as

the

solution

contains

excess

strong acid.

section of this topic.

14

NaOH

+

NaCl

intercept

(NaOH only)

Hp

7

pH drop

equivalence point

(NaCl

only)

NaCl

+ HCl

0 V(HCl)

p Figure9

(N aOH)

The

main

and

bases

features

are

Ty p i c a l

with

of

pH

a

pH

strong

curves

curve

acid

for

for

the

t i t ra t i o n

of

a

strong

base

(HCl)

titration

experiments

involving

strong acids

summarized in table6.

pH

Analyte

Titrant

y-axis equivalence

attening out

intercept

strong

strong

acid

base

p Table 6

strong

strong

S ummary

base

low

7.0

high

acid

high

7 .0

low

of

pH

curves

for

t i t ra t i o n s

i nv o l v i n g

strong

acids

and

bases

Conductometric acid–base titration

The

by

progress

This

is

ionic

in

of

measuring

possible

species

the

bec ause

present

the

water

are

acid–base

bec ause

fully

of

the

c an

concentration

the

reaction.

molecular,

of

the

and

solution

all

the

of

+ Cl

this

Tool

1:

•

Tool

2:

Use

•

Tool

3:

Sketch

•

Tool

3:

Extrapolate

•

Tool

3:

Interpret

Titration

sensors

example,

graphs

with

labelled

but

unsc aled

graphs

species features

of

graphs

+

Na

+

(aq)

+

OH

(aq)

→

Na

you

will

S afety (aq)

(aq)

+

+

H

2

O(l)

•

Wear

•

Dilute

eye

protection.

hydrochloric

acid

and

sodium

hydroxide

perform a conductometric

irritants. titration.

You

graphic al

of

558

the

will

form.

acid

axes

dissociated into ions:

(aq)

practic al,

Relevant skills

•

aqueous

decreases

+

(aq)

monitored

mixture.

For

other

Cl

In

be

reaction

hydrochloric acid and sodium

conductivity

+

H

the

during

between

is

reaction

conductivity

changes

reaction

hydroxide,

an

the

analyse

Then,

aects

the

and

you

interpret

will

shape

the

consider

of

the

resulting data in

how

titration

the

strength

curve.

•

Dispose

of

all

substances

appropriately.

are

Reactivity

3.1

Proton

transfer

reactions

Questions

Materials

3

•

0.01 mol dm

•

0.1 mol dm

•

250 cm

•

burette,

•

magnetic stir bar and stir plate

hydrochloric acid, HCl(aq)

1.

Plot

a

graph

NaOH(aq)

3

sodium

hydroxide,

showing conductivity

vs.

volume of

added.

NaOH(aq)

3

2.

beaker

Draw

two

lines

equivalence

of

best

point

fit

and

(one

for

another

the

points

before the

for the points after).

burette clamp and stand

Identify

the

equivalence

point

by

extrapolating the

two lines.

•

conductivity

probe

3.

Compare

Explain

Instructions

and

any

contrast

this

differences

graph

to

your initial sketch.

between the two plots.

3

1.

You

will

titrate

a

known

volume

of

4.

0.01 mol dm

Interpret

and

explain

the

shape

of

the

graph, noting

3

HCl(aq)

how

with

the

0.1 mol dm

volume

conductivity.

above,

as

the

of

With

explain

reaction

NaOH(aq)

NaOH(aq)

reference

why

the

added

to

the

the

and

measure

affects the

ionic

conductivity

approaches

and

equation

should

decrease

explaining

the

following:

a.

change in conductivity before the equivalence point

b.

conductivity

c.

change in conductivity after the equivalence point.

at

the

equivalence point

equivalence point. 5.

Interpret

your

graph

to

compare

and

+

2.

Predict

and

explain

conductivity

vs

the

volume

shape

of

of

the

conductivity of H

graph of

NaOH(aq)

added

that

explain the

+

(aq)

ions

and

Na

(aq) ions.

you 6.

The

total

volume

of

solution

in

the

beaker

was

expect to obtain. kept

3.

Use

the

conductivity

necessary

probe,

equipment,

to

and

hydroxide

any other

prepare

and

roughly

constant

concentration

hydrochloric

measure the

beaker

conductivityof:

by ensuring that the sodium

acid.

should

be

was ten times that of

Suggest

kept

why

constant

the

in

volume in the

this

practic al.

3

a.

0.1 mol dm

b.

0.01 mol dm

HCl(aq)

3

A

conductometric

titration

of

0.1 mol dm

ethanoic acid,

3

NaOH(aq)

3

CH c.

distilled

3

COOH(aq),

Using

make

the

any

0.1 mol dm

water

sodium

hydroxide,

3

NaOH(aq)

4.

with

measurements

necessary

you

changes

obtained in step 3,

and

The

data

added

in

recorded

2.0 cm

are

increments

shown

in

was

c arried out.

gure 11.

refinements to the 5

graph

you

sketched in step 2.

1

up

the

equipment

as

shown

in

figure 10:

mc

Set

Sm( / ytivitcudnoc

3

0.1 mol

)

5.

dm

NaOH

500

μS cm

4

3

2

1

3

100 cm

of

conductivity 0

3

0.01

probe

mol

dm

HCl

0

5

10

15

20

25

30

magnetic stir bar

35

3

volume

p  Figure 11

of

NaOH

Conductivity

added / cm

against

volume

of

sodium

3



p  Figure10

Experimental

a p p a ra t u s

for

hy d r ox i d e

conductometric

for

the

t i t ra t i o n

of

0.1 m o l d m

CH

3

CO O H ( a q )

3

acid–base

6.

Record

with

t i t ra t i o n

the

initial

conductivity.

Add

the

0.1 m o l d m

K.,

Ed i o nw e,

E.

11,

1217–1221

N aOH(aq).

and

Michel,

Source

B.,

J.

of

data:

Chem.

Smith,

Educ.,

C.

2 0 1 0,

87,

NaOH(aq) in

3

small

of

(~1 cm

NaOH(aq)

adding

so

you

before

)

increments.

added

NaOH(aq)

c an

and

later

after

and

well

Record

the

past

the

exact

conductivity.

the

volume

Continue

equivalence point,

compare the conductivity changes

the

7.

Compare

the

graph

and

you

contrast

the

obtained.

graph

Explain

in

figure 11 with

any

differences

you

observe.

equivalence point.

559

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

The pOH

The

a

pOH sc ale

broad

for

allows

range of OH

solutions

related

sc ale (Reactivity 3.1.9)

to

of

the

us

to

(aq)

Arrhenius

solution

represent

the

basicity

concentrations.

bases,

pOH

as

in

which

The

the

of

pOH

aqueous

solutions

over

sc ale is particularly useful

concentration

of

hydroxide ions is

follows:

pOH

pOH

Basic

The

equations

= −log[OH

solutions

with

]

[OH

high

]

=

10

concentrations of OH

(aq)

ions

have

low

pOH

values

for the while

acidic

solutions

with

low

concentrations of OH

(aq)

ions

have high pOH

+

interconversion of pH, pOH, [H

] values.

and

[OH

booklet.

]

are

given in the data

The

pOH

requires

sc ale

some

is

a

logarithmic

practice

but

sc ale,

greatly

like

the

simplies

pH

the

sc ale.

Working with logarithms

c alculations.

The

14

“pOH=14”

likely

use

to

is

c ause

addition

log(a

×

more

errors

and

b)

compact

=

when

written.

subtraction

log a

+

than

“[OH

Also,

instead

of

]=1.00×10

logarithms

and

expression

3

mol dm

”

p-numbers

and

less

allow us to

multiplic ation and division:

log b

a log

As

a

use

(

result,

in

=

)

b

log a

−

formulas

log b

with

logarithms

or

p-numbers

are

easier to memorize and

c alculations (table7).

Without p-numbers

+

[H

With p-numbers

14

][OH

]

=

1.00 × 10

pH

+ pOH

pH

=

=

14

14

1

+

[H

×

10

] =

[OH

14

−

pOH

]

14

1 ]

[OH

×

10

pOH

=

=

14

−

pH

+

[H

]

+

p Table7

Useful expressions involving K

, [H

w

],

[OH

], pH and pOH

Worked example 4

C alculate the pOH

values for the following solutions:

3

a.

0.025moldm

KOH(aq)

3

b.

0.025moldm

H

2

SO

4

(aq).

Solution

a.

First,

write

the

equation

for

the

dissociation

of

potassium

hydroxide:

+

KOH(aq)

The

→ K

(aq)

concentration

+

of

OH

(aq)

KOH

is

equal

to

the

concentration

of

hydroxide ions:

3

[OH

]

Then,

pOH

560

=

[KOH]

use

the

=

0.025 mol dm

expression pOH

= −log 0.025

=

1.60

= −log[OH

] to determine pOH:

Reactivity

Write

the

equation

for

the

H

2

SO

(aq)

→

one

mole

4

2H

(aq)

+

SO

13.

(aq)

4

of

sulfuric

acid

dissociates,

it

C alculate:

a.

the pOH of a

produces two moles of H 3

5.0 × 10

+

ions.

Therefore,

the

concentration of H

3

mol dm

ions is double that of sulfuric

solution

of

Ba(OH)

acid:

b. +

[H

3

]

=

2×[H

2

SO

reactions

2

+

When

transfer

Practice questions

dissociation of sulfuric acid:

+

Proton

LHA

b.

3.1

4

]

=

2

×

the

2

(aq)

concentration of

3

0.025 mol dm

=

0.050 mol dm hydroxide ions in a solution

+

You

c an

use

the

expression

K

=

w

[H

][OH

]

to

determine

[OH

with

]:

pOH

=

4.70.

14

K [OH

]

1.00

w

×

10

13

=

=

=

3

2.00 × 10

mol dm

+

[H

Then,

use

]

0.050

the

expression pOH

= −log[OH

] to determine pOH:

13

pOH

The

pH

= −log(2.00 × 10

same

answer

could

= −log0.050

pOH

=

14

−

1.30

)

≈

be

≈

1.30

=

12.70

12.70

obtained

using

the

formula pH

+

pOH

=

14:

Weak acids and bases (Reactivity 3.1.10 and

Reactivity 3.1.11)

As

with

by

an

any

other

equilibria,

equilibrium

constant.

dissociation constant,

K

a

the

dissociation

This

equilibrium

of

a

weak

constant

acid

is

c an

be

characterized

known as the

acid

:

+

[H

+

HA(aq)

⇌

H

(aq)

+ A

(aq)

K

a

][A

]

= [HA]

Bases

c an

be

characterized

by the

base dissociation constant,

K

b

:

+

[BH

+

B(aq)

+

H

2

O(l)

⇌

BH

(aq)

+

OH

(aq)

K

b

][OH

]

= [B]

Notice

that

the

equilibrium

Stronger

acids

dissociation

relative

concentration

constant

and

bases

constants.

strengths

of

of

the

solvent

(water)

is

not

included in the

expressions.

dissociate

Therefore,

K

to

a

a

greater

and

K

b

extent

values

c an

and

be

therefore

used

to

have

larger

compare the

dierent acids and bases.

+

Like [H

(their

],

[OH

negative

] and

K

w

decimal

,

the

values of

p

pK

a

= −log K

K

a

a

=

10

In

b

= −log K

contrast to

and

bases,

K

a

K

b

and

K

b

a

and

K

b

are

oen

expressed

as

p-numbers

=

10

Ka

p

pK

K

logarithms):

,

respectively

b

larger

Kb

values of pK

(table8).

stronger than ethanoic acid (pK

a

stronger base than ammonia (pK

For

=4.76),

b

a

and pK

b

correspond

to

example, methanoic acid (pK

a

while

methylamine (pK

b

weaker acids

=3.75) is

=3.34) is a

=4.75).

561

3

What

are

the

mechanisms

LHA

Reactivity

of

chemic al

K

Acid

change?

pK

a

K

Base

a

pK

b

4

HCOOH

1.78 × 10

4

3.75

5

CH

3

COOH

1.74 × 10

4.76

(CH

3

)

2

NH

5.37 × 10

p Table8

3

NH

4.57 × 10

2

9.21

Dissociation constants of

weak

dissociation

in

know

the

the

bases at

constants

by

1.78 × 10

we

c an

6

H

3

5

NH

4.75

7.41 × 10

2

9.13

298 K

for

measuring

dissociation

solution,

examples

C

acids and

experimentally

we

3.34

5

NH

10

6.17 × 10

The

3.27

4

CH

10

HCN

b

4

3.17

sesaercni

6.76 × 10

htgnerts

HF

weak

the

constant

c alculate

acids

pH

of

for

the

and

their

a

weak

pH

of

bases

c an

standard

acid

that

or

be

determined

solutions.

base

solution,

and

as

Conversely, if

its

shown

concentration

in

the

worked

below.

Worked example 5

3

A 0.0100 mol dm

solution of propanoic acid, CH

3

CH

2

COOH(aq), has a pH of 3.44. Determine the

pK

a

of propanoic

acid.

Solution

First

of

all,

we

need

to

consider

all

acid–base

equilibria in the solution:

+

[CH

+

CH

3

CH

2

COOH(aq)

⇌

CH

3

CH

2

COO

(aq)

+

H

(aq)

K

a

3

CH

[CH

+

H

2

O(l)

⇌

H

2

COO

] [H

]

=

3

CH

2

COOH]

+

(aq)

+

OH

(aq)

K

w

=

[H

14

][OH

]

=

1.00 × 10

+

Although H

(aq)

ions

are

formed

by

the

dissociation

of

both

propanoic

acid

and

water,

typic al

K

a

values

of

organic acids

+

(table8)

are much higher than

K

w

.

Therefore,

we

c an

assume

that

nearly all H

(aq)

ions

in

the

solution

are

produced

by

+

the

acid.

In

such

c ase,

[CH

3

CH

2

COO

+

[CH

Weak

3

CH

acids

2

COO

]

≈

dissociate

molecules, CH

approximately

3

CH

the

2

[H

to

≈

[H

],

which

3.44

]

a

=

10

very

as

its

3.63 × 10

extent,

Therefore,

initial

follows

4

≈

small

COOH(aq).

same

]

so

we

from

the

rst

equilibrium.

3

moldm

most

c an

concentration.

of

the

assume

In

that

propanoic

that

the

acid

in

the

equilibrium

solution

exists

as

undissociated

concentration of CH

3

CH

2

COOH(aq) is

c ase:

3

[CH

When

3

CH

2

COOH]

performing

approximations

quadratic

Substitute

the

valid.

will

−

K

a

×

10

The

be

values

(3.63

0.0100 mol dm

equilibrium

are

equations.

approximations

≈

In

c alculations,

many

use

of

c ases,

always

state

any

approximations

quadratic

equations

will

approximations

greatly

not

be

simplify

required

the

in

you

make

and

c alculations,

examination

explain

which

papers,

why these

otherwise

so

any

would

involve

reasonable

accepted.

into

the

expression

−

4

)(3.63

×

10

K

a

:

)

−

≈

=

for

4

5

1.32 × 10

0.0100

Then use pK

a

= −log K

a

to determine pK

−

pK

The

a

= −log(1.32 × 10

approximations

However,

arevalid.

562

the

nal

made

answer

a

:

5

)

≈

in

is

4.88.

this

example

could

potentially

very close to the actual pK

a

reduce

value

of

both

the

propanoic

accuracy

acid

and

(4.87),

so

precision

all

the

of

our

c alculations.

approximations

Reactivity

3.1

Proton

transfer

reactions

LHA

Worked example 6

3

Using the pK

value from the previous example,

a

c alculate the pH

of a 0.100 mol dm

solution of propanoic acid.

Solution

+

5

We

determined

K

to be 1.32 × 10

a

+

[CH

3

CH

2

COO

][H

+

]

+

[H K

≈

[H

in

the

previous

example.

We

c an

assume

that

[CH

3

CH

2

COO

]

≈

[H

], so

2

]

,

giving

the

following

expression

for

K

a

:

2

]

=

a

[CH

3

CH

2

COOH]

3

The

concentration

of

propanoic

acid

is

given

in

the

question

(0.100 mol dm

).

Substitute

the

values of

K

a

and

+

[CH

3

CH

2

COOH]

into

the

+

[H

5

1.32 × 10

above

expression to determine [H

]:

2

]

= 0.100

+

[H

6

2

]

=

1.32 × 10

+

[H

6

]

√ 1.32

=

×10

3

≈

1.15 × 10

3

mol dm

+

Then use pH

= −log [H

] to determine pH:

3

pH

= −log(1.15 × 10

)

=

2.94.

Worked example 7

3

A 0.0100moldm

solution of trimethylamine,

(CH

3

)

3

N(aq),

has a pH

of 10.90. Determine the pK

b

of trimethylamine.

Solution

Similar

to

worked

example5,

we

need

to

consider

all

acid–base

equilibria in the solution:

+

[(CH

3

)

3

N(aq)

+

H

2

O(l)

⇌

(CH

3

)

3

NH

) 3

+

(CH

(aq)

+

OH

(aq)

K

b

NH

[(CH

) 3

+

H

2

O(l)

Amines

⇌

are

H

][OH

]

3

= N] 3

+

(aq)

stronger

+

OH

(aq)

bases

than

c ase,

[(CH

K

=

w

water,

so

we

c an

assume that all OH

(aq)

[H

14

][OH

ions

in

]

the

=

1.00 × 10

solution

are

formed

by the ionization

+

of

the

amine.

In

such

3

)

3

NH

3

]

≈

[OH

].

Since

amines

are

weak

bases,

[(CH

3

)

3

N]

≈

0.0100 mol dm

pOH

Use pOH

=

14

−

pH

to

pH

=

work

out

the

value

of

pOH,

and

then

use

[OH

]

=

10

to

determine

the

concentration of

hydroxide ions:

pOH

=

14

−

14

−

10.90

3.10

[OH

Then

]

=

10

substitute

the

K

b

×

3.10

4

=

7.94 × 10

values

−

(7 .94

=

10

in

3

mol dm

the

expression

−

4

)

(7 .94

×

for

K

b

:

4

10

)

=

−

≈

5

6.31 × 10

0.0100

−

pK

Our

b

=

−log(6.31 × 10

5

)

≈

4.20

answer matches the actual pK

b

of

trimethylamine.

563

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

Worked example 8

Using the pK

value from the previous example,

b

c alculate the pH of a

3

0.100 mol dm

solution of trimethylamine.

Solution

5

K

b

was

determined to be 6.31 × 10

in

the

+

that

[(CH

3

)

3

NH

expression

for

previous

example.

+

]

K

≈

b

[OH

],

so

[(CH

3

)

3

NH

We

c an

assume

2

][OH

]

≈

[OH

]

,

giving

the

following

:

2

[OH K

b

]

= [(CH

Substitute

3

the

)

3

N]

values into the

K

b

expression:

2

[OH

]

–5

6.31 × 10

= 0.100

2

[OH

]

[OH

]

6

=

6.31 × 10

6

=

√ 6.31

3

×10

≈

2.51 × 10

3

mol dm

Then determine the pOH, and nally the pH:

3

pOH

pH

A

= −log(2.51 × 10

=

14

conjugate

−

2.60

=

acid–base

)

≈

2.60

11.40

pair

c an

be

characterized

by a single ionization constant,

Practice questions as the pK

14.

C alculate

the

pH

values

for the

the

of the acid and pK

a

acid–base

represented

following solutions:

equilibria

as

b

of

the

base

are

related

to

each

other.

For

involving the acid HA and its conjugate base A

example,

c an be

follows:

+

3

a.

[H

+

0.0200 mol dm

HA(aq)

⇌

H

(aq)

+ A

(aq)

K

a

][A

]

= [HA]

hydrogen

cyanide, [HA][OH A

HCN(aq)

(aq)

+

H

2

O(l)

⇌

HA(aq)

+

OH

(aq)

K

b

b.

5.00 × 10

6

H

5

NH

2

]

3

mol dm

phenylamine,

C

]

= [A

3

When

these

giving

the

two

equations

equation

for

the

are

added

ionic

together, HA(aq) and A

product

of

(aq)

c ancel out,

water:

(aq). +

H

+

O(l)

⇌

H

According

to

table1

2

(aq)

+

OH

(aq)

K

w

=

[H

][OH

]

Refer to table 8.

added

the

K

w

At

together,

product

=

K

a

pK

Note

a

b

any

The

b

w

=14,

−

pK

=

14

−

pK

these

values

Reactivity 2.3,

equilibrium

14

w

=

pK

constants

+

pK

b

of

of

two

the

the

chemic al

resulting

individual

K a

w

= K

b

K

w

is

equations.

are

equal to

As

a

result,

.

b

a

are

valid

non-conjugated

=

equations

equation

so:

equations

of

a

when

constant

only

acids

for

conjugate

and

bases

acid–base

are

not

way.

equation

K

the

from

equilibrium

and pK

=

that

and pK

564

of

b

298 K, pK

pK

in

×K

the

K

a

×

K

b

c an

be

rearranged

as

follows:

pairs, as the pK

a

related to one another

Reactivity

equation

shows

the

inverse

relationship

between

the

Proton

transfer

reactions

LHA

This

3.1

strengths of

Practice questions conjugates:

stronger

the

the

stronger

base,

the

the

acid,

the

weaker

its

conjugate

base, and the

weaker its conjugate acid.

15.

The pK

for

the

HCO

a

and pK

(aq),

3

b

expressions

hydrogenc arbonate ion,

are

10.32

and

7.64,

Acid–base equilibria in solutions of salts respectively.

(Reactivity 3.1.12) a.

As

you

already

reaction

an

solution

c an

therefore

aecting

on

the

salt

acid

c an

and

react

the

c alled hydrolysis

depend

any

between

aqueous

are

know,

a

reactions.

of

The

water

pH.

The

the

considered

base.

with

solution’ s

strengths

be

ions

and

The

and

a

product

produced

by

of

the

a

neutralization

reactions

and

base

between

salt in an

extent

that

form

of

the

salts

and

hydrolysis

salt.

The

K

reactions

of

parent

acids

and

bases

are

strong

acid–strong

base,

a

equations

equilibria

characterized

and

K

b

by

of the

hydrogenc arbonate ion.

possible b.

combinations

the

represent the

acid–base

water

four

Formulate

that

form conjugate acids and/or bases,

direction

acid

as

C alculate the pK

base,

weak

acid–strong

base

and

weak

acid–weak

a

value

for

strong c arbonic acid, H

acid–weak

2

CO

3

(aq),

base. and the pK

b

value

for the

2

c arbonate

ion,

CO

3

(aq),

S alts of strong acids and strong bases at

Both

and

the

c ation

acid,

chloride

and

the

anion

respectively),

dissociates

in

so

in

they

salts

do

aqueous

of

not

this

type

undergo

solutions

as

have

298 K.

strong conjugates (base

hydrolysis.

For

example, sodium

follows:

+

NaCl(aq)

→

Na

(aq)

+ Cl

(aq)

+

The

hypothetic al

NaOH(aq),

reaction

which

of

c annot

Na

exist

(aq)

in

with

water

aqueous

would

solutions

in

produce

the

strong base

undissociated

form:

+

Na

(aq)

Similarly,

HCl(aq),

Cl

the

is

(aq)

only

the

H

2

O(l)

no

reaction of Cl

which

Therefore,

The

+

+

also

H

2

does

O(l)

neither

ion

of

is

in

O(l)

⇌

H

The H

would

aqueous

produce

solutions

in

the

strong acid

molecular

solutions

in

of

acid–base

salts

equilibria.

formed

by

strong

acids

and

(aq)

formed

and OH

solution

by

a

strong bases

−

+

OH

(aq)

−

the

form:

water:

(aq)

+

and

water

in

reaction

involved

+

2

with

exist

no

equilibrium

dissociation

H

(aq)

not

reaction

+

(aq)

ions

remains

strong

acid

are

produced

neutral (pH=7).

and

a

strong

base

in

In

equal amounts, so [H

other

has

no

words,

eect

the

on

−

]

=

presence

the

[OH

of

a

],

salt

solution’s pH.

S alts of strong acids and weak bases

In

salts

of

this

conjugates

solutions

is

type,

a

the

hydrolysis

favourable

produces

the

4

Cl(aq)

→

NH

4

For

c ations

example,

only,

as

the

ammonium

formation

chloride

in

of

weak

aqueous

following ions:

+

NH

involves

process.

−

(aq)

+ Cl

(aq)

−

The conjugate of the Cl

(aq)

ion

is

a

strong acid, HCl(aq), so the chloride anion

+

does

weak

not

undergo

base, NH

3

hydrolysis.

(aq),

so

the

4

contrast, the conjugate of the NH

ammonium

+

NH

In

c ation

itself

behaves

as

a

4

(aq) ion is a

weak acid:

+

(aq)

⇌

NH

3

(aq)

+

H

(aq)

565

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

LHA

+

According to table8, pK

b

(NH

3

)

=

4.75,

so

pK

a

(NH

)

4

=

14

−

4.75

=

9.25.

+

Therefore, the acidity of NH

HCN(aq),

The

true

last

which

is

equation

nature

of

a

very

c an

the

(aq)

4

is

comparable

weak acid with a pK

be

expanded

hydrolysis

to

include

a

of

hydrogen

cyanide,

9.21.

molecule

of

water

and

show the

+

(aq)

4

of

a

that

process:

+

NH

to

+

H

2

O(l)

⇌

NH

3

(aq)

+

H

3

O

(aq)

+

As

so

mentioned

the

+

earlier, the H

hydrolysis

of

the

symbol

ammonium

is

used

ion

c an

as

be

a

shorthand

represented

+

two

equations.

In

IB

equivalent of H

3

O

,

by either of the

+

assessments, the use of H

or H

3

O

ions

will

be

equally

accepted.

+

The

excess H

+

(aq) or H

3

O

+

(aq)

ions

means that [H

]

>

[OH

], so the solution

becomes acidic (pH7).

Anions

Practice question

16.

Using

value

CH

3

that

table8,

c alculate the pK

for

polyprotic

weak

acids

accept

a

single

proton

to

form their conjugates,

2 b

CO

(aq)

3

+

H

2

O(l)

⇌

HCO

(aq)

3

+

OH

(aq)

for the ethanoate anion,

COO

this

(aq),

ion

is

a

and

HCO

show

weak

the

(aq)

+

H

2

O(l)

⇌

H

2

CO

3

(aq)

+

OH

(aq)

that

each

equation

involves

only

one

molecule

of

water

and

produces a

strength of this single

with

3

base. Notice

Compare

base

of

example:

other

basic

hydroxide anion.

species

listed in table8.

S alts of weak acids and weak bases

In

these

salts,

ammonium

both

the

ethanoate

c ation

and

produces

the

the

anion

undergo

hydrolysis.

For

following ions:

+

CH

Both

3

COONH

ions

CH

3

have

COO

4

(aq)

weak

(aq)

→

CH

3

COO

conjugates,

+

H

2

O(l)

⇌

(aq)

so

CH

3

they

+

4

COOH(aq)

+

(aq)

+

H

(aq)

⇌

2

O(l)

⇌

NH

3

(aq)

or

+

NH

566

4

+

NH

3

(aq)

+

H

(aq)

4

react

+

NH

NH

+

H

3

O

(aq)

(aq)

reversibly

+

OH

with

(aq)

water:

example,

Reactivity

3.1

Proton

transfer

reactions

+

reactions

basicity)

base.

base

will

In

If

solution

conjugate

the

salt

depends

acid

c ation,

for

the

(aq)

on

the

salt

solution

and OH

the

(aq)

relative

anion

will

be

is

ions,

so

strengths

slightly

slightly

the

of

acidity

the

(or

parent acid and

stronger than the conjugate

acidic

(pH7).

c ase

of

b

For

the

the

for

the

pK

of

produce both H

LHA

These

the

many

of

ammonium

parent

salts,

base

the

ethanoate, the pK

(4.75)

relative

are

almost

strengths

of

a

of

the

identic al,

their

parent

so

parent

the

acid

acid

(4.76) and the

solution

and

pH

base

will

be

7 .0.

are similar to

+

each

other, so [H

]

≈

[OH

] and

pH≈7.

Thinking skills

ATL

Hydrolysis of salts: summary

Certain chemistry topics, such as

acids The

pH

of

a

salt

solution

depends

on

the

relative

strengths

of

the

and

dierent base,

as

shown

in

types

involve

of

several

c alculations. This

table9.

task

Ions Parent acid

bases,

parent acid and

Parent base

will

help

you

to

organize the

c alculations covered in this chapter,

Hydrolysis

Solution pH produced

so

you

c an

better

understand

how

+

strong

strong

strong

weak

none

[H

]

=

[OH

]

7

>

[OH

]

< 7

only

[H

are

make

+

c ation

they

]

a

interconnected.

list

of

the

First,

various types of

c alculations covered in this chapter. +

weak

strong

weak

weak

anion

only

[H

]


7

]

≈

[OH

]

≈ 7

Two

examples

are

shown

below:

+

c ation

and

anion

[H

•

Converting

between

+

p Table9

Hydrolysis of

pH

salts and solution pH

and [H

]:

+

[H

This

table

c an

be

•

“hydrolysis

•

“the

summarized

is

for

the

by

the

following

pH

]

=

10

informal rules:

+

weak”

pH

]

[H

stronger wins”.

+

pH

= –log

[H

]

10

These

rules

relative

emphasize

strengths

of

the

the

fact

that

parent

the

acid

pH

and

of

a

base:

salt solution depends on the

if

the

acid

is

stronger, the solution

•

Converting

between

+

will

be

acidic,

and

if

the

base

is

stronger, the solution will be basic.

K

a

and

[H

]:

+

[H

]

=



K

[HA] a

Practice questions

17 .

Formulate

the

equations

for

acid–base

equilibria

in

aqueous solutions of

+

K

[H

a

the

following

]

salts:

a.

potassium

cyanide,

b.

potassium

sulfate, K

KCN

c.

methylammonium

+

[H K

2

]

= a

2

SO

[HA] 4

As methanoate,

HCOONH

3

CH

shown

you

d.

For

each

methylammonium

solution,

predict

bromide, CH

whether

it

will

3

NH

be

3

in

the

worked

examples,

3

will

oen

need

to

apply

several

Br. dierent

neutral, acidic or basic.

a

single

c alculations when solving

problem

and

bases.

that

shows

involving acids

Create a scheme

and

interconnects

pH curves of strong and weak acids and all

bases (Reactivity 3.1.13)

the

c alculations

listed.

answering The

shape

of

the

pH

curve

in

an

acid–base

titration

depends

on

the

the

acid

and

the

base.

There

are

four

distinct

shapes

that

practice

types

of

acid–base

pair:

strong

acid–strong

base,

strong

weak

acid–strong

base

and

weak

acid–weak

You

relationships and no

acid–weak

longer base,

questions.

eventually memorize these

correspond to the

quantitative following

have

strengths

will of

you

Refer to this scheme when

need

this

sc aold.

base.

567

3

What

are

the

mechanisms

of

LHA

Reactivity

chemic al

change?

pH curves involving strong acids and strong bases

The

pH

shown

curves

in

for

titration

gures8

and

9

experiments

in

the

SL

involving

section

of

this

strong

topic.

acids

At

and

typic al

bases

are

concentrations

3

of

the

the

•

analyte

following

y-axis

and

titrant

(approximately

0.1 mol dm

each),

these

pH

curves

have

features:

intercept at pH≈1 (when the analyte is an acid) or pH ≈13 (when the

analyte

is

a

base)

•

gradual

rise

•

sharp

•

equivalence at pH=7

•

attening

out

or pH≈1

(when

rise

or

or

fall

in

pH

at

drop

in

pH

near

at

the

(no

end

the

the

beginning

the

the

the

titration

equivalence point

hydrolysis

of

of

of

the

salt)

titration to pH≈13

(when

the

titrant

is

a

base)

titrant is an acid).

pH curves involving weak acids and strong bases

A

typic al

pH

gure12.

gure8,

•

The

curve

for

Although

there

curve

are

the

the

titration

overall

several

important

intercepts the

dissociates

only

of

partially

a

shape

y-axis

and

weak

of

this

acid

with

curve

is

a

strong

base

is

shown in

somewhat similar to that in

dierences:

at

a

higher

therefore

pH.

This

produces

a

is

bec ause

lower

the

weak acid

concentration of

+

H

•

Buer

solutions

will

be

(aq) ions.

There is a

buer

region

before

the

equivalence point. This is when the

discussed solution

contains

both

components

of

a

weak

conjugate

acid–base

pair.

later in this topic.

•

The

jump

in

•

The

equivalence

undergoes

pH

near

is

the

equivalence

achieved

hydrolysis

and

at

a

point

is

pHgreater

produces OH

smaller

than7,

than

as

the

that

in

gure8.

salt anion

(aq) ions.

14

1

CH

COONa

3

+

NaOH

equivalence point 1 (CH

COONa

3

only)

pH jump

 Hp

buffer region

6

4 CH

COOH

3

+ CH

COONa

3

 intercept

(CH

COOH only)

3







1

1

3

V(NaOH),

cm

3

p Figure12

pH

curve for the titration of 0.1 mol dm

3

0.1 mol dm

568

NaOH(aq) (strong base)

CH

COOH(aq) (weak acid) with

3



Reactivity

nal

c ases

curves

The

part

the

the

curve

in

contains

gure12

excess

is

very

strong

similar

base,

to

that

in

NaOH(aq).

transfer

reactions

gure8, as in both

Therefore, both

atten out at pH≈13.

stage

of

neutralized

is

of

solution

Proton

LHA

The

3.1

the

is

titration

equal to the pK

dissociates

at

which

known as the

as

of

a

the

exactly

one-half

of

the

half-equivalence point.

weak

acid.

In

our

acid

The

example,

the

pH

has

at

been

half-equivalence

weak ethanoic acid

follows:

+

CH

3

COOH(aq)

⇌

CH

3

COO

(aq)

+

H

(aq)

+

[CH K

a

3

COO

][H

]

= [CH

3

COOH]

+

At

half-equivalence,

[CH

3

COO

]

=

[CH

3

COOH], so

K

=

a

[H

] and pK

=

a

pH.

3

Figure12

shows

that

the

equivalence

is

achieved at

V(NaOH)

=

10 cm

, so the

3

half-equivalence

approximately

The

a

pH

curve

mirror

and

the

of

in

a

4.8,

for

image

occur

use

occurs at

the

of

the

which

is

acid

a

of

a

5 cm

strong

except

half

as

=

.

At this point, the solution pH is

very close to the pK

titration

gure12,

second

weak

V(NaOH)

of

that

the

titrant

base

the

curve.

has

no

a

of

with

buer

a

acid

(4.76, table8).

weak acid would be almost

region would be much longer

Titrations

practic al

ethanoic

of

this

type

are uncommon, as

value.

pH curves involving strong acids and weak bases

A

in

typic al

pH

gure13.

following

•

The

curve

The

for

the

overall

titration

shape

of

of

a

this

weak

curve

base

is

with

similar

a

to

strong

that

in

acid

is

shown

gure9, with the

dierences:

curve

intercepts the

basedissociates

ofOH

only

y-axis

partially

at

a

and

lower

thus

pH.

This

produces

is

a

bec ause

lower

the

weak

concentration

(aq) ions.

•

There

is

a

buer

•

The

drop

•

The

equivalence

in

pH

region

near

is

before

the

the

equivalence point.

equivalence

achieved

at

point

pH>pK

their

+1,

+

conjugate

lower thanpK

the

a

(aq)

acid

HInd(aq).

wavelengths,

H

conjugate

forms the conjugate base Ind

dierent

⇌

is

have

dierent

occurs

within

known as the

slightly

predominantly in its

loses

colours

a

a

certain

pH

or

broader

range,

of

the

transition

absorption

compounds

spectra

are

of

chemic al

discussed in

Structure 1.3

(gure15).

transition range

narrower

The

proton and

(aq) absorb visible light at

typic ally

indic ator.

ranges

(table10).

Colour

pH transition pK

Indic ator

a

range

methyl

orange

bromothymol

3.7

blue

phenolphthalein

p Table10

3.1–4.4

Acid

Base

red

yellow

7.0

6.0–7.6

yellow

blue

9.6

8.3–10.0

colourless

pink

Common acid–base indic ators

p Figure15

red

The

universal

overlapping

transition

0

to

14,

indic ator

transition

range,

as

and

is

a

mixture

ranges.

its

As

colour

a

of

several

result,

changes

acid–base

the

universal

gradually

over

and

indic ators with

indic ator

the

has

whole

no

pH

specic

range

from

shown in table1.

indic ators

are

oen

used

in

titration

experiments,

as

they

chemists

to

and

the

monitor

the

reaction

progress

common

when

the

colour

and

transition

acid–base

ranges

indic ators

by observing the solution colour are

titration

colours

allow of

stop

yellow in neutral

alkaline solutions

The

Acid–base

Methyl orange indic ator is

in acidic solutions and

changes.

This

moment,

given

in

section 18 of the data

known as the booklet.

end point

indic ator

of

the

titration, depends on the pK

approximately

a

of

the

indic ator. The pK

a

of the

corresponds to the pH at the end point.

571

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Identifying appropriate indic ators

(Reactivity 3.1.15)

A

suitable

pH

at

and

for

a

the

indic ator

strong

this

range

for

equivalence

type

base,

of

6.0–7.6

satisfactory

base

titration

in

a

should

pH

equivalence

titration

results

acid–strong

the

(table

a

point

is

due

is

However,

to

the

titrations

large

a

transition

When

a

range that includes the

titration

achieved at pH=7 ,

bromothymol

10).

have

curve.

all

the

a

strong acid

best

indic ator

blue, which changes colour within the pH

three

pH

involves

so

common

jump

at

the

indic ators

equivalence

would

point

produce

in

strong

(gure 16).

14

phenolphthalein

Hp

bromothymol

7

methyl

blue

orange

0 V(NaOH)

p Figure16

Titration of

a strong acid

with a strong base using

various indic ators

The

amount

result.

E ach

indic ator

the

you

the

titrations

at

pH>7,

try

involving

so

changes

Methyl

weak

colour,

it

to

to

you

makes

use

the

analysed

solution

Brønsted–Lowry

reacts

error

solution

always

added

a

the

titrant.

introduce.

it

just

with

dicult

enough

acid

On

to

The

the

see

or

c an

more

other

the

aect

the

indic ator

hand,

solution

you add,

adding

colour.

too

little

Therefore,

indic ator to make the colour change

the

colour

orange

a

best

weak

acid

indic ator

at

c annot

pH>7

be

and

for

a

this

strong

type

(gure17).

used,

as

it

of

base,

the

Bromothymol

would

equivalence

produce

a

blue

very

could

large

Hp

bromothymol

blue

methyl orange

0 V(NaOH)

p Figure17

Titration of a weak acid

various indic ators

achieved

also

be

systematic

phenolphthalein

7

is

titration is phenolphthalein, which

14

572

titration

base, so when the

visible.

In

also

is

systematic

to

should

clearly

indic ator

changes

greater

indic ator

of

indic ator

with a strong base using

used.

error.

Reactivity

titrations

at

pH

7,

such

as

phenol

red

(see

section hydroxide,

18 of the data

19.

Identify

an

indic ator

Titrations

involving

acid–base

the

whole

gradually,

point

in

a

weak

indic ators.

In

experiment

and

NaOH(aq).

booklet).

the

titrations

end

of

acid

such

and

a

weak

titrations,

the

base

c annot

change

in

be

pH

is

performed using

very

gradual during

(gure14), so the colour of the solution will also change

point

this

type

will

c an

be

impossible

only

be

to

determine.

determined

using

a

The

pH

equivalence

meter.

titration

appropriate

for an acid–base

that

ammonium

NH

4

Br(aq),

point.

produces

bromide,

at

the

equivalence

Refer to table10 and

section 18 of the data booklet.

573

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Buer solutions (Reactivity 3.1.16)

In

many

chemic al

constant

These

pH

of

and

the

solutions

biologic al

reaction

resist

experiments, it is important to maintain a

mixture.

changes

in

This

pH

c an

when

be

achieved using

small

amounts

of

buer solutions.

acids

or

bases

are

added to the solution.

E ach

in

buer

which

buer,

it

is

it

solution

both

is

the

contains

acid

neutralized

neutralized

by

the

buers

are

and

by

both

the

the

weak

components

base

weak

acid.

are

weak.

base.

As

a

of

Similarly,

result,

a

conjugate

When

the

a

strong

when

a

solution

acid–base

acid

strong

pH

is

pair,

added to a

base

is

added,

remains almost

unchanged.

Acid–base

extremely

ecient

in

maintaining

−

3

we

add

water,

a

single

the

pH

drop

of

the

(about

0.05 cm

resulting

)

solution

of

will

drop

by

their

pH.

For

example, if

3

0.1 mol dm

3

HCl(aq)

2.7,

from

to

7.0

100 cm

to

of

pure

approximately

3

4.3.

However,

phosphate

solution

will

laboratory

Buer

For

if

add

decrease

pH

are

the

simple

salt,

units

buer

sodium

following

by

very

pH

by

of

same

quantity

used

less

in

than

3

When

the

3

buer

ions,

solution

ethanoate.

a

⇌

COONa(aq)

strong

c an

be

is

kept

systems

c arbon

When

CH

acid,

conjugate

→

3

is

below

prepared

two

of

all

within

that

dioxide

these

COO

CH

such

base

when

neutralized

by

a

3

as

of

a

all

buer

c ases,

buer ’s

acid

574

living

involve

and

from

organisms.

narrow

range of

hydrogenc arbonate and

proteins.

the

weak ethanoic acid and its

compounds

are

dissolved

+

H

in

water, the

(aq)

COO

+

(aq)

HCl(aq),

the

+

is

buer

Na

(aq)

added

to

this

solution,

+

CH

3

COO

(aq)

→

CH

3

a

weak

weak

acid

conjugate base

conjugate acid

weak

base,

such

conjugate

as

NaOH(aq),

acid

of

the

is

by

system:

−

+

CH

3

COOH(aq)

→

CH

3

COO

(aq)

+

weak

weak

conjugate base

reactions

the

action

releases

neutralized

added to this solution, it is

buer

conjugate acid

strong

is

COOH(aq)

base

of

it

system:

strong

buer

is

a

the

reacts

acid,

the

of

strong

system

always

acid

conjugate

and

typic al

detection limit of most

strong

(aq)

neutralization

components

the

the

very

−

of

a

–

(aq)

strong

the

OH

The

of

+

(aq)

+

H

Similarly,

100 cm

processes take place:

COOH(aq)

weak

to

experiments, the pH of the

components

blood

−

CH

HCl(aq)

This

−

CH

of

biologic al

0.001.

important

human

several

hydrogenphosphate

A

the

meters.

solutions

example,

7.35–7.45

we

buer (pH≈7 .0)

a

acid

and

a

strong

known as the

same,

with

while

buer ’s

are

the

regardless

buer ’ s

strong

base

conjugate

base.

base

2

O(l)

by

buer action.

of

the

with

dierent

The

nature

buer components. In

conjugate

reacts

H

base

the

and

releases the

buer ’s conjugate

Reactivity

other

words,

replaced

extent,

An

with

they

are

usually

from

weak

be

a

used

weak

acid CH

is

replaced

Since

weak

with

acids

a

weak

and

acid,

bases

and

a

reactions

strong base is

dissociate only to a small

eect on the solution pH.

c an work eciently only if both components of its conjugate

in

compounds.

but

their

3

little

present

base

or

not

solutions

for

very

buer

acid

base.

transfer

the

solution

in

sucient

concentrations.

This

c an

achieved when the conjugate acid and the conjugate base originate

conjugate,

Buer

strong

weak

and

dierent

acid,

a

have

acid–base

pair

a

Proton

LHA

In

3.1

a

both

are

A

salt,

at

oen

solution

may

the

same

classied

preparation.

For

of

contain

a

a

single

high

compound,

such

as

a

weak

concentration of only one

time.

according

example,

COOH(aq) and its anion CH

3

the

to

the

acid

ethanoate

COO

(aq),

and

(or

base)

buer

the

and

the

salt

contains

the

weak

ammonia

buer

+

contains

as CH

3

buers

salt”,

the

weak base NH

COONa(s) or NH

are

sometimes

respectively.

4

3

(aq)

Cl(s),

c alled

Common

and

were

“a

its

used

weak

types

Type

c ation NH

of

for

acid

(aq).

4

preparing

and

its

acid–base

salt”

specic

the

and

buers

Example

If

salts, such

solutions,

“a

are

the

same

weak base and its

listed in table11.

Conjugate acid

pK

Conjugate base

a

weak acid ethanoate

buer

CH

3

COOH(aq)

CH

3

COO

(aq)

4.76

and its anion

weak base +

ammonia and

its

buer

NH

(aq)

4

NH

3

(aq)

9.25

c ation

anions of two

phosphate 2

H acid

anions

of

salts

an

2

PO

4

(aq)

HPO

4

(aq)

7.20

buer

acid

salt

c arbonate 2

HCO and

a

p Table11

In

a

normal

salt

3

(aq)

CO

3

(aq)

10.32

buer

Common acid–base buers

laboratory,

buer

solutions

c an

be

prepared

by

various

methods.

For

Practice question example,

the

ethanoate

buer

c an

be

made

as

follows:

20. 1.

solid

sodium

ethanoate

and

liquid

ethanoic

acid

are

dissolved

in

Explain,

the

2.

solutions

3.

excess

ethanoic

of

sodium

4.

excess

sodium

ethanoate

and

ethanoic

acid

are

using

ionic

equations,

water

mixed together

buer action of a solution

containing

ammonium

c ations,

+

NH acid

is

mixed

with

sodium

NH

ethanoate

solution

is

mixed

4

(aq), and ammonia,

hydroxide

with

3

(aq).

hydrochloric acid.

Regardless of the method, the resulting buer will contain the same two

components, the weak acid CH

In

method

CH

CH

In

3

3

3

the

conjugate

+

NaOH(aq)

COOH(aq)

+

OH

4,

the

COONa(aq)

(aq)

+

→

→

conjugate

HCl(aq)

COOH(aq) and its conjugate base CH

3

base

COOH(aq)

method

CH

3,

will

CH

CH

acid

→

3

3

3

as

(aq)

form

as

COO

3

(aq).

follows:

COONa(aq)

COO

will

CH

form

+

H

+

2

H

2

O(l)

O(l)

molecular equation

net

ionic

equation

follows:

COOH(aq)

+

NaCl(aq)

molecular equation

+

CH

3

COO

(aq)

+

H

(aq)

→

CH

3

COOH(aq)

net

ionic

equation

575

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

If

Activity

the

of

Suggest

three

preparation

of

buer

solutions

prepared

by

dierent

methods

contain

the

same

concentrations of both the conjugate acid and the conjugate base, the properties

methods

the

these

buers

will

also

be

the

same.

for the

ammonia

buer.

pH

The

of buer solutions (Reactivity 3.1.17)

pH

of

a

conjugate

The

buer

solution

acid–base

acid–base

pair

equilibrium

its conjugate base A

depends

and

(aq)

in

is

the

a

on

the

ratio of the components in the

dissociation

buer

solution

characterized

constant

of

containing

by the

K

a

of

the

the

a

weak

conjugate.

weak acid HA(aq) and

weak acid:

+

[H

+

HA(aq)

⇌

H

(aq)

+ A

(aq)

K

][A

]

=

a

[HA]

+

To

as

nd

the

concentration of H

(aq)

ions,

we

need

to

rearrange the

K

a

expression

follows:

[HA]

+

[H

]

=

K

a

× ]

[A

We

c an

then

take

the

logarithm

log[H

]

=

logK

a

+

a

every factor to determine the pH:

log [A

Add

of

[HA]

+

minus

sign

to

every

]

term

+

−log[H

in

the

equation:

[HA] ]

= −logK

−

a

log ]

[A

+

Finally, subsitute in pH

= −log[H

x rule that

−log

= log

pK

+

a

= −logK

a

, and use the

: x

[A =

a

y

y

pH

] and pK

]

log [HA]

The

to

Practice questions

21.

C alculate

the

pH

for

the

equation,

c alculate

c an

in

last

be

the

the

used

buer

to

known as the

pH

of

nd

any

the

solution

at

buer

ratio

a

of

Henderson–Hasselbalch

solution

the

of

known

equation,

composition.

c an

be

used

Alternatively, it

concentrations of the conjugate acid and base

known pH.

buer

solutions containing the An

important

consequence

from

the

Henderson–Hasselbalch

equation is that

following compounds: the

3

a.

a

0.50 mol dm

pH

of

certain

a

buer

factor, both [A

[A NaH

2

PO

4

(aq) and

solution

]

is

not

and

aected

[HA]

by

dilution.

decrease

by

the

If

the

same

solution

factor.

is

diluted

by

Therefore, the

]

value of log

remains

unchanged,

and

so

does

the

solution

pH.

However, a

[HA]

3

0.20 mol dm buer Na

2

HPO

4

solution

c annot

be

diluted innitely without changing its pH. An innitely

(aq) diluted

aqueous

solution

of

any

substance

at

298 K

will

have

a

pH

of

7.00.

3

b.

0.25 mol dm

NH

3

(aq)

3

and

The

also NH

ability

4

limited

are

a

values

given

are

buer

to

resist changes in pH on addition of acids and bases is

the

nite.

At

amounts

the

point

ceases

to

function,

table10. addition

576

buer

of

when

the

weak conjugate acid and base in the

either

of

the

weak

conjugates

is

used

up,

for conjugate acids the

in

a

bec ause

Cl(aq).

solution

The pK

of

0.50 mol dm

of

an

acid

or

base.

and

the

solution

pH

changes

signic antly on further

Reactivity

3.1

Proton

transfer

reactions

LHA

Worked example 10

3

C alculate the pH

CH

3

of an ethanoate buer containing 0.100 mol dm

3

CH

3

COOH(aq) and

0.200 mol dm

COONa(aq).

Solution

The conjugate acid is CH

3

COOH(aq) and the conjugate base is CH

3

COO

(aq),

which

is

produced

by

the

dissociation of

sodium ethanoate:

+

CH

3

COONa(aq)

→

CH

3

COO

(aq)

+

Na

(aq)

3

All

salts

are

strong

electrolytes,

According to table8, pK

a

(CH

3

so

sodium

COOH)

=

ethanoate

dissociates

completely,

and

[CH

3

COO

]

=

0.200 mol dm

4.76, so:

0.200 pH

=

4.76

+

log

=

5.06

0.100

Worked example 11

A buer solution with a pH

of 11.00 was prepared

by the reaction of methylamine, CH

3

NH

2

(aq),

with hydrochloric

acid, HCl(aq).

a.

Identify the conjugate acid–base pair in this buer and

b.

State,

c.

C alculate the mole ratio of the conjugate acid to the conjugate base in the solution.

with a reason,

which of

state the role of

each species.

the two reactants was in excess.

Solution

a.

The

reaction

between

CH

NH

(aq)

+

HCl(aq)

(aq)

+

H

3

2

hydrochloric

→

CH

3

NH

+

CH

3

The

NH

net

2

ionic

3

acid

and

methylamine

Cl(aq)

proceeds

molecular

as

follows:

equation

+

(aq)

→

equation

CH

3

NH

involves

(aq)

3

two

net

species

that

dier

ionic

by

a

equation

single

proton.

Therefore, the conjugate acid is

+

methylammonium, CH

3

NH

(aq),

3

and

the

conjugate

base

is

methylamine, CH

3

NH

2

(aq).

+

Note

that

the

solution

contains

two

more

conjugate

acid–base

pairs,

H

3

O

(aq)/H

2

O(l)

and

H

2

O(l)/OH

(aq).

+

However,

b.

A

none

of

these

buer must contain

methylamine

Therefore,

(weak

pairs

both

c an

a

components

conjugate

methylamine

form

was

base)

buer,

of

the

would

be

as

H

3

O

conjugate

(aq)

is

a

strong

acid–base

consumed,

and

the

pair.

acid,

If

while

OH

hydrochloric

solution

could

not

(aq)

acid

act

as

is

a

strongbase.

were

an

in

excess, all

acid–base

buer.

inexcess.

+

c.

According to table8, pK

[A pH

=

pK

a

+

b

(CH

3

NH

2

)

=

3.34,

so pK

a

(CH

3

NH

3

)

=

14

−

3.34

=

10.66.

Substituting pK

a

and pH into

]

log

gives: [HA]

[CH 11.00

=

10.66

+

3

NH

2

]

log +

[CH

3

NH

[CH

3

]

3

NH

2

]

Rearrange in terms of log

: +

[CH

[CH

3

NH

2

3

NH

3

]

]

log

=

0.34

+

[CH

Simplify

the

[CH

3

3

NH

3

]

expression

NH

2

]

by

making

each

term

the

exponential of 10:

0.34

=

10

=

2.19

+

[CH

3

NH

3

]

+

Therefore,

[CH

3

NH

3

]:[CH

3

NH

2

]

=

1 : 2.19. Note that

n

=

c

×

V,

so

the

mole

ratio

is

equal

to

the

ratio

of

concentrations.

577

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

End of topic questions

Extended-response questions

Topic review

5. 1.

Using

your

knowledge

from the

Reactivity 3.1

Identify

for answer

the

guiding

question

as

fully

as

the

chemic al

formulas

of

parent acids and bases

topic, the

following

salts:

possible:

a.

C a(NO

b.

Fe

c.

NH

What happens when protons are transferred?

2

(SO

3

4

)

)

2

3

Exam-style questions 4

HCO

3

Multiple-choice questions 6.

2.

A solution of acid

acidY

has

a

pH

A

[X] >

[Y]

B

Acid

X

X

of

has a

3.

pH of 1 while a

Which

statement

is

solution of

stronger than acid

each

one

correct?

Two

salt

from

molecular

parent

7.

is

For

acid

question

equation

and

unlabelled

5,

that

formulate and balance

produces

that

salt

from the

base.

bottles

contain

solutions

of

a

strong

Y monoprotic

acid,

HX(aq),

and

a

weak

monoprotic acid,

+

C

[H

] in the solution of

X

] in the solution of

Y

is

three times higher than

X

is

HY(aq),

of

the

same

concentration.

+

[H

a.

D

[OH

] in the solution of

lower

than

[OH

Suggest

how

these

acids

c an

be

distinguished

] in the using

solid

c alcium

c arbonate,

C aCO

(s). 3

solution of

Y b.

3.

Which

pair

of

Formulate

reaction

species is a conjugate acid–base pair?

molecular

of

each

and

acid

net

with

ionic

c alcium

equations

for the

c arbonate.

+

A

H

B

H

and

C

HCl

B

H

OH 8.

Deduce

the

formulas

of

conjugate

acids

and

bases

for

+

3

O

and H

and

2

O

NaOH

each

species

lled

for

you

listed

as

in

table12.

The

rst

two

rows

are

examples.

2–

2

CO

3

and

CO

3

Species

LHA

4.

Which

indic ators

ammonia, NH

3

c an

be

used

for

the

titration of

(aq), with sulfuric acid, H

2

SO

4

2

O

(aq)?

H

Bromocresol

green (pK

a

=

3

O

HCl

4.7) HF

II

Methyl

red (pK

III

Phenol

red (pK

a

=

5.1)

=

7.9)

NH

a

(CH

A

I and II only

B

I and III only

3

3

)

3

3

2

CO

578

−

3

II and III only

p Table12

D

N

−

HCO

C

I, II and III

Conjugate base

−

OH

−

Cl

I

Conjugate acid

+

H

Conjugate acids and bases

does

not

exist

Reactivity

9.

2-Amino

acids

exist as

zwitterions,

which

have

both

a

positive

and

a

negative

+

within

the

Proton

same

transfer

reactions

species:

+

H

+

charge

3.1

H

+

–

H

COOH

3

H

COO

3

H

+

+

H

+

R

The

K

w

conjugate

value

at

10 °C

acid–base

is

3.47

R

zwitterion

pairs

times

involved

in

these

lower than that at

anion

equilibria

16.

25 °C.

and

The pK

ion, H

a.

C alculate

b.

Discuss

the

pH

whether

of

pure

pure

water

water

at

at

10 °C.

a.

2

and pK

PO

role

b

(aq),

4

of

each

values

are

for

7.20

the

acid–base

equilibria

species.

the

and

equations

dihydrogenphosphate

11.88,

that

respectively.

represent the

characterized

by

K

10 °C is acidic, basic

the

a

and

K

b

dihydrogenphosphate ion.

neutral.

b.

LHA

11.

a

the

Formulate

of or

state

LHA

10.

two

H

R

c ation

Identify

COO

2

+

C alculate

the

pH

for

the

C alculate the pK

acid, H

0.015 mol dm

b.

0.010 mol dm

3

PO

4

value

for phosphoric

(aq), and the pK

b

value

for the

2

3

a.

a

following solutions:

HNO

hydrogenphosphate

(aq)

ion,

HPO

(aq).

4

3

3

H

SO 2

(aq) 4

17.

Formulate

the

equations

for

acid–base

equilibria in

3

c.

0.020 mol dm

KOH(aq)

aqueous

3

12.

A

0.100 dm

3

sample

of

0.020 mol dm

KOH(aq)

was

solutions

of

the

a.

sodium

b.

potassium

c.

ammonium

d.

trimethylammonium

following

methanoate,

salts:

HCOONa

3

mixed

the

with

nal

0.900 dm

solution.

of

water.

Assume

that

C alculate the pH of

solution

volumes

iodide, KI

are cyanide, NH

4

CN

additive.

13.

chloride,

(CH

3

)

3

NHCl.

C alculate: For

each

solution,

predict

whether

it

will

be

neutral,

3

a.

the

b.

the

pOH

of

hydroxide,

a

0.015 mol dm

acidic or basic.

solution of sodium

NaOH(aq)

The pK concentration

of

a

and pK

values

b

for

weak

acids

and

bases

are

given

hydroxide ions in a solution in table8.

with

pOH

=

9.50.

18.

C alculate

the

pH

values

for

buer solutions containing

3

14.

A

C

0.020 mol dm

6

H

5

COOH(aq),

solution of benzoic acid,

has

a

pH

of

the

following compounds:

2.95. 3

a.

a.

Determine the pK

b.

Using the pK

a

a

of benzoic acid.

value

from part

a,

0.25 mol dm

3

HCOOH(aq)

and

0.50moldm

HCOONa(aq);

3

c alculate the pH of

b.

0.50 mol dm

3

CH

3

NH

2

(aq)

and

0.20moldm

3

a

15.

0.10 mol dm

C alculate

the

pH

solution of benzoic acid.

values

for

the

following solutions:

CH

19.

0.010 mol dm

b.

2.0 × 10

Explain,

NH

methylamine, CH

3

3

NH

2

3

Cl(aq).

using

solutions

3

a.

3

from

ionic

the

equations,

the

buer action of the

previous question.

(aq)

3

mol dm

ethanoic acid, CH

3

COOH(aq). 20.

Refer to table8.

Using

table10, identify the most suitable acid–base

indic ator

a.

for

the

titration of:

methylamine, CH

3

NH

2

(aq),

with

hydrochloric acid,

HCl(aq);

b.

hydrogen

cyanide,

hydroxide,

HCN(aq),

with

potassium

KOH(aq).

579

Reactivity 3.2

Electron transfer reactions

What happens when electrons are transferred?

In

a

will

reaction

lose

dierent

species

reduced.

without

to as

where

electrons

electrons — the

It

is

will

redox

gain

of

transferred,

to

have

another.

one

species

oxidized — and a

electrons — the

impossible

reduction

are

species is

These

If

species is

oxidation

of

one

reactions

are

Some

the

made

species

referred

reactions.

cells

be

redox

to

or

transfer

occur

to

the

power

spontaneous

electroplating,

to

are

a

one

wire,

energy

the

require

and

exothermic.

species to another is

such

as

released

appliances.

and

or

spontaneous

from

through

batteries,

used

not

reactions

electron

Other

of

electrochemic al

the

redox

energy

reduction

in

in

to

process

c an

reactions

are

occur, such as in

aluminium

oxide, Al

2

O

3

,

produce aluminium metal.

Understandings

Reactivity 3.2.1 — Oxidation

described

oxidation

in

terms

state,

of

and

electron

oxygen

reduction

Reactivity 3.2.10 — Functional

c an be

compounds

transfer, change in

gain/loss

or

hydrogen

may

undergo

of

oxidation

and

reduction,

separate

showing

the

the

in

organic

loss/gain.

Reactivity 3.2.11 — Reduction

Reactivity 3.2.2 — Half-equations

groups

reduction.

processes

loss or gain of

by

the

addition

of

hydrogen

of

unsaturated compounds

lowers

the

degree of

unsaturation.

electrons.

relative

ease

of

oxidation and +

reduction

its

of

position

an

in

element

the

in

periodic

a

group

table.

c an

The

be

predicted

reactions

hydrogen half-cell H

(aq)

1 e

from

between

H

⇌

2

(g)

is

assigned

a

standard

electrode

2

potential

of

zero

by

convention.

It

is

used in the

⦵

metals

ease

and

of

aqueous

oxidation

of

metal

ions

demonstrate

the

relative

measurement

of

standard

electrode potential,

E

dierent metals. Reactivity 3.2.13 — Standard cell potential,

⦵

Reactivity 3.2.4 — Acids

react

with

reactive metals to

E

cell

,

c an

be

c alculated

from

standard

electrode

⦵

release

potentials.

hydrogen.

E

cell

spontaneous

Reactivity 3.2.5 — Oxidation

has

a

positive

value

for a

reaction.

occurs at the anode and ⦵

reduction

occurs

at

the

Reactivity 3.2.6 — A

electrochemic al

spontaneous

cell

redox

c athode

primary

that

in

electrochemic al cells.

(voltaic) cell is an

converts

reactions

to

energy

electric al

from

Reactivity 3.2.14 — The

shows

the

Gibbs

energy

a

relationship

and

equation

between

standard

redox

reactions

that

c an

be

using

cells

involve

electric al

energy.

electrochemic al

chemic al

cell

energy

by

electrolytic cell is an

that

converts

bringing

electric al

about

Reactivity 3.2.9 — Functional

compounds

may

undergo

non-spontaneous

groups

oxidation.

in

potential

anode

competing

and

c athode,

of

electrolysis

reactions

for

including

organic

electrolytic

thinlayer.

of

aqueous

c an occur at the

the

oxidation and

water.

Reactivity 3.2.16 — Electroplating

energy to

reactions.

580

solutions,

reduction

Reactivity 3.2.8 — An

cell

energy.

(rechargeable)

reversed

⦵

= −nFE

reaction.

Reactivity 3.2.15 — During

Reactivity 3.2.7 — Secondary

cell

ΔG

standard change in

coating

of

an

involves the

object with a metallic

LHA

+

Reactivity 3.2.12 — The

Reactivity 3.2.3 — The

Reactivity

3.2

Electron

transfer

reactions

Oxidation and reduction (Reactivity 3.2.1) TOK

Reduction

and

oxidation

c an

be

dened

in

several

ways:

In

1.

in

terms

of

the

loss

and

gain

of

chemistry,

c an

2.

in

terms

of

the

gain

and

3.

in

terms

of

electron

loss

of

be

terms

of

oxidation

state.

on

example, in

three

to

the

rst

denition,

oxygen.

of

metals

these

oxidation

Examples

of

this

is

a

type

reaction

of

4Fe(s)

During

and

6

aerobic

H

12

This

O

According

removed

the

O

2

+ 3O

water.

C

+

6

(g)

to

2

form

metal

to

(g)

reduction

→

2Fe

2

O

3

oxygen

a

of

also

6O

rst

a

Lewis

dierent

theories

are

aspect of

species’

chemistry.

What other

reaction include the in

chemistry

in

dierent

c an be

ways?

(s)

described

the

by

of

2MgO(s)

respiration,

+

E ach

oxides:

c an

(aq)

from

→

you

where a substance

oxidation

dened

2Mg(s)

Reactivity 3.1,

and

concepts combustion

by

separate denitions of

theories.

a with

ways

knowledge.

Brønsted–Lowry,

informed

combines

several

dierent

acids and bases: the Arrhenius,

1. Oxidation and reduction in terms of oxygen gain/loss

According

same concept

in

transfer met

in

dened

drawing

hydrogen

For

4.

the

oxygen

2

be

(g)

→

6CO

denition,

substance.

nickel(II)

2

reacts

(g)

+

with

an

6H

reduction

Examples

oxide

as

by

of

2

is

this

c arbon

glucose

oxidation

to

to

form

c arbon

dioxide

reaction:

O(l)

a

reaction

type

of

where

oxygen is

reduction

produce

pure

reaction include

nickel

and

c arbon

monoxide:

NiO(s)

In

all

+

redox

C(s)

→

Ni(s)

reactions,

+

one

CO(g)

species

is

oxidized,

and

another

is

reduced. If the

p Figure 1

substance

being

oxidized

gains

oxygen,

then

the

substance

being

iron(III)

oxygen.

In

oxygen,

and

the

CuO(s)

experiment

hydrogen

+

H

2

(g)

→

in

gas

gure

is

Cu(s)

2,

being

+

H

2

copper(II)

oxidized,

oxide

gaining

is

being

In

the

ox i d a t i o n

of

iron,

reduced loses ox i d e,

or

rust,

is

produce d

reduced, losing

oxygen:

O(g)

black copper(II) oxide hydrogen in

heat

p Figure 2

In

some

the

c ases,

same

chlorate,

this

while

one

reaction.

KClO

4KClO

In

Experimental

3

(s)

3

substance

This

is

3KClO

some

formula

4

(s)

+

formula

units

for

c an

the

be

known as

decomposes

→

reaction,

other

,

set-up

are

on

re duction

of

copper(II)

simultaneously

reduced

disproportionation.

heating

as

ox i d e

For

by

and

hy d r o g e n

oxidized in

example,

potassium

follows:

KCl(s)

units

of

oxidized

KClO

to

3

are

KClO

4

reduced

by

to

gaining

KCl

by

losing

oxygen

oxygen.

581

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

2. Oxidation and reduction in terms of hydrogen loss/gain

Oxidation

reaction

c an

hydrogen

to

4HCl(g)

Reduction

example

C

The

process

c atalysis

is

of

heterogeneous

detailed in

This

2

also

between

H

4

+

O

+

reaction

2

(g)

be

such

(g)

considered

as

a

chloride

loss

and

of

hydrogen.

oxygen,

For

example, in the

hydrogen chloride loses

form chlorine gas:

c an

of

be

hydrogen

H

2Cl

2

(g)

considered

a

2

→

reaction

(g)

→

typic ally

C

2

is

H

6

+

as

2H

the

the

2

O(g)

addition

of

hydrogen

to

a

species. An

hydrogenation of ethene:

(g)

requires

Ni(s)

as

a

heterogeneous

c atalyst.

Structure 3.1

3. Oxidation and reduction in terms of electron transfer

During

a

reaction,

electrons,

OIL:

RIG:

it

the

instead

as

2

(g)

Sodium

→

the

of

Gain

electrons,

mnemonic

it

for

is

oxidized,

and

if

a

remembering this is

species gains

OIL RIG:

electrons

of

electrons

between sodium metal and chlorine gas:

2NaCl(s)

be

described

are

not

oxidation

sodium

loses

loses

useful

elements

between

metal

A

Loss

Is

c annot

these

describe

electrons

species

reaction

+ Cl

reaction

a

Is

Reduction

2Na(s)

oxygen,

if

reduced.

Oxidation

Consider

This

is

and

and

electrons

in

terms

of

involved

in

reduction

the

gain

the

or

loss

reaction.

occurring

in

of

hydrogen and

However,

terms

of

we

the

c an

transfer of

chlorine.

to

form

sodium

c ations,

so

it

is

oxidized.

+

2Na(s)

These

Cl

The

→

2Na

electrons

2

(g)

+

sodium

2e

+

are

→

transferred

to

chlorine

gas,

reducing it to chloride anions:

2Cl

c ations

three-dimensional

2e

and

chloride

lattice

anions

structure,

are

NaCl(s)

held

together

by ionic bonds in a

(gure 3).

+

The

formation

structures

is

of

ionic

Na

Cl

Na

2,8,1

2,8,7

2,8

Cl

2,8,8

+

lattice

NaCl (Na

Cl

−

)

detailed in

p Figure 3

Sodium

atoms

are

ox i d i z e d

the

formation

(lose

e l e c t r o n s)

and

Structure2.1 re duce d

582

−

(g a i n

e l e c t r o n s)

in

of

sodium

chloride

chlorine

atoms

are

Reactivity

3.2

Electron

transfer

reactions

C ase study: Redox reactions in optometry

Optometrists

darken

a

in

redox

the

oen

prescribe

presence

of

glasses

ultraviolet

with

light

photochromic lenses. These lenses

(from

sunlight);

this

change

is

based on

reaction.

Ordinary

glass

copper(I)

oxidized

is

composed

chloride,

to

CuCl,

chlorine

of

and

atoms

silic ates

silver

on

while

photochromic lenses contain

chloride,

exposure

to

AgCl.

The

chloride

ions

are

ultraviolet light (hf ).

hf

Cl

→

Electron

Cl

+

e

transfer

then

takes

place,

reducing

the

silver

c ations

to

silver atoms.

+

Ag

The

+

silver

e

light,

Ag

atoms

darkening

become

→

turn

process

transparent

the

following

2

+

Cl

→

lenses

reversed

again.

dark,

by

inhibiting

copper(I)

When

the

the

transmission of light. The

chloride

lenses

are

no

allowing the lenses to

longer

exposed

to

ultraviolet

reaction takes place:

+

Cu

the

is

+

Cu

+ Cl

+

The

chlorine

atoms

formed

by

the

exposure

+

2

In turn, Cu

ions

are

to

light

are

+

2

oxidized to Cu

reduced

by the Cu

ions.

+

ions. These Cu

ions

then

oxidize

silver

+

atoms

to

2

Ag

Cu

As

a

ions:

+

+

+

result,

Ag

the

→

Cu

+

+

lenses

Ag

become

transparent

again

and

the

silver and chlorine

+

atoms

return

to

the

initial

species,

Ag

and Cl

4. Oxidation and reduction in terms of oxidation state

change

Some

redox

example,

c arbon

in

and

reactions

the

do

formation

sulfur

atoms,

not

of

involve

c arbon

forming

a

a

transfer

disulde,

covalently

of

electrons

electrons

bonded

are

between

shared

species.

For

between the

molecule:

p Figure 4

C(s)

C arbon

c an

state

the

2S(s)

of

describe

the

charge

so

CS

has

2

P hotochromic

lenses

(l)

the

following

Lewis

structure:

S

composed

ionic,

→

disulde

C

S

We

+

all

oxidation

atoms

that

of

in

an

atom

ions.

shared

the

In

and

reduction

reacting

would

other

have

words,

electrons

in

in

terms

species. The

in

all

each

a

of

compound

polar

bond

if

covalent

are

the

change

oxidation state

the

oxidation

compound

bonds

formally

in

represents

are

were

treated as

transferred

to

the

more

electronegative atom. You

learned

assigning A

compound

is

oxidized

if

the

and

reduced

if

the

rules

for

oxidation state of an atom in that compound

atoms increases,

the

oxidation states to

in

covalent compounds in

oxidation state of an atom in that compound

Structure3.1. decreases.

583

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

In

0

the

in

c arbon

the

electrons

with

Elemental

state

of

disulde

reactant,

the

sulfur

each

example,

C(s), to

more

has

sulfur

an

+4

in

the

the

oxidation

electronegative

oxidation

atom

state

product, CS

state

sulfur.

of

decreases to

0.

−2.

In

2

(l),

of

c arbon

where

C arbon

c arbon

Sulfur

is

is

increases

c arbon

shares

therefore

disulde,

therefore

the

from

four

oxidized.

oxidation

reduced.

TOK

The

are

underlying

transferred

not

real

bec ause

a

molecule.

If

you

study

Formal

shared

of

both

do

these

not

course,

are

also

we

simplifying

that

shared.

represent

you

the

assigning

not

learned

articial

charges,

tools

when

atoms,

between

useful

formulas,

they

HL

formal

equally

spite

are

the

charges

assigning

In

assumption

between

and

not

that

us

to

is

that

oxidation

electrons

states

are

charges on the atoms in

formal

charges in

represent

the

states

actual

electrons

in

Structure 2.2.

charges.

covalent

When

bonds

are

involved.

assumptions,

help

actual

about

do

assume

atoms

the

oxidation

However,

oxidation

explain

redox

states

and

reactions

formal

and

charges

Lewis

respectively.

Worked example 1

Sodium

chloride,

4NaCl(s)

+ 2H

2

sulfuric acid,

SO

4

(aq)

and

+ MnO

2

manganese(IV) oxide react

(s) →

2Na

2

SO

4

(aq)

+ MnCl

a.

Deduce the change in oxidation states for each atom.

b.

State which atom

c.

Identify the oxidizing agent and

is oxidized,

and

which atom

2

according to the following chemic al equation.

(aq)

+ 2H

2

O(l)

+ Cl

2

(g)

is reduced.

the reducing agent.

Solution

a.

First,

review

the

species

on

each

side

of

the

On the le-hand side

In

NaCl,

In H

2

SO

In MnO

Na:

4

2

, H:

+1, Cl:

+1,

, Mn:

S:

+4,

O:

In

O:

−2

Na

the

oxidation

chlorine,

b.

The

oxidation

chlorine

c.

As

is

NaCl(s),

states

the

−2

In H

state

stay

the

oxidation

of

assign

oxidation

states

to

each atom:

SO

2

same,

state

manganese

2

4

,

O, H:

2

Na:

, Mn:

+1,

S:

+2, Cl:

+1,

O:

+6,

O:

−2

−1

−2

, Cl: 0

except

changes

for

manganese,

from

decreases,

so

−1

to

where

the

oxidation

state

changes

from

+4 to

+2, and

0.

manganese

is

reduced.

The

oxidation

state

of

chlorine

increases, so

oxidized.

manganese

oxidizing

584

where

2

In MnCl

In Cl

All

and

On the right-hand side

−1

+6,

equation

has

agent.

the

been

reduced

Chlorine

has

reducing agent.

and

been

c aused

oxidized

chlorine

and

to

c aused

be

oxidized,

manganese

manganese(IV)

to

be

reduced.

oxide, MnO

That

makes

2

(s), is the

sodium

chloride,

Reactivity

3.2

Electron

transfer

reactions

Worked example 2

Consider the following balanced

Fe(s)

a.

+

2HBr(aq)

→

FeBr

equation:

2

(aq)

Deduce the oxidation states of

+ H

2

(g)

iron and

hydrogen in the reactants and

products.

b.

State which species is oxidized,

and

c.

Identify the oxidizing agent

the reducing agent.

and

which species is reduced.

Solution

a.

On the le-hand side

On the right-hand side

Fe: 0

H

in

Br

Fe

HBr:

in

+1

HBr:

in

FeBr

2

H in H

Br

−1

in

2

:

+2

: 0

FeBr

:

−1

2

b.

The

oxidation

state

c.

The

of

H

state

of

decreases,

oxidizing

agent

Fe

so

is

increases,

HBr(aq)

HBr(aq)

is

and

so

Fe(s)

is

oxidized.

The

oxidation

reduced.

the

reducing

agent

is

Fe(s)

An oxidizing agent causes another species to be oxidized, with the oxidizing The

use

of

Roman

numerals to

agent itself being reduced in the process. A reducing agent causes another represent

oxidation states of

species to be reduced, with the reducing agent itself being oxidized in the process. oxyacids

and

compounds

transition element

was

covered in

Practice questions Structure 3.1.

1.

Consider

Cl

2

(aq)

a.

+

Deduce

and

2.

the

the

→

balanced

2KCl(aq)

oxidation

+ I

2

states

equation:

(aq)

of

chlorine

and

iodine

in

the

reactants

products.

b.

State

c.

Identify

Identify

following

2KI(aq)

which

the

the

element

is

oxidizing

oxidizing

oxidized,

agent

agents

and

and

and

the

the

which

element

is

reduced.

reducing agent.

reducing

agents

in

the

following

reactions:

a.

+

b.

2CuO(s)

c.

Mg(s)

d.

2KI(aq)

When

not

PbO(s)

writing

aer

it.

In

H

+

+ Cl

+

2

(g)

C(s)

2

(g)

Br

2

→

→

(aq)

oxidation

worked

→

Pb(s)

CO

2

(g)

MgCl

→ I

2

states,

example

+

2

2

+

O(l)

2Cu(s)

(s)

(aq)

the

2,

H

+

2KBr(aq)

sign

the

is

always

oxidation

placed

state

of

before the number and

hydrogen

in

HBr is

+1 and

not1+.

Remember

individual

and

to

describe the

atoms

reducing

in

the

agents.

compounds

compounds.

In

worked

as

This

being

also

example

2,

oxidized

applies

HBr(aq)

to

is

and

reduced, not the

describing

the

oxidizing

reduced and is the

oxidizing agent, not H.

585

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Half-equations (Reactivity 3.2.2)

Consider

previous

the

reaction

between

2Na(s)

+ Cl

You

saw

two

equations:

that

2

the

processes

→

These

(g)

+

+

2e

it

writing

chlorine

gas,

discussed in the

easier

redox

1.

Identify

2.

Separate

oxidation

and

reduction

could

be

separated into

e

2Cl

are

helps

to

c alled

to

balance

the

in

species

the

half-equations.

show

reactions

the

of

−

→

equations

half-equations

make

and

−

Na

−

2

metal

(g) → 2NaCl(s)

+

Na(s)

Cl

sodium

section:

the

full

transfer

equation

aqueous

being

equation

an

Separating

electrons.

for

solutions

oxidized

into

of

and

the

a

redox

process into two

Half-equations

redox

reaction.

The

c an also

steps

for

are:

reduced.

oxidation

half-equation

and

reduction

half-equation.

3.

Balance

4.

For

the

side

of

any

oxidation

the

For

the

hand

of

magnitude

6.

Balance

equals

7.

Add

8.

If

the

the

of

half-equation,

The

in

The

in

the

so

write

that

electrons lost on the right-hand

electrons

state

the

number

electrons

half-equations

of

oxidation

half-equations

of

reduced.

oxidation

half-equation,

change

or

write

number

change

number

two

oxidized

equation.

the

these

the

the

the

of

reduction

side

being

equation.

magnitude

5.

atoms

the

together

should

the

the

number

in

and

equal to the

species.

gained on the le-

electrons

of

be

oxidized

electrons

of

state

gained

of

should

reduced

of

be

equal to the

species.

electrons

lost

in

oxidation

reduction.

c ancel

the

reaction is occurring in acidic solution, add H

2

electrons.

O(l)

to

balance

any

+

oxygen atoms and H

(aq)

to

balance

any

hydrogen atoms.

–

9.

For

H

10.

586

2

neutral or basic solutions, add OH

O(l)

Finally,

to

balance

add

up

(aq)

to

balance

oxygen atoms and

hydrogen atoms.

the

charges

and

check

that

the

sum

is

equal

to

zero.

Reactivity

3.2

Electron

transfer

reactions

Worked example 3

+

Iron metal,

Fe(s),

will react with a solution of

3

aqueous iron(III) ions,

Write the balanced

silver(I) ions,

Ag

(aq),

to form

+

Fe

(aq),

and

silver metal,

Ag(s).

equation for this redox reaction.

Solution

First,

write

the

unbalanced

+

Fe(s)

The

is

+

Ag

oxidation

therefore

3

(aq) →

state

of

equation

for

the

reaction:

+

Fe

(aq)

iron

+

Ag(s)

increases

from 0 to

+3,

so

Fe(s)

loses

electrons and

oxidized.

+

The

oxidation

and

is

Write

state

therefore

the

of

silver

decreases

from

+1

to

0,

so

Ag

(aq)

gains

electrons

reduced.

oxidation

and

reduction

half-equations,

ensuring

that

the

atoms

are

balanced:

3

oxidation:

Fe

→

+

Fe

+

reduction:

To

save

nal

time,

Ag

you

→

c an

Ag

omit

states

of

reacting

species

in

all

steps

except the

answer.

Add

the

electrons

equation

is

equal

3

Fe

→

to

to

make

the

sure

that

magnitude

the

of

number

the

of

change

electrons

in

in

each half-

oxidation state:

+

Fe

+

3e

+

Ag

Then,

+

e

→

multiply

electrons

in

→

the

each

3

Fe

Ag

reduction

half-equation

by

three,

so

that

there

are

three

half-equation:

+

Fe

+

3e

+

3Ag

Add

the

+

3e

two

→

3Ag

half-equations

+

Fe

+

3Ag

3

+

→

Finally,

three

+

check

(aq)

that

→

the

Fe

therefore

nal

charge.

the

So,

the

equation

is

c ancel

the

electrons:

+

3Ag

+

+

(aq)

charges

silver(I) ions with a 1+

with a 3+

the

3Ag

3

and

+

Fe

+

Fe(s)

together

+

are

balanced:

charge

total

3Ag(s)

each,

charge

balanced.

on

Do

on

and

each

not

the

on

le-hand

the

side

forget

right,

of

to

the

add

side,

there

is

there

one

are

iron ion

equation is 3+

states

to

all

and

species in

equation.

587

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Worked example 4

2

Iron(II) ions,

+

Fe

2

(aq),

and

dichromate(VI) ions, Cr

3

and

chromium(III) ions, Cr

Deduce the balanced

2

O

−

3

(aq),

7

react

in acidic solution to form iron(III) ions,

+

Fe

(aq),

+

(aq).

redox equation for this reaction.

Solution

First,

write

the

unbalanced

equation:

Add

the

two

half-equations

together

and

c ancel the

electrons: 2

+

Fe

2

(aq)

+ Cr

2

O

3

(aq)

7

→

+

Fe

3

(aq)

+

+

Cr

(aq)

2

+

2

6Fe

The

oxidation

2

therefore

state

of

iron

increases

from

+2 to

+

Fe

2

(aq)

loses

electrons

and

is

+

Cr

+

Cr

2

O

3

+

7

→

+

3

6Fe

+

+

2Cr

+

+3,

oxidized.

+

2

6Fe

2

O

3

→

7

+

6Fe

3

+

+

2Cr

2

In Cr

2

O

(aq),

7

oxidation

chromium

state

of

has

chromium

an

oxidation state of

decreases

from

+6. The

As

+3,

to

+6 to

this

reaction

balance

the

is

taking

oxygen

place

and

in

acidic

hydrogen

solution,

atoms.

we

There

need

are

2

therefore Cr

2

O

(aq)

7

gains

electrons

and

is

seven

reduced.

so

Write

the

oxidation

and

reduction

oxygen

add

seven

all

the

atoms

that

change

their

oxidation

states

to

2

2

Fe

3

→

balance

+

the

le-hand

side

of

the

equation,

oxygen atoms:

2

6Fe

oxidation:

the

water to the right-hand side of the

are

balanced:

+

on

of

half-equations, ensuring equation

that

atoms

moles

+

Cr

2

O

3

→

7

+

6Fe

3

+

+

2Cr

+

7H

2

O

+

Fe

There

are

now

14

hydrogen atoms on the right-hand side of

+

2

reduction: Cr

Add

the

2

electrons

O

3

→

7

such

the

equation, so add 14 moles of H

the

equation

that

number

of

electrons

in

each

2

is

state

balance

the

2

+

Cr

2

O

+

+

7

the

second

2

(remembering

3

Fe

→

Fe

Then,

there

2

O

+

that

there

are two Cr

ions

Finally,

multiply

are

six

hand

e

(6

3

+ 6e

7

→

the

check

side,

in

× 2)

On

oxidation

half-equation

each

+

3

6Fe

by six, so that

(6

the

O

7

+

+

7H

2

O

that

there

the

are

charges

six

are

balanced.

+

Fe

On the le-

2

, Cr

2

O

+

and

7

14

H

ions:

+

×

(−2)

+

(14 ×

3)

right-hand

+

the

and

6e

3

+ 6e

+

1)

=

24

(2

× 3)

side,

=

there

are

six

Fe

+

3

+

and two Cr

ions:

24

half-equation:

total

charge

therefore

forget

2

2Cr

on

each

side

of

the

equation

is

24+

+

2

Cr

3

+

3

2Cr

electrons

→

+

+

So,

2

6Fe

6Fe

+

2

Cr

→

+

half-equation):

+

3

14H

2

in

hydrogen atoms:

equal to the magnitude of the change in

3

oxidation

to

+

6Fe

half-equation

to the le-hand side of

+

2Cr

→

to

add

the

equation

states

to

all

is

balanced.

reacting

As usual, do not

species in the nal

+

2Cr equation:

2

+

6Fe

2

(aq)

+

Cr

2

O

+

(aq)

7

+

14H

3

6Fe

(aq)

→

+

3

(aq) +

2Cr

+

(aq)

+

7H

2

O(l)

Practice questions

3.

Write

balanced

equations

for

the

following

reactions that occur in acidic

solutions:

2

a.

Zn(s)

+

SO

2

(aq)

4

→

Zn

+

(aq)

2

b.

MnO

c.

I

d.

Cr

2

(s)

4

(aq)

+ OCl

+

Br

(aq)

(aq)→

→

IO

2

588

2

O

7

3

Mn

(aq)

+

(aq)

+ Cl

2

(aq)

+

C

2

O

4

SO

3

(aq)

→

2

(g)

+

Cr

+

BrO

3

(aq)

(aq)

+

(aq)

+

CO

2

(g)

Reactivity

3.2

Electron

transfer

reactions

Oxidation and reduction of metals and

halogens (Reactivity 3.2.3)

Relative ease of reduction of halogens

Halogens

halogens

c an

in

act

their

as

halogen anions. In

increases

going

increasing

F

Cl

2

This

the

Cl

2

Br

that

most

example,

Structure 3.1,

up

the

I

2

you

redox

reactions.

electrons,

learned

being

that

the

In

these

reduced

reactions,

to

singly

charged

reactivity of halogens

group.

uorine

easily

2Br

However,

in

gain

2

→

2

,

2Cl

chlorine

is

the

reduced,

chlorine, Cl

+

agents

state

reactivity

2

means

and

oxidizing

elemental

c an

+

Br

c annot

strongest

followed

oxidize

oxidizing agent among the halogens,

by

chlorine,

bromide

ions,

and

Br

then

bromine.

For

:

2

oxidize

uoride

ions

bec ause

uorine

is

a

stronger

is

possible:

oxidizing agent:

Cl

2

+ 2F

Instead,

F

2

Iodine

the

+

is

reduced

other

no

reverse

2Cl

the

by

→

reaction

+ Cl

weakest

the

strong

2F

reaction

between

and

chloride

ions

2

oxidizing

other

uorine

agent

halogens.

among

However,

the

halogens,

iodine

will

so

oxidize

it

c annot be

many metals and

reducing agents.

Relative ease of oxidation of metals

Group

1

electron

and

metals

easily.

therefore

c an

The

the

increasing

Li

For

a

Na

other

pure

then

If

no

the

reaction

reducing

of

ease

agents

group

of

1

bec ause

metals

oxidation

they

increases

increases

lose

their

going

going

valence

down

down

the

the

group,

group.

reactivity

into

pure

as

relative

K

metals,

metal

act

reactivity

Rb

you

a

c an

Cs

deduce

solution

metal

is

occurs,

more

then

of

easily

the

their

ions

of

a

relative

oxidized

metal

ease

dierent

and

of

metal.

it

comprising

is

a

the

oxidation

If

a

stronger

ionic

by placing

reaction

occurs,

reducing agent.

solution

is

more

easilyoxidized.

589

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Consider

zinc

the

reaction

between

zinc

metal

and

copper(II)

2

the

copper(II)

copper

zinc

nitrate

metal

reduced

nitrate

will

to

solution,

dissolve

copper

to

there

form

metal,

a

zinc

which

nitrate solution. In

+

are copper(II) ions, Cu

will

(aq).

In

this

reaction,

nitrate solution, and copper(II) ions will be

precipitate out as a solid:

solution

Zn(s)

p Figure 5

Therefore,

Zinc will displace copper in

solution, changing the colour of the solution

oxidized.

from

agent

blue to colourless and

forming a red

copper precipitate

zinc

is

a

are

in

more

this

3

)

2

(aq) →

stronger

Conversely,

and

electrons

+ Cu(NO

this

reducing

means

easily

reaction

Zn(s)

2

reduction: Cu

You

c an

repeat

series. In

least

a

easily

this

→

oxidized

3

)

2

(aq)

agent

+

than

copper(II)

than

Cu(s)

zinc

copper,

ions

ions.

are

You

and

a

is

more

stronger

c an

track

easily

oxidizing

the

transfer of

half-equations:

+

Zn

(aq)

+

2e

+

(aq)

+

2e

experiment

reactivity

that

reduced

using

2

oxidation:

Zn(NO

series,

metal

with

the

is

→

Cu(s)

several

most

easily

dierent metals to obtain a

oxidized

metal

is

listed

reactivity

rst and the

listed last.

Worked example 5

Strips of ve dierent

nitrate counterparts.

metals,

zinc,

iron,

magnesium,

copper and

The mixtures were observed for a period

silver,

were each added to solutions of their metal

of time to see whether a reaction has occurred or not.

These observations were recorded in table 1.

Zn(NO

) 3

(aq)

Fe(NO

2

) 3

(aq)

Mg(NO

2

) 3

(aq)

Cu(NO

2

) 3

(aq)

AgNO

2

Zn(s)

–

Yes

No

Yes

Yes

Fe(s)

No

–

No

Yes

Yes

Mg(s)

Yes

Yes

–

Yes

Yes

Cu(s)

No

No

No

–

Yes

Ag(s)

No

No

No

No

–

p Table 1

S ummary

of

re actions

between

metals

Use table 1 to deduce the reactivity series of

and

metal

ion

(aq) 3

solutions

the ve metals.

Solution

M agnesium

reacts

with

all

four

solutions

and

is

therefore

the

most

easily

oxidized

and

the

most

reactive.

Silver metal

+

does

not

making

react

AgNO

Completing

oxidized:

590

with

3

(aq)

the

Mg(s),

list

any

the

by

Zn(s),

solution

best

and

the

least

easily

oxidized.

However,

Ag

(aq) ions

are

the

most

easily

reduced,

oxidizing agent on the list.

inspection

Fe(s),

is

gives

Cu(s),

the

Ag(s).

following

activity

series,

from

the

most

easily

oxidized

to

the

least

easily

Reactivity

3.2

Electron

transfer

reactions

Redox reactions of acids and metals

(Reactivity 3.2.4)

Reactive

strong

dilute

In

metals,

acids,

solutions,

as

as

these

H

Zn(s)

+

2HCl(aq) →

these

the

SO

reactions

+

2

4

(aq) →

reactions,

state

of

following

magnesium,

zinc,

and

iron,

are

readily

oxidized

hydrochloric acid, HCl(aq), and sulfuric acid, H

Zn(s)

oxidation

by

such

such

the

ZnSO

ZnCl

2

produce

4

(aq)

(aq)

oxidation

hydrogen

+

+

H

state

changes

H

2

2

hydrogen

gas

and

a

metal

2

SO

by

4

(aq). In

salt:

(g)

(g)

of

zinc

from

changes

+1

to

0.

from 0 to

The

+2, and the

electron

transfer

is

shown

half-equations:

p Figure 6

2

oxidation:

Zn(s)

→

Zn

The reaction of metals with

+

(aq)

+

acids c an be detected

2e

the gas released

from

by the “pop” test:

the reaction mixture

+

reduction: 2H

(aq)

+

2e

→

H

2

(g)

is collected

in an inverted

test

tube, and a lit

splint is held close to the test tube opening.

Therefore,

in

the

reaction

between

a

metal

and

an

acid,

the

metal

is

the

reducing A small explosion (“pop”) suggests the

agent,

and

the

acid

is

the

oxidizing agent. presence of hydrogen gas that

reacts with

oxygen in the air

Copper

do

not

and

react

silver

with

are

less

dilute

easily

oxidized

solutions

of

than

common

potassium

magnesium,

acids

zinc

and

iron, so they

(gure 7).

most reactive

sodium

calcium

magnesium

aluminium

zinc

iron

tin

lead

(hydrogen)

copper

silver

gold

platinum

p Figure 7

oxidized.

least reactive

A reactivity series showing the most easily oxidized

Metals above hydrogen on the list c an react

metals to the least

easily

with common acids, those below

c an not

591

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Thinking skills

ATL

A

student

investigating

qualitative

the

reactivity

of

zinc

and

copper

noted

the

following

observations:

1.

Copper wire was placed in dilute sulfuric acid. No change was observed. A

2.

Some copper wire was wrapped around one end of the zinc strip. This

zinc strip was placed in dilute sulfuric acid. Bubbles appeared. p Figure 8

Gold

is at

the bottom of

the reactivity series of metals, so it is not

oxidized

easily.

end of the strip and the surrounding copper were placed in dilute sulfuric

Therefore, it is the most

likely to be found

in its reduced

form, with

acid. Bubbles evolved quickly on the surface of the copper.

zero oxidation state: elemental gold. It is

3. impossible to “pan for lithium”

Strips of copper and zinc were placed in dilute sulfuric acid and

bec ause it is

connected to each other. Bubbles evolved on the surface of the copper. at the top of the activity series

zinc

dilute

Explain

the

electron

Write

or

SO

2

student’ s

transfer

two

H

4

observations,

reactions,

three

copper

linking

using

your

knowledge

of

metal

reactivity,

reactions of acids and metallic bonding.

questions

relating

the

concepts

illustrated

by this

experiment.

Redox titration

Iron

with

supplements

a

solution

oxidizing

of

agents.

are

used

In

their

2

manganese(II) ions, Mn

592

to

potassium

treat

iron

deciency.

manganate(VII)

reaction

with

of

iron(II)

You

known

in

will

determine

concentration.

acidic

solution,

the

iron

content

purple

manganate(VII)

+

(aq).

In

this

process,

iron(II)

ions

are

in

iron

tablets

M anganate(VII) ions, MnO

oxidized

to

iron(III) ions.

ions

are

4

by

(aq),

titrating them

are

powerful

reduced to pale pink

Reactivity

Relevant skills

4.

In

the

meantime,

propagation •

Tool

1:

•

Tool

3:

General mathematics

•

Tool

3:

Record

•

Inquiry

review

the

sections in the

3.2

Electron

transfer

reactions

titration and uncertainty

Tools for chemistry

Titration chapter

before starting part B.

Part B Titration against potassium manganate(VII)

and

propagate uncertainties

5.

2:

Assess

reliability

and

validity

of

Filter

the

iron

tablet

extract.

Transfer

the

ltrate into

results

3

a

volumetric

with

ask

distilled

and

water.

make

Store

up

in

a

to

the

250 cm

labelled

mark

reagent

jar.

S afety

3

•

Wear

eye

6.

protection

Fill

the

burette

with

0.020 mol dm

manganate(VII), KMnO •

Sulfuric acid, H

•

Dilute

2

SO

4

4

potassium

(aq).

(aq), is an irritant

3

7. potassium manganate(VII), KMnO

4

Transfer

conic al irritant and will

25.0 cm

of

iron

tablet

solution

to

a

clean

(aq), is an

ask.

Place this ask on a white tile under the

stain skin and fabrics

burette.

Materials

8. 2

•

iron

tablets

(or

other

source

of

iron(II),

Fe

Perform

a

rough

titration

of

the

iron tablet solution,

+

(aq),

ions) stopping when the solution in the ask permanently

3

•

1.0 mol dm

•

distilled

sulfuric acid, H

2

SO

4

turns pale pink.

(aq)

water

9.

•

0.020 mol dm

KMnO

4

Repeat

the

titration

concordant

3

several

times

until

you obtain two

values.

potassium manganate(VII) solution,

(aq) 10.

•

pestle and mortar

•

top pan balance

Clear

up

according

to

the

directions

given

by

your

teacher.

Questions

3

•

two

•

100 cm

250 cm

•

250 cm

•

funnel

conic al asks

1.

Write

the

oxidation

and

reduction

half-equations

for

thisreaction.

3

measuring

cylinder

3

volumetric ask

2.

Deduce

3.

Using

the

your

redox

results,

equation

for

determine

this

the

reaction.

mass

of

iron in one

tablet. •

lter paper

•

reagent jar

4.

Propagate

the

•

burette

•

25 cm

5.

3

•

volumetric

the

measurement uncertainties to obtain

uncertainty

of

the

Compare

your

the

packaging.

result

mass

to

the

of

iron per tablet.

iron

content

reported on

pipette tablet

C alculate

the

percentage

error.

white tile

6.

Comment

7.

Describe

on

the

reliability

and

validity

of

your

result.

Instructions

Part A Preparation of acidified iron tablet extract

suggest

1.

Grind

four

iron

tablets

into

a

ne

and

Weigh

the

two

sources

of

systematic

error and

of

error.

mortar.

8.

2.

least

powder using a sources

pestle

at

improvements that would minimize these

iron

tablet

powder

and

Explain

why

not

one.

you

used

four

iron

tablets

in

this

analysis,

transfer it into a just

conic al ask.

9.

Add

100 cm

Explain

of

1.0 mol dm

sulfuric acid, H

2

SO

4

the

tablet

you

le

the

iron

tablet

powder in dilute

(aq), acid

to

why

3

3

3.

powder

and

leave

for

24

to

48

solution

for

24

to

48

hours.

hours.

10.

Suggest

be

why

redox

titrations

such

as

this

are

said to

“self-indic ating”.

593

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Electrochemic al cells (Reactivity 3.2.5 and

Reactivity 3.2.6)

An

electrochemic al cell

two

types

1.

In

of

primary (voltaic) cells,

energy

produced

electric al

2.

In

produce

are

and

chemic al

energy.

secondary (rechargeable) cells, and

spontaneous

There

are

chemic al

reactions

is

used

fuel cells

to

the

generate

the

irreversible

while

move

reduced.

energy

electric al

energy

is

used

to

drive

forward non-

reactions.

electrons

being

and

electric al

chemic al

reactions,

substance

exothermic,

to

by

electrolytic cells,

redox

the

electric al

energy.

spontaneous

In

interconverts

electrochemic al cell:

from

Nearly

released

energy.

Redox

secondary

the

all

in

substance

spontaneous

these

reactions

being

redox

chemic al

used

(rechargeable)

in

cells

oxidized to

reactions

changes

primary

utilize

c an

are

be

used

(voltaic) cells

reversible

redox

reactions.

In

the

that

by

to

a

18th

an

moist

twitch

Italian

He

century,

electric

the

current

substance.

by

touching

scientist,

showed

Italian

could

He

it

two

that

chemic al

Luigi

produced

noticed

with

doubted

that

scientist

be

that

he

animal

legs

c an

two

could

dissimilar

reactions

Galvani

by

c ause

metals.

were

discovered accidentally

dissimilar

an

Alessandro

integral

produce

metals

connected

amputated

to

produce

electricity

frog

leg

Volta, another

and

electricity.

made

the

rst

battery

p Figure 9

(a) Galvani’s frog legs experiment

of electric battery

594

(b) Volta’s voltaic pile,

the rst modern type

Reactivity

Any

two

one

higher

will

dissimilar

have

in

its

the

ions

metals

activity

and

their

series

reduced

to

ions

will

the

c an

oxidize

pure

participate

to

metal.

For

2

by

copper(II) ions,

to

ions

and

in

redox

the

example,

one

zinc

3.2

Electron

transfer

reactions

reactions. The

lower in the series

c an

be

oxidized

+

form zinc ions. The Cu

ions

get

reduced, and act as an

oxidizing agent.

2

Zn(s)

+

Separating

+

Cu

this

into

2

Zn(s)

2

→

+

Zn

two

(aq)

+

Cu(s)

half-equations

gives:

+

Zn

(aq)

+

2e

+

Cu

These

→

2

(aq)

(aq)

two

+

2e

→

processes

Cu(s)

c an

occur

in

separate

beakers,

c alled

Zn

Cu

2+

Zn

2+

(aq)

Cu

p Figure 10

The

In

the

in

two

half-cells

c an

electrochemic al

c athode.

this

c an

c ase,

be

RED CAT

Zn(s)

2

When

of

zinc

→

the

at

the

a

slight

on

a

slightly

As

therefore

becomes

polarized,

To

get

around

half-cells

and

solution,

such

the

positive

with

the

charged

this,

on

a

and

salt

complete

as

ions

the

and

half-cell

with

to

the

This

and

reduction at

CAThode.

Therefore,

electrochemic al cell

Cu(s)

the

produced

copper

loses

two

half-cell

oxidation

of

by

the

half-cell

electrons

gains

which

positively

2

to

zinc

two

is

repels

oxidation

to

reduce

and

now takes

electrons, taking

prevented

the

by the

electrons. The

charged zinc half-cell. The cell

connect the solutions in the two

circuit.

SO

4

toward

ions

to

half-cell,

used

Na

ow

electrochemic al cell.

reaction stops.

electric al

negative

→

copper

slightly

is

an

anode,

REDuction at

electrons

wire

further

redox

sulfate,

(c ations)

the

the

form

c athode.

2e

half-cell

copper

bridge

the

sodium

c athode

in

the

the

+

the

the

Any

to

−

zinc

while

the

remain

wire

occurs at the

equations:

(aq)

through

charge.

charge

a

+

connected,

charge

electrons

2

|| Cu

result,

negative

Cu(s)

following

2e

ow

a

positive

+

are

anode

negative

and

−

half-cells

on

the

with

always

useful mnemonic:

anode

by

(aq)

ions.

connected

is a

(aq)

The zinc and copper half-cells

oxidation

+

Zn

the

copper(II)

slight

is

represented

Zn(s)

be

cells,

half-cells.

,

or

the

(anions)

A

salt

bridge

potassium

slightly

to

ow

consists

of

nitrate, KNO

negatively

toward

the

an

ionic

.

allows

3

It

salt

charged half-cell

slightly

positively

anode.

595

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Consider

the

addition

of

a

sodium

sulfate

salt bridge to our zinc–copper cell. The

+

slightly

negatively

through

the

salt

charged

bridge

copper

and

half-cell

solution.

The

attracts

slightly

Na

2+

(aq) and Zn

positively

(aq)

c ations

charged zinc half-cell

2–

attracts SO

(aq)

anions

through

the

salt

bridge

and

solution.

This

neutralizes

4

the

charge

complete

cell,

in

each

primary

named

aer

half-cell,

cell

its

so

(gure

the

11).

inventor,

the

redox

This

reaction

kind

British

of

c an

primary

chemist

John

continue.

cell

is

This

known

Frederic

is

as

now a

the

D aniell

D aniell.

e

e V

–

NO

+

+

Na

3

Zn anode

Cu

cathode

+

–

NO

u Figure 11

NO

3

A primary cell consisting of

–

2+

a zinc metal anode (labelled

bec ause it

in ZnSO

4

is the source of

(aq),

NO

Zn

as negative

NO

electrons) dipped

3

2+

Cu

3

a copper metal c athode 2+

Zn(s) (labelled

3

positive as it

dipped in CuSO

4

connecting wire,

(aq),

→

Zn

2+

(aq)

+

2e

Cu

(aq)

+

2e

→

Cu(s)

attracts electrons)

movement of cations

an electric al

a voltmeter and a

movement of anions salt bridge

As

the

the

reaction continues, the blue colour of the copper(II) sulfate solution fades,

copper

bar

increases

in

size

as

it

becomes

coated

in

more

copper, and the

2

zinc

bar

gets

c athode

thinner.

side,

the

Cell diagrams

convention,

always

written

vertic al

lines.

cell

are

the

Once

on

used

the

c an

is

a

+

signic ant build-up of Zn

(aq) ions on the

ceases to function.

as

c athode

You

there

a

is

short-hand

always

le-hand

use

the

way

to

represent primary cells. In this

written on the right-hand side and the anode is

side.

The

following

salt

bridge

general

is

represented

template

to

write

by

cell

two

parallel

diagrams

for metal–ion primary cells:

anode

being

oxidized

|

product

of

Therefore,

the

cell

2

diagram

for

+

Zn(s) | Zn

2

(aq)

|| Cu

of

oxidation

species

on

the

le-hand

2

being

oxidized to Zn

the

D aniell

(aq)

|

596

represent

the

being

reduced

|

product

cell

would

be

written

as

follows:

Cu(s)

side,

Zn(s) | Zn

+

(aq),

represent

the

anode, with zinc

+

2

(aq).

The

species

2

Cu(s),

species

+

2

The

metal

||

reduction/c athode

c athode, with Cu

on

the

right-hand

side, Cu

+

(aq)

being

reduced

to

Cu(s).

+

(aq) |

Reactivity

3.2

Electron

transfer

reactions

Worked example 6

M anganese metal reacts with nickel(II) ions to form manganese(II) ions and

nickel metal.

a.

Write the redox reaction that

occurs between nickel(II) ions and

manganese metal.

b.

Assuming that this redox reaction occurs in a manganese–nickel primary

cell,

write the half-equations that

occur in each half-cell.

c.

Write the cell diagram to represent the primary cell for this redox reaction.

d.

Sketch a primary cell for this reaction and

direction of electron ow and

identify the anode,

c athode,

direction of ion ow.

Solution

2

a.

Mn(s)

+

+

Ni

2

(aq)

→

2

b.

+

Mn

(aq)

At

the

c athode: Ni

(aq)

At

the

anode:

→

+

You

c an

anode

use

Mn(s)

the

being

general

→

(aq)

+

template

oxidized

|

Ni(s)

+

Mn

of

2e

for

product

product

This

Ni(s)

2e

2

c.

+

+

of

cell

diagrams:

oxidation

||

species

being

reduced |

reduction/c athode

gives:

2

Mn(s) | Mn

+

2

(aq) || Ni

+

(aq)

|

Ni(s)

d. V cathode

anode

(

)

(+)

salt bridge

Ni(s)

Mn(s)

2+

Electrons

ow

2+

(aq)

Mn

from

Ni

le

2

the

electric al

also

The

the

ow

wire. Mn

from

anions

of

le

the

to

to

(from

the

anode

to

the

c athode)

through

+

(aq)

right

salt

right

(aq)

ow

and

the

through

from

c ations

the

right

salt

to

of

the

bridge

le

salt

in

toward

through

the

the

the

salt

salt bridge

c athode.

bridge

toward

anode.

597

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Secondary cells (Reactivity 3.2.7)

A

in

battery

a

materials

the

a

will

be

the

polarize

also

c ause

a

both,

or

of

cells

c an

satisfy

cells

that

c an

replacement

have

be

to

self-discharged

well

is

the

the

under

are

on

cell

current

a

batteries

using

during

an

demands

phone,

the

and

the

and

Either

will

c an

no

solution

reaction

the

than

in

to

surface

of

reduce

its

be

the

anode,

thrown

longer

and

stop.

the

reaction

be

away.

used.

salt bridge

Polarization

c an

anode. These

electric al output.

low-current household devices.

chemic al

and

to

cells

phones

energy.

to

reactions

current

primary

cell

need

storage

for

electric

electric al

you

battery

enclosed

the

high-current demands, such as in ash

suitable

applying

the

the

typic ally

ultimately

reversible.

through

chemic al

bubbles

of

not

or

cells,

cell,

oxidized

rechargeable cell,

by

is

replaced,

travelling

but

example,

for

reaction

electrode)

ions

c ars,

recharged

battery

be

resistance

reversed

For

the

c auses

operate

higher

self-discharge.

and

electrochemic al

electrochemic al

hydrogen

electric

be

the

which

internal

not

more

(negative

cell,

the

c an

or

primary

need

secondary cell, or a

electricity

of

a

discussed,

do

photography

In a

or

build-up

cells

two

In

consumed,

the

increase

Primary

of

anode

previously

c an

c an

series

container.

electrolyte,

Typic ally,

As

is

single

it

but

are

When

charge

the

have

made

you

that

cell.

of

a

generate

Secondary

higher

rate

secondary

purchase a

before

use

bec ause it will

transportation.

electron flow (current)

through external circuit

negative plate

positive plate

(anode)

(cathode)

electrolyte

oxidized oxidized

positive ions metal

metal

metal negative ions

metal or

lower oxide

electrolyte

case

p Figure 12

Structure of

an electrochemic al cell.

ow of ions c auses polarization.

be reversed

598

The negative anode is oxidized and the

This process c annot

in a secondary (rechargeable) battery

be reversed

in a primary cell,

but

it

c an

Reactivity

3.2

Electron

transfer

reactions

C ase study: Lead–acid battery

C ar

is

batteries

used

to

systems

the

in

the

and

will

the

in

discharge,

does

not

gradually

typic al

Pb(s),

made

c ar.

combustion

during

A

are

power

a

c ar

of

secondary

motor

This

the

is

that

known as

engine

therefore

provide

cells.

starts

is

the

electric al

discharge.

used

to

recharging

enough

The

engine,

energy

Some

reverse

the

to

and

the

battery.

recharge

energy

to

of

power

the

is

battery

chemic al

energy

from

reactions that occur

An

engine

the

idling

battery,

and

runs

so

the

slowly

battery

discharge.

uses

lead(IV)

a

the

electric al

chemic al

a

lead–acid

oxide

battery.

c athode,

PbO

2

This

battery

is

composed

(s), and sulfuric acid, H

strong acid, so it will ionize into H

2

of

SO

4

a

lead

(aq).

anode,

Sulfuric

−

+

acid

from

any

(aq) and HSO

(aq).

4

When

the

battery

−

is

powering

the

motor

and

the

c ar ’s

electric al

systems, HSO

4

(aq)

will

oxidize

+

Pb(s)

to

at

the

the

anode, and H

following

(aq)

discharge

will

reduce

PbO

2

(s)

at

the

c athode.

−

anode

(oxidation):

Pb(s)

+

HSO

+

(aq) →

4

PbSO

overall

cell

reaction:

2

(s) + 3H

Pb(s)

+

dissociation

together

of

and

gives rise

(s)

+

H

(aq)

PbO

2

+ HSO

(s)

+

2H

2

4

−

(aq)

+

2e

−

(aq) + 2e

SO

4

(aq)

→

→ PbSO

2PbSO

4

4

(s)

(s) + 2H

+

2H

2

2

O(l)

O(l)

–

+

Note that the H

4

−

+

cathode (reduction): PbO

This

reactions:

(aq)

and HSO

sulfuric

acid,

shown as H

2

so

SO

4

(aq)

4

in

ions

the

are

overall

ultimately

cell

produced

reaction

they

by the

are

combined

(aq).

Worked example 7

Determine the reactions that

cell,

and

occur at

the overall cell reaction,

the anode and

c athode in a c ar battery

when the engine is charging the battery.

Solution

The

charging

reactions

are

the

reverse

+

anode:

PbSO

4

(s) +

H

of

the

discharge

(aq)

+

2e

reactions:

−

−

→

Pb(s)

+ HSO

4

(aq)

−

+

c athode:

PbSO

4

(s) +

2H

2

O(l)

overall cell reaction: 2PbSO

The

continual

oxygen

topped

from

up

charging

water.

with

of

a

(s)

PbO

+ 2H

lead–acid

Therefore,

distilled

4

→

2

2

(s)

O(l)

battery

non-sealed

+

3H

→ Pb(s)

tends

c ar

(aq)

to

+

HSO

+ PbO

2

produce

batteries

4

−

(aq) +

(s) + 2H

2e

2

SO

4

(aq)

hydrogen and

occ asionally

need to be

water.

t Figure 13

of

a series of

A lead–acid

battery consists

secondary cells with lead(IV)

oxide plates,

lead plates, and sulfuric acid

electrolyte

599

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

C ase study: Lithium-ion batteries

Lithium-ion

graphite

is

a

lithium–cobalt

lithium

ions

During

and

are

through

graphite

lattice.

polymer

gel,

the

to

use

pure

complex,

opposite

battery

lithium

The

as

than

lithium

lithium

LiCoO

2

.

atoms

metal

The

embedded

for

the

lithium

in

anode.

atoms

are

a

lattice of

The

c athode

oxidized to

discharge.

the

reduced

batteries

rather

oxide

during

charging,

migrate

a

rechargeable

electrodes,

atoms.

battery

lithium

process

medium

occurs:

the

These

medium

reacts

to

the

anode,

atoms

must

be

vigorously

lithium

where

become

water.

in

the

complex

accept

electrons

embedded in the

completely

with

ions

they

non-aqueous, usually

Table 2 summarizes

thesereactions:

Electrode

Discharging reaction

+

anode

Li(s)

→

+

c athode

p Table 2

The

flow

of

Li

cell

+

e

Li

−

+

e

→

Li(s)

−

+

Anode and

overall

Charging reaction

−

+

Li

e

+

+

CoO

2

(s) →

LiCoO

2

(s)

LiCoO

2

(s) →

Li

−

+

e

+

CoO

2

(s)

c athode reactions in the lithium-ion battery

reaction

during

discharge

is

Li(s)

+

CoO

2

(s) →

LiCoO

2

(s).

electrons e

during

anode

discharge

Li Li

+

Li

Li

Li Li

Li

+

Li Li Li

Li

Li

Li

cathode

anode

cathode

p Figure 14

Structure of

a typic al

lithium-ion rechargeable battery. The

When

the

lithium-ion

battery

is

in

use,

electrons

ow

from the anode to the

battery consists of a series of secondary

c athode

cells composed

through

the

external

circuit

while

lithium

ions

ow

from the anode to

of c athodes and

the

c athode

are

le

through

the

polymer

gel

inside

the

cell.

When

no

more lithium ions

anodes with a layer of polymer (yellow)

on

the

anode,

the

battery

is

at.

To

recharge

separating them

transferring

600

lithium

ions

back

to

the

anode.

it,

the

process

is

reversed,

Reactivity

3.2

Electron

transfer

reactions

Practice questions

4.

During

the

discharge

of

a

nickel–c admium

battery,

the

following

reactions

occur in the cells:

anode:

Cd(s)

c athode:

+

2OH

NiO(OH)(s)

a.

Write

the

overall

b.

Write

the

cell

c.

Determine

(aq)

+

H

→

2

Cd(OH)

O(l)

equation

+

e

2

(s)

→

+

2e

Ni(OH)

2

(s)

+

OH

(aq)

for the cell.

diagram.

the

charging

reactions that occur in the cell.

Fuel cells

Fuel

or

cells

are

ethanol

pollution

but

and

unlike

reactions

For

a

type

and

are

very

primary

in

the

oxygen

electrochemic al

into

water,

ecient.

cells,

cell

example, in a

while

of

oxygen

c an

they

Like

a

is

supplied

to

convert

cells,

steady

and

they

supply

hydrogen, methanol,

heat.

are

of

They

not

fuel

c ause

little

rechargeable,

and

oxygen, so the

indenitely.

hydrogen fuel cell,

gas

that

dioxide

primary

require

continue

cell

c arbon

the

hydrogen

c athode.

gas

The

is

supplied to the anode

following

reactions occur in a

hydrogen fuel cell:

+

anode: H

2

(g)

→

2H

(aq)

+

2e

+

c athode: O

overall

In a

cell

2

(g)

+

4H

(aq)

equation: 2H

2

direct methanol fuel cell

following

reactions

occur

in

a

+

4e

(g)

+

→

O

2

2H

(g)

(DMFC),

2

→

O(l)

2H

2

O(l)

methanol

is

supplied

to

the

anode. The

DMFC:

+

anode: CH

3

OH(l)

+

H

(g)

+

6H

2

O(l) →

CO

2

(g)

+

6H

(aq)

+

6e

3 +

c athode:

O

2

(aq)

+

6e

→

3H

2

O(l)

2

3 overall

cell

equation: CH

3

OH(l)

+

O

2

(g)

→

CO

2

(g)

+

2H

2

O(l)

2

A

•

typic al

fuel

cell

has

the

following key components:

Electrolyte or separator: this keeps components from mixing. For example, a

+

proton exchange membrane

(PEM) is a polymer that allows H

ions to diuse

through but prevents the diusion of other ions, electrons or molecules.

•

•

Electrodes:

The

electrodes

chemic al

reactions

reducing

electrode

Bipolar plate:

ensures

This

to

are

occur.

made

There

is

of

an

a

c atalyst

oxidizing

that

allows

electrode

for the

(anode) and

(c athode).

conducts

the

electric al

current

from cell to cell and

uniform distribution of the fuel gas.

601

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

external

circuit

– – –

fuel H

O

2

2

(air)

+

+

heat

+

+

fuel air and H

O

2

recirculated

gas diffusion catalyst

gas diffusion

catalyst electrode (cathode)

electrode (anode)

proton exchange

membrane (PEM)

p Figure 15

In a hydrogen fuel cell, H

2

(g) is oxidized at the anode and O

2

(g) is reduced

at

the c athode.

The proton exchange membrane

+

(PEM) allows H

ions to diuse but

prevents the diusion of other ions,

Hydrogen

fuel

greenhouse

2H

be

2

O(l)

utilized,

from

c atalysts

use

The

the

used

for

expensive

metals,

sc ale.

hydrogen

The

c atalyst

main

1.

and

Clean

of

electrons

gas

formed

increasing

as

in

the

the

the

fuel.

These

exothermic

eciency

of

cells

do

not

reaction 2H

the

cell.

The

2

produce

(g)

+

oxygen

O

2

(g)

gas

→

c an

air.

the

gas

electrodes

makes

used

the

fuel

must

are oen made of platinum or other

cells

be

reduction

of

expensive

very

the

pure

cell

to

to

run

on

a

commercial

prevent the poisoning of the

electric al

output.

There

are two

hydrogen gas:

hydrogen

wind

hydrogen

heat

which

therefore

sources

or

5.

be

obtained

The

Practice question

c an

cells

gases.

molecules and

c an

generators

be

produced

provide

the

by

the

cleanest

electrolysis

form

of

of

energy

water. Solar cells

for

powering the

O utline the function of the electrolysis.

proton

exchange

(PEM) in fuel

membrane

cells.

2.

Hydrogen

especially

DMFC s The

eect

of

atmospheric

602

the

methanol

as

also

be

with

advantage

the

fuel.

temperatures is

is discussed in

have

c an

obtained

steam: CH

of

not

by

4

the

(g)

needing

+

to

reaction

H

2

O(g)

extract

of

⟶

hydroc arbons,

3H

2

(g)

hydrogen

+

gas

CO(g)

bec ause they

greenhouse gases

use on

gas

methane,

Structure 3.2

a

greenhouse gas.

However,

they

produce

c arbon

dioxide,

CO

2

(g), which

Reactivity

3.2

Electron

transfer

reactions

Electrolytic cells (Reactivity 3.2.8)

An

electrolytic

chemic al

cell

energy.

is

non-spontaneous,

the

An

chemic al

dipped

the

in

cell

c an

free-moving

so

be

they

This

of

a

of

and

external

converts

reactions

source

of

electric al

in

an

electrolysis.

single

container

lled with an

ionic

Two

salt,

or

a

molten

electrodes

direct

current

(the

(DC)

energy to

electrolytic

cell

are

electricity to bring about

known as

an

a

that

is

anions.

electrolyte,

cell

reduction

an

process

solution

and

and

require

consists

a

c ations

the

electrochemic al

oxidation

changes.

electrolytic

electrolyte

an

The

electrolyte. The

ionic

salt,

c athode

power

composed of

and

source

the

is

anode)

are

connected to

electrodes.

+

DC power

e

e

source

electrolyte

electrodes

p Figure 16

In

a

closed

terminal

is

of

circuit,

the

connected

reduce

anode

the

and

positive

electrons

DC

to

power

the

c ations

terminal

and

in

Consider

the

of

ions

ow

the

electrolyte.

DC

an

The

power

of

The

cell

anions

to

in

ow

the

of

to

the

the

DC

the

positive

power

source

c athode and

electrons

circuit.

electric

chloride,

to

the

electrolyte

The

complete

comprises

sodium

terminal

terminal

electrons.

source

molten

negative

electrons

releasing

electrolytic

electrolysis

the

negative

Therefore,

oxidation,

the

in

from

source.

c athode.

undergo

electrons

The structure of an electrolytic cell

ow to the

ow to the

The

ow of

current

NaCl(l),

shown

in

gure

17.

ammeter to DC power source measure current –

+

A

–

–

e

e

anode

cathode

oxidation

reduction

occurs here

–

occurs here

–

2Cl

+

+

2e

Na

–

+

e

2

–

Cl

+

Na t Figure 17

The electrolysis of molten sodium chloride

heat

603

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

In

the

electrolytic

cell,

molten

sodium

chloride

is

the

electrolyte, which contains

+

sodium

c ations,

reducing

the

reduction

Na

, and chloride anions, Cl

sodium

c ations

half-equation

is

as

in

the

.

electrolyte

Electrons

to

ow

to

the

c athode,

form molten sodium metal. The

follows:

+

Na

+ e

At

the

to

complete

anode,

2Cl

The

→

chloride

the

→ Cl

overall

Na(l)

2

anions

circuit.

(g)

+

The

are

oxidized,

oxidation

producing

half-equation

is

chlorine

as

gas

and

electrons

follows:

2e

equation

for

reactions

in

the

electrolytic cell is:

Practice question

2NaCl(l) 6.

Write

full

the

for

the

molten

2Na(l)

+ Cl

lead

2

(g)

electrolysis This

of

→

half-equations, and

equation,

bromide,

PbBr

electrolytic

cell

is

therefore

useful

for

the

production of sodium metal and

2

chlorine gas.

Reactive

metals,

obtained

the

by

such

as

lithium,

electrolysis

electrolysis

must

of

take

their

place

magnesium,

molten

in

an

salts.

inert

aluminium,

These

and

metals

sodium

react

with

are all

oxygen, so

atmosphere.

Oxidation of organic compounds

(Reactivity 3.2.9)

Some

functional groups

certain

conditions.

hydroxyl group

forming a

in

For

in

organic

example,

in

compounds

the

presence

secondary alcohols

c an

be

c an

of

undergo

an

oxidation under

oxidizing agent, the

oxidized to a

c arbonyl group,

ketone:

OH

O

[O]

R

C

C

R

+

R

H

O

2

R

H

secondary

The

of

symbol

oxygen

the

[O]

is

atoms.

compound,

additional

bond

the

rst

step,

In

to

this

forms

c an

the

indic ate

oxidation

forming

Primary alcohols

In

used

alcohol

water

be

the

an

group

is

two

oxygen

c arbon

oxidized to

hydroxyl

oxidizing

reaction,

with

between

ketone

and

agent,

which

hydrogen

atom

from

provides

atoms

the

are

a

lost

source

from

oxidizing agent. An

oxygen.

c arboxylic acids

oxidized

to

a

in

a

two-step

c arbonyl

group,

reaction.

forming an

aldehyde:

OH The

denitions

of

O

primary,

[O] secondary and tertiary alcohols R are

given in

C

C

H

+

H

O

2

Structure 3.2.

R Namingcompounds with

H

H hydroxyl,

c arboxyl

and

c arbonyl

primary functional

groups

is

also

alcohol

aldehyde

covered in

Structure 3.2

Like

in

the

additional

604

oxidation

bond

of

forms

a

secondary

between

alcohol,

c arbon

and

two

hydrogen

oxygen.

atoms

are lost and an

Reactivity

In

a

the

second

step,

the

c arbonyl

group

is

oxidized to a

c arboxyl group,

3.2

Electron

transfer

reactions

forming

c arboxylic acid:

O

O

water

out

[O]

C

C

condenser

R

H

aldehyde

In

this

oxidation

reaction,

the

c arboxylic acid

aldehyde

gains

an

oxygen atom.

water in

The

oxidation

to

ketone,

a

c an

mixture with

them

back

potassium

of

a

to

a

primary

be

to

accomplished

reux

the

alcohol

condenser,

reaction

reactionmixture.

For

the

by

2

Cr

2

c arboxylic

reux.

which

mixture

dichromate(VI), K

the

a

cools

(gure

O

7

,

and

Reux

any

18).

a

acid,

An

or

a

secondary alcohol

involves

vapours

heating

produced

the

reaction

and

returns

oxidizing agent, such as

concentrated

acid

are

also

added to

alcohol

is

oxidation

present

oxidation

in

to

of

excess

an

a

primary

to

ensure

aldehyde

alcohol

to

complete

aer

the

rst

a

c arboxylic

two-step

oxidation

acid,

the

oxidation,

oxidizing agent

rather than partial

+ excess

oxidizing agent

+ concentrated acid

step.

heat

The

oxidation

distillation.

of

a

primary

Distillation

alcohol

allows

the

to

an

aldehyde

aldehyde

to

be

c an

be

accomplished

isolated

before

it

by

undergoes

p Figure 18

The experimental set-up

for a reux reaction. Reux allows vapours

to condense back to the boiling reaction

further

oxidation

to

c arboxylic

acid

(gure

19).

In

this

c ase, the alcohol, not the

mixture for further oxidation

oxidizing

agent,

is

used

in

excess.

water out

condenser

primary alcohol

in excess

+

oxidizing agent water in +

concentrated

acid

anti-bumping granules heat

t Figure 19

The experimental set-up

for

the distillation of an aldehyde obtained

by

aldehyde the oxidation of a primary alcohol

605

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Worked example 8

Write the equations for the following oxidation reactions,

displayed

showing

formulas:

a.

oxidation of ethanol to an aldehyde

b.

oxidation of propan-1-ol to a c arboxylic acid

c.

oxidation of propan-2-ol.

Use [O] to symbolize the oxidizing agent.

reacting species is in excess and suggest

reux or distillation,

must

In each equation, state which

which experimental procedure,

be used.

Solution

a. H

H

H O

[O]

C

C

H

H

C

+

H

O

2

H H

ethanol

ethanol

in

ethanal

excess

distillation

b.

H

H

H

H

H O

[O]

H

C

C

C

C

OH H

H

H

propan-1-ol

oxidizing

agent

in

H

H

propanoic acid

excess

reux

c.

H

H

H

H

H

[O]

H

C

H

C

OH

H

propan-2-ol

oxidizing

reux

606

agent

in

excess

C

H

O

H

propanone

+

H

O

2

Reactivity

In

both

primary

c arbonyl

c arbon

and

group

with

secondary

involves

the

the

hydroxyl

alcohols,

removal

group.

of

oxidation

the

Tertiary

of

the

hydrogen

alcohols

do

hydroxyl

atom

not

3.2

Electron

group to a

so

alcohols

they

c annot

be

oxidized

in

the

same

way

as

reactions

H

connected to the

have

this

H

hydrogen

C

H

H atom,

transfer

primary

and

H

secondary

(gure 20). H

C

C

H

H

H

Reduction of organic compounds

C

OH

(Reactivity 3.2.10) p Figure 20

C arboxylic

acids

involving

an

alcohols.

The

c an

be

aldehyde

reduced

to

primary

intermediate,

reactions

are

the

and

opposite

alcohols

ketones

of

the

c an

via

a

be

two-step

reduced

corresponding

reaction

to

secondary

oxidation

2-methylpropan-2-ol is a

tertiary alcohol,

so it

c annot

be oxidized by

reux or distillation in the presence of an

oxidizing agent

reactions.

In

the

presence

reduced

to

a

of

a

reducing

hydroxyl

agent,

group,

the

forming

a

c arbonyl

group

in

a

ketone

c an be

secondary alcohol:

O

OH

[H] C

R

R

C R

R

H

ketone

The

symbol

atoms.

of

the

In

bonds

C arboxylic

the

[H]

this

rst

is

to

between

acids

step,

used

reduction

the

c an

secondary alcohol

indic ate

reaction,

c arbon

be

and

reduced

c arboxyl

the

the

group

oxygen

to

is

reducing

ketone

is

primary

reduced

agent,

gains

which

two

provides

hydrogen

hydrogen atoms and one

broken.

alcohols

to

a

in

a

c arbonyl

two-step

group,

reaction. In

forming an

aldehyde:

O

O

[H] C

C

R

OH

R

c arboxylic acid

In

this

In

the

reduction

second

reaction,

step,

the

the

c arboxylic

c arbonyl

H

aldehyde

group

acid

is

loses

reduced

an

to

oxygen atom.

a

hydroxyl

group,

forming a

primary alcohol:

O

OH

[H] C

R

R

C H

H

H

aldehyde

Like

in

the

reduction

of

c arbon–oxygen

bonds

In

of

the

reduction

product,

but

in

a

is

ketone,

two

hydrogen

atoms

are

gained and one of the

broken.

c arboxylic

most

primary alcohol

c ases

it

acids, the

c annot

be

aldehyde

is

produced

as

an

intermediate

isolated.

607

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

A

common

hydride,

reducing

LiAlH

reduced

by

4

.

agent

used

Aldehydes

sodium

and

in

all

the

ketones

borohydride,

NaBH

reactions

(but

not

above is lithium aluminium

c arboxylic

acids)

c an also be

4

Science as a shared endeavour

Vincent

van

Gogh

create

many

Redox

reactions

over

of

his

used

lead

colourful

have

and

and

c aused

chromium paints to

Chemists,

well-known paintings.

some

of

the

to

colours to fade

devise

time.

p Figure 21

used

historians

and

and

conservators work together

understand these changes, as well as to

possible

chemistry

The yellow paint

art

explore

preservation

methods.

Where else does

intersect with the world of art?

in Van Gogh’s Bedroom at Arles was probably made from

lead

chromate,

PbCrO

4

(right)

Reduction of alkenes and alkynes

(Reactivity 3.2.11)

Alkenes

double

and

compounds

In

the

alkynes

bond,

and

c an

presence

alkynes

c an

be

be

of

are

unsaturated

alkynes

a

have

reduced

suitable

reduced

by

by

a

compounds.

c arbon–c arbon

the

addition

c atalyst,

such

as

of

Alkenes

triple

have

bond.

a

c arbon–c arbon

Unsaturated

hydrogen to the multiple bond.

deactivated

palladium,

H

+

C

H

H

C

H

C

R

alkyne

In

this

equation,

hydrogenatom.

608

the

Pd(s),

hydrogen gas to alkenes:

symbol

hydrogen gas

R

represents

either

an

R

alkene

alkyl

group or a

Reactivity

Alkenes

are

reduced

by

hydrogen

gas

to

C

+

reactions

H

H

H

R

C C

R''

alkene

reaction

R

R'''

R''

same

transfer

R

C

This

Electron

alkanes:

H R

3.2

also

c atalysts

hydrogen gas

requires

c an

be

a

used

transition

to

reduce

R'''

alkane

metal

c atalyst,

alkynes

such

directly

to

as

Ni(s)

alkanes

or

Pt(s). The

using

excess

hydrogen gas:

R–C≡C–R’

+

alkyne

2H–H

→

R–CH

excess

2

–CH

2

–R’

alkane

hydrogen

Reduction

of

alkynes

and

alkenes

decreases

the

degree

of

unsaturation of these

compounds.

ATL

Thinking skills

In

this

task,

in

this

unit

transfer

notes.

known

•

will

as

create

identify

reactions.

This

task

is

You

the

will

a

concept

map

to

summarize

connections

between

need

sheet

based on a

a

large

of

the

various

paper

material

aspects

and

of

covered

electron

several sticky

Harvard Project Zero Visible Thinking Routine

Generate-Sort-Connect-Elaborate.

M ake

a

These

•

you

and

Write

list

are

the

of

the

the

title

key

nodes

words in this unit and write them on sticky notes.

of

your

“Electron

concept

transfer

map.

reactions”

at

the

centre

of

a

large

sheet of

paper.

•

Arrange

general

•

Draw

the

to

nodes

more

lines

(sticky

notes)

on

the

paper

around

the

title,

from

specic.

between

pairs

of

nodes

to

represent

the

connections

between

them.

•

Write

key

•

a

brief

words

Share

your

received

statement

are

along

each

connecting

line

to

describe

how the

linked.

concept

map

with

your

class

and

expand

it

once

you

have

feedback.

Practice questions

7 .

Predict

the

products

a.

propene

b.

pent-1-yne.

Write

the

full

of

the

equations,

reaction

including

of

excess

displayed

hydrogen gas with:

formulas,

for

these

reactions.

609

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

Standard electrode potentials

(Reactivity 3.2.12)

The

be

ease

of

oxidation

described

and

reduction

numeric ally using a

of

a

species

in

an

electrochemic al

standard electrode potential, or

cell

c an

standard

⦵

reduction potential,

hydrogen-based

.

Standard

where

electrode

the

1

+

H

E

half-cell

(aq)

+

e

potentials

following

reaction

are

dened

relative to a

occurs:

⦵

H

⇌

2

(g)

E

=

0 V

2

This half-cell is known as the standard hydrogen electrode

a

standard

SHE

bec ause

Species

of

in

with

oxidation,

the

will

reactivity series and the

corresponding

standard

the

are

They

greatest

data

given

booklet.

in

are

provided

reduced

more

series.

ease

will

ease

be

of

is

in

the

inert

platinum

(SHE) and is assigned

electrode

is

used in the

hydrogen gas and not a metal.

to

with

reduction

lower

An

standard

tendency

Species

of

zero.

species

negative

greater

of

a

electrode

reduce

more

and

potential

other

positive

greater

will

have

greater

standard

tendency

to

electrode potential

oxidize other

reactivity series. Of all the elements, lithium has

oxidation

and

potentials

are

uorine

has

the

greatest

ease

of

reduction.

electrode

measured

in

volts

(V).

They

are

correct

for

section 19 of temperature

and

pressure

(SATP)

conditions

(temperature

=

298 K,

=

100 kPa).

All

aqueous

species

present

in

the

half-cell

equation must

section 4. 3

also

have

a

concentration

of

1.0 mol dm

Practice question

8.

The

standard

electrode

potentials

of

four

metals

are

given in table 3.

⦵

Metal

tin,

E

Sn(s)

c alcium,

lithium,

/ V

−0.14

C a(s)

−2.87

Li(s)

aluminium,

−3.04

Al(s)

−1.66

⦵

p Table 3

Order

E

these

readily

610

ease

species. They will be higher

SATP conditions pressure

in

potential

electrode

standard the

a

and

greater

Standard potentials

the

reactivity

have

species.

The

electrode

values for selected metals

metals

oxidized

according

rst.

to

their

ease

of

oxidation, with the most

Reactivity

3.2

Electron

transfer

reactions

LHA

Standard cell potentials (Reactivity 3.2.12)

The

in

voltage

each

of

an

half-cell.

electrochemic al

The

further

apart

cell

the

depends

species

on

are

the

on

identity

the

of

the

electrodes

reactivity series, the

⦵

greater

the

voltage.

This

voltage,

known as the

standard cell potential,

E

, cell

c an

be

c alculated

by

nding

the

dierence

between

the

standard

electrode

potentials of the half-cells:

⦵

E

⦵

=

cell

E

⦵

(reduced

species) –

E

(oxidized

species)

⦵

For

a

a

reaction

positive

in

an

value.

In

electrochemic al

that

c ase,

the

cell

to

be

reduction

spontaneous,

will

occur

at

the

E

must

cell

have

electrode with

⦵

the

more

positive

value of

E

(the

c athode)

and

the

oxidation will occur at the

⦵

electrode

with

⦵

E

the

more

negative

⦵

=

cell

E

value of

E

(the

anode).

(c athode)

−

E

The

data

an

2

electrochemic al

booklet

electrode

states

potentials

+

Fe

2

as

the

cell

composed

half-cell

+

2e

Cu

+

+

Fe

2

and

copper

reduced

forward

to

standard

Fe(s)

E

= −0.45 V

⦵

2e

⇌

Cu(s)

electrode

Cu(s),

| Cu half-cells.

their

E

= +0.34 V

⦵

The

+

| Fe and Cu

reactions

follows:

−

(aq)

words:

⦵

⇌

+

of

reduction

−

(aq)

other

(anode)

2

Consider

In

⦵

and

has

the

the

more

positive

half-equation

at

value of

that

E

2

, so Cu

electrode

will

+

(aq) will

be

proceed in the

direction.

⦵

The

iron

2

to

electrode

(aq),

direction

to

and

you

⇌

c an

stated

more

negative

that

value of

E

cell

=

+

,

so

the

equations

will

Fe(s)

will

proceed

be

in

oxidized

the

reverse

2e

determine

E

electrode

above:

the

overall

standard potential of the cell:

⦵

=

nd

at

−

(aq)

⦵

(reduced

= (+0.34 V)

To

the

half-equation

+

Fe

⦵

E

the

that

2

Fe(s)

Now

has

+

Fe

−

species)

−

E

(oxidized

species)

(−0.45 V)

0.79 V

overall

equation

together,

for

c ancelling

the

electrochemic al cell, add the two half-

the

electrons

and

ensuring

that

the

equation is

balanced:

2

Fe(s)

This

+

+ Cu

reaction

is

2

(aq)

⇌

+

Fe

described

(aq)

+

Cu(s)

as

being

to

predict

spontaneous

in

the

forward

direction.

You

⦵

c an

therefore use

reversible

redox

E

data

reaction

in

an

which

direction

will

be

spontaneous

for a

electrochemic al cell.

611

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Worked example 9

2

+

An electrochemic al cell composed of a Zn

in gure 22.

+

2H

2

(aq)

| Zn half-cell and

a standard

hydrogen electrode (SHE) half-cell is shown

The redox reaction in the cell c an be written as follows:

+

Zn(s)

+

⇌ Zn

(aq)

+ H

2

(g)

voltmeter

e

H



e

zinc

gas

at 1 bar

salt bridge

ZnSO

(aq)

4

3

(1 mol

latinm

dm

)

electrode

Hl(aq)

3

(1 mol

standard

hydrogen

dm

)

electrode

standard zinc half-cell

2

p Figure 22

is set

up

The electrochemic al cell with a SHE half-cell and

a Zn | Zn

+

half-cell.

The voltmeter

to measure the overall cell potential

a.

State which half-cell contains the anode and

b.

Predict

c.

Show that

whether the forward

which contains the c athode.

or backwards reaction is spontaneous.

⦵

the standard cell potential,

E

cell

,

is +0.76 V using section 19 of the data booklet.

Solution

a.

First,

write

standard

2

the

half-cell

electrode

reduction

potentials

Before

(aq)

+

2e

⇌

Zn(s)

E

you

+

H

(aq)

+

e

half-equations

together,

you

rst

hydrogen

half-equation

of

electrons

in

to

ensure

each

that

there

is

an

equal

half-equation:

⦵

H

⇌

the

= −0.76 V number

1

add

need to double the stoichiometric coecients in the

⦵

+

Zn

reactions and

from the data booklet:

2

(g)

E

=

0.00 V

2

+

2H

−

(aq)

+

2e

⇌

H

2

(g)

⦵

The SHE half-cell has the more positive value of

E

, so it

Now will

involve

the

c athode.

a

reduction

reaction

and

add

the

half-equations

together,

c ancelling the

therefore contains

electrons:

⦵

The

so

it

zinc

will

half-cell

involve

has

an

the

more

oxidation

negative

reaction

value of

and

E

+

,

2H

2

(aq)

Therefore, contains

the

Reduction

equation

occurs at the SHE half-cell, so its half-

will

equation

proceed

in

the

forward

direction

as

occurs at the zinc half-cell, so its half-

will

612

(aq)

+

H

2

(g).

reaction

for

will

the

be

spontaneous in the

equation

proceed

in

the

reverse

Zn

+

−

(aq)

+

2e

given in the

given

E

⦵

cell

=

E

⦵

(reduced

(0.00 V)

−

direction:

=

⇌

+

Zn

species)

booklet.

2

Zn(s)

the

direction

⦵

Oxidation

→

question.

c. in the data

Zn(s)

anode.

forward

b.

+

therefore

0.76 V

(−0.76 V)

−

E

(oxidized

species):

Reactivity

3.2

Electron

transfer

reactions

LHA

Worked example 10

2

+

An electrochemic al cell comprises a Cu

+

| Cu half-cell and

an Ag

| Ag

half-cell.

a.

State which half-cell contains the anode, and which contains the

c athode.

b.

Write the spontaneous reaction that

occurs in the electrochemic al cell.

c.

C alculate the standard cell potential,

⦵

E

cell

.

Solution

a.

Copy

cell

the

half-equations

+

Ag

2

the

standard

reduction

potentials

for

each half-

silver

+

e

⇌

+

Ag(s)

E

+

2e

has

half-cell

= +0.80 V

⦵

−

(aq)

half-cell

copper

⦵

−

(aq)

Cu

The

and

from the data booklet:

⇌

the

Cu(s)

more

contains

E

positive

the

= +0.34 V

value,

so

it

contains

the

c athode, and

anode.

⦵

b.

The

in

silver

this

half-cell

half-cell

direction.

needs

to

you

⇌

add

stoichiometric

equal

number

Now

add

positive

in

the

is

the

+

copper

half-equations

of

in

the

in

each

electrons

together,

silver

It

is

E

=

+

2e

⇌

but

Cu(s)

the

species

(reduced

=

0.46 V

by

the

occurs

correct

half-equation

you

rst

need to double the

to

ensure

that

there is an

together,

c ancelling

→

2Ag(s)

+

Cu

the

electrons:

+

(aq)

⦵

(+0.80 V)

important

so

the

2Ag(s)

2

+

=

multiplied

half-cell,

in

−

(aq)

E

reduction

half-equation:

⦵

cell

so

shown

half-equation

+

c.

value,

2e

half-equations

2Ag

⦵

E

already

−

(aq)

coecients

(aq)

the

occurs

+

Cu

+

2Ag

more

half-equation

reversed:

2

Before

the

the

Oxidation

be

Cu(s)

has

and

to

−

note

two.

species)

E

(oxidized

species):

(+0.34 V)

that

The

−

the

half-cell

electrons

only

lost

voltage

and

potential

of

the

cell

involved

in

the

half-equation,

for

gained

depends

and

on

the

not

on

silver

need

electrode is not

to

be

chemic al

the

way

balanced,

nature of the

the

half-equation is

balanced.

613

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

Gibbs energy and standard cell potentials

(Reactivity 3.2.14)

⦵

The

This

equation

and

the

standard change in Gibbs energy,

reaction

the

F araday

constant

are

occurring

in

an

cell,

,

over

c an

the

be

course

of

determined

a

chemic al

from the

⦵

E

cell

:

sections 1 and 2 of the data

⦵

booklet.

Change

ΔG,

also

c an

be

in

Gibbs

ΔG

energy,

the

enthalpy

⦵

= −nFE

cell

dened in terms where

of

electrochemic al

given standard cell potential,

in

ΔG

value of

change,

n

is

the

number

of

electrons

transferred

in

and

the

balanced

redox

equation,

entropy 1

4

change

and

F

is

the

You

know

F araday

constant,

9.65 × 10

C mol

temperature, as

described in

Reactivity 1.4

⦵

(AHL). that

an

electrochemic al

As

the

right-hand

term

be

spontaneous if

ΔG

in

is

the

reaction

equation

will

above

be

has

spontaneous if

a

negative

sign,

E

a

cell

is

positive.

reaction will

negative.

1

F

has

units

of

coulombs

per

mole,

C mol

,

where coulomb is the SI unit of

⦵

charge.

E

⦵

cell

has

units

of

volts,

V.

When the

n,

F

and

E

terms

cell

are

multiplied

1

together,

unit

of

the

resulting

energy

per

unit

value has units of C V mol

charge;

in

other

1

1 J C

one

One

joule

volt

is

per

equivalent to one

coulomb:

1 V

=

1

.

Substituting

1

C J C

this

into

1

mol

C V mol

gives:

1

=

J mol

⦵

Therefore, the units of

by

words,

.

dividing

by

⦵

1

ΔG

are J mol

, or more oen

ΔG

1

is converted to kJ mol

1,000.

Worked example 11

⦵

In worked example 9,

E

+

cell

for the reaction 2H

2

(aq)

+

Zn(s)

→ Zn

+

(aq)

+

⦵

H

2

(g) was c alculated to be

+0.76 V.

C alculate

ΔG

for this reaction.

Solution

Two

electrons

are

⦵

ΔG

transferred

in

this

redox

reaction, so

n

=

2.

⦵

= −nFE

cell

1

4

= −2

×

(9.65 × 10

C mol

×

0.76 J C

1

5

= −1.47 × 10

1

)

J mol

1

or

−147 kJ mol

Practice questions

⦵

9.

The

standard

following

change

in

Gibbs

electrochemic al

Fe(s) + CuSO

a.

State

b.

C alculate

4

whether

(aq)

the

→

energy,

ΔG

1

, is

–152 kJ mol

for the

reaction:

FeSO

reaction

4

(aq)

will

+

Cu(s)

occur

spontaneously.

⦵

the

value of

E

cell

. Compare

the

value

obtained with the

2+

dierence

in

electrodes,

614

standard

which

are

reduction

given

in

potentials

of

Fe

2+

/Fe and Cu

section 19 of the data booklet.

/Cu

Reactivity

3.2

Electron

transfer

reactions

LHA

Measuring standard cell potentials

3

The

displacement

ions

is

as

reaction

2

Zn(s)

If

the

into

salt

+

oxidation

two

+

Cu

and

half-cells

bridge,

redox

between zinc and copper(II)

•

50

cm

•

filter paper

•

tweezers

•

high-resistance

•

crocodile clips

•

connecting

beakers

follows:

2

(aq)

is

Cu(s)

reduction

connected

chemic al

reaction

→

energy

converted

+

+

Zn

(aq)

processes

by

an

from

into

are

external

the

voltmeter

separated

wire and

wires

spontaneous

electric al

energy. In this

Instructions practic al,

you

copper cell

will

measure the cell potential of a zinc–

and use this to determine

∆G

for

the

reaction.

−3

1.

Prepare

a

1.0 mol dm

copper(II) sulfate solution and

3

transfer

5.0 cm

of

it

into

a

weighing

bottle.

3

You

will

be

using

1.0 mol dm

solutions,

which

require −3

2. large

quantities

of

the

corresponding

hydrated

salts.

Prepare

a

1.0 mol dm

zinc sulfate solution and

To 3

transfer minimize

be

used.

the

waste,

Once

electrolyte

they

c an

be

very

small

nished,

solutions

used

for

volumes

you

are

instead

other

of

the

encouraged

of

5.0 cm

of

it

to

a

second

weighing

bottle.

electrolytes will

to

recover

3.

disc arding them, as

S and

the

two

electrodes

to

remove

any surface

contaminants.

experiments. 4.

Prepare

a

small

volume

of

chloride solution. Cut a

saturated

potassium

strip of filter paper to use as

Relevant skills

a

•

Tool

1:

Measuring

potential

salt

into

•

Tool

1:

•

Tool

3:

Constructing

the

percentage

Construct

•

error

Tool

3:

the

connect

with •

between

the

two

half-cells.

Dip the strip

potassium chloride solution.

electrochemic al cells 5.

C alculate

bridge

difference

the

voltaic cell:

the

two

solutions

in

weighing

bottles

salt bridge

General mathematics

•

connect

the

crocodile

electrodes

clips

and

to

the

voltmeter using the

connecting

wires

S afety

• •

Wear

•

Copper(II)

eye

dip

the

electrodes

•

their

corresponding

protection solutions

irritant.

into

sulfate

Avoid

Copper(II)

and

contact

sulfate

and

zinc

with

zinc

sulfate

in

weighing

bottles.

are harmful and

6.

Measure

7 .

Clear

the

potential

difference.

eyes and skin

sulfate

are

up

according

to

the

directions

given

by

your

toxic to the

teacher. environment

and

must

be

disposed

of

safely

Questions

Materials

1.

•

copper(II)

sulfate

•

zinc

•

distilled

•

zinc

•

copper

•

potassium chloride

•

two

sulfate

pentahydrate, CuSO

heptahydrate, ZnSO

4

•7H

2

4

•5H

2

O

your

2.

electrode

3.

electrode

weighing

•

sandpaper

bottles

(or other small wide-mouth

theoretic al

cell,

Compare

your

value

c alculate

and

Suggest

value

containers)

the

zinc–copper

potentials in the data

O

water

C alculate

at

the

two

from

∆G from

cell

potential

standard

result

to

percentage

reasons

the

the

for

electrode

booklet.

experimental

least

differs

standard

using

why

theoretic al

your

the

theoretic al

error.

your

measured

value.

4.

C alculate

5.

Research the relationship between E

measured cell potential.

cell

and ∆G under

non-standard conditions. Use this to briefly outline two

research questions related to voltaic cells, E

cell

and ∆G

615

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Electrolysis of aqueous solutions

(Reactivity 3.2.15)

We

have

discussed

reduced

ionic

and

salts

compete

anode

The

the

anions

introduces

with

and

the

electrolysis

are

oxidized.

oxidation

redox

of

molten

The

and

reactions

ionic

electrolysis

reduction

of

the

salts

of

reactions

anions

and

O(l)

involving

c ations

of

water

c an

proceed

as

are

of

water, which

the

salt at the

follows:

1 2

c ations

c athode.

reduction

H

where

aqueous solutions of

+ e

⦵

→

H

2

(g)

+

OH

(aq)

E

= −0.83 V

2

If

the

standard

−0.83 V,

then

hydrogen

The

other

booklet,

electrode

water

gas

will

be

possible

the

will

at

competing

of

of

the

salt

preferentially

formed

reduction

1

potential

be

the

is

is

is

the

given

as

2

(g)

+ 2H

the

negative than

salt

c ation and

oxidation

of

water. In the data

follows:

⦵

+

O

more

over

c athode.

reaction

oxygen

c ation

reduced

(aq)

+

2e

→ H

2

O(l)

E

= +1.23 V

2

Reversing

the

reduction

potential

equation

has

gives

the

to

reversed

be

1 H

2

O(l)

half-equation

as

well,

for

the

giving

2

(g)

+

2H

of

water. The

oxidation potential:

⦵

+

O

→

oxidation

an

(aq)

+ 2e

E

= −1.23 V

2

If

the

oxidation

water

will

formed

For

at

be

the

example,

chloride,

potential

consider

NaCl(l),

the

electrolysis

in

the

reduction

of

the

salt

anion

oxidized

over

is

more

the

negative than

salt

anion

and

−1.23 V, then

oxygen gas will be

anode.

In

oxidation

of

preferentially

and

of

of

two

one

electrolytic

composed

NaCl(l),

sodium

chloride

ions

electric al

ions

to

to

cells:

of

an

one

energy

form

composed of molten sodium

aqueous

is

sodium

sodium

provided

metal

at

to

the

chloride,

the

cell,

NaCl(aq).

resulting

c athode and the

form chlorine gas at the anode:

+

c athode:

Na

+ e

→

Na(l)

1 anode: Cl

→

Cl

2

(g)

+

e

2

overall

In

the

salt,

equation:

electrolysis

oxidation

two

cell

and

species

and

of

aqueous

reduction

that

water.

c an

The

2NaCl(l)

sodium

reactions

potentially

two

→

2Na(l)

hydroxide,

involving

be

+ Cl

(g)

NaCl(aq), the two competing

water

reduced

competing

2

at

reduction

the

are

reactions

(aq)

+

e

→

Na(s)

E

1 H

2

are

There

sodium

as

ions

are

now

from the

follows:

⦵

+

Na

introduced.

c athode:

O(l)

+

e

→

H

2

(g)

+

OH

= −2.71 V

⦵

−

(aq)

E

= −0.83 V

2

The

reduction

more

easily

potential

reduced

of

than

water

the

is

less

sodium

negative

ions.

than

that

Therefore,

the

of

sodium,

only

so

product

water is

formed

+

at

the

c athode

will

be

hydrogen gas, H

2

(g).

The

sodium

−

hydroxide ions, OH

616

(aq),

will

stay in the solution.

ions,

Na

(aq), and

Reactivity

chloride

anode.

The

oxidation

ions

from

the

salt

half-equations

of

chloride

ions

and

and

and

water

standard

water

are

also

compete

electrode

shown

to

be

potentials

Electron

transfer

reactions

LHA

The

3.2

oxidized at the

involving the

below.

1

⦵

Cl

2

(g)

+

e

→ Cl

(aq)

E

chlorine gas

= +1.36 V

hydrogen gas

2

1

⦵

+

O

2

(g)

+

2H

(aq)

+

2e

→

H

2

O(l)

E

sodium

= +1.23 V

2 chloride

The

equations

need

to

be

reversed

to

reect

the

oxidation

1 Cl

of

these

species:

solution

⦵

Cl

(aq) →

2

(g)

+

e

E

= −1.36 V

2

1 H

2

⦵

+

O(l) →

O

2

(g)

+

2H

(aq)

+

2e

E

= −1.23 V

2

The

chloride

expect

the

ions

water

to

dierence

high

a

between

overall

cell

2

more

oxidized

the

concentration of Cl

chlorine gas, Cl

The

have

be

(g)

negative

two

(aq)

equation

+ Cl

to

than

water,

chloride

ions.

so

we

c an

However,

the

main

product

formed at the anode will be

(gure 23).

is

as

follows:

1 (aq)

potential

compared

oxidation potentials is small, so in solutions with

ions

+

Na

oxidation

preferentially

(aq)

+

H

2

O(l)

1 H

→

2

(g)

+

Cl

+

2

2

(g)

+

Na

(aq)

+

OH

(aq)

2

carbon rod

carbon rod

as positive

as negative

electrode ( +)

electrode (

+

or

1 NaCl(aq)

+

H

2

O(l)

)

1

→

H

2

(g)

+

Cl

2

2

(g)

+

NaOH(aq) p Figure 23

2

The electrolysis of aqueous

sodium chloride

Doubling

the

stoichiometric

2NaCl(aq)

In

is

dilute

the

only

2H

solutions

oxidized

chlorine

+

along

gas

is

of

2

O(l)

at

H

2

NaCl(aq),

with

the

at

(g)

+ Cl

the

chloride

produced

product

→

coecients

(g)

+

the

nal

In

that

anode.

In

equation:

2NaOH(aq)

concentration of Cl

ions.

the

2

gives

c ase,

very

a

(aq)

mixture

dilute

ions

of

is

low,

so

water

oxygen gas and

solutions,

oxygen gas will be

anode.

Worked example 12

Deduce the products of the electrolysis of

CuSO

4

(aq),

with inert

electrodes.

aqueous copper(II) sulfate,

Write the overall cell equation.

Solution

First,

write

reduction

the

of

2

half-equations

copper(II)

ions

and

and

standard

electrode

+

Cu

2

involving the

⦵

(aq)

+

2e

→

Cu(s)

E

1 H

potentials

water.

O(l)

+

e

H

→

= +0.34 V

⦵

2

(g)

+

OH

(aq)

E

= −0.83 V

2

The

so

reduction

copper(II)

potential

ions

will

of

be

copper(II)

reduced

to

ions

is

more

copper

positive

metal

at

the

than

that

of

water,

c athode.

2

The

two

(aq),

is

in

its

from

competing

and

water.

highest

sulfate

In

species

this

c ase,

for

the

oxidation

sulfate

oxidation state of

would

result

in

an

+6.

at

ions

the

Therefore,

impossible

anode

c annot

be

are sulfate ions, SO

oxidized

removing

electron

any

4

bec ause sulfur

more

conguration

for

electrons

sulfur.

617

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

This

means

released,

that

with

water

will

hydrogen

be

1 H

2

Deduce

the

the

of

overall

aqueous

2

2

Cr

2

O

7

O

2

(g)

cell

+

2H

equation

+

Cu

K

anode

and

oxygen gas will be

(aq)

+

2e

is

therefore:

products of

electrolysis

potassium

the

2

The 10.

at

staying in solution:

+

O(l) →

Practice question

oxidized

ions

1

2

(aq)

+

SO

(aq)

4

+ H

2

O(l)

→

Cu(s)

+

O

+

2

(g)

+

2H

2

(aq)

+

SO

(aq)

4

2

dichromate,

or

(aq), with inert

electrodes. 1 CuSO

4

(aq)

+

H

2

O(l)

→

Cu(s)

O

+

2

(g)

+

H

2

SO

4

(aq)

2

Doubling

the

2CuSO

4

stoichiometric

(aq)

+

2H

2

O(l)

coecients

→

2Cu(s)

+

gives

O

2

(g)

the

+

nal

2H

2

SO

equation:

4

(aq)

Electroplating (Reactivity 3.2.16)

The

electrolysis

uses

inert

made

will

from

add

a

are

example,

to

(aq),

right).

to

form

metal.

2

Cu

as

+

The

the

plating

an

with

sulfate

24,

le).

electrolysis

described

However,

of

and

take

and

eroding,

an

in

if

worked

the

aqueous

material

away

example12

electrodes

solution

from

the

of

are

an

ionic

salt

anode. These

respectively.

electrolytic cell containing a solution of copper

the

the

anode

and

electric

the

current

c athode

is

each made of copper metal

applied, the copper anode will

+

form Cu

Cu(s).

(gure

c athode

When

2

erode

to

the

known as

4

copper(II)

metal,

consider

sulfate, CuSO

(gure24,

aqueous

electrodes

reactive

material

processes

For

of

c arbon

(aq)

This

ions

process

impure

copper

while

c an

is

the

be

used

same

used

as

to

the

ions

purify

anode,

will

a

be

reduced

sample

which

of

will

at

the

c athode

impure copper

be

eroded

to

produce

+

(aq)

pure

ions.

These

copper

ions

metal.

will

The

then

be

impurities

readily

oxidized

than

copper

metal)

readily

reduced

than

copper(II)

reduced

will

or

at

either

remain

the

stay

in

c athode

on

the

the

and

anode

solution

(if

plated

(if

their

they

ions

there

are

are

less

less

ions).

+

A

A

inert electrodes

e

copper electrodes

e (Pt or graphite)

(+)

(

)

(+)

copper

oxygen

(

)

deposited

+ 2

evolved

Cu

+ 2e

copper

Cu

2

+ 4e

Cu

copper +

Cu

dissolves

2 + 2

(aq)

Electrolysis of

reduces copper(II) ions at

producing oxygen gas at

copper(II) sulfate with inert

the c athode and

(aq)

4

solution retains blue colour

electrodes

oxidizes water,

the anode. The blue colour of the solution

fades as copper(II) ions are replaced

CuSO

+ 2e

4

solution loses blue colour

618

+ 2e

O

CuSO

p Figure 24

deposited

+ 2

+

2H

with hydrogen ions.

Electrolysis using copper electrodes c auses the copper anode to

dissolve and

copper metal to deposit

on the c athode. The amount of

copper(II) ions in the solution remains constant, so the blue colour of

the solution does not fade

Reactivity

electrolysis.

travel

layer

For

A

involves

metal

through

of

metal

example,

c athode,

and

thin

of

the

on

in

coating

anode

solution

the

an

is

an

object

oxidized

to

the

to

with

form

c athode,

a

thin

layer

c ations

where

they

in

of

pure

the

are

metal

solution.

reduced

to

by

The

Electron

transfer

reactions

LHA

Electroplating

3.2

+

c ations

form a thin

c athode.

electrochemic al

copper(II)

sulfate

cell

comprising

solution,

the

steel

a

copper

c athode

anode,

will

be

a

steel

plated with a steel ring

layer

copper

copper

(gure 25).

cathode

anode

(

)

( +) to be plated

with copper

Observations copper(II) sulfate

solution Scientists make inferences from their observations. Observation involves use

of the senses. Our knowledge of the behaviour of matter then allows us to infer

conclusions from observed data. For example, you can observe gas bubbles

being generated at an electrode during electrolysis, and you can infer the

p Figure 25

In copper electroplating,

copper(II) ions are reduced

to form

at

the c athode

a thin layer of Cu(s) on the surface of

the c athode (in this c ase,

a steel ring)

identity of the gas from your knowledge of the composition of the electrolyte.

Similarly, you may observe a brownish-red solid being deposited at the cathode

during the electrolysis of aqueous copper(II) sulfate and, from that, infer that

copper is reduced at that electrode. What “counts” as an observation in science?

Worked example 13

Deduce the half-equations at each electrode in the electroplating of a steel

electrode with copper in a copper(II) sulfate solution, CuSO

4

(aq).

Solution

At

the

anode,

copper

2

Cu(s)

At

the

→

c athode,

2

Cu

metal

is

oxidized to copper(II) ions:

+

Cu

(aq)

the

+

2e

reverse

reaction

occurs:

+

(aq)

+

2e

→

Cu(s)

Practice question

11.

A nickel spoon is used in a silver electroplating experiment as shown below.

+

silver

nickel

spoon

to be plated

(cathode)

silver

a.

Write

i.

the

half-equation

c athode

for

ii.

nitrate solution

the

reaction occurring at the:

anode.

p Figure 26

b.

Describe

and

explain

what

would

happen

to

the

nitrate

the Berlin International Film

Describe

and

Festival, the

electrolyte. Golden Bear,

c.

The trophy for the top prize

concentration of the at

silver(I)

explain

the

mass

changes

at

each

of

the

two

electrodes.

is made of bronze coated with

a thin layer of gold

by electroplating

619

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

End of topic questions

7.

The

following

reaction

occurs

in

a

voltaic cell.

Topic review +

2Ag

1.

Using

your

answer

knowledge

the

guiding

from the

question

as

Reactivity 3.2

fully

as

topic,

What

2+

(aq)

reaction

+

Cu(s)

occurs

at

→

2Ag(s) + Cu

each

(aq)

electrode?

possible: C athode

Anode

What happens when electrons are transferred? +

A.

Ag

2+

(aq)

→

Ag(s)

+

e

Cu(s) → Cu

(aq)

+

e

→

Ag(s)

Cu(s)

+ 2e

(aq)

+

e

→

Ag(s)

Cu(s)

→ Cu

Exam-style questions

2+

B.

Ag

C.

Ag

D.

Cu(s)

→ Cu

+

Multiple-choice questions

Which

species

contains

nitrogen

with

an

(aq)

2+

2+

2.

(aq) + 2e

+

→

Cu

(aq) + 2e

+

(aq)

+

2e

Ag

(aq)

+

e

→

Ag(s)

oxidation state

of +5? 8.

A.

N

Which

B.

N

C.

NO

O 2

2

D.

HNO 3

3.

Which

element

2MnO

(aq)

+

is

Br

reduced

(aq) + H

4

in

the

O(l)

following

A.

Mn

B.

O

C.

Br

I

alkene

II

c arboxylic acid

III

aldehyde

A.

I and II only

+

BrO

D.

H

of

the

A.

2H

B.

CH

C.

C

O

D.

Na + Cl

following is

→

2H

2

compound

c an be

(aq) + 2OH

(aq)

B.

I and III only

C.

II and III only

D.

I, II and III

Which

class

not

a

redox

+

6

→ 2

→

→

CH

COOLi + H

C

H

A.

aldehyde

B.

ether

C.

ketone

D.

c arboxylic acid

3

6

10.

2

Hydrogen

C

H

2

Li

B.

Li and F

and

most

c an

(g) + H 4

Which

A.

formed

when

a

secondary

2

Br

2NaCl

reacts

is

O

2

pair

compound

reaction?

3

Br

of

oxidized?

2

LiOH

3

+

is

O + O 2

COOH

H

Which

of

3

alcohol

2

5.

classes

reaction?

9.

3

following

→

(s) 2

Which

the

2

2MnO

4.

of

reduced?

2

be

(g)

added

→

2

of

the

C

H 2

to

ethene

to

produce

ethane.

(g) 6

following

is

correct

for

this

reaction?

vigorously?

Br Degree of

2

Species that undergoes

unsaturation

reduction

2

C.

K

and

Br 2

D.

increases

ethene

ethene

K and F 2

6.

A.

What

are

lead(II)

the

products

bromide,

PbBr

of

the

electrolysis of molten

B.

decreases

C.

increases

hydrogen

D.

decreases

hydrogen

? 2

Anode product

11.

Consider

the

standard

reduction potentials:

2+

lead

bromine

Cd

⦵

(aq) + 2e

⇌

Cd(s)

E

3+

Cr B.

lead

Which lead(II)

bromine

D.

lead(II)

bromide

is

the

2+

A.

Cd

B.

Cr

C.

Cd

D.

Cr

3+

620

⇌

Cr(s)

E

bromide

C.

=

–0.40 V

=

–0.74 V

⦵

(aq) + 3e

strongest

oxidizing agent?

LHA

C athode product

A.

Reactivity

LHA

12.

What

are

the

major

concentrated

products

sodium

of

chloride

the

electrolysis of

solution,

16.

a.

Write

NaCl(aq)?

of

b. Negative electrode

an

equation

i.

state

of

c arbon in

the

[1]

mean

oxidation

state

of

c arbon

O

H 2

2

ethanol.

[1]

Cl 2

iii.

State

Cl

H 2

and

states,

2

Extended-response questions

between

ethanedioate ions, C

O 2

,

in

whether

ethanol

Ethanol

c an

with

reference

c arbon

is

to

oxidized

oxidation

or

reduced

undergoes complete

acidic

[1]

also

be

oxidized

when

reacted with

and 4

acidied manganate(VII) ions, MnO

explain,

combustion.

c.

2–

reaction

oxidation

dioxide.

Deduce

when

The

the

2

in

13.

for the complete combustion

O

Na

D.

reactions

[1]

Deduce

ii.

B.

transfer

Positive electrode

Na

C.

Electron

ethanol.

c arbon A.

3.2

solution

is

a

potassium

dichromate(VI).

redox

4

reaction.

shown

The

incomplete

equation

reaction is

of

+ MnO 4

→

CO

4

Deduce

the

oxidation

state

of

formulas and state the names

possible

organic

products of this

oxidation

State

the

[2]

methods

used

to

isolate

each of the

c arbon in the

substances

2–

O 2

The

two

2

ethanedioate ion, C

b.

the

structural

+ Mn

ii. a.

the

reaction.

2+

O 2

this

Draw

below:

2–

C

for

i.

.

in

your

answer

above.

[1]

[1]

4

half-equation is:

17.

A

student

c arries

out

the

following

reactions

between

2–

C

O 2

(aq)

→

2CO

4

(g) + 2e

three

2

Deduce

the

including

reduction

state

unknown

solutions.

half-equation,

symbols.

The

metals

(X,

following

Y

and

results

Deduce

the

full

redox

reaction.

+

Z(NO

) 3

Sodium

a.

chloride

Describe

is

how

found

in

table

bonding

Observations

(aq)

Y

is

a

clear

conductor

sodium

when

chloride

solid

but

is

c an

not

an

on

the

conduct

electricity

molten.

[1]

X

+

Y(NO

Z

+

X(NO

)

Molten

sodium

chloride

c an

be

the

half-equation

for

)

of

the

solution’s

solid

metal.

blue

(aq) is 2

reacted

appeared

O ver

colour

faded.

(aq)

No

(aq)

No

reaction

reaction

No

reaction

2

electrolysed. Z

Identify

red-brown

surface

the

When

2

) 3

c.

a

solution.

electric al

3

when

grey metal. Z(NO

blue

together,

occurs in sodium

[2]

why

shiny

3

a

time, Explain

obtained:

2

salt.

chloride.

b.

were

several dilute

[1] Y

14.

and

[2] Reactants

c.

Z)

the

+

HCl(aq)

reaction that takes a.

List

metals

X,

Y

and

Z,

in

order

of

increasing

place at the reactivity.

i.

ii.

c athode

anode.

b.

Suggest,

LHA

Deduce

c.

A

that

the

takes

half-equation

place

when

for

dilute

the

oxidation

is

metal,

standard

A

student

chloride

prepares

solution

phenolphthalein.

solution.

student

two

observations,

or

Z,

could

be

copper.

voltaic

cell

is

[2]

set

up.

Draw

the

cell

diagram

for

this

voltaic cell and

[1]

adds

The

the

Y

the

following:

concentrated sodium

and

Describe

during

a

X,

nickel–copper

aqueous sodium

electrolysed.

label

e.

reference to

reaction

i. chloride

with

[1] which

d.

[1]

[1]

a

few

student

two

-

ions

-

c athode

in

each solution

-

anode

-

direction

drops of

then

electrolyses the

observations

made

by the

electrolysis.

of

travel

of

electrons

of

travel

of

c ations

in

the

external

[2]

circuit

+

15.

a.

Li

+ e

→

Li(s)

electrodes

electrode

in

is

a

a

reaction occurring at one of the

lithium-ion

-

(anode/c athode)

this

reaction

whether

this

is

the

charging

or

Deduce

Explain

[2]

equation,

including

state

symbols,

the

at

the

opposite

electrode.

why

lithium-ion

batteries

must

be

this

spontaneous

redox

reaction

that

occurs

cell.

[2]

[1]

sealed. [1]

C alculate

redox

the

standard

reaction.

booklet.

Refer

cell

to

potential,

in

V,

LHA

reaction

d.

c.

the

[2] in

the

salt

discharging for

Identify

the

occurs

reaction.

b.

in

bridge.

ii.

and

direction

battery. State at which

for the

section 19 of the data

[1]

621

Reactivity 3.3

Electron sharing reactions

What happens when a species possesses an unpaired electron?

When

an

atoms

breaks

atoms.

atom

The

or

polyatomic

through

radic als

the

are

species

process

highly

of

has

an

unpaired

homolytic

reactive

and

c an

electron,

ssion,

the

combine

two

with

it

is

c alled

electrons

other

a

radic al.

involved

radic als

to

When

in

form

the

a

covalent

bond

more

move

stable

bond

onto

covalent

between two

the

separate

molecules.

Understandings

Reactivity 3.3.1 — A

an

unpaired

radic al

electron.

Reactivity 3.3.2 — R adic als

ssion,

e.g.

light

heat.

or

of

is

a

R adic als

halogens,

in

are

the

molecular entity that has

are

highly

produced

presence

Reactivity 3.2.3 — R adic als take part in substitution

reactive.

reactions

with

alkanes,

producing

a

mixture

of

products.

by homolytic

of

ultraviolet (UV)

Introduction to radic als (Reactivity 3.3.1)

Organic

reaction

reactants

are

broken

One

A

type

radic al

from

of

to

and

of

is

a

other

species

species.

and

entity

(such

In

how

electrons

that

contrast,

A

has

ions)

in

an

in

a

means

move

such

a

over

and

the

the

c an

anions

is

conversion of

explaining which bonds

course

electron.

radic al

species

of

of

mechanisms

unpaired

that

c ations

radic al

descriptions

are

are

of

the

reaction.

radic als.

R adic als

are

dierent

exist independently

will

always

indic ated

by

a

have a

dot

(•).

There

are

two

common

•

a single atom, such as a halogen radical, for example the chlorine radical, Cl•

•

a

polyatomic

of

detailed

involved

as

counter-ion.

types

are

mechanisms

species

chemic al

corresponding

These

formed,

chemic al

charged

any

mechanisms

products.

radic al:

species,

for

example

the

methyl

radic al, •CH

3

,

and

the

hydroxyl

radic al, •OH.

When

the

placed

radic al

next

radic al,

the

to

consists

the

dot

is

atom

of

several

with

placed

the

next

to

atoms,

unpaired

c arbon,

the

dot

in

electron.

as

c arbon

the

For

has

chemic al

example,

the

formula is

in

the

unpaired

methyl

electron:

H

C

H

H

Due

to

their

reaction,

react

with

unpaired

species

This

622

high

that

is,

each

reactivity,

they

other

electrons.

c an

also

process

is

are

to

the

form

This

react

radic als

not

new

process

with

known as

a

are

usually

ultimate

is

covalently

known as

non-radic al

propagation.

formed

reaction

as

intermediates in a

products.

Two

radic als

c an

bonded compound with no

termination.

species

to

create

However,

further

radic al

radic al

species.

Reactivity

3.3

Electron

sharing

reactions

Hypotheses

Scientists

universal

living

make

things

ageing

provisional

observation

deteriorate

(also

mid-20th

that

known

century

mitochondria

as

and

(gure

puzzled

physic ally

the

has

1)

explanations

has

as

free

been

a

over

radic al

for

the

humans

time.

theory

widely

by-product

patterns

The

of

free

radic al

ageing)

accepted.

of

they

observe. A

for a long time is ageing. All

was

R adic als

metabolic

hypothesis of

proposed in the

are

produced in

processes.

0.5 μm

p Figure 1

The

The structure of a mitochondrion

gradual

with

the

hypothesis

from

the

accumulation

oxidation

suggests

build-up

mitochondrial

However,

The

free

What

of

that

critics

of

ageing

is

this

of

radic al

nature

by

is

the

have

are

time

“oxidative

been

is

associated

The

free

radic al

stress” that arises

observed

between

age,

damage.

the

c ausality

therefore

science

over

biomolecules.

oxidative

challenge

ageing

of

species

other

c aused

and

hypothesis

of

and

Correlations

production

hypothesis

aspects

these

proteins

radic als.

radic al

radic al

other

of

of

DNA,

an

area

hypotheses

of

of

the

relationship.

ongoing

connected

research.

to?

Formation of radic als (Reactivity 3.3.2)

Curly

arrows

are

mechanisms,

as

movement

a

p Figure 2

A

of

used

pair

of

illustrate

are

or

electrons

sh

the

broken

The double-barbed

single-barbed,

electron

to

bonds

movement

and

made.

A

of

electrons

reaction

arrow

shows the

(gure 2).

arrow represents the movement

hook,

in

double-barbed

arrow

is

used

to

show

the

of

an electron pair

movement of a single

(gure 3).

p Figure 3

The sh hook

arrow represents the movement

of

a single electron

623

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

When

you

draw

•

The

base

•

The

arrowhead

•

The

arrow

region

Fish

is

hook

formed

covalent

each

of

mechanisms

of

arrows

bond

a

arrow

must

starts

the

when

have

halogen

the

are

a

an

split

Notice

at

at

the

or

other

diatomic

hook

the

destination

region

used

in

and

to

moving

of

ends

the

at

the

between

Halogen

halogen

reactions

the

following:

electrons.

electrons.

an

electron-poor

involving

homolytic ssion.

two

atoms

radic als

are

molecule

is

The

resulting

formed

broken

in

this

radic als.

two

two

way,

A

radic al

electrons of a

radic als that

where the

homolytic ally.

Cl

2

sh

of

attention

species.

Homolytic ssion of chlorine, Cl

how

pay

origin

exact

undergoes

evenly

arrows,

the

electron-rich

electron.

a

curly

start

commonly

+

p Figure 4

nish

molecule

are

in

must

molecule

single

bond

at

using

curly

arrows

are

,

to form

used

to

chlorine radic als, Cl•

show

the

path

of

each

electron that

make up the chlorine–chlorine bond.

For

to

homolytic

ultraviolet

halogen

For

this

ssion

(UV)

radic als

reason,

is

it

of

light,

the

is

halogens

or

to

heated.

rst

step

in

known as the

occur,

The

a

the

reaction

homolytic

series

of

initiation

mixture

ssion

chain

of

must

halogens

reactions

be

to

initiated

exposed

form

by

radic als.

step.

R adic al substitution reactions

(Reactivity 3.3.3)

TOK

One Organic

common

type

of

reaction

in

organic

chemistry

is

substitution

reactions.

mechanisms depict the Substitution

movement

of

molecule reaction.

In

is

the

replacement

of

an

atom

or

a

group

of

atoms

in

an

organic

electrons during a

this

section,

you

with

another

atom

or

group of atoms.

have

–1

encountered

sh

hook

arrows

Alkanes

are

relatively

inert

due

to

the

strength

of

the

c arbon–c arbon

(346 kJ mol

)

–1

(gure

3),

which

movement

of

a

single

Double-barbed

represent

electron

in

the

represent the

electron.

arrows

(gure2)

movement of

pairs.

These

are

discussed

and

c arbon–hydrogen

non-polar,

alkanes

replaced

using

which

into

makes

more

with

radic al

(414 kJ mol

them

reactive

polar

bonds.

unreactive

In

addition,

towards

polar

the

bonds

reagents.

in

To

alkanes

transform

One

way of achieving this is to

halogenate

the

alkane

reactions.

Reactivity 3.4

For

To

what

extent

mechanisms

explanatory

are

curly-arrow

descriptive,

or

example,

heat,

when

methane

chloromethane

example

of

a

radic al

and

reacts

with

hydrogen

substitution

chlorine

chloride

are

in

the

presence of UV light or

formed

(gure 5). This is an

reaction.

interpretative? H

H

UV light H

C

H

+

Cl

Cl

H or

H

p Figure 5

624

are

species, some of their non-polar bonds must be

bonds.

substitution

)

C

Cl

+

H

heat

H

R adic al substitution in methane to form

chloromethane

Cl

Reactivity

There

are

three

stages

involved

in

radic al

substitution

3.3

Electron

sharing

reactions

reactions: initiation,

propagation and termination.

Initiation

The initiation stage involves the homolytic ssion of a

species.

In

producing

below

the

presence

two

show

of

identic al

the

UV

light,

chlorine

movement

of

the

chlorine

radic als,

Cl•

(gure

step.

6).

splits

The

homolytic ally,

Lewis

structures

electrons.

+

Cl

p Figure 6

molecule to produce radic al

molecule

Cl

The homolytic ssion of chlorine is the initiation

This is the same as the reaction in gure 4

Propagation

The

propagation

species

this

to

c ase,

chlorine

form

the

stage

a

rst

radic al,

includes

dierent

pair

propagation

Cl•

reactions

of

a

of

a

radic al

non-radic al

step

occurs

between

H

this

step,

a

propagate.

C

H

propagation

The

methyl

desired

Cl•

methyl

step

radic al

is

formed,

substitution

•CH

3

,

halogenoalkane,

(gure

is

further

reacts

which

therefore

reactive

with

a

allows

a

chain

species. In

molecule and a

+ CH

Cl

chlorine

chloromethane,

the

reaction

reaction,

as

to

continue, or

reactions in the

radic als.

CH

3

molecule,

Cl,

and

producing

another

the

chlorine

radic al,

H

C

Cl

H

Cl

H

p Figure 8

of

the

chlorine

two

C +

H

The second

regenerated

cycle

non-radic al

8).

H

This

a

in the radic al substitution of methane

H

The

methane

H

Cl

propagation step

produce

radic al,

with

radic al

H

The rst

R adic al

a

H

H

In

a

and

(gure 7).

H

p Figure 7

species

species

propagation step

radic al

c an

take

in the radic al substitution of methane

part

again

in

the

rst

propagation

step.

propagation steps will continue until a termination step

occurs.

625

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Termination

The

termination

non-radic al

of

radic als

that

in

these

the

the

includes

The

radic al

mixture.

need

reactions

reactions

termination

reaction

reactions

termination

In

step

species.

a

slow

substitution

You

constant

down

of

c an

see

of

reaction,

methane,

two

therefore

supply

the

Cl

between

step

three

from

the

radic als

termination

Cl

Cl

a

result,

a

mixture

Ethane

is

also

but

C

has

organic

molecule

in

organic

halogenated

is

too

not

C

H

H

products

a

H

is

chlorine,

which

greater

direct

formed,

which

is

a

bonds

reactivity

and

completely.

are

possible:

ClC

bromine

the

In

in

of

has

the

is

be

H

H

C C

H

H

the

H

desired

recycled

the

too

which

been

same

for

the

of

low,

a

so

is

polar.

product

initiation

step.

Therefore, an

generated,

way,

presence

alkanes

formation

iodine

H

including

c an

bond,

reactivity

uorination

of

Therefore,

it

by-product.

reactions.

or

stopping

reactions

H

Cl

c arbon–chlorine

with

chlorine

so

c arbon–c arbon

the

C

chemistry

by

reactive,

contrast,

H

produced,

Chloromethane

other

of

continue.

H

H

also

form a

Cl

H

chloromethane,

to

concentration

H

H

As

species

the

propagation steps

to

eventually

H

H

radic al

reduces

oen

any

of

UV

leads

complex

radic al

which

other

light

to

or

the

mixture

be

used

c an be

heat. Fluorine

breaking of

of

iodination

c an

alkane

of

products. In

alkanes does

occur.

Molecular modelling

You

c an

simple

choice,

and

build

molecular

materials

model

diatomic

such

the

as

free

bromine,

models

using

plasticine

radic al

Br

2

soware,

and

mechanism

, under

specialized model kits, or

toothpicks.

of

the

Using

a

reaction

medium

of

your

between ethane

UV light.

Relevant skills

•

Tool

2:

Physic al

and

digital

molecular modelling

Instructions

1.

Start

by

modelling

the

initiation

step.

Then,

model

the

propagation steps.

Finally, model the termination steps.

2.

Share

may

your

something

626

model

decide

to

with

create

else

(for

a

your

class.

Choose

stop-motion

video

example, a ick book).

a

suitable

way

recording,

live

to

do

so.

You

explanation, or

Reactivity

3.3

Electron

sharing

reactions

End of topic questions

Extended-response questions

Topic review

1.

Using

your

answer

knowledge

the

guiding

from the

question

as

Reactivity 3.3

fully

as

topic,

5.

a.

Dene

possible:

homolytic

reaction

What happens when a species possesses an unpaired

b.

Write

electron?

an

equation

electrons

6.

Exam-style questions

ssion,

including

the

required

conditions.

during

This

question

a.

Ethane, C

is

2

6

,

show

the

homolytic

about

H

[2]

to

movement of

ssion

of

iodine.

[1]

c arbon and chlorine compounds.

reacts with chlorine in sunlight.

Multiple-choice questions State

2.

What

is

a

propagation

mechanism

of

ethane

step

with

in

the

radic al substitution

A

Cl

•C

C

•C

D

C

2

→

2

2

H

H

5

5

2Cl•

+

→

2

Cl•

C

→

2

C

H

2

5

H

Cl

5

+

H

6

+

Cl•

→

C

2

H

5

and

Methane

Cl

+

4

equations

it

occurs.

one

for

the

[1]

termination

from

two

step

propagation

in

the

formation of

ethane.

[3]

Chloromethylbenzene, C

in

synthetic

reactions

and

in

6

H

the

5

CH

2

Cl,

is

a

useful

reagent

manufacture of pesticides,

fragrances.

reacts with chlorine in sunlight.

(g) + Cl

2

(g)

→

CH

3

Cl(g)

+

Draw

the

type

of

reaction

structural

formula

of

methylbenzene, also

HCl(g)

known Which

reaction and the name of the

•H

a. CH

this

which

Cl

medicines 3.

of

by

chloroethane

Cl•

7 .

2

Formulate

steps

+ Cl

type

chlorine? b.

B

the

mechanism

as

toluene.

[1]

occurs?

b. A

radic al substitution

B

electrophilic substitution

Methylbenzene

in

the

presence

c an

of

undergo chlorination

UV

light

chloromethylbenzene.

C

nucleophilic substitution

D

electrophilic addition

in

the

initiation

c.

Formulate

d.

Write

e.

Write

the

to

produce

Explain

the

role of UV light

step.

[1]

equation

to

describe the initiation

step. 4.

Which

of

these

mechanism

in

reactions

the

proceeds

by

a

C

6

H

B

C

C

CH

D

CH

6

H

6

6

+ Cl

+

2

3H

→

2

equations

C

→

6

H

C

6

5

Cl

H

+

3

CH

CH

2

3

+

HBr

+ Cl

the

two

propagation steps and

termination

2

→

→

step.

[3]

HCl

an

equation

for

the

overall

reaction, using

12

structural

2

for

presence of UV light?

one A

[1]

radic al

CH

CH

3

3

CH

CH

2

2

formulas.

[3]

Br

Cl

+

HCl

627

Reactivity 3.4

Electron-pair sharing

reactions

What happens when reactants share their electron pairs with others?

When

a

heterolytic

nucleophile.

electrophile),

ssion

This

of

a

molecule

nucleophile

forming

a

new

is

occurs,

one

electron-rich

of

and

the

c an

two

share

fragments

an

receives

electron

pair

the

with

bonding

an

electron

pair,

electron-decient

forming

species (an

covalent bond.

Understandings

nucleophile

is

a

reactant

that

forms a Reactivity 3.4.6 — A

bond

to

its

reaction

partner

(the

electrophile)

acceptor both

bonding

Lewis

and

a

Lewis

base

Reactivity 3.4.2 — In

nucleophile

bond,

as

is

electron-pair

an

an

electron-pair

donor.

electrons.

Reactivity 3.4.7 — When

a

acid

is

by donating

a

donates

another

nucleophilic

an

bond

electron

breaks

substitution

pair

to

form

producing

a

a

reaction,

new

a

Lewis

acid,

a

Nucleophiles

leaving

a

Lewis

coordination

are

Lewis

base

bond

bases

and

is

reacts with

formed.

electrophiles

are

Lewis acids.

group. Reactivity 3.4.8 — Coordination

when Reactivity 3.4.3 — Heterolytic

ssion

is

the

ligands

one

of

bond

the

two

when

both

fragments

bonding

electrons

forms

a

bond

c ations,

to

its

reaction

forming

is

partner

a

reactant that

(the

nucleophile)

include

halogenoalkanes

the

both

bonding

electrons

are

to

formed

transition

complex ions.

reactions

and

from

that

between

nucleophiles.

by Reactivity 3.4.10 — The

accepting

pair

Reactivity 3.4.9 — Nucleophilic substitution

formed.

electrophile

bonds

electron

remain with

reactions

Reactivity 3.4.4 — An

an

breakage of a element

covalent

donate

rate of the substitution

reaction reactions

is

inuenced

by the identity of the

partner. leavinggroup.

Reactivity 3.4.5 — Alkenes

attack

bec ause

c arbon–c arbon

of

the

high

double

electrophilic addition.

are

susceptible

to

electrophilic

electron density of the

bond.

These

reactions

lead to

Reactivity 3.4.11 — Alkenes

electrophilic

addition

Reactivity 3.4.12 — The

c arboc ations

hydrogen

be

used

in

the

halides

to

undergo

relative stability of

addition

and

explain

readily

reactions.

the

reactions

unsymmetric al

reaction

between

alkenes

c an

mechanism.

Reactivity 3.4.13 — Electrophilic substitution

reactions

include

electrophiles.

628

the

reactions of benzene with

LHA

Reactivity 3.4.1 — A

Reactivity

3.4

Electron-pair

sharing

reactions

Nucleophiles (Reactivity 3.4.1)

In

Reactivity 3.3,

hydrogen

polar

It

that

it

is

nucleophile

c an

be

donate

This

you

with

saw

a

open

is

an

neutral

a

pair

forms

a

of

or

process

bond,

to

by

electron-rich

a

electrons

full

to

of

atom.

C–X,

attack

c arry

covalent

electrophile

the

halogen

c arbon–halogen

means

A

atom

a

so

the

species

c arbon

that

atom

contains

charge.

bond

alkane

A

by substituting a

halogenoalkane contains a

known as a

is

electron-decient. This

nucleophile.

a

lone

pair

nucleophile

electron-decient

coordination

an

resulting

species

negative

an

activating

The

species

between

the

of

electrons.

(reactant)

c alled an

c an

electrophile.

nucleophile and the

(gure 1).

Nu

E

[Nu

Nu

p Figure 1

E]

E

Nu

A negatively charged

E

or neutral nucleophile (Nu)

c an attack an electrophile (E) forming a coordination bond

Water, H

2

O,

is

an

example

of

a

neutral

nucleophile,

as

it

has

two

lone

pairs of The

electrons

on

the

oxygen

atom

and

no

charge.

The

hydroxide ion,

nature

formation negatively

charged

nucleophile,

with

three

lone

pairs

of

and

mechanism of the

OH, is a of

coordination bonds

electrons. are

discussed in

Structure2.2.

O O H

p Figure 2

H

H

Water and

the hydroxide ion are both

nucleophiles bec ause they contain at

least one lone pair of

electrons

Other

examples

halogen

neutral

The

an

ions

of

(Cl

,

nucleophiles

Br

, I

),

include

cyanide

ion

charged atoms and ions, such as the

(CN

molecules, such as ammonia (NH

strength

of

a

nucleophile

depends

3

)

on

)

and

and

its

hydrogensulde ion (HS

methylamine

ability

to

(CH

donate

its

3

NH

2

), and

).

electron pair to

electrophile.

Practice questions

1.

2.

What

must

be

present

A.

Negative

charge

B.

Lone

of

C.

Positive

D.

Symmetric al

Which

of

A.

CH

B.

(CH

C.

(CH

D.

(CH

3

pair

3

3

)

)

2

3

a

nucleophile?

electrons

charge

the

NH

in

distribution

following is

not

of

an

electrons

example

of

a

nucleophile?

2

NH

N

+

3

)

4

N

629

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Nucleophilic substitution reactions

(Reactivity 3.4.2)

In

most

pair

to

reactions

the

reactions,

involving

electrophile,

this

producing

a

also

small

nucleophiles,

forming

results

in

the

molecule or

An

example

of

with

CH

The

presence

chlorine

the

of

bond

c arbon

3

a

in

atom

CH

nucleophile

breaking

of

one

in

of

donates

an

electron

nucleophilic substitution

the

bonds

in

the

electrophile,

R

Mechanism

+

Nu

for a nucleophilic substitution

where Nu = nucleophile, R = electron-decient

in the electrophile and X = leaving group

nucleophilic

halogenoalkane,

the

However,

X

p Figure 3

reaction,

bond.

leaving group

+

Nu

atom

a

substitution

aqueous

2

Cl(g)

highly

+

sodium

OH

(aq)

is

→

electronegative

chloroethane.

The

electron-decient

the

reaction

of

chloroethane, a

hydroxide.

CH

CH

chlorine

resulting

and

3

2

OH(aq)

atom

partial

therefore

+ Cl

(aq)

polarizes

positive

the

c arbon–

charge makes

susceptible

to

attack

by

nucleophiles.

The

hydroxide

between

Practice questions

in

the

and

bond

Deduce

the

the

equations

reactions

in

nucleophile

oxygen

electrons

detailed 3.

the

the

and

c arbon–chlorine

creating

the

the

an

electron

atoms.

bond

leaving

At

the

moves

group,

a

pair

to

same

onto

the

chloride

form

a

new

covalent bond

time, the bonding pair of

chlorine

ion

(Cl

).

atom,

This

breaking

mechanism is

gure 4.

for

between the H

following

donates

c arbon

H

H

H

H

C C

H

H

reactants, identifying

nucleophile

and

훿–

훿+

leaving H

C

C

H

H

Cl

OH

group:

Cl a.

2-bromo-2-methylpropane

b.

bromopentane and the

and

potassium

OH

hydroxide nucleophile

p Figure 4

The nucleophilic substitution of chloroethane with

cyanide ion. a hydroxide ion nucleophile.

c arbon atom

is indic ated

The electron-decient

nature of the

with a partial positive charge(δ+)

Science as a shared endeavour

For

over

a

chemists

century, the journal

to

public ation

circulation.

replic ating

in

the

While

the

experts

the

manuscript.

The

to

the

who

not

author

that

all

There

all

the

check

many

that

is

is

then

that

widely

to

employing

journals

thoroughly

methods

either

must

been

experiment

submitted

for

peer-reviewed scientic journals in

be

and

claims

addressed

as

provided

in

a

are

reviewed.

accepted,

recognized

information

has

every

submitted

are

research

revisions

process

ensures

are

articles

manuscript

for

and

investigations

scrutinize

The

peer-review

method

journal.

Organic Syntheses

repeat

experiment,

are

back

630

independently

or

by

reviewers

described in

rejected, or sent

before

valuable

journal

veried

The

further

quality

articles

is

review.

control

reliable.

Reactivity

3.4

Electron-pair

sharing

reactions

Heterolytic ssion (Reactivity 3.4.3)

When

an

bonding

this

unsymmetric al

pair

process,

atom

are

one

receives

c ation

When

that

is

of

none

the

of

decient

drawing

cleavage

distributed

atoms

the

of

is

le

bonding

an

heterolytic

of

a

covalent

unevenly.

with

a

is

and

both

an

bond

occurs,

known as

bonding

electrons.

electron

ssion,

This

This

anion

double-barbed

electrons in the

electrons while the other

results

that

the

heterolytic ssion. In

in

has

curly

the

an

formation of a

extra

arrow

is

electron.

used

to

show +

A the

movement

of

the

electron

pair.

The

B

A

+

c ation

where

the

electrons

are

moving

to.

In

the

c ase

of

gure

5,

atom

B

and

atom

A

becomes

a

anion

becomes an

p Figure 5

anion

B

arrow starts at the bond and nishes

Heterolytic ssion of a

c ation.

diatomic molecule

Organic

at

the

compounds,

such

c arbon–halogen

c ations

with

the

as

bond

positive

halogenoalkanes,

to

form

charge

on

a

halogen

the

c arbon

c an

undergo

anion

atom

and

are

an

heterolytic

alkyl

c alled

c ation.

ssion

Alkyl

c arboc ations

Worked example 1

Draw the mechanism for the heterolytic ssion of

bromomethane and

hence deduce the nal products.

Solution

First,

draw

the

structure

of

bromomethane:

Then

from

draw

the

the

double-barbed

curly

arrow, originating

c arbon–bromine bond and nishing on the

H bromine atom:

H

C

Br

H

훿+

H

Identify

the

partial

charges

H

in

the

molecule:

C

훿–

Br

H

H The

훿+

H

훿–

C

products

are

therefore

a

methyl

c arboc ation and a

bromide anion:

Br

H

H

H

훿+

H

C

H

The

two

species

therefore

inan

have

overall

a

formed

short

during

lifespan.

heterolytic

This

means

ssion

that

are

they

usually

are

훿–

Br

+

H

C Br

H

unstable, and

usually

intermediates

reaction.

Linking question

What

and

is

the

the

dierence

between

bond-breaking

that

the

occurs

bond-breaking

in

nucleophilic

that

forms

a

substitution

radic al

reactions?

(Reactivity 3.3)

631

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Electrophiles (Reactivity 3.4.4)

F 훿–

We

훿+

dened

an

electrophile

as

an

electron-decient

species.

Electrophiles

B

readily

accept

a

pair

of

electrons

from

an

electron

donor,

a

nucleophile,

to

form

F

F 훿–

훿–

a

covalent

p Figure 6

Boron triuoride is an

neutral

electrophile,

with an electron-decient

charges

boron atom

bond.

Electrophiles

molecules

are

molecule

with

a

generated

are

partial

by

the

either

positive

positively

charged

charge (δ+)

presence

of

a

highly

on

one

ions

of

(c ations), or

the

atoms.

electronegative

Partial

species in the

resulting in the polarization of a bond.

+

The

methyl

charge.

atom

a

is

c ation,

Boron

susceptible

partial

positive

Compounds

CH

3

,

is

triuoride,

to

3

,

example

has

an

nucleophilic

charge

with

an

BF

of

an

electrophile

electron-decient

attack.

It

is

an

with

boron

example

of

a

full

positive

atom.

an

The

boron

electrophile with

(gure 6).

c arbonyl

or

c arboxyl

groups,

such

as

aldehydes, ketones and

c arboxylic acids, are also electrophiles. The electron-decient c arbon atom of the

c arbonyl

group

is

susceptible

to

nucleophilic

attack

(gure 7).

훿–

O Non-polar

molecules, such as

훿+

bromine,

Br

,

c an

also

behave as

2

OH electrophiles.

detail

in

the

This

AHL

is

covered in

section of this

p Figure 7

topic.

C arbon atoms in c arbonyl

groups or c arboxyl groups, such as that in

butanoic acid,

have a partial positive charge

Global impact of science

Alice

Ball

Hawai‘i.

was

At

treatment

method

to

for

for

DP

to

the

chemistry

concepts

chemist

it

leprosy,

was

but

to

and

researcher

known

it

was

that

not

suitable

chaulmoogra

patients.

This

working

oil

method

in

chaulmoogra

for

into

was

early 20th century

oil

was

injection.

ethyl

used

esters

to

treat

an

Ball

that

eective

developed a

were

leprosy

far

for

easier

dec ades

introduction of antibiotics.

chemistry

converting

US

time,

converting

administer

prior

a

the

covers

much

chaulmoogra

of

the

Ball

of

the

extract

Method”.

theory

into

You

underlying

ethyl

will

ester.

Ball’s

Search

recognize

many

procedure

online

of

the

for

for

“the

core

chemic al

involved in this method.

Electrophilic addition reactions in alkenes

(Reactivity 3.4.5)

In

p Figure 8

and

rst

Alice Ball was the rst woman

Afric an Americ an to earn a master ’s

degree from the University of

died

in 1916 at

Hawai‘i. She

the age of 24 before her

work on the Ball method

a

Structure 2.3,

we

c arbon–c arbon

alkenes

c arbon

more

reactive

double

electrophilic

bond

attack.

than

is

a

This

alkenes

bond.

the

as

The

unsaturated

corresponding

region

of

reactivity

hydroc arbons that contain

presence of the double bond makes

high

saturated

alkanes.

The

c arbon–

electron density that is susceptible to

means

that

alkenes

are

oen

used in industrial

was published

processes as the

between

632

dened

double

starting molecules

alkenes

and

electrophiles

for

are

synthetic

known

as

reactions.

The

electrophilic

reactions

addition

reactions.

Reactivity

3.4

Electron-pair

sharing

reactions

Electrophilic addition of halogens

An

example

diatomic

across

of

a

For

Br

2

of

electrophilic

halogen

the

electron-rich

disubstituted

example,

(aq),

the

yields

a

addition

molecule, X

2

.

In

reaction

H

the

reaction

reaction,

c arbon–c arbon

halogenoalkane

single

is

this

with

double

the

halogen

bond,

general

between ethene gas, C

product,

between an alkene and a

the

2

formula C

H

4

(g),

1,2-dibromoethane, C

2

and

H

4

Br

2

n

in

H

2n

the

X

2

is

added

formation

.

bromine

water,

:

H

H

H

C C

Br

Br

H

C

+

C

Br

Br

H

H

H

ethene

bromine

water

1,2-dibromoethane

(brown)

This

molecule

resulting

reaction

c an

also

be

used

to

test

(colourless)

for

the

presence

of

unsaturated compounds

Practice questions in

a

mixture

of

hydroc arbons,

as

the

bromine

water

will

turn

colourless in the

presence of alkenes or alkynes.

4.

a.

Determine

the

gas, C

Electrophilic addition of hydrogen halides

3

gas, Cl Electrophilic

addition

reactions

will

also

occur

between

alkenes

and

H

2

6

HX.

The

in

the

C

H

n

molecule

process

formation

2n+1

is

is

added

similar

of

a

across

to

the

the

addition

of

halogens:

c arbon–c arbon

monosubstituted

double

the

halogenoalkane

with

This

(g).

general

b.

formula

Iodine

consider

the

electrophilic

addition

reaction

between

4

H

8

(g),

and

aqueous

hydrogen

bromide,

HBr(aq).

But-2-ene

is

a

so

to

the

addition

product:

of

a

hydrogen

2-bromobutane, C

halide

4

H

9

molecule

will

many

colours,

brown

from

or

purple.

how an iodine

produce only one c an

be

used to

Br. detect

the

presence of

unsaturated

H

H

H

C

H

Br

hydroc arbons.

H

H

C

H

in

form solutions

symmetric al

solution

possible

to

various

Explain

alkene,

products.

but-2-ene, yellow

C

and

dissolves

solvents

X.

example,

Draw the

formulas of all

results

of

For

propene

hydrogen

bond.

the

product of

hydrogen

reactants halide

of

(g), with chlorine

displayed halides,

the

reaction

+

C

H

Br

C

3

C

CH

3

CH

C

3

3

but-2-ene

hydrogen

2-bromobutane

bromide

The

reaction

halide

will

of

yield

H

3

unsymmetric al

two

possible

alkene,

such

as

propene,

with

a

hydrogen

products:

H

H

H

C

H

Br

H

H

H

C

Br

H

H

C

H

an

+

C

C

propene

H

Br

3

C

C

H

and

3

C

C

H

H

hydrogen

1-bromopropane

2-bromopropane

bromide

633

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

Electrophilic addition of water

Electrophilic The

selectivity

of

addition

occurs involving

addition

reactions

also

occur

between

water and alkenes. This

reactions when

an

alkene

is

added

to

an

acidic

solution,

resulting

in

the

formation

unsymmetric al alkenes is of

discussed

in

the

AHL

an

alcohol.

The

reaction

involves

the

addition

of

a

water

molecule

across

section of this the

c arbon–c arbon

double

bond,

forming

an

alcohol

with

the

general

formula

topic. C

n

H

2n+1

OH.

This

reaction

is

also

known

as

a

hydration

reaction.

TOK

Chemists

interested

knowledge

synthetic

of

pathways

reasoning

all

limitations

of

What

and

For

H

2

are

the

example,

to

part

so

in

these

ways

in

6

H

12

one

innovation.

opens

thinking

up

Imagination

the

when

possibility

solving

transcends the

of

new

chemic al

ideas.

problems

situations?

undergoes

alcohol

product

+

C

C

CH

2

is

CH

2

H

electrophilic

hexan-3-ol, C

6

H

13

addition

OH(aq).

with

water,

Hex-3-ene is

formed.

with

OH

H

CH

3

C

2

3

hex-3-ene

As

(l),

and

of

novel

secondary

only

properties to determine suitable

scientic

knowledge

knowledge

the

physic al

H

H

C

C

H

OH

H

C

CH

of

and

hex-3-ene, C

H

3

synthesizing a particular compound use their

produce that compound. Imagination, intuition and

their

roles

form

symmetric al,

H

play

to

acquired

applying

O(l),

in

structure

water

hydrogen

halides,

two

CH

CH

2

hexan-3-ol

products

will

be

formed

in

the

electrophilic

Practice questions addition

5.

Deduce

the

reaction

between

alkenes

a.

and

equations

the

of

water

to

an

unsymmetric al

alkene.

for the

following

electrophiles:

Linking questions

2-methylbut-2-ene and

Why hydrogen

is

bromine

water

decolourized

in

the

dark

by

alkenes

but

not

by

alkanes?

bromide

(Reactivity 3.3)

b.

pent-2-ene and iodine

c.

ethene

d.

cyclohexene and hydrogen

e.

methylpropene and

Why

are

alkenes

sometimes

known

as

“starting

molecules” in industry?

(Structure 2.4) and

water

chloride

ATL

Social skills

hydrogen iodide Collaboratively

chapter.

as

M ake

others,

that

responses

contrast

the

common

634

in

a

develop

a

list

help

of

answers

the

you

answers

in

to

three of the linking questions in this

understandings

address

document

themes

key

the

shared

developed

everyone’ s

linking

with

by

the

the

from

rest

of

people

answers.

this

questions.

your

in

chapter, as well

Summarize

class.

your

your

Compare and

class.

Draw out the

3

Reactivity

Electron-pair

sharing

reactions

+

H

OH H

In

Reactivity 3.1,

we

dened a

Brønsted–Lowry base

as

a

substance

LHA

Lewis acids and bases (Reactivity 3.4.6)

3.4

that

H

c an + H

+

accept

a

proton

(a

hydrogen ion, H

).

The

presence

of

at

least one pair of H

electrons

in

Brønsted–Lowry

bases

allows

them

to

form a

coordination bond

+

H

N

H N

with

a

bases

proton.

The

hydroxide

ion

and

ammonia

are

examples

of

Brønsted–Lowry

H

(gure 9).

p Figure 9

A

H

H

H

Lewis acid

is

dened

as

an

electron-pair acceptor and a

Lewis base

The lone pair of

electrons on

as an

Brønsted–Lowry bases forms a coordination

electron-pair

donor.

The

Lewis

acid–base

theory

is

a

more

general denition

bond

when

range

Both

compared

of

ammonia

bases,

a

We

in

acid,

c an

use

reactions

between

BF

In

this

and

donating

Lewis

3

to

the

substances

as

triuoride,

3

B

proton

is

is

a

Brønsted–Lowry

to

c an

the

electron

theory

are

BF

not

3

with a proton

theories, enabling a wider

as

Brønsted–Lowry

The

and

Lewis

hydrogen ion acts as

pair.

identify

involved.

a

base

involved,

ion.

the

For

role

of

each

example,

3

reacting

consider

the

species

reaction

:

3

donates

Lewis

to

act

hydrogen

, and ammonia, NH

NH

ammonia

ammonia

no

acid–base

the

protons

F

ion

electrons

acid–base

3

and

included.

hydroxide

of

accepts

Lewis

reaction,

Arrhenius

be

the

pair

where

:NH

Therefore,

reaction,

it

boron

+

a

to

so

lone

and

it

pair

of

boron

c annot

be

electrons

triuoride

to

is

described

boron

a

triuoride.

Lewis acid. In this

using

Brønsted–Lowry

theory.

Activity

2+

Copy

CH

3

and

complete

COOH, OH

the

, NH

3

table

and

for

HF.

Species

BF

Brønsted–Lowry acid

No

each

The

of

the

example

following

of

BF

3

species: H

has

been

2

O, Cu

,

completed.

3

Brønsted–Lowry base

No

Lewis acid

Yes

Lewis base

No

Practice question

6.

Which

but

species

not

a

is

a

Lewis acid

Brønsted–Lowry acid?

2+

A

Cu

B

NH

C

Cu

D

CH

Linking question +

What

is

acids

and

the

relationship

between

Brønsted–Lowry

acids

and

bases

and

Lewis

4

bases? (Reactivity 3.1)

3

COOH

635

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

Lewis acid and base reactions (Reactivity 3.4.7)

You

have

least

(anion)

Lewis

An

learnt that a

one

lone

or

a

pair

neutral

base,

as

it

an

neutral

When

electron

is

Lewis

Consider

the

donate

an

An

base

orbital

resulting

of

an

electron-rich

be

c an

be

a

with

either

nucleophile

pair

a

boron

2

of

a

c an

species

negatively

therefore

that

possesses at

charged

be

species

described as a

electrons.

either

c an

species

a

that

positively

therefore

Lewis acid, a

triuoride

be

will

accept

charged

a

pair

species

described

as

a

coordination bond

with

ammonia.

Boron

of

electrons

(c ation) or

Lewis acid.

is

has

formed.

an

1

2s

in

is

c an

electrophile

2

orbitals,

It

reacts

reaction

conguration of 1s

a

A

It

electron-decient

donor.

molecule.

a

electrons.

molecule.

c an

electrophile

from

nucleophile

of

electron

2

2p

and

trigonal

in

boron

planar

triuoride

geometry

it

and

will

a

form

vac ant

three sp

hybrid

unhybridized 2p

z

(gure 10).

2

2s

2p

ground

p Figure 10

2s

state

2p

2sp

2p

z

excited state

Hybridization of boron in boron triuoride

The

lone

pair

of

electrons

on

the

nitrogen

atom

in

ammonia

is

donated to this

F vac ant 2p

H

F

N

H

orbital,

F

B

and

generating

F

H

a

coordination

bond

with

the

electron-decient

F

a

single

product (gure

11).

boron

Therefore, ammonia acts as a

N H nucleophile

F

forming

H atom

B

z

and

Lewis

base,

and

boron

triuoride

acts

as

an

electrophile and

H Lewis acid.

p Figure 11

Ammonia donates an

Coordination electron pair to boron triuoride,

or

as

an

arrow

Anhydrous

central

c an Coordination

orbitals

were

bonds

and

bonds

are

covalent

bonds,

so

they

c an

be

drawn

as

ordinary bonds

forming a

coordination bond

from

aluminium

aluminium

react

the

with

of

the

chloride, AlCl

atom

each

source

is

other

electron

to

form

a

lone

3

,

is

pair

to

another

decient.

dimer,

the

example

Two

where

electron-decient atom.

of

aluminium

the

lone

a

Lewis acid, as the

chloride

molecules

pairs on the chlorine

hybrid

atoms

form

atoms

(gure

coordination

bonds

with

adjacent

electron-decient aluminium

introduced in

12).

Therefore,

aluminium

chloride

c an

act

as

both

a

Lewis acid and

Structure 2.2.

a

Lewis

base.

Cl 훿–

훿–

Cl

Cl Cl

훿+

AI

Al

Al Cl

Cl

훿–

Cl

훿+

Cl

훿–

Cl

훿–

AI

Cl

훿–

Cl

p Figure 12

The formation of

the aluminium

chloride dimer, Al 2

636

Cl 6

Cl

Reactivity

3.4

Electron-pair

sharing

reactions

LHA

Practice questions

7 .

Which

8.

statements

are

I

Lewis

II

Electrophiles

bases

III

Lewis

acids

A

I and II only

B

I and III only

C

II and III only

D

I, II and III

Which

type

of

correct?

c an

act

are

are

as

nucleophiles.

Lewis acids.

electron

pair

acceptors.

Linking question

bond

is

formed

when

a

Lewis

acid

reacts

with

a

Lewis

base? Do

coordination

dierent

A

covalent

C

double

B

dipole–dipole

D

hydrogen

covalent

bonds

properties

have

any

from other

bonds? (Structure 2.2)

Coordination bonds and complex ions

(Reactivity 3.4.8)

Transition

elements

are

metals

that

c an

form

ions

with

a

partially

lled d-subshell.

4

An

example

Transition

bonds

is

In

the

Lewis

of

species

c an

chromium(II)

also

bases.

as

an

the

are

A

be

complex

are

with

Lewis

transition

and

ligands.

.

They

This

as

it

as

are

c an

has

the

an

acids,

electron

so

element

they

ion

surrounding

neutral,

such

therefore

electrophile,

ions,

normally

anions,

hydroxide ion, OH

electrons,

c ations

ion

conguration of [Ar] 3d

c an

form

bonded

several

to

.

coordination

several

Lewis bases

complex ion.

context

Ligand

but

a

the

element

with

c alled

is

a

the

such

species

considered

positive

is

Lewis

water, H

cyanide ion, CN

electron-rich

be

relationship

as

bases

and

in

c alled

, chloride ion, Cl

with

at

accepts

gure

are

ligands.

O, and ammonia, NH

nucleophiles.

charge

summarized

2

13,

3

,

, and the

least one lone pair of

The

metal

electron

where

c ation acts

pairs

arrows

from

represent

coordination bonds.

electron pair Lewis acid

Lewis base

electron pair electrophile

nucleophile

electron pair

transition element

ligand cation

p Figure 13

All Lewis bases are nucleophiles,

are electrophiles,

and vice versa.

transition element

and

all Lewis acids

All ligands are nucleophiles, and all

c ations are electrophiles,

but

not

vice versa

637

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

2+

LHA

2+

For

example, copper(II) ions in solution, Cu

(aq),

form

coordination bonds

OH 2 2+

H

with

OH

O 2

molecules

to

give

the

complex

ion

[Cu(H

ion

if

2

O)

6

]

.

It

has

octahedral

2

geometry

Cu

H

water

O

and

is

blue

in

colour

(gure 14).

OH

2

2

You

c an

deduce

the

charge

on

a

complex

you

know

the

charge on the

OH 2

transition

element

c ation,

the

charges

on

the

ligands

and

the

number

of

charged

ligands.

O verall

of

charge

of

complex ion

charged ligands

×

=

(charge

on

transition

element

c ation)

+

(number

charge on a ligand)

Worked example 2

Consider the equilibrium between two cobalt(II) complex ions,

n

[Co(H

2

O)

6

]

m

and

[CoCl

4

]

, in a solution containing chloride ions. Their

charges are unknown.

n

[Co(H

2

O)

6

]

m

(aq) + 4Cl

(aq)

⇌

[CoCl

pink

p Figure 14

The hexaaquacopper(II)

4

]

(aq) + 6H

2

O(l)

blue

Deduce the charges on each complex ion.

complex ion

Solution

In

each

complex

molecule,

ion,

therefore

the

the

cobalt

overall

c ation

charge

has

of

a

charge of 2+.

the

complex

ion

Water

with

is

a

neutral

water ligands is

2+

the

same

Chloride

second

sum

of

as

ions

of

a

cobalt

c ation:

charge of 1−,

ion.

The

overall

[Co(H

and

2

O)

there

charge

of

6

]

are

the

.

four chloride ligands in the

complex

ion

is

equal to the

charges:

charge

(number

the

have

complex

the

O verall

that

of

of

complex ion

charged ligands

×

=

(charge

on

transition

element

c ation)

+

charge on a ligand)

=

2

=

2−

+

4

×

(−1)

The identity of the ligands also aects

the

colour

of

the

complex ion. This

2

was

discussed in

Structure 3.1

Therefore,

You

c an

c ation,

use

that

the

the

is,

second

reverse

you

need

complex

process

to

ion

to

know

is

[CoCl

deduce

the

4

]

the

overall

charge

charge

of

on

the

the

transition metal

complex ion, and

charge and number of the ligands:

Charge

of

638

on

transition

charged ligands

×

element

c ation

=

(overall

charge on a ligand)

charge

of

complex ion)

−

(number

Reactivity

3.4

Electron-pair

sharing

reactions

LHA

Practice questions

Worked example 3

Deduce the charge on the transition element

c ation in the following 9.

Deduce

the

charge on the

complexes: metal

ion

in

the

following

2+

a.

[Fe(OH)(H

b.

[TiF

2

O)

5

complexes:

]

3+ 2−

6

a.

[Cr(H

b.

[NiBr

c.

[Pt(CN)

d.

[Fe(H

]

2

O)

6

]

2

4

]

Solution 2

a.

The

overall

charge

of

the

complex ion is 2+.

It

contains

ve

neutral

6

]

water 3+

ligands,

and

a

hydroxide ligand with a 1−

2

O)

2

(NH

3

)

4

]

charge:

2

e.

Charge

on

transition

element

c ation

=

(overall

charge

of

complex ion)

10. (number

of

charged ligands

×

[Pd(CN)

4

(NH

3

)

2

]

−

Deduce

the

total

charge (n)

charge on a ligand)

on

=

2

−

=

3+

1

× (−1)

the

complex ion in the

following

complexes:

n

a.

[Cr(H

b.

[Ni(OH)

c.

[Pt(CN)

d.

[Fe(H

2

O)

6

]

, Cr(III)

3+

Therefore,

the

the

reverse

metal

process

c ation

to

is

Fe

c alculate

.

You

the

c an

overall

check

your

charge

on

working

the

n

by doing

2

Br

2

]

, Ni(II)

complex ion n

and

checking

that

it

4

(H

2

O)

]

,

)

]

2

Pt(IV)

equals 2+

n

b.

The

overall

charge

of

the

2

O)

2

(NH

3

4

,

Fe(II)

complex ion is 2−. It contains six uoride n

e. ligands,

Charge

each with a 1−

on

(number

transition

of

[PdCl

charge:

element

charged ligands

c ation

×

=

(overall

charge

of

complex ion)

6

]

,

Pd(IV)

−

charge on a ligand)

=

(−2)

=

4+

−

6

×

(−1)

4+

Therefore,

the

metal

c ation is Ti

Element

Electronegativity,

c arbon

χ

2.6

uorine

4.0

chlorine

3.2

bromine

3.0

iodine

2.7

Nucleophilic substitution in halogenoalkanes

(Reactivity 3.4.9)

Halogenoalkanes

electronegativity

contain

of

the

a

c arbon–halogen bond, which is polar due to the high

halogen

atom

compared

to

that

of

c arbon (table 1).

p Table 1

The

electron-decient

c arbon

atom

is

susceptible

to

nucleophilic

Halogen atoms have high

attack.

electronegativity and

Halogenoalkanes

c an

therefore

undergo

nucleophilic

substitution

form polar bonds

reactions,

with c arbon

where

There

the

are

reactions:

halogen

two

S

N

reactant is a

1

atom

types

and

S

of

N

displaced

mechanism

2.

primary,

is

The

by

that

or

nucleophile.

occur

mechanism

secondary

the

that

in

훿+

nucleophilic substitution

훿–

occurs depends on whether the

tertiary halogenoalkane.

p Figure 15

Representation of the partial

charges within the polar c arbon–halogen

S

2

reaction mechanism

N

bond

Nucleophilic

mechanism.

substitution

This

in

primary

mechanism

is

an

halogenoalkanes

example

of

a

follows the S

concerted

N

reaction,

2

reaction

which

means

Primary, that

reactants

are

S

mechanism

converted

directly

into

products

in

a

single

step.

secondary and tertiary

Therefore, the

halogenoalkanes N

2

does

not

involve

an

were

dened in

intermediate.

Structure 3.2

639

LHA

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

The

‘2’

in

S

2

me ans

that

there

are

two

molecules

involved

in

the

rate-

N

Reaction

order

and

rate

equations determining

are

discussed in

step

(slow

step).

Therefore,

the

rate-determining

step

involves

Reactivity 2.2. both

on

the

the

halogenoalkane

concentrations

re action

rate

and

=

Consider

ions,

has

the

of

and

the

both

nucleophile,

re actants.

following

rate

is

the

rate

described

as

of

a

re action

second

depends

order

equation.

k[halogenoalkane][nucleophile]

the

reaction

between

bromoethane, C

OH(aq), which yields ethanol, C

group,

It

so

Br

(aq).

The

mechanism

is

2

H

shown

5

2

H

OH(aq)

in

5

Br(l),

and

a

and

HO

Br

C

Br

HO

+

C

Br

H

H

H

hydroxide

leaving

H

C

HO

ion

gure 16.

H

H

aqueous

bromide

CH CH

3

CH

3

3

transition state

p Figure 16

The S

N

2 mechanism

for the reaction between the primary

halogenoalkane bromoethane and

Transition

states

and

intermediates The

are

discussed

in

hydroxide ion

hydroxide

nucleophile

attacks

the

electron-decient

c arbon

atom,

forming a

greater detail in transition state

that

includes

the

halogen

and

the

hydroxyl

group.

This

transition

Reactivity 2.2 state

has

c arbon

a

partially

atom,

broken.

and

These

formed

a

covalent

weakened

partial

bonds

bond

between

the

nucleophile and the

c arbon–bromine bond that has not completely

are

represented

by

dotted lines.

Practice questions

A 11.

Halogenoalkanes

c an

transition

state

is

not

the

same

as

an

intermediate.

A

transition

state

exists

for

undergo

an

innitesimally

small

period

of

time

and

represents

the

structure

of

the

reacting

nucleophilic substitution

species reactions

with

bonds potassium

State

that

the

are

an

equation

of

the

reaction,

of

H

9

Cl(l),

with

hydroxide,

potassium

an

the

Using

and

the

for

the

reaction

formed,

but

it

pathway.

does

not

It

typic ally contains

represent

the

entire

reaction

occurs

in

a

single

step.

In

a

discrete

contrast, an

some

degree

product,

point

in

a

so

it

of

stability

and

represents

multistep

the

does

not

structure

immediately

of

the

transform

reacting

species at

reaction.

drawing S

N

2

reaction

mechanisms,

pay

attention

to

the

following:

formulas

arrows,

reaction

the

as

KOH(aq).

structural

curly

has

ultimate

intermediate

When b.

along

and

1-chlorobutane, into

4

energy

broken

for the intermediate

reaction

C

highest

partially

hydroxide.

step

a.

with

aqueous

draw

1

mechanism

The

curly

charge,

arrow

and

from

the

terminates

nucleophile

at

the

originates

from

electron-decient

its

lone

pair

or

negative

c arbon atom.

conversion of 2

The

curly

arrow

representing

the

halogen

leaving

group originates at the

1-chlorobutane to bond

between

the

c arbon

and

halogen

atoms.

This

c an

be

shown either in

butan-1-ol. the

3

H

reaction

Partial

substrate

bonds

in

the

or

in

the

transition

transition

state

are

state.

represented

by

dotted lines,

C 3

i.e.,

HO

C

The

transition

X

111.5°

C

4

charge

state

shown

is

enclosed

outside

the

in

square

brackets

with

a

single

negative

brackets.

H 5

p Figure 17

Both

the

nal

product

and

the

leaving

group

must

be

shown.

The tetrahedral arrangement

The S of bromoethane.

The bond

N

2

mechanism is

stereospecic,

which

means

that

the

product

formed

angle diers

will

have

a

specic

stereochemistry,

rather

than

be

a

mixture

of

isomers. In

slightly from the theoretic al value of

109.5° due to the presence of

two large

halogenoalkanes,

the

electron-decient

c arbon

atom

on

the

c arbon–halogen

3

substituents, CH

and

Br

bond is sp

hybridized,

and

therefore

it

has

a

tetrahedral

3

gure

640

17

shows

the

geometry

of

bromoethane.

geometry.

For

example,

Reactivity

N

position

2

reaction,

of

the

the

bromine

nucleophile

leaving

creates

steric hindrance,

c arbon

atom

c auses

inside

an

from

the

inversion

out

of

which

same

the

will

group.

attack

This

prevents

side

as

molecule

the

is

the

the

sharing

reactions

c arbon atom at 180° to the

bec ause

the

nucleophile

halogen

Electron-pair

LHA

In the S

3.4

atom.

conguration,

large halogen atom

from

attacking the

Therefore,

much

like

an

the

nucleophile

umbrella turning

(gure 18).

H

C

CH

3

3

inversion

H HO

Br

H

p Figure 18

H

Inversion of stereochemic al conguration in S

2 reactions

N

S

1 reaction mechanism

N

T ertiary halogenoalkanes undergo nucleophilic substitution in two steps. This is

known as the S

1 mechanism. In S

N

1 mechanisms, only one molecule is involved

N

in the rate-determining step. In the rst step of the S

1 mechanism, the bond to the

N

leaving group in the halogenoalkane breaks, forming an intermediate carbocation.

This is the rate-determining step, and it only involves the halogenoalkane. Therefore,

this is a rst order reaction, and the rate equation is as follows:

rate

For

=

example,

aqueous

the

k [halogenoalkane]

the

reaction

hydroxide

chloride

ion

ions

leaving

between

yields

group

the

via

2-chloro-2-methylpropane, C

product

a

c arboc ation

CH

intermediate

CH 3

4

H

2-methylpropan-2-ol, C

9

4

Cl, and

H

9

OH, and

(gure 19).

CH 3

3

OH +

H

Cl

C

C

H

3

C

C

H

3

OH

C

C

+

Practice questions Cl

3

12. CH

CH 3

Halogenoalkanes

S

1 mechanism

undergo

3

substitution

p Figure 19

c an

CH 3

for the reaction between a tertiary halogenoalkane

sodium

reactions with

hydroxide solution.

N

and

aqueous hydroxide ion

a.

State

the When

drawing S

N

1

reaction

mechanisms,

pay

attention

to

the

an

equation

methylbutane

1.

The

curly

2.

The

c arboc ation

3.

The

curly

bond

arrow

between

representing

the

the

halogen

leaving

clearly

show

a

positive

charge

on

the

c arbon atom.

Using

and

from

the

nucleophile

originates

at

its

lone

pair

or

H

11

I) and

terminates

at

the

positively

charged

structural

curly

formulas

arrows,

reaction

draw

mechanism

negative for

and

5

c arbon and the halogen atoms.

must

the

charge

(C

NaOH.

group originates at the

b.

arrow

for

reaction of 2-iodo-2-

following:

the

conversion of

c arbon. 2-iodo-2-methylbutane to

4.

Both

the

nal

product

and

the

leaving

group

must

be

shown.

2-methylbutan-2-ol.

641

3

What

LHA

Reactivity

are

the

mechanisms

of

chemic al

+

CH

Inductive eects

C

3

change?

The p Figure 20

An arrow is used

dierent

nucleophilic

halogenoalkanes the movement

substitution

mechanisms

of

reactions

involving

to represent

c an

be

explained

by

the

inductive

eects of substituents. In

of electron density c aused

the

C–H

bond,

the

c arbon

atom

has

a

slightly

greater

electronegativity than

by the inductive eect

hydrogen,

the

c arbon

donating 3 °

2°

1 °

R

R

R

atom

+

+

C

+

(gure

When

R

H

H

20).

several

weak

a

result,

density

This

alkyl

dipole

is

and

adjacent

and

are

slight

alkyl

reducing

known as a

groups

a

shi

groups

the

of

bonding

stabilize

positive

electrons

the

charge

on

towards

c arboc ation

the

central

by

c arbon

positive inductive eect

positive

inductive

bonded

eect

to

the

positively

increases.

charged

Therefore,

a

c arbon, their

c arboc ation

formed

H by

a

tertiary

halogenoalkane

halogenoalkane. p Figure 21

As

electron

combined

R''

atom.

a

C

C

R

creating

This

is

explains

more

why

stable

the

than

that

formed

by a primary

nucleophilic substitution of a tertiary

The inductive eect

halogenoalkane

is

more

likely

to

proceed

according to the S

decreases from tertiary to secondary to

The

stability

of

intermediate

Bond

N

1

mechanism

(gure21).

primary c arboc ations

Bond enthalpy

so

c arboc ations

between

secondary

formed

those

formed

halogenoalkanes

c an

from

secondary

from

primary

undergo

halogenoalkanes is

and

tertiary

nucleophilic

halogenoalkanes,

substitution

according

−1

/ kJ mol

C–F

to both the S

N

1 and S

N

2

mechanisms.

492

C–Cl

324

C–Br

285

Linking questions

C–I

228

What

S p Table 2

Bond

enthalpies of

N

2

dierences

would

be

expected

between

the

energy

proles

for S

N

1 and

reactions? (Reactivity 2.2)

c arbon–

halogen bonds

What

are

the

rate

How

equations

useful

are

for these S

N

1 and S

N

2

reactions? (Reactivity 2.2)

mechanistic models such as S

N

1 and S

N

2?

(Reactivity 2.2)

R ate of nucleophilic substitution reactions

Bond

enthalpy

c alculations in

data

are

used in

(Reactivity 3.4.10)

Reactivity 1.2 The

by

rate

the

of

a

nucleophilic

identity

mechanisms,

of

the

c arbon–halogen

electronegative

the

substitution

halogen

in

the

rate-determining

bond,

atom.

in

which

The

faster

reaction

leaving

step

the

this

in

involves

two

halogenoalkanes

group. In both S

the

bonding

step

is

N

heterolytic

electrons

completed,

the

is

1 and S

inuenced

N

2

ssion of the

move

to

higher

the

the

more

rate of

Practice question reaction.

13.

In

separate

samples

of

reaction

vessels,

2-iodopropane,

2-bromopropane

an

are

The rate of heterolytic ssion of the carbon–halogen bond depends on the strength

of that bond, which is characterized by the bond enthalpy (table 2). The higher the

2-chloropropane and

added to

bond enthalpy, the stronger the bond and therefore the slower the reaction.

aqueous solution of sodium Fluoroalkanes

are

virtually

inert

due

to

the

high

strength of the C–F bond

hydroxide. 1

(492 kJ mol

List

these

halogenoalkanes

according

from

642

to

highest

their

to

reaction

lowest.

halogen

rates

).

bond

As

anions

move

decreases

electronegativity

halide

you

of

also

the

as

down

the

halogen

increases

group

size

of

atom

down

the

17,

the

the

strength

halogen

decreases.

group.

atom

of

the

c arbon–

increases and the

Additionally, the stability of the

Reactivity

3.4

Electron-pair

sharing

reactions

LHA

Data-based questions

Under

certain

nucleophiles

alcohols.

To

conditions,

and

halide

substitute

investigate

the

how

ions

c an act as

hydroxyl

the

Alcohol

Halogenoalkane

Percentage of

reactant

products formed

products / %

groups in

position

of

the

hydroxyl pentan-1-ol

group

of

aects

the

experiments

pentan-1-ol,

2-ol.

E ach

chloride

identity

were

pentan-2-ol,

of

these

and

of

the

c arried

The

was

87

product, a series

alcohols

pentan-3-ol

alcohols

bromide

reaction

out.

1-bromopentane

and

reacted

used

2-methylbutan-

with

a

1-chloropentane

were

pentan-2-ol

13

2-bromopentane

50

2-chloropentane

22

mixture of

nucleophiles.

3-bromopentane

H

O, H

2

SO

2

19

4

3-chloropentane R–OH

R–Cl

NH 1°,

2°

or

Cl, NH

4

+

9

R–Br

Br

4

pentan-3-ol

3°

2-bromopentane

21

halogenoalkanes

2-chloropentane

Aqueous

and

ammonium

bromide

amounts

were

of

used

however,

and

ions

for

salts

the

ammonium

as

reactants

did

not

used

in

The

to

nucleophilic

chloride

contain

bromoalkanes.

were

each

and

substitution.

ammonium

experiment.

equal

amounts

relative

9

provide chloride

amounts

of

of

The

products,

chloroalkanes

the

3-bromopentane

50

3-chloropentane

20

2-methylbutan-

2-bromo-2-

53

2-ol

methylbutane

Equal

bromide

products

2-chloro-2-

47

1

were

determined

by

H

NMR

spectroscopy.

methylbutane

Q uestions

p Table 3

reactions.

1.

For

each

of

the

four

alcohols

used

in

the

Product

Source of

the

skeletal

secondary

or

Identify

In

certain

the

S aunders

solvents,

such

as

water,

bromide

ions

are

independent

and

dependent

variables in the

results that support this statement.

investigation.

8.

3.

R aza, P.

nucleophiles than chloride ions. Identify and

explain

the

R.

J. Chem. Ed.,2021,98(10), 3319–3325

tertiary.

better

2.

nucleophilic substitution

Herasymchuk,

formula and determine whether it is

7 . primary,

K.

experiments, and N. Merbouh,

draw

analysis of

data:

Answer

this

part

of

the

question

using

Using

your

knowledge

of

nucleophilic substitution

your reaction

mechanisms,

identify

and

explain which of

knowledge of chemistry and without looking at the the

results

which

ion

or

your

in

table

ion

is

3.

the

chloride

Formulate

stronger

ion.

hypothesis

Predict

will

be

a

hypothesis

nucleophile:

how

the

and

a.

S

b.

S

bromide

ideas behind

with

N

N

1

Again,

bromoalkane

Some

2

of

mechanism.

the

results

at

explains

taken

place,

table

3.

Formulate

a

mechanism, S

N

hypothesis that

1 and/or S

N

each

of

the

alcohols

used.

in

ideas

the

behind

amounts

produced

5.

Select

to

6.

and

in

of

the

your

alkyl

hypothesis

chloride

reactions

construct

a

with

suitable

will

and

Predict

be

alkyl

a

that

a

rearrangement

positive

atom

explain

the

of

within

formed

the

moves

charge of the

to

a

dierent

molecule. Identify and

results that support this statement.

Considering

the

results,

evaluate

the

hypotheses that

how

formulated

in

your

answers to questions 3 and 4.

reected

bromide

each of the alcohols.

type

initially

c arbon

you the

which

2, is

10. for

in

each of the alcohols.

which

favoured

suggest

produced in the

answer this part of the question without

looking

predominantly via an:

mechanism

c arboc ation

4.

reacted

reected in the amounts of

has reactions

have

explaining

the

9. chloroalkane

alcohols

graph or chart

11.

Formulate

a

conclusion

to

your

analysis, which

includes:

•

the

•

a

•

an

•

any

aim(s)

of

the

investigation

represent the data in table 3.

Describe

three

patterns,

trends

or

relationships

summary

of

the

outcomes

of

the

investigation

you appraisal

of

the

hypotheses

you

proposed

see in the data.

unanswered

questions

or

issues.

643

3

What

are

the

mechanisms

of

chemic al

change?

LHA

Reactivity

Linking question

Why

is

the

iodide

ion

a

better

leaving

group

than

the

chloride

ion?

(Structure 3.1)

Electrophilic addition mechanisms

(Reactivity 3.4.11)

Earlier in this topic, you saw that the electron-rich carbon–carbon double bond

in alkenes was susceptible to electrophilic attack. This type of attack leads to C arbon–c arbon

double

bonds

are electrophilic addition reactions. In this section, you will learn about the mechanism

electron-rich

due

to

the

presence of these reactions.

of

readily

accessible pi (π) bonds

either side of the bond axis. This is

Electrophilic addition of halogens to symmetric al alkenes discussed in

Structure 2.2 (AHL )

Consider

the

discussed

breaks,

reaction

earlier

and

two

in

between ethene gas, C

this

chapter.

One

c arbon–bromine

H

of

the

bonds

2

H

two

are

4

(g),

and

+

C

H

bromine

electrophile.

Br

Br

bromine

molecule

The

bromine

bromine

The

is

non-polar,

electrophilic

molecule

c arbon-c arbon

2.

double

molecule

are

water

charge,

and

is

so

(aq),

H

H

H

C C

Br

Br

H

1,2-dibromoethane

(colourless)

polarized

bond

bond

the

it

addition

of

repelled,

electron-rich C=C

positive

2

formed:

(brown)

The

Br

H

ethene

1.

water,

H

C

The

bromine

c arbon–c arbon bonds in ethene

is

the

must

as

it

polarized

via

the

approaches

alkene.

resulting

attacked

bromine

be

proceeds

The

in

by

an

the

molecule

before

it

c an act as an

following steps:

the

electron-rich

bonding

induced,

electrons within the

temporary

dipole.

bromine atom with a partial

splits

heterolytic ally

to

form a

bromide anion.

3.

When

of

the

the

produces

4.

Finally,

anion

positively

c arbon

a

the

charged

atoms,

another

c arboc ation

reaction

results

in

the

bromine

c arbon

atom

atom

forms

a

covalent bond with one

becomes

positively

charged. This

intermediate.

between

formation

the

of

unstable

the

c arboc ation

product,

and

the

bromide

1,2-dibromoethane.

+

H

H 2

2

C 2

CH

H 2

C

CH

2

2

훿+

Br induced

Br

dipole

Br

Br

Br

훿–

Br

p Figure 22

644

Mechanism for the electrophilic addition of bromine to ethene

Reactivity

1.

drawing

The

curly

c arbon

electrophilic

arrow

double

that

addition

shows

bond

and

the

mechanisms,

electrophilic

nishes

at

the

pay

attack

attention

to

the

following:

at

the

c arbon–

originates

Electron-pair

sharing

reactions

LHA

When

3.4

electron-decient atom of the

electrophile.

2.

3.

The

curly

and

nishes

The

last

charge

arrow

on

curly

on

for

heterolytic

the

arrow

the

leaving

originates

resulting

ssion

group

anion

at

to

the

and

originates

at

the

bond

being

broken

give an anion.

lone

pair

nishes

at

of

the

electrons

or

positively

the

negative

charged

c arbon

+

atom, C

4.

The

,

in

the

structural

c arboc ation.

formula

of

the

nal

product

must

be

shown.

Electrophilic addition of hydrogen halides to

symmetric al alkenes

For

is

the

electrophilic

similar

bond

is

to

that

for

already

addition

of

halogens.

polar,

as

the

hydrogen

The

only

halogen

halides

exception

atom

is

to

is

more

alkenes,

that

the

the

mechanism

halogen–hydrogen

electronegative than the

hydrogen atom.

Consider

aqueous

the

electrophilic

hydrogen

reaction

between

but-2-ene, C

4

H

8

(g), and

HBr(aq).

H

H

H

C

H

Br

H

H

C

H

addition

bromide,

+

C

H

Br

C

3

C

CH

3

CH

C

3

3

but-2-ene

hydrogen

2-bromobutane

bromide

The

electrophilic

1.

The

addition

electron-rich

atom,

and

the

proceeds

C=C

bond

hydrogen

is

via

the

following steps:

attacked

bromide

by

the

molecule

partially

splits

positive

hydrogen

heterolytic ally

to

form a

bromide anion.

2.

When

the

atoms,

hydrogen

another

c arboc ation

3.

The

reaction

results

in

the

atom

c arbon

forms

atom

covalent

bond

positively

with

one

charged.

of

the

This

c arbon

produces a

intermediate.

between

the

formation

of

unstable

the

CH

3

c arboc ation

product,

H C

a

becomes

C

and

the

bromide anion

2-bromobutane.

CH

3

H 3

C

CH

3

3

3

+

C

C

H

C

C

H

C C

H

Br

H

Br H

H H

훿+

H

훿–

2-bromobutane

p Figure 23

Mechanism for the electrophilic addition of hydrogen bromide to

but-2-ene

645

3

What

are

the

mechanisms

of

chemic al

LHA

Reactivity

change?

Electrophilic addition of water

to symmetric al alkenes

The

you

the

for

third

example

reaction

this

by

Consider

water

in

the

the

the

CH

2

involves

the

addition

of

electrophilic

and

protonation

a

water

+

C

CH

C

electrophilic

1.

The C=C

mixture,

2

CH

water

c arbon

bond

The

the

CH

3

C

CH

2

CH

2

3

is

H

In

one

this

molecule,

atom

in

resulting

ion

of

CH

3

alkene

and

to

nally

between

OH

the

by

the

c ase,

H

earlier

form

the

in

The

the

loss

the

topic

was

mechanism

c arboc ation,

of

a

hex-3-ene, C

6

proton.

H

12

(l) and

H

H

3

CH

2

C

C

C

H

OH

as

a

via

the

proton

proton

a

CH

2

CH

3

hexan-3-ol

following steps:

present

c arbon–c arbon

the

acting

is

acting

nucleophile,

in

the

bonds

as

an

attacks

acidied

and

reaction

producing a

electrophile.

the

positively

charged

c arboc ation.

oxonium

ion,

a

deprotonates,

protonated

forming

the

alcohol,

alcohol

is

strongly acidic. The

hexan-3-ol

and

regenerating

proton.

H H

the

reaction

proceeds

attacked

breaking

oxonium

H

of

at

solution.

3

addition

c arboc ation.

3.

looked

acidied

water

The

A

in

molecule

addition

hex-3-ene

2.

water,

H

C

3

alkenes

addition

presence of an acid.

H

H

electrophilic

between

reaction

followed

of

C

2

CH

CH

2

CH

3

C

CH

2

CH

2

3

H

H

3

CH

3

C

CH

H

C C

H

H

2

CH

2

3

+

H C

C

O

H

2

+

+

H

H

C C

H

H

H

C C

H

H

O

OH

H H

+

H

p Figure 24

Mechanism

for the electrophilic addition of water to hex-3-ene

A

proton

is

consumed

regenerated

at

the

at

end

the

of

beginning

the

reaction,

of

so

the

it

reaction,

acts

as

a

and

a

proton is

c atalyst.

Practice question

14.

646

Which

of

A.

CH

B.

C

C.

CH

D.

CH

3

3

H

3

4

the

CH

7

I

2

+

CH

following

CH

2

Cl

KCN

2

CH

+ Cl

2

2

→

→

C

reactions

CH

3

H

CHCH

→

CH

3

2

Cl

7

3

is

CHCH

CN

+

Br

+

2

+ HCl

2

an

+

example

of

electrophilic

HCl

KI

→

CH

3

CH

2

CH

2

CH(Br)CH

2

Br

addition?

Reactivity

3.4

Electron-pair

sharing

reactions

LHA

C arboc ations in electrophilic addition

reactions (Reactivity 3.4.12)

To

predict

an

unsymmetric al

potential

that

the

present

the

the

The

product

alkene,

c arboc ations

stability

in

the

c arboc ation

of

major

three

major

of

a

bec ause

produced

A

the

of

positive

the

alkenes

electrophilic

to

during

the

charge

the

the

You

involving

learned

inductive

has

density

reaction

relative stability of the

reaction.

on

c arboc ation

eects

previously

of

alkyl

groups

greater stability than a primary

is

oset

by

the

inductive

eects

(gure 21).

electrophilic

c an

addition

understand

depends

tertiary

substituents

products

unsymmetric al

an

need

c arboc ation

molecule.

alkyl

of

we

be

addition

predicted using

of

hydrogen halides to

Markovnikov’ s rule.

In an unsymmetrical alkene, there are two possible carbon atoms on the carbon–

carbon double bond that are susceptible to electrophilic attack. Markovnikov’ s rule

states that the electropositive part of the polarized electrophile will preferentially

bond to the carbon that has the least number of alkyl substituents. This results in a

carbocation with the positive charge centred on the most substituted carbon, so

the major reaction product will form via the more stable carbocation.

For

to

example,

the

consider

unsymmetric al

2-bromopropane

H

3

electrophilic

and

propene.

addition

This

reaction

reaction

has

of

hydrogen

two

possible

bromide

products:

1-bromopropane.

H

H

H

C

H

Br

H

H

H

C

Br

H

H

C

H

the

alkene

+

C

H

Br

3

C

C

H

and

3

C

C

H

H

C

propene

hydrogen

1-bromopropane

2-bromopropane

bromide

2-bromopropane

will

form

via

a

will

primary

2-bromopropane

will

form

via

a

secondary

c arboc ation.

be

the

major

The

c arboc ation

secondary

product

H

while

1-bromopropane

c arboc ation

is

more

stable, so

(gure 25).

CH

H

CH

3

H

C

C

3

H

H

C

C

H

Br

H

+

Br

H CH

H

3

C

H

C

secondary

major

c arboc ation

product

H

훿+

H

훿–

H

CH

CH

3

H

C C

3

H

H

C

C

Br

H

H

+

Br H

p Figure 25

primary

minor

c arboc ation

product

The major product

in this reaction is 2-bromopropane, as the

reaction proceeds preferentially via a more stable c arboc ation

647

LHA

Reactivity

ATL

3

What

are

the

mechanisms

of

chemic al

change?

Thinking skills

Organic synthesis converts a starting material via a series of

Summarize

reactions into the desired product. Each step produces an

of

intermediate product in quantities less than the theoretical

for

paper.

each

all

Use

of

the

reactions

your

the

from

summary

to

Reactivity 3

propose

a

on

one

synthetic

sheet

route

following:

yield, so an ecient synthetic pathway must involve the

a.

methanoic

b.

propanone

acid

c.

ethyl

from

bromomethane

smallest possible number of steps. Synthetic organic

chemists oen use a method referred to as retrosynthesis.

from

propene

Starting with knowledge of the structure and properties of ethanoate

from

ethene.

the target compound, they think “in reverse” to determine

possible synthetic pathways to that compound.

Electrophilic substitution in benzene

(Reactivity 3.4.13)

Benzene

The

structure

discussed in

of

benzene

was

Structure 2.2 (AHL )

its

does

six-electron

mechanism

nitration

of

not

readily

aromatic

undergo

ring.

electrophilic

addition

Instead,

it

reactions

undergoes

substitution

in

bec ause of the stability of

substitution

benzene

c an

be

reactions. The

illustrated

by the

reaction.

+

The

rst

which

of

step

acts

of

as

nitronium

benzene

the

ions,

concentration

of

nitration

electrophile

but

in

these

a

in

is

mixture

ions

the

this

of

formation

reaction.

sulfuric

increases

as

a

of

Pure

acid

result

the

nitric

and

of

nitronium ion, NO

acid

nitric

the

contains

acid

at

following

,

2

only

traces

50 °C the

reactions:

+

HNO

+ H

3

2

SO

4

+

H

In

2

NO

turn,

the

reaction,

1.

The

H

2

NO

+

3

HSO

4

+

⇌

3

⇌

NO

high

which

+

2

H

2

O

concentration

proceeds

nitronium

ion

as

of

nitronium

ions

increases

the

rate

of

the

nitration

follows:

electrophile

is

attracted

to

the

deloc alized

pi

electrons of

the benzene ring.

+

2.

Two

electrons

C–N

bond

moves

from

forms

onto

the

the

while

benzene

a

pi

ring

electron

are

donated to the NO

from

one

N–O

bond

in

2

ion,

the

so

a

new

nitronium ion

oxygen atom.

+

H

+

N

+

O

O

O

The

addition

ring.

This

is

represents

aromatic

step

648

of

of

the

ring

the

the

nitronium

depicted

by

the

deloc alization

in

benzene

reaction.

ion

to

benzene

incomplete

of

the

requires

breaks

dashed

positive

energy,

charge.

so

the

aromaticity of the

circle in the ring, which also

this

Breaking

process

is

of

the

the

very stable

rate-determining

Reactivity

Water

then

restoring

acts

the

as

a

base,

aromaticity

deprotonating

of

the

benzene

the

c arboc ation

ring,

which

Electron-pair

sharing

reactions

LHA

3.

3.4

intermediate and

gives

the

nal

product,

nitrobenzene.

+

+

H

H OH

NO

When

drawing

benzene,

1.

The

pay

curly

the

arrow

deloc alized

the

2.

3.

to

the

for

in

electrophilic

substitution

reaction

involving

following:

representing

electrons

an

2

the

electrophilic

benzene

and

attack originates at the ring of

terminates

at

the

positive

charge on

electrophile.

The

a

NO

2

mechanism

attention

O

3

2

structure

positive

The

curly

of

the

c arboc ation

must

show

an

incomplete

dashed

circle and

charge on the ring.

arrow

representing

the

hydrogen

ion

leaving originates at the bond

between the c arbon and hydrogen atoms and terminates at the benzene ring

Practice questions c ation.

15. 4.

The

last

curly

arrow

originates

at

a

lone

electron

pair

of

Benzene

is

an

aromatic

water and terminates

hydroc arbon. at

the

hydrogen

ion

leaving.

a. 5.

The

structural

formula

of

the

substituted

benzene

must

+

the

released

hydrogen ion, H

be

State

or

hydronium ion, H

3

O

the

typic al

reactions

shown along with

that benzene and

+

,

.

cyclohexene

with

will

undergo

bromine.

TOK b.

Explain

the

Arrows

and

have

many

chemistry

signify

uses

focuses

movement

or

in

on

chemistry.

the

Arrows

oen

transformations

imbalance.

There

are

of

represent

matter.

several

types

mechanism

of

for

benzene,

transformations,

Arrows

of

the

nitration

using

curly

show

the

arrows to

c an also

arrows,

movement of

each with electronpairs.

its

own

specic

meaning:

•

transformation

of

•

movement

of

a

•

movement

of

an

reactants

into

products

Linking questions

single

electron

What

electron pair

C

6

to •

reversible

H

6

,

are

the

that

features

make

undergo

it

not

addition

of

benzene,

prone

reactions,

reaction despite

being

highly

unsaturated?

(Structure 2.2) •

resonance

structures

•

coordination bond

•

bond dipole

Nitration of benzene uses a

mixture

sulfuric

of

concentrated nitric and

acids

to

generate

a

strong

+

electrophile, NO

acid/base

in

How

are

arrows

used

as

symbols

in

other

areas

of

knowledge?

this

2

.

How

c an the

behaviour of HNO

mixture

be

3

described?

(Reactivity 3.1)

649

Reactivity

3

What

are

the

mechanisms

of

chemic al

change?

End of topic questions

Which

is

A.

electrophile

an

example

of

a

Lewis

LHA

5.

base?

Topic review

1.

Using

your

answer

knowledge

the

guiding

from the

question

as

Reactivity 3.4

fully

as

an

topic, B.

BF

C.

CH

D.

a

3

possible:

4

What happens when reactants share their electron pairs nucleophile

with others?

6.

Exam-style questions

Which

of

attacking

species

is

matched

with

its

mechanism

reaction?

Multiple-choice questions

–

2.

Identify

c annot

and

act

explain

as

a

why

one

of

the

following

A.

species

OH

electrophilic substitution

+

nucleophile.

B.

Cl

C.

NH

nucleophilic addition

D.

NO

+

A.

H

B.

H

nucleophilic addition

4

+

C

B

N

electrophilic substitution

2

H

7.

H

Which

bromoalkane

is

most

likely

to

hydrolyse via a S

N

1

mechanism?

C.

H

H

C

H

H

H

C C

H

H

C

H

3.

D.

S

A.

CH

B.

(CH

C.

(CH

D.

CH

3

H

CHBrCH

3

3

)

CHBr

)

CBr

2

3

2

CH

3

H

Ethene, C

2

H

4

,

reacts

with

steam

in

the

presence of a

3

CH

2

CH

2

CH

2

Br

Extended-response questions

strong acid. 8.

Organic

chemistry

c an

be

used

to

synthesize

a

variety of

products. C

2

H

4

+

H

2

O

→

C

2

H

5

OH

a.

What

is

the

name

of

this

type

of

Draw

the

between

Nucleophilic substitution

B.

Neutralization

C.

Condensation

D.

Electrophilic addition

b.

Sketch

the

LHA

Which

statement

is

curly

c.

Electrophiles

B.

Nucleophiles

C.

Electrophiles

D.

Nucleophiles

product

for

reaction

for

with

the

[1]

reaction of

hydrogen

bromide using

why

in

[3]

the

part

major

(b)

is

organic

product of the

2-bromo-2-methylbutane and

correct?

are

are

are

2-bromo-3-methylbutane.

are

[2]

Brønsted–Lowry acids.

Brønsted–Lowry acids.

Lewis acids.

9.

Chlorine, Cl

a.

State

b.

Predict,

c.

Explain

2

,

the

reacts

undergoes

type

with

of

many

reactions.

reaction occurring when ethane

chlorine

to

produce

chloroethane.

[1]

Lewis acids.

giving

chloroethane

NaOH(aq),

movement

a

is

reason, whether ethane or

more

reactive.

mechanism

and

using

of

of

the

aqueous

curly

electron

[1]

reaction

sodium

arrows

pairs.

to

LHA

the

chloroethane

650

nal

water.

arrows.

Explain

not

A.

the

and

mechanism

2-methylbut-2-ene

reaction

4.

of

but-2-ene

LHA

A.

structure

reaction?

between

hydroxide,

represent the

[3]

Reactivity

Propene, C

3

H

6

,

is

an

important

starting

material

for

11.

Benzene

nitration

occurs

Electron-pair

when

benzene

sharing

reactions

reacts with the

+

many

a.

products.

Consider

nitronium ion, NO

the

conversion

halogenoalkane

with

LHA

i.

State

the

type

of

ii.

State

the

IUPAC

iii.

Outline

iv.

Write

of

the

propene to a

general

formula C

3

H

7

an

it

of

[1]

the

major

Write the equation for the production of the nitronium

b.

Explain

ion from concentrated sulfuric and nitric acids.

reaction.

name

is

the

equation

halogenoalkane

hydroxide

to

major

product.

for

the

product

using

[1]

product.

the

curly

aqueous sodium

12.

But-1-ene

series

of

formula C

3

H

8

O.

and

is

an

But-1-ene

aqueous

sodium

possible

was

Explain

between

the

rate

of

were

[4]

unsymmetric al

to

alkene

that

c an

undergo a

form an alcohol.

c an

with

undergo

an

hydrogen

electrophilic addition

products

of

iodide.

this

Deduce the two

reaction.

which

compound

is

the

[2]

major

product

of

the

reaction.

halogenoalkane.

The

for

[2]

Draw

and

explain

the

mechanism

for

the

reaction

following using

results

movement of

reaction and the c.

concentration

[1]

benzene,

c arried out to determine the the

relationship

the

of

hydroxide, an b.

experiment

indic ate

nitration

[1]

For the reaction between the major halogenoalkane

product

to

the

pairs.

reactions

reaction

b.

arrows

for

produce a compound with the a.

general

mechanism

[1]

reaction of the major

with

.

a.

Cl.

electron why

2

LHA

10.

3.4

curly

arrows.

[4]

obtained.

d.

The

major

product C

nucleophilic

potassium

4

H

I

substitution

hydroxide.

mechanism

9

for

the

c an

undergo a

reaction

Draw

reaction

and

with

aqueous

explain the

using

curly

arrows. [3]

etar

[halogenoalkane]

i.

State

the

ii.

the

order

Under

certain

independent

ions.

is S

iii.

of

the

reaction

with

respect to

halogenoalkane.

N

Deduce

1 or S

N

Sketch

the

arrows

to

electron

2.

[1]

conditions,

of

the

Explain

represent

pairs.

reaction

concentration

whether

reaction

the

the

your

of

reaction

rate is

hydroxide

mechanism

answer.

[2]

mechanism, using curly

the

movement of

[4]

651

Exam-style questions

Cross topic exam-style questions

DP

exam

explore

questions

the

links

may

be

between

topic-specic

various

or

concepts,

refer

as

to

well

content

as

from

aspects

of

across

NOS

dierent topics. These questions

and

the

skills

in

the

study

of

chemistry.

Below, three exam-style questions have been annotated to show their links to dierent parts of the course. Next

time

The

you

do

an

enhanced

Nitrous

exam-style

question,

greenhouse

oxide, N

O,

and

eect

c arbon

try

and

to

link

greenhouse

to

the

ozone-layer

dioxide,

CO

²

also

it

,

various

depletion

are

course

are

greenhouse

two

topics,

NOS

separate

gases.

and

skills

atmospheric

as

shown

below.

problems.

Chlorouoroc arbons

(CFC s)

are

²

gases,

but

they

are

primarily

known

for

their

ozone-depleting

properties.

Question 1

a.

Nitrous

oxide, N

O,

is

a

greenhouse gas.

2

i.

Draw

a

possible

Lewis

Structure

2.2

Lewis

[2]

Structure

2.2

VSEPR

[2]

Structure

3.2

IR

Structure

2.2

AHL

formula of N

formulas

2

ii.

State

iii.

LHA

iv.

and

Explain

explain

why

Deduce

the

nitrous

formal

molecular

oxide

is

charge

IR

of

geometry

of

nitrous

oxide.

active.

each

of

the

atoms

in

nitrous

oxide.

[1]

Formal

b.

C arbon

c arbon

dioxide

dioxide

Describe

is

is

a

greenhouse

shown

one

of

gas.

An

image

of

a

limitation

of

this

charge

molecular model of

below:

strength

representation

652

the

the

and

one

bonding

in

c arbon

model’ s

dioxide

molecules.

[2]

NOS

–

models

spectroscopy

Exam-style questions

Question 2

a.

C arbon

i.

dioxide

State

ii.

is

the

produced

balanced

Determine

bond

the

in

complete

equation

molar

enthalpy

the

for

enthalpy

the

of

combustion

combustion

combustion

of

of

of

organic compounds such as alcohols.

propan-1-ol.

propan-1-ol

[1]

Reactivity

Explain

why

the

Fuels

from

data.

[3]

Reactivity

iii.

1.3

value

you

obtained

above

diers

from

the

1.2

standard

Energy

cycles

–1

enthalpy

b.

A

student

using

the

of

combustion

determines

apparatus

the

of

propan-1-ol,

enthalpy

shown

of

which

combustion

is

of

.

–2021 kJ mol

propan-1-ol

by

[1]

c alorimetry

below:

thermometer

water c alorimeter

spirit burner

i.

Outline

the

experimental

method

employed

by the student, identifying

T ool

1:

Experimental

techniques ii.

The

student’ s

experimental

enthalpy

of

combustion

of

propan-1-ol

–

calorimetry

was

–1

.

–894 kJ mol

C alculate

the

percentage

error.

[1]

T ool

3:

skills Predict,

giving

a

reason,

the

sign

of

the

entropy

change

Mathematical

percentage LHA

iii.

–

for the

error combustion

of

propan-1-ol.

[1]

–1

iv.

The

standard

entropies

–1

and

205 J K

of

propan-1-ol

and

oxygen

are

193 J K

–1

mol

–1

mol

,

respectively.

Further

standard

entropy

values

are

Reactivity listed

in

AHL standard

entropy

change

for

the

combustion

of

v.

J K

Entropy

propan-1-ol,

and –1

in

1.4

section 13 of the data booklet. Using these data, determine the

spontaneity

–1

mol

Determine

. [1]

the

Gibbs

energy

for

the

combustion

of

propan-1-ol

at

298 K.

–1

Give

your

answer

in

kJ mol

.

[2]

653

Exam-style questions

Question 3

Ozone

The

is

a

gas

following

found

in

the

mechanism

Step 1

O

upper

has

+ Cl• →

atmosphere

been

→

proposed

2

A

student

and

b.

LHA

c.

Identify,

The

describes

suggest

the

giving

rate

Cl•

correct

a

harmful

UV

depletion

radiation

of

from

the

Sun.

ozone:

O

→

Cl• + 2O

as

for

²

a

“chloride

anion”.

Outline

the

student’s mistake

term.

reason,

equation

the

²

ClO• +

³

a.

absorbs

describe

ClO• + O

³

Step

which

to

the

the

[2]

species

reaction

is

in

the

found

mechanism

to

be

rate

=

that

k[O

is

a

c atalyst.

][Cl•].

Reactivity

3.3

Radicals

[1]

Reactivity

2.2

Catalysts

[1]

Reactivity

2.2

AHL

Identify,

³ giving

a

reason,

determining

which

step

in

the

mechanism

is

likely

to

be

the

rate

step.

Rate d.

Chlorouoroc arbons

(CFC s) such as CF

Cl ²

to

certain

below

to

wavelengths

show

the

of

are

of

a

Cl•

of

radiation.

sh-hook

arrows

and

the

when

species

Complete

from a CF

Cl ²

Include

Cl•

exposed

²

electromagnetic

formation

sources

equations

structural

formula

of

the

the

diagram

molecule. ²

missing

species.

[2]

Reactivity

sharing

3.3

Electron

reactions

F

hf

Cl

C

F

e.

CFC s

are

HFOs,

in

ozone-depleting

are

their

unsaturated

applic ation

decompose

is

faster

as

substances

organic

refrigerants.

and

have

used

compounds

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shorter

as

refrigerants.

with

HFOs

lifetimes

in

the

are

more

the

Hydrouoroolens,

potential

to

replace

reactive

atmosphere.

than

An

CFC s

CFC s, they

example

of

a

HFO

2,3,3,3-tetrauoropropene.

i.

Draw

the

skeletal

structure

of

2,3,3,3-tetrauoropropene.

[2]

Structure

3.2

Representation ii.

Suggest

why

HFOs

are

more

reactive

than

CFC s.

organic

654

of

[1]

Reactivity

3.4

Reactivity

of

compounds

alkenes

The inquiry process

Introduction

This

section

during

the

claries

DP

the

learning-through-inquiry

chemistry

course.

Tool 1: Experimental techniques,

during

lessons,

internal

Figure

1

support

experiments,

It

will

help

you

approach

to

Tool 2: Technology

the

collaborative

you will use

develop the skills in

and

sciences

Tool 3: Mathematics

project, and the

assessment (IA).

shows

the

how

inquiry

the

skills

detailed in the

Tools for chemistry

chapter

c an

process.

Background

Research

ypothesis

research

uestion

and prediction

Improvement

Identiying

and extensions

variables Exploring

Experimental

Evaluating

techniues

methodology

Evaluating

reen

esigning

Evaluating chemistry hypotheses

ontrolling

variables

Systematic and

inimising

random error

errors

Uncertainties

in results

ontrolling

Risk

variables

assessment

oncluding

Percentage

orking

error

saely

Identiying and Reliability collect data and validity

Interpreting

ollecting

results

data

Uncertainties

Accuracy

Use o

and precision

sensors Processing Trends and Uncertainty data patterns propagation

Interpreting raphing graphs

Identiying alculation outliers

▴

Figure 1

The inquiry cycle with examples of supporting skills

655

The

inquiry

process

The

are

skills

in

the

grouped

skills,

research

process

approaches

into

and

ve

skills

tools

to

learning

c ategories:

and

(ATL)

thinking

self-management

shown

in

framework

skills,

support

learning. They

communic ation skills, social

skills.

Where

do

ATL skills t into the

gure 1?

Theory of knowledge

You

will

into

further

of

notice

knowledge

themselves

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owns

owning

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want

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to

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process

The

cycle.

starting

knowledge?

What

dierent

ways.

covered

in

go

cyclic al:

Scientists

points

in

more

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for

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outcomes

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expansion of science as a body

use

their

rights

work

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previously

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research.

and

responsibilities come with

knowledge?

start

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for

also

chemistry.

to

on

others)

equipment

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inquiry.

subtopics

learn

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of

scientic

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cooking.

using

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its

your

through

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the

chemistry,

favourite

might

the

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such

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loc al

section,

inquiry

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want

to

Alternatively,

or

as

the

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cycle

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and

chemic al

issue

consider

identify

deeper into one or

may

materials

environmental

you

go

you

reactions

that

one

the

have

a

hobby and

chemistry

c an

used to

involved in

be

examined

c ase study that will

skills

required to nd

your questions.

C ase study 1: Ocean acidic ation

The

concentration

fuels

that

has

oceans

c arbon

of

increased

have

dioxide

atmospheric

dramatic ally

absorbed

dissolves

in

over

in

c arbon

recent

25%

seawater,

of

it

dioxide

times.

all

produced

Since

the

anthropogenic

forms

by

burning

1980s,

it

c arbon

c arbonic acid, H

is

dioxide.

CO 2

fossil

estimated

,

a

As

weak acid

3

+

that

dissociates

into

hydrogenc arbonate

ions,

HCO

,

and

hydrogen ions, H

. A

3

+

higher

concentration of H

ions

increases

ocean

acidity,

resulting

in

a

decrease

in pH.

An

the

inquiry

might

chemistry

may

want

to

involve

behind

create

a

testing

the

list

this

theory

of

hypothesis,

and

nd

preliminary

but

reliable

questions

rst

to

form

such as:

656

•

What

is

•

Is

a

•

Which

this

ocean

loc al

acidic ation?

or

a

evidence

global

issue?

supports

the

statement

you

need

information

above?

the

to

to

understand

support

basis

of

it.

You

your inquiry

The

The

next

step

supporting

to

ocean

1990,

to

do

claim

research

that

acidic ation.

government

to

is

the

data.

For

Figure

according

to

using

increased

this

2

the

reliable

c arbon

purpose,

shows

data

the

from

sources

dioxide

process

provide evidence

emissions

has

contributed

you could use journal articles or

change

the

that

inquiry

in

Global

pH

of

Ocean

the

sea

D ata

surface

Analysis

from

1700

Project

(GLODAP).

Δ sea–surface pH [–]

–0.12

▴

Figure 2

The

the

to

map

explain

better

you

this?

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to

know

dissolve

explains

dioxide

this

that

in

with

quantitative

this

data,

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at

learning

lower

energy

to

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–0.06

polarity

of

dioxide

such

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0

greater

extent

in

regions close to

from the subtopics in DP chemistry

their

gases

molecules.

will

dissolve

in

water

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covalent

is

you

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molecules

non-polar

may

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is

discussed.

molecule

conrm

and

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to

as

choose

molecular

WebMO.

you

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you

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should

polarity

Molecular modelling (Tool 2: Technology)

need

will

–0.02

1700 to 1990

temperatures,

of

–0.04

phenomenon?

However,

data.

has

of

kinetic

this

the

You

pH

any

–0.08

of the sea surface from

example,

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water.

purpose.

the

use

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Structure 2.2,

you

that

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due

chemistry

In

The change in pH

shows

poles.

–0.1

is

a

build

the

c arbon

has

a

net

that

dioxide

dipole

readily

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valuable

soware

dioxide

not

of

tool

for

produces

molecule in

moment

of

0 D.

657

The

inquiry

process

You

c an

also

check

experimentally.

shown

in

the

For

eect

of

example,

exposure

you

could

to

set

c arbon

up

the

dioxide

on

water pH

experimental

apparatus

gure 3.

01

C arbon

dioxide

collected

02

03

04

Hydrochloric

acid 05

Water

C alcium

c arbonate

▴

A

Figure 3

test

tube

containing

is

Experimental apparatus to collect

containing

a

measured.

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into

the

and

between

that

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of

the

have

yield

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pH.

aect

the

connected

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produced

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establish

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water.

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of

use

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in

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factors

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no

results.

may

aect

matter

select

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tube

the

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there

to

your

how

results

simple,

instruments

with

and

you

analyse

need

should

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to

list

on

sizes

be

the

and

will

is

a

a

the

be

vessel

water

c alcium

released

correlation

support

the

that

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to

of

you change the

this

based

tube

pH

containing

reaction

value.

data

a

initial

whether

water

critic ally

with

The

considered in Tool 1: Experimental techniques

reliable

experiment,

results

will

added

record

you

and

is

is

deionized

dioxide

then

water,

of

acid

c arbon

c an

temperature

temperature

external

658

the

You

c arbonate

volume

Hydrochloric

water.

temperature

c alcium

measured

c arbon dioxide

your

data

not

in

all

hypothesis

gure 2.

experiments

experiment: what

controlled?

variables

Before

aecting

precisions.

any

your

The

The

following

are

some

•

How

many

•

How

will

you

measure

•

How

will

you

weigh

trials

inquiry

process

relevant points to consider:

will

you

do?

the

the

volume

sample

of

of

water?

c alcium

c arbonate?

How

much

will

you

use?

•

What

•

Will

•

How

concentration

the

container

much

know

•

How

c an

•

How

will

•

What

You

should

results.

time

when

the

you

you

of

will

enter

completed

procedure,

or

water

be

ensure

record

the

the

pH

you

the

open

acid

or

for

will

you

closed

the

to

reaction

use?

the

to

occur?

c arbon

of

in

the

your

record

you

How

will

you

dioxide

dissolves

in

the

water?

water?

in

measurements?

a

table

and

produce

graph (Tool 3: Mathematics).

follow-up

surroundings?

ended?

there

experiment,

any

has

that

are

be

required

the

data

interpret

the

hydrochloric

reaction

uncertainties

Then,

of

should

think

experiments

how

you

you

may

c an

want

a

Once

to

graph

you

of

your

have

improve the

do

based

on

your

results.

You

may

you

are

water:

you

want

likely

sulfur

will

nd

reduces

forming

the

oxides,

extensions

found

impact

of

to

disrupts

positions

of

oxides

is

your

gases

and

in

shells

equilibrium

equilibria.

All

for

these

While

the

coastal

c arbonate

their

inquiry.

aect

ammonia.

signic ant

build

the

to

other

c alcium

organisms

also

that

nitrogen

their

availability

marine

the

consider

have

that

concentration

shiing

to

to

areas.

your

surface

more

research,

ocean

research,

Ocean

acidic ation

reefs and shell-

skeletons.

reactions

inquiry

doing

of

Doing

coral

and

pH

that

lines

Increased acid

occur

will

in

the

initiate

a

ocean,

new inquiry

cycle.

We

will

now

examine

the

individual

steps

of

the

inquiry

cycle

in

gure 1.

659

Exploring, designing and

controlling variables

In the

•

•

exploring

topic

step

Find

a

your

question

that

of

the

inquiry

interests

and

use

you.

critic al

cycle,

Find

you

should

information

thinking

to

consider

that

helps

the

following:

you

to

There are many sources for your background reading, and you

select

•

those

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a

•

out

a

will

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that

be

step,

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acquired

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are

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knowledge

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that

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you

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should

must c arefully

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help

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and

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interpret the data.

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use

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simulations

or

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modelling.

•

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that

•

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should

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you

•

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Could

•

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•

to

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independent

number

the

the

be

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will

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need

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•

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your questions.

variables

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the

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•

and

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your

storage

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reproduce

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always

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pilot

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improvement.

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controlling

•

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to

are

environmental

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most

you

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stable

temperature,

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in

consider

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following:

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terms

or

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of

their

monitor

sizes

them

if

and

you

precision?

are unable to

pressure and humidity).

containers

to

prevent

heat

loss or gain.

C ase study 2: The pH sc ale

At

times

now

For

your

that

you

example,

accepted

have you

660

inquiry

have

this

the

a

pH

range

may

be

better

sc ale

as

based

on

the

understanding

is

that

dened

was

how

as

it

content

of

values

was

you

chemistry,

are familiar with, but

it

between

covered

in

may look puzzling.

0

all

and

14.

previous

You

have

courses.

ever considered whether it is correct? In your initial research, you

But

might

The

nd

out

–1.1,

that

while

There

are

Russia,

naturally

contain

estimated

as

in

low

the

commercially

saturated

as

pH

US.

have

These

When

nding

under

which

are

this

as

low

the

as

most

results

–1.7.

well.

Extremely

acidic

your

were

as

a

next

At

of

sulfuric

acidic

mine

currently

step

obtained.

pH

springs

and

underground

waters

process

hydrochloric acid has a pH of

has

Hot

hydrochloric

encountered

information,

these

solution

examples

occurring

been

concentrated

hydroxide

occurring

naturally

values

–3.6

available

sodium

inquiry

near

Ebeko

acids,

waters

the

and

with

volc ano,

have

pH

values

Iron Mountain Mine

known.

would

times,

in

15.0.

be

to

check the conditions

you will nd it dicult to

reproduce them in the school laboratory. For example, it is unusual to nd buers

to

c alibrate

probes.

your

You

attempting

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prepare

pH

also

it,

as

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probe

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highly

acidic

simple

samples

with

at

the

necessary

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substances

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increasing

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acid

pH

the

are

values

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or

risks

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concentration

you

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need

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before

corrosive.

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and

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observe

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correlation

+

between

pH

determined

approach,

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then

values

value.

as

the

to

This

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to

the

[H

suggest

measured

possibly

hypothesis

c alculated as

values

] concentration

that

will

consider

DP

chemistry

deviate

molecular

deviation.

You

and

from

is

the

the

experimentally

using

a

simplied

c alculated

values.

You

interactions and suggest an initial

could

also

propose

a

new denition

+

of

pH

that

does

unreliable,

consider

not

which

involve

may

whether

you

also

[H

].

It

also

contribute

have

enough

worth

to

the

noting

that

deviation.

evidence

to

pH

meters

Therefore,

support

your

c an be

you should

hypothesis.

C ase study 3: The eect of temperature on

equilibrium

Equilibrium

in

is

open

investigating

K,

of

the

the

following

to

many

eect

of

investigations.

changes

in

Aer

some

by

serial

research,

this.

The

dilution,

maximum

on

you

the

may

be

interested

equilibrium constant,

2+

(aq) + SCN

investigate

example,

reaction:

3+

Fe

For

temperature

(aq)

you

⇌

will

[FeSCN]

nd

experiment

producing

absorbance

a

that

(aq)

colorimetry

would

involve

c alibration

is

a

suitable

preparing

technique to

standard solutions

curve, and then determining the

wavelength (Tool 1: Experimental techniques).

Due to You

time

constraints,

you

may

need

to

c arry

out

dierent

parts

of

the

will

need

to

use

your

investigation on mathematics skills to make

dierent

days. c alculations, and consider and

process uncertainties Here

are

some

of

the

challenges

you

may

face

in

designing

the

(Tool 3:

experiment and

Mathematics). controlling

variables:

Before •

Typic al

to

•

instruments

in

schools

only

provide

you

begin

any

experiment,

reliable data to absorbances up

you

must

and

the

identify

the

risks

involved

1.

The

optimum

wavelength

should

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risks.

ways to minimize these

You

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should

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involved

any

and

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2+

•

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661

Collecting data,

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C ase study 4: C aeine in painkillers

You

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lid

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▴

Figure 4

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663

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▴

664

Figure 5

TLC

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▴

Table 1 Suggested

improvements to the experiment

based

on the results in gure 5

C ase study 5: Antioxidant eects of vitamin C

When

considering

example,

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temperature,

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C ase study 6: Enthalpy of solution for metal chlorides

A

student

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666

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The

inquiry

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C ase study 7: Solubility of potassium bitartrate

A

student

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K

, of sp

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▴

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668

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Figure 1

Possible steps in developing your research design

669

The

internal

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When

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question.

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The

c ase

studies

below

how they link to the inquiry

process.

C ase study 1 (Inquiry 1: Exploring)

A

student

content

wants

of

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kale

to

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question:

a

as

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it

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water.

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temperature changes when cooking

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C ase study 2 (Inquiry 1: Exploring, Designing, The

Controlling variables)

a

skills

titration

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for conducting

discussed in

Tool 1:

Experimental techniques, in the A

student

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acid

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this

The

acid

describing

provides

research

the

tea.

of

the

6

to

4

and

for the

4–

ions 6

react

assay

realizes

suitable

compounds

3–

will

(DPPH)

tea, and nds it in the school

tea solutions.

student

assay

decides

the

of

the

Prussian blue test,

which

chemic al

equations

add

value to the

reactions that occur in

with

state

symbols.

671

The

internal

assessment

C ase study 3

(Inquiry 1: Designing, Controlling

variables)

A

of

student

oxalic

they

wants

acid

will

in

to

investigate

the

determine

dierent

times

by

spinach.

the

how

The

the

concentration

titrating

it

with

cooking

student

of

time

outlines

oxalic

aects

their

acid

in

the

concentration

methodology

the

cooking

as

follows:

water at

potassium manganate(VII), KMnO

. 4

Several

control

the

mass

of

the

source

spinach,

and

methodology

•

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•

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of

will

to

boiling.

example,

and

•

a

of

Should

the

or

should

for

the

sample.

the

be

this

investigation, such as

range of cooking times, and

The

spinach

used?

following

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is

reasonable

acid

provide

clear

line

be

aspects of the

will

of

varieties.

the

within

of

oxalic acid in

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information is

titrant.

the

context of the

decompose if the solution

sucient

of

used?

concentration

dierent

concentration

oxalic

will

a

all

between

suitable

shows

best

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and

with

reliable

the

data?

equation

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included

R².

spinach

be

stem

intervals

2

water,

considered:

example,

value of

the

a

considered

of

spinach

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temperatures

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sample

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signic antly

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high

the

to

volume

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to

prepare

investigation?

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of

spinach

vary

range

need

the

need

leaves,

type

required

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storage

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spinach

•

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blended

for?

in

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processor

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surface

rst?

area

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experiment

used

c an

instance,

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•

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hot

water

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they

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several

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and

c an

there

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672

when

their

could

to

used

be

to

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be

prepared?

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samples

process

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very

on

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the

small,

concentration

results.

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means that

lower.

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desired

temperature

remain

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at

each

samples

continue,

equipment

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research.

based

temperatures?

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monitored?

removed?

experimental

instruments

solution

sped

any

up

stable?

asks

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and

temperature? Should

are

not

aecting

immediately

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results.

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pipettes should

in

a

the

must

a

considered

reacts

if

data

with

are

collected

water, and this

presence of sunlight.

heating

pilot

decompose.

be

manganate(VII)

requirements

gentle

in

available.

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Potassium

specic

ascertained

will

be

more

spinach

on

experiments

needs

temperature

precision?

based

recorded

bath

be

preparing solutions or dilutions. The golden rule is to use the

therefore

oxalate

water

solution

solution? The student could run a pilot

future

titrant

cooking

sessions.

is

the

provide

the

the

for

the

down

precision

reaction

•

and

for

a

the

should

cooled

used

of

or

for

volumes

could

down,

highest

•

plate

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size

manganate(VII)

concentration

titrant

appropriate

their

be

the

long

be

cooled

•

a

optimized

bath

sample.

•

with

be

concentration

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potassium

concentration

is

for

the

titration?

advisable.

experiment.

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the

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most

is

a

slow

suitable

reaction,

temperature

temperature is too high, the

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•

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into

safety

the

disposed

always

must

issues

picture?

of ?

be

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worn

consider

possible

of

must

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there

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be

considered?

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ethic al

when

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associated

and

state

leover

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issues?

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experiment

with

c an

spinach

chemic al

not

a

and

few:

involve

reagents,

environmental

assessment

green chemistry come

samples

mention

does

internal

hazards

try

reagents be

safety

gear should

serious

to

use

as

risks.

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little as

involved.

0.48

g

0.43

/ el dica cilaxo fo ssam

0.38

0.33

y = –0.0211x + 0.5145

2

0.28

R

= 0.52

0.23

0.18

2

4

6

8

10

12

14

16

cooking time / minutes

▴

Figure 2

E ach

Graph of measured

decision

must

methodologic al

with

scissors

area

of

pilot

the

the

and

time

is

adds

The

and

same

value

to

student

also

ruler

sample

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on

is

the

to

the

each

a

solid

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rationale

student

measure

consistent.

sample

student

cool

for

the

their

a

shows

spinach.

the

planning

uses

spinach

thermometer

based

considerations.

experiment

chop

be

mass of oxalic acid in spinach vs cooking time

also

samples

sample.

gets

the

pieces

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to

shows

to

cut

aer

be

the

samples

for this method as a

using

the

a

food

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temperature using a

heating,

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justied in the

leaf

ensure that the surface

opts

when

monitor

immediately

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should

and

student

warmer

decides

and

decides

so

control

of

the

exposure

variables and

investigation.

checks

research

that

there

design,

are

high

therefore

precision

minimizing

instruments

of

available while

uncertainties

aecting the

results.

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student

goggles,

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every

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to

in

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should

damage

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conducts

mask

to

are

use

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experiment

gloves

briey

the

special

approach

good

and

be

due

minimum

bin

the

described

for

a

that

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in

the

quantities

through

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the

in

risks

company

disposed

environment.

shows

to

to

student

hood

using

potassium

a

lab

coat,

safety

manganate(VII)

presents.

research design. The student makes

and

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collect.

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leovers

fume

that

sink

of

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adheres

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this

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waste is

report claries that the

compound

spinach

safety,

are

used

ethic al

c an

c ause

for composting.

and

environmental

practice.

673

The

internal

assessment

C ase study 4 (Inquiry 1: Designing)

A

student

exposure

of

designs

to

sodium

UV

a

research

light

aects

thiosulfate,

Na

S 2

The

following

•

reagents

•

concentrations

•

chosen

•

distance

•

elimination

•

range

•

the

method

the

O 2

to

extent

,

and

investigate

of

the

starch

how the time of chlorine

photolysis

as

an

reaction,

using

a

mixture

indic ator.

3

methodologic al

considerations

need to be made:

involved

and

storage of solutions

wavelength of UV light

of

of

best

chlorine

of

sample

other

from lamp

sources of light

exposure times

way

to

establish

the

endpoint

of

the

titration and the number of

trials.

E ach

time

will

needs

a

require

periods

to

be

challenge

constant

system

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in

for

the

should

student’ s

and

buers

cover

gives

a

the

each

to

the

rate

required.

need

of

the

to

they

avoid

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of

the

be

a

with

would

is

aect

using

the

blank,

pH

out

the

you

for

a

of

to

24

the

the

temperature

avoid

to

hours,

a

that

extended

the

temperature

system. This is usually

c annot be kept

of

the

reaction

probe.

c alibrating

prevents

shows

example,

temperature

good

transparent

results

pilot

For

reaction

the

showing

the

the

temperature

details

This

make

stable

c arry

If

needed,

glass

photolysis.

the

pH

probe with two

experimental

systematic

wavelength

technique. The

errors. The student

of

365 nm, which

evaporation and changes in

unreliable.

so

the

experiment.

student

The

starch

prepares

solution

a

is

fresh solution

also

freshly

degradation.

errors

idea

However,

if

excluded

experimental

methodology.

674

of

will

if

results.

environment,

includes

optimum

that

processing

procedure.

this

especially

reliable

investigations.

prepare

container

that

as

obtain

monitored

thiosulfate

day

prepared

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be

report

concentration

Sodium

longer

to

experimental

they

high

decisions,

needed

controlled

buers

covers

several

are

is

must

to

any

and

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use

of

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the

more

set-up

considered

standard

trials

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deviation,

produce

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add

ve

outlier

necessary.

value

to

trials

results,

the

are the minimum

then

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these

or

results

photographs

description of the

D ata analysis

The data analysis for your scientic investigation requires you to:

•

communic ate

your

recording

and

processing

of

data

in

a

clear

and

precise

manner

•

consider

uncertainties

•

adequately

process

and

the

their

propagation

relevant

data

to

allow

you

to

answer

the

research

question.

Communic ation

is

essential

when

addressing

this

criterion.

M ake

sure

that

you

follow these rules:

1.

Present

clear

2.

Produce

3.

Include

4.

Use

tables

well

the

and

designed

graphs

tables

uncertainties

correct

symbols

for

of

with

that

the

adequate titles.

allow

easy comparison of data.

instruments

physic al

quantities

used.

and

their

units.

Remember to use

SI units.

5.

Show

your

processing

clearly,

but

do

not

add

unnecessary steps and

descriptions.

6.

Report

7 .

Do

8.

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9.

Conduct

not

decimal

use

sure

images

that

point

recorded

data.

without

c atalyst,

a

completion,

reaction.

be

in

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you

repeats

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to

places

8:

replace

report

for

any

your

and

are

example,

the

data

raw

if

data

data,

the

the

results

testing

be

would

in

inferences.

conclusions

were

might

be

not

where

you

inference

should

qualitative data.

raw

trials

inferences

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raw

to

consistently.

the

be

your

data

you

the

time

that

are inconsistent.

make

rate

of

taken

the

based

a

for

c atalyst

analysis,

and

on

your

reaction with and

the

reaction

increases

the

to

the

reach

rate of

inference should

your conclusion.

c ase

studies

investigations

below

and

how

give

examples

these

link

to

of

the

data

analysis

inquiry

in

three scientic

process.

675

The

internal

assessment

C ase study 5

(Inquiry 2: Collecting data, Processing

data)

A

student

the

rate

dierent

of

c arbon

glucose

account

CO

the

to

a

with

for

will

xed

a

the

be

eect

dioxide

concentrations

solutions

of

investigates

of

of

cooling

at

a

that

measured

changing

glucose

volume

hotplate

of

production

of

are

water

occurs

a

and

while

probe

concentration of glucose on

fermentation.

prepared

yeast.

temperature

with

the

during

for

by

The

slightly

data

700

samples with

dierent

student

above

are

Four

adding

the

plans

to

masses

heat the

expected one to

collected.

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concentration

seconds.

2

In

this

experiment,

monitored

points

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is

a

required

student’ s

pilot

the

temperature

thermometer.

for

report

the

To

must

independent

includes

four

be

establish

controlled

a

trend,

a

with

a

water bath and

minimum

of

ve data

variable.

graphs.

Figure

3

shows

the

results

from the

experiment.

Figure

3

respect

shows

a

with

shows

to

two

linear

that

time,

trials

so

c arbon

a

that

line

of

dioxide

best

include

the

t

concentration

is

not

other

changes

appropriate

glucose

for

samples,

non-linearly with

this

graph.

which

also

Figure 4

do

not

show

trend.

40000

y = 59.426x

35000

2

R

= 0.9954

mpp / noitartnecnoc

30000

25000

20000

15000

2

OC

10000

5000

0

0

100

200

300

400

time / s

▴

676

Figure 3

Graph of c arbon dioxide evolved vs time during fermentation for one glucose sample

500

600

The

internal

assessment

45000

40000

35000

mpp / noitartnecnoc

30000

25000

20000 0.10

g

0.25 g

OC

2

15000

0.50 g

1.00

g

10000

5000

0

100

0

200

300

400

500

600

time / s

70000

60000

mpp/ noitartnecnoc

50000

40000 0.10 g

0.25 g

30000 0.50 g

1.00 g

2

OC

20000

10000

0

0

100

200

300

400

500

600

time /s

▴

Figure 4

The

two

1.00

g.

Graphs showing two trials for the fermentation experiment

trials

In

this

show

c ase,

inconsistent

a

repeat

is

values

for

all

samples,

especially

0.10 and

required.

677

The

internal

assessment

In

the

nal

Menten

graph

plot

assessed

in

in

the

(gure

DP

report,

5).

The

the

ability

student

to

has

tried

to

produce a Michaelis–

produce a Michaelis–Menten plot is not

chemistry.

90

80

70

s

1–

60

)mpp(

50

/

40

V

0

30

20

10

0

0

0.01

0.02

0.03

0.04

glucose concentration /

Figure 5

▴

For

for

Michaelis–Menten plot

each

data

point,

other

data

points.

bar

ranges

More

data

student

that

are

also

the

include

for the fermentation experiment

of

the

example,

all

required

assumes

range

For

to

that

of

the

error

the

data

produce

the

0.05

M

a

graph

bars

rst

and

points.

overlaps

last

with

points

Therefore,

in

the

the

the

error

graph

data

are

bar

range

have

not

error

reliable.

reliable Michaelis–Menten plot. The

starts

at

the

origin

but

does

not

have data to

support this.

C ase study 6

(Inquiry 2: Collecting data, Processing

data)

A

student

length

the

of

standard

booklet

of

the

are

The

c arbon

You

used

data

should

aim

investigation

discourage

ways

to

reasons

678

for

were

as

you

of

from

your

selecting

in

at

least

are

if

three

you

the

pursuing

this

the

chosen

a

traditional

and

form

a

of

an

variable is the

dependent

hands-on

the

other

sc atter

databases

interesting

three

variable is

experiment.

contains

One

predicted

graph with the number

limited

investigations using

for a database

available, this should not

You should try to think of other

secondary

databases

in

idea

databases

investigation.

if

the

values,

reliable

than

hypotheses

and

variable.

have

fewer

independent

Two databases and the DP chemistry data

conducting

independent

the

alcohols,

experimental

However,

there

where

primary

processed

the

use

in

combustion.

instead

and

support

of

includes

to

data.

investigation

chain

enthalpy

atoms

secondary

an

c arbon

databases

values.

of

conducts

the

should

be

data

is

unavailable. The

included.

The

The

DP

papers

from

chemistry

is

also

not

secondary

data

a

booklet

good

data,

idea

and

this

is

as

is

reliable,

the

a

but

authors

task

you

it

is

not

have

must

a

internal

assessment

database. Using scientic

already

selected

the

values

perform.

M any databases do not include uncertainties. In this situation, an estimate

through

reported

reported

values

precision

in

dierent

is

too

simplistic.

databases

An

would

analysis

provide

a

of

the

more

dierences in

realistic

uncertainty.

rorre

20.0 ±

srab

aerae chane in mass of alcohol urin combustion

1.2

maximum slope:

y

1.1

= 0.1125x  0.4908

2

=

R

1

smar ni

pentanol 1

0.9 butanol

lohocla fo ssam

0.8 propanol

0.7

maximum slope:

ethanol

y = 0.0925x  0.5508

0.6

2

methanol

R

= 1

ni enahc earea

0.5

y = 0.11x  0.5066 0.4

2

= 0.9741

R

0.3

0.2

0.1

0

0

1

2

3

4

5

6

number of carbon atoms in alcohol

▴

Figure 6

Graph of change of mass in combustion of primary alcohol vs

number of c arbon atoms in chain

The

student

Sc atter

graphs

variable

is

is

number

of

the

line

c arbon

continuous

data

should

discrete

bec ause

for

processes

handling

their

not

and

of

skills

be

to

used

best

The

such

t

display

with

incorrectly

exist.

will

plotting

the

implies

shown

in

gure 6.

where the independent

intervals

that

sc atter

however

the

graph

data

regular

Therefore

student

as

produce

to

quantitative

atoms

data.

data

(as

in

alcohols

graphs

receive

gure 7). This

with

should

marks

for

a

fractional

only

be

used

demonstrating

graph, identifying the line of best t,

2

extracting the

when

the

R

value

and

independent

line

of

variable

is

best

t

equation.

discrete

and

You should use a bar chart

qualitative.

679

The

internal

assessment

C ase study 7

(Inquiry 2: Collecting data, Processing

data)

A

student

in

a

wants

voltaic

they

use

cell

to

investigate

aects

copper

and

the

how

potential

iron

the

temperature of copper(II) sulfate solution

dierence

electrodes,

and

of

the

sodium

cell.

In

nitrate,

their

investigation,

NaNO

,

for

the

salt

3

bridge.

In

their

the

the

the

want

shows

The

student

the

the

electrodes

will

be

not

student

the

reports

distance

distance

ow

for

temperatures

student

not

should

The

pressure

use

the

The

electrolyte

electrolyte.

but

that

beaker.

thermometers

nor

to

procedure,

electrolyte,

report

of

They

were

aected

has

the

20,

not

if

the

it

30,

overall

parallel

reported

the

A

and

but

that

of

will

rooms

the

prepares

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need

the

c alculations

the

to

student

trials.

for

ensure

voltmeter

The

that

Their

▸

Table 2

The

680

the

the

bottom

controlled, and

the

container. The

placed in the bulk of the

of

the

surroundings,

nd

of

the

•5H

their

are

molar

methodology.

cheap

mass

O but they use the molar 2

concentration. This is a common mistake

and

of

easy

water

in

M any

to

salts in school stock

store.

the

In

this

hydrated

c ase, the student

salt when making

concentrations.

a

large

also

enough

decides

to

volume

of

change

the

the

electrolyte to use in all

salt

bridge

and

sand the

each trial.

involving

the

electrochemic al

temperature

needs

includes

experiment.

process

for

reading

student

to

they

the

student

experiments

to

in

as

prepares

The

electrodes

In

factor

be

temperature

electrolyte with CuSO

salt

accuracy

hydrates,

will

The

student’s results

the

to

electrodes in

three trials.

anhydrous

aect

are

of

included in the

touched

touch

be

4

mass

area

picture

needs

electrodes

either

40, 50 and 60 °C.

surface

them.

electrodes

touching

not

10,

between

between

be

of

to

does

stabilize

qualitative

However,

not

cells,

a

change

before

data

water

while

is

bath

should

be

used

collecting data and the

recorded.

observations on the solutions at the start of the

qualitative

observations on changes that occur during the

are most useful.

results

are

shown

Temperature

in

table

2

below.

Potential dierence / mV ± 1 mV

/ K ± 0.1 K Trial 1

Trial 2

Trial 3

Trial 4

Trial 5

Mean

323

619

614

627

611

627

620

313

631

634

642

638

633

636

303

642

639

639

640

649

642

293

616

622

602

597

619

611

283

579

573

582

589

586

582

The

Always

trial.

reect

Also

thinking

to

rationale

The

the

equation.

theoretic al

and

an

exclude

decide

In

either

whether

the

nd

the

soware,

you

should

you

mean

c ase,

you

of

the

theoretic al

Nernst

experimental

equation

experimental

them.

If

c alculates

use

and

and

c alculating

you

standard

need

and

use

should

deviation

include

a

to

repeat

your

assessment

any

critic al

provide a brief

using

picture

of

your

the

graphic al

c alculations

used.

also

The

data

when

decision.

computer

values

student

outliers

or

your

or

collected

for

keep

for

c alculator

with

on

look

internal

for

data,

the

so

line

the

data

of

data

equation

are

best

student

t

is

for

the

not

voltaic cell using the Nernst

assessed

presented

in

a

(gure 7). The

c annot

report

R²

that

in

DP

chemistry. Both

graph with a line of best t

is

relatively

there

is

a

low

strong

for the

correlation.

900 hcae ta decudorp Vm

800 erutarepmet

= theoretical 700

600 pearo correlatio coefficiet = avg. eperimetal

value

500 2

= 0.67

R

y = 790 400

2

R y

=

1

= 0.045x 24.65

300

0

1

2

3

4

5

6

temperature, 1 = 283K, 5 = 323K

▴

Figure 7

You

must

when

Graph of potential dierence vs

remember

designing

an

to

consider

experiment.

the

limitations

The

–3

concentrations up to 10

temperature for an iron–copper voltaic cell

Nernst

of

the

theory

you plan to use

equation only applies to solutions of

–3

mol

dm

,

and

the

solutions

that

the

student

prepares

–3

have

the

in

a

concentration

error

your

report

bars

were

report.

should

of

too

0.1 mol dm

small

However,

include

if

the

to

they

error

be

are

bar

.

In

their

shown.

used,

report, the student mentions that

The

you

use

of

should

c alculations

and

error

use

a

bars is not mandatory

them

brief

correctly. The

analysis of their

signic ance.

681

Conclusion

The

conclusion

•

present

a

for

your

scientic

conclusion

that

is

investigation

relevant

to

the

requires

you to:

research

question

and

justied

by

the data

•

include

The

c ase

and

shows

scientic

study

below

how

this

C ase study 8

A

student

in

wants

antacids

The

student

antacids,

to

impact

Hydrochloric

context

acid

gives

links

to

to

an

example

the

inquiry

your conclusion.

of

conclusion

in

a

scientic

investigation

process.

(Inquiry 3: Concluding)

investigate

their

is

support

how

dierent

neutralizing

eect

ratios

on

of

active

ingredients

hydrochloric acid, HCl(aq).

used to simulate stomach acid.

chooses

to

magnesium

investigate

c arbonate,

mixtures

MgCO

of

(s),

ve

and

dierent

c alcium

mass

ratios of two

c arbonate,

C aCO

³

The

is

range

0.2 g.

of

The

mass

ratios

method

hydrochloric

acid

used

involves

and

is

1:1

to

adding

titrating

the

the

(s). ³

1:5,

and

the

antacid

resulting

overall

mass

mixtures

solution

with

to

of

each

mixture

0.1 mol dm

sodium

³

hydroxide,

NaOH(aq).

Their

results

are

shown

in

table

3

below.

3

Mass ratio

of MgCO

Volume of NaOH / cm

to Trial 1

Trial 2

1:1

27.6

30.8

24.1

27 .5

1:2

28.3

29.2

28.6

28.7

1:3

29.1

30.5

33.5

31.0

1:4

35.2

34.1

32.7

34.0

1:5

37.5

37.2

37 .2

37.3

3

C aCO

Trial 3

Mean

± 0.001 3

▴

Table 3

Table 3 The student’s results

The

student

The

choice

skills

in

682

as

sc atter

the

from

“amount”.

two

graph

data-handling,

information

to

includes

of

this.

such

is

graphs

a

as

shown

good

one

identifying

However,

the

as

in

it

the

gures

allows

line

of

volume on the

8

and

the

best

y-axis

9

in

their

student

t

is

and

to

report.

demonstrate

extracting

erroneously

referred

The

internal

assessment

40

2

R

= 0

35

) 3

mc( HOaN fo

30

25

20

tnuoma

15

10

5

0

1:1

1:2

1:3

1:4

1:5

mass ratio

▴

Figure 8

Graph of volume of sodium

with dierent

hydroxide required

to neutralize hydrochloric acid

mixtures of antacids

30

mc( desilartuen

3

)

25

20

2

R

= 0

15

lCH fo tnuoma

10

5

0

0

1:1

1:2

1:3

1:4

1:5

mass ratio

▴

Figure 9

neutralized

In

their

Graph of volume of hydrochloric acid

with dierent

mixtures of antacids

by sodium hydroxide

conclusion,

c arbonate

in

the

the

student

antacid

states

increases,

that

the

as

the

relative

neutralizing

mass

eect

of

of

the

c alcium

mixture

2

decreases.

However,

not

The

both

also

mentions

report

the

that

value

as

the

0

0.97

and

the

value of

origin

R

of

justies

the

0.97

this

trend.

value is

reported.

The

main

drawback

hydrochloric

also

not

antacid

As

student

graphs

acid

considered

mixture

presented,

Therefore,

to

this

mole

of

the

experimental

design

is

that

the

concentration

for

used poorly mimics stomach conditions. The student has

the

temperature

of

the

stomach,

and

the

time

given

for the

react with the acid.

investigation

ratios

rather

is

than

more

mass

related

ratios

to

stoichiometry

should

be

used

c alculations.

for the antacid

mixtures.

683

Evaluation

The

evaluation

•

identify

•

suggest

realistic

stage

in

the

that

minimal

you

considered

is

study

below

and

pilots,

the

and

against

will

an

how

not

to

this

the

you to:

addressing

your

before

example

links

you

while

of

to

to

an

the

so

any

marks.

suggestions

allow

previously

methodology.

criterion,

aect

that

this

signic ant mistakes

Identifying

collecting

such

reach

as

the

data

‘doing

issues at this

will

earn

more

trials’

are

second band.

evaluation in a scientic

inquiry

process.

(Inquiry 3: Evaluating)

the

eect

cyclohexene

student

will

addressed

gives

requires

improvements,

limitations

design

and

shows

investigates

cyclohexane

and

remember

simple

C ase study 9

student

relevant

have

Please

investigation

weaknesses and limitations

assessed

should

too

investigation

A

and

experimental

credit.

c ase

scientic

weaknesses

methodology

made

The

your

methodologic al

identied

Your

for

nds

that

of

on

the

changing

the

mole

the

volatility

ratios

of

1:0,

mole

the

3:1,

ratio

of

mixture.

1:1,

1:3

a

mixture of

Aer

and

several

0:1

produce

reasonable data.

The

student

then

allows

nature

of

the

monitored.

ensure

decides

the

that

to

sample

solvents.

The

the

soak

to

piece

of

lter

evaporate

in

a

The

student

surface

a

temperature

measures

area

is

the

the

paper

fume

and

for

the

pressure

length

same

in

solvent

cupboard

and

each

in

width

due

the

of

mixture.

to

mixture, and

the

fume

the

They

hazardous

hood

lter

are

papers to

decide

to

control

3

the

amount

graduated

The

and

mixture.

and

solvents

attaches

the

submerges

The

value

the

by

c arefully

measuring

their

volumes

with

a

2.00

cm

pipette.

student

band,

the

of

student

remains

record

the

lter

both

collects

stable

for

decreasing

paper

through

data

two

to

a

a

temperature

hole

until

the

minutes,

values

until

in

a

test

probe using a rubber

tube

temperature

they

they

remove

stabilize

lled

with

remains

the

the

solvent

constant.

When

probe with the paper

for two minutes using a data

logger.

The

student

estimate

random

Their

▸

Table 4

of

propagates the uncertainties ( Tool 3: Mathematics) to nd an

the

errors

results

systematic

range

are

from

shown

in

error.

3.38

table

The

to

4

overall

percentage

error is 8.96%, and

5.72%.

below:

The

Mole fraction

Average rate

Total random

Total

Total

Absolute

of cyclohexane

of cooling /

error / %

systematic

percentage

uncertainty

error / %

error / %

student’s results

–1

–1

°C s

684

/ °C s

1.00

0.1200

3.63

5.33

8.96

±0.0044

0.75

0.0840

5.72

3.24

8.96

±0.0048

0.50

0.1050

4.89

4.07

8.96

±0.0051

0.25

0.0970

5.60

3.36

8.96

±0.0054

0.00

0.1000

3.38

5.58

8.96

±0.0028

The

The

student

uses

their

results

to

create

the

graph

shown

in

gure

internal

assessment

10.

0.2

s

1-

C° / gnilooc fo

0.15

0.1

etar egareva

0.05

0

0

0.2

0.4

0.6

0.8

1

mole fraction of cyclohexane to cyclohexene

▴

Figure 10

solvent

Vapour

the

an

In

pressure

change

c an

Graph of rate of

use

in

their

example

the

surface

the

to

paper

another

a

result

an

the

introduces

same.

is

measure

vapour

of

evaluating

band

cooling of solvent

mixture vs

mole fraction of cyclohexane in

mixture

They

to

predict

the

is

volatility

of

proportional

the

relative

a

to

liquid.

the

volatility

of

The

student

change

the

in

knows that

temperature, so

dierent

mixtures. This is

inference.

results

a

in

their

systematic

suggest

minimize

needs

of

pressure

this

more

systematic

report,

error,

using

a

as

thin

the

student

they

c annot

contact

lm

considers that the rubber

conrm

strip

with

its

a

position

is

always

well-established

weakness in the design. The student also thinks that

contact

error.

The

with

the

student

temperature

suggests

probe,

using

which

introduces

cylindric al lter paper

instead.

The

the

student

same

professional

report

sensor

The

also

also

be

identify

to

ensure

to

errors

entire

that

measure

in

are

the

of

paper

so

they

same

vapour

and

may

not

always absorb

recommend

volume

measuring

the

valid

their

lter

mixture,

instead

weaknesses

useful

the

solvent

that

used

minimize

some

that

the

mentions

could

to

of

device

identied

eort

notes

volume

the

is

using

absorbed

evaporation

a

more

every

rate,

a

time. The

pressure

pressure.

deserve

methodology

credit as the student has made

earlier

in

the

process, and they

improvements.

685

Index

Index

3D

representation

of

molecules

129,

259, 260

addition

absolute uncertainty 355

ageing,

absolute

air

zero

absorbance

absorption

accident

accuracy

acetic

17, 389

76, 339

acid

pollution

spectra 34, 35, 39

311–12,

242, 433

boiling points 269

467

classic ation 281

(ethanoic acid) 526, 549

acid

632–4, 644–7

theory of 623

aqueous solubility 152

experiments 309

measurements

acetylsalicylic

electrophilic

radic al

alcohols 261

prevention,

of

reactions,

free

(aspirin)

combustion

472

achirality 286

naming

classic ation

systems 552

acid–base

equations, balancing of 555–6

primary

acid–base

equilibria,

reduction

acid–base

indic ators

acid–base

titration

oxidation 604–6

salt solutions 565–7

318–19,

468,

427

276

acid–base

in

398,

homologous series 264, 265

571–3

557–9,

alcohols

secondary

567–73

242, 554

tertiary

604–6,

alcohols

alcohols

604,

deposition

acid

dissociation constant 561–5

functional

acid

rain

homologous series 265

acid

residue

242–3, 554

547

reduction

anions 548

terminal

Arrhenius acids 538, 539

binary

acids,

periodic

Brønsted–Lowry

region 568

buer

solutions

conjugate

239,

isomers 283

metallic

574–7

acids

241,

character of 238

reactions 239

alkaline

541,

half-equivalence

276

periodic table 230

547

acids

607, 608

position

alkali metals 550

539, 546–7

571

point

569,

species 541

alkanes 262

dissociation constant 561–5

Lewis

276

group

aliphatic 262

trends 549

acids

buer

classic ation

boiling points 268, 269

571

branched-chain

635–6

alkanes

homologous

oxide

melting points 268

reactions

241–2

parent acids 556

naming

periodic

radic al

trends 549

pH

curves

pH

sc ale 543–4

reactions

redox

557–9,

568,

569–70

root

series

241–2, 552, 553–6, 591

see

263

272

substitution

names

also

624

272

straight-chain

537, 538, 546–50

alkanes 268

halogenoalkanes

alkenes 262

reactions 552, 591

addition polymerization 217–19

strong acids 548

pH

boiling points 269

curves

568,

569–70

cis–trans isomerism 284–5

theories 538–41

electrophilic

weak

homologous series 264

acids

530, 549–50

naming

dissociation constant 561–5

reduction 608–9

reversible ionization 530

activation

energy

addition

anions 552

titration

391,

271

combustion 425–6, 428

oxidation states 549

properties

572–3

492–4,

active metals 553

686

264,

607

oxidation 604–6

acids

addition

261,

605,

607

acid

aldehydes

607

607

polymers 217–19

507–8

632–4, 644–6

274

straight-chain alkenes 264

alkoxy

alkyl

group 261, 264, 283

group

260,

272

alkynes 264

homologous series 264

Index

reduction 608–9

allotropes,

weak acids 552

c arbon 135–6

anodes

437

alloys 205–9

cell

aluminium

electrochemic al cells 598

boiling

and

melting

points

192

electrolysis 618

recycling 190

aluminium

chloride

aluminium

nitrate

amides

amido

diagrams 596

electrolytic cells 603

204,

636

hydrogen fuel cell 602

107

lithium-ion

127, 261, 266

RED

group 261

CAT

batteries 600

mnemonic

595

antacids 555

amines

anthocyanins 318

classic ation 281–2

antibiotics

hydroc arbon substituents 551

aqueous

naming 282

127

solutions

13,

72, 465

electrolysis 616–18

primary amines 261, 266

area

secondary amines 266

aromatic

sux 282

Arrhenius acids 538, 539

tertiary amines 266

under

a

curve,

graphs

compounds

Arrhenius

bases

amino acids 222, 542

Arrhenius

equation 506–8

amino

Arrhenius factor 506

group 261

ammeters 319

Arrhenius,

ammonia 551

Arrhenius

Svante 538

Āruni,

combustion 413

aspirin

Haber

asymmetric

Uddālaka 5

472

centre 286

ionization 540

atmosphere,

reverse

atom

van

reaction 515

der

Waals

parameters 84

ammonium ion 104

ammonium

amphoteric

atomic

number

atomic

orbitals

d

241, 542

f

techniques

colorimetric

colorimetry

infrared

liquid

mass

p

chromatography 155

50, 53–4

47, 251, 252

orbitals

49, 50

47

chromatography 155

chromatography

nuclear

10, 11, 155, 156, 331

magnetic

diagrams 48–54

orbitals

45,

47

exclusion principle 48

s orbitals 44–7

293–6

resonance

spectroscopy

296–302

atomic

radii, periodicity 232–3

atomic

theory 3, 5–6, 46

atomic

volume,

atomization,

periodicity

237

enthalpy of 418

atoms/atomic

structure 3, 5–6, 20–6

Bohr model 40–2, 43

spectrophotometry

thin

orbitals

Pauli

spectroscopy

column

proton

orbitals

orbital

77, 339–40

spectrometry 29–32, 290–2

paper

176–8

principle

Hund’s rule 50

analysis 253

combining of 302–3

gas–liquid

24–6

44–54,

degenerate

analyte 335, 469

analytic al

475–6

Aufbau

species 542

species

E arth’ s 268, 429

economy

nitrate 104

amphiprotic

539, 560

theory 552

bond angle 129

process 525

538,

375–8

163, 164, 261

layer

titration

76, 339–40

chromatography

334–6,

400,

468,

157,

diameter of atoms 22

158–9, 331

469,

557–9,

567–73,

electron

congurations 34–62

empiric al

formula 68–71

592–3 excited state 41

Anastas,

Paul

475 ground state 41

anions

97, 98, 102, 105 ionization

energy 54–60

acids 548 isotopes 26–9

electric al conductivity 111 mole

oxyanions

unit

63–4,

66,

72

245 “plum pudding” model 21

primary

(voltaic) cells 596 quantum

mechanic al model 43–6

radii of 233–4 relative

atomic

charge 23

relative

atomic

mass 23, 65, 66

solubility 112, 113, 114

687

Index

Rutherford model 20–1, 22

Aufbau

aurora

principle,

borealis

atomic

orbital

(Northern

structure 162–7

lling

50, 53–4

beryllium

Lights) 35

electron

Avogadro’s constant 64

Avogadro’s

axial

law

78

bonds 169

titration 469

backward

ring

bimolecular

reactions 496

see

Ball,

Alice

also

reversible

biofuels

427,

632

models,

molecular

geometry

169,

170

biologic al

barium

bipolar

hydroxide 104

dissociation constant 561–5

blocks

alkalis 541

Bohr

538,

Brønsted–Lowry

region 568

buer

solutions

half-equivalence

bases

oxide

reactions

239,

539,

241,

model,

atomic

structure 40–2, 43

boiling points

542,

554,

635

and bonding type 198

ethanal 150

105,

ethanol 150

541,

point

571

graphic al

570

model

of

270

group 1 metals 191

635–6

halogens 141

241–2

and

homologous

series

parent bases 556

hydrides 145

pH

curves

methanoic acid 150

pH

sc ale 543–4

557–9, 568–71

properties

reactions

period

241–2, 552, 553–6

pH

269,

270

pentane 142

537, 538, 550–2

3

metals

192

propane 150

strong bases 558

straight-chain

curves 568–9

alkanes 268, 269

Boltzmann, Ludwig 443

theories 538–41

bomb

weak bases

bond

axis

dissociation constants 561–5

bond

dissociation

pH

bond

enthalpy

curves 569–71

c alorimeter 401

(internuclear

axis)

175

energy 391

125,

293, 405–7

reversible ionization 530

average

titration

c arbon–halogen bonds 642

573

batteries

bond

enthalpy 405–6

denition 405

lead–acid

batteries 599

lithium-ion

primary

batteries

112,

positive

191,

439, 600

(voltaic) cell 594, 598

enthalpy

value 405

bonding

axial bonds 169

recycling 111

bond angle 128, 129

secondary

bond-breaking

(rechargeable) cells 598

voltaic pile 338

bent

table) 229–31

microplastics in 216

574–7

bases

Lewis

539, 560

bases

buer

conjugate

region 436

plate, fuel cells 601

(periodic

blood,

bases

world

c arbon xation 434

bioplastics 216

Arrhenius

477

by

pros and cons of 435

bases

(V-shaped)

geometry,

1,

molecules 128, 129

2-disubstituted benzene compounds 166

c arbon–c arbon bonds in 165

electron density 165

electrophilic substitution 648–9

hydrogenation 166

isomers 166

resonance

and

bond-forming 404–7

bond length 125

benzene

688

433–6,

bar charts 365, 366

base

495

materials/products 213, 215, 216

production

ball-and-stick models 341

balloon

reactions

biodegradable

reactions

246

127

Gerd 23

biochemic al

reactions 514, 522

energy

beta-lactam

Binnig,

back

conguration 56

ionization

energy 165–6

bond

order

124–5

bonding

continuum

bonding

electrons 120

coordination

bonds

187,

126,

198–200, 201

251–2,

629,

635,

637–9

electron

domains

127–30

equatorial bonds 169

metallic

bonds

95,

metallic–covalent

96,

187, 191–3

bonding

continuum

187

636,

Index

multiple bonds 130

c alorimetry

pi

c ar

bonds

174,

175–6

polarity 132

sigma

valence

bonding

176

193–4, 251–2

641,

642,

647

pair

repulsion

model

groups 260

allotropes of 135–6

127–9, 131

biologic al xation 434

c arbon–c arbon

(diagrams) 198, 201–5

c atenation

cycles 418–21

chirality

boron

bonds

124, 125, 165

257

286,

287, 288

hybrid orbitals 177–8

electron

conguration 56

ionization

mass

orbital

Boyle,

Robert

law

hybridization

246

and

c arbon

also

food

87

207

hydroc arbons

271

concentration

industry

Lewis

alkanes

176–8

hardness

atmospheric

87, 539

82,

branched-chains

steel

dioxide 5

632

81,

81,

see

diagram 50

triuoride

Boyle’s

energy

spectrum 30

boron

brass

631,

alkyl

diagrams 198, 201–5

electron

399, 401

c arbon

covalent bonds; ionic bonds

triangles

Born–Haber

bonding

shell

also

174–5,

elements

triangular

see

c arboc ations

bonds

transition

333–4,

batteries 599

use

of

formula 121

molecular polarity 133

207

seawater

concentration 526

brittleness, and bonding type 198, 199

c arbon–halogen bonds 642

bromine

c arbon

monoxide 5, 428

electrophilic addition 644

c arbon nanotubes 135, 136

reversible

c arbonate ion 104, 122–3

bromopentane

reactions 512–14

279

c arbonates

Brønsted, Johannes 538

Brønsted–Lowry

acids

Brønsted–Lowry

bases

Brønsted–Lowry

theory

bronze

acid

239,

539, 546–7

239,

539,

542,

554,

635

540, 552

action

buer

region 568

buer

solutions

acid

c arbonyl

group 261

(buckyballs) 135, 136

of

of

reduction

c arboxyl

574–7

compounds

276–7

607

group

isomers 283

oxidation 604

c arboxylic acids 261, 265

uncertainty/error 352, 383

boiling points 269

274

naming

277–8

boiling and melting points 268

oxidation 604–6

chain

reduction

isomers

empiric al

enthalpy

in

group 261, 265

functional

576–7

measurement

273,

242, 526

oxidation 604

574

burettes 314, 335

butane

c arbonic

naming

buer

pH

reactions 554

aqueous solubility 114

207

buckminsterfullerenes

429, 430–1

477

and

279

molecular

change

molecular

structural

of

formula

formulas

formula 68

formation 413

sux

c atalysts

263

247, 494–5

and

263

607

277

clean

energy 434

equilibrium position 523

butanoic acid 294, 526

Haber

process

524

hydrogen fuel cell 602

c alcium 52, 53

c alcium

c atalytic

c arbonate

104,

469,

c alcium uoride 109

c alcium

hydroxide 550

473

converters

c atenation

c athodes

194,

247–8

257

437

cell

diagrams 596

c alcium nitride 106

electrolysis 618

c alcium

electrolytic cells 603

oxide 105

c alculator skills 533

hydrogen fuel cell 602

c alibration

lithium-ion

curves

c alorimeters

391,

76–8, 339

395, 401

RED

CAT

batteries 600

mnemonic

595

689

Index

c ations

29,

97 ,

98,

99, 101, 102

loc ating

electric al conductivity 111

paper

primary

retardation

(voltaic) cell 596

see

also

chromatography

thin

c arboc ations

layer

chromatography

diagrams 596

discovery

cell

phones 342, 598

electron

cellulose 211, 212

sc ale 17

rust

CERN 90

279

of

193

conguration 54, 101

cis–trans

oxidation

citations,

charge

climate change 431, 434

ionic

nuclear

charge

charge 233

closed

charge 109

also

systems

electron

mass

charge 234

coecient

see

collision

also

chaulmoogra

chemic al

oil

bonds

graphs; tables

colour

deduction

chemic al

equations

chemic al

equilibrium

chemic al

formulas

see

chemic al

industry,

reaction

chemic al

kinetics

chemic al

reactions

chemic al

shi

chemic al

symbols

chemic al

weapons 525

286,

of

243–5

equations

see

equilibrium

formulas

yields

see

column

determination

combined

analysis 253

77, 339–40

substances,

gas

combustion

474

reactions

law 88–9

427

enthalpies of 334

greenhouse

symbols

gas

emissions 429–31

incomplete combustion 428

metals

reactions 509

424–5

non-metals 425

287, 288

phlogiston

theory 426

primary alcohols 398

standard

aqueous solubility 114

complementary

lattice

complex

enthalpy 110

chlorine

ions

complexes,

homolytic

isotopes

ssion

26,

624, 625

27

reactions 582

297,

chloromethane

630

chromatograms

156,

248,

251,

equilibrium 520

157, 158

oxidation

concentrated

states,

solutions

molar

chromatography 155

of

curves

76–8

analysis 253

equilibrium position 519–20

mass

column

deduction

73

colorimetry 77

experimental

chromatography 155

251–3,

72–8

colorimetric

classic ation of 156

technique/apparatus 331–3

637–9

elements

637

compounds 6

c alibration

chromatography 155–60

gas–liquid

enthalpy change 398, 412–17

colours 340

transition

concentration/s

627

chromate–dichromate

liquid

157

complete combustion 425–7

chlorides

chloroethane

chromatography

424–8

alcohols

chloride ion 98

redox

379

487–8

chromatography 155

297, 298, 300

see

conguration 101

wheel 251–2, 340

colourless

see kinetics

chemiluminescent

of

colorimetry

chemic al compounds 6

see

warming

spectrum 31

theory

colorimetric

see bonding

states,

global

colorimeters 342

632

oxidation

practice 311

cobalt

notation of 102

nuclear

249, 250

387, 388

ions 98–100

charts 365–6

chirality

ac ademic

see

172–4

states

isomers 284–6

changes of state 13–14, 15

formal

158–9, 331

prevention 206

variable

eective

157,

in oil paints 608

temperature

isomers

10, 11, 155, 156, 331

157–8

chromium

cell

chain

factor

stationary phase 155

solubility 112, 113, 114

Celsius

157

mobile phase 155

radii of 233–4

690

agents

coordination bonds 251

concentration

74

concentration

73–5

problem-solving 465

reaction

rate 488, 491

243–5

Index

spectrophotometry

standard

titration

solutions,

electric al conductivity 151

values,

curve 504–5

volatility 151

polymers 220–4

crystallization

condensation

of

condensers

the

two

dierent

monomers 221

same monomer 222–3

9,

11,

327

crystals 96, 104

current,

224

structural

electric 603

cycloalkanes,

cis–trans isomerism 285

formula 258

324

d

atomic

orbitals

47, 251, 252

congurational

isomers 284

d-block elements 230

conformational

isomers 284

D alton, John 5

conjugate

acid–base

conjugate

acids

conjugate

bases

pairs 541, 564–5

data

105,

571

of

energy,

conservation

of

mass,

continuous

D aniell cell 596

571

conservation

constitutional

isomers

law of 404

law

and

bonds

of

decimal

470

Lewis

acid

measurement

atomic

deloc alization

251–2,

629,

635

in

637–9

base

636

162,

orbitals

molecule 164

hybridization 180

density, gases 468

ancient artefact 186

antimicrobial

electron

deposition 14, 15

properties

207

detergents 134, 152–3

conguration 54, 101

diamond 135–6

electroplating 619

diamorphine

mass

digital balance 313

spectrum 30

variable

oxidation

states

249, 250

copper(II) sulfate

crystals 96

relative

(heroin) 153

digital

pH

digital

sensors 342

meters 543

digitalin 260

electrolysis 618

molecular

dilute

mass 65, 66

solutions

dimensional

dipole–dipole

correlation,

dipole-induced

graphs

371

73

analysis 22

Cornell method, note-taking 490

forces

143,

forces

dipole moment 132, 133

cotton 212

dipoles,

bonds

95, 96, 117–34

diprotic

temporary

acids

direct-methanol

bond

direct

bond length 125

bond

order

electron

126,

636

127–30

of

of

dissolution

polarity 132

distillation

network

632

pair

repulsion

117, 135–8

covalent substances 149–54

safe

practice 309

water 545

(solvation) 11, 112

10, 11

experimental

electron

structures

chemic als,

dissociation constant, acid 561–5

multiple bonds 130

shell

438–9, 601, 602

371

ionization comparison 540

formulas 120–3

valence

(DMFC)

graphs

dissociation

formation of 118

polarization

cell

disproportionation 581

electronegativity 119

Lewis

fuel

proportionality,

disposal

bonds

domains

139, 140

discharge 599

124–5

coordination

147, 148

547 , 550

bond angle 128, 129

enthalpy 125

147 , 148, 149

142,

corrosion 198, 199

covalent

50, 251

Democritus 5

copper

covalent

49,

187, 188, 194

benzene

and

reactions

measurement 356–7

reasoning 38

degenerate

ions

and

see

places,

deductive

279

126,

complex

collection

databases 345–6

spectrum 35

coordination

strength of 149–50

crosslinking 213

between

condensed

relative

solubility 152

measurement 312

condensation

hydrolysis

forces,

properties 151–4, 198

condensation 15, 443

condensation

intermolecular

technique/apparatus 334–6

concentration–time

concordant

76

measurement uncertainty 77–8

model

127–9, 131

fractional

technique/apparatus

distillation

271,

324–5

271, 325, 329

oxidation of primary alcohols 605

simple

distillation

324

691

Index

DMFC

DNA

see

direct-methanol fuel cell

(deoxyribonucleic

double

bonds,

acid),

molecules

hydrogen

condensed

bonds

147

124

emission

full

drugs/medicines

isoelectronic 98

for

aspirin

noble

472

632

diamorphine

ibuprofen

278

impurities

472,

penicillamine

474

286,

287

dry ice 13, 17

samples

a

constant

metals

dynamic

equilibrium 512–16, 519–25

elements

decient

electron

domain

electron

domains,

electron

microscope 23

mass 326

ductility,

conguration 98, 99

diagrams 48–54

electron

electron-pair

to

gas

transition

(heroin) 153

congurations 52

nitrogen atom 56–7

orbital

clinic al trials 555

drying

electron

antacids 555

oil

congurations 52–3

spectra 34–40

doublets 300

chaulmoogra

189, 190

geometry

covalent

sharing

248

electrophilic

acids

169–71,

bonds

179–80

127–30

reactions 628–51

bonds

electrophiles

E arth

101,

molecules 123

coordination

Lewis

and

complex

ions

637–9

632

addition

and

bases

632–4

635–6

nucleophiles 629

atmosphere

nucleophilic

c arbon

dioxide

concentration

429, 430–1

oceans,

eective

nuclear

electron

sharing

substitution

electron shielding 233

ozone

electron

layer 161

c arbon

dioxide

transfer

reactions 580–621

electrochemic al cells 594–602

concentration 526

charge 233

electrolysis

of

aqueous solutions 616–18

electrolytic cells 603–4

Einstein, Albert 43

electroplating 618–19

elasticity 199

Gibbs

electric

current

electric

potential

electric al

319, 603

energy

and

standard cell potential 614

half-equations 586–8

dierence 320

conductivity

630

reactions 622–7

composition of 266

magnetic eld 35

oxidation 580–5

137

metals 589–90

and bonding type 198

organic compounds 604–7

covalent substances 151

redox

ionic compounds 111

reduction 580–5

reactions of acids and metals 591

measurement of 319

electric al

conductors 190

electric al

resistance, metals 189

alkenes and alkynes 608–9

halogens 589

organic

compounds

electrochemic al cells 336–8, 594–602

secondary cells 598–602

electrode

standard cell potential 611

potential,

electrodes

standard 610

437, 601

see

also

standard

anodes;

c athodes

electrolysis 603–4

electronegativity

electrode potential 610

covalent bonds 119

halogens

copper sulfate 618

ionic bonds 102–3, 104

chloride, molten 603–4, 617

water 438

20,

21,

24–5

atomic

electrolytic

bonding

electromagnetic

336–7, 594, 603–4

(EM)

radiation 34, 35, 38, 39

639

periodicity 236

electrons

electrolyte 601, 603

cells

607–8

187, 201

aqueous solutions 616–18

sodium

692

electron

for copper atom 54

orbitals

44–54,

176–8

electrons 120

deloc alization

164,

180,

187, 188, 194

electron anity 235–6, 418

electron anity 235–6, 418

electron

congurations 34–62

electron

for

beryllium atom 56

energy

for

boron atom 56

inner

for

c alcium atom 52, 53

ionization

for

chromium atom 54

relative

energy

levels

core

40, 41

37,

40–2,

44,

47, 50–2

electrons 52

energy 54–60

mass

and

charge 23

Index

spin

48–50,

valence

174

Gibbs

electrons 52

wave–particle duality 43

see

also

electrophiles

electrophilic

electron...

629,

in

air

647

oceanic

water 646

enzymes

attraction

charge 23

elementary

steps,

elements

EM

see

5,

6,

tests

formula

point,

286,

37

energy

of

c alculations

green chemistry

495

equations 555–6

equation 506–8

36,

ideal

gas

ionic

equations 553

net

57–8

total

equation 553

ionic

equation 43

equation 553

equatorial bonds 169

571

14,

prole

15,

109,

391,

392, 404, 442

equilibrium 442, 514–16

acid–base

492

chemic al

equilibrium position 522

equilibria,

in

electromagnetic

and

matter 3–4

energy

cycles 404–23

energy

density

energy

distribution

energy

levels,

energy

proles

energy

transfers

enrichment,

dynamic

radiation 39

439, 440

curves

electrons

391–3,

493–4

bond

and

of

the

equilibrium

519–24

of

reaction

Gibbs

conditions

on

524

energy change 454–6

law 516–18

equilibrium sign 512

standard enthalpies 394–401, 412–17

entropy

equilibrium 514

water 545

equilibrium

law 411

enthalpy of combustion 334, 398, 412–17

of

524–5

reversible ionization 530

447

hydrogenation 166

enthalpy

process

Châtelier ’ s principle 519–25

eect

cycles 418–21

denition 390

Hess’ s

temperature on 522–3

energy change 454–6, 531–2

c alculations 526–30

enthalpy 125, 406

entropy

of

position

combustion 334, 398, 412–17

and

pressure on 521

eect

equilibrium constant 516–18

energy 391

Born–Haber

of

model of 523

387–403

activation

concentration on 519–20

eect

Le

uranium 28

change

c atalysts on 523

of

heterogeneous

enthalpy of atomization 418

enthalpy

of

eect

Haber

492

387–90, 391

equilibrium 520

equilibrium 512–16, 519–25

eect

Gibbs

37, 40–2, 44, 48, 50–2

salt solutions 565–7

equilibrium 514–16, 523

chromate–dichromate

energy

of

576

equations 500–6

Schrödinger

257, 296

equation

equation 90–1

ionic

rate

37

287 , 289

reactions

energy

also

427, 430

Henderson–Hasselbalch

radiation

68–71,

titration

endothermic

waste 215–16

half-equations 586–8

observation

enantiomers

warming

plastic

Arrhenius

mechanism 496

24

ionization

empiric al

concentrations, atmospheric and

equations 26, 461, 462

187

spectra 34–40

ame

242, 433

dioxide

acid–base

reaction

electromagnetic

emission

end

102,

242–3, 554

global

see

electrophilic substitution, benzene 648–9

elementary

rain

pollution

429, 430–1, 526

hydrogen halides 645

electroplating 618–19

values 445, 446

issues

c arbon

halogens 644–5

electrostatic

entropy

acid

632–4, 644–7

c arboc ations

entropy change 445, 446

standard

environmental

632

addition

energy change 448, 452

standard

hydrogenation 166

equivalence point 335

eroding 618

error

bars,

graphs

369,

370

errors

389, 442–7

and

enthalpy

and

physic al changes 443–4

change

entropy change 443–6

c alculation of 445–6

447

and

graphs 385

and

processed

random

results 383–4

errors 354, 383, 385

systematic

errors

383,

385,

395, 399

693

Index

ester

group 261, 283

esteric ation

reproducibility

reaction

220, 526

risk

esters 220–1, 261, 266

safety 309–11

ethanal

variables,

150

ethane 262

see

boiling and melting points 268

molecular

formula

also

extrapolation,

control of 89

263

f

structural

f-block elements 230

formulas

263

atomic

orbitals

47

falsic ation 21, 426

ethanoic acid (acetic acid) 526, 549

femtosecond

ethanol

fermentation, glucose 435

aqueous solubility 152

fertilizers

biofuels

lter

427, 435

ltrate

equilibrium constants 526

ltration

intermolecular

rst

forces 150

formula 152

energy,

formal

formation,

charge

evaporation

formula units 65

exchangeable

excited

acid

condensed

hydrogen atoms 546–7

reactions

energy

empiric al

ionic

14,

prole

15,

101,

391,

392, 404, 442

octets

Lewis

120,

492

experimental

yield

for

167–8

accident

isolation

471

forward

a

mass 326

also

reversible

distillation

radic al

frequency

separation

fuel

of

formula

287–8

formula 258, 341

reactions 514, 522

recrystallization 329

of

257–9

163, 258

reactions

mixtures 326–9

cells

variables 311–20

mass of a gas, determination of 86–7

271, 325, 329

hypothesis of ageing 623

freezing 15

factor, particle collisions 506

437–9, 594, 601–2

direct-methanol fuel cell 438–9

melting point data 8

molar

formula

fragmentation 290

free

constant

257

skeletal

fractional

techniques

to

105–7, 108

167–8

fossil fuels 429–33

formula, determination of 71

drying

formula

compounds

see

integrity 311

measurement

hydrogen

fuels

fuel

cell

437–8

424–41

non-Newtonian uids 14

biofuels

planning

c arbon-neutral fuels 413

and

preparation

risk

assessments 11

techniques

dilution

of

distillation

reux

fossil fuels 429–33

results 312

477

complete combustion 425–7

energy density 440

324

of

433–6,

324–5

rate 509

repeatability

427 ,

standard solutions 322–3

standard solutions 321–2

reaction

413–16,

formula 258

104,

organic

structural

prevention 309

empiric al

of

257, 296

120–3,

stereochemic al

geometry 169–71

experiments 309–41

ac ademic

formulas

molecular

167–71

formulas

molecular

change

structural

formula

compounds

Lewis

equilibrium position 522

expanded

enthalpy

278

formulas

463

state, atoms 41

exothermic

57–8

172–4

standard

formic

395,

hydrogen

627

ethyl ethanoate 220

reactant

of

37

589,

ethic al

9, 11, 328

327

uoroalkanes 642

ethers 261, 265

spreadsheets 343

of

9, 11, 326–7

tests

uorine

issues 28

uting

9, 326

ionization

ame

addition polymerization 217–19

lasers 499

393

paper,

boiling point 150

ethene 262

excess

techniques

graph data 380–1

ethanoate anion 180

Excel

results 313

analytic al

molecular polarity 133

structural

694

of

assessments 309

incomplete combustion 428

specic

energy 432, 440

storage

and

fullerenes 136

transportation 440

419, 420

Index

functional

group

functional

groups

isomers 283

concentration–time

260–78

correlation

classes 261

direct

formulas 261

error

homologous

series

263–70

and

suxes 261

see

funnel,

also

curve 504–5

371

proportionality

bars

369,

371

370

errors 385

extrapolation 380–1

specic

functional

groups

gas

separating 329

laws 88

Gibbs

energy

change/temperature

relationship

457–8

gradient

gallium 231

373–4

374–5

interpolation 380–1

galvanizing 206

gas–liquid

(slope)

intercepts

Galvani, Luigi 594

interpretation

chromatography 155

inverse

gases 13

Boyle’s

law

combined

81,

gas

82,

line

87

gas

ideal

gases

curve

logarithmic

law 88–9

maximum

density of 468

ideal

or

outliers

80–93

molar

mass 86–7

molar

volume

of

best

sc ale

t

372

369–70

367

minimum

values

375–8

372

of

potential

ideal gas 85–7

371–2

367

plotting

an

of

and

non-linear

equation 90–1

of

proportionality

367

energy/distance

between

hydrogen

atoms

125 pressure–volume

relationship

81–2,

83,

85,

87–91

proportionality real

gases

Van

der

versus

371–2

ideal gases 82–5

rate–concentration Waals

reaction GDC

see

giant

covalent

graphic-display

order 503–5

c alculator

sketching structures

energy

change

of

363–4

117, 135–8

tangent Gibbs

curve 503–4

parameters 84

line

374

447–58, 531–2

see

also

charts;

tables,

quantitative data

c alculation of 448–51

gravimetric

analysis

313,

473

entropy change 452

gravity and

ltration

327

equilibrium 454–6, 531–2

green reversible

chemistry

216,

atom spontaneous

economy

495

standard cell potential 614

cost temperature

455,

of

476–7

457–8

principles glassware,

of

475

volumetric 321

greenhouse global

474–7

475–6

reactions 448

c atalysts and

310,

reactions 454–5

warming

eect

429, 430

427, 430

greenhouse gases 429–31 see also climate change

ground

state, atoms 41

glucose

group 1 metals, boiling and melting points 191 empiric al

and

molecular

formula 68

groups

(periodic

table) 54, 229–31

fermentation 435

molecular

structure

257

Haber,

Fritz

393

gold

Haber

process

524–5

electroplating 619

half-cells nuclear

595, 596, 612

symbol notation 25

half-equations 586–8 reactivity

of

592

half-equivalence point 569–71 gradient

halide ions graphs

373–4

halogen

reactions 239

tangent line 484

nucleophilic substitution 643 graphene 135, 136

halides, melting points 200 graphic-display

c alculator

(GDC) skills 533

halite 113 graphite

135,

136,

137

halogenate

624

graphs

halogenoalkanes area

under

a

curve

261,

263

375–8

classic ation 281 boiling

points

and

homologous

series

270

heterolytic coecient

of

determination

ssion

631

379

homologous series 264

695

Index

naming

275

combustion 412–13

nucleophilic

substitution

630,

639–42

complete combustion 425–6

halogens

fractional

electronegativity

steam

halide

ion

addition

633, 644–5

substituents 551

see

character of 238

also

alkanes; alkenes

hydrochloric acid 548

hydrogels 214–15

reduction 589

hydrogen

hardness

test,

275

Vickers

covalent bonding 118

207

electron distribution 140

symbols, chemistry labs 309

388,

390,

and

see

392, 394

temperature 389

also

electron

energy

electron

transitions 42

emission

enthalpy...

rst

Werner 43

ion

helium

40, 41

spectrum

exchangeable

water 28

Heisenberg,

hydroc arbons 262

reforming of 438

periodic table 230

substituents

heavy

639

reactions 239

non-metallic

heat

271

saturated/unsaturated

electrophilic

hazard

distillation

boiling points 141

ionization

40–3,

57

hydrogen atoms 546–7

energy

57–8

formation 99

isotopes of 26, 28

emission

Van

der

spectrum

Waals

37

nuclear

parameters 84

Henderson–Hasselbalch

equation

heroin

(diamorphine) 153

Hess’s

law 408–11

spin

297

“pop” test 591

576

potential

redox

of

hydrogen 543

reactions 582

specic

energy 440

enthalpy change of combustion 414–7

hydrogen bonding 143–6, 148, 149

enthalpy

change, determination of 411

hydrogen

enthalpy

change

hydrogen uoride 132

enthalpy

cycle

summation

heterogeneous

of

of

formation 414–17

diagram

method

equations

c atalysts

410, 416

method

247

409, 416

chloride

hydrogen

fuel

hydrogen

halides

cell

hydrogen

iodide

143,

293,

547, 548

437–8, 601, 602

633, 645

293

heterogeneous composition 6

hydrogenation, of benzene 166

heterogeneous

equilibrium 514

hydrogenc arbonates,

heterogeneous

mixtures 488–9

hydrolysis

heterolytic

hexane

ssion

631, 642

amides

274

molecular

formula

structural

formulas

structural

isomers

263

homologous

Lewis

hydrophilic

337

physic al

homolytic

Hund’ s

ssion

rule,

141,

trends

405,

molecules 152–3

formula 104

Lewis

268–70

formula 122–3

hydroxides

624

aqueous solubility 114

degenerate orbitals 50

properties 550–1

hybrid orbitals 177–9

hybridization

formula 122–3

hydroxide ion

263–70

in

extent 565–7

567

molecules 152–3

hydrophobic

reactions 489

series

and

solutions

coordination bond 126

273

homogeneous composition 6

homogeneous

reactions 554

hydronium ion 540–1

263

histograms 365, 366

voltameter

salt

acid

224

127

direction

boiling and melting points 268

Hofmann

reactions

water

176–80

reactions

241

hydroxonium 540

and

deloc alization 180

and

molecular

geometry

hydroxyl

179–80

group 261

cellulose 212

hydrates 65

functional

group

hydride anions 99

morphine

molecule 153

hydrides, boiling points 145

nucleophilic substitution 643

hydroc arbons

oxidation

696

604,

isomers 283

607

Index

hypotheses 499

lattice

structure 108

naming 104–5

ibuprofen

278

periodicity 103–4

ice

properties 110–14, 154, 198

ideal

changes of state 13, 15

redox

hydrogen bonding 145–6

solubility 112–14

molecular

standard

gases

structure 341

80–93, 139

assumptions

combined

ideal

gas

enthalpy

ideal gas model 80–2

equation 90–1

ionic

equations 553

ionic

lattices 108

mass 86–7

ionic

product

molar

volume 85–7

ionic

radii

ionic

salts, solubility of 114

real gases 82–5

immiscible liquids 152

impurities,

drug

indic ators,

acid–base

indigenous

of

water 545–6

ionization

production

472,

318–19,

474

of ammonia 540

571–3

dissociation comparison 540

peoples 431

ionization

energy

54–60, 101, 418

eect 642

c alculation

inductive

reasoning 38

data

initial

(IR)

spectroscopy

reaction

293–6

core

inquiry

rst

electrons 52

dipoles

instantaneous

reaction

intensive

139, 140

complex

traces

297, 298

graphs

specic

heat

c apacity 394

London

see

IR

forces

143,

forces

(dispersion)

147, 148, 149

142,

see

147, 148

forces

136,

139–42,

bonds

26,

175

95, 96, 102–7

electronegativity 102–3, 104

lattice

enthalpy 109–10

lattice

structure 108

non-directionality 108

periodic table position 103–4

polyatomic ions 104

ionic

charge 109

ionic

compounds

also

anions;

95, 96

electric al conductivity 111

104,

251,

637–9

solutions

567

c ations

infrared

electron

147, 148,

ion

spectroscopy

conguration 101

formation 100

iron

energies 101

disulde

(“fool’s

105–7, 108

enthalpy 109–10

gold”)

137

isoelectronic 98, 234

isolated

systems

387

isomers

benzene 166

cis–trans

isomers 284–6

congurational

isomers 284

conformational

isomers 284

functional

optic al

dissolution of 112

formulas

248,

salt

oxidation 581

graph data 380–1

lattice

also

ionization

axis)

of

iron 206

assessment 668

(bond

58–9, 60

spectrum 294

269, 432

axis

energies

spectator ions 553

hydrogen bonding 143–6, 148, 149

ionic

57–8

polyatomic ions 104, 122

forces 138–50

dipole-induced

interpolation,

246

hydronium ion 540–1

374–5

dipole–dipole

internuclear

in

energy

ionization

ions

hydrolysis

intermolecular

internal

57–8

charge 98–100

intermediate compounds 494

149,

data

alloys 205

rate 484–5, 486

measurement uncertainty 351–2

properties,

intercepts,

spectral

95, 96–102

70

integration

ionization

successive

ions

instantaneous

integers

from

periodicity 234–5

process 665

instruments,

of

collection 60

discontinuities

rate 484, 485, 486

initiation 625

inner

419, 420

109, 233–4

inductive

infrared

formation

ionic–covalent bonding continuum 201

law 88–9

molar

and

of

volatility 111

of

gas

reactions 98

group

isomers 283

isomers 286–90

stereoisomers 284–90

structural

isomers

273,

279–83

isotope labelling 28

isotopes 26–9

relative

atomic

mass

30, 31

697

Index

IUPAC

nomenclature

271

liquids

10, 13, 152

lithium 191

journals,

scientic

630

lithium-ion

batteries

112,

439, 600

litmus 538, 539

Kekulé,

Kelvin

Friedrich

August

temperature

von 164

loc ants

sc ale 15–17

ketones 261, 264

group

isomers 283

London

lone

oxidation 604

reduction

(dispersion)

pairs

Lowry,

607, 608

LPG

277

standard

energy

forces

(LDFs)

136,

139–42,

147 ,

127

see

liqueed

3D

petroleum gas

structure of 223

measurement 16

189,

389,

487 ,

m/z

493–4

ratio

29,

30, 31

macromolecules 209

kinetics 480–511

activation

energy

Arrhenius

equation 506–8

magnesium 6

492–4

boiling

nuclear

c atalysts 494–5

collision

theory

magnesium

487–8

M axwell–Boltzmann

multistep

energy

distribution

curves

melting

points

192

symbol notation 25

hydroxide 555

493–4

magnesium

oxide 26

magnesium sulde 6

reactions 496–9

magnetic

equations 500–6

reaction

and

magnesium iodide 204

femtochemistry 499

rate

367

470

M artin 538

lysozyme,

Kevlar 211

kinetic

graphs

Mikhail

269, 432

homologous series 265

kilogram,

sc ale,

Lomonosov,

functional

sux

272

logarithmic

rate 480–6, 488–91, 509

eld,

malleability,

Kwolek, Stephanie 211

E arth’s 35

magnetite 69

metals

189, 190

manganese 101

laboratory

work

see

lattice

enthalpy

lattice

structure 108

L avoisier,

analytic al

techniques;

109–10, 204, 418, 420

Antoine

drying

conservation

of

energy 404

law

of

conservation

of

mass

Le

London

lead

mole

470

(dispersion)

chromate 113

group

length,

measurement of 316

leprosy

Lewis

630

241,

bases

acid

Lewis

formulas

Lewis

theory 552

reactions

126,

limiting

reactant

251–2,

636–7

of

or

line

graphs 365, 366

698

concentration

mass

percentage 69

mass

spectra 29–32, 290–2

mass

spectrometry 29–32, 290–2

science

74

197–227

geometry,

428,

t,

brittleness 199

green chemistry 216

463–8

graphs

369–70

hydrogels 214–15

magnesium iodide 204

plasticity 199

molecules

128,

130, 169

polymers

product

petroleum

column

biodegradable materials 213, 215, 216

corrosion 199

167–8

lipids 153

liquid

mass 141

elasticity 199

395,

best

636–7

637

line

liqueed

63–4, 66

bonding continuum 198–200

reactions

120–3,

ligands

linear

mass 326

72, 313

aluminium chloride 204

635–6

base

635–6

Lewis

curve

470

constant

alloys 205–9

Lewis

Lewis

of

a

mass

materials

632

acids

unit

mass

to

subatomic particles 23

batteries 599

leaving

of

samples

molecular

forces

chromate 608

lead(II)

647

measurement

Châtelier ’s principle 519–25

lead–acid

rule

mass

conservation

of

see

M arkovnikov’ s

470

law

LDFs

experiments

gas

(LPG) 426

chromatography 155

209–24

life

cycle 208

silicon 203

mathematics 350–85

148,

149,

Index

experimental

graphs

and

error,

tables

sources of 383–5

silver halides 200

363–82

straight-chain

SI units 350

memory metals 206

uncertainties 351–62

Mendeleev,

matter

meniscus,

changes of state 13–14, 15

metal

characteristics 4

metallic

oxides

measurement 314, 315, 383

241–3

bonds

95,

96,

187

non-directionality 189

and

strength of 191–3

energy 3–4

pure

of

37

transition

and

mixtures 6–12

metallic–covalent

matter 13–15

M axwell–Boltzmann

values,

of

substances

states

energy

metallic

distribution

curves

elements

bonding

493–4

ductility

measurement 362

311–12,

concordant

467

189, 190

resistance 189

malleability

values 312

properties

decimal places 356–7

decimal

189, 190

188–90, 198

superconductors 189

prexes 64

thermal

conductivity

189, 190

electric

current 319

metalloids, periodic table 230

electric

potential

metals

dierence 320

electric al conductivity 319

boiling

length 316

combustion

mass

electric al

72, 313

mean

values 362

ame

and

melting

37

precision

periodicity

311–12,

467

reactions

product

395

redox

230, 238–40

cycle 208

reactions 591

standard

SI units 15, 16, 66, 355

see

gures 356–7

229,

life

reliability 312–13

signic ant

group 1 metals 191

resistance 189

tests

oxidation 589–90

rate 482

points,

424–5

pH of solution 318–19

reaction

187, 201

electric al conductivity 190

electric al

accuracy

193–4

continuum

structures 186–91

measurement

also

electrode potentials 610

alkali

metals;

transition elements

methane

standard solutions 77–8

boiling and melting points 268

temperature

bond angle 129

316–17, 351, 352

time 315

molecular

uncertainties 351–62

radic al

error

bars

369,

370

formula

structural

formulas

tetrahedral

human

Van

reaction time 353

instrument uncertainty 351–2

methanoic

mean

methanol

values 362

der

acid

276,

random

microbeads 216

mixtures 353

279, 526

microplastics 216

Milley–Urey

experiment 268

validity 312–13

mirrors 188

volume 314–15

miscible liquids 10

Lise 28

mitochondria 623

melting 15

parameters 84

278

438–9, 526

methylpropane

value uctuation 353

mixtures 6–12

melting points

alloys 205

and bonding type 198

heterogeneous

determination of 8, 330

separation

group 1 metals 191

period

624, 625, 626

263

geometry 177

Waals

150,

propagation of 358–61

errors 354

263

substitution

expression of 355–6

reaction

Meitner,

Dmitri 231

and

composition 3–12

observations

mean

alkanes 268

3

metals

192

potassium halides 200

mixtures 488–9

techniques 9–12, 326–9

mnemonics

OIL RIG 582

RED

CAT

595

699

Index

models/modelling

see

also

22,

molar

concentration

molar

mass

molar

volume

mole

ratio

mole

unit

66–7,

of

81,

(molarity)

86–7,

an

68–9,

63–4,

46,

347–8

neutralization

molecular models

neutrons

73–5

461–2,

formula

molecular

geometry

68,

468,

mass

477 , 481

nicotine

70,

257

nitrates,

127–31

179–80

peak 291

95

283–4,

303,

representation

electron

decient

intermolecular

Lewis

341,

348–9, 626

of

129,

259, 260

molecules 123

forces 138–50

dioxide 162

nitrogen

oxides

nitrogen

trichloride

NMR

see

mole

242

nuclear

121,

172

magnetic

electron

resonance

congurations 98, 99

mass 66–7

unit

63–4,

theory

non-metal

66,

72

oxides

241–3

non-metals

176

combustion 425

periodic table 229

polarizability of 141

non-Newtonian uids 14

relative

non-polar

molecular

mass 65, 66

solvents 112

resonance

structures 160–7

non-spontaneous

temporary

dipoles

Northern

139, 140

M ario 161

Lights

reactions 442

(aurora

borealis) 35

note-taking method 490

monomers 209

nuclear

charge 234

nuclear

ssion 28

nuclear

magnetic

morphine 153

nuclear

reactors 28

multiple bonds 130

nuclear

spin

multiplicity 300

nuclear spin quantum number 296

multistep

nuclear

monoprotic

acids

547

monosaccharides

224

reactions 496–9

naming

symbol

nucleophilic

of

alcohols

276

of

alkanes

272

of

alkenes

274

in

nylon

c arbonyl

of

c arboxylic

of

halogenoalkanes

group

compounds

acids

277

octet

rule

paints 608

271

compounds

271–8

245

isotopes

systems

optic al activity 286

optic al

26,

27

isomers 286–90

optometry,

natural compounds 260

orbital

natural

organic

net

polymers

ionic

209, 212

equation 553

387, 388

opiates 153

245

abundance,

OIL RIG mnemonic 582

open

245

molecules

120, 123

industry 269

symbols

639–42

rate 642

geometry,

99,

oil

table

630

220, 221

oil

nomenclature

oxyanions

24–6

276–7

octahedral

275

nanotechnology 136

natural

substitution

IUPAC

periodic

spectroscopy 296–302

637

halogenoalkanes

of ionic compounds 104–5

in

(NMR)

nucleus, atomic 20–6

of

of

notation

629,

reaction

of amines 282

organic

resonance

297

nucleophiles

of

spectroscopy

periodic table 230

polarity 133–4

700

246

noble gases

formulas 120–3

orbital

393

energy

nitrogen

mass 141

molar

conguration 56–7

process

ionization

molecules

3D

conguration 101

aqueous solubility 114

nitrogen

Haber

hybridization

models

537

287

and

molecular

468,

nitrate ion 104

electron

ion

241,

24–5

spectrum 31

expanded octets 169–71

molecular

Molina,

23,

electron

72

molecular

reactions

21,

nickel

463

ideal gas 85–7

66,

20,

redox

reactions 583

diagrams 48–54

chemistry

257

organic compounds

170, 171

Index

3D

models

of

classic ation

259, 260

penicillamine

260, 281–2

pentane

boiling and melting points 142, 268

formulas

molecular

257–9

groups

260–78

naming

structural

271–8

formulas

percentage composition 69

percentage uncertainty 355

607–8

organic

synthesis 648

graph

mechanisms 622

data

reaction

percentage

yield

470

period 3 elements

367

acid–base

order 500

properties

properties

oxidation 98, 580–5

period

denitions 581–4

3

metals,

242

187

boiling

and

melting

acids 549

electron

alkali metals 230

transfer 582

half-equations 586–8

alternative

hydrogen

atomic

radii 232–3

atomic

volume

loss/gain 582

424, 589–90

representations

blocks 229–31

oxidation state change 583–5

electron anity 235–6

oxygen

electronegativity 236

gain/loss 581

oxidation

states

series

590,

(oxidation

592

number)

variable

oxidizing

oxoacids

agents

101,

243–5, 583–5

and

elements

oxidation

585,

248–50

states

ionic

ionic

248–50

charge 99

radii 233–4

ionization

589, 590

energy

54–7, 234–5

discontinuities

in

245

metal

oxides

241–3

metalloids 230

bonding 160

metals

covalent bonding 118

naming

empiric al

and

hybridization

ionization

molecular

ozone

redox

formula 68

229,

bonds

oxidation

160, 161

states

243–5

transition elements

complexes 251–3

160, 161

properties

variable

atomic

orbitals

45,

47

see

p-block elements 230

periods

painkillers

personal

153,

278

chromatography

245

241–3

periods 229–31

160, 161

oxygen)

oxides

non-metals 229

246

reactions 581

(diatomic

230, 238–40

conventions

non-metal

179

energy

oxygen–oxygen

10, 11, 155, 156, 331

PET

also

equipment (PPE) 309

terephthalate) 209

petrochemic al industry 269

pH

pasc al unit 81

of

Pasc al’s triangle 301

measurement of 318–19

peak

sc ale 102

ratio 301

peer-review

PEM

see

process

proton

630

exchange

membrane

pH

curves

248–50

table) 229–31

protective

parent bases 556

exclusion principle 48

states

group 1 metals, boiling and melting points

(periodic

(polyethene

247–8

oxidation

parent acids 556

Pauling

246

Mendeleev’s work on 231

oxygen

Pauli

240

ionic compounds 103–4

547 , 549

oxyanions

of

groups 229–31

98,

acids 549

transition

192

237

organic compounds 604–7

reactivity

points

periodic table/periodicity 229–55, 698

electrolytic cells 603–4

metals

paper

263

peptide bonds 222

reduction

reaction

p

263

molecular model 142

oxidation 604–7

organic

ozone

formula

space-lling

incomplete combustion 428

overall

287

complete combustion 425–7

functional

outliers,

286,

273

buer

solutions

576–7

557–9

of

strong

acids

and

strong bases 568

of

strong

acids

and

weak

of

weak

acids

and

strong bases 568–9

of

weak

acids

and

weak

bases

bases

569–70

570–1

701

Index

pH

sc ale 543–4

pharmaceutic al

phases,

phenyl

“pop”

drugs

reaction

see

drugs/medicines

rate 488–9

positive

group 261

phlogiston

test,

positional

hydrogen 591

isomers

inductive

279

eect 642

potassium

theory 426

boiling and melting points 191

phosphate ion 104

electron

orbital

lling

photochromic lenses 583

potassium

photons

potassium uoride 104, 109

pi

37 ,

bonds

40, 43

174,

175–6

potassium halides, melting points 200

pie charts 365, 366

potassium

pipettes 314, 321

potential

dierence,

potential

of

placebo

eect 555

plane-polarized

planetary

plant

light,

model,

pigments,

rotation of 289–90

atomic

thin

structure 21

layer

chromatography 159

PPE

see

permanganate

protective

precipitation

equipment

reactions

plastics 213

gravimetric

bioplastics 216

precision

microplastics 216

prexes,

pollution

pressure

plating 618

“plum

76

precipitate 114

aspirin

issue 215–16

73,

measurement of 320

hydrogen 543

personal

plasticity 199

platinum

diagram 51

bromide 420

of

production

472

analysis

measurements

473

311–12,

467

decimal 64

equilibrium position 521

247,

248

pudding”

pasc al unit 81

model,

atomic

structure 21

reaction

pOH

sc ale 560–1

pressure–volume

polar

covalent bonds 132

primary alcohols

polar

solvents 112

rate 488

relationships,

gases

81–2,

83,

85,

87–91

aqueous solubility 152

polarity

combustion 398

bond polarity 132

oxidation 604–6

molecular polarity 133–4

polarization

polarized

595,

light,

598,

632

reduction

rotation of 289–90

polyamides 221

primary compounds 281–2

primary

(voltaic)

polyatomic ions 104, 122

polyatomic

607

primary amines 261, 266

cell

338,

molecules 294

principal

quantum

number

poly(chloroethene) 218

product

polyester 221

propagation 622, 625

polyethene 211

propanal 258

polyethene

propane

terephthalate (PET) 209

poly(isoprene)

(natural rubber) 211

life

cycle

boiling

208,

and

melting

intermolecular

polymers

molecular

209–24

polymers 217–19

condensation

and

the

polymers 220–4

environment 215–16

examples of 211

natural

polymers

structural

40,

47, 230

477

polymerization 209

addition

437, 596

batteries 594, 598

points

150, 268

forces 150

formula

formulas

263

258,

263

propanoic acid 526

propanone 258

propene 258

209, 212

proteins 222

properties 212–13

proton

acceptors 539

repeating units 209–11

proton

donors 539

synthetic

proton

exchange

proton

nuclear

polymers

209, 212–13

membrane

(PEM)

437, 601

1

polypeptides 222

polypropene

210, 218

magnetic

resonance (

H

NMR)

spectroscopy

296–302

1

H NMR 300–1

high-resolution

polyprotic acids 552

1

polysaccharides

proton

polystyrene 213

polyvinyl

702

low-resolution

224

chloride

(PVC) 218

transfer

H

reactions

acid–base

NMR

297–9

537–79

equilibria

in

salt solutions 565–7

Index

acid–base

indic ators

571–3

graphic al

acids and bases

properties 546–52

reaction

systems

reactions 553–6

reaction

yield

theories 538–41

reactions 4

weak acids and bases 561–5

amphiprotic

buer

conjugate

ionic

of

pH

curves

protons

public

pure

20,

amphoteric

buer

of

576–7

567–71

23,

24–5

energy

proles 391–3

energy

transfers

excess

extent

387–90

change

387–403

389, 442–7

reactant

experimental

covalent bonds 132

of

a

395,

yield

forward

PVC

Gibbs

energy

Hess’ s

law 408–11

quantization

chloride) 218

40, 43

limiting

reactions 514, 522

mechanic al atomic model 43–6

metal

quantum

numbers 40

multistep

137

change

reactant

quantum

quartz

oxides

395,

radic al

oxides

substitution

percentage

reactions

624–6

reversible

radic als 622–7

sc attering

random

yield

241–2

Gibbs

errors 354, 383, 385

curve 503–4

standard

rate

equations 500–6

rate

of

497

theoretic al

average

rate

reactions 496–7

yield

unimolecular

reaction 480–6

eect 453

enthalpy change 394–401

termolecular

step

520, 530

energy change 448

temperature

rate constant 500

rate-determining

512–16,

reactions 442

entropy change 443

eect 298

rate–concentration

reactions 442

470

reactions

spontaneous

formation 623–4

R aman

463–8

reactions 496–9

non-metal

mixtures 289

447–58

241–2

non-spontaneous

racemic

463

471

reaction 464

pure substances 6, 7

(polyvinyl

bond-forming 404–7

cycles 404–23

entropy

understanding of science 440

and

energy

enthalpy

sc ale 560–1

21,

reactions 496

bond-breaking

water 545–6

557–9,

reactions 514, 522

bimolecular

pairs 541

solutions

387

470–4

backward

species 542

574–7

acid–base

product

pH

pOH

and

solutions

representations of 503–5

reaction quotient 525–6

463,

471

reactions 496

water 239

of

concentration

reaction 481

eect 488, 491

see

also

electron

reactions;

electron-pair

reactions;

exothermic

sharing

sharing

reactions;

electron

transfer

reactions; endothermic

reactions;

kinetics;

rate

of

reaction

denition 480

reactivity

series

590,

591,

592, 610

experiments 509

real gases 82–5, 139 factors

aecting 488–91

reasoning, types of 38 heterogeneous

mixtures 488–9

rechargeable homogeneous

cells

437, 594, 598–602

reactions 489

recrystallization 329 initial

reaction

rate 484, 485

recycling instantaneous

reaction

rate 484–5, 486

batteries 111 measurement 482

metals overall

rate

of

190, 208

reaction 481, 483

plastics 215 phases 488–9

RED pressure

CAT

mnemonic

595

eect 488

redox surface

area

of

reactions

424, 580–5

reactants 489

acids 591 temperature

eect 489

electron

transfer 594–7

units 486

half-equations 586–8 reaction

coordinate

492

reaction

kinetics

reaction

mechanism

reaction

order 500

ionic compounds 98 see kinetics

metals 591 496,

497

in optometry 583

oxidizing

agents

585,

589, 590

703

Index

reducing

redox

titration

reducing

agents

335,

agents

585,

589,

590, 608

592–3

585,

salt

bridge

589,

590, 608

acid–base

reduction 98, 580–5

equilibria 565–7

parent acids and bases 556

alkenes 608–9

of

strong

acids

and

strong bases 565

alkynes 608–9

of

strong

acids

and

weak bases 565–6

denitions 581–4

of

weak

acids

and

strong bases 566

weak

acids

and

weak bases 566–7

electrolytic cells 603–4

electron

of

transfer 582

S aruhashi,

K atsuko 526

half-equations 586–8

saturated

halogens 589

saturated solutions 514

hydrogen

organic

loss/gain 582

compounds

sc ale

607–8

resolution,

sc anning

oxygen

sc atter

gain/loss 581

referencing

style,

hydroc arbons 262

practice 311

324, 605

measurement 316

tunnelling

microscope

equation 43

scientic

journals

scientic

knowledge 218

630

relative abundance of isotopes 29

falsiability of 21, 426

relative

atomic

charge 23

public

relative

atomic

mass

relative

molecular

relative

(or

reliability

26,

27,

30, 31, 65, 66

mass 65, 66

renewable

measurements 312–13

repeatability,

reproducibility,

470

scientic models 22, 46, 81

theories 46, 539

seawater

experimental

units,

of

laws 444

scientic

energy 433–6

understanding of 440

sharing

scientic

fractional) uncertainty 356

of

repeating

23,

results 312

c arbon

polymers 209–11

experimental

dioxide

concentration 526

reverse osmosis 11

results 313

secondary

alcohols,

oxidation

residue 326

secondary amines 266

resonance

energy, benzene 165–6

secondary compounds 281–2

resonance

structures 160–7

secondary

benzene 162–7

seesaw

deloc alization 162

retardation

factor,

(rechargeable)

geometry,

157–8

serial

reactions 530

dynamic

Gibbs

RGB

risk

SHE

equilibrium 520

equilibrium 512–16

energy change 454–5

analyser, smartphones 342

assessments,

experiments 12, 309

Rohrer, Heinrich 23

root

names,

rotary

alkanes

272

hydrogen

electrode

SI

system, dening constants 355

SI

units

sigma

15,

16,

bonds

signic ant

66,

350, 355

174–5,

gures,

silicon

137, 203

silicon

dioxide

skeletal

prevention

model,

78, 323

standard

176

measurement 316, 356–7

(silic a)

137

silver sulde 104

rulers 316

Rutherford

dilution

see

silver halides 200

199, 211

rusting/rust

437, 594, 598–602

silver chloride 204

evaporation 328

rubber

199, 206

atomic

structure 20–1, 22

formula

163, 258

smartphones 342, 598

S

1

reaction

mechanism 641

N

Rydberg constant 40

S

2

reaction

mechanism

639–41

N

snowakes 146

s atomic orbitals 44–7

boiling and melting points 191

safety,

emission

salicylic

704

sodium

s-block elements 230

experiments 309–11

acid

472

607

sensors 342

separating funnel 329

seawater 11

chromate–dichromate

cells

605,

semiconductors 203

chromatography

reverse

reversible

604,

molecules 169

retrosynthesis 648

osmosis,

(STM) 23

graphs 365, 366

Schrödinger

ac ademic

length

sc andium 101, 250

oxidation state change 583–5

reference plane 285

reux

595–6, 612

salts

reactions

spectrum 36

239, 582

Index

successive

sodium

c arbonate

ionization

106,

energies 59

equilibrium

473

concentrations, determination of 528–9

functions/operators 344

sodium chloride

modelling

crystals 96

square

planar

dynamic

square

pyramidal

equilibrium 514

347–8

geometry,

molecules

geometry,

170

molecules

170

electrolysis 603, 617

standard cell potential 611, 614, 615

lattice

enthalpy 204

standard

change

lattice

structure 108

standard

electrode potential 610

standard

enthalpy change of combustion 398, 412–17

standard

enthalpy

change

of

sodium uoride 109

standard

enthalpy

change

for

sodium

mass

concentration

molar

74

concentration

74

hydroxide 616–17

in

Gibbs

energy

see

formation

a

standard

entropy change 445, 446

standard

entropy

solubility 514

standard

hydrogen

and bonding type 198

standard

reduction potential 610

covalent substances 152

standard

solubility rules 114

solute

72–8

starting

solutions

concentrated

dilute

72, 465, 616–18

solutions

solutions

73

states

of

73

76, 77–8

entropy change 443

symbols 462

stereocentre

dilution

78, 323

solutions

solutions

632

matter 13–15

serial

75,

76, 77–8, 321–3

75

206,

207

(asymmetric

stereochemic al

formula

centre) 286

287–8

stereoisomers 284–90

(dissolution) 11, 112

cis–trans

72

isomers 284–6

stereospecic 640

chromatography 155, 156, 158

hybrid

610, 612

technique 322–3

molecules

steels

stock

sp

75,

(SHE)

state function 389

measurement 318–19

standard

solvent

13,

saturated/unsaturated solutions 514

solvation

electrode

starch 212

aqueous

pH

419, 420

preparation of 321–2

72, 514

solutions

413–16,

values 445, 446

solutions

dilution

energy change

reaction 394–401

solids 13

ionic compounds 112–14

Gibbs

orbitals

steric

178

STM

hindrance 641

see

sc anning

tunnelling

microscope

2

sp

hybrid

orbitals

178

stock

solutions

75

3

sp

hybrid orbitals 177

space-lling

models

stoichiometric coecient 65, 461

127, 341

stoichiometry 461–2

specic

energy 432, 440

straight-chain

specic

heat

straight-chain alkenes 264

c apacity 394–5

spectator ions 553

strong acids 548

spectrophotometers 339

spectrophotometry

spectroscopes

36,

pH

76, 339–40

proton

37

pH

spectroscopy

nuclear

293–6

magnetic

vibrational

resonance

293

speed of light 39

spin,

spin

electrons

resonance

isomers

chain

substituents

48–50,

174

substitution

spectroscopy 296

temperature

eect 453

isomers

279

secondary and tertiary compounds 281–2

271,

272,

275, 551

reactions

electrophilic substitution 648–9

nucleophilic

substitution

reaction

entropy change 443

energy change 448–51

279–83

279

sublimation 13, 15

reactions 442

Gibbs

273,

isomers

primary,

spin–spin coupling 300

spontaneous

curves 568–9

structural

spectroscopy

spectroscopy 296

spectroscopy

569–70

formula 258, 341

positional

resonance

568,

structural

296–302

spin

curves

strong bases 558

spectroscopy

infrared

alkanes 268, 269

radic al

639–42

rate 642

substitution

624–6

sucrose 68

sulfate

ion

104,

173

spreadsheets 343–5

705

Index

sulfates 114

thermography 388

sulfur 6

thermometric

sulfur

dioxide

sulfur

hexauoride

sulfuric

Sun,

acid

127,

242

167–8

thin

73, 425, 548

absorption

spectrum 40

agents

layer

time,

states

conventions

of

matter

polymers

systematic

errors

back

245

end

symbols 462

383,

385,

tables

(periodic

tables,

395, 399

pH

molecules 169

of

table) 54

tangent

line

also

charts;

point

graphs

TLC

technology 342–9

see

total

databases 345–6

curves

of

weak

acid

of

weak

base

ionic

layer

592–3

strong

347–9

with

strong

with

and

change

455,

251–3,

states

101,

101,

248

248–50

periodic table 230

457–8

properties

variable

316–17, 351, 352

rate 489

reactions 453

transition

range,

transition

state

triangular

247–8

oxidation

acid–base

bonding

temperature

sc ales 15–17

trigonal

bypyramidal

276

trigonal

planar

termination 622, 626

trigonal

pyramidal

termolecular

triiodide

607

ion

248–50

indic ators

571

diagrams 198, 201–5

trichloromethane 133

reactions 496–7

states

493, 640

gradient 389

alcohols

637

congurations

temperature

position

geometry,

geometry,

molecules

triprotic

tertiary compounds 281–2

tritium 26

geometry,

molecules 129

167–8

tests

acids

547, 550

tungsten 194

tests,

“pop”

test,

Vickers

tetrahedral

tetravlent

metals

37

hydrogen 591

hardness

geometry,

test

uncertainties,

207

molecules

129,

theoretic al

yield

130, 177

471

thermal

conductivity/conductors

thermal

energy 389

thermochemistry

390, 399

thermodynamics 404

law of 444

bars

369,

370

expression of 355–6

theories, scientic 46, 539

second

decimal places 356–7

error

463,

measurement 351–62

absolute uncertainty 355

176

human

189,

190, 198

reaction time 353

instrument uncertainty 351–2

least count 351

mean

169–70, 171

molecules 128, 130

triplets 300

tertiary amines 266

ame

573

193

oxidation

87–91, 139

spontaneous

572–3

atomic orbital lling 53, 55

heat 389

reaction

acid

equation 553

electron

measurement

572

chromatography

coordination bonds 126

85,

base

strong

spreadsheets 343–5

energy

base

titration 400

complexes

and

tertiary

335,

with

transition elements 100–1

equilibrium position 522–3

706

acid

bonding

83,

567–73

557–8

sensors 342

Gibbs

557–9,

technique/apparatus 334–6

titration

strong

thin

temperature

terminal

468,

and science 218

gases

259, 260

571

thermometric

374, 484

modelling

129,

half-equivalence point 569–71

quantitative data 364–5

see

titration

experimental

209, 212–13

geometry,

158–9, 331

molecules

titration 469

redox

T-shaped

of

titration

symbols 309

naming

157,

titanium 101

(surfactants) 152–3

acid–base

hazard

(TLC)

representation

measurement of 315

symbols 6

synthetic

chromatography

three-dimensional

superconductors 189

surface-active

titration 400

Thiele tube 330

values 362

percentage uncertainty 355

propagation of 358–61

Index

reaction

relative

mixtures 353

(or

signic ant

hydrogen bonding 145–6

fractional) uncertainty 356

intermolecular

gures 356–7

ionic

value uctuation 353

uncertainty

principle,

unimolecular

universal

Heisenberg’s 43

reactions 496

metal

oxide

non-metal

uranium 28

as

bond

valence

electrons 52

valence

shell

validity

of

327

theory

174,

176

as

electron

pair

repulsion

model

(VSEPR)

127–9, 131

der

Waals

forces

van

triangular

bonding

der

Waals

parameters, gases 84

diagrams 198, 201–5

147

states

vaporization 15

states

248–50

Nicolas-Louis

light,

molecular

model

Waals

parameters 84

aqueous

solutions;

test

293

electromagnetic

radiation 34, 35, 39

293

anions 552

207

dissociation constant 561–5

wavelengths 34, 35, 38, 39

reversible ionization 530

572–3

weak bases

covalent substances 151

dissociation constant 561–5

ionic compounds 111

pH

Alessandro 594

338,

hydrolysis

Schrödinger ’ s 44, 45

titration

cell

127

matter 13

wave–particle duality 43

and bonding type 198

voltaic

der

also

functions,

volatility

Volta,

of

see

wavenumber

193

reaction 239

weak acids 549–50

spectroscopy

hardness

243

solvent 112

Van

wavelengths,

c aliper 317

Vickers

visible

wave

measurement of 311–20

vibrational

state

space-lling

vanadium 101

vernier

polar

sodium

van

Vauquelin,

241

reverse osmosis 11

measurements 312–13

oxidation

reactions

reduction 616

Arkel-Ketelaar

variables,

oxide

percentage composition 69

van

variable

241–2

nucleophile 629

oxidation

valence

reactions

formula 68

molecular polarity 133

unsaturated solutions 514

ltration

water 545–6

formula 121

molecular models 348, 349

hydroc arbons 262

vacuum

forces 138

of

Lewis

molecular

indic ator 543

unsaturated

product

curves 569–71

reversible ionization 530

437, 596

titration

batteries 594, 598

573

work 389

voltaic pile 338

voltmeters 320

X-ray

volume

xenon 84

measurement of 314–15

diraction 165

xenon

trioxide

167–8

problem-solving 465

volumetric

analyses 314

Zewail,

volumetric

glassware 321

zinc

VSEPR

see

Warner,

waste

valence

John

shell

electron

pair

Ahmed 499

247, 590

repulsion model

475

disposal,

laboratory

chemic als 309

water

alkali

metal

reactions 239

bond angle 129

dissociation 545

electrolysis 438

electrophilic

empiric al

addition

634, 646

formula 68

equilibrium constant 545

heating

heavy

curve

graph 15

water 28

707

The periodic table

alkali

metals

group

s-block

1

1

Non-metals 1

H

1.01

3

2

Metalloids

4

Metals 2

Li

6.94

11

3

n

=

4

5

6

7

Be

d-block

9.01

12

Na

Mg

22.99

24.31

19

20

3

21

4

5

6

7

8

22

23

24

25

26

K

Ca

Sc

Ti

V

Cr

Mn

Fe

39.10

40.08

44.96

47.87

50.94

52.00

54.94

55.85

37

38

Rb

Sr

39

Y

40

Zr

Nb

41

Mo

42

Tc

43

Ru

85.47

87.62

88.91

91.22

92.91

95.96

[98]

101.07

55

56

57

Cs

Ba

132.91

137.33

87

88

Fr

Ra

[223]

[226]

72

73

74

Hf

Ta

W

Re

Os

138.91

178.49

180.95

183.84

186.21

190.23

89

104

105

106

107

Rf

Db

Sg

Bh

[267]

[268]

[269]

[270]

La

Ac

†

‡

[227]

†lanthanoids

75

44

61

76

108

Hs

[269]

58

59

60

Ce

Pr

Nd

Pm

Sm

62

140.12

140.91

144.24

[145]

150.36

90

91

92

93

Th

Pa

U

Np

232.04

231.04

238.03

f-block

‡actinoids

708

[237]

94

Pu

[244]

Atomic

number

Element

Relative

atomic

mass noble

gases

p-block

halogens

18

2

He

13

12

14

15

16

17

4.00

5

6

7

8

9

10

B

C

N

O

F

Ne

10.81

12.01

14.01

16.00

19.00

20.18

13

14

15

16

17

18

Al

Si

P

S

Cl

Ar

39.95

9

10

11

26.98

28.09

30.97

32.06

35.45

27

28

29

30

31

32

33

34

35

36

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

58.93

58.69

63.55

65.38

69.72

72.63

74.92

78.96

79.90

83.80

45

46

47

48

49

50

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

102.91

106.42

107.87

112.41

114.82

118.71

121.76

127.60

126.90

131.29

77

78

79

80

81

82

83

84

85

86

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

192.22

195.08

196.97

200.59

204.38

207.20

208.98

[209]

[210]

[222]

109

110

111

113

114

115

116

117

118

Mt

Ds

Rg

Cn

Nh

Fl

Mc

Lv

Ts

[278]

[281]

[281]

[285]

[286]

[289]

[288]

[293]

112

51

69

52

70

53

[294]

63

64

65

66

67

68

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

151.96

157.25

158.93

162.50

164.93

167.26

168.93

173.05

174.97

95

96

97

98

99

100

101

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

[243]

[247]

[247]

[251]

[252]

[257]

[258]

[259]

[262]

102

54

Og

[294]

71

103

709

Oxford

Resources

for

IB

Diploma Programme

2 0 2 3

E D I T I O N

C H E M I S T RY

CO U R S E

CO M PA N I O N

Written by expert and

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