UNIT 1 - HCC Applied Sciencehccappliedscience.weebly.com/.../2017_2018_u1_chem_notes_full.pdf · 4.8 Uses and applications of the substances in this unit 52 5. Quantitative ... videos

Level 3 Applied Science 2017-2018 Unit 1 (Chemistry) 1

Level 3 Applied Science

UNIT 1:

Principles & Applications

of Science I

CHEMISTRY SECTION

Name: ………………………………………………………………………………………..

Teacher: ……………………………………………………………………………………..


Contents 1. The Periodic Table, Atoms, & Ions Page Vid # Done revised

PCP & PCC Tracker 3

1.1 Introduction to the atom 11 1

1.2 Elemental symbols in the Periodic Table 12 2

1.3 Introduction to the Periodic Table of elements 13 3

1.4 Making ions 17 4

1.5 Trends in ionic radius 18 5

1.6 Ionisation energy 19 6

1.7 Electron affinity 21 7

2. Compounds, Bonding & Intermolecular Forces

2.1 Introduction to bonding in compounds 23 8

2.2 Ionic bonding 25 9

2.3 Covalent bonding 27 10

2.4 Metallic bonding 28 11

2.5 Electronegativity 29 12

2.6 Introduction to intermolecular forces 31 13

2.7 Permanent dipole-dipole forces 32 13

2.8 Temporary dipole-induced dipole forces 32 13

2.9 Hydrogen bonding 33 14

2.10 Trends in melting and boiling points across a period 34 15

2.11 Trends in melting and boiling points down a group 37 15

3. Orbital Theory

3.1 Sub-shells and orbitals 38 16

3.2 Electron in box diagrams and electron configurations 40 16

3.3 Blocks in the periodic table 41 17

3.4 Ionisation energies re-visited 42 17

4. Balanced Equations & Chemical Reactions

4.1 balancing equations 43 18

4.2 Reactions of period 2 and 3 elements with oxygen 44 19

4.3 Reaction of metals with oxygen, water, and dilute acids 46 20

4.4 Redox 48 21

4.5 Oxidation numbers for transition metals and oxyanions 50 22

4.6 Reactivity series 51 23

4.7 Displacement reactions 52 23

4.8 Uses and applications of the substances in this unit 52

5. Quantitative Chemistry

5.1 Moles and masses 53 24

5.2 Moles and solutions 54 24

5.3 Moles and equations 55 25

5.4 Percentage yield 57 26

Periodic Table 58

Videos can be found at hccappliedscience.weebly.com under ‘unit 1’ and ‘chemistry’


PCP & PCC Tracker – Unit 1 Chemistry

Watch the videos on EDpuzzle. Ask your teacher to re-set your video if you achieve below 80%.

Answers to questions will be available on the website – make sure you check and mark your answers.

At level 3 you should be doing a MINIMUM of 5 hours independent study per week per teacher.

Week Lesson Date Objectives / checklist Pre-class preparation Post-class consolidation

1

1

State the relative charge, mass and position in

a Bohr atom of protons, neutrons and electrons

Define atomic and mass number

Determine the number of protons, electrons

and neutrons in an atom or ion from given

atomic and mass numbers

Define relative atomic mass

Calculate relative atomic mass

Watch chem videos 1 and

2

Read sections 1.1 and 1.2

in the notes booklet

Make summary notes on

videos 1 and 2

Complete questions 1-15

in the questions booklet.

Check answers online and

mark your own work.

Revise theory for

upcoming chemistry test

1.

2

Identify the Groups, Periods, metals and non-

metals in the Periodic Table

State that the elements are arranged in order of

increasing atomic number in the Periodic

Table

Explain why elements in the same Group have

similar chemical properties

State patterns seen across a period and define

Periodicity

Explain how the atomic radius changes across

a Period and down a Group using CARS

Watch chem video 3

Read section 1.3 in the

notes booklet


video 3




mark your own work.

Revise theory for


1.

3

State what is meant by cation and anion

Explain the octet rule

Predict the charge on an ion of an element

from its position in the Periodic Table

Explain trends across a Period and down a

Group for cations and anions

Compare and explain the size of a neutral

atom with its cation / anion

Explain what is meant by the term

isoelectronic, giving examples


5




videos 4 and 5




mark your own work.

Revise theory for


1.


4

Define ionisation energy

Write equations showing ionisation energies

Explain why energy is required to remove an

electron from an atom

Explain trends in ionisation energy across a

Period and down a Group using CARS

Define electron affinity

Write equations showing electron affinities

Explain why energy is released when adding

an electron to an atom for the first electron

affinity

Explain why energy is required for the second

electron affinity

Explain trends in electron affinity values

across a Period and down a Group using

CARS


7




videos 6 and 7




mark your own work.

Revise theory for


1.

5

Determine the type of bonding present from

the combination of elements in a compound

Use the cross-over method to work out the

formula of ionic compounds

Correctly draw dot & cross diagrams for ionic

and covalent compounds

State what is meant by the term ionic bond

Use the cross-over method and draw dot &

cross diagrams for ionic compounds

Calculate the relative formula mass for ionic

compounds

Explain the two factors that affect the strength

of an ionic bond

Recall the common molecular ions.


9




videos 8 and 9

Come prepared for

chemistry test 1.




mark your own work.

Revise theory for


2.







2 1

Explain what is meant by a covalent bond and

a dative covalent bond

Draw dot & cross diagrams for covalently

bonded compounds (single, multiple and

dative)

Know how the ammonium ion is made

Calculate the relative molecular mass for

covalently bonded molecules

State that organic molecules such as methane

have a tetrahedral shape

Explain that shorter covalent bonds (double

and triple) are stronger bonds

State what is meant by the term metallic

bonding

State why metals conduct heat and electricity

State what is meant by the terms malleable and

ductile

Watch chem videos 10

and 11




videos 10 and 11




mark your own work.

Revise theory for


2.

2

Explain what is meant by the terms

electronegativity, polar, non-polar and dipole

Explain the trend in electronegativity values in

the Periodic Table

Explain what factors affect electronegativity of

an element using CARS

Explain how electronegativity values can be

used to predict the type of bonding present

Explain that ionic bonds can also show

polarisation

Watch chem video 12


notes booklet


video 12




mark your own work.

Revise theory for


2.


3

State what is meant by the term intermolecular

force

State what the three intermolecular forces are

in order of strength

Explain that permanent dipole-dipole forces

exist between polar molecules

Explain how temporary dipole-induced dipole

forces arise

Know that temporary dipole-induced dipole

forces exist between all molecules

Explain that temporary dipole-induced dipole

forces increase in strength with more electrons

Watch chem video 13

Read sections 2.6, 2.7 and

2.8 in the notes booklet


video 13




mark your own work.

Revise theory for


2.

4 Explain that hydrogen bonding occurs when

there is an O-H, N-H or F-H bond present

Draw labelled diagrams showing hydrogen

bonding, including lone pairs and dipoles

Watch chem video 14


notes booklet


video 14




mark your own work.

Revise theory for


2.

5 Explain the trends in melting/boiling points

across periods 2 and 3, in terms of structure

and forces

Explain the trend in boiling points down

Groups 1, 2 and 7

Watch chem video 15

Read sections 2.10 and



video 15

Come prepared for

chemistry test 2.




mark your own work.

Revise theory for


3.







3

1

Determine the number of electrons in each

shell using the formula 2n2

State what is meant by the term orbital and

know the shapes of s and p orbitals.

Know the order of filling the orbitals and sub-

shells when writing out electron configurations

Draw electron in box diagrams and know the

rules for filling the boxes with electrons

Watch chem video 16




video 16




mark your own work.

Revise theory for


3.

2

Identify s, p and d blocks in the Periodic Table

Explain the anomalies in ionisation energy

trends (Be/B and N/O)

Watch chem video 17




video 17




mark your own work.

Revise theory for


3.

3

Be able to balance chemical equations with the

use of state symbols

Explain the trend in melting/boiling points for

the oxides of Period 2 and 3 elements

Explain the properties of the oxides of Period

2 and 3 elements

Construct balanced equations for the formation

of Period 2 and 3 oxides


and 19




videos 18 and 19




mark your own work.

Revise theory for


3.


4 Construct balanced equations for the reaction

of metals with O2, H2O, HCl and H2SO4

Watch chem video 20


notes booklet


video 20

Complete question 81 in

the questions booklet.


mark your own work.

Revise theory for


3.

5

Recall the rules for oxidation numbers and

assign these to species in a chemical equation

Define oxidation, reduction and redox

Determine whether a reaction is REDOX from

a chemical equation

Know that transition metals have variable

oxidation states, shown by roman numerals

Know that roman numerals can also be used

with oxyanions


and 22




videos 21 and 22

Come prepared for

chemistry test 3.




mark your own work.

Revise theory for


4.







4

1

Explain the relative position of metals in the

reactivity series and order of reactivity

Be able to construct displacement reactions for

the metals and the halogens

Be able to explain why the displacement

reactions happen for metals and the halogens

Identify uses for substances covered in this

unit

Watch chem video 23

Read sections 4.6, 4.7 and



video 23




mark your own work.

Revise theory for


4.

2

Define the terms mole and molar mass

Use GMR to calculate grams/moles/molar

masses for solids

Be familiar with the units for concentration

and volumes

Use MCV to calculate

moles/concentrations/volumes for liquids

Watch chem video 24




video 24




mark your own work.

Revise theory for


4.

3

Understand molar ratios in balanced equations

(stoichiometry)

Use GMR, MCV and stoichiometry to solve

calculations in chemistry

Watch chem video 25


notes booklet


video 25

Complete questions 96-

103 in the questions

booklet. Check answers

online and mark your own

work.

Revise theory for upcoming

chemistry test 4.


4

Recall the formulae for percentage yield

calculations

Calculate the percentage yield of a reaction

with use of GMR and stoichiometry

Watch chem video 26


notes booklet


video 26

Complete questions 104-

105 in the questions

booklet. Check answers

online and mark your own

work.

Revise theory for


4.

5 Revision of unit 1 chemistry content Come prepared for

chemistry test 4.

Revise all theory for

upcoming mock exam.


1. The Periodic Table, Atoms & Ions

1.1 Introduction to the atom

In the nucleus of an atom are found:

Protons (+1 charge)

Neutrons (neutral)

In shells around the nucleus you find electrons (-1 charge).

Relatively, protons and neutrons have a similar mass. So if we say the mass of a proton is 1 then the mass

of a neutron is also 1. Relative to protons and neutrons, the mass of an electron is very small (1/2000th the

mass).

You must be able to recall the relative mass and charge for each of the sub-atomic particles in the table

below. You should be able to see that most of the mass is found in the nucleus of the atom.

Particle Relative mass Relative charge Position in atom

Proton 1 +1 Nucleus

Neutron 1 0 Nucleus

Electron 1/2000 -1 Electron shells

Note that we use dots (or crosses) to show electrons in shells. Up to two electrons are found in the first

shell outside the nucleus and eight electrons in the next shell. For a neutral atom (one that has no charge

and therefore is not an ion), the number of positive protons must equal the number of negative electrons.

The atom shown above has 7 protons and 7 electrons and is nitrogen. We can tell how many protons,

neutrons and electrons are in an atom by looking at the elemental symbols in the periodic table (see next

section).


+ +

+ +

+ +

+

Electron

Shell

Nucleus

Neutron

Proton


1.2 Elemental symbols in the Periodic Table

You can work out how many protons and neutrons are in the nucleus of a particular element when given its

elemental symbol along with its atomic number and mass number:

N

Atomic number: the number of protons in the nucleus of an atom

Mass number: the number of protons and neutrons in the nucleus of an atom

For a neutral atom (one that has no charge and therefore is not an ion), the number of positive protons must

equal the number of negative electrons. Therefore the atomic number also tells you how many electrons

are present in a neutral atom. The number of neutrons can be found by subtracting the atomic number from

the mass number. E.g. nitrogen has 14 – 7 = 7 neutrons.

Note that in the periodic table the average of the mass numbers, the relative atomic mass, is given, not the

mass number.

Cl

For example, in nature two isotopes of chlorine are found; 35Cl and 37Cl. Both isotopes have an atomic

number of 17, telling us that chlorine has 17 protons and also 17 electrons when a neutral atom. The only

difference between any isotope of the same element is the number of neutrons and therefore the mass

numbers. The two isotopes of chlorine react the same despite having a different number of neutrons and

different mass numbers because they still have the same number of electrons. Chemical reactivity depends

on electron movement.

You might think to yourself that the mean of 35 and 37 is (35 + 37) / 2 = 36. But the relative atomic mass

(R.A.M.) is not worked out in this simple way – it considers the relative contribution of each isotope. The

symbol in the Periodic Table shows an average of 35.5. Because the average is closer to 35 this must mean

there are more 35Cl atoms present in a sample of chlorine found in nature and less 37Cl. So to work out the

number of neutrons in an atom, you do not look at the relative atomic mass, as this is just the average mass,

but instead you look at the mass number given for that pure isotope.

Relative atomic mass: the average mass of an atom of an element compared to 1/12th of the mass of an

atom of 12C

14

7

Mass number

Atomic number

Elemental symbol

35.5

17 Atomic number

Elemental symbol

Relative atomic mass


Look at the definition for R.A.M. above – if you take an atom of pure 12C, only 1/12th of that atom is used

as the standard for weighing atomic masses. Atomic masses are weighed relative to 1/12th of 12C. The

relative atomic masses do not have any units because they are only relative numbers.

To calculate the relative atomic mass for a sample of an element found in nature, simply multiply the mass

number by the percentage for each isotope and then divide by 100 (see the example calculation below).

Remember to use brackets on your calculator and give your answer to one decimal place.

WORKED EXAMPLES

1. A sample of chlorine contains 25% 37Cl and 75% 35Cl. Calculate the relative atomic mass?

(25 x 37) + (75 x 35) = 35.5

100

2. A sample of magnesium contains 78.6% 24Mg, 10.1% 25Mg and 11.3% 26Mg. Calculate the relative

atomic mass?

(78.6 x 24) + (10.1 x 25) + (11.3 x 26) = 24.3

100

1.3 Introduction to the Periodic Table of elements


Period 2

Period 3

Period 4

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Period 1

METALS

NON-METALS


The Periodic Table shows all of the chemical elements arranged in order of increasing atomic number.

Elements are organised into vertical columns called groups and horizontal rows called periods. Period 1

contains the elements H and He. Chemical properties are similar for elements that are in the same group

because they all have the same number of electrons in the outer shell.

The atomic number increases as you move from left to right across a period because each element has one

more proton than the element before it in the same period. There are trends (or patterns) that repeat

themselves each time you go across a period. For example, each time you go from left to right across a

period you go from metals to non-metals and the atomic radius also decreases. This repeating pattern seen

by the elements across a period is called periodicity.

Periodicity: the repeating pattern seen by the elements in the periodic table

The diagram below shows how the atomic radius decreases across a period but increases down a group.

The two electrons in the first shell have been shown. Add the missing electrons to the other electron shells

using the atomic numbers for the neutral atoms to prove: (i) electrons are added to the same shell across a

period but (ii) a new shell of electrons is added down a group.

To explain trends in atomic radius seen above (and other trends that you shall meet in this unit), the follow

pneumonic shall help you to re-call all of the factors that are responsible for the trends:

Nuclear Charge

Nuclear Attraction

Radius

Shielding

3Li 4Be 5B 6C 7N 8O 9F 10Ne

11Na


Across a period, the atomic number increases as the number of positive protons in the nucleus

increases by one each time, so the nuclear charge increases (e.g. +3 to +4 to +5). The increased

positive nuclear charge each time you go across a period means even stronger attraction on the

electrons; the electrons experience stronger nuclear attraction which draws them to be closer to the

nucleus.

Because the electron shells are closer to the nucleus across a period due to the increased nuclear

charge and attraction, this means that the atomic radius has decreased.

One last factor is shielding. The inner shell of electrons ‘shields’ the outer shell of electrons from

the positive nuclear charge. The inner shell of electrons also repel the outer shell of electrons to be

further from the nucleus. Hence, the more inner shells of electrons there are in an atom the greater

the shielding. Across a period electrons are added to the same shell, not a different shell. This

means that there is the same number of inner shells and therefore the same shielding across a period

(and is not a big factor affecting the radius).

+3 +4 +5

The electron experiences

increased nuclear attraction as

the nuclear charge increases

- - -

+3

Attraction

Repulsion.

Shielding increases as the number of

inner shells of electrons increases

- - -

Radius decreases across a

period as the nuclear charge and

attraction increases


Each time you go down a group however, there is an additional electron shell added, resulting in an

increase in radius. The greater number of inner shells also results in increased shielding. Therefore

the outermost electrons are less strongly attracted to the nucleus with them being further away with

the larger radius, so there is weaker nuclear attraction on the electrons. Note that the nuclear charge

does also increase down a group, and you may think this would draw the electron shells inwards to

be closer to the nucleus. However, the increased shielding and larger radius from more electron

shells outweigh the increase in nuclear charge, resulting in an overall larger atomic radius.

MODEL ANSWERS

1. Explain the difference in atomic radius between C and N.

N is to the right of C in the same period (period 2)

N has more protons so a greater nuclear charge than C

The electrons in N experience greater nuclear attraction than those in C

As N and C are in the same period, shielding is the same

N therefore has a smaller radius than C

2. Explain the difference in atomic radius between Mg and Ca.

Ca is below Mg in Group 2

Ca has more shells so a larger radius

Ca has more shells so more shielding

The electrons in Ca experience weaker nuclear attraction

Ca does have a greater nuclear charge than Mg, but this is outweighed by the increased radius and

shielding in Ca (resulting in a weaker overall nuclear attraction)


More shells down a group:

Larger radius

Increased shielding,

Weaker nuclear attraction on electrons


1.4 Making ions

Cation: ion with positive charge

Anion: ion with a negative charge

So far we have only looked at neutral atoms, where the number of positive protons equals the number of

negative electrons. However, atoms can gain or lose electrons during chemical reactions to form charged

particles called ions in order to satisfy the octet rule.

The octet rule states that elements gain or lose electrons in order to have eight electrons in the outermost

shell, like the noble gases. The noble gases already have eight electrons in their outer-most shell and they

are very stable, existing as atoms only and do not form ions. E.g. Na+ has lost one negative electron to

leave behind a positive charge; Cl- has gained one negative electron and therefore has a negative charge;

N3- has gained three negative electrons.

We can predict the charge that an ion of a given element shall form by looking at its position in the periodic

table. You must learn the charges that elements from each group form.

E.g. sodium (Na) is in Group 1 of the periodic table and therefore only has one electron in its outer shell.

To have a complete outer shell containing eight electrons, the sodium atom must lose this one outer electron

(electrons are negatively charged), which would result in a sodium ion with a charge of +1 (written as Na+).

All group 1 elements form +1 ions for the same reason. Note that group 4 elements do not usually form

ions; they have four electrons in their outer-most shell and it takes too much energy to gain or lose four

more electrons in order to complete the octet. Group 0 (also called Group 8) elements already have eight

electrons in their outer-most shell and are very stable atoms, so they do not form ions.

+1 +2 +3 -3 -2 -1

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6

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0

+2 common

but varied

Transition metals



1.5 Trends in ionic radius

You have seen previously that for NEUTRAL atoms there are trends in radius across a period and down a

group. The trends can be explained using the pneumonic CARS. The same patterns also exist with the

cations/anions – radius decreases across a period and increases down a group due to more shells.

But you must also know how the size of a cation/anion compares with the neutral atom across a period.

Across a period for cations: across a period, the radius of neutral atoms and cations decreases. But

compared with neutral atoms, the cations have lost electrons to gain a positive charge. This means that the

cations attract the remaining electrons even more strongly than the neutral atoms because there is still the

same number of protons in the nucleus now attracting less electrons in the cation. This means that the

electron shells are drawn to be even closer to the nucleus in a cation compared with the neutral atom,

resulting in the cations being smaller than the neutral atoms.

There is an interesting property with cations across a period – they all have the same number of electrons

despite being different elements, so they are isoelectronic. This is shown below with the cations of the first

three elements in Period 3 as an example:

Neutral atom electron structure Cation electron structure

Na: 2,8,1 Na+: 2,8

Mg: 2,8,2 Mg2+: 2,8

Al: 2,8,3 Al3+: 2,8

Isoelectronic: having the same number of electrons

Na Mg Al

Na+ Mg2+ Al3+

Neutral atom

Cation (smaller)

Across a period, the radius decreases

for both cations and the neutral atom

– CARS!


Across a period for anions: across a period, the radius of neutral atoms and anions decreases. But compared

with neutral atoms, the anions have gained electrons to complete with a negative overall charge. The added

electron(s) when you form the anion cause extra repulsion, with the electrons pushing themselves to be

further apart. This results in a larger radius for the anion than the corresponding neutral atom.

Across a period, the anions are also isoelectronic, as shown with the first three anions in Period 3 below:

Neutral atom electron structure Anion electron structure

P: 2,8,5 P3-: 2,8,8

S: 2,8,6 S2-: 2,8,8

Cl: 2,8,7 Cl-: 2,8,7

But how are the cations/anions made in the first place? There are two ways of forming ions; removing

electrons is the ionisation energy and adding electrons is the electron affinity (see next section).

1.6 Ionisation energy

First ionisation energy: the energy required to remove one mole of electrons from one mole of gaseous

atoms to form one mole of gaseous +1 ions

Because there is a massive number of atoms even in 1 gram of a substance, it is not appropriate to count

atoms in millions or even billions. We need a bigger quantity to count atoms - chemists use the mole:

One million: 1 000 000

One billion: 1 000 000 000

One mole: 602300000000000000000000 or 6.023 x 1023 in standard form

P S Cl

P3- S2- Cl-

Neutral atom

Anion (larger)


Across a period, the radius decreases

for both anions and the neutral atom

– CARS!


When we remove electrons from atoms, we measure the energy required to remove one mole of electrons

from one mole of atoms in the gaseous phase. We can write equations to show the process. Removing an

electron leaves a positively charged ion after the arrow and we also show the electron that has been

removed. The state symbol for the gaseous phase must be shown.

Na (g) → Na+ (g) + e first ionisation energy = +496 kJmol-1

You do not need to remember the numerical values, only the trends shown across a period and down a

group (see below). For the example shown above, 496 kJ of energy are required per mole of electrons

removed from one mole of sodium atoms in the gaseous phase.

The positive sign before the numerical value shows that energy is required for this process; energy is

required to break the attraction between the electron and the positively charged nucleus.

More than one electron can be removed in a stepwise process. For example, if you want to form a 2+ ion

then you initially remove the first mole of electrons from the neutral atom in the gaseous phase to form a

+1 ion and then go back and remove the second mole of electrons from the +1 ion to form the +2 ion. Hint:

if writing the equation for the first ionisation energy then you are going to form the +1 ion after the arrow.

If writing the equation for the 7th ionisation energy then you are going to form the +7 ion after the arrow.

A few examples are shown below:

Mg+ (g) → Mg2+ (g) + e second ionisation energy = +1450 kJmol-1

Al2+ (g) → Al3+ (g) + e third ionisation energy = +2740 kJmol-1

There are periodic trends in the first ionisation energy which can be explained using the pneumonic CARS

which we met earlier. The answer is the exact same as used for explaining trends in atomic radius. Across

a period the first ionisation energy increases (there are some anomalies e.g. between Be and B and also N

and O – the reason for this shall be explained later in section 3.4). Down a group the ionisation energy

decreases.

Ionisation energies increase across a

Period left to right.

There are some anomalies which will

be discussed in section 3.4 later

Ionisation energies decrease down a

Group


MODEL ANSWERS

1. Explain why the first ionisation energy for F (1680 kJmol-1) is greater than the first ionisation energy

for C (1090 kJmol-1).

F is to the right of C in the same period (period 2)

F has more protons so a greater nuclear charge than C

F has a smaller radius than C

As F and C are in the same period, shielding is the same

The electrons in F experience greater nuclear attraction than those in C

This means more energy is required to overcome the attraction and remove an electron in F

2. Explain why the first ionisation energy for K (418 kJmol-1) is less than the first ionisation energy for Na

(494 kJmol-1).

K is below Na in Group 1

K has more shells so a larger radius

K has more shells so more shielding

The electrons in K experience weaker nuclear attraction

This means less energy is required to overcome the attraction and remove an electron from K

K does have a greater nuclear charge than Na, but this is outweighed by the increased radius and

shielding in K (resulting in a weaker overall nuclear attraction)

1.7 Electron affinity

Electron affinity: the change in energy when one mole of a gaseous atom gains one mole of electrons to

form one mole of gaseous negative ions

Cl (g) + e → Cl- (g) first electron affinity = -349 kJmol-1

When adding electrons to atoms, the negative sign before the numerical value indicates that energy is

released (the opposite of ionisation energy). There are also the same general periodic trends in electron

affinity values as were seen with ionisation energy values. These can also be fully explained using the

pneumonic CARS.

Across a period the electron affinity values increase as more energy is released when nuclear attraction for

the electron is stronger. Down a group when nuclear attraction for the electron to be captured is weaker,

less energy is released. There are some anomalies. Look at the table of electron affinity values for group

6 and group 7 elements below. You should notice that O and F do not fully fit the trend going down a

group. This is because these atoms are very small and placing an extra electron in a crowded area is difficult



and there is significant repulsion. This repulsion lessens the attraction for the incoming electron resulting

in a lower than expected electron affinity value.

Group 6 element First electron

affinity/kJmol-1 Group 7 element

First electron

affinity/kJmol-1

O -141 F -328

S -200 Cl -349

Se -195 Br -324

Te -190 I -295

Note that Group 6 elements can also have a second electron affinity to form the 2- ions. Adding a second

electron to the 1- ion shall require energy to overcome the repulsive force, as this time the negative electron

to be captured shall be repelled by the negative 1- ion. The energy change shall therefore be positive as

energy is required to force the second electron into the 1- ion.

E.g. values and equations for oxygen:

O (g) + e → O- (g) first electron affinity = -141 kJmol-1 (energy released)

O- (g) + e → O2- (g) second electron affinity = +844 kJmol-1 (energy required)



2. Compounds, Bonding & Intermolecular Forces

2.1 Introduction to bonding in compounds

Atoms or ions can come together to form compounds. There are three main types of bonding; ionic,

covalent and metallic. You can only determine the type of bonding present if you know where the metals

and non-metals are located in the Periodic Table of elements and also what the charges are on ions from

the position of the element in the Periodic Table.

Metallic bonding occurs when there are only metals present; either a metal on its own or a mixture of

metals (an alloy). The negative delocalised electrons are attracted to the positively charged metal ions.

Metallic bonding exists in e.g. Cu, Fe and the alloy brass (mixture of Zn & Cu).

Covalent bonding occurs between two non-metals. Ions are not involved; instead the two non-metals

come together to share one electron each via a single covalent bond. More than one electron may be

shared (a multiple bond) and there are also cases where the pair of electrons in the covalent bond has

come from only one of the non-metals (dative covalent bond). Examples of covalently bonded

molecules are O2, F2 and NH3.

Ionic bonding occurs between metals and non-metals (and also when there are molecular ions such as

NH4+ present in a compound). Two oppositely charged ions attract each other in the ionic bond. Use

the cross-over method to find the formula of the ionic compound.

Period 2

Period 3

Period 4

Gro

up 8

Gro

up 7

Gro

up 6

Gro

up 5

Gro

up 4

Gro

up 3

Gro

up 2

Gro

up 1

Period 1

METALS

NON-METALS

Charge on ions: +1 +2 +3 -3 -2 -1


The cross-over method for deducing ionic formulae

(i) Write the charge above each ion by checking the position of the element in the Periodic Table.

(ii) Make the simplest ratio for the charges.

(iii) Swap numbers and use them as a subscript for the other ion.

Example 1: what is the formula for aluminium oxide?

Example 2: What is the formula for calcium chloride?

Example 3: what is the formula for magnesium oxide?

Example 4: what is the formula for aluminium carbonate (carbonate is the molecular ion CO32-)?

The octet rule states that atoms with eight electrons in their outer-most shell are extremely stable (apart

from the first shell, where two electrons are required). The Noble gases (group 8 elements) exist as single

atoms with 8 electrons in their outer-most shell (except Neon, which has 2 outer electrons in the first shell

only). They are therefore extremely stable (and extremely un-reactive).

What about atoms that do not have eight electrons in their outer-most shell? These atoms can satisfy the

octet rule by either:

- Sharing electrons with other elements until both have eight electrons (covalent bonding), or

- Transferring electrons until both have eight electrons (to make ions, ionic bonding).

We can use dot & cross diagrams to show the number and source of the electrons in a compound.

(i) Draw a circle to represent the outermost shell only.

(ii) Use dots to represent the electrons from one of the elements and crosses to represent electrons from the

other element. In cases where there is more than two different types of element involved, use a third symbol

(and therefore a symbol key to tell the examiner what your chosen third symbol represents e.g. a triangle,

a square, the letter ‘e’ etc).

(iii) Covalent compounds: draw two overlapping circles and put the electrons being shared in the

overlapping region. Include those electrons not involved in bonding (lone pairs).

(iv) Ionic compounds: show two separate circles as ions, with square brackets around the circle and the

ionic charge to the top right outside the brackets. The charge should be the same as the Group number and

you should also check the formula with the cross-over method so that both opposite charges cancel out

overall due to the correct number of each type of ion.

Al3+ O2- → Al2O3

Ca2+ Cl1- → CaCl2

Mg2+ O2- → Mg1 O1 → MgO

Al3+ CO32- → Al2(CO3)3

*Use brackets for

molecular ions


Example 1: Draw a dot & cross diagram for chlorine, Cl2?

Example 2: Draw a dot & cross diagram for Sodium chloride, NaCl?

Note: All ionic compounds actually exist as a giant ionic lattice – see later. We do not have small molecules of NaCl; each ion

attracts the oppositely charged ion in all directions to form a giant ionic lattice.

2.2 Ionic bonding

Ionic bonding: the electrostatic force of attraction between two oppositely charged ions

Ionic bonding occurs when you have a mixture of metals and non-metals bonded together. The Metal loses

electrons to form a positively charged ion and the non-metal gains electrons to form negative ions. After

electrons have been transferred to give two ions with a stable octet these oppositely charged ions then attract

each other. Electrostatic attraction is the force experienced by oppositely charged particles and holds

particles strongly together. You must also use the cross-over method to find the formula of the ionic

compound.

When drawing dot and cross diagrams you must draw separate circles as ions; the charge on each ion can

be predicted by looking at the position of the element in the Periodic Table. Also ensure that you have the

correct number of each type of ion drawn out by using the cross-over method to work out the correct

formula.

Cl X

Cl

Two non-metals, so the bonding is covalent.

Draw two overlapping circles.

Each Cl atom has 7 electrons in its outer-most shell (group 7). If

each Cl atom contributed one electron to the shared area then

each Cl atom would now have access to 8 electrons, satisfying the

octet rule.

Use dots to represent the electrons in one Cl atom and crosses to

represent electrons in the second Cl atom.

X X

X X

X X

Na X

Cl

- + A metal and a non-metal, so the bonding is ionic.

Draw two separate circles with square brackets.

Na is a Group 1 metal. It forms +1 ions. Cl is a Group

7 non-metal. It can accept the electron that the Na

wants to lose so that both now have 8 electrons to

satisfy the octet rule. The Cl now has a charge of -1.

In the example, the dots represent the 7 electrons in Cl

and the cross represents the electron gained from Na.

Complete question 41


You may also be asked to calculate the relative formula mass for an ionic compound. The relative formula

mass is the sum of the relative atomic masses for each element in the formula. There are no units for

relative formula mass; the numbers are simply relative to 12C.

WORKED EXAMPLE

Draw a dot & cross diagram for NaCl. Find its relative formula mass?

Na is a metal, in Group 1, so forms +1 ions. Cl is a non-metal in Group 7, so forms -1 ions. Draw separate

circles with brackets and charges for ionic compounds

Remember that ionic compounds do not exists as simple molecules. Each ion attracts the oppositely

charged ion in all directions resulting in a giant ionic lattice, which is a regular arrangement of positive and

negative ions. E.g. the giant ionic lattice structure of NaCl:

The strength of the electrostatic force and therefore the ionic bond depends on:

(i) the ionic charge: a bigger ionic charge results in a stronger force of attraction

(ii) ionic radius: a larger radius (e.g. when you have more shells of electrons) means the ionic charge is

spread over a larger surface area, resulting in a weaker attraction for the oppositely charged ion compared

with a smaller radius.

Finally, it is a good idea to learn the following ions which exist as molecules (molecular ions):

+1 charged ions -1 charged ions -2 charged ions

Ammonium NH4+ Hydroxide OH-

Nitrate NO3-

Carbonate CO32-

Sulfate SO42-

Na+ Na+ Na+

Na+ Na+ Na+

Na+ Na+ Na+

Na+

Na+ Na+

Na+

Cl – Na+ Cl –

Cl –

Cl –

Draw in an exam as:


Na X

Cl

- +

Relative formula mass:

23 + 35.5 = 58.5


2.3 Covalent bonding

Covalent bond: a shared pair of electrons

Dative covalent bond: a shared pair of electrons and both electrons have come from the same atom

Covalent bonding occurs between two non-metals where the atoms come together to share a pair of

electrons to form a molecule. This is drawn with overlapping circles and the shared pair of electrons in the

covalent bond shown in the overlapped area. One pair of electrons in total in the shared area results in a

single bond, two pairs of electrons a double bond and three pairs of electrons a triple bond. The term

‘relative molecular mass’ is used for covalently bonded compounds as these do exist as molecules.

‘Relative formula mass’ was used for ionic compounds which do not exist as molecules.

Where both electrons have come from the same atom, a dative covalent bond is formed. An example is in

the reaction of NH3 with a H+ ion to form the molecular ion NH4+.

WORKED EXAMPLE

Draw a dot & cross diagram for CH4. Find its relative molecular mass?

Both C and H are non-metals. Draw over-lapping circles for the covalent bonds. Both atoms contribute

one electron each, so single covalent bonds are formed (not a dative bond).

Dative covalent bond.

Two dots to show that both of the

electrons have come from the N atom.

Because the H+ ion is now part of the

molecule, the entire molecule now has

a charge of +1 (NH4+).

Note: only draw the final product, on

the right, in an exam, unless asked for

the entire reaction.

→

Relative formula mass:

12 + (4 x 1) = 16


Consider the dot & cross diagram drawn above for methane. The diagram would suggest that this is a flat

molecule. This is not the case. Organic molecules (compounds that contain one or more carbons in a

carbon chain) have a 3D shape. In the 3D diagram below, the bold wedge shows the bond that comes out

of the plane of the paper towards you. The dashed wedge represents the bond that goes into the plane of

the paper, away from you. The shape around each carbon atom in the alkanes is tetrahedral with a bond

angle of 109.50.

Methane propane

Finally, you should know that single bonds are longer than double bonds and double bonds are longer than

triple bonds. The shorter the bond the stronger it is and therefore would require more energy to break.

Bond Length (pm) Energy (kJ mol-1)

C-C 154 347

C=C 134 612

C=C 120 820

2.4 Metallic bonding

Metallic bonding: the electrostatic force of attraction between positive metal ions and negative delocalised

electrons.

Metallic bonding exits between metals only. Metals exist in a giant metallic lattice, which is a 3-D structure

of positive metal ions surrounded by negative delocalised electrons. There is a very strong attraction

between the positively charged metal ions and the negative delocalised electrons, so metallic bonding is

strong. However, the attraction is not usually as strong as in ionic or covalent bonding.

e.g. diagram showing metallic bonding in Na:

Na+ Na+ Na+

Na+ Na+ Na+

– – –

– – –

delocalised

electrons

Positive

metal ions

Na+ Na+ Na+

– – –

Shorter bonds

Stronger bonds



Because the delocalised electrons are free to move and carry charge, metals conduct electricity, when solid

or molten. This is why copper is used in electrical cables and wires. The delocalised electrons can also

absorb heat energy, which gives them kinetic energy, hence metals are also good thermal conductors.

Copper and aluminium are examples of metals used in saucepans, heat sinks in computers and radiators.

Two other properties of metals are:

(i) malleable – they can be hammered into shape without breaking.

(ii) ductile – they can be stretched out into wires without breaking.

These two properties can be explained by the fact that the metal ions are in layers, and they can roll and

slide over each other without breaking the metallic bonding, as shown in the diagram below. Aluminium

is very malleable and along with its thermal conductivity makes it suitable for use in aluminium foil.

2.5 Electronegativity

Consider a H2 molecule. Both atoms are identical and each has an equal share of the electron pair in the

covalent bond. The electron pair is equally distributed between both atoms.

Now consider a molecule of HCl. Both atoms are different. One atom is likely to attract the electron pair

in the covalent bond more strongly than the other atom (and therefore the electron pair will be closer to this

atom). We say that the atom with the greater attraction for the pair of electrons in the covalent bond is

more electronegative than the other atom.

Electronegativity: the tendency of an atom to attract a bonded pair of electrons in a molecule

Polar molecule: molecule with a partial negative charge on one end and a partial negative charge on another

end due to an uneven distribution of electrons

Non-polar molecule: a molecule where the electrons are equally distributed throughout the molecule

H H H Cl X X



In the HCl molecule, Cl is more electronegative than H. This means that Cl has the greater attraction for

the pair of electrons in the covalent bond and therefore the pair of electrons is closer to the Cl atom than

the H atom. This difference in electronegativity between the two atoms results in a small charge difference

(because electrons are negatively charged) across the H – Cl bond called a permanent dipole, which we

show with a δ- and a δ+. These are partial charges only, not full ionic charges; the Cl has not captured the

electron pair to form a Cl- ion. Quite simply, the electrons are just closer to the Cl atom. ‘Permanent’

dipole means the dipole is always present.

H – Cl

If we have permanent dipole’s across a bond, because the two atoms have a difference in electronegativity,

then we say that the bond is polar. As a result, the molecule may also be polar. H – Cl is a polar molecule

but H – H is a non-polar molecule; in H2, both atoms are identical, there is no difference in electronegativity.

How do we work out which atom is the more electronegative atom? We use the Pauling scale to compare

the relative electronegativity of atoms; the bigger the number, the more electronegative the atom.

Do not learn these numbers, they shall be provided to you in an exam. But you should know the general

trend; electronegativity increases across a period left to right and also up a group. F is the most

electronegative atom and Group 0 elements do not have electronegativity that can reliably be determined –

Group 0 elements exist as atoms not molecules.

You can use the pneumonic CARS to help you remember the factors that affect the electronegativity of an

element. Electronegativity depends on the number of prtons in the nucleus, the distance from the nucleus

of the bonding pair of electrons and how much shielding there is from inner electrons.

The electronegativity of an element can be used to predict the type of bonding in a compound. It is actually

rare to have a wholly ionic or a wholly covalent compound; bonding is a spectrum from ionic to covalent

bonding with most compounds sitting somewhere between the two. If the electronegativities are similar

between both atoms then a covalent bond forms. As the difference in electronegativity increases the

Increasing electronegativity

Incr

easi

ng e

lect

ron

egati

vit

y

δ- δ+


covalent bond will become more polar. If the difference in electronegativity is very large then the bond

becomes ionic as one of the atoms has captured the electron pair completely.

It is not just covalent bonds that can be polar. Ionic bonds can also show polarity. The extent of polarisation

shall depend on whether:

either ion is highly charged

the cation is relatively small

the anion is relatively large

e.g. a small cation that is highly charged can draw electrons towards it. A large anion that is highly charged

has an electron cloud that is easily distorted. If these two anions attract then the small cation can share

some of the negative charge on the anion. This gives the ionic bond some covalent characteristics.

2.6 Introduction to intermolecular forces

Intermolecular force: the attractive or repulsive force between molecules

When we have covalently bonded molecules, we can also have weaker attractive forces (compared with

covalent and ionic bonds) that exist between the molecules. These are known as intermolecular forces or

van der waals forces and there are three types:

Temporary dipole-induced dipole force weakest

Permanent dipole-dipole force

Hydrogen bonding strongest

Molecule with covalent

bonds between atoms

Intermolecular force

between molecues




2.7 Permanent dipole-dipole forces

Polar molecules, such as HCl, have permanent dipoles. The permanent dipole of one polar molecule can

attract the opposite permanent dipole in a neighbouring molecule. When this happens, we have a weak

permanent dipole-dipole force between the neighbouring molecules.

H – Cl H – Cl H – Cl

2.8 Temporary dipole-induced dipole forces

Consider a H2 molecule. The molecule is non-polar as there is no difference in electronegativity, no

permanent dipoles and therefore no permanent dipole-dipole force between neighbouring molecules.

However, the electrons are constantly moving within the molecules. At any one moment in time, the

electrons may be temporarily closer to one of the H atoms in the H2 molecule. This would instantaneously

lead to a temporary dipole on that H atom (not permanent; if the electrons continue moving, this temporary

dipole disappears, but can reappear again, temporarily). The temporary dipole can influence and induce a

neighbouring molecule into also forming a temporary dipole; these temporary dipoles attract each other

leading to a temporary dipole-induced dipole force of attraction (also called London dispersion forces).

London dispersion forces exist in all molecules, whether polar or non-polar. A non-polar molecule will

only have London dispersion forces between neighbouring molecules. A polar molecule with permanent

dipoles shall have permanent dipole-dipole forces and also London dispersion forces between neighbouring

molecules.

H – H H – H H – H

You need to be able to explain how temporary dipole-induced dipole forces arise (model exam answer):

Due to the movement of electrons, there is an uneven distribution of electrons throughout the molecule.

This causes temporary dipoles on the molecule.

The temporary dipole induces a dipole in a neighbouring molecule.

The temporary dipoles and induced dipoles attract each other to form weak intermolecular forces called

temporary dipole-induced dipole forces (also called London dispersion forces).

δ+ δ+ δ+ δ- δ- δ-

δ+ δ+ δ+ δ- δ- δ-

Permanent dipole-dipole force

this molecule has temporary

dipoles which were induced

by the first H2 molecule


dipoles which were induced

by the second H2 molecule

Temporary dipole-induced dipole force


dipoles caused by the

movement of electrons

Permanent dipoles


Temporary dipole-induced dipole forces increase with an increase in the number of electrons. More

electrons results in larger temporary and induced dipoles which results in stronger attractive London

dispersion forces between the molecules. Stronger attractive forces between the molecules results in an

increase in boiling point.

WORKED EXAMPLE

Explain the following trend in the boiling points of the halogens listed?

F2 Cl2 Br2 I2

-1880C -350C 590C 1840C

The boiling point of the halogens increases as you go down group 7.

This is because I2 has more electrons than Br2, which has more electrons than Cl2, which has more

electrons than F2; more electrons results in larger temporary dipoles.

This means there would be stronger temporary dipole-induced dipole forces between molecules of I2

followed by Br2 followed by Cl2 followed by F2.

Note: temporary dipole-induced dipole forces can also exist between atoms, such as the noble gases.

2.9 Hydrogen bonding

This is a special type of permanent dipole-dipole force when O – H, N – H or F – H bonds are present in a

molecule. Because there is a large difference in electronegativity between H and the O/N/F atoms, these

bonds are highly polar. The permanent dipole-dipole forces between these bonds in neighbouring

molecules are particularly strong and are given a special name: hydrogen bonding.

In hydrogen bonding, the Hδ+ in one molecule attracts the lone pair of electrons on the O δ-, N δ- or F δ- in

the neighbouring molecule. E.g. a diagram showing hydrogen bonding between neighbouring water

molecules is shown below. You must include dipoles, lone pairs and a label for the hydrogen bond.

δ-

δ-

δ+ δ+

δ+ δ+

Hydrogen bond



MODEL ANSWER – comparing intermolecular forces present

Explain why HCl has a higher boiling point than H2?

HCl is a polar molecule. Due to the difference in electronegativity between H and Cl, there are

permanent dipoles. This means permanent dipole-permanent dipole forces between the molecules in

addition to temporary dipole-induced dipole forces.

H2 is a non-polar molecule. There is no difference in electronegativity between the atoms. The only

intermolecular force present is temporary dipole-induced dipole forces.

Permanent dipole-dipole forces are stronger than temporary dipole-induced dipole forced. This means

the intermolecular forces between HCl molecules are stronger than those in H2. The stronger

intermolecular forces in HCl require more energy to break resulting in a higher boiling point.

2.10 Trends in melting and boiling points across a period

There are periodic trends in melting/boiling points across a period which can be explained in terms of the

bonding or intermolecular forces present. Melting (solid to liquid) and boiling (liquid to gas) points depend

on the strength of the forces between the atoms or molecules that you want to separate to go to a liquid or

gaseous phase.

During melting, energy is required to overcome the attractive forces between the atoms or molecules and

boiling usually means that most of the rest of the attractive forces are broken. The stronger the forces

between the atoms/molecules the more energy is required to break them and the higher the melting or

boiling point. In general there is an increase in melting/boiling points across periods 2 & 3 (left to right)

up to Group 4. There is then a sharp decrease in melting/boiling points between Groups 4 and 5. E.g. the

trends in boiling points across period 3 is shown in the graph below:

3000

2500

2000

1500

1000

500

0

boiling point

/K

11 12 13 14 15 16 17 18

atomic number

Na Mg

AlSi

P

S

ArCl

general increase

groups 1 to 4

sharp decrease

groups 4 to 5

generally low for

groups 5 to 8



The table below summarises the reasoning for the trend in melting & boiling points across periods 2 & 3:

Period 2 Li Be B C N2 O2 F2 Ne

Period 3 Na Mg Al Si P4 S8 Cl2 Ar

boiling

point

explained

Giant metallic lattice

with strong electrostatic

forces of attraction

between positive metal

ions and delocalised

electrons to break

Giant covalent

lattice with strong

covalent bonds to

break between the

atoms

Simple

molecular lattice with only weak

temporary dipole-induced dipole

forces of attraction between the

molecules (atoms for Ne and Ar)

to break

As you go across Periods 2 & 3, left to right, the melting/boiling points are higher for the metals (Li,

Be, Na, Mg & Al) due to the strong metallic bonding present between positive metal ions and

delocalised electrons in the giant metallic lattice. There is also an increase in melting points for the

metals each time you go across the period (e.g. Na → Mg → Al) because there is one more proton so a

greater nuclear charge and also more delocalised electrons with the next metal across the period. This

results in a stronger attraction between the positive metal ions and the delocalised electrons (stronger

metallic bonding) which then requires more energy to break for the melting/boiling point.

A giant metallic lattice

Across a period the nuclear charge and number of delocalised electrons increases in the metal resulting in a higher melting point

The melting point increases again as you move from the metals to the giant covalent lattices across

Periods 2 & 3. Covalent compounds usually exist as small molecules. However, C (in the form of

diamond and graphite), B, Si and SiO2 exist as a giant covalent lattice structure. A small cross section

of the giant covalent lattice of carbon in the form of diamond is shown below. To melt/boil a giant

covalent lattice even more energy is required because you have to break stronger covalent bonds to

achieve this.

Na+ Na+ Na+

– – –

Mg2+

– – – – – –

Mg2+ Mg2+ Al3+ Al3+ Al3+

– – – – – –

– – –

Na+ Na+ Na+

Na+ Na+ Na+

– – – delocalised

electrons

Positive

metal ions

Na+ Na+ Na+

– – – strong

metallic

bonding


There is then a sharp decrease in melting/boiling points between Groups 4 and 5, and the boiling points

for Groups 5-8 are relatively low. This is because the elements now exist as small covalently bonded

molecules between Groups 5-8, with only weaker temporary dipole-induced dipole forces of attraction

to break between these non-polar molecules. Look at the table above; you must know how the

molecules exist, e.g. chlorine exists as Cl2 and phosphorus as P4 molecules. The structure when you

have small molecules organised by attractive intermolecular forces is referred to as a simple covalent

lattice or simple molecular lattice.

Note that across Period 3, Groups 5-8, S8 has the highest boiling point followed by P4 followed by Cl2

and then Ar. This is because the strength of temporary dipole-induced dipole forces depends on the

number of electrons in the molecule. More electrons means larger temporary and induced dipoles

resulting in stronger temporary dipole-induced dipole forces between molecules which require more

energy to break.

MODEL ANSWER

Explain why Mg has a higher boiling point than P?

Mg exists as a giant metallic lattice with stronger metallic bonds to break.

P exists as P4 molecules in a simple molecular lattice with weaker temporary dipole-induced dipole

forces between the molecules.

The electrostatic attraction between positive metal ions and delocalised electrons in Mg is stronger

than the weaker temporary dipole-induced dipole forces between P4 molecules. Breaking the metallic

bonds in Mg requires more energy hence the higher boiling point.

Weak temporary

dipole-induced dipole

forces between

molecules to break

Strong covalent

bonds between C

atoms to break in the

giant covalent lattice

in diamond

Cl2 molecules:


You also need to know the trend in boiling points for the oxides of the elements in Periods 2 and 3. The

formulae of the oxides can be worked out using the cross-over method for the ionic compounds. Some

important properties of the oxides are also listed.

Period 2 Li2O BeO B2O3 CO or CO2

NO,

NO2,

N2O5

O3

Period 3 Na2O MgO Al2O3 SiO2 P4O6

P4O10

SO2,

SO3

Properties

of oxides

Metal oxides are ionic

compounds, existing in a

giant ionic lattice with

strong ionic bonds to

break – the strong

electrostatic force of

attraction between

oppositely charged ions.

They form alkaline

solutions. Al2O3 is

amphoteric.

SiO2 is giant covalent,

strong covalent bonds

to break. CO and

CO2 are simple

molecules with

weaker intermolecular

forces to break. CO

and CO2 are also

acidic oxides.

Simple molecular structures and

gases with weaker intermolecular

forces to break. Nitrogen forms a

range of oxides (acidic) with

different oxidation states. O2 and

O3 are allotropes. Ignore Groups 7

and 8.

2.11 Trends in melting and boiling points down a group

Melting points decrease down Groups 1 & 2 as there are more shells down a group meaning a larger radius

and more shielding, resulting in weaker forces of attraction between the particles.

The melting points increase as you go down group 7 however. This is because the Group 7 elements exist

as non-polar molecules and as you go down the Group the diatomic molecules have more electrons resulting

in larger temporary dipoles and therefore stronger temporary dipole-induced dipole forces to break between

the molecules which require more energy to break.




3. Orbital theory

3.1 Sub-shells and orbitals

So far we have assumed that electrons orbit the nucleus of an atom in a similar way the planets orbit the

sun. We have also assumed that two electrons are found in the first shell followed by eight electrons in the

next shells. This was a simplified model of the atom and is not the true picture for the structure of the atom!

However, this Bohr model that you have come across at GCSE and earlier in this course is a very useful

simplified model and is still widely used. You shall continue to use this model for the previous topics met

in this unit. We shall now look at the true model of the atom that was developed from the 1900’s. You

shall use this new model when writing out electronic structures for atoms or ions.

The actual number of electrons per shell can be found by using the formula 2n2, where n is the shell number.

Shell Electrons

1 2

2 8

3 18

4 32

Each shell is then broken down into sub-shells within which are found orbitals. There are four types of

orbitals; s, p, d & f. Electrons are found in these orbitals

Shell 1 is made up of the sub-shell 1s. Shell 2 is made up of the sub-shells 2s and 2p. The number for the

sub-shell tells you which shell you are in. The letter, s, p, d or f, tells you the type of orbital where the

electrons are located. Electrons are actually found in orbitals and each orbital can hold up to two electrons.

You should notice that in shell 1 there is only an s orbital. In shell 2, p orbitals come in. In shell 3, d

orbitals come in and in shell 4, f orbitals come in. The table below summarises how many of each type of

orbital is found in sub-shells.

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

Shell 1 contains the sub-shell 1s

Shell 3 contains the sub-shells 3s, 3p, 3d

Shell 2 contains the sub-shells 2s, 2p

Shell 4 contains the sub-shells 4s, 4p, 4d, 4f

Nucleus


Orbital: a region of space where there is a 95% probability of locating an electron. Each orbital can hold

a maximum of two electrons.

Type of

orbital

Shape Number of these

orbitals per sub-shell

Total number of electrons

held in all of these orbitals

s Sphere 1 1 x 2 = 2

p Dumb-bell

3 3 x 2 = 6

d

5 5 x 2 = 10

f

7 7 x 2 = 14

The table below summarises where the electrons are found in each shell in an atom.


s

s

s

p

p

d

2

6

2

2

6

6

10

10

14

8

18

32


3.2 Electron in box diagrams and electron configurations

In the Bohr model, we used dots or crosses to show electrons in shells. How do we show electrons in this

new model of the atom where there are also subshells and orbitals? We use boxes to represent the orbitals

and arrows to represent the electrons in the orbitals. Because p orbitals come in threes we use three boxes

for p orbitals. In a similar fashion, we use five boxes for d orbitals and seven boxes for f orbitals. Each

box (orbital) can hold up to two electrons. Electrons are negatively charged and repel each other; we show

this by having one arrow pointing up and one pointing down. For the same reason, each orbital is filled

singularly before pairing starts for p, d and f sub-shells.

Note that in the electron in box diagram below, the 4s sub-shell is filled with electrons before the 3d

subshell. This is because the sub-shells in the 3rd and 4th shell are very close together and they can overlap,

which results in the 4s sub-shell coming below the 3d sub-shell.

Simply writing out how many electrons appear in each subshell is called the electron configuration. This

is the most important point and summarises this topic; remember the following order for writing electron

configurations: 1s2 2s2 2p6 3s2 3p6 4s2 3d104p6. Be very careful when writing out electron configurations

for ions – remove electrons from the highest filled subshell first.

The electron in box diagram below is for Mn, which has an atomic number of 25 and so 25 electrons when

a neutral atom. Note how in the 3d sub-shell the orbitals are filled singularly before pairing starts.

4p Each orbital holds up to two electrons with

opposite spins. Show this with one arrow

pointing upwards and one pointing downwards

Remember to fill from the lowest sub-shell

upwards.

Each energy level must be full before the next one

higher up is filled (Aufbau principle)

Each orbital is filled singularly before pairing

starts.

The electron configuration of Mn is:

1s2 2s2 2p6 3s2 3p6 4s2 3d5 1s

3d

2s

2p

3s

3p

4s


WORKED EXAMPLE

1. Write out the electron configuration for P and P3-?

P has an atomic number of 15, so 15 electrons: 1s2 2s2 2p6 3s2 3p3

P3- has gained 3 electrons so now has 18 electrons: 1s2 2s2 2p6 3s2 3p6

2. Draw an electron in box diagram for P and explain how the subshells are filled with electrons?

3.3 Blocks in the periodic table

If you write out the electronic structure (using s, p, d notation) for any element in Groups 1 or 2, you’ll find

that the outer electron is always in a s orbital. Similarly, if you write out the electronic configuration for

any element in groups 3-8, you’ll find that the highest energy electron is always in a p-orbital. For this

reason, we can assign certain blocks of the Periodic Table as being either s, p or d blocks.

1 2 3 4 5 6 7 0

1 H

2

3

4

5

6

7

E.g. C is a p-block element because it has the electronic structure 1s22s22p2. The highest energy electron

is in a p orbital.

s d p


1s

2s

2p

3s

3p Lowest energy sub-shell is filled first.

Each energy level must be full before the next one

higher up is filled (Aufbau principle)

Each orbital is filled singularly before pairing

starts. Single pairing in the 3p for this example.



3.4 Ionisation energies re-visited

We have seen previously seen that there are trends in ionisation energies across a period; in general there

is an increase in first ionisation energy across a period. However, small decreases can also be seen when

going across a period from the graph above. For example, across Period 2 (Li → Ne), there is a decrease

between Be and B and also between N and O. These slight anomalies are linked to the filling of s and p

sub-shells, which shall be explained using the electron in box diagrams below.

Comparing Be and B:

Beryllium Boron

Comparing N and O:

Nitrogen Oxygen

1s

2s

2p

1s

2s

2p 2p is empty for Be Electron in 2p for B, which

is higher in energy and also

experiences shielding from

the 2s, so it is easier to

remove.

So B has a lower first

ionisation energy than Be

1s

2s

2p

1s

2s

2p 2p electrons

are unpaired

for N

A paired 2p electron in O is

removed. The repulsion

between the paired electrons

means it is easier to remove.

So O has a lower first

ionisation energy than N


4. Balanced Equations and Chemical Reactions

4.1 balancing equations

A chemical equation for a reaction is always written:

REACTANTS PRODUCTS

A chemical equation may also include the state symbols below:

(s) solid (l) liquid (g) gas (aq) aqueous

In a chemical reaction, atoms are never created or destroyed; an equal number of the same atoms must

appear on both sides of an equation. Equations are balanced by putting numbers before the formulae to

ensure that there is the same number of atoms of each element on both sides of the equation.

WORKED EXAMPLES

1. Write an equation for the reaction of hydrogen gas with nitrogen gas to make gaseous ammonia.

Nitrogen exists and N2, hydrogen as H2. Put these before the arrow with a small (g) to show they are

in the gaseous state. After the arrow write the formula for ammonia:

N2 (g) + H2 (g) → NH3 (g)

The equation must now be balanced. There are two nitrogen atoms on the left hand side of the equation

but only one on the right hand side. Putting a ‘2’ before NH3 (g) balances the number of nitrogen atoms.

NEVER CHANGE THE FORMULA OR ANY OF THE SUBSCRIPT NUMBERS, ONLY WRITE

NUMBERS IN FRONT OF A FORMULA! THE MULTIPLYING NUMBER MULTIPLIES

EVERYTHING THAT APPEARS IN THE FORMULA. IF YOU HAVE BRACKETS, THE

SUBSCRIPT APPLIES TO EVERYTHING INSIDE THE BRACKETS.

N2 (g) + H2 (g) → 2 NH3 (g)

There are now 2 x 3 = 6 H atoms on the right hand side of the equation, but only 2 H atoms on the left

hand side. Putting a ‘3’ before H2 (g) balances the hydrogen’s:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

There is now an equal number of each atom on each side of the equation; it is fully balanced.

2. Write an equation for the reaction between sodium and oxygen to form sodium oxide.

Sodium exists as Na, oxygen as O2. Sodium is a metal so must be a solid at room temperature, oxygen

is a gas. Sodium oxide is an ionic compound (which are usually solids) because it is made up of a metal

and a non-metal; use the cross-over method to work out its formula. ALWAYS DO THIS FOR IONIC

COMPOUNDS!

Na (s) + O2 (g) → Na2O (s)


Balance the oxygens on the right hand side of the equation:

Na (s) + O2 (g) → 2 Na2O (s)

The oxygens are balanced but there are now 2 x 2 = 4 Na atoms on the right but only one on the left.

Placing a number 4 before Na in the equation shall balance it:

4 Na (s) + O2 (g) → 2 Na2O (s)

4.2 Reactions of period 2 and 3 elements with oxygen

You have previously seen how as you move across periods 2 and 3, left to right, there is an increase in

melting/boiling points from Groups 1 to 4 followed by a sharp decrease with relatively low boiling/melting

points for elements in Groups 5-8. This is because you move from giant metallic lattice (stronger metallic

bonds to break) to giant covalent (stronger covalent bonds to break) to simple molecular lattice (weaker

intermolecular forces to break).

As you move across periods 2 and 3 from left to right, the products formed when these elements react with

oxygen tend to be giant ionic lattices to giant covalent structures to simple molecular structures (see the

table on the next page). You get giant ionic when a metal reacts with the non-metal O2 to form a metal

oxide. The bonding present is responsible for the properties of the products formed. The products made

change from solids with the giant ionic and giant covalent lattices to gases for the simple molecules. The

products made also change from being alkali (metal oxides are examples of bases) to amphoteric to acidic.



WHEN WRITING OUT EQUATIONS FOR THE FORMATION OF IONIC COMPOUNDS, YOU

MUST USE THE CROSS OVER METHOD TO ENSURE YOU HAVE THE CORRECT FORMULA

FOR THE IONIC PRODUCTS!

A summary table:

Period 2 Li2O BeO B2O3 CO or CO2

NO,

NO2,

N2O5

O3

Period 3 Na2O MgO Al2O3 SiO2 P4O6

P4O10

SO2,

SO3

Properties

of oxides

Metal oxides are ionic

compounds and form

alkaline solutions. Al2O3

is amphoteric.

CO from incomplete

combustion and CO2

from complete

combustion. SiO2 is

giant covalent.

Acidic oxides.

Simple molecular structures and

gases. Nitrogen forms a range of

oxides (acidic) with different

oxidation states. O2 and O3 are

allotropes. Ignore Groups 7 and 8.

Alkaline solution: a solution with a pH value above 7

Acidic solution: a solution with a pH value below 7

Amphoteric: a substance that can act as both an acid and a base

Oxidation: loss of electrons

Allotrope: two or more different physical forms that an element can exist in.

Si and SiO2 are examples of giant covalent lattices

Oxides are acidic



4.3 Reaction of metals with oxygen, water, and dilute acids

(i) Oxygen:

Metals react with the non-metal oxygen to form ionic metal oxides. The Group 1 and 2 metal oxides are

used as bases because they form alkali solutions. It is very important that you check the formula of the

metal oxide formed using the cross-over method and then fully balance the equation. When using the cross-

over method, the positive charge on the metal ion is the same as its Group number and the charge on the

oxide ion is -2. E.g:

4Na + O2 → 2Na2O (the general formula for Group 1 metals is 4M + O2 → 2M2O)

2Mg + O2 → 2MgO (the general formula for Group 2 metals is 2M + O2 → 2MO)

4Al + 3O2 → 2Al2O3 (the general formula for Group 3 metals is 4M + 3O2 → 2M2O3)

Further reading:

Group 1 metals react rapidly with oxygen and are stored under oil to prevent contact with air and the more

reactive Group 1 metals are stored in small sealed glass tubes. Be and Al form BeO and Al2O3 coatings

respectively, which makes them resistant to further oxidation and makes them behave as unreactive metals.

The Group 4 metals lead (Pb) and Tin (Sn) also form oxides. The transition metals are much less reactive

with oxygen than Group 1 or 2 metals. When transition metals (also called d-block metals) react with

oxygen the oxides are usually brittle; for example, when iron reacts with oxygen to form rust (iron oxide).

The transition metals form a range of oxides with different oxidation states. Some d-block metals such as

titanium are resistant to corrosion because they quickly form an outer unreactive oxide layer which

prevents any further oxidation.

We have so far assumed that oxygen only forms the oxide ion, O2-. However, the peroxide ion, O22- and the

super-oxide ion O2- also exist. These molecular ions contain a covalent bond between the two O atoms; O-

O2- and O-O-. These two negative molecular ions are only stable next to a large positive cation, so they

only form compounds with metals the further down a Group you go where the larger positive metals ions

are found.

For example, Li is a small positive ion and when next to a large negative ion such as O2- or O2-, the electrons

in the covalent bond between the two O atoms are strongly attracted to the small positive metal ion; the

ionic bond becomes polarised. This results in the covalent bond between the oxygen atoms breaking.

However, as you go down Group 1, you shall find peroxides are formed such as Na2O2 and K2O2 because

the positive metal ions become larger and therefore the ionic bond is less polarised. Further down the

Group super-oxides are formed such as KO2, RbO2 and CsO2.


(ii) Water:

Metals react with water to form solutions of a metal hydroxides as well as hydrogen gas. The hydroxide

molecular ion exists as OH-. Once again ensure you use the cross-over method when working out the

formula for the metal hydroxide. Where there is more than one hydroxide molecular ion, use brackets

around it with the small multiplying number outside the brackets. E.g.:

2Na + 2H2O → 2NaOH + H2 (general formula for Group 1 metals: 2M + 2H2O → 2MOH + H2)

Mg + 2H2O → Mg(OH)2 + H2 (general formula for Group 2 metals: M + 2H2O → M(OH)2 + H2)

Note that when in solution, the hydroxides exist as separate M+/M2+ and OH- ions. The hydroxide ions are

responsible for the alkalinity of the solutions formed. Group 1 metals are called the alkali metals because

they form a basic solution when reacted with water. The reactivity of Group 1 and 2 metals inceases down

the Group. This can be explained by CARS. In Group 2, Be does not react with water, Mg only reacts with

steam but the metals further down the Group react increasingly easier with water. Group 3 metals are not

very reactive with water and aluminium appears not to react at all due to the outer oxide layer. Group 4, 5

and 6 metals do not react with water. Some transition metals do react with water but only very slowly.

(iii) Dilute HCl and H2SO4

In a similar reaction to that with water, metals (those above copper in reactivity series) react with dilute

acids to form salts in a neutralisation reaction as well as H2 gas. Chloride salts form with HCl and sulfate

salts form with sulfuric acid. The sulfate molecular ion exists as SO42-. Once again remember to use the

cross-over method when working out the formula of the ionic salt. Note that the reaction of Ca, Sr and Ba

with H2SO4 leads to a protective sulfate layer that is insoluble, preventing them from reacting any further.

2Na + 2HCl → 2NaCl + H2

2Na + H2SO4 → Na2SO4 + H2

Mg + 2HCl → MgCl2 + H2

Mg + H2SO4 → MgSO4 + H2



4.4 Redox

Oxidation numbers (not to be confused with ionic charge) are a measure of the number of electrons that an

atom uses to bond with atoms of a different element. We can assign oxidation numbers to both ionic and

covalent compounds. The rules, which you shall use when looking at redox reactions, are listed below.

The sign of the oxidation number must be placed before the number.

Element Ox. No. Examples Exceptions

Un-combined element 0 O2, H2, Fe, S8

Combined H +1 H2O, NH3 -1 in metal hydrides

e.g. NaH, CaH2

Combined O -2 H2O, CaO

-1 in peroxides

e.g. H-O-O-H

+2 when bonded to F

e.g. F2O

Combined F -1 HF

Ions Charge on ion Na+ = +1; Mg2+ = +2

The sum of the oxidation numbers for each atom must equal the overall charge

Assign oxidation numbers to each individual atom that is in the molecule/ion when looking at redox

reactions. All of the oxidation numbers when added together must equal any charge on an ion.

Before we look at redox reactions, you need to know the difference between oxidation and reduction.

Oxidation: the loss of electrons (resulting in an increase in oxidation number)

Reduction: the gain of electrons (resulting in a decrease in oxidation number)

Remember this using: OILRIG

X3+ X2+ X1+ X0 X1- X2- X3-

Now consider the oxidation numbers in the following reaction:

2 Fe + 3 Cl2 → 2 FeCl3

Fe has gone from Fe = 0 to Fe = +3; it has been oxidised.

Cl has gone from Cl = 0 to Cl = -1; it has been reduced.

A reaction in which both oxidation and reduction is taking place is called a REDOX reaction.

redox reaction: one which involves both oxidation and reduction.

X is losing electrons – it is being oxidised

X is gaining electrons – it is being reduced


In the above reaction, Fe is a reducing agent; it reduced Cl (Fe itself was oxidised).

Cl is an oxidising agent; it oxidised Fe (Cl itself was reduced).

WORKED EXAMPLE

Assign oxidation numbers to the equation below and prove that it is a REDOX reaction.

Sr + 2 HCl → SrCl2 + H2

Oxidation state for Sr: un-combined element, so 0.

Oxidation state for H in HCl: combined H, so +1.

Oxidation state for Cl in HCl: sum of oxidation numbers must equal the charge. The charge is 0, and

the oxidation state of H is +1. So Cl must be -1 as +1-1 = 0.

Oxidation state for Sr in SrCl2: SrCl2 is an ionic compound (mixture of metal with non-metal). The

charge on any ion is its oxidation number. Sr is in Group 2 so forms 2+ ions. Oxidation state is +2.

Oxidation state of Cl in SrCl2: ionic compound and each Cl has a 1- charge. Oxidation numbers must

be assigned to each individual ion. Also the sum of the oxidation numbers equals the overall charge of

0. Cl is -1.

Oxidation state of H in H2: un-combined element so 0.

The oxidation numbers should be written above the equation as seen below. Either state what has been

oxidised and reduced or use arrows such as in the diagram below. Careful when you have more than one

of the same element in a compound as is the case with Cl in SrCl2. You must not write -2 above the ‘Cl2’

part. Oxidation numbers are assigned to EACH individual atom.

-1

0 +1 -1 +2 -1 0

Sr + 2 HCl → SrCl2 + H2

Sr has gone from Sr = 0 to Sr = +2, so it has been oxidised

H has gone from H = +1 to H = 0, so it has been reduced

Both oxidation and reduction have taken place in this reaction so it is a REDOX reaction

Oxidation also means the gain of oxygen. Questions may ask which substance in an equation has been

oxidised or reduced. In the equation below, CO has been oxidised to CO2.

Fe2O3 + 3 CO → 2 Fe + 3 CO2


Oxidised

Reduced


4.5 Oxidation numbers for transition metals and oxyanions

Transition elements can have variable oxidation numbers, e.g. Fe can exist with oxidation numbers of +2

and +3. To differentiate between the different states of +2 and +3 in Fe, the oxidation number of the

transition element is given as a roman numeral in brackets. The variable oxidation state is what makes the

transition metals useful as catalysts in many important industrial reactions.

1 2 3 4 5 6 7 8 9 10 I II III IV V VI VII VIII IX X

Formula Name Oxidation no. of

transition element

FeCl2 Iron(II) chloride Fe = +2

FeCl3 Iron(III) chloride Fe = +3

CuO Copper(II) oxide Cu = +2

Cu2O Copper(I) oxide Cu = +1

Oxyanions are:

negative molecular ions

that contain oxygen

combined with a second element

e.g. SO42-, CO3

2-, NO3-. The name of the oxyanion usually ends in –ate (e.g calcium sulphide, CaS vs

calcium sulfate, CaSO4).

The second element has several oxidation states, with the oxidation number of the second element given in

brackets after the name of the oxyanion.

The oxidation numbers given in brackets after the name of the oxyanion allows us to distinguish between

similar molecular ions. For example, SO42- is not the only oxyanion called sulfate; SO3

2- is another sulfate

ion. The oxidation number of the second element, in this case sulfur, written after the name allows us to

distinguish between the two; SO42- is referred to as sulfate(VI) and SO3

2- is referred to as sulfate(IV)!

Formula Name Oxidation no. of

second element

SO42- Sulfate(VI) S = +6

SO32- Sulfate(IV) S = +4

NO3- Nitrate(V) N = +5

NO2- Nitrate (III) N = +3

KMnO4 (K+ MnO4

-) Potassium manganate(VII) Mn = +7

NaNO3 Sodium nitrate(V) N = +5

NaNO2 Sodium nitrate (III) N = +3

CaSO4 Calcium sulfate (VI) S = +6



4.6 Reactivity series

The reactivity series lists metals in order of reactivity with oxygen, water and acids. The most reactive

metals are at the top of the list. Most reactivity series have potassium at the top as it is very reactive and in

Group 1 and gold and platinum at the bottom as these transition metals are so unreactive. It is useful to

place carbon and hydrogen in the list. Carbon is used to extract (or displace) metals from their ores; only

metals above carbon in the list can be extracted from their ores with the more reactive carbon. In a similar

way, only metals above hydrogen will react with acids or water to displace the hydrogen. This leads us

onto the topic of displacement reactions, which also tend to be redox reactions. We shall look at redox in

the next section before moving on to displacement reactions.

The order of reactivity for the metals is Group1, Group2, Group3, Group4, Transition metals.

The metals higher up are more reactive as they have a greater tendency to lose an electron to form a

complete outer shell. In general, for the metals, reactivity decreases across a period and increases down a

Group. Across a period nuclear charge and attraction increase, making it more difficult to lose an electron

during a chemical reaction, but down a Group nuclear attraction decreases making it easier to lose an

electron during a chemical reaction. The full explanation is CARS!



4.7 Displacement reactions

A more reactive metal shall displace a less reactive metal in a metal salt solution. You can predict when a

metal can displace another metal from its salt by using the reactivity series. Displacement reactions

involving metals are also usually redox reactions – you can prove this by assigning oxidation numbers to

each element in the equation. E.g. the displacement reaction below is a redox reaction:

0 +2 +6 -2 +2 +6 -2 0

Fe (s) + CuSO4 (aq) → FeSO4 (aq) + Cu (s)

Displacement reactions can also happen with the Halogens. The halogens become more reactive as you go

up the Group. A more reactive halogen shall displace a less reactive halogen from its metal salt:

Cl2 + 2 KBr → Br2 + 2 KCl

Note that the potassium ion does not actually take part in the reaction – it is a spectator ion. We can simplify

the equation by removing the potassium so we can see more clearly what is happening to the electrons in

the reaction:

Cl2 + 2 Br- → Br2 + 2 Cl-

The halogens are oxidising agents – this means that they oxidise (remove electrons from) another species.

In the above reaction, chlorine is a stronger oxidising agent than bromine, so it removes electrons from the

bromide ions to form the chloride ions.

4.8 Uses and applications of the substances in this unit

You may be asked for the uses of some of the substances that you have come across in this unit. Research

a few substances and summarise their uses below.


Oxidised

Reduced


5. Quantitative Chemistry

5.1 Moles and masses

Mole: a unit of substance equivalent to the number of atoms in 12g of carbon-12. 1 mole of a compound

has a mass equal to its relative atomic mass expressed in grams.

Molar mass: the mass of one mole of a substance.

We previously introduced the concept of one mole; one mole of molecules is 6.023 x 1023 molecules.

Where has this number come from? We have also previously seen how the isotope 12C is used as the

standard for weighing atoms when we introduced the idea of relative atomic mass, which is the average

mass of an atom compared with 1/12th of 12C.

If you weigh out exactly 12 grams of the isotope 12C, you shall have 6.023 x 1023 atoms – this number was

then used as the quantity for one mole of anything. It is also called Avogadro’s constant.

We can also introduce the idea of molar mass, which is the mass in grams of one mole of any substance. It

has units of g mol-1 (grams per one mole). To work out how many grams of any element you need to have

one mole, simply find it relative atomic mass. For a compound, it is the same as the relative formula mass

or relative molecular mass.

The molar mass for 12C is 12 gmol-1 (12 grams are required for you to have one mole, or 6.02 x 1023 atoms).

The molar mass for CO2 is 44 gmol-1 (44 grams are required for you to have one mole, or 6.02 x 1023

molecules.

The molar mass (R), the number of moles (M) and the mass in grams of the substance are related by the

following triangle:

amount in Moles = mass in Grams / Relative molar mass

WORKED EXAMPLE

1. What is the mass of 0.8 moles of CO2?

We have moles, M = 0.8 and the molar mass, R, of CO2 = 12 + (16 x 2) = 44

Using the above triangle, mass in grams, G = M x R

= 0.8 x 44 = 35.2 g

G = mass of compound (g)

M = number of moles of compound (mol)

R = relative molar mass of compound (g mol-1)


2. Calculate the number of moles of NaOH in 0.2 g of the solid?

We have G = 0.2 and R of NaOH = 23 + 16 + 1 = 40

Using the above triangle, moles, M = G / R

= 0.2 / 40 = 0.005 mol

5.2 Moles and solutions

The concentration of a solution tells you how much solute (solid) is dissolved in the solvent. Concentrations

are measured in moles per cubic decimetre (units mol dm-3). A decimetre is the same as one litre. To go

from cm3 to dm3, divide by 1000. To go from dm3 to cm3, multiply by 1000.

So a solution of NaOH with concentration of 1 mol dm-3 (sometimes written as 1 M) contains 1 mole of

NaOH dissolved per 1 dm3 of the solution ( = 1 mole NaOH per 1000 cm3).

A 0.5 mol dm-3 NaOH solution contains 0.5 moles of NaOH in every 1 dm3.

2 dm3 of a 0.5 mol dm-3 NaOH solution contains1.0 mol of dissolved NaOH.

The amount in moles (M), the concentration (C) and the volume of a solution (V) are linked via the

following equation.

amount in Moles = Concentration of solution (in mol dm-3) x Volume of solution (in dm3)

M = amount of solute (mol)

C = concentration of solution (mol dm-3)

V = the volume of solution (dm3)



WORKED EXAMPLES

1. Calculate the amount, in mol, of HCl in 100 cm3 of a solution with a concentration of 0.005 mol dm-3.

Using the above triangle, M = C x V (in dm3)

= 0.005 x 0.1 = 0.0005 mol

2. Calculate the concentration of a NaOH solution when 0.025 mol of NaOH is dissolved in 250 cm3 of

water.

Using the above triangle, C = M / V (in dm3)

= 0.025 / 0.25 = 0.1 mol dm-3

5.3 Moles and equations

We shall now link the number of moles to balanced chemical equations. We can use chemical equations

to work out the masses of the products we expect to make. Looking at a chemical equation, we do not think

about how many molecules are reacting but rather how many moles are reacting.

The balancing numbers in a balanced equation gives you the ratio of moles reacting in a given reaction.

Stoichiometry is the molar relationship between the relative quantities of substances taking part in a

reaction. Take for example the following equation:

4 Na + O2 → 2 Na2O

We can use these molar ratios to calculate the number of moles of reactants needed or the number of moles

of products formed, e.g., using the molar ratios in the above balanced equation;

2 moles of Na reacts with 0.5 moles of O2 to form 1.0 mole of Na2O.

4 moles of Na2O can be formed by reacting 2 moles of O2 with 8.0 moles Na.

One mole of Na was reacted with excess oxygen to form 0.5 moles of Na2O.

Stoichiometry is usually used together with the two triangles met earlier to help you work out quantities in

chemical reactions. You shall be using this idea in unit 2 when you perform titrations. In a titration, a

standard solution (who’s concentration you know) is used to determine the concentration of a second

solution where the concentration is unknown.

4 moles of Na reacts with ... 1 mole of O2 molecules ... to form 2 moles of Na2O



WORKED EXAMPLES

1. In a reaction, 0.14 g of calcium oxide, CaO, is reacted with excess hydrochloric acid.

CaO + 2 HCl → CaCl2 + H2O

(i) Calculate the amount, in mol, of CaO used in the reaction.

For CaO, we have G = 0.14 and R = 40.1 + 16 = 56.1

So we can calculate M = G / R

= 0.14 / 56.1 = 0.0025 mol

(ii) Determine the amount, in mol, of CaCl2 produced in the reaction.

From the stoichiometry in the equation, 1 mole of CaO produces 1 mole of CaCl2.

So 0.0025 mol CaO would also produce 0.0025 mol of CaCl2.

(iii) Calculate the mass, in grams, of CaCl2 produced in the reaction.

For CaCl2, we now have M = 0.0025 and R = 40.1 + (35.5 x 2) =111.1

So we can calculate G for CaCl2 = M x R

= 0.0025 x 111.1 = 0.2778 g

2. In an experiment, 0.10 g of Li metal was added to 200 cm3 of water to make a solution of LiOH.

2 Li + 2 H2O → 2 LiOH + H2

(i) Calculate the amount, in mol, of Li metal used in this reaction.

For Li, G = 0.10, R = 6.9

So we can calculate M = G / R

= 0.10 / 6.9 = 0.0145 mol

(ii) Calculate the concentration of the LiOH solution formed, in mol dm-3.

Molar ratio of Li to LiOH in the equation is 2:2 = 1:1, so M = 0.0145

C = M / V = 0.0145 / 0.2 = 0.0725 mol dm-3



5.4 Percentage yield

Once you work out the moles of one reactant in a chemical equation, you can calculate the moles of product

expected (also called the theoretical moles) using the molar ratios in the chemical equation. Once you know

the moles of product expected you can then calculate the mass in grams of product expected (also referred

to as the theoretical mass) using GMR.

However, you rarely produce the expected (or the theoretical) mass of product when you do a chemical

reaction. Some of the product could have been lost when transferring the product from one vessel to

another. Some of the reactant chemical may react with impurities. The reaction may be also be reversible.

Chemists need to know how efficient their reaction process is so they calculate the percentage yield. You

can calculate the percentage yield based on the number of moles or the mass in grams.

% yield = actual no. of moles x 100

expected no. of moles

% yield = actual mass x 100

expected mass

WORKED EXAMPLES

1. We are expecting 10g of product from a reaction but only produce 8g. What % is this?

% yield = actual mass x 100 = 8 x 100 = 80%

expected mass 10

2. We are expecting 0.5mol of product but only produce 0.2mol. What % is this?

% yield = actual no. of moles x 100 = 0.2 x 100 = 40%

expected no. of moles 0.5

3. 2.3 g of ethanol is oxidised to form 2.4 g of ethanoic acid. Calculate the percentage yield.

C2H5OH + 2 [O] → CH3COOH + H2O

Moles of ethanol used = G / R

= 2.3 / 46 = 0.05 mol

Expected moles of ethanoic acid = 0.05 mol also (1:1 molar ratio in equation)

Actual moles of ethanoic acid produced = G / R

= 2.4 / 60 = 0.04 mol

Percentage yield = (0.04 / 0.05) x 100 = 80%


Extension: what is empirical

formula and how do you calculate

this using moles?


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