UNIT 1 - HCC Applied Science -...

Level 3 Applied Science 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

1.1 Introduction to the atom 3 1

1.2 Elemental symbols in the Periodic Table 4 2

1.3 Introduction to the Periodic Table of elements 5 3

1.4 Making ions 8 4

1.5 Trends in ionic radius 10 5

1.6 Ionisation energy 11 6

1.7 Electron affinity 13 7

2. Compounds, Bonding & Intermolecular Forces

2.1 Introduction to bonding in compounds 15 8

2.2 Ionic bonding 17 9

2.3 Covalent bonding 19 10

2.4 Metallic bonding 21 11

2.5 Electronegativity 22 12

2.6 Introduction to intermolecular forces 24 13

2.7 Permanent dipole-dipole forces 25 13

2.8 Temporary dipole-induced dipole forces 25 13

2.9 Hydrogen bonding 26 14

2.10 Trends in melting and boiling points across a period 27 15

2.11 Trends in melting and boiling points down a group 29 15

3. Orbital Theory

3.1 Sub-shells and orbitals 30 16

3.2 Electron in box diagrams and electron configurations 32 16

3.3 Blocks in the periodic table 34 17

3.4 Ionisation energies re-visited 35 17

4. Balanced Equations & Chemical Reactions

4.1 balancing equations 36 18

4.2 Reactions of period 2 and 3 elements with oxygen 37 19

4.3 Reaction of metals with oxygen, water, and dilute acids 39 20

4.4 Redox 41 21

4.5 Oxidation numbers for transition metals and oxyanions 43 22

4.6 Reactivity series 44 23

4.7 Displacement reactions 45 23

4.8 Uses and applications of the substances in this unit 45

5. Quantitative Chemistry

5.1 Moles and masses 46 24

5.2 Moles and solutions 47 25

5.3 Moles and equations 48 26

5.4 Percentage yield 50 27

Questions 51

Checklist 54

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


1. The Periodic Table, Atoms & Ions

1.1 Introduction to the atom

The atom consists of a central nucleus where p__________ (positively charged) and neutrons (neutral) are

found. In s__________ around the nucleus are electrons (negatively charged). Relatively, protons and

neutrons have a s__________ mass. So if we say the mass of a proton is 1 then the mass of a neutron is

also __________. Relative to protons and neutrons, the mass of an electron is so small that it is negligible.

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

Particle Relative mass Relative charge Position in atom

Proton 1

Neutron

Electron

Most of the mass is found in the n__________ of the atom. Note that we use dots (or crosses) to show

electrons in shells. Up to t__________ electrons are found in the first shell outside the nucleus and

e__________ 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 e__________ 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).

+ +

+ +

+ +

+

Complete questions 1-2


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

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

e__________ are present in a neutral atom. The number of neutrons can be found by s__________ the

atomic number from the mass number. Nitrogen has __________ - __________ = __________ 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 i__________ of chlorine are found; 35Cl and 37Cl. Both i__________ 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 i__________ of the same element is the number of n__________ and

therefore the m__________ 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 e__________.

Chemical reactivity depends on e__________ movement.

You might think to yourself that the mean of 35 and 37 is (35 + 37) / 2 = __________. 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 m__________ 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 p__________ 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 s__________ for weighing atomic masses. Atomic masses are weighed r__________ to 1/12th of

12C. The relative atomic masses do not have any units because they are only relative numbers.

Examples for calculating the number of sub-atomic particles when given pure isotopes:

Pure

Isotope

Atomic no.

(protons)

Mass no. (protons

+ neutrons) No. of electrons No. of neutrons

R.A.M in periodic

table

16O

32S

10B

24Mg

1.3 Introduction to the Periodic Table of elements

Identify the groups and periods. Remember that period 1 contains the elements H and He and is not the

row from Li to Ne (a common mistake)! Also identify how you can find the metals and non-metals.

Period __________

Period __________

Period __________

Period __________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Gro

up _

_________

Complete questions 3-5 Research how you calculate

relative atomic mass, with at least

one example calculation shown

+


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

Elements are organised into vertical columns called g__________ and horizontal rows called p__________.

Period 1 contains the elements H and H__________. Chemical properties are similar for elements that are

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

The atomic number i__________ as you move from l__________ to r__________ across a period because

each element has o__________ more proton than the element before it in the same period. There are trends

(or patterns) that r__________ 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-__________ and the atomic radius also

decreases. This repeating pattern seen by the elements across a period is called p__________.

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

The diagram below shows how the atomic radius d__________ across a period but i__________ down a

group. Add the missing electrons to the electron shells using the atomic numbers for the neutral atoms.

To explain trends in atomic radius (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:

__________ C__________

__________ A__________

R__________

S__________

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

11Na


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

by o__________ each time, so the nuclear charge increases. 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 c__________ to the nucleus. Because the electron shells

are closer to the nucleus this means that the atomic radius has d__________ across the period. 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 r__________ 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 s__________ 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).

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

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

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

with the larger radius, so there is w__________ 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

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

e.g. MODEL ANSWER: explain the difference in atomic radius between C and N.

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e.g. MODEL ANSWER: explain the difference in atomic radius between Mg and Ca.

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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 p__________ protons e__________ the

number of n__________ electrons. However, atoms can gain or lose electrons during chemical reactions

to form charged particles called i__________ in order to satisfy the octet rule. The octet rule states that

elements gain or lose electrons in order to have e__________ 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 o__________ negative electron to leave

behind a positive charge; Cl- has g__________ o__________ negative electron and therefore has a negative

charge; N3- has gained t__________ negative electrons.

Below are examples of ions – careful when working out the number of electrons!

Atom/Ion Atomic no. Mass no. No. of protons No. of neutrons No. of electrons

14N3-

16O2-

23Na+

27Al3+

We can predict the charge that an ion of a given element shall form by looking at its p__________ 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 o__________ electron in its

outer shell. To have a c__________ outer shell containing e__________ electrons, the sodium atom must

l__________ this one outer electron (electrons are negatively charged), which would result in a sodium

i__________ with a charge of +1 (written as Na+). All group 1 elements form +1 ions for the same reason.

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

Gro

up

1

Gro

up

2

Gro

up

3

Gro

up

4

Gro

up

5

Gro

up

6

Gro

up

7

Gro

up

0

+2 common

but varied

Transition metals


Note that group 4 elements do not usually form ions; they have f__________ 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 e__________ electrons in their outer-most shell and are very

s__________ atoms so they do not form ions.

The ions below are missing charges – predict the charges expected – remember to always check the position

of the element in the Periodic Table!

Mg2+ F N Al P Br Sr O Cs K

Li Cl S Ca Rb Ba I Cu Fe Mn

Draw diagrams to show the formation of the following ions:

i) Mg to Mg2+

ii) K to K+

iii) O to O2-



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 same patterns also exist with the cations/anions. But you must also know the comparison of

the radius of a neutral atom to the radius of its cation/anion across a period.

Down a group for cations and anions: the same trend is seen as with the neutral atoms; as more electron

shells are added down a group the ionic radius i__________.

Across a period for cations: the same trend is seen as with neutral atoms; across a period the radius of a

cation d__________. The cations have l__________ electrons to have the positive charge and across a period

there is a greater nuclear charge and ionic charge on the cation, attracting the remaining electrons even

more s__________ resulting in the cations becoming smaller.

But you must also know that the ionic radius of a cation compared with the neutral atom is s__________

because there is still the s__________ number of protons in the nucleus of that atom now attracting

l__________ electrons (electrons were removed to form the cation).

Note that across a period the cations are isoelectronic – they have the s__________ number of electrons.

Prove this by completing the electronic structures of the first three cations in period 3 of the Periodic Table:

Neutral atom electron structure Cation electron structure

Na: Na+:

Mg: Mg2+:

Al: Al3+:

Isoelectronic: having the same number of electrons

Na Mg Al

Na+ Mg2+ Al3+

Neutral atom

Cation (smaller)


Across a period for anions: the same trend is seen as with neutral atoms; across a period the radius of an

anion d__________. The anions have g__________ electrons to complete the octet with a negative overall

charge, but with the extra protons across the period radius still decreases.

But you must also know that anions have a l__________ radius than the corresponding neutral atom because

the added electron(s) when you form the anion cause extra repulsion.

Across a period the anions are also isoelectronic – once again prove this by completing the electronic

structures of the first three anions in period 3 of the Periodic Table in the table below.

Neutral atom electron structure Anion electron structure

P: P3-:

S: S2-:

Cl: Cl-:

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: 1000000

One billion: 1000000000

One mole: 602300000000000000000000 or 6.023 x 1023 in standard form

P S Cl

P3- S2- Cl-

Neutral atom

Anion (larger)



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

electrons from one mole of atoms in the g__________ phase. We can write equations to show the process.

Removing an electron leaves a p__________ 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 g__________ phase. The p__________ sign before the

numerical value shows that energy is r__________ for this process; energy is required to break the

a__________ between the electron and the positively charged n__________.

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 n__________ 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 +__________

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

C__________ 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 i__________ (there are some anomalies e.g. between Be

and B and also N and O – the reason for this shall be explained in the later section on orbitals). Down a

group the ionisation energy d__________.


e.g. MODEL ANSWER: explain why the first ionisation energy for F (1680 kJmol-1) is greater than the

first ionisation energy for C (1090 kJmol-1).

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e.g. MODEL ANSWER: explain why the first ionisation energy for K (418 kJmol-1) is less than the first

ionisation energy for Na (494 kJmol-1).

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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

r__________ (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 be fully explained using the

pneumonic C__________.

Across a period the electron affinity values i__________ as more energy is r__________ when nuclear

attraction for the electron is s__________. Down a group when nuclear attraction for the electron to be

captured is w__________ less energy is r__________. 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 d__________ a group. This is because these atoms are very s__________ and placing

an extra electron in a c__________ area is difficult and there is significant r__________. This repulsion

lessens the attraction for the incoming electron resulting in a l__________ 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 s__________ electron affinity to form the 2- ions. Adding a

second electron to the 1- ion shall r__________ energy to overcome the r__________ force, as this time the

negative electron to be captured shall be r__________ by the negative 1- ion. The energy change shall

therefore be p__________ 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)

e.g. MODEL ANSWER: explain why group 6 elements have a lower electron affinity than group 7 elements

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e.g. MODEL ANSWER: explain why less energy is released for electron affinities down a Group.

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2.1 Introduction to bonding in compounds

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

c__________ and m__________. 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 – show these on the Periodic Table below:

Ionic bonding occurs between m__________ 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.

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-)?

Al3+ O2- → Al2O3

Ca2+ Cl1- → CaCl2

Mg2+ O2- → Mg1 O1 → MgO

Al3+ CO32- → Al2(CO3)3

*Use brackets for

molecular ions


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

metals come together to s__________ 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 o__________ of the non-metals (dative covalent bond). Examples

of covalently bonded molecules are O2, F2 and NH3.

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

mixture of metals (an alloy). The negative d__________ electrons are attracted to the p__________

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

The octet rule states that atoms with e__________ electrons in their outer-most shell are extremely

s__________ (apart from the first shell, where two electrons are required). The Noble gases (group 8

elements) exist as single a__________ 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:

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

- T__________ 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 c__________ to represent the outermost shell only.

(ii) Use d__________ to represent the electrons from one of the elements and c__________ 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 o__________ circles and put the electrons being shared in the

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

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

and the ionic c__________ 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.


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 m__________ and non-metals bonded together. The

Metal l__________ electrons to form a positively charged ion and the non-metal g__________ electrons to

form n__________ ions. After electrons have been transferred to give two ions with a stable octet these

oppositely charged ions then a__________ 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.

Name of ionic

compound

Sodium fluoride Magnesium

fluoride

Aluminium

chloride

Magnesium

phosphide

Cross-over

working

Formula of

ionic compound

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 30


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.

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.

Draw dot & cross diagrams for the following and also calculate the relative formula mass for each

compound given below.

(i) sodium fluoride, NaF

R.F.M ……………

(ii) magnesium chloride, MgCl2

R.F.M ……………

(iii) aluminium oxide, Al2O3

R.F.M ……………

(iv) Barium oxide, BaO

R.F.M ……………

(v) Calcium sulphide, CaS

R.F.M ……………

(vi) Barium nitride, Ba3N2

R.F.M ……………


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

charged ion in all d__________ resulting in a giant ionic l__________, which is a regular arrangement of

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

Finally, the strength of the electrostatic force and therefore the i__________ bond depends on:

(i) the ionic charge: a bigger ionic charge results in a s__________ 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 l__________ surface area, resulting in a w__________ attraction for the oppositely charged ion

compared with a smaller radius.

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 p__________

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 s__________ bond, two pairs of electrons a d__________ bond and three pairs of electrons a t__________

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

m__________. ‘Relative formula mass’ was used for ionic compounds which do n__________ exist as

molecules.

Where b__________ electrons have come from the s__________ atom, a d__________ covalent bond is

formed. An example is in the reaction of NH3 with a H+ ion to form the molecular ion NH4+.

Na+ Na+ Na+

Na+ Na+ Na+

Na+ Na+ Na+

Na+

Na+ Na+

Na+

Cl – Na+ Cl –

Cl –

Cl –

Draw in an exam as:



Draw dot & cross diagrams for the following and also calculate the relative molecular mass (covalent

compounds exit as MOLECULES) for each compound given below. Remember to include all lone pairs.

(i) hydrogen, H2

R.M.M ……………

(ii) chlorine, Cl2

R.M.M ……………

(iii) methane, CH4

R.M.M ……………

(iv) oxygen, O2

R.M.M ……………

(v) nitrogen, N2

R.M.M ……………

(vi) H3O+ (formed from the reaction of H2O with H+)

R.F.M ……………

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.

→


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 t__________ with a bond

angle of __________0.

Methane propane

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

__________ than triple bonds. The shorter the bond the s__________ it is and therefore would require

m__________ 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 m__________ only. Metals exist in a giant metallic lattice, which is a 3-D

structure of p__________ metal ions surrounded by negative d__________ electrons. There is a very strong

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

bonding is s__________. 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

Complete questions 33-36 Find the formula of the other

molecular ions: hydroxide, nitrate,

sulfate, carbonate

+


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

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

also a__________ heat energy, which gives them kinetic energy, hence metals are also good t__________

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 h__________ into shape without breaking.

(ii) ductile – they can be hammered thin or s__________ out into wires without breaking.

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

r__________ over each other without breaking the metallic bonding. Aluminium is very malleable and

along with its thermal conductivity makes it suitable for use in aluminium foil.

Diagram showing how metals are malleable/ductile:

2.5 Electronegativity

Consider a H2 molecule. Both atoms are identical and each has an e__________ 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 m__________ strongly than the other atom (and therefore the electron pair will be

closer to this atom). We say that the atom with the greater a__________ for the pair of electrons in the

covalent bond is more e__________ 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 c__________ 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

d__________, which we show with a δ- and a δ+. These are p__________ 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

c__________ 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 d__________ in

electronegativity, then we say that the bond is p__________. 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 r__________ 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 i__________ across a period left to right and also up a g__________. F is the most

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

Group 0 elements exist as a__________ not molecules. Show any dipoles on the examples below and state

whether the bond is polar or non-polar:

H – F

C – Cl

O – H

S – F

N – H

H – Cl

C – O

N – O

Cl – F

Br – Br

Increasing electronegativity

Incr

easi

ng e

lect

ron

egati

vit

y

δ- δ+


You can use the pneumonic C__________ to help you remember the factors that affect the electronegativity

of an element. Electronegativity depends on the number of p__________ in the nucleus, the d__________

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

s__________ between both atoms then a c__________ bond forms. As the difference in electronegativity

increases the covalent bond will become more p__________. If the difference in electronegativity is very

large then the bond becomes i__________ 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 w__________ attractive forces (compared

with covalent and ionic bonds) that exist b__________ the molecules. These are known as intermolecular

forces or v__________ d__________ w__________ 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

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

can a__________ the opposite p__________ 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 d__________ in electronegativity, no

p__________ dipoles and therefore no permanent dipole-dipole force between neighbouring m__________.

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

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

lead to a t__________ 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

i__________ a neighbouring molecule into also forming a temporary dipole; these temporary dipoles

a__________ each other leading to a temporary dipole-induced dipole force of attraction (also called

L__________ d__________ forces). London dispersion forces exist in all molecules, whether polar or non-

polar. A non-polar molecule will o__________ 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 i__________ with an increase in the number of electrons. More

electrons results in l__________ temporary and induced dipoles which results in s__________ attractive

London dispersion forces between the molecules. Stronger attractive forces between the molecules results

in an increase in boiling point.

e.g., we can explain the following trend in the boiling points of the four halogens listed:

F2 Cl2 Br2 I2

-1880C -350C 590C 1840C

The boiling point of the halogens i__________ as we go down group 7.

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

which has m__________ electrons than F2; more electrons results in l__________ temporary dipoles.

This means there would be s__________ 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. Can

you predict the trend in boiling points for the noble gases and also explain how the forces arise between the

atoms?

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 l__________ difference in electronegativity between H and the O/N/F atoms,

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

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

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

F δ- a the neighbouring molecule. E.g. a diagram showing hydrogen bonding between neighbourng 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



Draw diagrams to show the hydrogen bonding between:

(i) neighbouring NH3 molecules

(ii) neighbouring HF molecules

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

b__________ or i__________ forces present. Melting (solid to liquid) and boiling (liquid to g__________)

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 e__________ is required to break them and

the higher the melting or boiling point. In general there is an i__________ in melting/boiling points across

periods 2 & 3 (left to right) up to Group 4. There is then a sharp d__________ 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 h__________ for the metals

(Li, Be, Na, Mg & Al) due to the s__________ metallic bonding present between positive metal ions

and delocalised electrons in the g__________ metallic lattice. There is also an increase in melting points

for the metals each time you go a__________ the periods (see the graph above) because there is one

more p__________ so a greater n__________ charge and also more d__________ electrons with the next

metal across the period. This results in a s__________ 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. For example, the melting point i__________ going from Na to Mg to Al across

Period 3.

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 i__________ 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 and Si exist as a g__________ c__________ 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 s__________ c__________

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 d__________ in melting/boiling points between Groups 4 and 5, and the boiling

points for Groups 5-8 is relatively l__________. This is because the elements now exist as small

covalently bonded m__________ between Groups 5-8, with only w__________ 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 c__________ lattice or simple m__________ lattice. Note that across Period 3, Groups 5-

8, S8 has the h__________ 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 e__________ in the

molecule. More electrons means l__________ temporary and induced dipoles resulting in s__________

temporary dipole-induced dipole forces between molecules which require more energy to break.

2.11 Trends in melting and boiling points down a group

Melting points d__________ down Groups 1 & 2 as there are m__________ shells down a group meaning a

l__________ radius and m__________ shielding, resulting in w__________ forces of attraction between the

particles.

The melting points i__________ 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 m__________

electrons resulting in l__________ temporary dipoles and therefore s__________ temporary dipole-induced

dipole forces to break between the molecules which require more energy to break.

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



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 e__________ 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

3

4

Each shell is then broken down into sub-shells within which are found o__________. There are f__________

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 2__________. The number

for the sub-shell tells you which s__________ 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 o__________ and each orbital can hold up

to t__________ 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 Number of these orbitals per

sub-shell

Total number of electrons held

in all of these orbitals

s 1 1 x 2 = 2

p 3 3 x 2 = 6

d 5

f 7

Complete the table below to summarise where the electrons are found in each shell in an atom.


What are the shapes of s and p

orbitals?

+


3.2 Electron in box diagrams and electron configurations

In the Bohr model, we used dots or c__________ 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 b__________ to represent

the orbitals and a__________ to represent the electrons in the orbitals. Because p orbitals come in threes

we use t__________ boxes for p orbitals. In a similar fashion, we use f__________ boxes for d orbitals and

s__________ boxes for f orbitals. Each box (orbital) can hold up to t__________ electrons. Electrons are

n__________ charged and r__________ each other; we show this by having one arrow pointing up and one

pointing d__________. For the same reason, each orbital is filled s__________ before pairing starts for p, d

and f sub-shells.

Note that in the electron in box diagram below, the 4__________ 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

o__________, 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 e__________ 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 h__________ filled subshell first.

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

when a neutral atom. Note how in the 3d sub-shell the orbitals are filled s__________ 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


Draw electron in box diagrams and below each write out the electron configuration for:

(i) H

(ii) Be

(iii) O

(iv) K+

(v) Br

(vi) Br-


Write out electron configurations only for the following atoms and ions:

He

N

F

Al3+

P3-

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 o__________ 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.

Prove this fact by writing out the electron configurations for the following elements.

s-block

Li

Na

Ca

p-block

B

P

Cl

s d p


Prove that the cations Na+ and Mg2+ are isoelectronic by writing out their electronic configurations.

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 i__________ 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 s__________ 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 b__________ sides of an equation. Equations are balanced by putting n__________ before the

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

Worked Example 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 o__________ 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) b__________ the hydrogen’s:

N2 (g) + 3 H2 (g) → 2 NH3 (g)

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

Example 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 m__________ so must be a solid at room temperature,

oxygen is a gas. Sodium oxide is an i__________ compound (which are usually solids) because it is

made up of a m__________ and a non-metal; use the c__________-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)

Now have a go at balancing the following equations.

1. Mg + N2 → Mg3N2

2. Ba + HCl → BaCl2 + H2

3. C + O2 → CO2

4. C2H6 + O2 → CO2 + H2O

5. Fe + Cl2 → FeCl3

6. Mg + HNO3 → Mg(NO3)2 + H2

7. CuCO3 + HCl → CuCl2 + + CO2 + H2O

4.2 Reactions of period 2 and 3 elements with oxygen

You have previously seen how as you move across periods 2 and 3 l__________ to r__________, there is an

i__________ 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 m__________

lattice (stronger metallic bonds to break) to giant c__________ (stronger covalent bonds to break) to simple

m__________ lattice (weaker intermolecular forces to break).

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

with o__________ tend to be giant i__________ lattices to giant c__________ structures to simple

m__________ structures (see the table on the next page). You get giant ionic when a m__________ 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

g__________ for the simple molecules. The products made also change from being a__________ (metal

oxides are examples of bases) to a__________ to a__________.


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 i__________ metal oxides. The Group 1 and 2 metal oxides

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

of the metal oxide formed using the c__________-over method and then fully b__________ the equation.

When using the cross-over method, the positive charge on the metal ion is the same as its G__________

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)

Group 1 metals react rapidly with oxygen and are stored under o__________ 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 o__________ and makes them behave as

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

much l__________ 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 p__________ ion, O22-

and the super-oxide ion O2- also exist. These molecular ions contain a c__________ bond between the two

O atoms; O-O2- and O-O-. These two negative molecular ions are only stable next to a l__________ 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 s__________ positive ion and when next to a large negative ion such as O2- or O2-, the

the electrons in the covalent bond between the two O atoms are strongly attracted to the small positive

metal ion; the ionic bond becomes p__________. 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.

Write an equation for the reaction of magnesium with oxygen:


(ii) Water:

Metals react with water to form solutions of a metal h__________ as well as hydrogen g__________. 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

b__________ 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 a__________ metals

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

i__________ down the Group. 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 o__________ layer. Group 4, 5 and 6

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

Write an equation for the reaction of calcium with water:

Write an equation for the reaction of lithium with water:

(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 s__________ in a neutralisation reaction as well as H2 gas. C__________ salts form with HCl

and s__________ 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

Write equations for the reaction of lithium with HCl and H2SO4:



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 d__________ 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 charge

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

reactions. All of the oxidation numbers when added together must e__________ 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 r__________ agent; it reduced Cl (Fe itself was oxidised).

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

Worked examples: Assign oxidation numbers and say whether or not these are REDOX reactions.

i)

Mg + 2 HCl → MgCl2 + H2

ii)

2 Ca + O2 → 2 CaO

iii)

SrO + H2O → Sr(OH)2

iv)

Mg + H2O → MgO + H2

v)

BaO + 2 HCl → BaCl2 + H2O

vi)

Ca + 2 HCl → CaCl2 + H2



4.5 Oxidation numbers for transition metals and oxyanions

Transition elements can have v__________ 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 r__________ numeral in brackets. The variable oxidation state is what makes

the transition metals useful as c__________ 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

Cu2O Cu = +1

Copper(I) chloride

CuCl2

Oxyanions are:

n__________ molecular ions

that contain o__________

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 s__________ 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- N = +3

KMnO4 (K+ MnO4

-) Potassium manganate(VII) Mn = +7

NaNO3 Sodium nitrate(V) N = +5

NaNO2

CaSO4


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 t__________ of the list. The order of reactivity for the metals is Group1, Group2, Group3,

Group4, Transition metals. The metals higher up are m__________ reactive as they have a greater tendency

to lose an e__________ to form a complete outer shell. In general, for the metals, reactivity decreases across

a period and i__________ down a Group. This is because across a period the n__________ charge increases,

making it harder to lose the outer electron during a chemical reaction.

The most reactive metals are in group 1 and these are found towards the top of the reactivity series. The

group 1 metals become more reactive as you go down the group, so Fr is the most reactive metal but it is

not usually seen on lists as it is so radioactive and unstable.

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 first before moving on to

displacement reactions.


4.7 Displacement reactions

A more r__________ 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 r__________ series. Displacement

reactions involving metals are also usually r__________ reactions – you can prove this by assigning

oxidation numbers to each element in the equation. E.g. Prove that the displacement reaction below is also

a redox reaction:

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

Displacement reactions can also happen with the Halogens. The halogens become m__________ 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 s__________ 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 o__________ agents – this means that they oxidise (remove electrons from) another

species. In the above reaction, chlorine is a s__________ 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.




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 __________ molecules,.

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

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

a__________ 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 g__________ 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 o__________ 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, 6.02 x 1023 atoms).

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

m__________).

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

following triangle:

amount in Moles = mass in Grams / Relative molar mass

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)


Example 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, d__________ by 1000. To go from dm3 to cm3, m__________ 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 __________ moles of NaOH in every 1 dm3.

2 dm3 of a 0.5 mol dm-3 NaOH solution contains __________ 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)



Example 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

Example 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 b__________ numbers in a balanced equation gives you the ratio of moles reacting in a given reaction.

Stoichiometry is the m__________ 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 r__________ 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 __________ mole of Na2O.

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

One mole of Na was reacted with excess oxygen to form __________ 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



Example 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

Example 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 r__________ 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 t__________ 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 example: We are expecting 10g of product from a reaction but only produce 8g. What % is this?

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

worked example: 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%


What is empirical formula and

how do you calculate this using

moles?

+


QUESTIONS

1. State the relative mass and charge for protons, neutrons

and electrons.

2. Draw a fully labelled diagram of an atom, stating where

each sub-atomic particle is found.

3. Calculate the number of protons, neutrons and electrons

in the neutral element that has an atomic number of 34.

4. Si has isotopes with a mass numbers of 28 and 29, yet

the relative atomic mass is 28.1. What does this tell you

about the relative amounts of each isotope in a sample of

Si found in nature?

5. Complete the following table:

Pure

Isotope

Atomic

no.

Mass

no.

No. of

electrons

No. of

neutrons

R.A.M in

periodic

table

14N

15N

12C

13C

14C

6. Explain why Mg and Ca have similar chemical

properties.

7. Identify the element in (i) Group 3, Period 3 (ii) Group

8, Period 1 (iii) Group 7, Period 5.

8. Explain how you find the metals in the Periodic Table.

9. Explain how the elements are arranged in the Periodic

Table

10. Explain what is meant by the term Periodicity. Use

examples in your answer.

11. Compare and explain the atomic radius of F and Cl.

12. Compare and explain the atomic radius of Li and F.

13. Explain what is meant by cation and anion.

14. Explain the octet rule.

15. How many electrons are there in ions of the elements

(i) Ca (ii) P (iii) I (iv) S (v) K (vi) Al.

16. Compare, and explain, the radius of Ca with its cation.

17. Compare, and explain, the radius of P with its anion.

18. Explain what is meant by the term isoelectronic.

19. Identify any ions that are isoelectronic with Ca.

20. Explain what is meant by the term first ionisation

energy.

21. Explain why energy is required to remove electrons

from atoms. Is this shown by a positive or negative sign?

22. Write equations to show the following ionisation

energies (i) first for Li (ii) third for Ga (iii) ninth for Na

(iv) tenth for Al (v) sixth for C.

23. Explain why the first ionisation energy for Be is

greater than the first for Li.

24. Explain why the first ionisation energy for Na is less

than the first for Li.

25. Explain what is meant by first electron affinity.

26. Write equations for the first and second electron

affinities for (i) O (ii) N (iii) P (iv) S

27. Explain why energy is released for the first electron

affinity but energy is required for the second electron

affinity. Is this shown by a positive or negative sign?

28. Explain why less energy is released for the first

electron affinity of Br than the first for Cl.

29. Explain why less energy is released for the first

electron affinity of Se than the first for Br.

30. State, and explain, whether the following show ionic,

covalent or metallic bonding (i) Ca (ii) O2 (iii) NaBr (iv)

NH3 (v) Mg (vi) MgCl2 (vii) CO2 (viii) CH4 (iv) AlCl3.

31. Draw dot and cross diagrams for the following and

also calculate the relative formula mass (i) potassium

chloride (ii) aluminium chloride.

32. What two factors affect the strength of an ionic bond?

Explain your answer.

33. State what is meant by a covalent bond and draw dot &

cross diagrams for the following. Also calculate the

relative molecular mass for each compound (i) F2 (ii) CF4

(iii) CO2 (iv) HCN.

34. State what is meant by a dative covalent bond, using a

dot & cross diagram of the ammonium ion as your

example.

35. Draw a 3D diagram of methane. State the name of the

shape and the bond angle.

36. Compare the strength and length of the bonds in Cl2

and in O2.

37. State what is meant by the term metallic bonding,

using a labelled diagram to illustrate your answer. You

should also give examples of substances which have

metallic bonding.


38. Explain the following properties of metals: (i)

conduction of heat and electricity (ii) malleability and

ductility.

39. State what is meant by the term electronegativity.

Explain the trends in electronegativity values in the

Periodic Table.

40. Explain what factors affect the electronegativity of an

element.

41. Explain how polar and non-polar molecules arise.

42. State what is meant by the term intermolecular force.

List the three types of intermolecular forces in order of

increasing strength. How do intermolecular forces

compare with ionic, covalent and metallic bonds?

43. Explain how temporary dipole-induced dipole forces

arise.

44. Explain why ICl has a higher boiling point than Cl2.

45. Draw a diagram showing hydrogen bonding between

one molecule of water and one molecule of ammonia.

46. Identify the intermolecular forces present in the

following (i) O2 (ii) N2 (iii) F2 (iv) HF (v) CH3CH2NH2

(vi) CH3CH2OH (vii) HCl (viii) H2S

47. Explain why O2 has a lower melting point than Li.

48. Explain why Cl2 has a lower melting point than Si.

49. Explain why Al has a higher melting point than Mg,

which has a higher melting point than Na.

50. Explain why S has a higher melting point than P,

which has a higher melting point than Cl, which has a

higher melting point than Ar.

51. Explain the trend in boiling points down Groups 1, 2

and 7.

52. Explain what is meant by the term orbital. What are

the different types of orbital and how many of each are

there?

53. Draw an electron in box diagram and also write out the

electron configuration for P and P3-.

54. Draw an electron in box diagram and also write out the

electron configuration for Ca and Ca2+.

55. To which block in the Periodic Table do the following

elements belong (i) Mg (ii) Si (iii) I (iv) Mn (v) Ag.

56. Construct balanced equations for the reaction of the

following elements with oxygen. Explain any properties

of the oxides produced (i) Na (ii) Al (iii) C (iv) Si.

57. Explain why Na2O has a higher melting point than

CO2.

58. Construct balanced equations for the reactions of Na

and Al with (i) oxygen (ii) water (iii) hydrochloric acid

(iv) sulphuric acid.

59. The reaction between lithium metal and hydrochloric

acid is a redox reaction. Write an equation for this

reaction and prove that it is a redox reaction.

60. The element strontium forms a nitrate, Sr(NO3)2, which

decomposes on heating as shown below.

2 Sr(NO3)2 (s) → 2 SrO (s) + 4 NO2 (g) + O2 (g)

Using oxidation numbers, explain why the reaction involves

both oxidation and reduction.

61. Magnesium and copper(II) sulfate, CuSO4, take part in

a displacement reaction. Write the equation for this

reaction. Is the reaction a redox reaction? Why does the

reverse reaction not take place?

62. Iron and copper(II) sulfate, CuSO4, take part in a

displacement reaction. Write the equation for this reaction.

Is the reaction a redox reaction? Why does the reverse

reaction not take place?

63. Calculate the molar mass of the following:

1. CuO 2. Cl2 3. HF 4. CaCO3

5. NH4NO3 6. Na2SO4 7. Al2(SO4)3 8. Zn(OH)2

64. Calculate the mass, in grams, for the following.

1. 2 mol of

Na

2. 0.5 mol

of Zn

3. 10 mol

of Pb

4. 0.25 mol

of SO2

5. 5 mol of

Na2CO3

6. 0.2 mol

of Br2

7. 0.01 mol

of Ag2O

8. 0.02 mol

of KMnO4

65. Calculate the amount, in mol, for the following.

1. 2.3 g of

Na

2. 58.5 g of

NaCl

3. 50.0 g of

CaCO3

4. 40.0 g of

CH4

5. 0.98 g of

H2SO4

6. 8.0 g of

Br2

7. 160.0 g of

NaOH

8. 303.0 g

of KNO3

66. Find the amount, in moles, of solute dissolved in water

to form the following solutions.

1. 1 dm3 of 0.2

mol dm-3 H2SO4

2. 100 cm3 of 1

mol dm-3 KCl

3. 25 cm3 of 2.5

mol dm-3 NaCl

4. 22.5 cm3 of

0.2 mol dm-3 HCl

5. 2 dm3 of 2 mol

dm-3 H2SO4

6. 5 cm3 of 5 mol

dm-3 HNO3

67. Find the concentration of the following solutions in

mol dm-3.

1. 1 mol of HCl

in 1 dm3

2. 5 mol of HNO3

in 250 cm3

3. 0.1 mol of

HNO3 in 25 cm3

4. 2.25 x 10-3

mol in 15 cm3

5. 4 mol in 20

dm3

6. 8.8 x 10-2 mol

in 200 cm3


68. 12.7 g of Cu reacts with excess O2 to produce CuO.

2 Cu + O2 → 2 CuO

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

(ii) Calculate the amount, in mol, of CuO produced in the

reaction.

(iii) Calculate the mass, in grams, of CuO produced in the

reaction.

69. 6.0 g of Mg reacts with an aqueous solution of CuSO4

to form MgSO4 and Cu.

CuSO4 + Mg → MgSO4 + Cu

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

(ii) Calculate the amount, in mol, of Cu produced in the

reaction.

(iii) Calculate the mass, in grams, of Cu produced in the

reaction.

70. In a reaction, SrCO3 is used to neutralise 25 cm3 of a

0.15 mol dm-3 HCl solution.

SrCO3 + 2 HCl → SrCl2 + H2O + CO2

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

(ii) Calculate the mass, in grams, of SrCO3 used to

neutralise the HCl in the reaction.

71. In an experiment 0.50 g of K metal was added to 500

cm3 of water making a solution of KOH.

2 K + 2 H2O →2 KOH + H2

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

(ii) Calculate the concentration, in mol dm-3, of the KOH

solution formed.

72. An excess amount of Ba is treated with 200 cm3 of 0.05

mol dm-3 HCl to form an aqueous solution of barium

chloride.

Ba + 2 HCl → BaCl2 + H2

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

(ii) Calculate the concentration, in mol dm-3, of the BaCl2

solution formed.

73. 1 Kg of MgO is used to neutralise a 0.04 mol dm-3

solution of HCl.

MgO + 2 HCl → MgCl2 + 2 H2O

(i) Calculate the amount, in mol, of MgO used in the

reaction.

(ii) Calculate the volume, dm3, of the HCl solution used.

74. 50 kg of pure sulfuric acid was accidently released into

a lake when a storage vessel leaked. It was proposed that

CaCO3 was used to neutralise it.

Calculate the mass of CaCO3 required to neutralise the 50

kg of H2SO4.

H2SO4 + CaCO3 → CaSO4 + H2O + CO2

75. A student dissolved 2.794 g of an acid HY in 100 cm3

of water. Once HY was fully dissolved, the student then

added a further 150 cm3 of water to make 250 cm3 of a

solution of HY in water. 25cm3 of this HY solution was

then titrated against a 0.0614 mol dm-3 of Na2CO3 solution,

requiring 23.45 cm3 of the Na2CO3 solution for complete

neutralisation.

2 HY + Na2CO3 → 2 NaY + H2O + CO2

(i) Calculate the amount, in mol, of Na2CO3 used in the

titration.

(ii) Calculate the amount, in mol, of HY used in the titration.

(iii) Calculate the amount, in mol, of HY in the original 250

cm3 solution.

(iv) Calculate the molar mass of HY.

76. A chemist reacts 1.00g of Cu with excess oxygen. After

purification, 1.00g of CuO is isolated from the reaction.

2 Cu + O2 → 2 CuO

(i) Calculate the amount, in moles, of Cu reacted.

(ii) Determine the theoretical moles of CuO.

(iii) Calculate the actual moles of CuO made.

(iv) Determine the percentage yield of the reaction.

77. A chemist reacts 1.00g of methanol, CH3OH, with an

oxidant. After purification, 0.98g of methanal, CH2O, is

isolated from the reaction. Determine the percentage yield.

CH3OH + [O] → CH2O + H2O


Checklist for Unit 1 (Chemistry Section)

1. The Periodic Table, Atoms & Ions

1.1 Introduction to the atom

State the relative charge and masses of protons, neutrons and electrons

State the position of protons, neutrons and electrons in a Bohr atom

1.2 Elemental symbols in the Periodic Table

State what is meant by atomic and mass number

Calculate number of neutrons and electrons in atoms and ions from atomic and mass numbers

State what is meant by relative atomic mass

1.3 Introduction to the Periodic Table of elements

Identify the Groups and Periods 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

1.4 Making ions

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

1.5 Trends in ionic radius

Explain trends across a Period and down a Group for cations and anions

Compare and explain the size of neutral atom with its cation

Compare and explain the size of neutral atom with its anion

Explain what is meant by the term isoelectronic

1.6 Ionisation energy

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

1.7 Electron affinity

Define electron affinity

Write equations showing electron affinties

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


2.1 Introduction to bonding in compounds

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

2.2 Ionic bonding

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

2.3 Covalent bonding

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

2.4 Metallic bonding

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

2.5 Electronegativity

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

2.6 Introduction to intermolecular forces

State what is meant by the term intermolecular force

State what the three intermolecular forces are in order of strength

2.7 Permanent dipole-dipole forces

Explain that permanent dipole-dipole forces exist between polar molecules

2.8 Temporary dipole-induced dipole forces

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

2.9 Hydrogen bonding

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

2.10 Trends in melting and boiling points across a period

Explain the trends in melting/boiling points across periods 2 and 3, in terms of structure and forces

2.11 Trends in melting and boiling points down a group

Explain the trend in boiling points down Groups 1, 2 and 7

3. Orbital theory

3.1 Sub-shells and orbitals

Determine the number of electrons in each shell using the formula 2n2

State what is meant by the term orbital

3.2 Electron in box diagrams and electron configurations

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

3.3 blocks in the periodic table

Identify s, p and d blocks in the Periodic Table

3.4 Ionisation energies re-visited

Explain the anomalies in ionisation energy trends using electron in box diagrams (Be/B and N/O)

4. Balanced Equations and Chemical Reactions

4.1 balancing equations

Be able to balance chemical equations with the use of state symbols

4.2 Reactions of period 2 and 3 elements with oxygen

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

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

Construct balanced equations for the reaction of metals with O2, H2O, HCl and H2SO4


4.4 Redox

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

4.5 Oxidation numbers for transition metals and oxyanions

Know that transition metals have variable oxidation states, shown by roman numerals

Know that roman numerals can also be used with oxyanions

4.6 Reactivity series

Explain the relative position of metals in the reactivity series

4.7 Displacement reactions

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

4.8 Uses and applications of the substances in this unit

Identify uses for substances covered in this unit


5.1 Moles and masses

Define the terms mole and molar mass

Use GMR to calculate grams/moles/molar masses for solids

5.2 Moles and solutions

Be familiar with the units for concentration and volumes

Use MCV to calculate moles/concentrations/volumes for liquids

5.3 Moles and equations

Understand molar ratios in balanced equations (stoichiometry)

Use GMR, MCV and stoichiometry to solve calculations in chemistry

5.4 Percentage yield

Recall the formulae for percentage yield calculations

Calculate the percentage yield of a reaction with use of GMR and stoichiometry

UNIT 1 - HCC Applied Science -...

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